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3700 | https://en.wikipedia.org/wiki/Remainder | Jump to content
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Contents
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1 Integer division
1.1 Examples
2 For floating-point numbers
3 In programming languages
4 Polynomial division
5 See also
6 Notes
7 References
8 Further reading
Remainder
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From Wikipedia, the free encyclopedia
Amount left over after computation
For other uses, see Remainder (disambiguation).
In mathematics, the remainder is the amount "left over" after performing some computation. In arithmetic, the remainder is the integer "left over" after dividing one integer by another to produce an integer quotient (integer division). In algebra of polynomials, the remainder is the polynomial "left over" after dividing one polynomial by another. The modulo operation is the operation that produces such a remainder when given a dividend and divisor.
Alternatively, a remainder is also what is left after subtracting one number from another, although this is more precisely called the difference. This usage can be found in some elementary textbooks; colloquially it is replaced by the expression "the rest" as in "Give me two dollars back and keep the rest." However, the term "remainder" is still used in this sense when a function is approximated by a series expansion, where the error expression ("the rest") is referred to as the remainder term.
Integer division
[edit]
Given an integer a and a non-zero integer d, it can be shown that there exist unique integers q and r, such that a = qd + r and 0 ≤ r < |d|. The number q is called the quotient, while r is called the remainder.
(For a proof of this result, see Euclidean division. For algorithms describing how to calculate the remainder, see Division algorithm.)
The remainder, as defined above, is called the least positive remainder or simply the remainder.
In some occasions, it is convenient to carry out the division so that a is as close to an integral multiple of d as possible, that is, we can write
: a = kd + s, with |s| ≤ |d/2| for some integer k.
In this case, s is called the least absolute remainder. As with the quotient and remainder, k and s are uniquely determined, except in the case where d = 2n and s = ±n. For this exception, we have:
: a = kd + n = (k + 1)d − n.
A unique remainder can be obtained in this case by some convention—such as always taking the positive value of s.
Examples
[edit]
In the division of 43 by 5, we have:
: 43 = 8 × 5 + 3,
so 3 is the least positive remainder. We also have that:
: 43 = 9 × 5 − 2,
and −2 is the least absolute remainder.
These definitions are also valid if d is negative, for example, in the division of 43 by −5,
: 43 = (−8) × (−5) + 3,
and 3 is the least positive remainder, while,
: 43 = (−9) × (−5) + (−2)
and −2 is the least absolute remainder.
In the division of 42 by 5, we have:
: 42 = 8 × 5 + 2,
and since 2 < 5/2, 2 is both the least positive remainder and the least absolute remainder.
In these examples, the (negative) least absolute remainder is obtained from the least positive remainder by subtracting 5, which is d. This holds in general. When dividing by d, either both remainders are positive and therefore equal, or they have opposite signs. If the positive remainder is r1, and the negative one is r2, then
: r1 = r2 + d.
For floating-point numbers
[edit]
When a and d are floating-point numbers, with d non-zero, a can be divided by d without remainder, with the quotient being another floating-point number. If the quotient is constrained to being an integer, however, the concept of remainder is still necessary. It can be proved that there exists a unique integer quotient q and a unique floating-point remainder r such that a = qd + r with 0 ≤ r < |d|.
Extending the definition of remainder for floating-point numbers, as described above, is not of theoretical importance in mathematics; however, many programming languages implement this definition (see Modulo operation).
In programming languages
[edit]
Main article: Modulo operation
While there are no difficulties inherent in the definitions, there are implementation issues that arise when negative numbers are involved in calculating remainders. Different programming languages have adopted different conventions. For example:
Pascal chooses the result of the mod operation positive, but does not allow d to be negative or zero (so, a = (a div d ) × d + a mod d is not always valid).
C99 chooses the remainder with the same sign as the dividend a. (Before C99, the C language allowed other choices.)
Perl, Python (only modern versions) choose the remainder with the same sign as the divisor d.
Scheme offer two functions, remainder and modulo – Ada and PL/I have mod and rem, while Fortran has mod and modulo; in each case, the former agrees in sign with the dividend, and the latter with the divisor. Common Lisp and Haskell also have mod and rem, but mod uses the sign of the divisor and rem uses the sign of the dividend.
Polynomial division
[edit]
Main article: Euclidean division of polynomials
Euclidean division of polynomials is very similar to Euclidean division of integers and leads to polynomial remainders. Its existence is based on the following theorem: Given two univariate polynomials a(x) and b(x) (where b(x) is a non-zero polynomial) defined over a field (in particular, the reals or complex numbers), there exist two polynomials q(x) (the quotient) and r(x) (the remainder) which satisfy:
: a(x) = b(x)q(x) + r(x)
where
: deg(r(x)) < deg(b(x)),
where "deg(...)" denotes the degree of the polynomial (the degree of the constant polynomial whose value is always 0 can be defined to be negative, so that this degree condition will always be valid when this is the remainder). Moreover, q(x) and r(x) are uniquely determined by these relations.
This differs from the Euclidean division of integers in that, for the integers, the degree condition is replaced by the bounds on the remainder r (non-negative and less than the divisor, which insures that r is unique.) The similarity between Euclidean division for integers and that for polynomials motivates the search for the most general algebraic setting in which Euclidean division is valid. The rings for which such a theorem exists are called Euclidean domains, but in this generality, uniqueness of the quotient and remainder is not guaranteed.
Polynomial division leads to a result known as the polynomial remainder theorem: If a polynomial f(x) is divided by x − k, the remainder is the constant r = f(k).
See also
[edit]
Chinese remainder theorem
Divisibility rule
Egyptian multiplication and division
Euclidean algorithm
Long division
Modular arithmetic
Polynomial long division
Synthetic division
Ruffini's rule, a special case of synthetic division
Taylor's theorem
Notes
[edit]
^ Smith 1958, p. 97
^ Ore 1988, p. 30. But if the remainder is 0, it is not positive, even though it is called a "positive remainder".
^ Ore 1988, p. 32
^ Pascal ISO 7185:1990 6.7.2.2
^ "6.5.5 Multiplicative operators". C99 specification (ISO/IEC 9899:TC2) (PDF) (Report). 6 May 2005. Retrieved 2018-08-16.
^ "Built-in Functions – Python 3.10.7 documentation". 9 September 2022. Retrieved 2022-09-10.
^ Larson & Hostetler 2007, p. 154
^ Rotman 2006, p. 267
^ Larson & Hostetler 2007, p. 157
^ Weisstein, Eric W. "Polynomial Remainder Theorem". mathworld.wolfram.com. Retrieved 2020-08-27.
References
[edit]
Larson, Ron; Hostetler, Robert (2007), Precalculus: A Concise Course, Houghton Mifflin, ISBN 978-0-618-62719-6
Ore, Oystein (1988) , Number Theory and Its History, Dover, ISBN 978-0-486-65620-5
Rotman, Joseph J. (2006), A First Course in Abstract Algebra with Applications (3rd ed.), Prentice-Hall, ISBN 978-0-13-186267-8
Smith, David Eugene (1958) , History of Mathematics, Volume 2, New York: Dover, ISBN 0486204308 {{citation}}: ISBN / Date incompatibility (help)
Further reading
[edit]
Davenport, Harold (1999). The higher arithmetic: an introduction to the theory of numbers. Cambridge, UK: Cambridge University Press. p. 25. ISBN 0-521-63446-6.
Katz, Victor, ed. (2007). The mathematics of Egypt, Mesopotamia, China, India, and Islam : a sourcebook. Princeton: Princeton University Press. ISBN 9780691114859.
Schwartzman, Steven (1994). "remainder (noun)". The words of mathematics : an etymological dictionary of mathematical terms used in english. Washington: Mathematical Association of America. ISBN 9780883855119.
Zuckerman, Martin M (December 1998). Arithmetic: A Straightforward Approach. Lanham, Md: Rowman & Littlefield Publishers, Inc. ISBN 0-912675-07-1.
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ArticlesAnatomyBasicsIntroduction to the musculoskeletal systemMusculoskeletal system development
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Overview
Bone
Joints
Axial skeleton
Skull
Vertebral column
Ribs
Sternum
Limbs and appendicular system
Limbs
Appendicular skeleton
Skeletal muscle
Clinical aspects
Skull malformations
Vertebral malformations
Rib malformations
Sternal malformations
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ArticlesAnatomyBasicsIntroduction to the musculoskeletal systemMusculoskeletal system development
Musculoskeletal system development
Author:
Danny Ly, MSc
•
Reviewer:
Francesca Salvador, MSc
Last reviewed: August 14, 2023
Reading time: 22 minutes
Musculoskeletal anatomy is fascinating since it gives us insights as to how our body utilizes our muscles, bones, and joints to give us the ability to navigate in the world. If this article peaks your interest, you probably have a solid foundation in this topic and are ready to take your knowledge to the next level. By studying the embryological development of the musculoskeletal system, you will achieve a better understanding of how different types of congenital anomalies can occur.
Contents
Overview
Bone
Joints
Axial skeleton
Skull
Vertebral column
Ribs
Sternum
Limbs and appendicular system
Limbs
Appendicular skeleton
Skeletal muscle
Clinical aspects
Skull malformations
Vertebral malformations
Rib malformations
Sternal malformations
Limb malformations
Skeletal muscle malformations
Sources
Show all
Overview
The musculoskeletal system develops from three sources:
the paraxial mesoderm
the parietal layer of the lateral plate mesoderm
the neural crest cells
The development of bone and muscle begins at the fourth gestational week, when the paraxial mesoderm differentiates into somites; the latter gives rise to sclerotomes and dermomyotomes. Sclerotomes form the vertebra and the ribs, whereas myotomes form the majority of the muscular system.
Bone formation can occur either by intramembranous ossification or endochondral ossification. Although different, the occurrence of both processes first require the condensation of mesenchymal cells - the loosely organized embryonic connective tissue. Intramembranous ossification underlies the formation of the cranial vault, many bones of the face, and the clavicle. Endochondral ossification underlies the formation of the base of the skull , some bones of the face, the bones of the limbs and girdles, the vertebral column, the ribs, and the sternum.
Development of the limbs involves the inductive influences of the apical ectodermal ridge, the formation of circular constrictions to separate parts of the limbs, and opposite rotations of the upper and lower limbs. Development of the skeletal muscle involves the differentiation of myotome cells into myoblasts. This article will discuss the embryological development of the axial skeleton, the appendicular skeleton, and the skeletal muscle, as well as the associated malformations that may occur.
Bone
The first stage of any type of bone formation involves a mesenchymal condensation, where cells become densely packed together. From this point on, there are two ways osteogenesis can occur: intramembranous ossification and endochondral ossification. The process in which mesenchymal cells ensheathed in membranous tissue directly undergo ossification is known as intramembranous ossification. The process in which mesenchymal cells first differentiate into cartilage models before undergoing ossification is known as endochondral ossification.
Formation of the cranial vault, most bones of the face, and the clavicle occur by intramembranous ossification, whereas formation of the rest of the axial and appendicular skeleton occur by endochondral ossification. In other words, the base of the skull, some bones of the face, the vertebral column, the ribs, the sternum, and the bones of the limbs and girdles form by a two-step process: chondrification and ossification.
Joints
During the sixth gestational week, joints begin to develop with the formation of condensed mesenchyme in the interzone, the region between two bone primordia. Joints are classified as:
fibrous
cartilaginous
synovial
The development of fibrous joints involves mesenchymal cells in the interzone to differentiate into dense fibrous tissue (i.e. sutures of the skull). The development of cartilaginous joints involves mesenchymal cells in the interzone to differentiate into hyaline cartilage (i.e. costochondral joints) or fibrocartilage (i.e. pubic symphysis). The development of synovial joints involves a more extensive process: the central mesenchymal cells in the interzone undergo apoptosis to form the synovial joint cavities, whereas the peripheral cells differentiate into ligaments and dense fibrous tissue. Sequentially, the dense fibrous tissue forms the articular cartilage that covers the ends of the adjacent bone primordia. The remaining mesenchymal cells surrounding the interzone differentiate into chondrocytes to form the joint capsules and the synovial membrane.
Axial skeleton
The axial skeleton includes the:
skull
vertebral column
ribs
sternum
The skull consists of a neurocranium and a viscerocranium, with each having membranous and cartilaginous components. The bones that make up the skull thus form either by intramembranous ossification or endochondral ossification. The bones that make up the vertebral column, the ribs, and sternum form only by endochondral ossification. The vertebral column develops from a resegmentation process of the somites, while the ribs develop as extensions from the thoracic vertebrae. The sternum develops as two independent bands of mesenchymal cells before fusing and ossifying as one.
Skull
The skull can be divided in two parts: the neurocranium that forms a protective case around the brain, and the viscerocranium that forms the skeleton of the face. The neurocranium itself is divided into two other parts: the membranous part that surrounds the brain as a vault, and the cartilaginous part (chondrocranium) that forms the base of the skull. Both the neurocranium and the viscerocranium have distinct components that are formed either by intramembranous ossification or endochondral ossification.
Neurocranium
The membranous part of the neurocranium forms the calvaria (skullcap). It is derived from two sources: the paraxial mesoderm and the neural crest cells. Mesenchymal cells from these two sources surround the brain at various sites, form primary ossification centers, and undergo intramembranous ossification. This results in the formation of membranous flat bones that are characterized by needle-like bone spicules. Bone spicules progressively radiate from the primary ossification centers toward the periphery. Structures derived from the membranous neurocranium include the parietal bones, part of the temporal bones, and the occipital bone.
At birth, the membranous bones are separated from each other by dense connective tissue membranes that form fibrous joints, known as the cranial sutures (coronal, sagittal, and lambdoid). The site at which more than two bones meet are called the fontanelles (anterior, posterior, and two posterolateral).
The anterior fontanelle is the most prominent one and is found where the parietal and frontal bones meet.
The posterior fontanelle is found where the parietal bones and the occipital bone meet.
The posterolateral (mastoid) fontanelles are found where the parietal, occipital, and temporal bones meet.
As the brain and the skull continue to grow after birth, many of these sutures and fontanelles will remain membranous and open postnatally. Generally, the posterior fontanelle closes first by 2 months of age, the mastoid fontanelle by 6 months, the anterior fontanelle by 18 months, and the cranial sutures by 36 months.
The cartilaginous part of the neurocranium forms the base of the skull. It initially consists of a number of separate cartilages that eventually fuse together. Similar to the membranous neurocranium, the cartilaginous neurocranium is derived from the same sources. The neural crest cells form the prechordal chondrocranium anterior to the center of the sella turcica, whereas the paraxial mesoderm form the chordal chondrocranium posterior to the center of the sella turcica. The development of the base of the skull is complete when these cartilaginous structures fuse and undergo endochondral ossification. Structures derived from the chondrocranium include components of the occipital bone, the sphenoid bone, and the ethmoid bone, specifically the:
posterior half of the cribriform plate
lesser wings of the sphenoid
greater wings of the sphenoid
sella turcica
petrous part of the temporal bones and the adjacent parts of the occipital bone
clivus
condyles of the occipital bone
Viscerocranium
The viscerocranium is mainly formed by the first two pharyngeal arches. The first pharyngeal arch undergoes intramembranous ossification to give rise to the:
zygoma
maxilla
squamous part of the temporal bone
mandible
The dorsal tip of the mandibular process and the second pharyngeal arch undergo endochondral ossification to give rise to the malleus, the incus, and the stapes. The ossicles are the first bones to become fully ossified, with their ossification beginning in the fourth month of gestation.
Vertebral column
The vertebral column develops from the sclerotomes, the ventromedial part of the somite. By the fourth gestational week , sclerotome cells surround the neural tube and the notochord to merge with cells derived from the opposing somite. Each sclerotome then undergoes resegmentation, a process that involves the caudal half of each sclerotome to fuse with the cranial half of each adjacent sclerotome; this forms the centrum, the primordial vertebral body. Thus, each vertebra develops from two adjacent sclerotomes rather than from one sclerotome.
Not all cells in the caudal half of each sclerotome undergo resegmentation. Instead, some migrate cranially and contribute to the formation of the intervertebral disc. As development continues, the notochord completely degenerates in the centrum, but where it persists, it enlarges as a gelatinous center. This forms the nucleus pulposus, which is later surrounded by circularly arranged fibers known as the annulus fibrosis. Combined, these two structures form the intervertebral discs.
By the sixth gestational week, the sclerotome cells surrounding the neural tube form a cartilaginous vertebral arch, and fuse with the cartilaginous vertebral body. The spinous, transverse, and costal processes develop as extensions from this newly assembled cartilage model. In the lumbar region, the costal processes of the first sacral vertebrae fuse and form the lateral sacral mass, known as the ala of the sacrum. The process of chondrification continues until a cartilaginous vertebral column is fully formed.
Ossification of the vertebrae begins at the seventh gestational week, but only ends during the second decade of adulthood. By the eighth week, three primary ossification centers develop: one at the center of the cartilaginous vertebral body and one on each side of the cartilaginous vertebral arch. At puberty, five secondary ossification centers appear in the vertebrae: one at the tip of the spinous process, one at the tip of each transverse process, and one on both the superior and inferior rim of the vertebral body.
Ribs
Ribs develop from the costal processes of the thoracic vertebrae. They are cartilaginous during the embryonic period and undergo ossification during the fetal period. The original site where the costal process is connected to the vertebra becomes replaced by costovertebral synovial joints. The first seven pairs of ribs attach to the sternum through their own cartilages. The subsequent five pairs of ribs attach to the sternum through the cartilage of the seventh rib. The last two pairs of ribs do not attach to the sternum. Respectively, this forms the true ribs, the false ribs, and the floating ribs.
Sternum
The sternum develops from a pair of separate vertical, condensed bands of mesenchymal cells, known as the sternal bars. These sternal bars form independently lateral to the midline of the ventral body wall. Chondrification occurs while the sternal bars migrate medially. By the tenth gestational week , they fuse in cranial-to-caudal sequence at the midline and form the cartilage model of the manubrium, the sternal body, and the xiphoid process.
Limbs and appendicular system
Limbs
The appendicular skeleton includes the bones of the limbs and girdles. The formation of these structures begin by the end of the fourth gestational week, where limb buds become visible as outpocketings from the ventrolateral body wall. They consist of a core of mesenchymal cells - derived from the somatic layer of the lateral plate mesoderm - covered by a layer of ectoderm. At the distal border of the limb, the ectoderm forms the apical ectodermal ridge (AER). The AER exerts an inductive influence on the core of mesenchymal cells to remain undifferentiated and to rapidly proliferate; this region is known as the progress zone. As the limbs continue to grow, cells farther from the influence of the AER begin to differentiate into cartilage and muscle. Development of the limbs thus proceed proximodistally.
By the sixth gestational week, a circular constriction separates the terminal and proximal portions of the limb buds. Later, a second circular constriction separates the proximal portion into two additional segments; the familiar parts of the limbs thus become recognizable. Meanwhile, the terminal portion becomes flattened to form the handplates and footplates. Further formation of fingers and toes depends on three factors: their continued outgrowth under the influence of the AER, mesenchymal condensation to form cartilaginous digital rays, and apoptosis of intervening tissue between the rays. Cell death in the AER creates separate ridges for each digit forming webbed fingers and toes. Further cell death in the interdigital spaces are what creates the separation of the digits. By the end of the eight week, digit separation is complete while the fingers develop distal swellings known as tactile pads, which are what create patterns for fingerprints.
The structural development of the upper limbs and lower limbs are similar but with two exceptions: the development of the lower limb is approximately 1 to 2 days behind that of the upper limb, whereas the upper and lower limbs rotate in opposite directions. By the seventh gestational week, the upper limbs rotate 90° laterally, placing the extensor muscles on the lateral and posterior surface and the thumb laterally. On the other hand, the lower limbs rotate 90° medially, placing the extensor muscles on the anterior surface and the big toe medially.
Appendicular skeleton
While the external shape of the limbs becomes established, the bones of the limbs and girdles (with the exception of the clavicle) form by a two-step process: chondrification and endochondral ossification. In contrast, the clavicle is a membrane bone: it forms directly by intramembranous ossification. Chondrification involves the condensation and differentiation of mesenchymal cells into chondrocytes (cartilage cells). By the sixth gestational week, these chondrocytes differentiate into hyaline cartilage models, foreshadowing the prospective bones. While the process of forming these cartilage models is initiated, synovial joints form between the two chondrifying bone primordia at the interzone.
At the center of the cartilage model (diaphysis), primary ossification centers form where chondrocytes increase in size, calcify the matrix, and eventually die. Concurrently, blood vessels invade the diaphysis. This results in the recruitment of osteoblasts, the differentiation of certain invading cells into hematopoietic cells (blood cells of the bone marrow), and the restriction of proliferating chondrocytes towards the distal ends of the cartilage model (epiphyses). Endochondral ossification thus begins from these primary ossification centers at the diaphysis and proceeds toward the epiphyses. However, this process only starts by the end of the embryonic period.
At birth, the diaphysis of long bones is usually completely ossified, whereas the epiphyses are still cartilaginous. Only after birth, secondary ossification centers develop in the epiphyses, which will also undergo the same ossification and vascularization processes that took place in the diaphysis. However, a layer of epiphyseal cartilage plate, known as the growth plate, persists between the epiphyses and the diaphysis. Continued proliferation of the chondrocytes in the growth plate is what allows the diaphysis to lengthen and thus what maintains the growth of bones. Only at approximately 20 years of age are when the epiphyses and diaphysis fuse, indicating that skeletal growth is complete.
Skeletal muscle
Skeletal muscle is derived from the mesoderm. Recall that the paraxial mesoderm forms segmented series of tissue blocks on each side of the neural tube, the somites. Cells in the ventromedial part of the somite form the sclerotome. Cells in the dorsal part form the dermatome and two edges, the ventrolateral lip and the dorsomedial lip. Cells from these two edges migrate ventral to the dermatome and proliferate to form muscle cell precursors. Collectively, these structures form the dermomyotome. In turn, the dermomyotome will differentiate into dermatome cells forming the dermis of the back and the neck, and myotome cells forming the skeletal muscles.
Before developing into skeletal muscles, myotome cells first differentiate into myoblasts (embryonic muscle cells) through elongation of their nuclei and cell bodies. Myoblasts fuse to form elongated, multinucleated, and cylindrical muscle fibers. During or after fusion, myofilaments and myofibrils develop in the cytoplasm. As development continues, the muscle cells become invested with the external laminae, segregating them from the surrounding connective tissue. Fibroblasts form the epimysium and perimysium layers of the muscle, whereas the external lamina and reticular fibers form the endomysium. In limbs, myoblasts migrate to the limb buds and surround the primordial limb bones. The pattern of muscle formation is dictated by the same mesenchymal cells that give rise to the bones.
Clinical aspects
Skull malformations
Malformations of the skull include cranioschisis and craniosynostosis. Cranioschisis involves the failure of the cranial vault to form, thus exposing the brain tissue to amniotic fluid, resulting in anencephaly. Craniosynostosis involves the premature closure of one or more sutures of the skull. Premature closure of the sagittal suture can result in a long and narrow skull due to frontal and occipital expansions. Premature closure of the coronal suture can result in a short skull. As such, premature unilateral closure of sutures can result in an asymmetrical skull.
Vertebral malformations
Malformations of the vertebra include Klippel-Feil sequence and spina bifida. Klippel-Feil syndrome involves the fusion of cervical vertebrae, which results in reduced mobility, short neck, and low hairline. Spina bifida involves the failure of vertebral arches to fuse, thus generally exposing the spinal cord in the sacral region. In spina bifida occulta, there are minimal neurological deficits; the spinal cord is intact and is covered by skin. In spina bifida cystica, the meninges and/or the neural tissue protrude through the skin at the sacral region to form a cyst-like sac.
Rib malformations
Malformations of the ribs include accessory ribs and fused ribs. Accessory ribs are usually rudimentary and unilateral or bilateral; they develop from the costal processes of cervical or lumbar vertebrae. Cervical ribs are usually attached to the seventh cervical vertebrae. Lumbar ribs are usually clinically insignificant, whereas cervical ribs may impinge on the brachial plexus or subclavian vessels, resulting in varying degrees of anesthesia of the upper limbs. Fused ribs occur posteriorly when two or more ribs arise from a single vertebra.
Sternal malformations
Malformations of the sternum include cleft sternum, pectus excavatum, and pectus carinatum. Cleft sternum is the result of a complete or partial midline fusion of the sternal bars. The heart and its major vessels are covered only by skin and soft tissue and thus are unprotected. Pectus excavatum (hollow chest) involves a concave depression of the sternum. Pectus carinatum (keel-shaped chest) involves an anterior projecting sternum. Both congenital deformities are often asymptomatic, but may impair cardiac and respiratory function depending on the severity.
Limb malformations
Malformations of the limbs vary greatly and can include defects in the entirety of the limb, the hand or the foot, and the digits. Malformation of the entire limbs include amelia, meromelia, phocomelia, and micromelia. Amelia (no limb) involves the complete absence of one or more limbs, whereas meromelia (part limb) involves a partial absence. Phocomelia (seal limb) involves the absence of long bones, resulting in rudimentary hands and feet attached to the trunk and pelvis. Micromelia involves abnormally small limbs.
Malformation of the hands and feet is known as cleft hand andcleft foot, which consist of an abnormal cleft between the second and fourth metacarpal or metatarsal bones and soft tissues. The third phalangeal and metacarpal or metatarsal bones are almost always absent, resulting in the possible fusion of the adjacent digits.
Malformations of digits include brachydactyly, syndactyly, polydactyly, and ectrodactyly. Brachydactyly involves shortened digits. Syndactyly involves the fusion of two or more digits. Polydactyly involves the presence of extra digits. Ectrodactyly involves the absence of a digit.
Skeletal muscle malformations
Malformations of skeletal muscle can result in certain conditions such as Poland sequence, prune belly syndrome and muscular dystrophy. Poland sequence involves the absence of the pectoralis minor, partial absence of the pectoralis major, the absence or displacement of the nipple and areola, and the accompanying presence of digital defects. Prune belly syndrome involves the partial or complete absence of abdominal muscles; this results with a very thin abdominal wall, making the internal organs visible and easy to palpate. Muscular dystrophy involves a group of inherited muscle diseases that cause progressive muscular atrophy and weakness.
Sources
All content published on Kenhub is reviewed by medical and anatomy experts. The information we provide is grounded on academic literature and peer-reviewed research. Kenhub does not provide medical advice. You can learn more about our content creation and review standards by reading our content quality guidelines.
References
G.C. Schoenwolf, S.B. Bleyl, P.R. Braeur & P.H. Francis-West: Larsen’s Human Embryology, 5th edition, Churchill Livingstone (2015), p. 172-196, 501-523
K.L. Moore, T.V.N. Persaud & M.G. Torchia: The Developing Human: Clinically Oriented Embryology, 10th edition, Elsevier (2016), p. 355-378
T.W. Sadler: Langman’s Medical Embryology, 12th edition, Wolters Kluwer, Lippincott Williams & Wilkins (2012), p. 133-161
Illustrators:
Skull (ventral view) - Yousun Koh
Skull (cross-sectional view) - Paul Kim
Lamina of the vertebral arch (cranial view) - Liene Znotina
Transverse process (cranial view) - Liene Znotina
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3702 | https://www.youtube.com/watch?v=LVpatsvV48I | DMID1843 Thermofluids || Momentum Flow Analysis : Fire Hose Nozzle || S1_Wan_Momentum
Amsyar Ismail
1 subscribers
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Description
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Posted: 18 Jun 2025
Unpack the physics of fire hoses! This video breaks down the momentum flow analysis of a fire hose nozzle, a key thermofluids concept. Discover how high-velocity water jets create a powerful recoil force (around 89.6 N!) , and learn about nozzle design, common issues like hose kinking , and innovative solutions for firefighter safety and performance.
Thermofluids #FluidMechanics #Firefighting #FireHose #Engineering
External Video Credit:
Portions of this video feature content from "How an Akron nozzle works" by Mike Feldon and used for educational purposes. Watch the original here:
Transcript:
Hi and good day everyone. We are from session one and group one also known as the group of momentum. Uh cuz today we're going to present our assignment two for DMID thermoflates and our topic is momentum flow analysis. So first of all our group consists of five members. I'm Darian and my teammates are Amara, Mozamin, Hamiro and also one. Moving on, we've chosen this fire hose nozzle as a application which is a critical tool in firefighting. Specifically, we studied the Acron brass adjustable nozzle which is 35 cm long with inlet and outlet diameters of 4.5 cm and 2.5 cm respectively. It is designed efficiently to convert pressurized water into a high-speed jet. The sudden change in momentum generates a strong recoil force that firefighters must be able to control. These nozzles are used across municipal, industrial, and military sectors and typically caused between RM500 to 2000. We chose this system because it clearly demonstrates fluid mechanics principle in real world context. Thank you. Thank you Dian. Now I'll be explaining the mechanism behind the fire host nozzle and how it relates to momentum flow analysis. Firstly the working principle. The fire hose operates based on the conservation of momentum. Pressurized water flows from a hydrant or pump into the hose and then exits through a narrow nozzle at a much higher speed. Due to the continuity equation, when the nozzle gets narrower, the water speeds up. This fast moving jet helps deliver water over long distances. But according to Newton's third law, this creates an opposite reaction force, a recoil, which pushes back on the firefighter. Now, let's look at the working fluid. In this case, we're dealing with water, which has a density of 1,00 kg per cubic meter and a viscosity of about 0.001 pascal seconds. The speed of sound in water is around 1,500 m/s, which we use later when calculating the Mac number. Next, we apply the momentum flow equation, which actually tells us how to calculate the recoil force. We first find the mass flow rate which actually depends on the water's density, the cross-sectional area of the outlet and the velocity of the water exiting the nozzle. Let's go through a sample calculation. We assume that the inlet diameter is 4.5 cm which gives an area of about 0.159 square m. The outlet diameter is smaller 2.5 cm which results in a smaller area. We also assume that the water enters the nozzle at 5 m/s. Because the outlet is narrower, the water speeds up. Using the ratio of the inlet and outlet areas, we find that the exit velocity increases to around 16.22 m/s. Then by multiplying the water's density, the outlet area, and the outlet velocity, we calculate the mass flow rate to be approximately 7.95 kg/ second. Now to find the reaction force, we take that mass flow rate and multiply by the difference in velocity between the outlet and the inlet. This actually gives us a recoil force about 89.6 newtons, which is roughly 90 newtons pushing back on the firefighter. Next, we are going to apply Bernoli's principle to understand how the pressure changes as the water speeds up. It shows that the pressure at the inlet is higher than at the outlet. After plugging in the numbers, we calculate a pressure drop of around 119 kilopascals. Meaning a large amount of energy is converted from pressure to velocity. Now let's look at some dimensionless numbers used to describe the type of flow. First, the Reox number. This compares inertial forces to viscous forces. Our value comes out to over 400,000 which confirm that the flow is turbulent. Next the MAC number which compares the speed of the flow to the speed of sound with a flow velocity of 16.22 and a speed of sound of,500. The MAC number is about 0.011. Since this number is much lower than 0.3, we know the flow is incompressible which is typical for liquids like water. Lastly, the FR number. This actually compares inertia to gravity. Using the velocity and representative length, we calculate the FR number to be around 10.3, which tells us the flow is initiated. Meaning gravity has little effect on the motion. In summary, the fire hose nozzle clearly shows how fluid mechanics applies in real life from velocity changes and pressure loss to force generation and flow behavior. Encounter problems. Although the nozzle is effective, there are several real world problem during use. First, the recall force we just calculate almost 90 newtons can distabilize the firefighter without proper stand or support. This may lead to injured or poor aim. Second, horse can bend especially when quickly deployed. This disturb flow, reduce pressure and cause turbulence inside the horse. Third, during extended use, especially under high pressure, the nosal can experience localized overheating. Although it can degrade performance. Lastly, holding a high pressure noser for long period cause muscle fatigue reducing the effectiveness and safety of the firefighter. To improve safety and efficiency, we propose upgrade. First we use counter recoil stabilizer like tripod or recoil absorption mount. This reduce the force on the user so the user not get dragged back. Second install flow rate regulator to maintain consistent output even when supply pressure varies. So we know that the flow rate of the water is still same even though we use different shape different nozzle. Third asend the design with agonomic pet grip to reduce risk fat fatigue and improve manual variability. So when we grip the host so it we can easily to control it. And finally consider smart nozzle to keep the system with sensors to monitor pressure, temperature and flow rate in real time. This can aid both training and life operations. That's in conclusion, the fire host nosal is a powerful and practical demonstration of momentum flow analysis. This study of a fire hose nosle under the lens of momentum flow analysis demonstrates the practical application of fluid mechanics principle through calculation and theoretical understanding. It is clear how high velocity fluid flow can generate substantial reaction force. Addressing this through design improvement ensure safety, performance and reliability in firefighting operation. Understanding this mission allow us to design better system that are safer, more efficient and easy to use under extreme condition. Thank you for your attention. Now a video will be play to educational educational purpose. [Music] You've probably seen it before, an outofcrol fire hose and someone trying to get a hold of it. Some parents may say that it's similar to what they have to deal with when their kids are on a sugar high. But the incredible power seen in a loose fire hose is due to the high velocity and flow rate of the water exiting the nozzle. It's the velocity that allows the stream of water to travel a long distance so the firefighters can maintain a safe distance from the fire. And most of us know that using more water at higher flow rates will be faster at putting a fire out. There are two factors that affect the velocity and flow rate from the fire nozzle. The pressure of the water entering and the nozzle design itself. Akran Brass is a company that designs a variety of equipment used in firefighting, including nozzles. We spoke with Ajit Singh, a senior engineer there to understand the operation and design of the assault nozzle. Ajit explained that some pass nozzles were designed to ensure a constant pressure water flow no matter how much water was being pumped through. That nozzle design is called automatic. They have a spring-loaded baffle mechanism at the exit to restrict the water flow. The spring allows the baffle to open more if there's an increase in water supply to the unit, maintaining the pressure while allowing more water to flow through. Their main advantage is that the adjustable opening provides a steady stream flow pattern with varying water supply conditions. But fire crews encounter different water flow conditions at each fire scene. A traditional automatic nozzle may shoot a water spray that looks good, but could severely restrict the amount of water flow in case of a pressure drop. In this case, the amount of water in the spray may actually be very different, leading to some risk for the firefighter. That's why Akran brass designers came up with a fixed baffle design. This ensures that the controlling gap at the baffle remains constant and a pressure drop will not cause a severe decrease in the amount of water flow to allow a fire crew to use the nozzle for a variety of locations. The baffle is interchangeable so the fire crews can match the nozzle to their desired water flow to get an optimum stream quality. Ajet and his team used Creodirect modeling to quickly develop different geometries for their baffle designs. They could then import the designs into a computational fluid dynamics program to simulate the flow and determine which ones provided the best stream quality. This saved them a lot of time and expense compared to creating many prototypes and testing each one. Modeling also helped in ensuring that the unit was robust enough to handle water pressure of up to 250 PSI and flow rates of up to 350 gall per minute. This while keeping the unit light and easy to operate since some fires can take hours to extinguish. You can see Creole modeling in action. It's the same software that Akran Brass used to design the assault fire nozzle seen in this episode. Just go to ptc.com/go/modelingpe. If you like the show, please give it a like on Facebook, subscribe on YouTube, or give us a rating on iTunes. We'll see you next week for more great design engineering. [Music] |
3703 | https://www.chemteam.info/Bonding/Electroneg-Bond-Polarity.html | Electronegativity
Electronegativity: Classifying Bond Type
Return to Bonding Menu
The modern definition of electronegativity is due to Linus Pauling. It is:
The power of an atom in a molecule to attract electrons to itself.
Pauling was able to develop a numerical scale of electronegativities. Below are ten common elements with their values. Your textbook should have a more complete list.
F 4.0 C 2.5
O 3.5 S 2.5
Cl 3.0 H 2.1
N 3.0 Na 0.9
Br 2.8 K 0.8
When you examine a periodic table, you will find that (excluding the noble gases) the electronegativity values tend to increase as you go to the right and up. The reverse statement is that the values tend to decrease going down and to the left. This pattern will help when you are asked to put several bonds in order from most to least ionic without using the values themselves.
Electronegativity values are useful in determining if a bond is to be classified as nonpolar covalent, polar covalent or ionic.
What you should do is look only at the two atoms in a given bond. Calculate the difference between their electronegativity values. Only the absolute difference is important.
I. Nonpolar Covalent: This type of bond occurs when there is equal sharing (between the two atoms) of the electrons in the bond. Molecules such as Cl 2, H 2 and F 2 are the usual examples.
Textbooks typically use a maximum difference of 0.2 - 0.5 to indicate nonpolar covalent. Since textbooks vary, make sure to check with your teacher for the value he/she wants. The ChemTeam will use 0.5.
One interesting example molecule is CS 2. This molecule has nonpolar bonds. Sometimes a teacher will only use diatomics as examples in lecture and then spring CS 2 as a test question. Since the electronegativities of C and S are both 2.5, you have a nonpolar bond.
II. Polar Covalent: This type of bond occurs when there is unequal sharing (between the two atoms) of the electrons in the bond. Molecules such as NH 3 and H 2 O are the usual examples.
The typical rule is that bonds with an electronegativity difference less than 1.6 are considered polar. (Some textbooks or web sites use 1.7.) Obviously there is a wide range in bond polarity, with the difference in a C-Cl bond being 0.5 -- considered just barely polar -- to the difference the H-O bonds in water being 1.4 and in H-F the difference is 1.9. This last example is about as polar as a bond can get.
III. Ionic: This type of bond occurs when there is complete transfer (between the two atoms) of the electrons in the bond. Substances such as NaCl and MgCl 2 are the usual examples.
The rule is that when the electronegativity difference is greater than 2.0, the bond is considered ionic.
So, let's review the rules:
If the electronegativity difference (usually called ΔEN) is less than 0.5, then the bond is nonpolar covalent.
If the ΔEN is between 0.5 and 1.6, the bond is considered polar covalent
If the ΔEN is greater than 2.0, then the bond is ionic.
That, of course, leaves us with a problem. What about the gap between 1.6 and 2.0? So, rule #4 is:
If the ΔEN is between 1.6 and 2.0 and if a metal is involved, then the bond is considered ionic. If only nonmetals are involved, the bond is considered polar covalent.
Here is an example: Sodium bromide (formula = NaBr; EN Na = 0.9, EN Br = 2.8) has a ΔEN = 1.9. Hydrogen fluoride (formula = HF; EN H = 2.1, EN F = 4.0) has the same ΔEN. We use rule #4 to decide that NaBr has ionic bonds and that HF has a polar covalent bond in each HF molecule.
As you might expect, NaBr and HF are very different substances. NaBr exhibits the classic "lattice structure" of ionic substances whereas HF is a gas a room temperature.
A warning: rule #4 may not exist in your textbook. Often, the 1.6 value is used and the 1.6-2.0 range is lumped into the ionic category.
Return to Bonding Menu |
3704 | https://www.expii.com/t/multiplication-and-division-properties-of-radicals-4566 | Expii
Multiplication and Division Properties of Radicals - Expii
Multiplying or dividing radicals lets us combine two radicals. We can multiply the two numbers then put the product under one radical.
Explanations (4)
Caroline K
Text
6
Multiplication and Division Properties of Radicals
The Multiplication and Division Properties of Radicals can help you simplify expressions!
Image by Caroline Kulczycky
We can have radicals of any integer root, known as the nth root. Most problems will have n=2 or n=3. In other words, you will mostly be dealing with square roots (√9 or 2√9) and cube roots (3√9).
Let's do an example to practice!
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Related Lessons
Subtracting Radicals - Examples & Practice
Simplifying Radicals - Examples & Practice
Adding Radicals - Examples & Practice
Solve by Radical: xⁿ=c
View All Related Lessons
Zora Gilbert
Text
4
Combine the Terms Under One Radical
Unlike adding and subtracting radicals, we can multiply two radicals and change the number underneath. Here's the rule:
√a×√b=√a×b
Multiplying two separate radicals with numbers in them is the same as multiplying those numbers under one radical. For example:
√16×√9=√16×94×3=√14412=12
Although this example used perfect squares, no matter what expression is under the radical, either way we solve the problem, the answer comes out the same.
Does this property seem familiar? Maybe. This is kind of like the product rule for exponents.
The Steps
Sometimes it's easier to keep difficult numbers in radical form. In this case, you can:
Multiply the numbers from within the radicals together
Place the product under one radical symbol.
If the radicals were already multiplied by something, multiply those together and place them in front of the radical.
Step 1: Multiply the numbers from within the radicals together
Let's work with the expression 2√7×3√13. We're going to multiply each part of each term together separately, then combine them again. Let's start with the numbers inside the radicals.
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Zora Gilbert
Text
3
Combine terms under one radical
Radical division follows the same rule as radical multiplication. Dividing two radicals means we can instead divide the two numbers under one radical. For a≥0 and b>0:
√a√b=√ab
So, for example, we could evaluate the expression √64√16 in two ways.
√64√16=√641684=√42=2
The left side of each equation shows that you can simplify the radicals before dividing. The right side shows you can divide before you simplify the radicals.
It's always important to be mindful of possible factors in the radicand that can help simplify expressions quicker.
The steps
It's good to keep messy numbers within their radicals for as long as possible. To do that while dividing, follow these steps:
Divide the numbers from within the radicals.
Place the result (fraction or not) under one radical symbol.
If the radicals were already multiplied by something (in other words, if they had a multiplier as part of their term), divide those numbers and place them in front of the radical. If only one radical has a multiplier, pretend the other is multiplied by 1.
Let's try it with the expression, √223√11.
Step 1: Divide the numbers from within the radicals
The numbers in this expression won't have nice square roots at all. We're going to try to keep everything within the radicals for as long as possible.
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Alex Federspiel
Video
1
(Video) Multiplying and Dividing Radical Terms
This video by mathgirl915 shows you how to multiply and divide radical terms.
Summary
When multiplying radicals, first make sure that the degree of the radical is the same. Then, unlike addition and subtraction, the terms underneath the radical do not need to be the same! You can multiply the outer numbers, then put the product of the inner radical terms underneath the radical.
Suppose we want to multiply two radicals: √3×√12. All we have to do is multiply the two numbers and put them under one radical.
√3×√12=√3×12=√36=6
Let's work through one that has numbers both inside and outside a radical.
8√8×6√6
We're going to multiply each part separately.
8×8=48√8×√6=√48
We need to simplify √48
√48√16×34√3
Now we combine these two sections. Anything outside of the radical needs to be multiplied.
48×4√3192√3
You've reached the end
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3705 | https://stackoverflow.com/questions/75333224/how-to-implement-a-wrap-around-index | algorithm - How to implement a wrap around index? - Stack Overflow
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How to implement a wrap around index?
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Asked 2 years, 7 months ago
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I want a wrap around index like 1232123...., and the frame size is 3. How to implement it? Does it has a term?
rust
for i in 1..100 {
let idx = loop_index(i);
print!("{} ", idx);
}
Expected output for frame 3:
1 2 3 2 1 2 3 2 1...
Expected output for frame 4:
1 2 3 4 3 2 1 2 3 4 3 2 1...
algorithm
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asked Feb 3, 2023 at 8:31
PurkylinPurkylin
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1 Can you think of how the modulo might help you achieve this?Nelewout –Nelewout 2023-02-03 08:33:38 +00:00 Commented Feb 3, 2023 at 8:33
Modulo can get a loop like 1 2 3 1 2 3 Purkylin –Purkylin 2023-02-03 08:36:22 +00:00 Commented Feb 3, 2023 at 8:36
Cound up and than count down and than repeat the sequence every 2n-2 indexes,MrSmith42 –MrSmith42 2023-02-03 08:37:06 +00:00 Commented Feb 3, 2023 at 8:37
1 It's not called wrap-around (that would just be modulo). I've seen it referred to as ping-pong but there probably isn't an official term.Thomas –Thomas 2023-02-03 08:50:57 +00:00 Commented Feb 3, 2023 at 8:50
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For a size of 3, notice that the sequence 1232 has length 4, and then it repeats. In general, for size n, the length is 2(n-1). If we take the modulo i % (2(n-1)), the task becomes simpler: turn a sequence 0123..(2(n-1)-1) into 123..(n-1)n(n-1)..321. And this can be done using abs and basic arithmetic:
n = 3
r = 2 (n - 1)
for i in range(20):
print(n - abs(n - (i % r)) - 1))
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edited Feb 3, 2023 at 9:50
answered Feb 3, 2023 at 8:50
ThomasThomas
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When you reach the top number, you start decreasing; when you reach the bottom number, you start increasing.
Don't change direction until you reach the top or bottom number.
```python
def count_up_and_down(bottom, top, length):
assert(bottom < top)
direction = +1
x = bottom
for _ in range(length):
yield x
direction = -1 if x == top else +1 if x == bottom else direction
x = x + direction
for i in count_up_and_down(1, 4, 10):
print(i, end=' ')
1 2 3 4 3 2 1 2 3 4
```
Alternatively, combining two ranges with itertools:
```python
from itertools import chain, cycle, islice
def count_up_and_down(bottom, top, length):
return islice(cycle(chain(range(bottom, top), range(top, bottom, -1))), length)
for i in count_up_and_down(1, 4, 10):
print(i, end=' ')
1 2 3 4 3 2 1 2 3 4
```
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edited Feb 3, 2023 at 9:00
answered Feb 3, 2023 at 8:46
StefStef
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3706 | https://nntdm.net/papers/nntdm-29/NNTDM-29-2-241-259.pdf | Notes on Number Theory and Discrete Mathematics Print ISSN 1310–5132, Online ISSN 2367–8275 2023, Volume 29, Number 2, 241–259 DOI: 10.7546/nntdm.2023.29.2.241-259 Representations of positive integers as sums of arithmetic progressions, I Chungwu Ho1, Tian-Xiao He2 and Peter J.-S. Shiue3 1 Department of Mathematics, Southern Illinois University at Edwardsville and Evergreen Valley College Current Address: 1545 Laurelwood Crossing Ter., San Jose, California 95138, United States e-mail: cho@siue.edu 2 Department of Mathematics, Illinois Wesleyan University Bloomington, Illinois, United States e-mail: the@iwu.edu 3 Department of Mathematical Sciences, University of Nevada, Las Vegas Las Vegas, Nevada, United States e-mail: shiue@unlv.nevada.edu Received: 21 October 2022 Revised: 10 March 2023 Accepted: 26 April 2023 Online First: 27 April 2023 Abstract: This is the first part of a two-part paper. Our paper was motivated by two classical papers: A paper of Sir Charles Wheatstone published in 1844 on representing certain powers of an integer as sums of arithmetic progressions and a paper of J. J. Sylvester published in 1882 for determining the number of ways a positive integer can be represented as the sum of a sequence of consecutive integers. There have been many attempts to extend Sylvester Theorem to the number of representations for an integer as the sums of different types of sequences, including sums of certain arithmetic progressions. In this part of the paper, we will make yet one more extension: We will describe a procedure for computing the number of ways a positive integer can be represented as the sums of all possible arithmetic progressions, together with an example to illustrate how this procedure can be carried out. In the process of doing this, we will also give an extension of Wheatstone’s work. In the second part of the paper, we will continue on the problems initiated Copyright © 2023 by the Authors. This is an Open Access paper distributed under the terms and conditions of the Creative Commons Attribution 4.0 International License (CC BY 4.0). by Wheatstone by studying certain relationships among the representations for different powers of an integer as sums of arithmetic progressions.
Keywords: Arithmetic progressions, Sylvester theorem, Complementary factors.
2020 Mathematics Subject Classification: 11B25, 11A41.
1 Introduction Motivated by the well-known fact that the square of a natural number can be represented as the sum of consecutive odd integers, Sir Charles Wheatstone, a fellow of the Royal Society of London and the inventor of the Wheatstone’s bridge for measuring electrical resistance, published a paper in 1844, showing various ways of representing the power 𝑁𝑘of a positive integer 𝑁as sums of arithmetic progressions for 𝑘≤4. In 1882, J. J. Sylvester stated in [11, Section 17, pp. 265-266] that the number of ways a positive integer 𝑁may be represented as the sum of consecutive positive integers is equal to the number of odd factors of 𝑁that exceed 1 (see also [6, Vol. 2, Chapter 3, p. 139]). This result is known as the Sylvester Theorem (see ).
A special consequence of this theorem is that a power of 2 cannot be represented as the sum of any sequence of consecutive positive integers. In this two-part paper, we will deal with the problems initiated in both Wheatstone’s and Sylvester’s work in the same framework of arithmetic progressions, and extend both of their results.
Since the outset of the 20th century, many authors tried to extend Sylvester Theorem in different ways. For instance, Thomas E. Mason in his paper of 1912 , allowed the consecutive integers to include zero and negative terms. In 1930, Laurens E. Bush studied the problem of representing a positive integer as sums of arithmetic progressions of a given common difference, for both progressions of positive terms and for progressions including zero and negative terms (see ).
In particular, he extended Sylvester Theorem on the impossibility of representing a power of 2 as the sum of any arithmetic progressions of an odd common difference ( [4, the corollary on top of p. 356]). For more recent publications see [1–3,5,8], and . However, in all these extensions, the authors considered, as Bush did, the arithmetic progressions of a given common difference 𝑑. In Part I of this paper, we will try to build up a procedure for finding the number of ways a positive integer 𝑁can be represented as the sum of an arithmetic progression of positive terms of any common difference 𝑑≥1. Our results require somewhat different methods. We believe that our extension might be more in line with Sylvester’s original intention, since almost all the 80 pages of his paper were devoted to the problem of partitioning of positive integers. To investigate the ways a positive integer can be partitioned into terms of arithmetic progressions, we should consider arithmetic progressions of all possible common differences.
To prepare for our investigations, we will first determine in Section 2, for a given positive integer 𝑁, the values of 𝑟and 𝑑for which 𝑁can be represented as the sum of an arithmetic progression consisting of 𝑟terms that has 𝑑as its common difference (Theorem 2.1). With this theorem, we will extend Wheatstone’s work in Section 3 (see Theorems 3.1, 3.2 and 3.3) and lay the foundation for our general procedure for determining the number of ways a positive integer 𝑁 242 can be represented as the sums of different arithmetic progressions.The general procedure will be described in Section 5, and an example for illustrating the procedure will be given in Section 5.
In part II of this paper , we will continue on the problems initiated by Wheatstone, by studying certain relationships among the representations of different powers of an integer.
2 Generating the sum of an arithmetic progression As indicated in the introductory section, the problem of representing an integer as sums of arithmetic progressions has a long history. Some of our results in this section, though appear in a new form, may have already been covered by, or can be proved easily from, the existing publications. However, we will still give a self-contained account here since it would be easier for the reader if all the relevant facts are collected in one place.
It is also because the results of the existing publications are not stated in a form suitable for our purpose: We need to look for the sums of arithmetic progressions for which the parities of the number of the terms 𝑟and the common difference 𝑑are important. The existing publications are either too general or not going far enough.
In the following, by a representation of an integer 𝑁, we mean a representation of 𝑁as the sum of an arithmetic progression. We will let 𝑆(𝑎, 𝑟, 𝑑) to represent the sum of an arithmetic progression, beginning with the term 𝑎≥1, consisting of 𝑟terms, and with a common difference 𝑑. We will require 𝑑> 0 and 𝑟> 1 to rule out the trivial cases. We will call a pair of positive factors 𝑟and 𝑠of 𝑁complementary if 𝑟𝑠= 𝑁.
Theorem 2.1. Let 𝑁be a positive integer. The existence of a representation 𝑁= 𝑆(𝑎, 𝑟, 𝑑), depending on whether 𝑟|𝑁, can be characterized by exactly one of the following two cases: 1).
𝑟|𝑁.
In this case, either 𝑑is even or 𝑟is odd and 1 2(𝑟−1)𝑑< 𝑠, where 𝑠is the complementary factor of 𝑟in 𝑁. Conversely, if 𝑁= 𝑟𝑠for some integers 𝑟and 𝑠such that 𝑟> 1 and 1 2(𝑟−1)𝑑< 𝑠, then 𝑁= 𝑆(𝑎, 𝑟, 𝑑) for some integer 𝑎≥1. Furthermore, in this case, the first term 𝑎of the representation 𝑁= 𝑆(𝑎, 𝑟, 𝑑) is 𝑎= 𝑠−1 2(𝑟−1)𝑑.
2). 𝑟∤𝑁. In this case, 𝑑is odd and 𝑟is even. Write 𝑟= 2𝑟0, then 𝑟0|𝑁. Let 𝑠0 be the complementary factor of 𝑟0, then 𝑠0 is an odd integer with 𝑠0 > (2𝑟0 −1)𝑑. Conversely, if 𝑟0 and 𝑠0 are a pair of complementary factors of 𝑁satisfying the following conditions: a) 𝑟= 2𝑟0 does not divide 𝑁and b) 𝑠0 is an odd integer satisfying 𝑠0 > (2𝑟0 −1)𝑑, then 𝑁= 𝑆(𝑎, 𝑟, 𝑑) for some integer 𝑎≥1.
In this case, the first term of the representation 𝑁= 𝑆(𝑎, 𝑟, 𝑑) is 𝑎= 1 2[𝑠0 −(2𝑟0 −1)𝑑].
Proof. We first show that if 𝑁= 𝑆(𝑎, 𝑟, 𝑑) for some positive integers 𝑎, 𝑟, and 𝑑, then depending on whether 𝑟|𝑁or 𝑟∤𝑁, there are two possibilities as described in the theorem.
Suppose 𝑁= 𝑆(𝑎, 𝑟, 𝑑), i.e., 𝑁= 𝑎+ (𝑎+ 𝑑) + · · · + [𝑎+ (𝑟−1)𝑑] = 𝑟𝑎+ 1 2𝑟(𝑟−1)𝑑 = 𝑟[𝑎+ 1 2(𝑟−1)𝑑].
243 Note that 𝑟|𝑁if and only if [𝑎+ 1 2(𝑟−1)𝑑] is an integer. Since 𝑎is an integer, 𝑟|𝑁if and only if either 𝑟is odd or 𝑑is even, and 𝑟∤𝑁if and only if 𝑟is even and 𝑑is odd.
In the case when 𝑟is odd or 𝑑is even, the complementary factor 𝑠of 𝑟is given by 𝑠= [𝑎+ 1 2(𝑟−1)𝑑]. Since 𝑎≥1, we must have 1 2(𝑟−1)𝑑< 𝑠.
Conversely, let 𝑟> 1 be a factor of 𝑁with 𝑠as its complementary factor, then for any integer 𝑑≥1 such that 𝑎) either 𝑟is odd or 𝑑is even and 𝑏) 1 2(𝑟−1)𝑑< 𝑠. We will show that there exists a representation of 𝑁= 𝑆(𝑎, 𝑟, 𝑑) with the given integers 𝑟and 𝑑where 𝑎is some integer greater than or equal to 1. Since by the condition 𝑎), 1 2(𝑟−1)𝑑is a positive integer. By condition 𝑏), 𝑎= 𝑠−1 2(𝑟−1)𝑑is also a positive integer. Then 𝑎+ (𝑎+ 𝑑) + (𝑎+ 2𝑑) + · · · + (𝑎+ (𝑟−1)𝑑) = 𝑟𝑎+ 𝑟(𝑟−1) 2 𝑑 = 𝑟[𝑎+ 1 2(𝑟−1)𝑑] = 𝑟[𝑠−1 2(𝑟−1)𝑑+ 1 2(𝑟−1)𝑑] = 𝑟𝑠 = 𝑁.
Thus, 𝑁= 𝑆(𝑎, 𝑟, 𝑑). In this case, the first term of 𝑆(𝑎, 𝑟, 𝑑) is 𝑎= 𝑠−1 2(𝑟−1)𝑑.
Now, consider the case when 𝑁= 𝑆(𝑎, 𝑟, 𝑑) for some positive integers 𝑎, 𝑟and 𝑑such that 𝑟∤𝑁. From the proof above, 𝑟is even and 𝑑is odd. In this case, let 𝑟= 2𝑟0 for some positive integer 𝑟0, and let 𝑀and 𝑀+ 𝑑be the two middle terms in the sum 𝑆(𝑎, 𝑟, 𝑑). These two middle terms add up to 2𝑀+ 𝑑. The two terms closest to these two middle terms, 𝑀−𝑑and 𝑀+ 2𝑑, again have a sum of 2𝑀+ 𝑑. If we collect all the terms of the sum in pairs, one preceding the terms already considered and one succeeding them, there are 𝑟0 such pairs, each of which has a sum of 2𝑀+ 𝑑. Thus, 𝑁= 𝑟0(2𝑀+ 𝑑), and hence, 𝑟0|𝑁, with its complementary factor 𝑠0 = (2𝑀+ 𝑑). From this, we conclude that 𝑠0 is odd and 𝑀= 1 2(𝑠0 −𝑑). Since 𝑀is the 𝑟0th term in the progression, 𝑀= 𝑎+ (𝑟0 −1)𝑑or 𝑎= 𝑀−(𝑟0 −1)𝑑. Since 𝑎> 0, we must have 𝑀> (𝑟0 −1)𝑑. Now substitute 1 2(𝑠0 −𝑑) for 𝑀in this inequality and also in the equality for 𝑎 and simplify, we will get both (2𝑟0 −1)𝑑< 𝑠0 and 𝑎= 1 2[𝑠0 −(2𝑟0 −1)𝑑].
Conversely, let positive integers 𝑟= 2𝑟0 and 𝑑be given such that 𝑑is odd and 𝑁= 𝑟0𝑠0, where 𝑠0 is an odd integer and (2𝑟0 −1)𝑑< 𝑠0. Note that in this case, 𝑟∤𝑁. This is because that since 𝑁= 𝑟0𝑠0 = 𝑟(𝑠0/2) but 𝑠0 is an odd integer. Let 𝑎= 1 2[𝑠0 −(2𝑟0 −1)𝑑]. Note that 𝑎is a positive integer and then 𝑎+ (𝑎+ 𝑑) + (𝑎+ 2𝑑) + · · · + (𝑎+ (𝑟−1)𝑑) = 𝑟𝑎+ 𝑟(𝑟−1) 2 𝑑 = 𝑟 [︀ 𝑎+ 1 2(𝑟−1)𝑑 ]︀ = 2𝑟0 [︀1 2(𝑠0 −(2𝑟0 −1)𝑑) + 1 2(2𝑟0 −1)𝑑 ]︀ = 𝑟0𝑠0 = 𝑁.
Again, 𝑁= 𝑆(𝑎, 𝑟, 𝑑), and in this case, 𝑎= 1 2[𝑠0 −(2𝑟0 −1)𝑑].
244 As a corollary of this theorem, we will give a stronger form of Bush’s extension of the forbidden representations for the powers of 2, not only for odd common differences, but also for odd numbers of terms. This result will be extended further in our Corollary 3.1, which will impose more restrictions on the arithmetic progressions whose sums can be a power of 2.
Corollary 2.1. None of the representations for a power 2𝑘, with 𝑘≥1, can have an odd common difference d or can consists of an odd number of terms.
Proof. Consider a power 2𝑘with 𝑘≥1. If 2𝑘= 𝑆(𝑎, 𝑟, 𝑑), we claim that 𝑟and 𝑑must both be even. First note that 𝑟cannot be an odd integer. This is because by Theorem 2.1, if 𝑟is odd, then 𝑟|2𝑘. Since we do not allow the trivial case of 𝑟= 1, 𝑟= 2𝑗for some positive integer 𝑗, and thus, 𝑟is even. Say, 𝑟= 2𝑟0. Now, if 𝑑is odd, by the proof of Theorem 2.1, 2𝑘= 𝑟0𝑠0, for some odd integer 𝑠0 > (2𝑟0 −1)𝑑. This is impossible. We conclude that 𝑑is even.
3 Representing the power of a prime as the sum of an arithmetic progression For a positive integer 𝑁, we will let #(𝑁, 𝑟) be the number of ways that 𝑁may have a representation consisting of 𝑟terms, and let #(𝑁) be the sum of #(𝑁, 𝑟) for all possible 𝑟> 1.
In this section, we will first show how Wheatstone’s work in can be extended. We will then compute, for a given power 𝑝𝑘of a prime 𝑝, the possible values of 𝑟for which 𝑝𝑘can be represented as the sum of an arithmetic progression consisting of 𝑟terms. We will then determine both #(𝑝𝑘, 𝑟) for all such 𝑟and #(𝑝𝑘). These computations can also be considered as extensions of Wheatstone’s work, and they will also be used in building our general procedures.
The main result in is the observation that for a positive integer 𝑁, any power 𝑁𝑘, 𝑘≥2, can always be represented as the sum of an arithmetic progression consisting of 𝑁terms. However, instead a proof, Wheatstone only showed some examples for 𝑘≤4: with 2 examples for 𝑘= 2 and 4, and 3 examples for 𝑘= 3 (there is a second example for 𝑘= 2, but it involves fractions).
No mentioning of the number of ways this can be done. In our next theorem, we will prove Wheatstone’s Theorem, together with a computation for the number of ways this can be done and for the construction method for these representations.
Theorem 3.1. Let 𝑁and 𝑘be any two integers ≥2. 𝑁𝑘always has a representation consisting of 𝑁terms. In fact, #(𝑁𝑘, 𝑁) = ⌊︁ 2𝑁𝑘−1 𝑁−1 ⌋︁ . Each integer 1, 2, . . . , ⌊︁ 2𝑁𝑘−1 𝑁−1 ⌋︁ gives rise to a value of 𝑑for such an expression, and for each such 𝑑, the arithmetic progression can be constructed by letting the initial term 𝑎be 𝑎= 𝑁𝑘−1 −1 2(𝑁−1)𝑑.
Proof. Let 𝑁and 𝑘be two integers specified by the theorem. Since 𝑁|𝑁𝑘, the cofactor for 𝑁is 𝑠= 𝑁𝑘−1. By the equivalence 1 2(𝑁−1)𝑑< 𝑁𝑘−1 ⇔𝑑< 2𝑁𝑘−1 𝑁−1 and the fact that 𝑑is an integer, we may claim that, by Theorem 2.1, there exists a representation of 𝑁𝐾consisting of 𝑁terms if and only if 𝑑≤ ⌊︁ 2𝑁𝑘−1 𝑁−1 ⌋︁ . But for an 𝑁and 𝑘≥2, it is certainly 245 true that ⌊︁ 2𝑁𝑘−1 𝑁−1 ⌋︁ ≥1. Thus, for each 𝑑= 1, 2, . . . , ⌊︁ 2𝑁𝑘−1 𝑁−1 ⌋︁ , there is a representation of 𝑁𝑘of 𝑁terms. By Theorem 2.1 again, for each such 𝑑, the initial term of the representation is given by 𝑎= 𝑁𝑘−1 −1 2(𝑁−1)𝑑. The arithmetic progression can easily be constructed by keeping adding 𝑑to each successive term beginning with 𝑎until all the 𝑁terms of the progression are obtained.
We now return to our computations of #(𝑝𝑘, 𝑟) and #(𝑝𝑘). First note that 1 and 2 cannot be represented as a sum of any arithmetic progression consisting of more than one term, and thus, #(𝑁) = 0 for 𝑁= 1 or 2. The following lemma shows that any integer 𝑁≥3 can always be represented as such a sum.
Lemma 3.1. Any integer 𝑁≥3 always has a representation consisting of two terms, and #(𝑁, 2) = ⌊︀1 2(𝑁−1) ⌋︀ . Furthermore, the difference 𝑑between these two terms is of the same parity as that of 𝑁: If 𝑁is odd, 𝑑may take the values of 1, 3, . . . , 𝑁−2. If 𝑁is even, 𝑑may take the values of 2, 4, . . . , 𝑁−2.
Proof. Let an integer 𝑁≥3 be given. 𝑁has a representation consisting of two terms if and only if 𝑁= 𝑎+ (𝑎+ 𝑑), where 𝑎is the first term of the progression and 𝑑is the common difference.
This condition is equivalent to 𝑑= 𝑁−2𝑎for some integer 𝑎≥1. From this, we conclude that a necessary and sufficient condition for 𝑑to be an integer ≥1 is 𝑎< 𝑁/2. Thus, 𝑎can be any integer 1, 2, . . . , ⌊︀1 2(𝑁−1) ⌋︀ and 𝑑= 𝑁−2𝑎. Consequently, #(𝑁, 2) = ⌊︀1 2(𝑁−1) ⌋︀ .
Since 𝑑= 𝑁−2𝑎, the common difference 𝑑is of the same parity as that of 𝑁. Thus, if 𝑁is odd, 𝑑may take the values of 1, 3, . . . , 𝑁−2. If 𝑁is even, 𝑑may take the values of 2, 4, . . . , 𝑁−2.
We might observe in passing that since we do not allow 𝑟= 1, for any positive integer 𝑁≥3, the shortest length 𝑟for 𝑁= 𝑆(𝑎, 𝑟, 𝑑) is 𝑟= 2. We now study the problem of determining the number of ways the power of a prime can be represented as sums of arithmetic progressions. We now begin with the prime 𝑝= 2. As noted above that 2 cannot be represented as a sum of any arithmetic progression consisting of more than one term. We now consider the representations 2𝑘= 𝑆(𝑎, 𝑟, 𝑑) for 𝑘> 1.
Theorem 3.2. For 𝑘> 1, 2𝑘= 𝑆(𝑎, 𝑟, 𝑑) if and only if 𝑟= 2𝑗for some integer 𝑗, with 1 ≤𝑗≤⌊𝑘/2⌋. For each such 𝑗, let 𝑙𝑗be the integer such that 0 ≤𝑙𝑗< 𝑗and 𝑘≡𝑙𝑗(mod 𝑗).
There are 2𝑘−𝑗−2𝑙𝑗 2𝑗−1 different ways for 2𝑘= 𝑆(𝑎, 2𝑗, 𝑑). Each of these representations corresponds to an even integer 𝑑= 2𝑑0 with 𝑑0 being one of the integers 1, 2, . . . , 2𝑘−𝑗−2𝑙𝑗 2𝑗−1 , and the initial term of the representation for such a 𝑑0 is given by 𝑎= 2𝑘−𝑗−(2𝑗−1)𝑑0. The arithmetic progression can then be found by adding repeatedly 𝑑(= 2𝑑0) to 𝑎until we obtain all the 2𝑗terms of the progression. In particular, #(2𝑘) = ⌊𝑘/2⌋ ∑︁ 𝑗=1 2𝑘−𝑗−2𝑙𝑗 2𝑗−1 .
(1) Furthermore, the longest 𝑟in the representation of 2𝑘= 𝑆(𝑎, 𝑟, 𝑑) is for 𝑟= 2⌊𝑘/2⌋.
246 Proof. Consider a power 2𝑘with 𝑘≥2. If 2𝑘= 𝑆(𝑎, 𝑟, 𝑑), by Corollary 2.1, 𝑑must be even.
Write 𝑑= 2𝑑0 for some positive integer 𝑑0. By Theorem 2.1, 𝑟|2𝑘. Since 𝑟> 1, 𝑟= 2𝑗for some positive integer 𝑗, and 𝑠= 2𝑘−𝑗, where 𝑠is the complementary factor of 𝑟in 2𝑘. The condition 1 2(𝑟−1)𝑑< 𝑠now becomes (2𝑗−1)𝑑0 < 2𝑘−𝑗. In particular, (2𝑗−1) < 2𝑘−𝑗. This implies that 1 ≤𝑗≤⌊𝑘/2⌋. Now, let 𝑙𝑗be the integer such that 0 ≤𝑙𝑗< 𝑗and 𝑘≡𝑙𝑗(mod 𝑗), then 0 ≤𝑙𝑗< 𝑗≤⌊𝑘/2⌋. From (2𝑗−1)𝑑0 < 2𝑘−𝑗, we have 𝑑0 < 2𝑘−𝑗 2𝑗−1 = 2𝑘−𝑗−2𝑙𝑗+ 2𝑙𝑗 2𝑗−1 = 2𝑙𝑗 (︂2𝑘−𝑗−𝑙𝑗−1 2𝑗−1 )︂ + 2𝑙𝑗 2𝑗−1.
(2) We contend that the first term, 2𝑙𝑗 (︁ 2𝑘−𝑗−𝑙𝑗−1 2𝑗−1 )︁ , is an integer and the second term, 2𝑙𝑗 2𝑗−1, is less than or equal to 1. This is because that since 𝑘≡𝑙𝑗(mod 𝑗) and 𝑗≤⌊𝑘/2⌋, 𝑘−𝑗−𝑙𝑗is a positive multiple of 𝑗, and thus, 2𝑗−1 is a factor of 2𝑘−𝑗−𝑙𝑗−1. As for the term 2𝑙𝑗 2𝑗−1, we first consider the case that 𝑙𝑗= 0. In this case, 2𝑙𝑗 2𝑗−1 = 1 2𝑗−1 ≤1. Now, assume that 𝑙𝑗> 0 and 2𝑙𝑗> 1. Since 𝑗> 𝑙𝑗, 2𝑗−𝑙𝑗−1 ≥1 and 2𝑗−2𝑙𝑗= 2𝑙𝑗(2𝑗−𝑙𝑗−1) ≥2𝑙𝑗> 1, or, 2𝑗−1 > 2𝑙𝑗.
(3) Thus, 0 < 2𝑙𝑗 2𝑗−1 < 1 when 𝑙𝑗> 0. Now, since both 𝑑0 and 2𝑙𝑗 (︁ 2𝑘−𝑗−𝑙𝑗−1 2𝑗−1 )︁ are integers, and 0 < 2𝑙𝑗 2𝑗−1 ≤1, from the inequality (2), we may conclude that 𝑑0 ≤2𝑙𝑗 (︁ 2𝑘−𝑗−𝑙𝑗−1 2𝑗−1 )︁ = 2𝑘−𝑗−2𝑙𝑗 2𝑗−1 .
In fact, it is not difficult to show that 1 2(𝑟−1)𝑑< 𝑠⇔𝑑0 ≤2𝑘−𝑗−2𝑙𝑗 2𝑗−1 .
(4) According to Theorem 2.1, 2𝑘= 𝑆(𝑎, 𝑟, 𝑑) for some 𝑟= 2𝑗if and only if 𝑑0 is less than or equal to 2𝑘−𝑗−2𝑙𝑗 2𝑗−1 , and each of these values of 𝑑0 will give rise to a different representation of 2𝑘= 𝑆(𝑎, 2𝑗, 2𝑑0). Thus, there are exactly 2𝑘−𝑗−2𝑙𝑗 2𝑗−1 different ways to write 2𝑘= 𝑆(𝑎, 2𝑗, 𝑑) with this 𝑗, or #(2𝑘, 2𝑗) = 2𝑘−𝑗−2𝑙𝑗 2𝑗−1 for 1 ≤𝑗≤⌊𝑘/2⌋. The formula for the first term 𝑎of the progression follows from Theorem 2.1. Collecting #(2𝑘, 2𝑗) for all such 𝑗, we have the Formula (1). This finishes the proof.
Corollary 3.1. Let 𝑘≥2 be an integer. If 2𝑘= 𝑆(𝑎, 𝑟, 𝑑) for some arithmetic progression, then both 𝑟and 𝑑are even. Furthermore, all the term of 𝑆(𝑎, 𝑟, 𝑑) are of the same parity, and these terms are even if and only if the common difference 𝑑≡0 (𝑚𝑜𝑑4).
Proof. By Corollary 2.1, if 𝑘≥2 and 2𝑘= 𝑆(𝑎, 𝑟, 𝑑), then both 𝑟and 𝑑are even. All terms of an arithmetic progression with an even 𝑑must be of the same parity. Thus, if 2𝑘= 𝑆(𝑎, 𝑟, 𝑑), all the terms of 𝑆(𝑎, 𝑟, 𝑑) are all even or are all odd, depending on whether its first term 𝑎is even or odd.
But 𝑎= 2𝑘−𝑗−(2𝑗−1)𝑑0. Thus, 𝑎is even if and only if 𝑑0 is even. But 𝑑= 2𝑑0. Consequently, all the terms of 𝑆(𝑎, 𝑟, 𝑑) are even if and only if 𝑑≡0(𝑚𝑜𝑑4).
Remark 3.1. Our Theorem 3.2, together with Corollary 3.1, may be considered as an extension of Sylvester Theorem for the powers of 2 since they specify not only what kind of arithmetic progressions can have a sum which is a power of 2, but also how many of them can there be for a given power of 2. We now consider the case of 𝑝𝑘= 𝑆(𝑎, 𝑟, 𝑑), and our next theorem can be considered as an extension of Sylvester Theorem for powers of an odd prime.
247 Theorem 3.3. Let p be an odd prime and 𝑘an integer ≥1. Depending on whether 𝑟|𝑝𝑘or not, there are two types of representations for 𝑝𝑘= 𝑆(𝑎, 𝑟, 𝑑): 1. 𝑟|𝑝𝑘. The only way for 𝑝𝑘= 𝑆(𝑎, 𝑟, 𝑑) with 𝑟|𝑝𝑘is when 𝑘> 1 and 𝑟= 𝑝𝑗for some 𝑗 such that 1 ≤𝑗≤⌊𝑘/2⌋. For each such 𝑗, let 𝑙𝑗be the integer such that 0 ≤𝑙𝑗< 𝑗 and 𝑘≡𝑙𝑗(mod 𝑗). There are 2 (︁ 𝑝𝑘−𝑗−𝑝𝑙𝑗 𝑝𝑗−1 )︁ many ways for 𝑝𝑘= 𝑆(𝑎, 𝑝𝑗, 𝑑). Each of the integers 1, 2, 3, . . . , 2 (︁ 𝑝𝑘−𝑗−𝑝𝑙𝑗 𝑝𝑗−1 )︁ gives rise to a value of 𝑑for a different arithmetic progression whose sum equals to 𝑝𝑘, and the initial term 𝑎of the progression for this 𝑑is given by 𝑎= 𝑝𝑘−𝑗−1 2(𝑝𝑗−1)𝑑.
2. 𝑟∤𝑝𝑘. In this case, 𝑟= 2𝑝𝑗for some 𝑗with 0 ≤𝑗< ⌊𝑘/2⌋. Each odd integer ≤ ⌊︁ 𝑝𝑘−𝑗−1 2𝑝𝑗−1 ⌋︁ gives rise to a value of 𝑑for a distinct representation 𝑝𝑘= 𝑆(𝑎, 2𝑝𝑗, 𝑑). These are the only possible values for 𝑑for this 𝑟= 2𝑝𝑗. For each of these representations, the initial term 𝑎is given by 𝑎= 1 2(𝑝𝑘−𝑗−(2𝑝𝑗−1)𝑑). Thus, #(𝑝𝑘, 2𝑝𝑗) = ⌊︁ 1 2 (︁ 𝑝𝑘−𝑗−1 2𝑝𝑗−1 + 1 )︁⌋︁ , where 0 ≤𝑗< ⌊𝑘/2⌋.
Collecting all the terms above, we have #(𝑝𝑘) = ⌊︂1 2(𝑝𝑘−1) ⌋︂ + ⌊𝑘/2⌋ ∑︁ 𝑗=1 (︂ 2 (︂𝑝𝑘−𝑗−𝑝𝑙𝑗 𝑝𝑗−1 )︂ + ⌊︂1 2 (︂𝑝𝑘−𝑗−1 2𝑝𝑗−1 + 1 )︂⌋︂)︂ .
(5) where the summation is zero if 𝑘= 1.
Proof.
1. Consider a power 𝑝𝑘for an odd prime 𝑝and an integer 𝑟> 1 such that 𝑟|𝑝𝑘. Then 𝑟= 𝑝𝑗for some 𝑗≥1, we claim that if 𝑝𝑘= 𝑆(𝑎, 𝑟, 𝑑) for such an 𝑟, then 𝑘> 1 and 𝑟= 𝑝𝑗 for an integer 𝑗with 1 ≤𝑗≤⌊𝑘/2⌋. This is because that if 𝑠be the complementary factor of 𝑟in 𝑝𝑘, then 𝑠= 𝑝𝑘−𝑗, and by Theorem 2.1, 𝑠> 1 2(𝑟−1)𝑑. Thus, 2𝑝𝑘−𝑗> (𝑝𝑗−1), or 2𝑝𝑘−𝑗≥𝑝𝑗, or 𝑝𝑘−2𝑗≥1 2. Since 𝑝is an odd prime, we must have 2𝑗≤𝑘or 𝑗≤⌊𝑘/2⌋. In particular, 𝑘> 1 and 1 ≤𝑗≤⌊𝑘/2⌋.
Now, let 𝑗be an integer such that 1 ≤𝑗≤⌊𝑘/2⌋. For 𝑟= 𝑝𝑗and 𝑠= 𝑝𝑘−𝑗. 𝑝𝑘= 𝑆(𝑎, 𝑟, 𝑑) if and only if 1 2(𝑟−1)𝑑< 𝑠, or (𝑝𝑗−1)𝑑< 2𝑝𝑘−𝑗. Let 𝑙𝑗be the integer such that 0 ≤𝑙𝑗< 𝑗 and 𝑘≡𝑙𝑗(mod 𝑗). We have 𝑑< 2𝑝𝑘−𝑗 𝑝𝑗−1 = 2(𝑝𝑘−𝑗−𝑝𝑙𝑗+ 𝑝𝑙𝑗) 𝑝𝑗−1 = 2𝑝𝑙𝑗 (︂𝑝𝑘−𝑗−𝑙𝑗−1 𝑝𝑗−1 )︂ + 2𝑝𝑙𝑗 𝑝𝑗−1.
(6) As in the proof of Theorem 3.2, 2𝑝𝑙𝑗 (︁ 𝑝𝑘−𝑗−𝑙𝑗−1 𝑝𝑗−1 )︁ is an integer. We now show that 0 < 2𝑝𝑙𝑗 𝑝𝑗−1 ≤1. Since 𝑝is an odd prime and 0 ≤𝑙𝑗< 𝑗, we have 𝑝𝑗−2𝑝𝑙𝑗= 𝑝𝑙𝑗(𝑝𝑗−𝑙𝑗−2) ≥1 or 𝑝𝑗−1 ≥2𝑝𝑙𝑗. This shows that 2𝑝𝑙𝑗 𝑝𝑗−1 ≤1. From (6), we conclude that 𝑑≤2𝑝𝑙𝑗 (︁ 𝑝𝑘−𝑗−𝑙𝑗−1 𝑝𝑗−1 )︁ . It is not difficult to show that 𝑑≤2𝑝𝑙𝑗 (︁ 𝑝𝑘−𝑗−𝑙𝑗−1 𝑝𝑗−1 )︁ is equivalent to the condition 1 2(𝑟−1)𝑑< 𝑠.
Thus, there are exactly 2𝑝𝑙𝑗 (︁ 𝑝𝑘−𝑗−𝑙𝑗−1 𝑝𝑗−1 )︁ , or 2 (︁ 𝑝𝑘−𝑗−𝑝𝑙𝑗 𝑝𝑗−1 )︁ , many ways for 𝑝𝑘= 𝑆(𝑎, 𝑝𝑗, 𝑑) with each of the integers 1, 2, . . . , 2 (︁ 𝑝𝑘−𝑗−𝑝𝑙𝑗 𝑝𝑗−1 )︁ giving rise to a value of 𝑑for a different arithmetic progression, and the corresponding initial term 𝑎of the arithmetic progression for this 𝑑is 𝑎= 𝑝𝑘−𝑗−1 2(𝑝𝑗−1)𝑑. This finishes the proof of Part 1 of the theorem.
248 2. Now assume that 𝑟is an integer such that 𝑟∤𝑝𝑘. If 𝑝𝑘= 𝑆(𝑎, 𝑟, 𝑑) for such an 𝑟, then by Theorem 2.1, 𝑟= 2𝑟0 for some positive factor 𝑟0 of 𝑝𝑘and 𝑑is an odd positive integer.
Thus, 𝑟0 = 𝑝𝑗for some integer 𝑗with 0 ≤𝑗≤𝑘, and its complementary factor is given by 𝑠0 = 𝑝𝑘−𝑗. By Theorem 2.1 again, 𝑁= 𝑆(𝑎, 𝑟, 𝑑) for 𝑟= 2𝑝𝑗if and only if (2𝑟0−1)𝑑< 𝑠0, or (2𝑝𝑗−1)𝑑< 𝑝𝑘−𝑗. From this we may conclude that 𝑗< ⌊𝑘/2⌋. Thus, for each such 𝑗, there is a representation for 𝑝𝑘= 𝑆(𝑎, 2𝑝𝑗, 𝑑) for each positive odd integer 𝑑satisfying 𝑑< 𝑝𝑘−𝑗 2𝑝𝑗−1 or 𝑑≤ ⌊︂𝑝𝑘−𝑗−1 2𝑝𝑗−1 ⌋︂ .
Furthermore, for each such 𝑑, the beginning term 𝑎= 1 2[𝑝𝑘−𝑗−(2𝑝𝑗−1)𝑑]. Note that the number of the odd integers less than or equal to ⌊︁ 𝑝𝑘−𝑗−1 2𝑝𝑗−1 ⌋︁ is given by ⌊︁ 1 2 (︁ 𝑝𝑘−𝑗−1 2𝑝𝑗−1 + 1 )︁⌋︁ .
Hence, #(𝑝𝑘, 2𝑝𝑗) = ⌊︂1 2 (︂𝑝𝑘−𝑗−1 2𝑝𝑗−1 + 1 )︂⌋︂ for each 𝑗with 0 ≤𝑗< ⌊𝑘/2⌋.
Note that when 𝑗= 0, the above formula becomes #(𝑝𝑘, 2) = ⌊1 2(𝑝𝑘−1)⌋, which agrees with the value given by Lemma 3.1.
The above two cases exhaust all the possible values of 𝑟. We can collect all the terms from the above two parts and get a formula for #(𝑝𝑘). Note that the formula in Part 1 is for 1 ≤𝑗≤⌊𝑘/2⌋, but for that in Part 2 is for 0 ≤𝑗< ⌊𝑘/2⌋. We may combine the two formulas under the same summation sign for 𝑗ranging from 1 to ⌊𝑘/2⌋by leaving out the term in Part 2 for 𝑗= 0 from the summation, and noting that the formula in Part 2 for 𝑗= ⌊𝑘/2⌋is automatically zero.
4 Representing a positive integer as sums of arithmetic progressions. I: A procedure A similar procedure for computing the number of representations for a general positive integer 𝑁 can also be established. This procedure is based on the prime factors of 𝑁since if 𝑁= 𝑆(𝑎, 𝑟, 𝑑), either 𝑟is a factor of 𝑁, or 𝑟= 2𝑟0 and 𝑟0 is a factor of 𝑁. Given a positive integer 𝑁, we will factor 𝑁into a product of powers of primes: 𝑁= 𝑝𝑘1 1 𝑝𝑘2 2 . . . 𝑝𝑘𝑛 𝑛. We will show how to compute #(𝑁, 𝑟) for all the possible values of 𝑟, for which either 𝑟is a product of the powers of the prime factors of 𝑁, or 𝑟= 2𝑟0 and 𝑟0 is a product of the powers of the prime factors of 𝑁. Collecting all such #(𝑁, 𝑟), we will get the number #(𝑁).
We start with the cases for #(𝑁, 𝑟) when 𝑟= 𝑝𝑖𝑗or 𝑟= 2𝑝𝑖𝑗, for each prime factor 𝑝𝑖of 𝑁 (Theorem 4.1 and 4.2). We will then build a computational procedure for 𝑟or 𝑟0 being the product of powers of two or more prime factors of 𝑁. This allows us to find #(𝑁, 𝑟) for all the possible values of 𝑟. Since the procedure is similar when 𝑟or 𝑟0 is the product of the powers of two or more prime factors of 𝑁, we will describe, as the typical case, the product of powers of three distinct prime factors of 𝑁(Theorem 4.3).
We now begin with powers of a single prime factor. As noted before, 1 and 2 cannot be written as the sum of any arithmetic progression. We now determine whether an integer 𝑁≥3 can be written as a sum 𝑆(𝑎, 𝑟, 𝑑) when 𝑟is a power of 2, and if so, how many ways can this be done.
249 Theorem 4.1. Let 𝑁≥3 be an integer. Depending on whether 𝑁is odd or even, we consider two cases: 1. If 𝑁is odd, the only way for 𝑁= 𝑆(𝑎, 2𝑗, 𝑑) for a positive integer 𝑗is when 𝑗= 1 and #(𝑁, 2) = 1 2(𝑁−1).
2. If 𝑁is even, say 𝑁= 2𝑘𝑁0, for some positive integers 𝑘and 𝑁0 such that 𝑁0 is odd, we need to consider two further cases: A) For each integer 𝑗such that 1 ≤𝑗≤𝑘, there are representations 𝑁= 𝑆(𝑎, 2𝑗, 𝑑) if and only if 𝑑(= 2𝑑0) is even and ⌊︁ 2𝑘−𝑗𝑁0−1 2𝑗−1 ⌋︁ ≥1. In such a case, each positive integer from 1 to ⌊︁ 2𝑘−𝑗𝑁0−1 2𝑗−1 ⌋︁ gives rise to a value of 𝑑0 for a representation 𝑁= 𝑆(𝑎, 2𝑗, 2𝑑0), and #(𝑁, 2𝑗) = ⌊︂2𝑘−𝑗𝑁0 −1 2𝑗−1 ⌋︂ for each 1 ≤𝑗≤𝑘.
(7) For each 𝑑0 specified above, the beginning term 𝑎= 2𝑘−𝑗𝑁0 −(2𝑗−1)𝑑0. The progression itself can then be found from these 𝑎and 𝑑(= 2𝑑0).
B) 𝑁= 𝑆(𝑎, 2𝑗, 𝑑) can also happen when 𝑗> 𝑘. For this to happen, we must have 𝑗= 𝑘+ 1 and ⌊︁ 𝑁0−1 2𝑘+1−1 ⌋︁ ≥1. In such a case, each odd positive integer from 1 to ⌊︁ 𝑁0−1 2𝑘+1−1 ⌋︁ gives rise to a value of 𝑑for a different representation of 𝑁, and #(𝑁, 2𝑘+1) = ⌊︂1 2 (︂𝑁0 −1 2𝑘+1 −1 + 1 )︂⌋︂ .
(8) For each of these arithmetic progressions, 𝑎= 1 2 (𝑁0 −(2𝑗−1)𝑑).
Proof.
1. Let 𝑁≥3 be an odd integer and 𝑟= 2𝑗for a positive integer 𝑗. We first show that 𝑁= 𝑆(𝑎, 𝑟, 𝑑) then 𝑗= 1. This is because that since 𝑁is odd, 𝑟∤𝑁. The only possibility for 𝑁= 𝑆(𝑎, 𝑟, 𝑑) is, by Theorem 2.1, for 𝑟= 2𝑟0 and 𝑟0|𝑁. But 𝑟0 = 1 2𝑟= 2𝑗−1. Thus, 𝑟0|𝑁implies that 𝑗= 1, and 𝑟= 2. The rest follows from Lemma 3.1.
2. Let 𝑁≥3 be an even integer. Write 𝑁= 2𝑘𝑁0 for some odd integer 𝑁0. We now consider the case that 𝑁= 𝑆(𝑎, 𝑟, 𝑑) for 𝑟= 2𝑗for some positive integer 𝑗.
A) Suppose that 1 ≤𝑗≤𝑘. In this case 𝑟= 2𝑗divides 𝑁. Since 𝑟is even, by Theorem 2.1, if 𝑁= 𝑆(𝑎, 𝑟, 𝑑), then 𝑑must be even. Write 𝑑= 2𝑑0 for some positive integer 𝑑0. By Theorem 2.1 again, 𝑁= 𝑆(𝑎, 2𝑗, 2𝑑0) if and only if (2𝑗−1)𝑑0 < 𝑠, where 𝑠= 2𝑘−𝑗𝑁0 is the complementary factor of 2𝑗in 𝑁. Thus, there is a representation 𝑁= 𝑆(𝑎, 2𝑗, 2𝑑0) for each integer 𝑑0 satisfying the inequality 𝑑0 < 𝑠 2𝑗−1, or 𝑑0 < 2𝑘−𝑗𝑁0 2𝑗−1 . In particular, ⌊︁ 2𝑘−𝑗𝑁0−1 2𝑗−1 ⌋︁ ≥1. In such a case, we may conclude that each positive integer from 1 to ⌊︁ 2𝑘−𝑗𝑁0−1 2𝑗−1 ⌋︁ gives rise to a value of 𝑑0 for a representation of 𝑁, and Formula (7) follows. The formula 𝑎= 2𝑘−𝑗𝑁0 −(2𝑗−1)𝑑0 also follows from Theorem 2.1.
Note that we do not need to worry about whether the condition ⌊︁ 2𝑘−𝑗𝑁0−1 2𝑗−1 ⌋︁ ≥1 in (7), for if it is not satisfied, this term in (7) would automatically be zero. We also note that when 𝑗= 1, ⌊︁ 2𝑘−𝑗𝑁0−1 2𝑗−1 ⌋︁ = ⌊2𝑘−1𝑁0 −1⌋= ⌊𝑁 2 −1⌋. Since 𝑁= 2𝑘𝑁0 is even, 250 𝑁−1 is odd, and ⌊1 2(𝑁−1)⌋= ⌊1 2(𝑁−2)⌋= ⌊𝑁 2 −1⌋. Thus, Formula (7) reduces to #(𝑁, 2) = ⌊︀1 2(𝑁−1) ⌋︀ , which agrees with Lemma 3.1. This finishes the proof of 2(A).
B) Consider the case when 𝑗> 𝑘. In this case 𝑟= 2𝑗does not divide 𝑁. By Theorem 2.1, we must have 𝑟= 2𝑟0 and 𝑟0 = 2𝑗−1 is a factor of 𝑁= 2𝑘𝑁0. Consequently, we must have 𝑗= 𝑘+ 1. Since 𝑟∤𝑁, by Theorem 2.1, 𝑁= 𝑆(𝑎, 𝑟, 𝑑) if and only if 𝑑is odd and (2𝑟0 −1)𝑑< 𝑠0, where 𝑠0 = 𝑁0 is the complementary factor of 𝑟0 in 𝑁. Thus, for each odd integer 𝑑satisfying the inequality 𝑑< 𝑁0 2𝑘+1−1, there is a representation of 𝑁= 𝑆(𝑎, 𝑟, 𝑑). Thus, we must require ⌊︁ 𝑁0−1 2𝑘+1−1 ⌋︁ ≥1. If this is the case, each odd integer from 1 to ⌊︁ 𝑁0−1 2𝑘+1−1 ⌋︁ gives rise to a value of 𝑑for an arithmetic progression, and Formula (8) follows.
The rest of the assertions in (B) are then immediate. Again, we do not need to worry about the condition ⌊︁ 𝑁0−1 2𝑘+1−1 ⌋︁ ≥1 for Formula (8), for if the condition is not satisfied, i.e., suppose 0 ≤𝑁0 −1 2𝑘+1 −1 < 1 ⇒1 ≤𝑁0 −1 2𝑘+1 −1 + 1 < 2.
Consequently, ⌊︁ 1 2 (︁ 𝑁0−1 2𝑘+1−1 + 1 )︁⌋︁ = 0. This finishes the proof of the theorem.
Theorem 4.2. Consider a representation 𝑁= 𝑆(𝑎, 𝑟, 𝑑) for a positive integer 𝑁= 𝑝𝑘𝑁0 where 𝑝is an odd prime, and 𝑘and 𝑁0 are positive integers such that 𝑁0 is relatively prime to 𝑝.
1. The only case for 𝑟= 𝑝𝑗for some integer 𝑗is for 1 ≤𝑗≤𝑘and ⌊︁ 2𝑝𝑘−𝑗𝑁0−1 𝑝𝑗−1 ⌋︁ ≥1. For such a 𝑗, #(𝑁, 𝑝𝑗) = ⌊︁ 2𝑝𝑘−𝑗𝑁0−1 𝑝𝑗−1 ⌋︁ . Each of the integers from 1 to ⌊︁ 2𝑝𝑘−𝑗𝑁0−1 𝑝𝑗−1 ⌋︁ gives rise to a value of 𝑑for a different representation of 𝑁. For each such 𝑑, the initial term is given by 𝑎= 𝑝𝑘−𝑗𝑁0 −1 2(𝑝𝑗−1)𝑑. The progression itself can then be found from these 𝑎and 𝑑.
2. Now consider the case for 𝑟= 2𝑝𝑗. For this part of the theorem, we may assume that 𝑁is an odd integer since the case 𝑁is even can be covered in the Part 1 of Theorem 4.3 (see also Remark 4.1) when 𝑟is a product of the powers of two or more primes, including 2.
In this case, 𝑟= 2𝑝𝑗for some integer 𝑗can happen if and only if 0 ≤𝑗≤𝑘 and ⌊︁ 𝑝𝑘−𝑗𝑁0−1 2𝑝𝑗−1 ⌋︁ ≥1. If these conditions are satisfied, for each such 𝑗there will be a representation 𝑁= 𝑆(𝑎, 𝑟, 𝑑), with 𝑟= 2𝑝𝑗, and #(𝑁, 2𝑝𝑗) = ⌊︁ 1 2 (︁ 𝑝𝑘−𝑗𝑁0−1 2𝑝𝑗−1 + 1 )︁⌋︁ . Each of the odd integers from 1 to ⌊︁ 𝑝𝑘−𝑗𝑁0−1 2𝑝𝑗−1 ⌋︁ gives rise to a value of 𝑑for such a representation, and for each of these 𝑑, the beginning term 𝑎= 1 2[𝑝𝑘−𝑗𝑁0 −(2𝑝𝑗−1)𝑑].
The above two cases exhaust all the possibilities for 𝑝𝑘𝑁0 = 𝑆(𝑎, 𝑟, 𝑑) for 𝑟= 𝑝𝑗or 𝑟= 2𝑝𝑗for all possible integer 𝑗.
Proof. Consider a representation 𝑁= 𝑆(𝑎, 𝑟, 𝑑) for a positive integer 𝑁= 𝑝𝑘𝑁0, where 𝑝, 𝑘, and 𝑁0 are as specified in the theorem.
1. If 𝑟= 𝑝𝑗for some integer 𝑗. Since 𝑟is odd, 𝑟|𝑁by Theorem 2.1, and hence, 1 ≤𝑗≤𝑘.
By Theorem 2.1 again, 𝑁= 𝑆(𝑎, 𝑟, 𝑑) for 𝑟= 𝑝𝑗if and only if the complementary factor 𝑠 of 𝑟satisfies 𝑠> 1 2(𝑟−1)𝑑, or 2𝑝𝑘−𝑗𝑁0 > (𝑝𝑗−1)𝑑. This condition is satisfied if and only if 𝑑is a positive integer such that 251 𝑑< 2𝑝𝑘−𝑗𝑁0 𝑝𝑗−1 or 𝑑≤ ⌊︂2𝑝𝑘−𝑗𝑁0 −1 𝑝𝑗−1 ⌋︂ .
In particular, ⌊︁ 2𝑝𝑘−𝑗𝑁0−1 𝑝𝑗−1 ⌋︁ ≥1. Thus, 𝑁= 𝑆(𝑎, 𝑝𝑗, 𝑑) for 𝑑= 1, 2, . . . , ⌊︁ 2𝑝𝑘−𝑗𝑁0−1 𝑝𝑗−1 ⌋︁ , and consequently, #(𝑁, 𝑝𝑗) = ⌊︂2𝑝𝑘−𝑗𝑁0 −1 𝑝𝑗−1 ⌋︂ .
Note that for certain values of 𝑗, 𝑘, 𝑝, and 𝑁0 the inequality 2𝑝𝑘−𝑗𝑁0 > (𝑝𝑗−1)𝑑is not satisfied for any positive integer 𝑑(for instance, when 𝑝= 3, 𝑗= 𝑘= 2 and 𝑁0 = 1), but this does not invalidate the formula for #(𝑁, 𝑝𝑗) since in such cases (2𝑝𝑘−𝑗𝑁0 −1) < (𝑝𝑗−1) and #(𝑁, 𝑝𝑗) is zero, and thus, when (2𝑝𝑘−𝑗𝑁0 −1) < (𝑝𝑗−1), or when 𝑠≤1 2(𝑟−1)𝑑, there is no such representation for 𝑁. The formula for #(𝑁) is still valid.
2. Now assume that 𝑁is an odd integer. If 𝑟= 2𝑝𝑗, then 𝑟∤𝑁. In this case, 𝑟0 = 𝑟 2 = 𝑝𝑗does divide 𝑁, and hence, 0 ≤𝑗≤𝑘. The complementary factor for 𝑟0 is then 𝑠0 = 𝑝𝑘−𝑗𝑁0.
By Theorem 2.1 again, 𝑁= 𝑆(𝑎, 𝑟, 𝑑) for 𝑟= 2𝑝𝑗if and only if (2𝑟0 −1)𝑑< 𝑠0, or (2𝑝𝑗−1)𝑑< 𝑝𝑘−𝑗𝑁0. Thus, there is a representation 𝑁= 𝑆(𝑎, 2𝑝𝑗, 𝑑) for each positive odd integer 𝑑satisfying 𝑑< 𝑝𝑘−𝑗𝑁0 2𝑝𝑗−1 or 𝑑≤ ⌊︂𝑝𝑘−𝑗𝑁0 −1 2𝑝𝑗−1 ⌋︂ .
In particular, ⌊︁ 𝑝𝑘−𝑗𝑁0−1 2𝑝𝑗−1 ⌋︁ ≥1. Again as before, the number of such representations is #(𝑁, 2𝑝𝑗) = ⌊︂1 2 (︂𝑝𝑘−𝑗𝑁0 −1 2𝑝𝑗−1 + 1 )︂⌋︂ .
The rest of the theorem is immediate.
Remark 4.1. We now consider a general positive integer 𝑁= 𝑝𝑘1 1 𝑝𝑘2 2 . . . 𝑝𝑘𝑛 𝑛𝑁0, where 𝑁0 is relatively prime to all the prime factors 𝑝𝑖’s. We will describe a way that 𝑁can be represented as sums of arithmetic progressions 𝑆(𝑎, 𝑟, 𝑑). The computations are all similar when 𝑟or 𝑟0 is a product of powers of two or more prime factors of 𝑁. In the theorem, we will describe only a typical case: when 𝑁= 𝑝𝑘1𝑞𝑘2𝑡𝑘3𝑁0, where 𝑝, 𝑞, 𝑡, are three distinct primes, all of which are relatively prime to 𝑁0, and 𝑟or 𝑟0 also involves these three primes with positive powers. The same procedure can be used when more, or fewer, number of prime factors of 𝑁are involved.
This procedure allows us to compute the number #(𝑁, 𝑟) for any factor 𝑟of 𝑁. By collecting all these #(𝑁, 𝑟), we can then get #(𝑁) itself.
Theorem 4.3. Consider a positive integer of the form 𝑁= 𝑝𝑘1𝑞𝑘2𝑡𝑘3𝑁0, where 𝑝, 𝑞, 𝑡are three distinct primes, each 𝑘𝑖> 0, and 𝑁0 is a positive integer relatively prime to 𝑝, 𝑞and 𝑡. Our procedure depends on whether 𝑁is even or odd: 1. 𝑁is an even integer. In this case, let 𝑝= 2 and write 𝑁= 2𝑘1𝑞𝑘2𝑡𝑘3𝑁0. The number of ways for the representation 𝑁= 𝑆(𝑎, 𝑟, 𝑑), where 𝑟of the form 𝑟= 2𝑗1𝑞𝑗2𝑡𝑗3, depends on two further cases: whether 𝑟|𝑁or 𝑟∤𝑁: 252 1A. If 𝑟|𝑁, there is a representation for each even integer 𝑑≤ ⌊︂2𝑘1−𝑗1+1𝑞𝑘2−𝑗2𝑡𝑘3−𝑗3𝑁0 −1 2𝑗1𝑞𝑗2𝑡𝑗3 −1 ⌋︂ .
For each of such 𝑑, the first term 𝑎is 𝑎= 2𝑘1−𝑗1𝑞𝑘2−𝑗2𝑡𝑘3−𝑗3𝑁0 −1 2(2𝑗1𝑞𝑗2𝑡𝑗3 −1)𝑑.
1B. If 𝑟∤𝑁, then 𝑗1 = 𝑘1 + 1 and there is a representation for each odd integer 𝑑≤ ⌊︂𝑞𝑘2−𝑗2𝑡𝑘3−𝑗3𝑁0 −1 2𝑘1+1𝑞𝑗2𝑡𝑗3 −1 ⌋︂ .
For each of such 𝑑, the first term is 𝑎= 1 2[𝑞𝑘2−𝑗2𝑡𝑘3−𝑗3𝑁0 −(2𝑘1+1𝑞𝑗2𝑡𝑗3 −1)𝑑].
Thus, the number of ways when 𝑁is an even integer and for 𝑟of the form 𝑟= 2𝑗1𝑞𝑗2𝑡𝑗3 for some nonzero powers 𝑗1, 𝑗2 and 𝑗3 is 𝑘1 ∑︁ 𝑗1=1 𝑘2 ∑︁ 𝑗2=1 𝑘3 ∑︁ 𝑗3=1 ⌊︂1 2 (︂2𝑘1−𝑗1+1𝑞𝑘2−𝑗2𝑡𝑘3−𝑗3𝑁0 −1 2𝑗1𝑞𝑗2𝑡𝑗3 −1 )︂ ⌋︂ + 𝑘2 ∑︁ 𝑗2=1 𝑘3 ∑︁ 𝑗3=1 ⌊︂1 2 (︂𝑞𝑘2−𝑗2𝑡𝑘3−𝑗3𝑁0 −1 2𝑘1+1𝑞𝑗2𝑡𝑗3 −1 +1 )︂ ⌋︂ . (9) 2. 𝑁is still even, but 𝑟= 𝑝𝑗1𝑞𝑗2𝑡𝑗3 is a product of powers of three odd primes. Since 𝑟is odd, 𝑟|𝑁and 𝑁= 𝑝𝑘1𝑞𝑘2𝑡𝑘3𝑁0, where 𝑁0 is even and 𝑘𝑖≥𝑗𝑖for each 𝑖= 1, 2 or 3. In this case, there a representation for each integer 𝑑≤ ⌊︂2𝑝𝑘1−𝑗1𝑞𝑘2−𝑗2𝑡𝑘3−𝑗3𝑁0 −1 𝑝𝑗1𝑞𝑗2𝑡𝑗3 −1 ⌋︂ .
For each of such 𝑑, the first term is 𝑎= 𝑝𝑘1−𝑗1𝑞𝑘2−𝑗2𝑡𝑘3−𝑗3𝑁0 −1 2 (𝑝𝑗1𝑞𝑗2𝑡𝑗3 −1) 𝑑, and consequently, for 𝑟= 𝑝𝑗1𝑞𝑗2𝑡𝑗3, #(𝑁, 𝑟) = ⌊︂2𝑝𝑘1−𝑗1𝑞𝑘2−𝑗2𝑡𝑘3−𝑗3𝑁0 −1 𝑝𝑗1𝑞𝑗2𝑡𝑗3 −1 ⌋︂ .
(10) 3. 𝑁= 𝑝𝑘1𝑞𝑘2𝑡𝑘3𝑁0 is an odd integer. In this case, 𝑝, 𝑞, 𝑡and 𝑁0 are all odd. There are again two possible ways for the representation 𝑁= 𝑆(𝑎, 𝑟, 𝑑) either for 𝑟of the form 𝑟= 𝑝𝑗1𝑞𝑗2𝑡𝑗3 or for 𝑟of the form 𝑟= 2𝑝𝑗1𝑞𝑗2𝑡𝑗3, depending on whether 𝑟|𝑁or 𝑟∤𝑁: 3A) 𝑟|𝑁. Then 𝑟= 𝑝𝑗1𝑞𝑗2𝑡𝑗3, where 1 ≤𝑗𝑖≤𝑘𝑖for each 𝑖= 1, 2 or 3. There is a representation for each integer 𝑑≤ ⌊︂2𝑝𝑘1−𝑗1𝑞𝑘2−𝑗2𝑡𝑘3−𝑗3𝑁0 −1 𝑝𝑗1𝑞𝑗2𝑡𝑗3 −1 ⌋︂ .
For each of such 𝑑, the first term is 𝑎= 𝑝𝑘1−𝑗1𝑞𝑘2−𝑗2𝑡𝑘3−𝑗3𝑁0 −1 2 (𝑝𝑗1𝑞𝑗2𝑡𝑗3 −1) 𝑑.
3B) 𝑟∤𝑁. In this case, 𝑟= 2𝑝𝑗1𝑞𝑗2𝑡𝑗3, where 1 ≤𝑗𝑖≤𝑘𝑖for each 𝑖. In this case, there is a representation for each odd integer 𝑑≤ ⌊︂𝑝𝑘1−𝑗1𝑞𝑘2−𝑗2𝑡𝑘3−𝑗3𝑁0 −1 2𝑝𝑗1𝑞𝑗2𝑡𝑗3 −1 ⌋︂ .
For each of such 𝑑, the first term is 𝑎= 1 2[𝑝𝑘1−𝑗1𝑞𝑘2−𝑗2𝑡𝑘3−𝑗3𝑁0 −(2𝑝𝑗1𝑞𝑗2𝑡𝑗3 −1)𝑑].
253 Combining these two cases, we have for an odd integer 𝑁= 𝑝𝑘1𝑞𝑘2𝑡𝑘3𝑁0 the number of possible representation for 𝑟= 𝑝𝑗1𝑞𝑗2𝑡𝑗3, or of the form 𝑟= 2𝑝𝑗1𝑞𝑗2𝑡𝑗3, is given by 𝑘1 ∑︁ 𝑗1=1 𝑘2 ∑︁ 𝑗2=1 𝑘3 ∑︁ 𝑗3=1 (︂⌊︂2𝑝𝑘1−𝑗1𝑞𝑘2−𝑗2𝑡𝑘3−𝑗3𝑁0 −1 𝑝𝑗1𝑞𝑗2𝑡𝑗3 −1 ⌋︂ + ⌊︂1 2 (︂𝑝𝑘1−𝑗1𝑞𝑘2−𝑗2𝑡𝑘3−𝑗3𝑁0 −1 2𝑝𝑗1𝑞𝑗2𝑡𝑗3 −1 + 1 )︂ ⌋︂ )︂ .
(11) Proof. In each case, for a given 𝑟or 𝑟0, we use the condition 1 2(𝑟−1)𝑑< 𝑠or (2𝑟0 −1)𝑑< 𝑠0 to determine the allowable values for 𝑑for the number of possible representations for 𝑁= 𝑆(𝑎, 𝑟, 𝑑).
In carrying out the computations, we need to consider the restrictions, as described in Theorem 2.1, that when 𝑟|𝑁, then either 𝑟is odd or 𝑑is even, but when 𝑟∤𝑁, 𝑟is even and 𝑑is odd.
Finally, when we compute the number of ways, if 𝑑is an odd integer less than or equal to a given number we have to add an 1 before taking the floor function of 1 2 of that number, but it the number is for 𝑑to be an even integer less than or equal to that number, we do not have to add an 1. The arguments are all similar to what we did before and will be skipped here.
5 Representing a positive integer as sums of arithmetic progressions. II: An example To show how our method can be carried out, we now sketch a computation of #(𝑁), for 𝑁= 233252 = 1800. We will first compute #(𝑁, 𝑟) for 𝑟or 𝑟0 being the power of a single prime factor of 𝑁, and then, for 𝑟or 𝑟0 being the product of powers for pairs of the prime factors of 𝑁, and so on, until all the factors of 𝑁are accounted for.
1. #(1800, 𝑝𝑗) for some positive integer 𝑗.
1A. First consider the case for 𝑝= 2 and 𝑟= 2𝑗for some integer 𝑗≥1.
(a) Suppose 2𝑗|𝑁. In this case 1 ≤𝑗≤3. In the form of 𝑁= 2𝑘𝑁0 of Theorem 4.1-Part 2A, 𝑁0 = 225. Thus, each integer from 1 to ⌊︁ 2𝑘−𝑗𝑁0−1 2𝑗−1 ⌋︁ gives rise to a value of 𝑑0 for an arithmetic progression. These are the only possible ways for 𝑁= 𝑆(𝑎, 2𝑗, 2𝑑0) for 1 ≤𝑗≤𝑘.
Hence, #(𝑁, 2𝑗) = ⌊︁ 2𝑘−𝑗𝑁0−1 2𝑗−1 ⌋︁ = ⌊︁ 23−𝑗225−1 2𝑗−1 ⌋︁ .
For each of these 𝑑0, the initial term of the progression is given by 𝑎= 2𝑘−𝑗𝑁0 −(2𝑗−1)𝑑0 = 23−𝑗225 −(2𝑗−1)𝑑0.
∙𝑗= 1. #(𝑁, 2) = ⌊899⌋= 899. To find these progressions, we can let 𝑑0 be any of the integers 1, 2, . . . , 899, and for each of the 𝑑0, let 𝑎= 22×225−𝑑0 = 900−𝑑0.
The progression will then consists of the two terms 𝑎and 𝑎+ 2𝑑0.
∙𝑗= 2. In this case, #(𝑁, 22) = ⌊︀2×225−1 3 ⌋︀ = 149. By a similar process, we can find all these 149 progressions: for each 𝑑0 = 1, 2, . . . , 149 we let 𝑎= 2 × 225 −3𝑑0 = 450 −3𝑑0, then keep adding 𝑑= 2𝑑0 to 𝑎until we obtain all 22 = 4 terms of the progression.
∙𝑗= 3.
⌊︁ 2𝑘−𝑗𝑁0−1 2𝑗−1 ⌋︁ = ⌊︀225−1 7 ⌋︀ = 32. Thus, 𝑑0 = 1, 2, . . . , 32 and #(𝑁, 22) = 32.
254 (b) For 𝑟= 2𝑗∤𝑁, then since 𝑟 2 = 𝑟0 divides 𝑁. We must have 𝑟0 = 23 and 𝑟= 2𝑗= 24, and 𝑠0 = 𝑁0 = 225. By asserton 2(B) of Theorem 4.1, each odd positive integer from 1 to ⌊︀𝑁0−1 2𝑗−1 ⌋︀ = ⌊︀224 15 ⌋︀ = 14 gives rise to a value of 𝑑for a distinct arithmetic progression, consisting of 2𝑟0 = 16 terms. Thus, 𝑑= 1, 3, . . . , 11, 13, and #(𝑁, 24) = 7.
Combining all the cases, we have #(1800, 2𝑗) = 899 + 149 + 32 + 7 = 1087. We stress again that if needed, we can construct any of these 1087 arithmetic progressions by using the values of 𝑟, the common difference 𝑑, and the first term 𝑎. This is also the case for any of the arithmetic progressions counted below.
1B. The case for #(1800, 3𝑗) is similar.
If 𝑟= 3𝑗, then 3𝑗|𝑁(= 233252 = 32𝑁0), where 𝑁0 = 200 and 𝑗= 1 or 2. By Theorem 4.2, each of the integers from 1 to ⌊︁ 2𝑝𝑘−𝑗𝑁0−1 𝑝𝑗−1 ⌋︁ = ⌊︁ 400×32−𝑗−1 3𝑗−1 ⌋︁ gives rise to a value of 𝑑for a distinct arithmetic progression.
(a) For 𝑗= 1, #(1800, 31) = ⌊︀1199 2 ⌋︀ = 599. There is a representation for 𝑟= 3 for each 𝑑= 1, 2, 3, . . . , 599.
(b) For 𝑗= 2, #(1800, 32) = ⌊︀399 8 ⌋︀ = 49. There is a representation for 𝑟= 9 for each 𝑑= 1, 2, . . . , 49.
Combining the above two cases, we have that #(1800, 3𝑗) = 599 + 49 = 648. Note that we need not compute #(𝑁, 2𝑝𝑗) described in Part (2) of Theorem 4.2 since here 𝑁0 is not an odd integer and it is not the case that 𝑟= 2𝑝𝑗∤𝑁.
1C. The case for #(1800, 5𝑗) is also similar. If 𝑟= 5𝑗, then 5𝑗|𝑁(= 233252 = 52𝑁0 = 52 ×72), and 𝑗= 1 or 2. Each of the integers from 1 to ⌊︁ 144×52−𝑗−1 5𝑗−1 ⌋︁ gives rise to a value of 𝑑for a different arithmetic progression.
(a) For 𝑗= 1, #(1800, 5) = ⌊︀719 4 ⌋︀ = 179.
(b) For 𝑗= 2, #(1800, 52) = ⌊︀143 24 ⌋︀ = 5.
Combining the above two cases, we have that #(1800, 5𝑗) = 179 + 5 = 184.
2. We now consider #(1800, 𝑟) for 𝑟= 2𝑗13𝑗25𝑗3, where at least two of 𝑗1, 𝑗2 and 𝑗3 are greater than zero.
2A. For 𝑗1 = 0, 𝑟= 3𝑗25𝑗3. This is the case described in Equation (10), except that we have only two prime powers instead of three. In this case 𝑟|𝑁, and each of 𝑗2 and 𝑗3 can be 1 or 2. The complementary factor 𝑠of 𝑟will then be 𝑠= 2332−𝑗252−𝑗3 and the condition 1 2(𝑟−1)𝑑< 𝑠or 𝑑< 2𝑠 𝑟−1 = 2432−𝑗252−𝑗3 3𝑗25𝑗3−1 . Thus, there is a representation for 𝑟= 3𝑗25𝑗3 for each value of 𝑑= 1, 2, . . . , ⌊︁ 2432−𝑗252−𝑗3−1 3𝑗25𝑗3−1 ⌋︁ , and for each such 𝑑, the corresponding initial term is 𝑎= 𝑝𝑘1−𝑗1𝑞𝑘2−𝑗2𝑡𝑘3−𝑗3𝑁0 −1 2 (𝑝𝑗1𝑞𝑗2𝑡𝑗3 −1) 𝑑.
(a) (𝑗2, 𝑗3) = (1, 1), 𝑟= 3𝑗25𝑗3 = 15, and ⌊︁ 2432−𝑗252−𝑗3−1 3𝑗25𝑗3−1 ⌋︁ = ⌊︀239 14 ⌋︀ = 17.
(b) (𝑗2, 𝑗3) = (2, 1), 𝑟= 3𝑗25𝑗3 = 45, and ⌊︁ 2432−𝑗252−𝑗3−1 3𝑗25𝑗3−1 ⌋︁ = ⌊︀80 44 ⌋︀ = 1.
255 (c) (𝑗2, 𝑗3) = (1, 2), and (𝑗2, 𝑗3) = (2, 2); 2432−𝑗252−𝑗3 < 3𝑗25𝑗3 −1, and hence, ⌊︁ 2432−𝑗252−𝑗3−1 3𝑗25𝑗3−1 ⌋︁ = 0. There are no representations for these cases.
Combining all the cases in 2A, we have #(1800, 3𝑗25𝑗3)= 17 + 1 = 18.
In the following, we will consider the cases when 𝑗1 > 0. Depending on whether 𝑟= 2𝑗13𝑗25𝑗3 is a factor of 𝑁= 233252 or not, there are two possibilities: i) If 𝑟|𝑁then 1 ≤𝑗1 ≤3, 0 ≤𝑗2 ≤2 and 0 ≤𝑗3 ≤2, but at least one of 𝑗2 and 𝑗3 is nonzero. In this case, 𝑟is an even integer, and by Theorem 4.3, 𝑑can only be an even integer less than or equal to ⌊︁ 2𝑘1−𝑗1+1𝑞𝑘2−𝑗2𝑡𝑘3−𝑗3𝑁0−1 2𝑗1𝑞𝑗2𝑡𝑗3−1 ⌋︁ .
ii). If 𝑟∤𝑁, then 𝑗1 = 𝑘1 + 1 = 4, and 𝑑will be an odd integer less than or equal to ⌊︁ 𝑞𝑘2−𝑗2𝑡𝑘3−𝑗3𝑁0−1 2𝑘1+1𝑞𝑗2𝑡𝑗3−1 ⌋︁ .
2B. Consider the possibility i), when 𝑟|𝑁. In this case 𝑟= 2𝑗13𝑗25𝑗3 with 1 ≤𝑗1 ≤3, 1 ≤𝑗2 ≤ 2 and 1 ≤𝑗3 ≤2, and there is a representation for each 𝑑≤ ⌊︂2𝑘1−𝑗1+1𝑞𝑘2−𝑗2𝑡𝑘3−𝑗3 −1 2𝑗1𝑞𝑗2𝑡𝑗3 −1 ⌋︂ = ⌊︂24−𝑗132−𝑗252−𝑗3 −1 2𝑗13𝑗25𝑗3 −1 ⌋︂ .
We now compute all the cases for 𝑟|𝑁: (a) 𝑗1 = 1 ∙(𝑗2, 𝑗3) = (1, 0). 𝑟= 2 × 3 = 6, and 𝑑is an even integer ≤ ⌊︁ 24−𝑗132−𝑗2 52−𝑗3−1 2𝑗13𝑗25𝑗3−1 ⌋︁ = ⌊︀599 5 ⌋︀ = 119. Thus, #(1800, 𝑟) = 118 2 = 59.
∙(𝑗2, 𝑗3) = (0, 1). 𝑟= 2 × 5 = 10, and 𝑑is an even integer ≤ ⌊︀360−1 9 ⌋︀ = 39, or 𝑑= 2, 4, . . . , 38. Thus, #(1800, 𝑟) = 19.
∙(𝑗2, 𝑗3) = (1, 1). 𝑟= 2 × 3 × 5 = 30, and 𝑑is an even integer ≤ ⌊︀120−1 29 ⌋︀ = 4, or 𝑑= 2 or 4. Thus, #(1800, 𝑟) = 2.
∙(𝑗2, 𝑗3) = (2, 0). 𝑟= 2 × 9 = 18 and 𝑑, is an even integer ≤ ⌊︀200−1 17 ⌋︀ = 11, or 𝑑= 2, 4, 6, 8, 10. Thus, #(1800, 𝑟) = 5.
∙(𝑗2, 𝑗3) = (0, 2), (1, 2), (2, 1), or (2, 2). In each of these cases, 24−𝑗132−𝑗2 52−𝑗3−1 2𝑗13𝑗25𝑗3−1 < 2, and ⌊︁ 24−𝑗132−𝑗2 52−𝑗3−1 2𝑗13𝑗25𝑗3−1 ⌋︁ ≤1. There cannot be any even integer for 𝑑in these cases, and consequently, and there are no representations in these cases.
Combining all the above for 𝑗1 = 1, we have #(1800, 𝑟) = 59 + 19 + 2 + 5 = 85.
(b) 𝑗1 = 2 ∙(𝑗2, 𝑗3) = (1, 0). 𝑟= 22×3 = 12, and𝑑isaneveninteger≤ ⌊︁ 24−𝑗132−𝑗2 52−𝑗3−1 2𝑗13𝑗25𝑗3−1 ⌋︁ = ⌊︀299 11 ⌋︀ = 27, or 𝑑= 2, 4, . . . , 26. Thus, #(1800, 𝑟) = 13.
∙(𝑗2, 𝑗3) = (0, 1). 𝑟= 22 × 5 = 20, and 𝑑is an even integer ≤ ⌊︀180−1 19 ⌋︀ = 9, or 𝑑= 2, 4, 6, 8. Thus, #(1800, 𝑟) = 4.
∙(𝑗2, 𝑗3) = (2, 0). 𝑟= 22 × 9 = 36, and 𝑑is an even integer ≤ ⌊︀100−1 35 ⌋︀ = 2, or 𝑑= 2. Thus, #(1800, 𝑟) = 1.
256 ∙(𝑗2, 𝑗3) = (1, 1), (0, 2), (1, 2), (2, 1) or (2, 2).
⌊︁ 24−𝑗132−𝑗2 52−𝑗3−1 2𝑗13𝑗25𝑗3−1 ⌋︁ = 0.
No representations in such cases.
Combining all the above for 𝑗1 = 2, we have #(1800, 𝑟) = 13 + 4 + 1 = 18.
(c) 𝑗1 = 3 ∙(𝑗2, 𝑗3) = (1, 0). 𝑟= 23×3 = 24 and 𝑑is an even integer ≤ ⌊︁ 24−𝑗132−𝑗2 52−𝑗3−1 2𝑗13𝑗25𝑗3−1 ⌋︁ = ⌊︀149 23 ⌋︀ = 6, or 𝑑= 2, 4, 6. Thus, #(1800, 𝑟) = 3.
∙(𝑗2, 𝑗3) = (0, 1). 𝑟= 23 × 5 = 40 and 𝑑is an even integer ≤ ⌊︀90−1 39 ⌋︀ = 2 or 𝑑= 2. Thus, #(1800, 𝑟) = 1.
∙(𝑗2, 𝑗3) = (1, 1), (0, 2), (2, 0), (1, 2), (2, 1) or (2, 2) .
⌊︁ 24−𝑗132−𝑗2 52−𝑗3−1 2𝑗13𝑗25𝑗3−1 ⌋︁ = 0.
No representations in such cases.
For 𝑗1 = 3, #(1800, 𝑟) = 3 + 1 = 4.
Summing up all the results for 2B for 𝑗1 = 1, 2, and 3: #(1800, 𝑟) = 85 + 18 + 4 = 107.
2C. Now, consider the possibility ii) when 𝑟∤𝑁. Then 𝑟= 2𝑟0 and 𝑟0|𝑁(= 233252). From these we may conclude that 𝑗1 = 4 and 𝑟= 243𝑗25𝑗3 with 1 ≤𝑗1 ≤3, 1 ≤𝑗2 ≤2, and 𝑠0 = 32−𝑗252−𝑗3. Furthermore, from Theorem 4.3, we know that there is a representation for each odd integer 𝑑≤ ⌊︁ 32−𝑗252−𝑗3−1 243𝑗25𝑗3−1 ⌋︁ .
We can check quickly that the only way for #(1800, 𝑟) ̸= 0 in this case is for 𝑟= 243150 = 48.
All other combinations of 𝑟= 243𝑗25𝑗3 with 1 ≤𝑗1 ≤3, 1 ≤𝑗2 ≤2, will make #(1800, 𝑟) = 0.
For the case that 𝑟= 48, 𝑟0 = 24, 𝑠0 = 3 × 52 = 75, and 𝑑= ⌊︂32−𝑗252−𝑗3 −1 243𝑗25𝑗3 −1 ⌋︂ = ⌊︂3 × 52 −1 243𝑗25𝑗3 −1 ⌋︂ = ⌊︂74 47 ⌋︂ = 1.
Summing all the results of above for 1A, 1B, 1C, 2A, 2B, and 2C, we may conclude that #(1800) = 1087 + 648 + 184 + 18 + 107 + 1 = 2045.
This is the total number of ways that 1800 can be written as sums of arithmetic progressions.
We wish to emphasize again that for any of the 2045 ways of representing 1800 as sums of arithmetic progressions, it is easy to find the arithmetic progressions themselves. All we need is to find the first term 𝑎of the progression for the given 𝑟and 𝑑, using either Theorem 4.3 or one of the earlier theorems. then keep adding 𝑑’s to 𝑎until all 𝑟terms of the progression are found. For instance, for the representation of 1800 as the sum of an arithmetic progression of 48 terms, since in this case 𝑑= 1, 𝑟0 = 𝑟 2 = 24, and 𝑠0 = 1800 24 = 75, by Theorem 4.3, (or even Theorem 2.1), the first term of the progression is 𝑎= 1 2[𝑠0 −(2𝑟0 −1)𝑑] = 1 2[75 −47] = 14, and the progression is 14 + 15 + · · · + 60 + 61 = 1800.
257 6 Concluding remarks J. J. Sylvester showed in his original paper that the number of ways a positive integer 𝑁can be represented as sums of consecutive integers is equal to the number of odd factors of 𝑁that is greater than 1. We now see that the number of ways a positive integer 𝑁can be represented as sums of arithmetic progressions 𝑆(𝑎, 𝑟, 𝑑) is also closely related to the factors of 𝑁, but in a more complicated way: we need to find among the factors of 𝑁the admissible values for 𝑟or 𝑟0 and use the conditions on 𝑟and 𝑟0 to find these representations. With some patience, we may find all the possible representations for any positive integer 𝑁with our procedure. For instance, from our computations, we found that the integer 1800 can be written as sums of arithmetic progressions consisting of as few as 2 terms and as many as 48 terms, and our method also allows us to produce any of these arithmetic progressions.
In the second part of our paper, we will take up the problem initiated by Wheatstone again, by using as the main tool, an extension of a method recently introduced by Junaidu, Laradje and Umar, to study the relationships among various representations for different powers of an integer.
Though these relationships provide us certain new insights for the representations studied in this part of the paper, each part of the paper can essentially be read independently from the other part.
Acknowledgements The authors would like to thank the editor of NNTDM for her painstaking comments and thoughtful suggestions on a preliminary draft of this paper. These comments and suggestions drastically improved both the organization and the presentation of the paper.
References Andrushkiw, J. W., Andrushkiw, R. I., & Corzatt, C. E. (1976). Representations of positive integers as sums of arithmetic progressions. Mathematics Magazine, 49, 245–248.
Apostol, T. M. (2003). Sum of consecutive positive integers. The Mathematical Gazette, 87(508), 98–101.
Bonomo, J. P. (2014). Not all numbers can be created equally. The College Mathematics Journal, 45(1), 3–8.
Bush, L. E. (1930). On the expression of an integer as the sum of an arithmetic series.
American Mathematical Monthly, 37, 353–357.
Cook, R., & Sharpe, D. (1995). Sums of arithmetic progressions. The Fibonacci Quarterly, 33, 218–221.
Dickson, L. E. (1923), (1966). History of the Theory of Numbers, Vol. II, Carnegie Institution of Washington, Washington, D.C., Vol. I 1919; Vol. II, 1920, Vol. III, 1923, reprinted by Chelsea, New York, 1966.
258 Ho, C., He, T.-X., & Shiue, P. J.-S. (2023). Representations of positive integers as sums of arithmetic progressions, II. Notes on Number Theory and Discrete Mathematics, 29(2), 260–275.
Junaidu, S. B., Laradje, A., & Umar, A. (2010). Powers of integers as sums of consecutive odd integers. The Mathematical Gazette, 94, 117–119.
Mason, T. E. (1912). On the representation of an integer as the sum of consecutive integers.
American Mathematical Monthly, 19, 46–50.
Robertson, J. M., & Webb, W. A. (1991). Sieving with sums. The Mathematical Gazette, 75, 171–174.
Sylvester, J. J. (1882). A constructive theory of partitions, arranged in three acts, an interact and an exodion. American Journal of Mathematics, 5(1), 251–330.
Wheatstone, C. (1854-1855). On the formation of powers from arithmetical progressions.
Proceedings of the Royal Society of London, 7, 145–151.
259 |
3707 | https://en.wiktionary.org/wiki/yuan | yuan - Wiktionary, the free dictionary
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Beginning
1 EnglishToggle English subsection
1.1 Alternative forms
1.2 Pronunciation
1.3 Etymology 1
1.3.1 Noun
1.3.1.1 Synonyms
1.3.1.2 Translations
1.3.2 See also
1.4 Etymology 2
1.4.1 Noun
1.4.1.1 Derived terms
1.5 Anagrams
2 FrenchToggle French subsection
2.1 Etymology
2.2 Pronunciation
2.3 Noun
2.4 Further reading
3 MandarinToggle Mandarin subsection
3.1 Romanization
3.1.1 Usage notes
4 Norwegian NynorskToggle Norwegian Nynorsk subsection
4.1 Noun
4.2 References
5 PortugueseToggle Portuguese subsection
5.1 Alternative forms
5.2 Etymology
5.3 Noun
6 RomanianToggle Romanian subsection
6.1 Etymology
6.2 Noun
6.2.1 Declension
7 SpanishToggle Spanish subsection
7.1 Etymology
7.2 Pronunciation
7.3 Noun
7.4 Further reading
yuan
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From Wiktionary, the free dictionary
See also:Yuan, yuán, yuàn, Yuán, yuān, Yuăn,andyuǎn
English
[edit]
English Wikipedia has an article on:
yuan
Wikipedia
English Wikipedia has an article on:
Chinese yuan
Wikipedia
Alternative forms
[edit]
yüan(Wade–Giles)
Pronunciation
[edit]
(UK)IPA(key): /ˌjuːˈæn/, /ˈwæn/
Audio (Southern England):Duration: 2 seconds.0:02(file)
(General American)IPA(key): /juˈɑn/, /ˈwɑn/, /ˈjuən/Homophone: won
Rhymes: -æn, -ɑːn
Etymology 1
[edit]
From Mandarin元 (yuán), from 圓／圆 (yuán, “circle > round coin”) referring to the piece of eight. Compare Japanese円(えん)(en), Korean원(圓)(won). Doublet of yen and won.
Noun
[edit]
yuan (pluralyuanoryuans)
(numismatics) The basic unit of money in China. quotations▼
1940, Evans Fordyce Carlson, “Test Tube for New China”, in Twin Stars of China: A Behind-the-Scenes Story of China's Valiant Struggle For Existence By A U. S. Marine who Lived & Moved with The People, New York: Dodd, Mead & Company, →OCLC, page 219:Sung paused to show me some of the notes. They were in issues of five yuan (a Chinese dollar, equal at this time to a little less than one American dollar), one yuan, twenty and ten cent denominations.
1978 [1955 June 24], Yenpei Prefectural Party Committee, “A Stern Struggle Must be Waged Against Graft and Theft”, in General Office of the Central Committee of the Communist Party of China, editor, Socialist Upsurge in China's Countryside [中国农村的社会主义高潮（选本）], First edition, Peking: Foreign Languages Press, →OCLC, page 481:The Nanchiaoshan Co-operative in Kuangling County originally had 14 households. But they became dissatisfied and 7 withdrew because nothing was done when the production-team leader (a rich peasant) and the bookkeeper made away with some 60 yuan of public money.
1985 April 29, Daniel Southerl, “Saturday Night Fever in Peking”, in The Washington Post, →ISSN, →OCLC, archived from the original on 27 August 2023:A ticket to these Saturday and Sunday night dances costs 15 yuan, or $5.35, which is a lot when one considers that the average Peking worker makes 60 to 70 yuan ($21 to $25) a month.
1998, Dru C. Gladney, “Getting Rich is Not So Glorious: Contrasting Perspectives on Prosperity Among Muslims and Han in China”, in Robert W. Hefner, editor, Market Cultures: Society and Morality in the New Asian Capitalisms, Westview Press, →ISBN, →LCCN, →OCLC, page 110:Donations to the mosque come from a village considered fairly poor by regional standards, with an average annual income of 300 yuan (about U.S. $100) per household. The 1982 average per capita annual income in Yongning County was substantially higher, 539 yuan according to the Population Census Office (1987:206).
2002, Kellee S. Tsai, “Creative Capitalists in Henan”, in Back-Alley Banking: Private Entrepreneurs in China, Cornell University Press, →ISBN, →LCCN, →OCLC, →OL, page 185:The provincial government has designated thirty-four counties in Henan as officially impoverished (pinkun xian) according to the provincial criterion: an average annual per capita income below 500 yuan and total household assets valued at less than 8,000 yuan, based on 1990 prices.
2021 April 5, Lily Kuo, Lyric Li, “China’s covid vaccine drive is lagging. Free food could help turn things around.”, in The Washington Post, →ISSN, →OCLC, archived from the original on 05 April 2021, Asia & Pacific:In Pinggu, a suburb of Beijing, residents were told that they would receive 50-yuan ($7.60) prizes in cash or merchandise once they had been fully vaccinated.
2023 September 4, Berry Wang and Michelle Toh, “Moutai coffee, anyone? Luckin is adding the fiery liquor to its lattes”, in CNN Business:The popular Chinese coffee chain rolled out the so-called “sauce-flavored latte” with a jolt of liquor for 38 yuan ($5.20) on Monday. Customers who order with an online coupon will be able to receive 50% off for a limited time, it said.
2024 February 20, Jessie Yeung and Hassan Tayir, “Pork flavored coffee is Starbucks’ newest China pitch”, in CNN Business:The drink is priced at 68 yuan ($9.45), according to the app.
For more quotations using this term, see Citations:yuan.
Synonyms
[edit]
kuai, kwai(informal)
Translations
[edit]
show ▼±basic unit of money in China
[Select preferred languages] [Clear all]
Arabic: يُوَانm(yuwān)
Armenian: յուան(hy)(yuan)
Belarusian: юа́нь(juánʹ)
Bulgarian: юа́нm(juán)
Burmese: ယွမ်(my)(ywam)
Catalan: iuan(ca)m
Chinese: Cantonese: 元(jyun 4), 蚊(yue)(man 4-1)(colloquial)Dungan: йүан(yüan)Hokkien: 箍(zh-min-nan)(kho͘)Mandarin: 圓/ 圆(zh)(yuán)(formal); 元(zh)(yuán)(less formal); 塊/ 块(zh)(kuài)(colloquial); 人民幣/ 人民币(zh)(rénmínbì)(Chinese currency in general)
Czech: jüan(cs)
Esperanto: juano
Estonian: jüaan
Finnish: juan(fi), yuan(fi)
French: yuan(fr)m
German: Yuan(de)m
Hebrew: יוּאָןm(yuán)
Hindi: युआन(yuān)
Hungarian: jüan(hu)
Italian: yuan(it)
Japanese: 元(ja)(げん, gen), ユアン(yuan), 人民元(じんみんげん, jinmingen)(Chinese currency in general)
Kazakh: юән(üän), қуай(quai)
Khmer: យ័ន(yuon)
Korean: 원(圓)(ko)(won), 유안(yuan), 위안(ko)(wian)
Latvian: juaņsm
Lithuanian: juanism
Macedonian: ју́анm(júan)
Malay: yuan(ms)
Maori: wāna
Marathi: युआन(yuān)
Mongolian: юань(mn)(juanʹ)
Persian: یوان(yuân)
Polish: juan(pl)m
Portuguese: yuan(pt)m
Russian: юа́нь(ru)m(juánʹ)
Serbo-Croatian: Cyrillic: јуанm Roman: juanm
Slovak: jüanm
Spanish: yuan(es)
Thai: หยวน(th)(yǔuan)
Ukrainian: юа́нь(uk)m(juánʹ)
Uyghur: يۈەن(yüen), كوي(koy)
Vietnamese: nguyên(vi)
Zhuang: maenz
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)
See also
[edit]
renminbi
CNY
RMB
Etymology 2
[edit]
From Mandarin院 (yuàn).
Noun
[edit]
yuan (pluralyuanoryuans)
(Taiwan,politics) One of the five branches of the government of the Republic of China.
Derived terms
[edit]
Control Yuan
Examination Yuan
Executive Yuan
Judical Yuan
Legislative Yuan
show more ▼
Anagrams
[edit]
Äynu, anyu
French
[edit]
Etymology
[edit]
Borrowed from Mandarin元 (yuán).
Pronunciation
[edit]
IPA(key): /jɥan/; /ɥan/
Noun
[edit]
yuanm (pluralyuans)
yuan(Chinese currency)
Further reading
[edit]
“yuan”, in Trésor de la langue française informatisé[Digitized Treasury of the French Language], 2012.
Mandarin
[edit]
Romanization
[edit]
yuan
nonstandard spelling of yuān
nonstandard spelling of yuán
nonstandard spelling of yuǎn
nonstandard spelling of yuàn
Usage notes
[edit]
Transcriptions of Mandarin into the Latin script often do not distinguish between the critical tonal differences employed in the Mandarin language, using words such as this one without indication of tone.
Norwegian Nynorsk
[edit]
Noun
[edit]
yuanm (definite singularyuanen, indefinite pluralyuanar, definite pluralyuanane)
currency used in China (since 1971)
References
[edit]
“yuan” in The Nynorsk Dictionary.
Portuguese
[edit]
Alternative forms
[edit]
iuane
Etymology
[edit]
From Mandarin元 (yuán).
Noun
[edit]
yuanm (pluralyuans)
yuan(currency unit of China)
Romanian
[edit]
Etymology
[edit]
Borrowed from GermanYuan.
Noun
[edit]
yuanm (pluralyuani)
yuan
Declension
[edit]
show ▼Declension of yuan| | singular | | plural |
---
| | indefinite | definite | indefinite | definite |
| nominative-accusative | yuan | yuanul | yuani | yuanii |
| genitive-dative | yuan | yuanului | yuani | yuanilor |
| vocative | yuanule | yuanilor |
Spanish
[edit]
Etymology
[edit]
Borrowed from Mandarin元 (yuán).
Pronunciation
[edit]
more ▼
IPA(key): /ˈʝwan/[ˈɟ͡ʝwãn]
IPA(key): /ˈʝwan/ˈɟ͡ʝwãn and Uruguay)
IPA(key): /ˈjwan/ˈjwãn and Uruguay)
Rhymes: -an
Syllabification: yuan
Noun
[edit]
yuanm (pluralyuanes)
yuan
Further reading
[edit]
“yuan”, in Diccionario de la lengua española [Dictionary of the Spanish Language] (in Spanish), online version 23.8, Royal Spanish Academy [Spanish: Real Academia Española], 10 December 2024
Retrieved from "
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How Microbiology started
Peter Collins1
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Correspondence: Peter Collins, pmd.collins@btinternet.com
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2025-08-13
ISSN: 1350-0872, Online ISSN: 1465-2080
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Volume 168, Issue 2
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First, launch the Society
Then, launch the journal
General microbiology?
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How Microbiology started
Peter Collins
1
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The Royal Society, 6–9 Carlton House Terrace, St. James’s, London, UK
Correspondence: Peter Collins, pmd.collins@btinternet.com
First, launch the Society
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First, launch the Society
Then, launch the journal
General microbiology?
Acknowledgements
References
From the outset, the founders of the Society for General Microbiology (as it was called until 2015) intended to launch a journal. The journal was a key resource for promoting their vision of microbiology as a coherent and wide-ranging scientific discipline. They would soon come to realise that it was also key to financing the new Society and to carrying its message to colleagues worldwide.
First they had to get the Society itself up and running. Many of the founders had worked directly or indirectly on the second congress of the International Society for Microbiology, held in London in 1936 – the bacteriologist John Ledingham, for example, had been President of the congress and Ralph St John-Brooks had been Secretary-General. Others were clustered around the Lister Institute for Preventive Medicine in Elstree, other medical settings, various commercial laboratories, or the Society of Agricultural Bacteriologists (SAB). That 1936 congress was, in Sid Elsdon’s assessment1, the first time that British microbiologists of all persuasions had come together and worked hard for common purpose. The idea of microbiology as a scientific discipline distinct from its diverse fields of application began to take root in the UK at that time.
The outbreak of war naturally disrupted progress. The next occasion when ideas about the future development of the discipline could be discussed with a relatively large number of those involved was not until September 1943, at the annual conference of the SAB. The SAB was worried about its long-term future. One option appeared to be turning itself into the learned society for this wider vision of microbiology. However, influential figures such as Ledingham and St John-Brooks felt that the SAB, with its roots firmly in the world of dairying, would be too narrow a starting point, and that a wholly new body would be more likely to deliver the wider vision. Marjory Stephenson, at her ‘forceful best’ in Sid Elsdon’s recollection, persuasively argued the case for building a new society around microbiology as a whole rather than around bacteriology, so that it could include ‘parasitologists and those studying yeasts, fungi and protozoa’ . She, Ledingham and St John-Brooks carried the day.
The upshot of that AGM was that Leslie Allen, newly elected President of the SAB, was deputed by the SAB to explore the possibilities of forming a new society with St John-Brooks (his near neighbour in Bushey, Hertfordshire). Together they set about the delicate process of building consensus. As a first step, they convened a meeting on 16 November 19432 with representatives of the Association of Applied Biologists, the Biochemical Society, the British Mycological Society, the Pathological Society and the SAB, and with individuals working in virology, animal parasitology, dairying, type cultures and various industrial settings3. The meeting was chaired by John Ledingham, with Allen and St John-Brooks acting as Joint Secretaries4. It was agreed that there should indeed be a new society, and a new journal, its name and scope mirroring those of the society itself. This would be one of the new society’s core activities, along with holding scientific meetings, enabling microbiologists of different backgrounds to interact, and seeking to influence the teaching of microbiology. The meeting also agreed to Marjory Stephenson’s suggested name, the ‘Society for General Microbiology’.
The subcommittee5 appointed to take it all forward held its first meeting a month later, chaired on that occasion by Marjory Stephenson6. The main focus then was on how to recruit members. That involved clarifying just what was covered by the term ‘general microbiology’. Muriel Robertson thought it should include ‘the biology of diatoms, protozoa, and microscopic algae’. John Ledingham opted for ‘the bionomics of bacteria, viruses, micro-fungi, protozoa, and microscopic algae’. Arthur Stableforth went for ‘the more fundamental aspects of … taxonomy, systematics, dissociation, physiology, nutrition and ecology’. It was left to Leslie Allen and Ralph St John-Brooks to sort it out, in consultation with BCJG (‘Gabe’) Knight who had been brought into the original discussions at the suggestion of Marjory Stephenson. Their task was encapsulated in the agreed, though never used, subtitle for the society: ‘The Society for the Establishment and Extension of Common Ground between all forms of Microbiology’.
The subcommittee was clear that the new Society, and its journal, should focus on the fundamental aspects of microbiology. As a concession, it accepted the case for including papers in applied areas where there was no suitable English journal, provided always that they contained results of ‘general scientific interest’.
Leslie Allen reported these developments to the SAB in January 1944. He explained that extensive consultations had confirmed support for a society dealing with the more fundamental aspects of microbiology. He defined these aspects as ‘the biology of bacteria, viruses, micro-fungi, protozoa, and microscopic algae’, with the study of these entities to include ‘taxonomy, systematics, dissociation, variation, physiology (including oxidation-reduction potentials, vitamins, etc.), and the study of germicides, nutrition, and ecology’. The proposed society would not challenge the SAB or similar societies within their own existing spheres. The SAB welcomed these moves and, in a parallel initiative, broadened its own scope, and its name, from Agricultural Bacteriologists to the more wide-ranging Society for Applied Bacteriology, an approximate analogue in the applied sphere of the Society for General Microbiology .
The subcommittee, then led by Ashley Miles, set about the business of establishing formal contacts with existing learned societies that might be affected by its plans, and developing a list of potential original members, to an initial total of about 250. It also had the fun of drafting formal rules for the Society. All these were debated, tweaked and approved by the full committee7 on 20 March 1944. The letters to potential members included a commitment that ‘as soon as circumstances permit’, the new Society would launch a journal for papers within its disciplinary scope. Importantly, the journal would accept any suitable papers, including those that had not first been read at its meetings.
The full committee met again on 25 May 1944 to review progress and to agree a slate of initial office-holders; but it did not at that stage address the matter of who would edit the journal. Given his leadership role among British microbiologists, John Ledingham would have been the obvious choice for the inaugural presidency, but had already effectively ruled himself out. Allen and St John-Brooks therefore wrote to David Keilin, who declined through pressure of work; then to Marjory Stephenson, who declined because she felt she was not the most suitable person, though she did later accept a subsequent invitation; and finally to Alexander Fleming, who accepted. The inaugural meeting, originally scheduled for 29 September 1944, was postponed till 16 February 1945 to allow time to get all the arrangements in place. The Second World War, after all, was still in full swing, London was subject to frequent heavy air-raids and flying bombs, and ‘the difficulties of travelling resulting from evacuation and the movement of troops’8 were considerable.
The formal rules approved at the inaugural meeting in February 1945 included a commitment ‘as soon as practicable to issue a journal known as The Journal of General Microbiology’; but the post of Editor still remained unfilled. A major problem for the intended journal was wartime restrictions on the availability of paper, which effectively put stop to such new initiatives9. The annual subscription was set initially at one guinea (£1.05), pending an increase to cover the costs of the journal when known; all members would be entitled to receive copies of all the Society’s publications.
Muriel Robertson told Allen and St John-Brooks that, in guiding the embryonic Society to its starting point they ‘had done a noble and arduous piece of work, and we should all be grateful to you’. ‘Praise from her is praise indeed!’ wrote St John-Brooks to Allen (St John-Brooks to Allen, personal communication 20 November 1944). Their reward at the inaugural meeting was to be voted in as Joint Secretaries, so that the arduous work could continue.
| |
| Allen and St John-Brooks [3, 4] Leslie Allen (1903–64) and the older Ralph St John-Brooks (1884–1963) became good friends through their shared experience in getting the Society off the ground. The fact that they happened to live near each other during the key period was a big help. It was natural that they should be appointed as the first Joint Secretaries of the new Society when it was formally launched in February 1945. In July 1946, St John-Brooks stepped back on medical advice from his role as Joint Secretary (he had suffered from pulmonary tuberculosis 20 years previously, and had never fully recovered), to be succeeded by Kits van Heyningen. In November 1946 he was made the Society’s first Honorary Member, which came as ‘a tremendous but very agreeable surprise’. Towards the end of that year he also retired from his long-standing role as Curator of the National Collection of Type Cultures at the Lister Institute. He then had stints at the American Type Culture Collection in Washington and the equivalent Swiss body at Lausanne. Allen wrote warmly to him in 1946 recalling fondly ‘the long fireside chats we used to have in the evenings when the project was taking shape’. ‘I feel we did a great work together’, he added. ‘The success of the new Society shows how much can be accomplished quietly by a few men of good will who work for a common aim, and who do not have to protect their personalities from the public gaze by putting on cloaks of prestige, or aggressiveness, or shyness, which make people misunderstand them. One wonders how much could be accomplished in a similar way in international affairs.’ Indeed! Allen, then Chief Microbiologist at the DSIR Water Pollution Research Laboratory, retired from his role as Joint Secretary of the Society in March 1948, citing pressure of other commitments, to be succeeded by JG Davis. St John-Brooks enjoyed sending Allen food parcels from America in 1947, which arrived safely, to their mutual relief. Allen would eventually write St John-Brooks’ obituary for the Journal of General Microbiology (JGM), but himself died before it was published; his own obituary was written by the Society’s inaugural Treasurer, Bill Bunker. |
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The Society’s newly elected Council10, at its first meeting on 16 March 1945, was keen to push ahead with the journal rather than dawdle until paper rationing was lifted. So it appointed a subcommittee comprising Gabe Knight, Alexander Mattick, Muriel Robertson and the two Secretaries (Allen and St John Brooks) to get things moving. By September 1945, opinion in the trade was that it should be possible to get permission from the Ministry of Production to launch a new scientific journal. By December, Council decided, on the advice of Cambridge University Press (CUP), that an annual subscription of 35/- (£1.75) would cover the additional costs of the journal with a sufficient margin for error. In February 1946, Council appointed CUP as publishers of the journal.
In February 1946, Council also, finally, addressed the issue of the Editor. It decided, pragmatically if a bit confusingly, to have two Joint Editors ‘of equal status’, one for ‘chemical’ papers and one for ‘biological’ papers11. It also approved the idea of a small Editorial Board, whose members (‘Associate Editors’) were to be selected primarily for their practical editorial experience and only secondarily, if at all, as covering a reasonable breadth of disciplines. This was in keeping with the Society’s aspiration of getting away from the prevailing disciplinary tribalism within microbiology. The Associate Editors would act as referees alongside the Editors (with help in peer reviewing from additional external experts as needed), and would prepare accepted papers for the press – though in practice most of that burden seems to have fallen on the two principal Editors.
Council elected Gabe Knight ahead of Marjory Stephenson’s former research student Donald Woods12 as the ‘chemical’ Joint Editor, but could not decide on the ‘biological’ Joint Editor; Knight invited Ashley Miles to fill that role after the meeting. Knight and Miles together selected the initial set of Associate Editors: Geoffrey Ainsworth, William Brierley, Tom Gibson, Alexander Mattick, Kenneth Smith, Arthur Stableforth13 and Donald Woods14. Brierley, Mattick and Smith were then also members of the Society’s Council, as were Knight and Miles. Both Knight and Miles were 43 years old in 1947; the others ranged from 35 years old (Woods) to 58 years old (Brierley). The group of nine – all men – included one current (Smith) and two future (Woods, Miles) Fellows of the Royal Society.
Muriel Robertson had, in November 1943, prepared a briefing on other people’s experiences in launching new learned societies and new journals, in which she commented: ‘The Editor must be hardworking and intelligent but also have some sufficient serpentine wisdom’. Beginnings, she noted, were easy: the real challenge was dealing with loss of momentum after the initial enthusiasm had worn off. Gabe Knight himself clearly had both the serpentine wisdom and the stamina required: he served as the most senior Editor until 1970, longer than anyone in the journal’s history.
The Editors and Associate Editors functioned mostly as a loose group of individuals rather than as a Board. They held formal meetings at the Linnean Society’s comfortable premises in Burlington House in September 1946, and again in December 1949, to go over various practical matters, but their plan to meet annually after that did not last long: in 1970 John Postgate was complaining that the Board had met only once in the previous 14 years. Ainsworth stepped down from his role in 1951; Miles moved to being an Associate Editor in 1951 and served in that capacity, with five of the original Associate Editors, until 1960; Gibson stayed on till 1969; and Knight outlasted them all. From 1951 to 1970, Arthur Standfast was the second Joint Editor.
The administrative office for the journal was established initially at the National Institute for Medical Research in London, where Ashley Miles was then Deputy Director and had access to secretarial support. At that early stage, the Society of course had no paid staff, but Council agreed in December 1946 that Miles’s secretary at NIMR could be paid an honorarium of £10 for each issue of the journal. The office moved to Standfast’s Department at Elstree, and his ‘indefatigable’ secretary Linda Peerless , when he succeeded Miles in 1951.
A notice advertising the new journal mentioned the ambition of getting the first issue of the journal published before the end of 1946, and called for papers on both specialised and general topics within the Society’s scope15. The Editors urged authors to make specialised papers accessible to readers from other disciplines. Twenty-seven papers were submitted by the end of 1946, and a further thirty-nine during 1947.
In November 1946, Council had the first of very many detailed discussions about the finances of the journal. Its costs greatly outweighed all other Society expenditure, so any miscalculation would have very serious consequences. On the other hand, Council quickly discovered that its profitability would make it the dominant source of Society income, outstripping members’ subscriptions by a considerable margin. Either way, finance would loom large in all policy discussions about the journal. At this early stage, the decision was simply to accept an element of advertising in the journal to mitigate any potential losses.
The caution about finance was understandable, indeed the only responsible course of action. But the journal proved to be a financial success from the outset. With the Society’s membership already approaching 600 and CUP estimating ‘trade’ sales (e.g. to academic libraries and industrial companies with relevant interests) of 350, the print run of the first issue was originally set at 1000. Allen got that increased to 1200, and the same for the second and third issues. The three issues of volume one (1947), with a total of 39 papers, cost the Society about £500 net (costs £1300, income £800); the three issues of volume two (1948), also with 39 papers, cost under £200 net because of increased sales. The print run for volume three (1949) was pushed up to 1500 after increased paper supplies had been secured. By 1955 the print run was 2750 (121 papers in six issues), and 4000 (137 papers also in six issues) in 1959. Already by the end of 1949, the journal had secured 500 external (i.e. non-member) subscriptions , which meant a degree of financial security.
| |
| Three Joint Editors of JGM Three individuals essentially ran JGM for its first 24 years. The longest-serving was Gabe Knight, who, with Ashley Miles, got it off the ground from 1946 onwards. Miles stepped down in 1951 and was succeeded by Arthur Standfast. Knight and Standfast both retired in 1970, and were elected Honorary Members of the Society in recognition of their long and effective service. Knight’s research was mainly on bacterial growth physiology, and he published important work on the nutritional requirements of bacteria including Staphylococcus aureus . His research career started with RK Cannan and then Paul Fildes on how the redox environment of the growth medium influences spore germination. In 1934 he moved with Fildes to a new MRC Unit on Bacterial Chemistry in the Middlesex Hospital. This was followed by four years at the Lister Institute at Elstree and, from 1943, eight years with the Wellcome Laboratories at Beckenham. In 1951 he was appointed to the first Chair in Microbiology in the first UK university Department of Microbiology, at Reading, where one of the buildings is now named after him. He retired from Reading in 1969. An avid and learned Francophile, he translated books on and by Diderot and Stendhal, and an internet search is as likely to connect him with that activity as with microbiology. Miles, ‘essentially a medical microbiologist of the classical type’, started in medicine and then held posts in bacteriology and pathology before taking the Chair in Bacteriology at University College Hospital in 1937 . This was soon combined with a range of war-related responsibilities. There followed a stint as Deputy Director at NIMR, with a particular focus on biological standards, during 1946–52, and then appointment as Director of the Lister Institute and Professor of Experimental Pathology at London. He retired from both these posts in 1971. He was elected FRS in 1961 and served as Biological Secretary of the Royal Society 1963–68. Miles’s wife Ellen, with whom he co-published several research papers, was a half-sister of Roald Dahl. Standfast (1905–90) joined the Elstree Laboratories of the Lister Institute in 1945, working on the preparation and study of bacterial vaccines (based on annual reports of the Lister Institute. Muriel Robertson was working in protozoology at the Institute when he arrived. At first Standfast seems to have been a one-man band, but he was gradually able to build up a team and came to serve as Head of the Bacterial Vaccine Department for 26 years, until his retirement in 1972. For the whole of this period his research was mainly on pertussis, cholera, typhoid and tetanus vaccines, and he organised their commercial production at Elstree. |
General microbiology?
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Given the Society’s origins and aspirations, it was inevitable that the disciplinary distribution of the papers published in its journal should attract close attention. Analyses presented at successive AGMs highlighted the fact that the bacteriological heritage of the Society was fading too slowly, to the occasional frustration of the Editors. Already at the September 1947 AGM, Ashley Miles remarked: ‘The Editors realise that the subjects covered so far do not represent all branches of microbiology, but hope that appropriate papers from all disciplines will be submitted to make the Journal more representative of general microbiology’. A few months later, in March 1948, Gabe Knight reiterated the message: ‘The Editors are aware that the Journal does not yet fully reflect in a balanced way all the different disciplines which the Society was formed to help unify. The remedy lies in the hands of the members’. And Ashley Miles again in April 1949: ‘The Society will notice that bacteriological papers still predominate. … The attainment of the desirable end of making the Journal more representative of general microbiology … lay in the hands of Members of the Society’.
These messages about the ‘great preponderance of bacteriological papers’ continued year after year. Finally, at the 1954 AGM, there was a slight change of tone: ‘The distribution of papers in 1953 showed an increase over 1952 in papers in Microbial Genetics, Mycology and Viruses, which the Editors feel is a good sign’. But only a slight change. By the 1956 AGM progress had stalled, and the Editors commented languidly: ‘It would seem that the Society gets the journal it writes, if not the journal it deserves’. But it was at least attracting about a third of its papers from outside the UK: ‘The journal can hardly be accused of being too insular’.
Yet the project to embed the distinct discipline of microbiology in the consciousness of the British scientific community did succeed over time. Academic appointments, university departments, undergraduate courses did gradually come to reflect that recognition. By 1994, when the journal dropped the word ‘general’ from its name and graduated to plain Microbiology, it was not giving up on its mission but rather signalling that the message had been accepted and no longer needed to be spelt out. But that’s another story.
Acknowledgements
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My warm thanks to Peter Cotgreave and Gavin Thomas for the opportunity to write this paper, and for interesting discussions during the writing of it.
Notes
1The sources for this paper are largely to be found within the uncatalogued archives of the Microbiology Society, held at the Wellcome Collection, and have not been itemised individually here. These archives include some correspondence, random Council papers, and valuable typescript histories of the Society including a 34-page essay by Sid Elsdon (President, 1969–72) on the foundation of the Society, and a 64-page piece by Doug Watson (Treasurer, 1979–87) on the first 50 years, based on analysis of Council minutes. I have also had access to a full set of Council minutes. Other sources are referenced below as relevant.
2The minutes wrongly give the date as Tuesday 17 November, but it was actually Tuesday 16 November.
3Those present on 16 November were: Leslie Allen, Christopher Andrewes, Bill Bunker, DHF Clayson, CE Coulthard, JT Duncan, Alexander Fleming, HB Hutchinson, Gabe Knight, John Ledingham, Walter Macleod, EW Mason, Alexander Mattick, Ashley Miles, Walter Morgan, Muriel Robertson, Ralph St John-Brooks, Kenneth Smith, Arthur Stableforth, Marjory Stephenson, Henry Thornton, Samuel Wiltshire and WR Wooldridge, with Sam Bedson, WJ Dowson and Paul Fildes sending apologies.
4Ledingham subsequently took more of a back seat as he found ‘the burden of presiding over our gatherings too exacting for him’, and he died on 4 October 1944 after a brief illness .
5Its members were: HJ (‘Bill’) Bunker, Gabe Knight, John Ledingham, Alexander Mattick, Ashley Miles, Muriel Robertson, Arthur Stableforth, Marjory Stephenson, Samuel Wiltshire; with Ralph St John-Brooks and Leslie Allen as Joint Secretaries .
6At her own request, Marjory Stephenson remained an active participant but was spared the burden of chairing subsequent meetings, which were taken instead by Ashley Miles.
7Essentially, the group that had met on 16 November 1943.
8Undated circular from Allen and St John-Brooks postponing the inaugural meeting, circa October 1944.
9By 1945, for example, newspapers were restricted to 25 % of their pre-war consumption of paper.
10It was, somewhat unhelpfully, termed the ‘Committee’ until July 1962, but for clarity I have used ‘Council’ throughout.
11In practice, they were always identified simply as ‘Editors’.
12Woods instead served as an Associate Editor, from 1946 to 1960. He was later nominated to succeed David Henderson as SGM President for the 1965–7 term, but died before he could assume office. His place was taken by Percy Brian.
13Stableforth, Senior Research Officer and later Director of the Weybridge Central Veterinary Laboratory, complained that the initial slate of six Associate Editors lacked any veterinary microbiologists, and thereupon was added to the team.
14Woods, then Reader in Microbiology at the Oxford Department of Biochemistry, would in 1955 be appointed to the country’s first Chair in Chemical Microbiology, in the same department.
15In fact, the first issue, labelled January 1947, appeared at the end of March 1947 – the first of many missed deadlines. Leslie Allen to SE Jacobs (original member of the SGM), 27 March 1947: ‘The first issue of the Journal of General Microbiology is expected to be out this week. It has been continually delayed for one reason and another, recently because of the coal shortage’.
References
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First, launch the Society
Then, launch the journal
General microbiology?
Acknowledgements
References
Society of Agricultural Bacteriologists
Minutes of the AGM (part ii) 9 September; 1943
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Hopton JW.
Sixty years of the Society for Applied Bacteriology: a story of growth, enterprise and consolidation. J Appl Microbiol 1991; 71:2–8 [View Article]
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Ralph St John-Brooks, 1884–1963. J Gen Microbiol 1966; 42:165–167 [View Article] [PubMed]
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Leslie Alfred Allen, 1903–1964. J Gen Microbiol 1966; 42:169–170 [View Article]
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Fifty years on (Society for General Microbiology); 1995; 7
6. Society for General Microbiology
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7. Zatman LJ.
BCJG Knight, 1904–1981. J Gen Microbiol 1983; 129:1261–1268 [View Article]
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8. Neuberger A
Arnold Ashley Miles, 1904–1988, Biographical Memoirs of Fellows of the Royal Society
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Round off the decimal number to the nearest one: 8.4
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In order to round off a decimal to the nearest ones, you need to look at the tenths digit. For the number 8.4, because 4 is less than 5, the number is rounded down to 8.
Explanation
In the system of mathematics, when you want to round off a decimal number to the nearest ones, you need to look at the digit in the tenths place. In this case, the decimal number to be rounded off is 8.4. You look at the number in the tenths place, which is 4. Since 4 is less than 5, the digit in the ones place remains as is. So, the rounded off number would be 8.
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To round off the decimal number 8.4 to the nearest whole number, you look at the tenths digit, which is 4. Since 4 is less than 5, you round down, and the result is 8. Therefore, 8.4 rounded to the nearest one is 8.
Explanation
To round off the decimal number to the nearest whole number, you will need to look at the digit in the tenths place. The given number is 8.4.
Identify the tenths digit: In 8.4, the tenths digit is 4.
Determine whether to round up or down: The rule for rounding is simple:
If the tenths digit is 5 or greater, then you round up the digit in the ones place.
If the tenths digit is less than 5, then you keep the ones place digit as it is.
Apply the rule: Since 4 is less than 5, you should round down the number. Therefore, the digit in the ones place remains 8.
The rounded number is 8.
In summary, when rounding 8.4 to the nearest one, the result is 8.
Examples & Evidence
For example, if you have the number 7.6, you would look at 6 in the tenths place which is greater than 5, so you round up to 8. Another example is 9.3; here, 3 is less than 5, and you would round down to 9.
The rounding rules are based on the common method used in mathematics, where numbers are rounded to make them easier to work with while maintaining a reasonable level of accuracy.
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Simplify the following expression by combining like terms: 5 x 2+6 x−3+4 x 2−8 x+2 a. A term is a single number, variable, or product of numbers and variables. How many terms are in the original expression? How many are in the simplified expression? b. When variable(s) are multiplied by a number, the number is called a coefficient. What is the coefficient of x in your simplified expression?
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2018 AIME II Problems/Problem 10
Contents
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1 Problem
2 Solution 1
3 Solution 2
4 Solution 3
5 Note (fun fact)
6 See Also
Problem
Find the number of functions from to that satisfy for all in .
Solution 1
Just to visualize solution 1. If we list all possible , from to in a specific order, we get different 's. Namely:
To list them in this specific order makes it a lot easier to solve this problem. We notice, In order to solve this problem at least one pair of where must exist.In this case I rather "go backwards". First fixing pairs , (the diagonal of our table) and map them to the other fitting pairs . You can do this in way. Then fixing pairs (The diagonal minus ) and map them to the other fitting pairs . You can do this in ways. Then fixing pairs (The diagonal minus ) and map them to the other fitting pairs . You can do this in ways. Fixing pairs (the diagonal minus ) and map them to the other fitting pairs . You can do this in ways. Lastly, fixing pair (the diagonal minus ) and map them to the other fitting pairs . You can do this in ways.
So
Solution 2
We perform casework on the number of fixed points (the number of points where ). There must be at least one fixed point, given has some value and is a fixed point. To better visualize this, use the grid from Solution 1.
Case 1: 5 fixed points
Obviously, there must be way to do so.
Case 2: 4 fixed points
ways to choose the fixed points. For the sake of conversation, let them be .- The last point must be or .- There are solutions for this case.
Case 3: 3 fixed points
ways to choose the fixed points. For the sake of conversation, let them be .- Subcase 3.1: None of or map to each other - The points must be and , giving solutions.- Subcase 3.2: maps to or maps to - The points must be and or and , giving solutions.- There are solutions for this case.
Case 4: 2 fixed points
ways to choose the fixed points. For the sake of conversation, let them be .- Subcase 4.1: None of , , or map to each other - There are solutions for this case, by similar logic to Subcase 3.1.- Subcase 4.2: exactly one of maps to another. - Example: - ways to choose the 2 which map to each other, and ways to choose which ones of and they map to, giving solutions for this case.- Subcase 4.3: two of map to the third - Example: - ways to choose which point is being mapped to, and ways to choose which one of and it maps to, giving solutions for this case.- There are solutions.
Case 5: 1 fixed point
ways to choose the fixed point. For the sake of conversation, let it be .- Subcase 5.1: None of map to each other - Obviously, there is solution to this; all of them map to .- Subcase 5.2: One maps to another, and the other two map to . - Example: - There are ways to choose which two map to each other, and since each must map to , this gives .- Subcase 5.3: One maps to another, and of the other two, one maps to the other as well. - Example: - There are ways to choose which ones map to another. This also gives .- Subcase 5.4: 2 map to a third, and the fourth maps to . - Example: - There are ways again.- Subcase 5.5: 3 map to the fourth. - Example: - There are ways to choose which one is being mapped to, giving solutions.- There are solutions.
Therefore, the answer is
~First
Solution 3
We can do some caseworks about the special points of functions for . Let , and be three different elements in set . There must be elements such like in set satisfies , and we call the points such like on functions are "Good Points" (Actually its academic name is "fixed-points"). The only thing we need to consider is the "steps" to get "Good Points". Notice that the "steps" must less than because the highest iterations of function is . Now we can classify cases of “Good points” of .
One "step" to "Good Points": Assume that , then we get , and , so .
Two "steps" to "Good Points": Assume that and , then we get , and , so .
Three "steps" to "Good Points": Assume that , and , then we get , and , so .
Divide set into three parts which satisfy these three cases, respectively. Let the first part has elements, the second part has elements and the third part has elements, it is easy to see that . First, there are ways to select for Case 1. Second, we have ways to select for Case 2. After that we map all elements that satisfy Case 2 to Case 1, and the total number of ways of this operation is . Finally, we map all the elements that satisfy Case 3 to Case 2, and the total number of ways of this operation is . As a result, the number of such functions can be represented in an algebraic expression contains , and :
Now it's time to consider about the different values of , and and the total number of functions satisfy these values of , and :
For , and , the number of s is
For , and , the number of s is
For , and , the number of s is
For , and , the number of s is
For , and , the number of s is
For , and , the number of s is
For , and , the number of s is
For , and , the number of s is
For , and , the number of s is
For , and , the number of s is
For , and , the number of s is
Finally, we get the total number of function , the number is
~Solution by (Frank FYC)
Note (fun fact)
This exact problem showed up earlier on the 2011 Stanford Math Tournament, Advanced Topics Test. This problem also showed up on the 2010 Mock AIME 2 here:
See Also
2018 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 9Followed by
Problem 11
1•2•3•4•5•6•7•8•9•10•11•12•13•14•15
All AIME Problems and Solutions
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.
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3711 | https://mathoverflow.net/questions/220030/normal-approximation-of-tail-probability-in-binomial-distribution | reference request - Normal approximation of tail probability in binomial distribution - MathOverflow
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Normal approximation of tail probability in binomial distribution
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My problem: From the Berry--Esseen theorem I know, that
sup x∈R|P(B n≤x)−Φ(x)|=O(1 n−−√),sup x∈R|P(B n≤x)−Φ(x)|=O(1 n),
where B n B n has the standardized binomial distribution and Φ Φ is the standard normal distribution function. I can prove this for x≈0 x≈0 with Stirling's formula and in a way similar to this proof.
Unfortunately Stirling's approximation becomes worse the bigger |x||x| is, so that I can only show that
sup|x|≤c|P(B n≤x)−Φ(x)|=O(1 n−−√)sup|x|≤c|P(B n≤x)−Φ(x)|=O(1 n)
for a fixed c>0 c>0.
My question: Is there a direct proof for sup|x|>c|P(B n≤x)−Φ(x)|=O(1 n√)sup|x|>c|P(B n≤x)−Φ(x)|=O(1 n), based on Stirling's formula? How can I estimate the error of the difference in the tail probability of the binomial distribution to the normal distribution?
Note:Chebyshev's inequality does not provide the right convergence speed. As zhoraster shows in his answer to Finding an error estimation for the De Moivre–Laplace theorem, also Chernoff's inequality does not provide the right convergence speed, too. That's why I am looking for another method. I also want to use more easy and direct approximations than Stein's method which is used in the proof of the Berry--Esseen theorem.
I already asked this question on MSE (see this thread), but I haven't got an answer there (even after adding a bounty). That's why I want to ask this question here.
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edited Apr 13, 2017 at 12:19
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asked Oct 5, 2015 at 9:55
Stephan KullaStephan Kulla
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The needed observation here is that, whereas the relative error of the Stirling approximation is worse for large |x||x|, it gets multiplied by a fast decreasing normal density function, and then the product gets integrated to produce the desired result. Here are the details.
Let X X be a binomial random variable with parameters n∈N n∈N and p∈(0,1)p∈(0,1), where n→∞n→∞ and p p is fixed. For any integer k k, let
z k:=k−n p n p q−−−√,q:=1−p.z k:=k−n p n p q,q:=1−p.
Let
K:={k=0,…,n:|z k|≤n 1/4},K:={k=0,…,n:|z k|≤n 1/4},
so that k/n→p k/n→p uniformly over k∈K k∈K. Let
p^:=k/n,q^:=1−p^=(n−k)/n,p^:=k/n,q^:=1−p^=(n−k)/n,
ϵ:=ϵ k:=z k/n−−√,λ:=q/p−−−√,ϵ:=ϵ k:=z k/n,λ:=q/p,
whence
p^p=1+ϵ λ,q^q=1−ϵ/λ.p^p=1+ϵ λ,q^q=1−ϵ/λ.
Uniformly over k∈K k∈K one has k∼n p k∼n p and n−k∼n q n−k∼n q, so that 1 k=O(1 n)1 k=O(1 n) and 1 n−k=O(1 n)1 n−k=O(1 n). Here and in what follows, the constant in O(⋅)O(⋅) depends only on p p. So, by Stirling's formula, over all k∈K k∈K
P(X=k)=n!k!(n−k)!p k q n−k=1 2 π n p^q^−−−−−−√1 R(1+O(1 n))P(X=k)=n!k!(n−k)!p k q n−k=1 2 π n p^q^1 R(1+O(1 n))
=1 2 π n p q−−−−−√1 R(1+O(|ϵ|+1 n))=δ 2 π−−√e−n H(1+O(|ϵ|+1 n))=1 2 π n p q 1 R(1+O(|ϵ|+1 n))=δ 2 π e−n H(1+O(|ϵ|+1 n))
where
R:=(p^p)k(q^q)n−k,R:=(p^p)k(q^q)n−k,
δ:=1 n p q−−−√,H:=H k:=ln(R 1/n)=p^ln p^p+q^ln q^q:=p ψ(ϵ λ)+q ψ(−ϵ/λ),δ:=1 n p q,H:=H k:=ln(R 1/n)=p^lnp^p+q^lnq^q:=p ψ(ϵ λ)+q ψ(−ϵ/λ),
ψ(s):=(1+s)ln(1+s).ψ(s):=(1+s)ln(1+s).
Since ψ(s)=s+s 2/2+O(|s|3)ψ(s)=s+s 2/2+O(|s|3) for |s|≤1/2|s|≤1/2 and |z k|3 n√+|ϵ|=|z k|3 n√+|z k|n√=O(|z k|3+1 n√)|z k|3 n+|ϵ|=|z k|3 n+|z k|n=O(|z k|3+1 n), we find that
(1)P(X=k)=δ ϕ(z k)(1+O(|z k|3 n−−√))(1+O(|ϵ|+1 n))(1)P(X=k)=δ ϕ(z k)(1+O(|z k|3 n))(1+O(|ϵ|+1 n))
=δ ϕ(z k)(1+O(|z k|3+1 n−−√))=δ ϕ(z k)(1+O(|z k|3+1 n))
over all k∈K k∈K, where ϕ ϕ is the standard normal density. Next, for u=O(n 1/4)u=O(n 1/4) and |h|≤δ/2|h|≤δ/2,
ϕ(u+h)+ϕ(u−h)2=ϕ(u)cosh(h u)e−h 2/2=ϕ(u)(1+O(h 2 u 2+h 2))=ϕ(u)(1+O(1 n−−√)).ϕ(u+h)+ϕ(u−h)2=ϕ(u)cosh(h u)e−h 2/2=ϕ(u)(1+O(h 2 u 2+h 2))=ϕ(u)(1+O(1 n)).
So, for k∈K k∈K
δ ϕ(z k)=1 2∫δ/2−δ/2[ϕ(z k+h)+ϕ(z k−h)]d h(1+O(1 n−−√))δ ϕ(z k)=1 2∫−δ/2 δ/2[ϕ(z k+h)+ϕ(z k−h)]d h(1+O(1 n))
=∫z k+δ/2 z k−δ/2 ϕ(z)d z(1+O(1 n−−√)).=∫z k−δ/2 z k+δ/2 ϕ(z)d z(1+O(1 n)).
Also, for any k∈K k∈K and any z∈[z k−δ/2,z k+δ/2]z∈[z k−δ/2,z k+δ/2] one has |z k|3−|z|3≤3 δ(|z k|+δ/2)2=O(1)|z k|3−|z|3≤3 δ(|z k|+δ/2)2=O(1), whence |z k|3/n−−√=O((|z|3+1)/n−−√)|z k|3/n=O((|z|3+1)/n). So, by (1)(1), for some positive real c c depending only on p p and for all k∈K k∈K
J k−≤P(X=k)≤J+k,J k−≤P(X=k)≤J k+,
where
J±k:=∫z k+δ/2 z k−δ/2 ϕ(z)(1±c(|z|3+1)n−−√)d z.J k±:=∫z k−δ/2 z k+δ/2 ϕ(z)(1±c(|z|3+1)n)d z.
Hence, for any x∈[0,n 1/4]x∈[0,n 1/4], letting K x:={k∈Z:0<z k≤x}K x:={k∈Z:0<z k≤x}, one has
P(0<B n≤x)=∑k∈K x P(X=k)≤∑k∈K x J+k P(0<B n≤x)=∑k∈K x P(X=k)≤∑k∈K x J k+
≤∫x+δ/2−δ/2 ϕ(z)(1+c(|z|3+1)n−−√)d z≤∫−δ/2 x+δ/2 ϕ(z)(1+c(|z|3+1)n)d z
≤∫x 0 ϕ(z)d z+A δ+B n−−√≤∫x 0 ϕ(z)d z+C n−−√,≤∫0 x ϕ(z)d z+A δ+B n≤∫0 x ϕ(z)d z+C n,
where A,B,C A,B,C depend only on p p. Quite similarly one obtains the corresponding lower bound on P(0<B n≤x)P(0<B n≤x), so that
(2)|P(0<B n≤x)−(Φ(x)−1/2)|≤C n−−√(2)|P(0<B n≤x)−(Φ(x)−1/2)|≤C n
for x∈[0,n 1/4]x∈[0,n 1/4], where Φ Φ is the standard normal cumulative distribution function.
On the other hand, for real x>n 1/4 x>n 1/4 one has 1−Φ(x)=O(1/n−−√)1−Φ(x)=O(1/n) and, by Bernstein's inequality,
P(B n>x)≤exp(−x 2 2+x/n p q−−−√)≤e−x 2/4+e−x n p q√/2=O(1/n−−√).P(B n>x)≤exp(−x 2 2+x/n p q)≤e−x 2/4+e−x n p q/2=O(1/n).
It follows that (2)(2) holds for all real x≥0 x≥0 (perhaps with a greater constant C C). Similarly, |P(B n≤0)−1/2)|≤C n√|P(B n≤0)−1/2)|≤C n. Thus, for all real x x
|P(B n≤x)−Φ(x)|≤2 C n−−√,|P(B n≤x)−Φ(x)|≤2 C n,
as desired.
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edited Oct 6, 2015 at 19:21
answered Oct 6, 2015 at 16:39
Iosif PinelisIosif Pinelis
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I have added details on the use of Stirling's formula and filled the gap on the estimation of 2 π n p^q^2 π n p^q^ by 2 π n p q 2 π n p q.Iosif Pinelis –Iosif Pinelis 2015-10-06 18:15:50 +00:00 Commented Oct 6, 2015 at 18:15
1 In the beginning of the answer, I have inserted a brief description of the main idea used in the proof.Iosif Pinelis –Iosif Pinelis 2015-10-06 19:23:18 +00:00 Commented Oct 6, 2015 at 19:23
Thanks a lot for your proof! It helped me a lot to find a good proof for my thesis (and I am sorry for my late response). I would like to cite your answer in my master thesis. Therefore I have the following question: You have found the proof by yourself, right? Shall I cite also someone else?Stephan Kulla –Stephan Kulla 2015-10-21 13:59:33 +00:00 Commented Oct 21, 2015 at 13:59
2 I am glad this proof has been helpful. I have not seen it elsewhere, even though the few tools used here are quite common: Stirling's formula, Taylor's expansion, summation via integration, Bernstein's inequality, and a few elementary estimates.Iosif Pinelis –Iosif Pinelis 2015-10-21 15:24:00 +00:00 Commented Oct 21, 2015 at 15:24
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3712 | https://www.youtube.com/watch?v=DOEXysLTgpo | Physics Summary. Chapter 18: Electric Charge and Electric Field
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I'm working through chapter summaries for introductory physics (algebra-based). I'm using the Openstax online (free) textbook College Physics. You can access this book here
Here is the playlist with all the videos and problems for this course.
In this chapter:
- Fundamental charges
- Conductors vs. Insulators
- conservation of charge
- Coulomb force
- Superposition of forces
- Vector addition review
- Electric field
3 comments
Transcript:
chapter summary for 18 chapter 18 using Open Stacks I'm going to start off with the demo I have here a little stick I have here a piece of myar let's see what happens just Watch What Happens magic stuff happens so this is a tiny little uh electrostatic generator and this is a piece of myar that I can charge and if they have the same charge they repel and this is an example of electrostatics so we got a bunch of stuff to do here I'm not going to solve any problems uh I'm just doing a chapter summary problems will be in a different video okay so let's start off with the fundamental idea of of charges because if you look around you and there's a lot of stuff in here that's more conceptual and I'm going to focus more on the on the mathematical stuff but I want to talk about some of the conceptual stuff too if I look at this chalkboard the chalk my face which I can't see the camera which I can see there are three things and all of those are made up of just three things a neutron a proton an electron and it's kind of amazing to think about that right there are other things there are other fundamental particles neutrons and protons are not technically fundamental we can break those up into other things but everything that you see around you is made up of those three things and by combining them in different combinations we make things like carbon and oxygen water bananas chalk do I have chalk on my head okay so the other important thing to realiz is that these Char these fundamental particles are charged the neutron has no charge but the proton electron have equal and opposite charges and we use this e is 1.6 109th kums I'll spell it out Kum so this is our fundamental unit of charge the electron has a charge of negative e this has a charge of plus e they're equal and opposite charges now I do want to say one other thing because the book has it in there too it says oh look it's a there's a proton there's a neutron um there's another proton there's another Neutron and then there's electrons that are orbit around there this would be this would be helium right and that's a fine model I just I want to point out it's not real okay we do not have atoms as tiny little solar systems although that was one of the original ideas it's way more complicated than that okay but there is an electric interaction between these charges just like there's an electric interaction between the myar and and the little charge generator okay so those are our fundamental charges I do want to point out that there is a positive charged electron there is a negatively charged Neutron these we call antimatter so the negatively charged proton is called an antimatter protron the neg the positively charged electron is called antimatter electron or a positron because it sounds cool that's a positron but it does EX this now I do want to point out one other thing because you can't create or destroy charge if I if I take a neutron all by itself this is a great example suppose I have a neutron all by itself it's right here it turns out that that's not stable and it will Decay into two things actually three it will Decay into a positive proton and a negative electron so the charge before this was Zero the charge afterward zero there's actually another particle produce a neutrino but that's not important right now so you cannot just create a proton by itself but if I create a proton and electron together then the net charge is constant and the net charge is conserved the same thing's true if I have uh an electron and a positron they're equal and opposite charges they can annihilate and create energy and that's fine they're gone they're gone but I didn't destroy any church what's next let's talk about uh conductors and insulators so we have two types of materials I have a material like this is a conductor and in a conductor we have positive core is the nucleus of the I mean the of the material and they all have you know an extra an electron and these electrons can actually move around and that's what makes it a conductor so conductors are things like aluminum uh iron copper gold silver metals are conductors some other things like graphite can be a conductor um but in those cases we have free charges that can move now in an insulator we also have charges but in this case those are bound together and they can do interesting things like separate a little bit but the charges can't move around so insulators would be things like glass plastic wood styrofoam rubber okay okay now if you add if I want to charge two objects if I want to get something charged I can can do that rubbing things together two different types of material together can cause them to be charged that's what's happening in here there's if you hear this there's a there's a motor in there that's rubbing things together and it's building up a charge and it and where do that charge come from it actually comes from me because the button right here is conducting so I can get charg to this um but when if I want to have something that's charged negatively charged it has just extra negative charges on there if I want to make something that's positively charged then I don't add extra positives because those would be in the nucleus I can't really move those around so it would just be fewer electrons I take some of them away now that one's positively charged okay now uh so I can charge things by rubbing uh or by contact and that's how you get things to be excessively charged okay there's a lot of interesting stuff about matter in there but but I want to get into the to the more important things and that is the force between two charges and this is what we call kum's law suppose I have a positive charge q1 and I have another positive charge over here Q2 and they're separated by distance R the two charges will repel each other so I could say this is F2 on one it technically is a vector and this is F1 on two is technically a vector I can calculate the magnitude of that Vector that Force either one of them they're the same right when I have two objects interacting the force that a pushes on B is equal to the force that b pushes on a that's Newton's third law forces come in pairs I can calculate the magnitude of this F is K q1 Q2 absolute value over R 2 where K is the coolum constant 8.99 109 Newtons m s perum s and coolum is a unit of charge I said that earlier right so this is what we call cool law it says that as those charges get further away the force gets smaller as you get them closer together the force gets greater and as you change the magnitude of their charges the force also changes and this is a very important con uh constant uh and actually we can write this as 1 over 4 Pi Epsilon KN uh and it turns out that that Epsilon not shows up in other places so it's useful but we'll always call K that's what the book does so I'll do that way too now yes that does look like the uh gravitational force between two objects if you remember that from the last semester where that was G Mass 1 Mass 2 over R 2 what's different is that the gravitational force between two masses is always attractive this could be attractive or repulsive if I change one of these to a negative charge then this would change directions both of these would change directions and it' be an attractive Force this equation isn't the best way to write kum's law because it's not a vector or equation but in this curriculum writing it as a vector equation becomes a little bit more complicated and we'll deal with it things in the in the easiest way that we could now what if I have more than two charges uh well forces obey the superposition principle so let's say that there are three charges so I have right here I'm just going to make up something q1 here's another one Q2 and then I'll put a negative charge on here Q3 and suppose I know all those distances and I want to find the chart the force on q1 Q2 so these two are going to repel so I'll write this as F uh one two and these two are going to attract and I write this this is F32 so I have two forces from the two interactions these two interact and those two interact and then the net force is just the vector sum of those and it'd be like this fnet okay so but you have to treat those as vectors because you would have to know the direction and the magnitude and it's not always a trivial problem but that's what we have now the next thing I want to talk about is the electric field this is you know there's two really big ideas in this semester the electric field and the magnetic field and how they relate they're they're huge huge okay uh I want to take a step back and talk about the gravitational field I just SP that with a capital G I don't know why I did capital r g gravitational field so suppose I have the Earth and I have a ball right there of mass m i can calculate the downward Force F I'll call it FG as m g so G is the gravitational field that has a magnitude of 9.8 Newtons per kilogram it's the force per unit Mass since the force depends on the mass and this other constant that constant is the force per unit Mass if I double the mass I double the force and that's the gravitational field it is technically a vector you didn't always write it as a vector but it's a vector okay it's the force per unit Mass how could I find the gravitational field well if I take some mass and I put it there and I C and I measure the force and I divide by its mass I get it so G the magnitude would be the force divided by the mass and that's the gravitational field now what if you move further away from the earth well in that case I have to use a different expression for the gravitational force here's the Earth here's my test Mass here's the vector R and then I had F the force on that mass is g m times the mass of the Earth over r squared if I divide that I can say G is f m that's G mass of the Earth over r s so still the gravitational field G is the gravitation field it's the force per unit Mass electric field so we have the same thing with the electric field so remember that if I have two charges q1 Q2 f I'm going to it's it's the magnitude I'm just going to write it as this q1 Q2 over R 2 what if I take e as F over Q let's say Q2 right I want to find the electric field due to this one so I'm going to divide by that and I get the electric that is the electric field it is the force per unit mass now that's a vector and that's a vector so electric field is a vector uh again it's a limitation of the course that they want to write these forces as scalar values uh but the but it is a vector and so is electric field so electric field is the force per unit charge and it would have units of Newtons Force per charge Kum that's the electric field and if I since I know the expression for the the force due to one charge and divide by that charge I get the max magude of electric field due to a single point charge is k q r 2 it is a vector but that's just the magnitude the electric field so let's just think about what this would look like the electric field we can we can visualize electric field uh in around some object so imagine that I have a positive charge and I I'm over here I want to calculate the electric field well I'll put a positive charge there and you'll experience a force this way I divide by the value of that force and I get the electric fields that way so over here it's this way there it's that way and so this is what I'm drawing is What's called the uh the electric field field a field so this is e so it points away from positive charges if I have a negative charge well if I I take that same charge and put it right there it's going to experience Force pulling it towards there so the a negative charge would have an electric field pointing towards it and as it get further away the the value of that charge decreases what if I have more than one charge well again the electric field obeys the superposition principle so suppose I have two charges near each other I can find the total electric field due to those two charges suppos I have a plus right there and a minus right there and I want to find the electric field down here well this is going to make an electric field going that way this is going to make an electric field going that way and when I add those two together let's call this e+ e minus and this would be the total electric field e so you still have to add them up now how do you add vectors that's something that we need to remember and I'm going to give you a quick Vector review uh in just a second um superp position let me give you uh let me say this just because I I put it in my notes and it is in there and it is fun there's this thing called uh if you apply a large enough electric field in the air then that air turns from an insulator into a conductor and when that happens you get a spark so you've seen Sparks before right that happens when the electric field exceeds 3 10 6 Newtons perum and we call that the breakdown electric field of err and it's just a useful problem that we can use in different situations okay let's review adding vectors and I do not think the book does a great job here um but you know I'm biased and I like to do things my own way but I try to do it along with the book um so let's just say vector versus a scaler so you hopefully have looked at these in your previous Physics course but if you didn't this is a quick review uh so a scalar is a thing that only has one thing to go with it so scalers are things like time Mass charge energy just one thing uh a vector has multiple Dimensions so it doesn't just matter what it is but what direction is going so you've seen this before with velocity position acceleration force and electric field so but how do you represent a vector let's just let's just draw a vector okay let's use a position Vector because it's easier so here's the XY AIS suppose I have some point right here I can represent that as the vector R and it has some uh distance from here to there and it has some angle with respect to the x-axis data I'm reviewing here not the full details okay we can represent this Vector as a sum of a vector in the X Direction plus a vector in the y direction why do we want to do that well if you break a vector into these X and Y components it's easier to add them so we'll call this RX X hat r y y hat and so this is just the X hat just means in the X Direction and that's a scalar value Y is a scalar value and that's in the y direction so I can write r as RX X hat plus r y y hat how can I find a relationship between the magnitude of the vector R RX and r y I can use trig so remember that we have these cosine sign functions so if I take the s of the angle Theta s is defined as opposite over hypotenuse well the opposite side has a length r y the hypotenuse has a magnitude R so I can solve this and I get r y is R sin Theta I can do the same thing with cosine cosine Theta is the adjacent over the hypotenuse so that's RX over R so RX is R cosine thet so this means that if I have if I have uh a vector and I know the angle with respect to the x or y AIS I can find these X and Y components now be careful two things what if I had this well in that case I have this is my right triangle so in this case the opposite side is the X component the adjacent side's the Y so don't think this is always our y you have to look at the picture next what if I have a vector over here and I have this is the angle well again I have the X's sign but it's negative right it's going in the negative X Direction and the Y component will be positive so you need to look at your picture and look at your angles and determine if your components are positive or negative why do we even care suppose I have two uh electric field vectors E1 is E1 x x hat plus E2 y y hat E2 is e2x X Plus E2 y y if I want to add those two vectors together E1 plus E2 I can add their components because the X's are in the same direction so I can just add them the Y's are in the same direction I can just add them now they might be negative and that's fine but I can do that so this would be E1 x + e2x x hat plus E2 1 y plus E2 y y hat now what if I want to find the magnitude of this total Vector I'll call it just e the magnitude of e now that I know the components I can go go back the other way uh I I know the adjacent side I know the opposite side I can find the hypotenuse using the Pythagorean theorem so this would be e1x I just call it e x right that's a total e x^2 plus e y^2 and that's that if you need to find the angle you can say the tangent of the angle Theta is opposite which is most likely ey over ex okay that's that I'm going to post some problems we're going to do some problems doesn't really make sense until we do problems um oh I did want to show you um there is a really great uh fat simulator but I'll I'll maybe play with that later in the if you read through the book there's some simulators embedded play with those they're really great I really I think they're awesome um let's see can I pull this up real quick let's see yeah okay let's do physics okay let's switch to the computer so this is this is fat um and I'm going to just turn off these don't want motion sound work energy light I want that one okay and let's go down here charges and Fields you can't see it make this a little bit smaller I should have fix this all beforehand but I didn't okay so here we can just move some charges onto our playing field there's a positive charge it shows the electric field uh at all these different locations uh as a vector and then I can put it in a negative charge right here and you'll see that I get this completely different shape of patterns because for each one of these locations this Vector points in the vector sum of the positive and the negative uh what's that do values no uh don't worry about that I can put another chart here so you can build up a whole bunch of things and play with it um I thought that was going be more interesting than it was but you should play with that anyway okay so that's that for chapter 18 I'll post some homework problems and some practice problems and we'll do that but other than that that's that and I don't know how to stop there it is stop button right there |
3713 | https://web.mst.edu/bohner/papers/tbhqde.pdf | Journal of Biological Dynamics, 2013 Vol. 7, No. 1, 86–95, The Beverton–Holt q-difference equation Martin Bohner and Rotchana Chieochan Department of Mathematics and Statistics, Missouri University of Science and Technology, Rolla, MO 65409-0020, USA (Received 13 December 2012; final version received 1 May 2013) The Beverton–Holt model is a classical population model which has been considered in the literature for the discrete-time case. Its continuous-time analogue is the well-known logistic model. In this paper, we consider a quantum calculus analogue of the Beverton–Holt equation. We use a recently introduced concept of periodic functions in quantum calculus in order to study the existence of periodic solutions of the Beverton–Holt q-difference equation. Moreover, we present proofs of quantum calculus versions of two so-called Cushing–Henson conjectures.
Keywords: Beverton–Holt equation; Cushing–Henson conjecture; time scale; dynamic equation; logistic equation; Jensen inequality AMS Subject Classifications: 39A10; 39A11; 39A12; 39A20; 34C25 1.
Introduction The Beverton–Holt difference equation has wide applications in population growth and is given by x(t + 1) = νK(t)x(t) K(t) + (ν −1)x(t), t ∈N0, (1) where ν > 1, K(t) > 0 for all t ∈N0, and x(0) > 0. We call the sequence K the carrying capacity and ν the inherent growth rate . The periodically forced Beverton–Holt equation, which is obtained by letting the carrying capacity be a periodic positive sequence K(t) with period ω ∈N, i.e., K(t + ω) = K(t) for all t ∈N0, has been treated with the methods found in [8–10]. For the Beverton–Holt dynamic equation on time scales, one article has been presented by Bohner and Warth . In , a general Beverton–Holt equation is given, which reduces to Equation (1) in the discrete case and to the well-known logistic equation in the continuous case. The approach given in opened a new path to the study of the discrete Beverton–Holt equation, which was pursued Corresponding author. Email: bohner@mst.edu Author Email: rckv9@mst.edu © 2013 The Author(s). Published by Taylor & Francis.
This is an Open Access article distributed under the terms of the Creative Commons Attribution License (
org/licenses/by/3.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. The moral rights of the named author(s) have been asserted.
Journal of Biological Dynamics 87 by Bohner et al. in . The crucial idea in [6,7] was to rewrite Equation (1) as x(t) = αx(t + 1) 1 −x(t) K(t) , t ∈N0, where α := ν −1 ν , (2) and thus identify the continuous version of the discrete Beverton–Holt equation (2) as x′(t) = αx(t) 1 −x(t) K(t) , t ∈R, (3) which turned out to be the usual logistic equation.1 This approach was generalized to any so-called dynamic equation of the form x(t) = αxσ(t) 1 −x(t) K(t) , t ∈T, (4) hence accommodating both the continuous and discrete equations (2) and (3). However, the restriction on the time scale T was that it should be periodic. Hence, N0 and R (and also hN0 with h > 0) were allowed, but qN0 for q > 1 was not.
In this paper, we are filling this gap by studying a quantum calculus version of the Beverton– Holt equation, namely, a Beverton–Holt q-difference equation. This became possible by using a new definition of periodic functions in quantum calculus which was introduced by the authors in [3, Definition 3.1] (see also ). Using this concept, we are interested in seeking ω-periodic solutions of the Beverton–Holt q-difference equation given by x(t) = a(t)xσ(t) 1 −x(t) K(t) , t ∈qN0, (5) where a is 1-periodic and K is ω-periodic, and x(t) = x(qt) −x(t) (q −1)t , xσ(t) = x(σ(t)), σ(t) = qt, t ∈qN0.
Using this notation and also our Assumptions (7) below, we can easily rewrite Equation (5) as x(qt) = νx(t)K(t) K(t) + (ν −1)x(t), t ∈qN0, where ν := 1 1 −(q −1)α .
(6) One can now observe the similarity of the discrete (additive) recursion (1) and the quantum (multiplicative) recursion (6).
The set-up of this paper is as follows. Section 2 contains some preliminaries on quantum calculus. We approach the periodic solutions of the Beverton–Holt q-difference equation (5) by some strategies presented in Section 3. In Sections 4 and 5, we formulate and prove the first and the second Cushing–Henson conjectures for the q-difference equations case, respectively.
2.
Some auxiliary results Definition 2.1 We say that a function p : qN0 →R is regressive provided 1 + μ(t)p(t) ̸= 0 for all t ∈qN0, where μ(t) = (q −1)t.
The set of all regressive functions will be denoted by R. Moreover, p ∈R is called positively regressive and we write p ∈R+ provided 1 + μ(t)p(t) > 0 for all t ∈qN0.
88 M. Bohner and R. Chieochan Definition 2.2 Exponential function Let p ∈R and t0 ∈qN0. The exponential function ep(·, t0) on qN0 is defined by ep(t, t0) = s∈[t0,t) [1 + (q −1)sp(s)] for t ≥t0, t ∈qN0.
Remark 2.3 See [5, Theorem 2.44] If p ∈R+, then ep(t, t0) > 0 for all t ≥t0, t ∈qN0.
Definition 2.4 See [3, Definition 3.1] A function f : qN0 →R is called ω-periodic if f (t) = qωf (qωt) for all t ∈qN0.
Theorem 2.5 See [5, Theorem 2.36] If p ∈R, then (i) e0(t, s) = 1 and ep(t, t) = 1; (ii) ep(t, s) = 1/ep(s, t); (iii) ep(t, s)ep(s, r) = ep(t, r); (iv) ep(σ(t), s) = (1 + μ(t)p(t))ep(t, s); (v) (1/ep(·, s))(t) = −p(t)/ep(σ(t), s).
The integral on qN0 is defined as follows.
Definition 2.6 Let m, n ∈N0 with m < n. For f : qN0 →R, we define qn qm f (t)t := (q −1) n−1 k=m qkf (qk).
Theorem 2.7 Integration by parts, see [5, Theorem 1.77] For a, b ∈qN0 and f , g : qN0 →R, we have b a f σ(t)g(t)t = f (b)g(b) −f (a)g(a) + b a f (t)g(t)t and b a f (t)g(t)t = f (b)g(b) −f (a)g(a) + b a f (t)gσ(t)t.
Theorem 2.8 Jensen’s inequality, see [12, Theorem 2.2] Let a, b ∈qN0 and c, d ∈R. Suppose g, h : ([a, b] ∩qN0) →(c, d) and b a |h(s)|s > 0. If F ∈C((c, d), R) is convex, then F b a |h(s)|g(s)s b a |h(s)|s ≤ b a |h(s)|F(g(s))s b a |h(s)|s .
If F is strictly convex, then ‘≤’can be replaced by ‘<’.
Journal of Biological Dynamics 89 3.
The Beverton–Holt equation Throughout this paper, we use the following assumptions and notation: ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ q > 1, t0 ∈qN0, K is positive and ω-periodic, ω ∈N, a(t) = α t , t ∈qN0, 0 < α < 1 q −1, λ := 1 −(q −1)α, β := 1 qωλ−ω −1.
(7) Note that Assumption (7) implies that 0 < λ < 1, −a ∈R+, and e−a(t, s) = λlogq(t/s) for all t, s ∈qN0.
(8) Note also that β is well defined as q/λ ̸∈{−1, 1} since λ ̸∈{−q, q} due to 0 < λ < 1.
In the q-difference equation (5), we assume x(t) > 0 for all t ∈qN0 and substitute u := 1 x .
Then, using the quotient rule [5, Theorem 1.20 (v)], Equation (5) becomes u(t) = −a(t)u(t) + a(t) K(t).
(9) ThegeneralsolutionofEquation(9)isgivenbyapplyingvariationofparameters[5,Theorem2.77] twice as u(t) = e−a(t, t0)u(t0) + t t0 e−a(t, σ(s)) a(s) K(s)s (10) = e−a(t, qωt0)u(qωt0) + t qωt0 e−a(t, σ(s)) a(s) K(s)s, (11) where t ∈qN0. Now, we require an ω-periodic solution ¯ x of Equation (5).This means that ¯ x satisfies ¯ x(t) = qω¯ x(qωt) for all t ∈qN0. This in turn means that a solution ¯ u = 1/¯ x of Equation (9) satisfies qω¯ u(t) = ¯ u(qωt) for all t ∈qN0.
(12) Lemma 3.1 Assume Assumption (7). If Equation (9) has a solution ¯ u satisfying Equation (12), then ¯ u(t0) = β qωt0 t0 e−a(t0, σ(s)) a(s) K(s)s.
Proof Assume Equation (9) has a solution ¯ u satisfying Equation (12). Then, ¯ u(t0) = q−ω¯ u(qωt0) = q−ωe−a(qωt0, t0)¯ u(t0) + q−ω qωt0 t0 e−a(qωt0, σ(s)) a(s) K(s)s 90 M. Bohner and R. Chieochan = q−ω 1 −q−ωe−a(qωt0, t0) qωt0 t0 e−a(qωt0, t0)e−a(t0, σ(s)) a(s) K(s)s = q−ωλω 1 −q−ωλω qωt0 t0 e−a(t0, σ(s)) a(s) K(s)s = 1 qωλ−ω −1 qωt0 t0 e−a(t0, σ(s)) a(s) K(s)s.
Thus, ¯ u satisfies the required initial condition.
■ 4.
The first Cushing–Henson conjecture Now we state and prove the first Cushing–Henson conjecture for the Beverton–Holt q-difference equation (5).
Conjecture 4.1 First Cushing–Henson conjecture The Beverton–Holt q-difference model (5) with an ω-periodic carrying capacity K has a unique ω-periodic solution ¯ x that globally attracts all solutions.
Using Equation (10) and Lemma 3.1, the solution ¯ u of Equation (9) can be written as ¯ u(t) = e−a(t, t0)¯ u(t0) + t t0 e−a(t, σ(s)) a(s) K(s)s = β qωt0 t0 e−a(t, σ(s)) a(s) K(s)s + t t0 e−a(t, σ(s)) a(s) K(s)s.
(13) Theorem 4.2 Assume Assumption (7) and let ¯ u be given by Equation (13). Then, ¯ x := 1/¯ u is an ω-periodic solution of the Beverton–Holt q-difference equation (5).
Proof By Equation (11), we have ¯ u(t) = e−a(t, qωt0)¯ u(qωt0) + t qωt0 e−a(t, σ(s)) a(s) K(s)s so that q−ω¯ u(qωt) = e−a(qωt, qωt0)q−ω¯ u(qωt0) + qωt qωt0 e−a(qωt, σ(s)) a(s) qωK(s)s = e−a(t, t0)¯ u(t0) + t t0 e−a(t, σ(s)) a(s) K(s)s = ¯ u(t) since by putting t0 = qm and t = qn, we have qωt qωt0 e−a(qωt, σ(s)) a(s) qωK(s)s = (q −1) ω+n−1 k=ω+m qke−a(qω+n, qk+1) a(qk) qωK(qk) = (q −1) n−1 k=m qk+ωe−a(qω+n, qk+ω+1) a(qk+ω) qωK(qk+ω) Journal of Biological Dynamics 91 = (q −1) n−1 k=m qke−a(qn, qk+1) a(qk) K(qk) = t t0 e−a(t, σ(s)) a(s) K(s)s.
Hence, ¯ u satisfies Equation (12) and thus ¯ x is indeed ω-periodic.
■ Now we are ready to prove the validity of the first Cushing–Henson conjecture.
Theorem 4.3 Assume Assumption (7). The solution ¯ x of Equation (5) given in Theorem 4.2 is globally attractive.
Proof First note that K is bounded. Indeed, define ˜ K := max 0≤k≤ω−1 K(qk).
For any m ∈N0, there exist ℓ∈N0 and 0 ≤k ≤ω −1 such that m = ℓω + k, and thus K(qm) = K(qℓω+k) = q−ℓωK(qk) ≤K(qk) ≤˜ K.
Now let x be any solution of Equation (5) with x(t) > 0 for all t ∈qN0. We have |x(t) −¯ x(t)| = 1 e−a(t, t0)u(t0) + t t0 e−a(t, σ(s))a(s)/K(s)s − 1 e−a(t, t0)¯ u(t0) + t t0 e−a(t, σ(s))a(s)/K(s)s = 1 e−a(t, t0)/x(t0) + t t0 e−a(t, σ(s))a(s)/K(s)s − 1 e−a(t, t0)/¯ x(t0) + t t0 e−a(t, σ(s))a(s)/K(s)s = 1 ¯ x(t0) − 1 x(t0) |e−a(t, t0)| e−a(t, t0) x(t0) + t t0 e−a(t, σ(s)) a(s) K(s)s e−a(t, t0) ¯ x(t0) + t t0 e−a(t, σ(s)) a(s) K(s)s ≤ |1/¯ x(t0) −1/x(t0)|e−a(t, t0) t t0 e−a(t, σ(s))a(s)/K(s)s 2 ≤ ˜ K2|1/¯ x(t0) −1/x(t0)|e−a(t, t0) t t0 e−a(t, σ(s))a(s)s 2 ≤˜ K2 |1/¯ x(t0) −1/x(t0)|e−a(t, t0) (1 −e−a(t, t0))2 , which due to [2, Theorem 2] tends to zero as t →∞.
■ 92 M. Bohner and R. Chieochan 5.
The second Cushing–Henson conjecture Now we state and prove the second Cushing–Henson conjecture for the Beverton–Holt q-difference equation (5).
Conjecture 5.1 Second Cushing–Henson conjecture The average of the ω-periodic solution ¯ x of Equation (5) is strictly less than the average of the ω-periodic carrying capacity K times the constant (q −λ)/(1 −λ).
In order to prove the second Cushing–Henson conjecture, we use the following series of auxiliary results.
Lemma 5.2 Assume Assumption (7). Then, for any t, u, v ∈qN0, we have v u e−a(t, σ(s))s = ve−a(t, v) −ue−a(t, u) 1 + α .
(14) Proof Using Theorems 2.5 and 2.7, we get v u e−a(t, σ(s))s = v u es −a(t, s) a(s) s = 1 αq v u σ(s)es −a(t, s)s = 1 αq ve−a(t, v) −ue−a(t, u) − v u e−a(t, s)s = 1 αq ve−a(t, v) −ue−a(t, u) − v u λe−a(t, σ(s))s = ve−a(t, v) −ue−a(t, u) αq + λ = ve−a(t, v) −ue−a(t, u) 1 + α , which shows Equation (14).
■ Lemma 5.3 Assume Assumption (7). Then, for any s, u, v ∈qN0, we have v u e−a(t, s) t2 t = q 1 + α e−a(u, s) u −e−a(v, s) v .
(15) Proof Using Theorems 2.5 and 2.7, we get v u e−a(t, s) t2 t = −1 α v u −a(t)e−a(t, s) t t = −1 α v u et −a(t, s) t t = −1 α e−a(v, s) v −e−a(u, s) u − v u e−a(σ(t), s) −1 tσ(t) t Journal of Biological Dynamics 93 = 1 α e−a(u, s) u −e−a(v, s) v −λ q v u e−a(t, s) t2 t = q α + 1 e−a(u, s) u −e−a(v, s) v , which shows Equation (15).
■ Now note that Equation (13) implies that for t0 ≤t < qωt0, we have ¯ u(t) = qωt0 t0 h(t, s) sK(s) s, (16) where h(t, s) := αe−a(t, σ(s))(β + χ(t, s)) with χ(t, s) := 1 ifs < t 0 ifs ≥t.
(17) Lemma 5.4 Assume Assumptions (7) and (17). Then, for t0 ≤s < qωt0, we have qωt0 t0 h(t, s) t2 t = α (α + 1)s.
(18) Proof Using Lemma 5.3 and βqωλ−ω −β −1 = 0, we obtain qωt0 t0 h(t, s) t2 t = α β qωt0 t0 e−a(t, σ(s)) t2 t + qωt0 σ(s) e−a(t, σ(s)) t2 t (15) = αq α + 1 β e−a(t0, σ(s)) t0 −e−a(qωt0, σ(s)) qωt0 + 1 qs −e−a(qωt0, σ(s)) qωt0 (8) = αq α + 1 1 qs + e−a(qωt0, σ(s)) qωt0 βqωλ−ω −β −1 = α (α + 1)s, which shows Equation (18).
■ Lemma 5.5 Assume Assumptions (7) and (17). Then, for t0 ≤t < qωt0, we have qωt0 t0 h(t, s)s = αt 1 + α .
(19) Proof Using Lemma 5.2 and βqωλ−ω −β −1 = 0, we obtain qωt0 t0 h(t, s)s = α β qωt0 t0 e−a(t, σ(s))s + t t0 e−a(t, σ(s))s (14) = α β qωt0e−a(t, qωt0) −t0e−a(t, t0) 1 + α + t −t0e−a(t, t0) 1 + α (8) = α 1 + α t + t0e−a(t, t0) βqωλ−ω −β −1 = αt 1 + α , which shows Equation (19).
■ 94 M. Bohner and R. Chieochan Now we are ready to prove the validity of the second Cushing–Henson conjecture.
Theorem 5.6 Let ¯ x be the unique ω-periodic solution of Equation (5). If ω ̸= 1, then 1 ω qωt0 t0 ¯ x(t)t < q −λ 1 −λ 1 ω qωt0 t0 K(t)t .
(20) Proof Since K is ω-periodic with ω ̸= 1, tK(t) cannot be a constant. In addition, F(x) = 1/x is strictly convex. Thus, we may use Jensen’s inequality (Theorem 2.8) for the single inequality in the forthcoming calculation to obtain qωt0 t0 ¯ x(t)t = qωt0 t0 1 ¯ u(t)t = qωt0 t0 1 qωt0 t0 h(t, s)/sK(s)s t = qωt0 t0 F qωt0 t0 h(t, s)/sK(s)s qωt0 t0 h(t, s)s 1 qωt0 t0 h(t, s)s t < qωt0 t0 qωt0 t0 h(t, s)F(1/sK(s))s qωt0 t0 h(t, s)s 2 t = qωt0 t0 qωt0 t0 h(t, s)sK(s)s qωt0 t0 h(t, s)s 2 t (19) = qωt0 t0 qωt0 t0 h(t, s)sK(s)s (αt/(1 + α))2 t = 1 + α α 2 qωt0 t0 qωt0 t0 h(t, s)sK(s) t2 st = 1 + α α 2 qωt0 t0 sK(s) qωt0 t0 h(t, s) t2 ts (18) = 1 + α α 2 qωt0 t0 sK(s) α (α + 1)ss = 1 + α α qωt0 t0 K(s)s, which shows Inequality (20). The proof is complete.
■ Theorem 5.7 If K is 1-periodic, then Inequality (20) becomes an equality, i.e., 1 ω qωt0 t0 ¯ x(t)t = q −λ 1 −λ 1 ω qωt0 t0 K(t)t .
(21) Journal of Biological Dynamics 95 Proof Since K is 1-periodic, we have K(t) = C t for some C > 0.
Now it is easy to check that ¯ x given by ¯ x(t) := 1 + α α K(t) = (1 + α)C αt is 1-periodic and satisfies ¯ x(t) = a(t)¯ xσ(t) 1 −¯ x(t) K(t) for all t ∈qN0.
Hence, ¯ x is the unique 1-periodic solution of Equation (5). Thus, (21) holds.
■ Remark 5.8 Note that the factor q −λ 1 −λ (22) is not present in the statements of the Cushing–Henson conjectures for the continuous and the discrete cases. However, in q-calculus, the presence of such quantities is common, and when replacing q by 1 (i.e., letting q →1+) in these terms, the continuous case is usually recovered.
Note that replacing q by 1 in the factor (22) yields the corresponding ‘continuous’ (and also ‘discrete’) factor 1.
Notes 1.
Note that, as Jim Cushing points out, the terminology ‘discrete logistic equation’ differs, due to Robert May , in a slight but essential way, from the discrete model (2).
References R.J.H. Beverton and S.J. Holt, On the Dynamics of Exploited Fish Population, Fishery Investigations (Great Britain, Ministry of Agriculture, Fisheries, and Food), Vol. 19, H. M. Stationery Office, London, 1957.
M. Bohner, Some oscillation criteria for first order delay dynamic equations, Far East J. Appl. Math. 18(3) (2005), pp. 289–304.
M. Bohner and R. Chieochan, Floquet theory for q-difference equations, Sarajevo J. Math. 8(21)(2) (2012), pp.
1–12.
M. Bohner and R. Chieochan, Positive periodic solutions for higher-order functional q-difference equations, J.Appl.
Funct. Anal. 8(1) (2013), pp. 14–22.
M. Bohner and A. Peterson, Dynamic Equations on Time Scales: An Introduction with Applications, Birkhäuser Boston, Boston, MA, 2001.
M. Bohner, S. Stevi´ c, and H. Warth, The Beverton–Holt difference equation, in Discrete Dynamics and Difference Equations, S. Elaydi, H. Oliveira, J. Ferreira, and J.Alves, eds.,World Scientific, Hackensack, NJ, 2010, pp. 189–193.
M. Bohner and H. Warth, The Beverton–Holt dynamic equation, Appl. Anal. 86(8) (2007), pp. 1007–1015.
J.M. Cushing and S.M. Henson,A periodically forced Beverton–Holt equation, J. Difference Equ.Appl. 8(12) (2002), pp. 1119–1120.
S. Elaydi and R.J. Sacker, Global stability of periodic orbits of nonautonomous difference equations in population biology and the Cushing–Henson conjectures, in Proceedings of the Eighth International Conference on Difference Equations and Applications, S. Elaydi, G. Ladas, B. Aulbach, and O. Dosly, eds., Chapman & Hall/CRC, Boca Raton, FL, 2005, pp. 113–126.
S. Elaydi and R.J. Sacker, Nonautonomous Beverton–Holt equations and the Cushing-Henson conjectures, J.
Difference Equ. Appl. 11(4–5) (2005), pp. 337–346.
R.M. May, Simple mathematical models with very complicated dynamics, Nature 261 (1976), pp. 459–467.
F.-H. Wong, C.-C. Yeh, and W.-C. Lian, An extension of Jensen’s inequality on time scales, Adv. Dyn. Syst. Appl.
1(1) (2006), pp. 113–120. |
3714 | https://www.youtube.com/watch?v=gw-4wltP5tY | Normal vector from plane equation | Vectors and spaces | Linear Algebra | Khan Academy
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Figuring out a normal vector to a plane from its equation
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Transcript:
What I want to do in
this video is make sure that we're good at
picking out what the normal vector
to a plane is, if we are given the
equation for a plane. So to understand that, let's
just start off with some plane here. Let's just start off--
so this is a plane, I'm drawing part
of it, obviously it keeps going in every direction. So let's say that is our plane. And let's say that this is a
normal vector to the plane. So that is our normal
vector to the plane. It's given by ai
plus bj plus ck. So that is our normal
vector to the plane. So it's perpendicular. It's perpendicular
to every other vector that's on the plane. And let's say we have
some point on the plane. We have some point. It's the point x sub p. I'll say p for plane. So it's a point on the plane. Xp yp zp. If we pick the origin. So let's say that
our axes are here. So let me draw our
coordinate axes. So let's say our coordinate
axes look like that. This is our z-axis. This is, let's say
that's a y-axis. And let's say that
this is our x-axis. Let's say this is our
x-axis coming out like this. This is our x-axis. You can specify this
is a position vector. There is a position vector. Let me draw it like this. Then it would be behind the
plane, right over there. You have a position vector. That position vector would
be xpi plus ypj plus zpk. It specifies this
coordinate, right here, that sits on the plane. Let me just call that something. Let me call that
position vector, I don't know-- let
me call that p1. So this is a point on the plane. So it's p-- it is p1
and it is equal to this. Now, we could take another
point on the plane. This is a particular
point of the plane. Let's say we just say, any
other point on the plane, xyz. But we're saying that
xyz sits on the plane. So let's say we take this
point right over here, xyz. That clearly, same logic, can
be specified by another position vector. We could have a position
vector that looks like this. And dotted line. It's going under the
plane right over here. And this position
vector, I don't know, let me just call it p, instead
of that particular, that P1. This would just be
xi plus yj plus zk. Now, the whole reason
why I did this set up is because, given some
particular point that I know is on the plane, and any other
xyz that is on the plane, I can find-- I can construct--
a vector that is definitely on the plane. And we've done this
before, when we tried to figure out what the
equations of a plane are. A vector that's
definitely on the plane is going to be the difference
of these two vectors. And I'll do that in blue. So if you take the yellow
vector, minus the green vector. We take this position,
you'll get the vector that if you view
it that way, that connects this point
in that point. Although you can
shift the vector. But you'll get a vector that
definitely lies along the plane So if you start
one of these points it will definitely
lie along the plane. So the vector will
look like this. And it would be lying
along our plane. So this vector lies
along our plan. And that vector is p minus p1. This is the vector p minus p1. It's this position vector minus
that position vector, gives you this one. Or another way to view
it is this green position vector plus this blue vector
that sits on the plane will clearly equal this
yellow vector, right? Heads to tails. It clearly equals it. And the whole reason
why did that is we can now take the dot product,
between this blue thing and this magenta thing. And we've done it before. And they have to be equal to 0,
because this lies on the plane. This is perpendicular
to everything that sits on the
plane and it equals 0. And so we will get the
equation for the plane. But before I do
that, let me make sure we know what the components
of this blue vector are. So p minus p1, that's
the blue vector. You're just going to subtract
each of the components. So it's going to be x minus xp. It's going to be x minus
xpi plus y minus ypj plus z minus zpk. And we just said,
this is in the plane. And this is, this
right, the normal vector is normal to the plane. You take their dot product--
it's going to be equal to zero. So n dot this vector is
going to be equal to 0. But it's also equal to this
a times this expression. I'll do it right over here. So these-- find some good color. So a times that, which is ax
minus axp plus b times that. So that is plus by minus byp. And then-- let me make
sure I have enough colors-- and then it's going to
be plus that times that. So that's plus cz minus czp. And all of this is equal to 0. Now what I'm going to do is,
I'm going to rewrite this. So we have all of these
terms I'm looking for, right? Color. We have all of the x terms-- ax. Remember, this is any
x that's on the plane, will satisfy this. So ax, by and cz. Let me leave that on
the right hand side. So we have ax plus by
plus cz is equal to-- and what I want
to do is I'm going to subtract each of
these from both sides. Another way is, I'm going
to move them all over. Let me do it-- let me
not do too many things. I'm going to move them
over to the left hand side. So I'm going to add
positive axp to both sides. That's equivalent of
subtracting negative axp. So this is going
to be positive axp. And then we're going to have
positive byp plus-- I'll do that same green-- plus byp,
and then finally plus czp. Plus czp is going
to be equal to that. Now, the whole reason why
did this-- and I've done this in previous videos, where we're
trying to find the formula, or trying to find the
equation of a plane, is now we say, hey, if
you have a normal vector, and if you're given a point
on the plane-- where it's in this case is
xp yp zp-- we now have a very quick way of
figuring out the equation. But I want to go the other way. I want you to be
able to, if I were to give you a equation for
plane, where I were to say, ax plus by plus cz, is equal to d. So this is the general
equation for a plane. If I were to give
you this, I want to be able to figure out the
normal vector very quickly. So how could you do that? Well, this ax plus by
plus cz is completely analogous to this part
right up over here. Let me rewrite all this over
here, so it becomes clear. This part is ax
plus by plus cz is equal to all of this stuff
on the left hand side. So let me copy and paste it. So I just essentially
flipped this expression. But now you see this, all of
this, this a has to be this a. This b has to be this b. This c has to be this thing. And then the d is all of this. And this is just
going to be a number. This is just going to
be a number, assuming you knew what the
normal vector is, what your a's, b's
and c's are, and you know a particular value. So this is what d is. So this is how you could get
the equation for a plane. Now if I were to give
you equation or plane, what is the normal vector? Well, we just saw it. The normal vector, this a
corresponds to that a, this b corresponds to that b, that
c corresponds to that c. The normal vector to
this plane we started off with, it has the
component a, b, and c. So if you're given
equation for plane here, the normal vector to this
plane right over here, is going to be ai
plus bj plus ck. So it's a very easy thing to do. If I were to give you
the equation of a plane-- let me give you a
particular example. If I were to tell you that
I have some plane in three dimensions-- let's say it's
negative 3, although it'll work for more dimensions. Let's say I have negative 3
x plus the square root of 2 y-- let me put it this
way-- minus, or let's say, plus 7 z is equal to pi. So you have this crazy--
I mean it's not crazy. It's just a plane
in three dimensions. And I say what is a normal
vector to this plane? You literally can just pick
out these coefficients, and you say, a normal
vector to this plane is negative 3i plus
the square root of 2 plus 2 square root
of 2 j plus 7 k. And you could ignore
the d part there. And the reason why
you can ignore that is that will just
shift the plane, but it won't fundamentally
change how the plane is tilted. So a this normal vector, will
also be normal if this was e, or if this was 100, it would be
normal to all of those planes, because all those
planes are just shifted, but they all have
the same inclination. So they would all kind of
point the same direction. And so the normal vectors would
point in the same direction. So hopefully you found
that vaguely useful. We'll now build on this
to find the distance between any point in three
dimensions, and some plane. The shortest distance that
we can get to that plane. |
3715 | https://fun2dolabs.com/predecessors-and-successors/ | Predecessors and Successors - Fun2Do Labs
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Predecessors and Successors
Predecessors and successors are important concepts in mathematics that are often used in various fields such as computer science, engineering, and business. Understanding these terms is essential for building a strong foundation in mathematics.
These terms are used for a series of whole numbers. The number that appears just before the given number in a series is called Predecessor.
The number that appears just after the given number in a series is called Successor.
Predecessor is a number that comes before another number, and it is one less than that number. The predecessor of any given number is obtained by subtracting 1 from the given number.
For example: The Predecessor of 68 is 68 – 1 = 67, similarly predecessors of the numbers 10, 15, and 20 are 9, 14, and 19, respectively.
After kids understand the concept of a predecessor, introduce the concept of the successor. A successor is a number that comes after another number, and it is one more than that number. The successor of any given number is obtained by adding 1 to the given number.
For example : The Successor of 68 is 68 + 1 = 69. Similarly, successors of the numbers 10, 15, and 20 are 11, 16, and 21, respectively.
Kids can easily relate to real-life examples. For instance, if a student is born in the year 2020, then their successor would be the year 2021, which comes after 2020 and the predecessor of the year 2020 would be the year 2019.
Posters
Teaching predecessors and successors with kid friendly, clear, and easy to understand posters from Uncle Math by Fun2Do Labs :
Activities
Learning predecessors and successors can be made enjoyable by incorporating interactive games and activities.
Jigsaw Puzzle :
There are puzzle pieces of 2-digit numbers along with P and S pieces.
The level of the game is decided as per the age of the child; initially, 2-digit numbers can be taken.
Kids can be instructed to make sets of predecessors and successors arranging numbers with the correct P or S piece.
Worksheets
Help your kids to practise predecessors and successors with interesting and engaging funworksheets and solutions from Uncle Math by Fun2do Labs.
Worksheet 066 : Successor And Predecessor
Solution 066 : Successor And Predecessor
Worksheet 067 : Write Successor And Predecessor
Solution 067 : Write Successor And Predecessor
Worksheet 068 : Predecessor
Solution 068 : Predecessor
Worksheet 069 : Successor
Solution 069 : Successor
Worksheet 070 : Successor And Predecessor
Solution 070 : Successor And Predecessor
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3716 | https://www.britannica.com/science/Earth-sciences/Geologic-time-and-the-age-of-Earth | SUBSCRIBE
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Geologic time and the age of Earth
in
Earth sciences
in
The 19th century
Written by
Claude C. Albritton
Hamilton Professor of Geology, Southern Methodist University, Dallas, Texas, 1955–78. Coauthor and editor of The Fabric of Geology; Uniformity and Simplicity.
Claude C. Albritton
,
Brian Frederick Windley
Professor of Geology, University of Leicester, England. Author of The Evolving Continents.
Brian Frederick Windley
•All
Fact-checked by
The Editors of Encyclopaedia Britannica
Encyclopaedia Britannica's editors oversee subject areas in which they have extensive knowledge, whether from years of experience gained by working on that content or via study for an advanced degree. They write new content and verify and edit content received from contributors.
The Editors of Encyclopaedia Britannica
Article History
By mid-century the fossiliferous strata of Europe had been grouped into systems arrayed in chronological order. The stratigraphic column, a composite of these systems, was pieced together from exposures in different regions by application of the principles of superposition and faunal sequence. Time elapsed during the formation of a system became known as a period, and the periods were grouped into eras: the Paleozoic (Cambrian through Permian periods), Mesozoic (Triassic, Jurassic, and Cretaceous periods), and Cenozoic (Paleogene, Neogene, and Quaternary periods).
Charles Darwin’s Origin of Species (1859) offered a theoretical explanation for the empirical principle of faunal sequence. The fossils of the successive systems are different not only because parts of the stratigraphic record are missing but also because most species have lost in their struggles for survival and also because those that do survive evolve into new forms over time. Darwin borrowed two ideas from Lyell and the uniformitarians: the idea that geologic time is virtually without limit and the idea that a sequence of minute changes integrated over long periods of time produce remarkable changes in natural entities.
The evolutionists and the historical geologists were embarrassed when, beginning in 1864, William Thomson (later Lord Kelvin) attacked the steady-state theory of Earth and placed numerical strictures on the length of geologic time. The Earth might function as a heat machine, but it could not also be a perpetual motion machine. Assuming that Earth was originally molten, Thomson calculated that not less than 20 million and not more than 400 million years could have passed since Earth first became a solid body. Other physicists of note put even narrower limits on Earth’s age ranging down to 15 million or 20 million years. All these calculations, however, were based on the common assumption, not always explicitly stated, that Earth’s substance is inert and hence incapable of generating new heat. Shortly before the end of the century this assumption was negated by the discovery of radioactive elements that disintegrate spontaneously and release heat to Earth in the process.
Concepts of landform evolution
The scientific exploration of the American West following the end of the Civil War yielded much new information on the sculpture of the landscape by streams. John Wesley Powell in his reports on the Colorado River and Uinta Mountains (1875, 1876) explained how streams may come to flow across mountain ranges rather than detour around them. The Green River does not follow some structural crack in its gorge across the Uinta Mountains; instead it has cut its canyon as the mountain range was slowly bowed up. Given enough time, streams will erode their drainage basins to plains approaching sea level as a base. Grove Karl Gilbert’s Report on the Geology of the Henry Mountains (1877) offered a detailed analysis of fluvial processes. According to Gilbert all streams work toward a graded condition, a state of dynamic equilibrium that is attained when the net effect of the flowing water is neither erosion of the bed nor deposition of sediment, when the landscape reflects a balance between the resistance of the rocks to erosion and the processes that are operative upon them. After 1884 William Morris Davis developed the concept of the geographical cycle, during which elevated regions pass through successive stages of dissection and denudation characterized as youthful, mature, and old. Youthful landscapes have broad divides and narrow valleys. With further denudation the original surface on which the streams began their work is reduced to ridgetops. Finally in the stage of old age, the region is reduced to a nearly featureless plain near sea level or its inland projection. Uplift of the region in any stage of this evolution will activate a new cycle. Davis’s views dominated geomorphic thought until well into the 20th century, when quantitative approaches resulted in the rediscovery of Gilbert’s ideas.
Gravity, isostasy, and Earth’s figure
Discoveries of regional anomalies in Earth’s gravity led to the realization that high mountain ranges have underlying deficiencies in mass about equal to the apparent surface loads represented by the mountains themselves. In the 18th century the French scientist Pierre Bouguer had observed that the deflections of the pendulum in Peru are much less than they should be if the Andes represent a load perched on top of Earth’s crust. Similar anomalies were later found to obtain along the Himalayan front. To explain these anomalies it was necessary to assume that beneath some depth within Earth pressures are hydrostatic (equal on all sides). If excess loads are placed upon the crust, as by addition of a continental ice cap, the crust will sink to compensate for the additional mass and will rise again when the load is removed. The tendency toward general equilibrium maintained through vertical movements of Earth’s outer layers was called isostasy in 1899 by Clarence Edward Dutton of the United States.
Evidence for substantial vertical movements of the crust was supplied by studies of regional stratigraphy. In 1883 another American geologist, James Hall, had demonstrated that Paleozoic rocks of the folded Appalachians were several times as thick as sequences of the same age in the plateaus and plains to the west. It was his conclusion that the folded strata in the mountains must have accumulated in a linear submarine trough that filled with sediment as it subsided. Downward crustal flexures of this magnitude came to be called geosynclines.
Hydrologic sciences
Darcy’s law
Quantitative studies of the movement of water in streams and aquifers led to the formulation of mathematical statements relating discharge to other factors. Henri-Philibert-Gaspard Darcy, a French hydraulic engineer, was the first to state clearly a law describing the flow of groundwater. Darcy’s experiments, reported in 1856, were based on the ideas that an aquifer is analogous to a main line connecting two reservoirs at different levels and that an artesian well is like a pipe drawing water from a main line under pressure. His investigations of flow through stratified beds of sand led him to conclude that the rate of flow is directly proportional to the energy loss and inversely proportional to the length of the path of flow. Another French engineer, Arsène-Jules-Étienne-Juvénal Dupuit, extended Darcy’s work and developed equations for underground flow toward a well, for the recharge of aquifers, and for the discharge of artesian wells. Philip Forchheimer, an Austrian hydrologist, introduced the theory of functions of a complex variable to analyze the flow by gravity of underground water toward wells and developed equations for determining the critical distance between a river and a well beyond which water from the river will not move into the well.
Surface water discharge
A complicated empirical formula for the discharge of streams resulted from the studies of Andrew Atkinson Humphreys and Henry Larcom Abbot in the course of the Mississippi Delta Survey of 1851–60. Their formula contained no term for roughness of channel and on this and other grounds was later found to be inapplicable to the rapidly flowing streams of mountainous regions. In 1869 Emile-Oscar Ganguillet and Rudolph Kutter developed a more generally applicable discharge equation following their studies of flow in Swiss mountain streams. Toward the end of the century, systematic studies of the discharge of streams had become common. In the United States the Geological Survey, following its establishment in 1879, became the principal agency for collecting and publishing data on discharge, and by 1906 stream gauging had become nationwide.
Foundations of oceanography
In 1807 Thomas Jefferson ordered the establishment of the U.S. Coast Survey (later Coast and Geodetic Survey and now the National Ocean Survey). Modeled after British and French agencies that had grown up in the 1700s, the agency was charged with the responsibilities of hydrographic and geodetic surveying, studies of tides, and preparation of charts. Beginning in 1842, the U.S. Navy undertook expansive oceanographic operations through its office of charts and instruments. Lieut. Matthew Fontaine Maury promoted international cooperation in gathering meteorologic and hydrologic data at sea. In 1847 Maury compiled the first wind and current charts for the North Atlantic and in 1854 issued the first depth map to 4,000 fathoms (7,300 metres). His Physical Geography of the Sea (1855) is generally considered the first oceanographic textbook.
The voyage of the Beagle (1831–36) is remembered for Darwin’s biological and geologic contributions. From his observations in the South Pacific, Darwin formulated a theory for the origin of coral reefs, which with minor changes has stood the test of time. He viewed the fringing reefs, barrier reefs, and atolls as successive stages in a developmental sequence. The volcanic islands around which the reef-building organisms are attached slowly sink, but at the same time the shallow-water organisms that form the reefs build their colonies upward so as to remain in the sunlit layers of water. With submergence of the island, what began as a fringing reef girdling a landmass at last becomes an atoll enclosing a lagoon.
Laying telegraphic cables across the Atlantic called for investigations of the configuration of the ocean floor, of the currents that sweep the bottom, and of the benthonic animals that might damage the cables. The explorations of the British ships Lightning and Porcupine in 1868 and 1869 turned up surprising oceanographic information. Following closely upon these voyages, the Challenger was authorized to determine “the conditions of the Deep Sea throughout the Great Ocean Basins.”
The Challenger left port in December of 1872 and returned in May 1876, after logging 127,600 kilometres (68,890 nautical miles). Under the direction of Wyville Thomson, Scottish professor of natural history, it occupied 350 stations scattered over all oceans except the Arctic. The work involved in analyzing the information gathered during the expedition was completed by Thomson’s shipmate Sir John Murray, and the results filled 50 large volumes. Hundreds of new species of marine organisms were described, including new forms of life from deep waters. The temperature of water at the bottom of the oceans was found to be nearly constant below the 2,000-fathom level, averaging about 2.5 °C (36.5 °F) in the North Atlantic and 2 °C (35 °F) in the North Pacific. Soundings showed wide variations in depths of water, and from the dredgings of the bottom came new types of sediment—red clay as well as oozes made predominantly of the minute skeletons of foraminifera, radiolarians, or diatoms. Improved charts of the principal surface currents were produced, and the precise location of many oceanic islands was determined for the first time. Seventy-seven samples of seawater were taken at different stations from depths ranging downward to about 1.5 kilometres. The German-born chemist Wilhelm Dittmar conducted quantitative determinations of the seven major constituents (other than the hydrogen and oxygen of the water itself)—namely, sodium, calcium, magnesium, potassium, chloride, bromide, and sulfate. Surprisingly, the percentages of these components turned out to be nearly the same in all samples.
Efforts to analyze the rise and fall of the tides in mathematical terms reflecting the relative and constantly changing positions of Earth, Moon, and Sun, and thus to predict the tides at particular localities, has never been entirely successful because of local variations in configuration of shore and seafloor. Nevertheless, harmonic tidal analysis gives essential first approximations that are essential to tidal prediction. In 1884 a mechanical analog tidal prediction device was invented by William Ferrel of the U.S. Coast and Geodetic Survey, and improved models were used until 1965, when the work of the analog machines was taken over by electronic computers.
Atmospheric sciences
Composition of the atmosphere
Studies of barometric pressure by the British chemist and physicist John Dalton led him to conclude that evaporation and condensation of vapour do not involve chemical transformations. The introduction of vapour into the air by evaporation must change the average specific gravity of the air column and, without altering the height of that column, will change the reading of the barometer. In 1857 Rudolf Clausius, a German physicist, clarified the mechanics of evaporation in his kinetic theory of gases. Evaporation occurs when more molecules of a liquid are leaving its surface than returning to it, and the higher the temperature the more of these escaped molecules will be in space at any one time.
Following the invention of the hot-air balloon by the Montgolfier brothers in 1783, balloonists produced some useful information on the composition and movements of the atmosphere. In 1804 the celebrated French chemist Joseph-Louis Gay-Lussac ascended to about 7,000 metres, took samples of air, and later determined that the rarefied air at that altitude contained the same percentage of oxygen (21.49 percent) as the air on the ground. Austrian meteorologist Julius von Hann, working with data from balloon ascents and climbing in the Alps and Himalayas, concluded in 1874 that about 90 percent of all the water vapour in the atmosphere is concentrated below 6,000 metres—from which it follows that high mountains can be barriers against the transport of water vapour.
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3717 | https://www.youtube.com/watch?v=W1m036J7qXU | Lecture 12 : Transient heat Conduction
NPTEL IIT Kharagpur
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[Music] so far we have us we have been studying transient conduction and several assumptions associated with it we have seen that incorporating lumped capacitance model that is if we can assume that the temperature is going to be space wise Isis isothermal as the object is getting cold that means temperature inside the solid will remain invariant with position it's a function of time only in that case significant simplification of the entire problem is possible and we are going to get a closed form solution for the variation of temperature with time we have also seen that this assumption can only be used when the Biot number which gives us some idea about the convection the conduction resistance and convection resistance it's the ratio of convection resistance to conduction resistance when the Biot number is small ideally when it is less than 0.1 then we can safely use the trap the lumped capacitance model in transient conduction so the significance of back number and that of another dimensionless number which is called Fourier number is clear to to all of us so what I would do in this class is L I'm going to solve another problem on transient conduction where you see that the lumped capacitance model is valid but it's slightly different than the one that we have attempted so far so far we have what we have done is when I write the energy equation it's rate of energy in minus rate of energy out plus any generation of energy rate of any generation of energy inside the control volume must be equal to the rate of energy stored in the control volume the the formula the methodology the modeling which we have adopted so far assumes that there is no energy which comes in it's let's say it's a quenching problem so there's no energy in to the control volume energy goes out principally by convection by a convection convective process and there's also no energy generated in the system and therefore the energy going out is simply going to be equal to the rate at which the energy is stored inside or stored or depleted inside the control volume and the equations which we have obtained essentially assumes that only two terms one term on the left hand side and the term on the right hand side they will remain in the physical description of the situation but what happens in the case where we have energy generation as well in the control volume so the whenever you are going to use any formula be careful to identify whether the basis on which the physical system based on which the formula has been derived whether or not it's identical with the problem that you are solving so the first tutorial problem of today is when an electric current passes through an electric wire and the temperature obviously will rise as a result of the heat generation and let's also assume that the wire is exposed to a conviction environment so as the temperature of the well Rises it's going to lose heat by convection and part of the heat which is which is generated in the world is going to be used to raise the energy the energy stored the capacitive energy stored in the system so rate of energy out - of rate of energy out + rate of energy generation must be equal to the rate of energy stored so this way the inclusion of the generation term makes the problem different then that then those we have already discussed and solved in this so the first problem today and I hope you can read this which it says is that a long way of diameter D equals 1 millimeter so you have a diameter aware of diameter one millimeter is submerged in an oil bath so it's it has oil everywhere in the oil bath is at a temperature of 25 degree centigrade and the word has electrical resistance per unit length so I call it as our a prime to be 0.01 ohm per meter and the current which flows through this that I is equal to hundred ampere the oil bath creates a convective environment where each can be eight the value of H is equal to 500 watt per meter square Kelvin so there are three parts or few parts to the problem the first one is what is the what is the steady state temperature of the word okay the second part is from the time the current is applied how long does it take for the work to reach a temperature which is within 1 degree centigrade of the steady state fellow so these are the and there are some properties of the wire are given for example Rho C and K thermal conductivity all these are provided so but essentially the problem is that of heat transfer by convection to the oil heat generation because of the flow of current current in it and since it is a transient process it would require some amount of heat it would it would also denote that some amount of heat is going to be stored in it so the equation if I write that rate of heat in - out plus-or-minus generation or depletion should be equal to rate of energy stored in the system so the first part of the problem we can write that at steady state when the steady state is there the right hand side of this equation is zero because at at at steady state it's the the the temperature is not going to be a function of time since the temperature it could still be a function it could still be a function of position but at steady but at steady state the there would be no change in in the in the energy stored inside the system there's no heat which comes into the system so out should be equal to generation so at steady state the amount of heat generated because of because of the resistance of the where and the current which is flowing through it must be equal to the energy which is going out and this out is by a convective process we know the value of H and so on so the energy generated must be equal to energy out so if I is the current which is flowing through it and R is the resistance of the entire or where then it should be Newton's law of cooling which it should give us just H a delta T and by H what we mean as the convective heat transfer coefficient times pi D L where L is the entire length and this is the T of the where at steady state minus T infinity so at steady state this equality must hold in order to obtain in order to obtain what is the steady state temperature now we we it's a simple equation which can be where you can substitute the values and get the final value of the temperature of the wire which is going to be a constant which is which will not change with location and therefore the value of T W can be obtained from this relation one point to note here is that this the value of the resistance per unit length is provided in the problem so if I divide the left hand side this side by re by L then I have all the all the way very all the numbers with with me so the a is hundred ampere square into R by L is 0.01 ohm per meter and then Y sorry I missed a D over here pi times D is 1 into 10 to the power of minus 3 into H is 500 plus T infinity which is given as 25 so this should be about eighty eight point seven degree centigrade so the steady state temperature steady state temperature of the wire eighty eight point seven degree centigrade the next part of the problem tells us that we need to find out what is the what is the what time does it take for the were to reach a temperature which is within one degree centigrate of this steady-state value so starting at 25 degree centigrade how much time has to has to has to after how much time the temperature will be one degree centigrade of the steady state value that means what is the time required for the way to reach a temperature of eighty seven point seven degree centigrade one thing we have to keep in mind here is that we are using the lumped capacitance assumption whenever we use lumped capacitance capacitance assumption or come lumped capacitance model it is imperative that we find out whatever we have done whether or not it's going to be valid so the way to check the the validity of lump capacitance model is to find out what's the value of the bad number so once the value of the mach number is calculated and if it is less than 0.2 0.1 then whatever we have done so far or whatever we are going to do by incorporating the lamp capacitance model would be correct so the next step or the one of the major steps of solving transient conduction problem using lumped capacitance is to show that your assumption is valid by calculating what is the value of the Biot number so that's what we should do next so this value of the bad number I can calculate right now which would be H times R 0 by K and see I have used R 0 ideally this length scale is simply volume by area so if it is volume by area for a cylindrical one it's not going to be R 0 it's going to be most likely R 0 by 2 but as they have mentioned in order to be on the conservative side of calculations on the conservative side the length scale is always chosen as the dimension across which we are getting the maximum temperature drop so for the case of a cylindrical system you would expect the maximum drop in temperature in the solid between the center line and that of the surface so to be really to be conservative about the number the magnitude of the Biot number will always use R 0 for the case of cylinders and for the case of spheres as well whereas for the case of a plane wall we will use half the thickness of the wall when it is heated uniform when it is heated from both sides uniformly from both sides heated or cold uniformly from both sides so that's the length scale that we should use for calculating the value of the Biot number so we put the values in here what you would get is the value of H is 500 the value of Arc zero can be can be substituted in here and the value of so this is by 2 and the value of K which would give you the value of bite number to be really really small definitely less than 0.1 and therefore the major this thing is LC lumped capacitance more as a model is valid in here so once we know that lump capacitance is valid then we should be able to able to calculate what is the time required for for this this thing to reach 1 within one degree of the steady state temperature and we know that the steady state temperature is 88 point seven so essentially we have to find out how long does it take for the way to reach a temperature of eighty seven point seven degree centigrade which is one degree of the steady state temperature in order to do that I am going to first write physically what is going to happen what is happening in this case so in this case the governing equation takes the form that a dot generation generation minus a dot out by convection is equal to that rate of change of energy stored in the system so this e dot G is the additional term which we are getting for this specific problem now what is e dot G if we express everything in terms in terms of energy only so this is re prime which is a resistance per unit length times the length of the length of the wire and a dot out would simply be H times a which is PI D times l t minus T infinity this T is the time variation time varying temperature of the system but since bite number is less than 0.1 the T inside the where will remain a constant okay so it's not going to vary in space but it's going to vary in time in the Y dot store would simply be equal to D DT the time rate of change of energy contained within the control volume so energy should always be expressed in terms of a reference but since that reference is a constant I can simply write it in this specific form which is Rho PI d square times L so that's that's the volume times C times DT by D time so this is this this is this is the M CP the rate of change of time with the rate of change of temperature with time so this I am going to put in my physical equation the understanding here and the length can be canceled from all sides so what we end up with is the variation of temperature with time so this pi will cancel D&D from here or cancel now if you look at this part this is nothing but a constant let's call this as C 1 and this is also a constant for H by Rho CP times D so this is another constant let's call it as C 1 so my C 1 is 4-h by Rho CP D and C 1 is simply 4 I square r-e prime by Rho CP by d square so that's straightforward so we what we got is a straightforward expression in terms of the variation of temperature with time which is so this is the governing equation which we need to solve and it requires one boundary condition so but before that I think we can also define that t minus T infinity to B equals theta it's just a it I am defining it in this way so my governing equation would therefore would become D theta DT to be equal to minus c1 theta minus c2 so this would be my more compact form of the same governing equation okay and at the boundary condition would be at theta equals sorry at time T equals 0 theta is Theta I which is the initial temperature difference between the wire and the surrounding medium so this equation has to be solved with this initial condition to obtain the variation of theta with with time now the solution of this is pretty simple and I would only show you some of these steps and I think you should do it on your own and what you would get this thing then is D theta DT would be if I take C 1 outside theta minus C 2 by C 1 in on integration you get ln theta minus C 2 by C 1 equals minus C 1 time plus C 3 where we have we can see that from our previous expression for C 2 and C 1 C 2 by C 1 simply bi square re prime by PI DL and this c3 is the constant of integration which can be evaluated if we supply the appropriate boundary conditions so the appropriate boundary condition as I said at T equals 0 theta is equal to theta I which would give rise to c3 to be ln of theta I minus C 2 by C 1 so 1 after everything is said and done what you get is t minus T infinity minus C 2 by C 1 you should check it on your own to do it or to be to ensure that the numbers are correct minus C 2 by C 1 would be equal to exponential minus C 1 T so if you look once carefully to this equation once again what you what you it should notice is that we are getting similar form to that what we have obtained when we did not have any energy generation but the presence of the energy energy generation term because of the Joule heating of the of the Joule heating present in the world creates additional introduces additional terms into the governing equation so which must be taken into account so the entire purpose of any modeling is for you to ensure that you did not miss out on any of the physical processes which are taking place in the system so once you identify the physical processes then it would it would be easy to substitute the corresponding form corresponding mathematical form of the physical process in your equation and when you do the algebraic sum of the contribution of all physical processes which are contributing to the total amount of energy which is stored in the system then what you have is your governing equation that's all they're there to is so just identify the processes put the terms in the conservation equation and the sum must be equal to the right hand side which is the time rate of change of energy stored inside the system it's that simple and once you do that then it's going to be a case of integration of an ordinary defer stored ordinary differential equation and you require a boundary condition in this case an initial condition at time T equals zero so what you see is if the time at T equal to zero that is initially you know what is the temperature of the solid to start with then that provides you with the boundary condition as is the case in this specific problem so what remains here is to put the values of Ti the values of the thermo physical purpose our physical properties for example Rho CP K Rho and CP Rho CP and gape in you use the value of convective heat transfer coefficient H which is a parameter which depends on many things including the operating conditions and what you what you are going to get is the unknown temperature and we need to calculate the unknown time I am sorry there you need to calculate the time required for the work to reach 1 degree centigrade of the steady state temperature so your T the temperature that you are that you are shooting for is one degree less than the steady state temperature and if that is known since the steady state temperature is known the only unknown in that equation is the time okay so that time you can calculate easily so when you put the values in here what you would get is that your T infinity is 25 you know the other other values that you would get the values of C 1 you can calculate for H by Rho CP D which should turn out to be 0.5 second inverse and C 2 by C 1 once you put the values should be 60 3.7 and your TI initial temperature when there is no current flowing through the wire the initial temperature must also be equal to T infinity so T is the Infinity and T is within one centigrade of steady state temperature so this must be equal to eighty seven point seven degree centigrade so you put all these values in there what you should get as the time required is eight point three one second so this gives you some some idea about how to look at a problem which is which is different than what is given what you have done what you have seen in your textbook okay so you should be prepared you should be prepared to modify the equation given in your text depending on what you have what what you have in the present scenario so this is all about long capacitance that I wanted to I wanted to teach in this class what there is still a vast number of situations in which the lumped capacitance would not be valid so how do you treat such a system the fundamental equation the fundamental conduction equation the differential equation will be valid but how we were going to solve them is the question so the problem shifts from that of physical understanding and heat transfer to the problem of solving partial differential equations where you may have in the in the in in the simplest case the temperature is a function of time and it's also a function of one of the dimensions let's say X so your governing equation would be K times delta T del X square is e to alpha is equal to is equal to Rho CP times Del temperature by Del x so this kind of equation can be solved in a number of ways the PD one of the ways of solving the PD is the method of combination of variables the method of separation of variables so these different types of methods are available in your text which I would not cover in this in this specific course but in you in most of the cases when you solve a two dimensional heat transfer problem you are going to end up with a series solution and that series solution can be truncated for some special cases in those truncated solution set of this series solution can be expressed in graphical form so that's one way of analytically finding out what is the how does the temperature vary as a function of X Y and time there is another method which is the numerical method and you have probably heard about the finite difference method if not you are going to read it in your numerical in any in your course on numerical computation where each equation can that differential equation can be approximated as a difference equation so you can define definite nodes in the control volume and write the difference equation across that node and what you would end up with is as is a series of equations series of algebraic equations which can then be solved to obtain what is the temperature of each of these nodes so the finite difference is a powerful tool of numerically solving differential equations the differential the heat diffusion equation in order to obtain the temperature at every node which essentially is a numerical come a problem so I I will not teach that in this in this course but it would it should have Remy it should you should be you should be familiar with how to convert a governing akan vert a differential equation to a difference equation and then solve the resulting algebraic equation to obtain the temperature at every node now in your textbook if you are interested you can see how the difference equation can be written there are different different ways by which you can write the difference equations how what you are going to what would be the form of the equation if one node is surrounded by other nodes in the free space or what if the node is essentially sitting on the surface based on that your the your difference equation would be would be slightly different but there is a specific methodology which one can adopt which is there in the in the textbook and if you are interested you can take a look at that by which you can convert the temp the difference equation the differential equation into difference equation now what would be the spacing between the nodes or so to say what is going to be the size of the grid and other numerical numerical considerations in order to in order to increase the accuracy of your method in order to ensure that your solution remains stable all these are part of a numerical solution of differential equations course which either you have done or you are going to do so we if you are interested take a look at the finite difference method of solution of heat diffusion equations so the major part which we have covered is the the lumped capacitance model which gives you a which gives you a closed form solution but there is a there is a vast literature available for analytical solution as well as for numerical solution in some special cases these numerical solutions or the reduced form of the analytical solutions can be expressed in graphical form which is very convenient to use from an engineering standpoint so in the next class we are going to see what is going to happen if we can we can truncate the series solution arising from the arising from the analytical solution of the heat diffusion equation and express them graphically and use them to solve for temperature variation with time and temperature variation with space for systems in which the biota number is not less than 0.1 such that the the lumped capacitance model is not valid so we will see the graphical solution of transient conduction problems where lumped capacitance assumption cannot be made in the next class |
3718 | https://gogeometry.com/geometry/perpendicular_chords_theorems_problems_high_school_college_index.html | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | |
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| | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- | | | | | | | | | Perpendicular Chords, Theorems and Problems - Table of Content | | | | | | Sagitta, Arc, Chord. | | | Geometry Problem 680. Concentric Circles, Radii, Chords, Perpendicular, Metric Relations. | | | Geometry Problem 651. Archimedes Book of Lemmas: Proposition 11 Perpendicular chords, Arcs, Radius. | | | Geometry Problem 650. Tangent Circles, Diameter, Radius, Chord, Perpendicular. | | | Geometry Problem 649. Archimedes Book of Lemmas: Proposition 10 Tangent, Chord, Perpendicular, Parallel, Midpoint. | | | Geometry Problem 648. Archimedes Book of Lemmas: Proposition 9 Perpendicular chords, Arcs. | | | Problem 578. Circle, Perpendicular Chords, Midpoints, Curvilinear Triangle, Area, 90 Degrees. | | | Proposed Problem 437. Tangent Circles, Diameter, Chord, Perpendicular, Congruence, Measurement. | | | Proposed Problem 436. Tangent Circles, Diameter, Chord, Perpendicular, Congruence, Measurement. | | | Proposed Problem 373. Square, Inscribed Circle, Diagonal, Perpendicular, Chord, Angle. | | | Proposed Problem 334. Cyclic Quadrilateral, Perpendiculars to Diagonals, Chord. | | | Proposed Problem 313. Circle, Chord, Tangent, Perpendicular, Geometric Mean. | | | Proposed Problem 297. Intersecting Circles, Chord, Secant, Radius, Angle, Perpendicular. | | | Proposed Problem 296. Intersecting Circles, Chord, Radius, Angle, Perpendicular. | | | Proposed Problem 293. Inscribed Quadrilateral, Perpendicular lines, Chord, Rectangle, Isosceles Right triangle, Area, Similarity, Parallel lines. | | | Arithmetic Mean, Geometric Mean, Harmonic Mean, Root Mean Square. Perpendicular Chord, Diameter. | | | | Home | Search | Geometry | Circles | Chord | Perpendicular lines | Email | Post a comment | By Antonio Gutierrez | | | | | | | | |
3719 | https://sges.sysu.edu.cn/article/297 | 测绘学堂：测绘基础知识简介讲解 | 中山大学遥感科学与技术学院
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测绘学堂：测绘基础知识简介讲解
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测绘学堂：测绘基础知识简介讲解
稿件来源：51GIS在线课堂 发布人：高心雨 发布日期：2019-12-05 阅读次数：3083
测绘基础知识简介
01.名词解释
1、测绘：是指对自然地理要素或者地表人工设施的形状、大小、空间位置及其属性等进行测定、采集的活动。
2、地图：就是依据一定的数学法则，使用制图语言、通过制图、综合在一定的载体上，表达地球上各种事物的空间分布、联系及时间中的发展变化状态的图形。
3、等高线：是地面上高程相等的各相邻点连成的闭合曲线。等高线又分为首曲线、计曲线、间曲线。
4、地物：是指的是地面上各种有形物（如山川、森林、建筑物等）和无形物（如省、县界等）的总称。现行图式将所有地图地物要素分为测量控制点、水系、居民地及设施、交通、管线、境界、地貌和植被与土质8个一级大类。一级地物要素共有8种，二级地物要素296种，三级地物要素520种和558个四级地物要素。
5、地貌：地球表面各种形态的总称。主要分为：山地、盆地、丘陵、平原等。
6、地形：地表的形态，是指地势高低起伏的变化。
7、1954北京坐标系：将我国大地控制网与前苏联1942年普尔科夫大地坐标系相联结后建立的我国过渡性大地坐标系。
8、1956年黄海高程系：是在1956年确定的。它是根据青岛验潮站1950年到1956年的黄海验潮资料，求出该站验潮井里横按铜丝的高度为3.61米，所以就确定这个铜丝以下3.61米处为黄海平均海水面。从这个平均海水面起，于1956年推算出青岛水准原点的高程为72.289米。我国测量的高程，都是根据这一原点推算的。
9、1980西安坐标系：采用1975年国际大地测量与地球物理联合会第十六届大会推荐的地球椭球基本参数设立的坐标系。该坐标系原点在陕西省泾阳县永乐镇，位于西安西北方向约60km，故称为1980年西安坐标系。基准面采用青岛大港验潮站1952-1979年确定的黄海平均海水面（即1985国家高程基准）。
10、1985国家高程基准：采用青岛验潮站1952-1979年的验潮数据确定的黄海平均海水面所建议的高程基准。水准原点起算高程为72.260米。
11、WGS-84坐标系：一种国际上采用的地心坐标系。坐标原点为地球质心，其地心空间直角坐标系的Z轴指向BIH （国际时间服务机构）1984.O定义的协议地球极（CTP)方向，X轴指向BIH 1984.0的零子午面和CTP赤道的交点，Y轴与Z轴、X轴垂直构成右手坐标系，称为1984年世界大地坐标系统。目的：在世界上建立一个统一的地心坐标系。
12、CGCS2000坐标系：是（中国）2000国家大地坐标系的缩写，该坐标系是通过中国GPS 连续运行基准站、 空间大地控制网以及天文大地网与空间地网联合平差建立的地心大地坐标系统。以ITRF 97 参考框架为基准, 参考框架历元为2000.0。
13、独立坐标系：任意选定原定和坐标轴的直角坐标系。
14、国家大地控制网：主要包括2000国家大地控制网和全国大地网成果。 国家施测的有：高精度A、B级（GPS网），全国GPS一、二级网以及“中国地壳运动观测网络”大约2500个；全国天文大地网（含三角点、导线点）48000多个。“十二五”期间将建设360个GNSS连续运动基准站和4500个卫星大地控制点。
15、国家高程控制网：包括国家一、二期水准网和1998年建成的国家二期一等水准网点测水准路试，长约31.5万千米。“十二五”期间一期工程将完成12.2万千米一等水准观测。还包括全国似大地水准面成果。
02.测绘专业简介
1、大地测量：确定地面点位置，地球形状大小和地球重力场的精密测量。
内容：三角测量、精密导线测量、水准测量、天文测量、卫星大地测量、重力测量、大地测量计算；
作用：
（a）为地形测图和大型工程测量提供平面和高程控制；
（b）为空间科学和国防建设提供精确的点位坐标，距离、方 位角和地球重力场数值；
（c）为研究地球形状大小，地壳运动和地震预报等提供资料；
2、地形测量：是指测绘地形图的作业。即对地球表面的地物、地形在水平面上的投影位置和高程进行测定，并按一定比例缩小，用符号和注记绘制成地形图的工作。
平板仪测图法、经纬仪测绘法、小平板仪与经纬仪联合测图法已经淘汰。
目前主要采用：
（1）航空摄影测量
（2）遥感
（3）无人机遥感
（4）全站仪极坐标法
（5）GPS-RTK
（6）三维激光扫描
（7）移动测量系统（车载）
3、摄影测量与遥感：属于地球空间信息科学的范畴，它是利用非接触成像和其他传感器对地球表面及环境、其他目标或过程获取可靠的信息，并进行记录、量测、分析和表达的科学与技术。
发展分为3个阶段：
（1）模拟测图
（2）解析测图
（3）数字摄影测量及3S技术
任务：
（1）测绘大面积的地形图
（2）获取4D测绘成果
DLG----数字线划图
DEM----数字高程横型
DOM----数字影像地图
DRG----数字栅格地图
（3）遥感解译：以多光谱、多分辨率和多时相卫星影像为数据源，研究地表变迁及地质调查的新方法。可以监测土地利用变化，环境污染监测、自然灾害监测，农业监测、大范围地表形状探测和气象变化监测等。
4、工程测量：按其工作顺序和性质分为：勘测设计阶段的工程控制测量和地形测量；施工阶段的施工测量和设备安装测量；竣工和管理阶段的竣工测量、变形观测及维修养护测量等。按工程建设的对象分为：建筑工程测量、水利工程测量、铁路测量、公路测量、桥梁工程测量、隧道工程测量、矿山测量、城市市政工程测量、工厂建设测量以及军事工程测量、海洋工程测量等等。因此，工程测量工作遍布国民经济建设和国防建设的各部门和各个方面。
5、地籍测量：是在权属调查的基础上运用测绘科学技术测定界址线的位置，计算面积，绘制地籍图，为土地登记、核发证书提供依据，为地籍管理服务。地籍测量是土地管理工作的重要基础，它是以地籍调查为依据，以测量技术为手段，从控制到碎部，精确测出各类土地的位置与大小、境界、权属界址点的坐标与宗地面积以及地籍图，以满足土地管理部门以及其它国民经济建设部门的需要。
为满足地籍管理的需要，在土地权属调查的基础上，借助仪器，以科学方法，在一定区域内，测量每宗土地的权属界线、位置、形状及地类等，并计算其面积，绘制地籍图，为土地登记提供依据而进行的专业测绘工作。它是土地管理的技术基础。要求分级布网、逐级控制，遵循“从整体到局部，先控制后碎部”的原则。地籍测量是在权属调查的基础上运用测绘科学技术测定界址线的位置，计算面积，绘制地籍图，为土地登记、核发证书提供依据，为地籍管理服务。
地籍测量任务：
（1）地籍平面控制测量
（2）地籍细部测量
（3）地籍原图绘制
（4）面积量算与汇总统计
（5）成果的检查与验收
6、房产测量：主要是测定和调查房屋及其用地状况，为房产产权、房籍管理、房地产开发利用、征收税费以及城镇规划建设提供测量数据和资料，是常规的测绘技术与房产管理业务相结合的测量工作。它是研究城镇的建成区和建成区以外的工矿企业、事业单位及其相毗连居民点的房产测绘的理论，仪器和方法的应用技术。房产测量与城市地形测量、地籍测量有相同之处，但由于服务对象不同，内容和要求又有所不同。房产测量的基本内容包括：房产平面控制测量、房产调查、房产要素测量、房产图绘制、房产面积测算、变更测量、成果资料的检查与验收等。
7、地图制图：是研究地图及其编制和应用的一门学科。它研究用地图图形反映自然界和人类社会各种现象的空间分布，相互联系及其动态变化。
工作内容：地图投影、地图编制、地图整饰和地图制印等部分组成。
主要包括：制图资料的选择、分析和评价,制图区域的地理研究,图幅范围和比例尺的确定，地图投影的选择和计算，地图内容各要素的表示法，地图制图综合的原则和实施方法，制作地图的工艺和程序，以及拟定地图编辑大纲等。
地图符号和色彩设计，地貌立体表示，出版原图绘制以及地图集装帧设计等。
地图制印包括地图复照、翻版、分涂、制版、打样、印刷、装帧等工艺技术。
8、海洋测绘：海洋水体和海底为对象所进行的测量和海图编制工作。主要包括海道测量、海洋大地测量、海底地形测量、海洋专题测量，以及航海图、海底地形图、各种海洋专题图和海洋图集等的编制.
发展阶段：
海洋测绘大致可分3个阶段：①20世纪30～50年代中期，开始对海洋进行地球物理测量，包括海洋地震测量、海洋重力测量等。这阶段利用回声探测数据绘制海底地形图，揭示了海洋底部的地形地貌；利用双折射地震法获取大洋地壳的各种地球物理性质，证明大洋地壳与大陆地壳有显著的差异。②1957～1970年，实施了国际地球物理年（1957～1958）、国际印度洋考察（1959～1965）、上地幔计划（1962～1970）等国际科学考察活动，发现了大洋中条带磁异常，为海底扩张说提供了强有力的证据，揭示了大洋地壳向大陆地壳下面俯冲的现象，观测了岛弧海沟系地震震源机制。③70年代以后，广泛应用电子技术和计算机技术于海洋测绘中
主要方法：
测量方法主要包括海洋地震测量、海洋重力测量、海洋磁力测量、海底热流测量、海洋电法测量和海洋放射性测量。因海洋水体存在，须用海洋调查船和专门的测量仪器进行快速的连续观测，一船多用，综合考察。
基本测量方式包括：
①路线测量。即剖面测量。了解海区的地质构造和地球物理场基本特征。
②面积测量。按任务定的成图比例尺，布置一定距离的测线网。比例尺越大，测网密度愈密。在海洋调查中，广泛采用无线电定位系统和卫星导航定位系统。
03.测绘新技术新产品简介
1、3S：目前测量界最先进的理论和技术
（1）GPS（Global Positioning System）
全球卫星定位系统（北斗系统）
（2）GIS（Geographic Information System）
地理信息系统
（3）RS（Remote Sensing）
遥感
2、GNSS系统：GNSS的全称是全球导航卫星系统（Global Navigation Satellite System），它是泛指所有的卫星导航系统，包括美国的GPS，俄罗斯的Glonass、欧洲的Galileo、中国的北斗卫星导航系统以及相关的增强系统。
3、定位系统连续运行基准站
国家的GNSS系统
贵州省的参考系统（在建）
4、车载移动测量系统
三维扫描+软件系统
国内有：中海达、北斗星空等
5、无人机遥感信息系统：它以无人驾驶飞行器为平台、以高分辨率数字遥感设备为传感器，以获取高分辨率遥感数据为应用目标，具有快速、实时对地观测、调查监测能力，因此在土地利用动态监测，矿产资源勘探，地质环境与灾害调查，海洋资源与环境监测，地形图更新，应急测绘等领域都将有广泛应用。
6、三维激光扫描技术：又被称为实景复制技术，是测绘领域继GPS技术之后的一次技术革命。它突破了传统的单点测量方法，具有高效率、高精度的独特优势.三维激光扫描技术能够提供扫描物体表面的三维点云数据，因此可以用于获取高精度高分辨率的数字地形模型。三维激光扫描技术是上世纪九十年代中期开始出现的一项高新技术，是继GPS空间定位系统之后又一项测绘技术新突破。它通过高速激光扫描测量的方法，大面积高分辨率地快速获取被测对象表面的三维坐标数据。可以快速、大量的采集空间点位信息，为快速建立物体的三维影像模型提供了一种全新的技术手段。由于其具有快速性，不接触性，穿透性，实时、动态、主动性，高密度、高精度，数字化、自动化等特性，其应用推广很有可能会像GPS一样引起测量技术的又一次革命。
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3720 | https://www.youtube.com/watch?v=cfq760WvLIY | Multiplying Fractions & Simplify - Fast & Easy Method
Math and Science
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Posted: 20 Aug 2024
Multiplying fractions is a vital math skill that is essential for various applications in both academic and real-life contexts. This video provides a comprehensive guide to understanding and mastering the multiplication of fractions, making the process clear and accessible.
The video starts by explaining the basic concept of fractions and the importance of multiplication in solving fraction problems. You will learn how to identify the numerator and denominator in each fraction and understand their roles in the multiplication process.
Step-by-step instructions demonstrate how to multiply fractions accurately. You will see examples of multiplying proper fractions, improper fractions, and mixed numbers. The video emphasizes simplifying fractions both before and after multiplication to ensure the final answer is in its simplest form.
Visual aids and practical exercises help reinforce the concepts, allowing you to follow along and practice independently. The video also addresses common mistakes, such as incorrectly multiplying numerators or denominators, and provides tips to avoid these errors.
Ideal for students, teachers, and anyone looking to enhance their math skills, this video simplifies the concept of multiplying fractions. By the end, you will be confident in your ability to multiply fractions accurately and efficiently, improving your overall proficiency in mathematics.
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so let's take probably the simplest multiplication problem I can come up with and that will be the following one-half the fraction one-half and it's multiplied by that's what the dot means uh one half and we don't use X's for multiplication anymore because we're going to be using variables letters and so we use x a lot so we're not using X for multiplication anymore when you see a dot it means multiply the same as before using an X so but first before we get started I want to represent what these fractions actually mean the first fraction fraction here is one half so if you have a whole circle and you take only cut it into two pieces and take one then we say we have a half and we're multiplying this fraction one-half by this other fraction so I'll put like a DOT between them because what's really happening is you're taking the first fraction and you're multiplying it by the second fraction now before we actually talk about what this means let's calculate the answer when we multiply fractions also when we divide fractions we do not need a common denominator you know common denominator we have to use it when we add and subtract fractions but actually we don't need to do any of that stuff when we're multiplying or dividing fractions it's actually much easier to multiply fractions all you do is you multiply the numerators together and then you multiply the denominators together it doesn't really matter that this is 2 and 2. we do not have to have a common denominator so we multiply on the in the numerator 1 times 1 and on the denominator it's 2 times 2. so I'm just showing you here what you're doing in each position 1 times 1 is 1 and 2 times 2 is 4. so the answer to this problem of one-half times one-half is the fraction one-fourth and we always ask ourselves can we simplify this fraction but we actually can't in this case 1 4 is the simplest way we can write this and of course the fraction 1 4 as you know you can put it any which way you want but this is what the fraction 1 4 is so what we want to actually do is understand why is it when we multiply one-half times one-half we get one-fourth all right so mechanically we multiply the tops we get an answer we multiply the bottoms we get an answer simplify and then that is our answer now what I want you to think about when it comes to multiplying fractions is I want you to think of an X you know the thing you come out to would cut a tree down you chop a tree down right what you're doing is you're starting with the first fraction you have one half that's what you're starting with half of a pizza think of it as half of a pizza and when you multiply by whatever it is you're multiplying by the second fraction is just representing how much you're going to chop down the first fraction so what it's basically saying here is you're going to start out with half of a pizza that's what this represents and you're going to chop it into a into one half of what it once was so you started out with this amount and this fraction just means if you think of a whole circle this fraction is a whole uh if you can think of a whole circle and this fraction is half of that that's what the fraction half means and it's telling you to take whatever you start with the whole which is what you've been given to start start with and you're cutting it in half of what you want started with so this this Size Doesn't the size doesn't matter at all it's just telling you start with this cut it into half because this fraction one half is one half of its own full circle so if you were to start with this and you cut it in half because this fraction is telling you how much to chop it by then you can see right away that one-fourth is the answer because if I start with this and cut it in half the answer that I get is 1 4 that is half of what I started with if I change this fraction to chop it by a different amount maybe I chop it by more than one half then I will start with this and I'll end up with a little bit more than half as my answer if I chop it by a fraction less than a half then I'm going to take what I start with and I'm going to cut it so that I have less than one half of what I started with so when you think of fraction multiplication try not to look at the sizes of the fractions too much what it's really doing is it's telling you start with this much stuff and then chop it by whatever the second fraction represents the second fraction in this case is a half so all we do is we start with this much stuff and we cut it in half and half of that of course is equal to 1 4. so that is what we're doing every time we multiply fractions so let's move on to problem two and I think we'll get the hang of it even further let's say we have again the fraction one-half is what we're starting with but instead of multiplying by a half we'll multiply by one-fourth so we're starting with half of a pizza but we're not cutting it in half we're cutting it to one-fourth of the size that it originally was so let's first calculate how to actually uh get the answer here what we're going to do to get the answer is we multiply the tops 1 times 1 is 1 and we multiply the bottoms 2 times 4 is 8. so to get the numerator one times one is one two times four is eight and we ask can we simplify this fraction we can't because 1 8 I cannot uh simplify by dividing top and bottom by something to make it any simpler so the answer is 1 8. let's try to understand what this actually means so again it's the same sort of thing we start out with a half of the pizza this is what half of a pizza looks like but we're not going to cut it in half we're going to cut it into one-fourth of the size this fat action you have to think of the full circle the full circle is a whole circle here you have one segment another segment another segment and then this would be the last fourth so what you're doing is you're asking yourself take this pizza cut it into four equal pieces and only take one of them so if we take this pizza and cut it into here's taking it cutting into half and then cutting it into half again cutting it into half again and take only one piece what's it going to look like it's going to look like 1 8 because that would be one piece the next piece would fit in something like this the third piece would fit in something like this and the fourth piece would be something like this so take what you start with and to chop it down by one-fourth of its original size you cut it into four equal pieces and only take one of them four equal pieces and only take one of them means the answer has to be 1 8. same thing here we started with a half and this is telling us to cut it in half which means to to cut it into two equal pieces and only take one of them and that's why the fraction that we got as an answer was one fourth so we're chopping it down in this case by something less than half of its original size so this is much smaller than half of its original size it is one-fourth of its original size all right this will be the last fraction where we will uh solve the problem and use the magnets after that we'll just be calculating let's take again one half times two-thirds you might be asking why am I using one-half so much mostly it's because it I want to choose a fraction large enough so you can see what's happening it this process this idea of chopping it it works for any multiplication problem but when we start with a half of a pizza everybody can visualize that so it makes it easier to talk about one half times two-thirds two-thirds is is a larger fraction so we're going to start with this one half and we're going to cut it down by this amount two-thirds or by this fraction two-thirds to get the answer what do we do we multiply the tops one times two is what two and we multiply the bottoms two times three is six so we get an answer of 2 6. and we always ask ourselves can we simplify this one we actually can simplify because they're both even numbers so we can divide the top fraction by two and we can divide the bottom fraction by two and we get 2 divided by 2 is one six divided by two is three and we get an answer of 1 3. so again we can calculate by multiplying the tops multiplying the bottoms and then simplifying most people can do that once I teach it to you but let's go a step further and try to understand what it means what it means is we start again with half of a pizza and we're not chopping it into a half of its original size we're not chopping it into fourths of its original size what we're doing is we're trying to represent and chop it by something called two-thirds think of the whole pizza being one piece two pieces three pieces in two thirds of the pizza is two out of three pieces so what it's telling us is take what you start with and chop it so when we chop it we're going to cut this P this piece that we started with into three equal pieces and we're going to keep two of them then we're going to chop it down by cutting it into three equal pieces and uh and only keeping two of them just as we have here for our whole circle two out of three pieces the answer we got is one-third let's see if this makes sense if we were to cut this thing into three equal pieces and keep two of them what would that actually look like well if we cut it into three equal pieces this would be one piece here this piece would be one and then we'd cut it right here this would be in the next piece and the third slice would be down here so if we cut it into one two three you gotta imagine another cut here one two three equal pieces and keep two of them then this whole orange wedge is what we actually end up with which is one third of a pizza and that's why it's one third same as two-sixths these are the exact same thing so we're going to solve more problems but in every case we start with what we are given one half we chop it by one half meaning we cut it into two pieces what we start with and we only keep one of them that's why it became one-fourth we started again with a half of a pizza we cut that and chopped it into four equal pieces and we only kept one of them so the answer was 1 8 we chopped it down in this case we started with one half of a pizza we cut it and chopped it down into three equal pieces would be one and then two one two three and we kept two of them and when we did that we got a fraction of one-third is the answer so in every case fraction multiplication is just taking what you start with and chopping it down by whatever the second fraction is telling you to chop it by so you cut it down by that by that fractional amount so for the rest of the problems we're not going to use any magnets we're just going to calculate and simplify let's take a look at two-fifths the fraction two-fifths and we're going to multiply it by two-thirds so it's the same problem as before we're just starting with a different amount of pizza two-fifths of a pizza will be a different amount but what we're going to do is we're going to cut it into three equal pieces and only take two of them how much will we have when we chop it by that amount we multiply the tops 2 times 2 is 4. remember multiply and 5 times 3 multiply to get 15. notice that you're not getting any kind of common denominator multiply the tops multiply the bottoms we cannot simplify this any further so we say 4 15 is the final answer all right great moving right along let's take a look at four fifths and we're going to multiply that by 3 8. again what does this mean it means you start with four-fifths of a pizza and you chop it by the fraction 3 8 that means whatever you start with you cut it into eight equal pieces and you only keep three of them that's what chopping does how do we get the answer we multiply the numerators 4 times 3 is 12. multiply the denominators 5 times 8 is 40. we have 12 40 and that is the correct answer but we can simplify this further 12 40. now we know that they're even they're even numbers so we could divide by two but then I recognize I can also divide by four so I'm going to do that 12 divided by 4 is 3 and 40 divided by 4 is 10. so the answer that you get is 310. so if you start with this much pizza and chop it down by by whatever this fraction is indicating in other words cut it into eight slices and keep three the answer I'm going to get is three tenths of a pizza when I line all of those pieces up together three tenths all right we're almost done halfway point we're past the halfway point let's take a look at 11 12 and we're going to multiply that by two-thirds so again take the first amount and we're chopping it down by two thirds so we're going to cut whatever this is into three equal pieces and only keep two of them how do we find the final answer multiply eleven times two which is 22 for the numerator and 12 times 3 which is 36 for that denominator so that's the final answer 22 well that's not the final answer but it's an intermediate answer because we can see that these are even numbers 22 36 so I know I can simplify it by dividing by 2 dividing by 2. 22 divided by 2 is 11 and 36 divided by 2 is 18. 18 times 2 is 36 you can grab a sheet of paper to make sure that you agree with that and 11 18 is the final answer I want you to pay attention to the final two problems they're both important let's take a look at Four Thirds and we're going to multiply that by four-fifths now what do you notice that's different about this one the main difference is you can tell that in all of the other problems 11 12 four fifths one-half two-fifths it they were improper but this fraction four-thirds is a uh improper fraction so all of the fractions before may have misspoke I don't know for sure but these are all uh proper fractions before this one is an improper fraction that means that this fraction is bigger than one whole pizza it's cutting a pizza into thirds but having four of them so you actually have one and a third pizzas here when you convert this to mixed number so you're starting with something larger than a whole pizza and then you're chopping it down sometimes you can end up with an answer that's bigger than a whole pizza you'll see what I'm saying here let's go ahead and just continue to multiply as usual we do the same thing 4 times 4 is 16 and 3 times 5 is 15. so we have 16 15. notice we got an an improper fraction there how do we convert this to a mixed number 16 divided by 15 how many times does it go one whole time and how many is left over 16 minus 15 is only one left over and and these are in terms of 15. so the fraction 16 15 is the same as the fraction 1 and 1 15. so converting fractions uh improper fractions to mixed numbers if you need practice for that go back to the previous lessons on that we've covered that in the past these two ways of writing the fraction are exactly the same thing cut cut a pizza into 15 into 15 slices but you have 16 of them which means you have more than one pizza or you can represent it as one whole pizza and 1 15 left over all right so I can really Circle either of them or both of them they're both fine but why do we get something bigger than one it's because we actually started with a pizza larger than a whole pizza we started with one whole pizza convert this to a mixed number four thirds you divide it in it'll be one and one-third so this is already bigger than a um a whole pizza and then we're cutting it down by four-fifths which is also really close to one so sometimes you can actually get an answer bigger than one because you start with an amount of pizza bigger than one which is what we have done here all right let's take a look at our last problem let's take a look at three-fourths and we're going to multiply that by five halves all right so let's just crank through it and then we'll try to understand more about what's going on in this problem multiply the numerators 3 times 5 15 multiply the denominators 2 times 4 is 8. we get a fraction that is again improper 15 is bigger so we have to convert 8 times 1 is 8 8 times 2 is 16. so it cannot go two times it only goes one time and then eight times one is eight so if we find the remainder 15 minus 8 what are we going to have we're going to have 7 left over out of eight and everything's in terms of eight slices so one and seven eighths is exactly the same thing as fifteen eight so these are correct again we got a fraction bigger than one y it's because this is less than one whole pizza three-fourths but this we're chopping it down by five halves we're chopping it down by something bigger than one think about what that means if you were to convert this to a mixed number two times two is four so it goes two times and then five minus four is one out of two so it really if you convert this it would go two and one half two and a half so what you're doing is you're taking this and you're multiplying it by two and a half which means you're multiplying it by something bigger than two you're more than doubling what you started with so of course the answer if you start with three-fourths and you more than double it you're going to get something bigger than one that's why these fractions are coming out to be uh larger than one so for most of these problems we are multiplying fraction times a fractions very small fractions together so most of the answers we're getting are just these small fractions but if you take a fraction that's larger than one and start multiplying it by another fraction you can get an answer bigger than one or an improper fraction however you want to write it same thing here this fraction was bigger than one so sometimes you can get a fraction larger than one there as well so here we've conquered multiplying fractions incredibly crucial material I'd like you to practice all of these yourself when you feel that you're comfortable follow me onto part two will get more skills and practice with multiplying fractions together learn anything at mathonscience.com |
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Boyle's Law Calculator
This Boyle's law calculator is a great tool when you need to estimate the parameters of a gas in an isothermal process. You will find the answer to "What is Boyle's law?" in the text, so read on to find out about the Boyle's law formula, see some practical examples of Boyle's law exercises and learn how to recognize when a process satisfies Boyle's law on a graph.
In case you need to work out the results for an isobaric process, check our Charles' law calculator.
Boyle's law definition
Boyle's law (also known as Boyle-Mariotte law) tells us about the relationship between the pressure of a gas and its volume at a constant temperature and mass of gas. It states that the absolute pressure is inversely proportional to the volume.
Boyle's law definition can also be phrased in the following way: the product of the pressure and the volume of a gas in a closed system is constant as long as the temperature is unchanged.
Boyle's law describes the behavior of an ideal gas. We can characterize this gas by the ideal gas equation, which you can read more about in our ideal gas law calculator. Boyle's law tells us about an isothermal process, which means that the temperature of the gas remains constant during the transition, as does the internal energy of the gas.
Boyle's law formula
We can write the Boyle's law equation in the following way:
p1 × V1 = p2 × V2
where p1 and V1 are initial pressure and volume, respectively. Similarly, p2 and V2 are the final values of these gas parameters.
We can write Boyle's law formula in various ways depending on which parameter we want to estimate. Let's say we change the volume of a gas under isothermal conditions, and we want to find the resulting pressure. Then, the equation of Boyle's law states that:
p2 = p1 × V1 / V2 or p2 / p1 = V1 / V2
As we can see, the ratio of the final and initial pressure is the inverse of the ratio for volumes. This Boyle's law calculator works in any direction you like. Just insert any three parameters, and the fourth one will be calculated immediately!
We can visualize the whole process on Boyle's law graph. The most commonly used type is where the pressure is a volume function. For this process, the curve is a hyperbola. The transition can progress in both ways, so both compression and gas expansion satisfy Boyle's law.
🔎 If a transition is an isochoric process (the constant volume), you'll find Omni's Gay-Lussac's law calculator helpful.
Boyle's law examples
We can use Boyle's law in several ways, so let's take a look at some examples:
Imagine that we have an elastic container that holds a gas. The initial pressure is 100 kPa (or 105 Pa if we use scientific notation), and the volume of the container equals 2 m3. We decide to compress the box down to 1 m3, but we don't change the overall temperature. The question is: "How does the pressure of the gas change?". We can use Boyle's law formula:
p2 = p1 × V1 / V2 = 100 kPa × 2 m3 / 1 m3 = 200 kPa
After halving the volume, the internal pressure is doubled. This is a consequence of the fact that the product of the pressure and the volume must be constant during this process.
2. The next Boyle's law example concerns a gas under 2.5 atm pressure while occupying 6 liters of space. It is then decompressed isothermally to the pressure of 0.2 atm. Let's find out its final volume. We have to rewrite Boyle's law equation:
V2 = p1 × V1 / p2 = 2.5 atm × 6 l / 0.2 atm = 75 l
You can always use our Boyle's law calculator to check if your evaluations are correct!
Where is Boyle's law applied?
Boyle's law describes all processes for which temperature remains constant. In thermodynamics, temperature measures the average kinetic energy that atoms or molecules have. In other words, we can say that the average velocity of gas particles doesn't change during that transition. Boyle's law formula is valid for a wide range of temperatures.
By opening the Additional parameters section of the calculator, you can choose any temperature you like, and we will calculate the number of molecules contained in the gas. You only have to ensure that the substance remains in the gas form (e.g., neither condensates nor crystallizes) at this temperature.
There are a few areas where Boyle's law is applicable:
Carnot heat engine — Consists of four thermodynamic processes, two of which are isothermal ones, satisfying Boyle's law. This model can tell us what the maximal efficiency of a heat engine is.
Breathing also can be described by Boyle's law. Whenever you take a breath, your diaphragm and intercostal muscles increase the volume of your lungs, which decreases gas pressure. As air flows from an area of higher pressure to a place of lower pressure, air enters the lungs and allows us to take in oxygen from the environment. During exhalation, the volume of the lungs decreases, so the pressure inside is higher than outside, so the air flows in the opposite direction.
Syringe — Whenever you have an injection, a doctor or a nurse draws a liquid from the small vial first. To do so, they use a syringe. By pulling the plunger, the accessible volume increases, which decreases pressure and, according to Boyle's law formula, causes the suction of the fluid.
Other thermodynamic processes
Boyle's law, along with Charles's and Gay-Lussac's law, are among the fundamental laws describing the vast majority of thermodynamic processes.
Other than working out the values of specific parameters like pressure or volume, it is also possible to discover something about the heat transfer and the work done by the gas during these transitions, as well as the internal energy change. We have gathered them all in our combined gas law calculator, where you can choose whichever process you like and evaluate the outcomes for real gases.
FAQs
Why is Boyle's law also called isotherm?
Boyle's law is one of the three fundamental thermodynamic processes. In each of them, we study a variation of two out of three quantities:
The pressure;
The temperature; and
The volume.
The third quantity remains constant during the process. In the case of Boyle's law, we don't change the temperature, thus we call the process isothermal.
How much will a balloon with initial volume 1000 cm³ expand at cruising altitude?
Follow these steps:
Find the initial pressure. We will take atmospheric pressure at sea level: Pi = 1 atm = 101,325 Pa.
Find the final pressure. In a cruising plane, the cabin is usually pressurized at about Pf = 0.8 atm = 81,060 Pa.
Calculate the final volume with Boyle's law: Vf = Pi · Vi/Pf = (101,325 Pa · 0.001 m3)/81,060 Pa = 0.00125 m3.
Find the expansion by subtracting the final and initial volumes : ΔV = Vf - Vi = (0.00125 - 0.001) m3 = 0.00025 m3 = 250 cm3.
How do I calculate Boyle's law?
To calculate Boyle's law, we need to perform a few simple steps, which depend on the initial data we know. To calculate the final pressure given the initial volume and pressure and the final volume:
Compute the product of the initial volume and pressure: Vi · Pi.
Divide the result by the final volume. The final pressure is Pf = (Vi · Pi)/Vf.
You can invert final and initial values freely (reversible transformation).
To find the final volume, switch it with the final pressure on the right-hand side of the formula.
What is the final pressure if the volume reduces by half with intial pressure 1 atm?
The final pressure in a process where the volume reduces by half, starting from Pi = 1 atm, is 2. To find this result:
Write Boyle's law for the final pressure: Pf = (Vi · Pi)/Vf.
In this formula, identify the ratio of the volumes. Since we know that the volume reduces by half, we can write Vi = 2 · Vf, hence Vi/Vf = 2.
The final pressure is, then: Pf = 2 · Pi = 2 atm.
Initial parameters
psi
cu ft
Final parameters
psi
cu ft
Additional parameters
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Section 1.4 Inverse Functions
Learning Objectives.
Determine the conditions for when a function has an inverse.
Use the horizontal line test to recognize when a function is one-to-one.
Find the inverse of a given function.
Draw the graph of an inverse function.
Evaluate inverse trigonometric functions.
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An inverse function reverses the operation done by a particular function. In other words, whatever a function does, the inverse function undoes it. In this section, we define an inverse function formally and state the necessary conditions for an inverse function to exist. We examine how to find an inverse function and study the relationship between the graph of a function and the graph of its inverse. Then we apply these ideas to define and discuss properties of the inverse trigonometric functions.
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Subsection 1.4.1 Existence of an Inverse Function
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We begin with an example. Given a function and an output we are often interested in finding what value or values were mapped to by For example, consider the function Since any output we can solve this equation for to find that the input is This equation defines as a function of Denoting this function as , and writing we see that for any in the domain of Thus, this new function, “undid” what the original function did. A function with this property is called the inverse function of the original function.
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Definition 1.97.
Given a function with domain and range its inverse function (if it exists) is the function with domain and range such that if In other words, for a function and its inverse
for all in and for all in
🔗
Note that is read as “f inverse.” Here, the is not used as an exponent and Figure 1.98 shows the relationship between the domain and range of and the domain and range of
🔗
Recall that a function has exactly one output for each input. Therefore, to define an inverse function, we need to map each input to exactly one output. For example, let’s try to find the inverse function for Solving the equation for we arrive at the equation This equation does not describe as a function of because there are two solutions to this equation for every The problem with trying to find an inverse function for is that two inputs are sent to the same output for each output The function discussed earlier did not have this problem. For that function, each input was sent to a different output. A function that sends each input to a different output is called a one-to-one function.
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Definition 1.99.
We say a is a one-to-one function if when
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One way to determine whether a function is one-to-one is by looking at its graph. If a function is one-to-one, then no two inputs can be sent to the same output. Therefore, if we draw a horizontal line anywhere in the -plane, according to the horizontal line test, it cannot intersect the graph more than once. We note that the horizontal line test is different from the vertical line test. The vertical line test determines whether a graph is the graph of a function. The horizontal line test determines whether a function is one-to-one (Figure 1.101).
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Note 1.100. Rule: Horizontal Line Test.
A function is one-to-one if and only if every horizontal line intersects the graph of no more than once.
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Example 1.102. Determining Whether a Function Is One-to-One.
For each of the following functions, use the horizontal line test to determine whether it is one-to-one.
Solution.
Since the horizontal line for any integer intersects the graph more than once, this function is not one-to-one.
Since every horizontal line intersects the graph once (at most), this function is one-to-one.
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Checkpoint 1.103.
Is the function graphed in the following image one-to-one?
Hint.
Use the horizontal line test.
Solution.
No.
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Subsection 1.4.2 Finding a Function’s Inverse
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We can now consider one-to-one functions and show how to find their inverses. Recall that a function maps elements in the domain of to elements in the range of The inverse function maps each element from the range of back to its corresponding element from the domain of Therefore, to find the inverse function of a one-to-one function given any in the range of we need to determine which in the domain of satisfies Since is one-to-one, there is exactly one such value We can find that value by solving the equation for Doing so, we are able to write as a function of where the domain of this function is the range of and the range of this new function is the domain of Consequently, this function is the inverse of and we write Since we typically use the variable to denote the independent variable and to denote the dependent variable, we often interchange the roles of and and write Representing the inverse function in this way is also helpful later when we graph a function and its inverse on the same axes.
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Note 1.104. Problem-Solving Strategy: Finding an Inverse Function.
Solve the equation for
Interchange the variables and and write
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Example 1.105. Finding an Inverse Function.
Find the inverse for the function State the domain and range of the inverse function. Verify that
Solution.
Follow the steps outlined in the strategy.
Step 1. If then and
Step 2. Rewrite as and let
Therefore,
Since the domain of is the range of is Since the range of is the domain of is
You can verify that by writing
Note that for to be the inverse of both and for all in the domain of the inside function.
🔗
Hint.Solution.
🔗
Subsubsection 1.4.2.1 Graphing Inverse Functions
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Let’s consider the relationship between the graph of a function and the graph of its inverse. Consider the graph of shown in Figure 1.106 and a point on the graph. Since then Therefore, when we graph the point is on the graph. As a result, the graph of is a reflection of the graph of about the line
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Example 1.107. Sketching Graphs of Inverse Functions.
For the graph of in the following image, sketch a graph of by sketching the line and using symmetry. Identify the domain and range of
Solution.
Reflect the graph about the line The domain of is The range of is By using the preceding strategy for finding inverse functions, we can verify that the inverse function is as shown in the graph.
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Checkpoint 1.108.
Sketch the graph of and the graph of its inverse using the symmetry property of inverse functions.
Hint.
The graphs are symmetric about the line
Solution.
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Subsubsection 1.4.2.2 Restricting Domains
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As we have seen, does not have an inverse function because it is not one-to-one. However, we can choose a subset of the domain of such that the function is one-to-one. This subset is called a restricted domain. By restricting the domain of we can define a new function such that the domain of is the restricted domain of and for all in the domain of Then we can define an inverse function for on that domain. For example, since is one-to-one on the interval we can define a new function such that the domain of is and for all in its domain. Since is a one-to-one function, it has an inverse function, given by the formula On the other hand, the function is also one-to-one on the domain Therefore, we could also define a new function such that the domain of is and for all in the domain of Then is a one-to-one function and must also have an inverse. Its inverse is given by the formula (Figure 1.109).
🔗
Example 1.110. Restricting the Domain.
Consider the function
Sketch the graph of and use the horizontal line test to show that is not one-to-one.
Show that is one-to-one on the restricted domain Determine the domain and range for the inverse of on this restricted domain and find a formula for
Solution.
The graph of is the graph of shifted left 1 unit. Since there exists a horizontal line intersecting the graph more than once, is not one-to-one.
On the interval is one-to-one. The domain and range of are given by the range and domain of respectively. Therefore, the domain of is and the range of is To find a formula for solve the equation for If then Since we are restricting the domain to the interval where we need Therefore, Interchanging and we write and conclude that
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Checkpoint 1.111.
Consider restricted to the domain Verify that is one-to-one on this domain. Determine the domain and range of the inverse of and find a formula for
Hint.
The domain and range of is given by the range and domain of respectively. To find solve for
Solution.
The domain of is The range of is The inverse function is given by the formula
🔗
Subsection 1.4.3 Inverse Trigonometric Functions
🔗
The six basic trigonometric functions are periodic, and therefore they are not one-to-one. However, if we restrict the domain of a trigonometric function to an interval where it is one-to-one, we can define its inverse. Consider the sine function. The sine function is one-to-one on an infinite number of intervals, but the standard convention is to restrict the domain to the interval By doing so, we define the inverse sine function on the domain such that for any in the interval the inverse sine function tells us which angle in the interval satisfies Similarly, we can restrict the domains of the other trigonometric functions to define inverse trigonometric functions, which are functions that tell us which angle in a certain interval has a specified trigonometric value.
🔗
Definition 1.112.
The inverse sine function, denoted or arcsin, and the inverse cosine function, denoted or arccos, are defined on the domain as follows:
if and only if and if and only if cos and
The inverse tangent function, denoted or arctan, and inverse cotangent function, denoted or arccot, are defined on the domain as follows:
if and only if tan and if and only if cot and
The inverse cosecant function, denoted or arccsc, and inverse secant function, denoted or arcsec, are defined on the domain as follows:
if and only if csc and if and only if sec and
🔗
To graph the inverse trigonometric functions, we use the graphs of the trigonometric functions restricted to the domains defined earlier and reflect the graphs about the line (Figure 1.113).
🔗
Note 1.114.
Go to the following site for more comparisons of functions and their inverses.
🔗
When evaluating an inverse trigonometric function, the output is an angle. For example, to evaluate we need to find an angle such that cos Clearly, many angles have this property. However, given the definition of we need the angle that not only solves this equation, but also lies in the interval We conclude that
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We now consider a composition of a trigonometric function and its inverse. For example, consider the two expressions and For the first one, we simplify as follows:
🔗
For the second one, we have
🔗
The inverse function is supposed to “undo” the original function, so why isn’t Recalling our definition of inverse functions, a function and its inverse satisfy the conditions for all in the domain of and for all in the domain of so what happened here? The issue is that the inverse sine function, is the inverse of the restricted sine function defined on the domain Therefore, for in the interval , it is true that However, for values of outside this interval, the equation does not hold, even though is defined for all real numbers
🔗
What about Does that have a similar issue? The answer is no. Since the domain of is the interval we conclude that if and the expression is not defined for other values of To summarize,
if
🔗
and
if
🔗
Similarly, for the cosine function,
cos if
🔗
and
cos if
🔗
Similar properties hold for the other trigonometric functions and their inverses.
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Example 1.115. Evaluating Expressions Involving Inverse Trigonometric Functions.
Evaluate each of the following expressions.
tan
cos
cos
Solution.
Evaluating is equivalent to finding the angle such that and The angle satisfies these two conditions. Therefore,
First we use the fact that Then tan Therefore, tan
To evaluate cos first use the fact that cos Then we need to find the angle such that cos and Since satisfies both these conditions, we have cos cos
Since cos we need to evaluate That is, we need to find the angle such that and Since satisfies both these conditions, we can conclude that cos
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Note 1.116. Project: The Maximum Value of a Function.
In many areas of science, engineering, and mathematics, it is useful to know the maximum value a function can obtain, even if we don’t know its exact value at a given instant. For instance, if we have a function describing the strength of a roof beam, we would want to know the maximum weight the beam can support without breaking. If we have a function that describes the speed of a train, we would want to know its maximum speed before it jumps off the rails. Safe design often depends on knowing maximum values.
This project describes a simple example of a function with a maximum value that depends on two equation coefficients. We will see that maximum values can depend on several factors other than the independent variable .
Consider the graph in Figure 1.117 of the function cos Describe its overall shape. Is it periodic? How do you know? Using a graphing calculator or other graphing device, estimate the - and -values of the maximum point for the graph (the first such point where \gt 0). It may be helpful to express the -value as a multiple of \pi.
Now consider other graphs of the form cos for various values of and . Sketch the graph when = 2 and = 1, and find the - and -values for the maximum point. (Remember to express the -value as a multiple of \pi, if possible.) Has it moved?
Repeat for = 1, = 2. Is there any relationship to what you found in part (2)?
Complete the following table, adding a few choices of your own for and :
Table 1.118.
| | | | | | | | | |
--- --- --- ---
| 0 | 1 | | | | | 1 | | |
| 1 | 0 | | | | 1 | | | |
| 1 | 1 | | | | 12 | 5 | | |
| 1 | 2 | | | | 5 | 12 | | |
| 2 | 1 | | | | | | | |
| 2 | 2 | | | | | | | |
| 3 | 4 | | | | | | | |
| 4 | 3 | | | | | | | |
5. Try to figure out the formula for the -values.
6. The formula for the -values is a little harder. The most helpful points from the table are (Hint: Consider inverse trigonometric functions.)
7. If you found formulas for parts (5) and (6), show that they work together. That is, substitute the -value formula you found into cos and simplify it to arrive at the -value formula you found.
🔗
Subsection 1.4.4 Key Concepts
For a function to have an inverse, the function must be one-to-one. Given the graph of a function, we can determine whether the function is one-to-one by using the horizontal line test.
If a function is not one-to-one, we can restrict the domain to a smaller domain where the function is one-to-one and then define the inverse of the function on the smaller domain.
For a function and its inverse for all in the domain of and for all in the domain of
Since the trigonometric functions are periodic, we need to restrict their domains to define the inverse trigonometric functions.
The graph of a function and its inverse are symmetric about the line
🔗
Subsection 1.4.5 Key Equations
Inverse functions for all in and for all in
🔗
🔗
This book is a custom edition based on OpenStax Calculus Volume 1. You can download the original for free at
🔗
Additional practice exercises are available in at the bottom on this section in OpenStax Calculus Volume 1: |
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3724 | https://statisticsbyjim.com/basics/outliers/ | 5 Ways to Find Outliers in Your Data - Statistics By Jim
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5 Ways to Find Outliers in Your Data
By Jim Frost36 Comments
Outliers are data points that are far from other data points. In other words, they’re unusual values in a dataset. Outliers are problematic for many statistical analyses because they can cause tests to either miss significant findings or distort real results.
Unfortunately, there are no strict statistical rules for definitively identifying outliers. Finding outliers depends on subject-area knowledge and an understanding of the data collection process. While there is no solid mathematical definition, there are guidelines and statistical tests you can use to find outlier candidates.
In this post, I’ll explain what outliers are and why they are problematic, and present various methods for finding them. Additionally, I close this post by comparing the different techniques for identifying outliers and share my preferred approach.
Outliers and Their Impact
Outliers are a simple concept—they are values that are notably different from other data points, and they can cause problems in statistical procedures.
To demonstrate how much a single outlier can affect the results, let’s examine the properties of an example dataset. It contains 15 height measurements of human males. One of those values is an outlier. The table below shows the mean height and standard deviation with and without the outlier.
Throughout this post, I’ll be using this example CSV dataset: Outliers.
With OutlierWithout OutlierDifference
2.4m (7’ 10.5”)1.8m (5’ 10.8”)0.6m (~2 feet)
2.3m (7’ 6”)0.14m (5.5 inches)2.16m (~7 feet)
From the table, it’s easy to see how a single outlier can distort reality. A single value changes the mean height by 0.6m (2 feet) and the standard deviation by a whopping 2.16m (7 feet)! Hypothesis tests that use the mean with the outlier are off the mark. And, the much larger standard deviation will severely reduce statistical power!
Before performing statistical analyses, you should identify potential outliers. That’s the subject of this post. In the next post, we’ll move on to figuring out what to do with them.
There are a variety of ways to find outliers. All these methods employ different approaches for finding values that are unusual compared to the rest of the dataset. I’ll start with visual assessments and then move onto more analytical assessments.
Let’s find that outlier! I’ve got five methods for you to try.
Use my free online Outlier Calculator to find outliers in your dataset!
Sorting Your Datasheet to Find Outliers
Sorting your datasheet is a simple but effective way to highlight unusual values. Simply sort your data sheet for each variable and then look for unusually high or low values.
For example, I’ve sorted the example dataset in ascending order, as shown below. The highest value is clearly different than the others. While this approach doesn’t quantify the outlier’s degree of unusualness, I like it because, at a glance, you’ll find the unusually high or low values.
Graphing Your Data to Identify Outliers
Boxplots, histograms, and scatterplots can highlight outliers.
Boxplots display asterisks or other symbols on the graph to indicate explicitly when datasets contain outliers. These graphs use the interquartile method with fences to find outliers, which I explain later. The boxplot below displays our example dataset. It’s clear that the outlier is quite different than the typical data value.
You can also use boxplots to find outliers when you have groups in your data. The boxplot below shows a different dataset that has an outlier in the Method 2 group. Click here to learn more about boxplots.
Histograms also emphasize the existence of outliers. Look for isolated bars, as shown below. Our outlier is the bar far to the right. The graph crams the legitimate data points on the far left.
Click here to learn more about histograms.
Most of the outliers I discuss in this post are univariate outliers. We look at a data distribution for a single variable and find values that fall outside the distribution. However, you can use a scatterplot to detect outliers in a multivariate setting.
In the graph below, we’re looking at two variables, Input and Output. The scatterplot with regression line shows how most of the points follow the fitted line for the model. However, the circled point does not fit the model well.
Interestingly, the Input value (~14) for this observation isn’t unusual at all because the other Input values range from 10 through 20 on the X-axis. Also, notice how the Output value (~50) is similarly within the range of values on the Y-axis (10 – 60). Neither the Input nor the Output values themselves are unusual in this dataset. Instead, it’s an outlier because it doesn’t fit the model.
This type of outlier can be a problem in regression analysis. Given the multifaceted nature of multivariate regression, there are numerous types of outliers in that realm. In my ebook about regression analysis, I detail various methods and tests for identifying outliers in a multivariate context.
For another advanced multivariate method for detecting outliers, consider using principal component analysis (PCA). This approach is particularly helpful when you have many variables in a high-dimensional dataset. Learn more in, Principal Component Analysis Guide and Example.
For the rest of this post, we’ll focus on univariate outliers.
Using Z-scores to Detect Outliers
Z-scores can quantify the unusualness of an observation when your data follow the normal distribution. Z-scores are the number of standard deviations above and below the mean that each value falls. For example, a Z-score of 2 indicates that an observation is two standard deviations above the average while a Z-score of -2 signifies it is two standard deviations below the mean. A Z-score of zero represents a value that equals the mean.
To calculate the Z-score for an observation, take the raw measurement, subtract the mean, and divide by the standard deviation. Mathematically, the formula for that process is the following:
The further away an observation’s Z-score is from zero, the more unusual it is. A standard cut-off value for finding outliers are Z-scores of +/-3 or further from zero. The probability distribution below displays the distribution of Z-scores in a standard normal distribution. Z-scores beyond +/- 3 are so extreme you can barely see the shading under the curve.
In a population that follows the normal distribution, Z-score values more extreme than +/- 3 have a probability of 0.0027 (2 0.00135), which is about 1 in 370 observations. However, if your data don’t follow the normal distribution, this approach might not be accurate.
Z-scores and Our Example Dataset
In our example dataset below, I display the values in the example dataset along with the Z-scores. This approach identifies the same observation as being an outlier.
Note that Z-scores can be misleading with small datasets because the maximum Z-score is limited to (n−1) / √n.
Indeed, our Z-score of ~3.6 is right near the maximum value for a sample size of 15. Sample sizes of 10 or fewer observations cannot have Z-scores that exceed a cutoff value of +/-3.
Also, note that the outlier’s presence throws off the Z-scores because it inflates the mean and standard deviation as we saw earlier. Notice how all the Z-scores are negative except the outlier’s value. If we calculated Z-scores without the outlier, they’d be different! Be aware that if your dataset contains outliers, Z-values are biased such that they appear to be less extreme (i.e., closer to zero).
For more information about z-scores, read my post, Z-score: Definition, Formula, and Uses.
The z-score cutoff value is based on the empirical rule. For more information, read my post, Empirical Rule: Definition, Formula, and Uses.
Related posts: Normal Distribution and Understanding Probability Distributions
Using the Interquartile Range to Create Outlier Fences
You can use the interquartile range (IQR), several quartile values, and an adjustment factor to calculate boundaries for what constitutes minor and major outliers. Minor and major denote the unusualness of the outlier relative to the overall distribution of values. Major outliers are more extreme. Analysts also refer to these categorizations as mild and extreme outliers.
The IQR is the middle 50% of the dataset. It’s the range of values between the third quartile and the first quartile (Q3 – Q1). We can take the IQR, Q1, and Q3 values to calculate the following outlier fences for our dataset: lower outer, lower inner, upper inner, and upper outer. These fences determine whether data points are outliers and whether they are mild or extreme.
Values that fall inside the two inner fences are not outliers. Let’s see how this method works using our example dataset.
Click here to learn more about interquartile ranges and percentiles.
Calculating the Outlier Fences Using the Interquartile Range
Using statistical software, I can determine the interquartile range along with the Q1 and Q3 values for our example dataset. We’ll need these values to calculate the “fences” for identifying minor and major outliers. The output below indicates that our Q1 value is 1.714 and the Q3 value is 1.936. Our IQR is 1.936 – 1.714 = 0.222.
To calculate the outlier fences, do the following:
Take your IQR and multiply it by 1.5 and 3. We’ll use these values to obtain the inner and outer fences. For our example, the IQR equals 0.222. Consequently, 0.222 1.5 = 0.333 and 0.222 3 = 0.666. We’ll use 0.333 and 0.666 in the following steps.
Calculate the inner and outer lower fences. Take the Q1 value and subtract the two values from step 1. The two results are the lower inner and outer outlier fences. For our example, Q1 is 1.714. So, the lower inner fence = 1.714 – 0.333 = 1.381 and the lower outer fence = 1.714 – 0.666 = 1.048.
Calculate the inner and outer upper fences. Take the Q3 value and add the two values from step 1. The two results are the upper inner and upper outlier fences. For our example, Q3 is 1.936. So, the upper inner fence = 1.936 + 0.333 = 2.269 and the upper outer fence = 1.936 + 0.666 = 2.602.
Using the Outlier Fences with Our Example Dataset
For our example dataset, the values for these fences are 1.048, 1.381, 2.269, and 2.602. Almost all of our data should fall between the inner fences, which are 1.381 and 2.269. At this point, we look at our data values and determine whether any qualify as being major or minor outliers. 14 out of the 15 data points fall inside the inner fences—they are not outliers. The 15 th data point falls outside the upper outer fence—it’s a major or extreme outlier.
The IQR method is helpful because it uses percentiles, which do not depend on a specific distribution. Additionally, percentiles are relatively robust to the presence of outliers compared to the other quantitative methods.
Boxplots use the IQR method to determine the inner fences. Typically, I’ll use boxplots rather than calculating the fences myself when I want to use this approach. Of the quantitative approaches in this post, this is my preferred method. The interquartile range is robust to outliers, which is clearly a crucial property when you’re looking for outliers!
Related post: What are Robust Statistics?
Finding Outliers with Hypothesis Tests
You can use hypothesis tests to find outliers. Many outlier tests exist, but I’ll focus on one to illustrate how they work. In this post, I demonstrate Grubbs’ test, which tests the following hypotheses:
Null: All values in the sample were drawn from a single population that follows the same normal distribution.
Alternative: One value in the sample was not drawn from the same normally distributed population as the other values.
If the p-value for this test is less than your significance level, you can reject the null and conclude that one of the values is an outlier. The analysis identifies the value in question.
Let’s perform this hypothesis test using our sample dataset. Grubbs’ test assumes your data are drawn from a normally distributed population, and it can detect only one outlier. If you suspect you have additional outliers, use a different test.
Grubbs’ outlier test produced a p-value of 0.000. Because it is less than our significance level, we can conclude that our dataset contains an outlier. The output indicates it is the high value we found before.
If you use Grubbs’ test and find an outlier, don’t remove that outlier and perform the analysis again. That process can cause you to remove values that are not outliers.
Challenges of Using Outlier Hypothesis Tests: Masking and Swamping
When performing an outlier test, you either need to choose a procedure based on the number of outliers or specify the number of outliers for a test. Grubbs’ test checks for only one outlier. However, other procedures, such as the Tietjen-Moore Test, require you to specify the number of outliers. That’s hard to do correctly! After all, you’re performing the test to find outliers! Masking and swamping are two problems that can occur when you specify the incorrect number of outliers in a dataset.
Masking occurs when you specify too few outliers. The additional outliers that exist can affect the test so that it detects no outliers. For example, if you specify one outlier when there are two, the test can miss both outliers.
Conversely, swamping occurs when you specify too many outliers. In this case, the test identifies too many data points as being outliers. For example, if you specify two outliers when there is only one, the test might determine that there are two outliers.
Because of these problems, I’m not a big fan of outlier tests. More on this in the next section!
My Philosophy about Finding Outliers
As you saw, there are many ways to identify outliers. My philosophy is that you must use your in-depth knowledge about all the variables when analyzing data. Part of this knowledge is knowing what values are typical, unusual, and impossible.
I find that when you have this in-depth knowledge, it’s best to use the more straightforward, visual methods. At a glance, data points that are potential outliers will pop out under your knowledgeable gaze. Consequently, I’ll often use boxplots, histograms, and good old-fashioned data sorting! These simple tools provide enough information for me to find unusual data points for further investigation.
Typically, I don’t use Z-scores and hypothesis tests to find outliers because of their various complications. Using outlier tests can be challenging because they usually assume your data follow the normal distribution, and then there’s masking and swamping. Additionally, the existence of outliers makes Z-scores less extreme. It’s ironic, but these methods for identifying outliers are actually sensitive to the presence of outliers! Fortunately, as long as researchers use a simple method to display unusual values, a knowledgeable analyst is likely to know which values need further investigation.
In my view, the more formal statistical tests and calculations are overkill because they can’t definitively identify outliers. Ultimately, analysts must investigate unusual values and use their expertise to determine whether they are legitimate data points. Statistical procedures don’t know the subject matter or the data collection process and can’t make the final determination. You should not include or exclude an observation based entirely on the results of a hypothesis test or statistical measure.
At this stage of the analysis, we’re only identifying potential outliers for further investigation. It’s just the first step in handling them. If we err, we want to err on the side of investigating too many values rather than too few.
In my next post, I’ll explain what you’re looking for when investigating outliers and how that helps you determine whether to remove them from your dataset. Not all outliers are bad and some should not be deleted. In fact, outliers can be very informative about the subject-area and data collection process. It’s important to understand how outliers occur and whether they might happen again as a normal part of the process or study area.
Read my Guidelines for Removing and Handling Outliers.
If you’re learning about hypothesis testing and like the approach I use in my blog, check out my eBook!
Reference
Ronald E. Shiffler (1988) Maximum Z Scores and Outliers, The American Statistician, 42:1, 79-80, DOI: 10.1080/00031305.1988.10475530
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Filed Under: BasicsTagged With: analysis example, conceptual, graphs
Reader Interactions
Comments
Sheikh says
February 22, 2024 at 9:31 am
Mr. Frost, I salute you for taking the time to diligently answer queries from readers.
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2. Bart says
October 10, 2022 at 9:34 pm
Thanks! That’s very helpful. And, actually, it helps to confirm my intuition about the need for control charts, which I was able to get somewhat. It’s not perfect because any one school can different in meaningful ways from others, but it’s the best approximation of what’s “normal.”
Thanks again for your feedback. Your willingness to provide such a thorough response goes above and beyond what I had anticipated. It was exceptional.
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Jim Frost says
October 10, 2022 at 9:40 pm
You’re very welcome!
I think pretty much any approach you use, whether its control charts or something else, you’ll need to use for individual schools. Unless you have reason to suspect all schools should follow the same distribution.
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Bart says
October 4, 2022 at 10:04 pm
Jim, I really appreciate the thorough (and timely) response. If you’ll allow me to impose on you just a bit more, I’ll reframe it.
Imagine there’s a class held annually that has 25-30 students. The grade on the mid-term does an excellent job of predicting the grade on the final, say, the mean difference from mid-term to final is -2 and the st.dev. is 4. Over a ten-year period, that mean & st.dev. have been consistent; however for the first 3 years, there were zero “outliers,” say, final exam scores that were 3 or more st.dev. from the mean; however, over the last 7 years, there was always at least one student and as many as three who received grades on the final that were 5 st.dev. greater than their mid-term grades.
What I’m wondering is how to determine if 7 consecutive years with 1-3 outliers is reasonable to expect or if the 3 years with zero outliers are actually the “odd” years.
My gut tells me that the reoccurrence of extreme outliers suggests there’s some sort of explanation, something “systematic” that can explain the consistency of those outliers; however, maybe it’s simply random variation. Also, there is no control because it wouldn’t be appropriate to compare the results from this class with a different one in a different subject or with a different instructor, and it’s not possible to see if the outliers made similar jumps in other courses (e.g., perhaps they become highly motivated after mid-terms) because they’ve only taken that one course.
Thanks again for your input!
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Jim Frost says
October 10, 2022 at 8:44 pm
Hi Bart,
Even though you reframed the scenarios, it’s the same basic idea. You’re trying to determine whether the outcomes over time reflect one underlying distribution, or has that distribution changed. That’s what control charts are designed to detect. So, I’d still recommend using a control chart in that scenario. In fact, I’ve written a post a while ago urging people outside manufacturing to use these charts more often. To learn more about them, read my post about Using Control Charts. If I remember correctly, I actually use changing test scores in an education setting as an example scenario.
Outside of control charts, you’d need to know the distribution of test scores from the years you’d consider normal. Then you could determine the likelihood of producing the outliers in later years if the distribution stayed constant. For example, if the scores follow a normal distribution, then you’d expect that 0.3% or 1 out of every 370 observations will fall more than +/- 3 standard deviations from the mean. You could see if the outliers in later years exceed those expected unusual values. By the way, control charts basically have those tests built into them, which is why I recommend them. They have a variety of tests including whether those points all fall on one side or both sides of the mean. And more. But that’s the type of think you need to look for an evaluate.
As for your question about which years are the normal or odd years, you won’t be able to answer that using statistics. Statistics can help you determine whether those two sets of years are different or not, but not which one is normal. That takes subject area knowledge.
It does seem like something has changed just reading your description. Although, it’s possible that the first three years were just flukes that happened to not have outliers. The difference between a year without outliers and a year with just one outlier might not be so large.
However, five standard deviations from the mean is extremely large and a single score with that value is almost definitely a red flag all by itself. You’d expect less than 1/100,000 scores to be that many standard deviations away from the mean. That’s got to be a red flag. It’s really off the charts. And if you have more than one that is 5 standard deviations from the mean, there’s really no question.
I don’t know how many students scores are being considered here. But there’s a massive difference between the 3 standard deviations (1 in 370) vs 5 standard deviations (
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Bart says
October 4, 2022 at 7:03 pm
If you have multiple samples, how can you tell if outliers are “expected.” What I mean is this: imagine a machine that churns out cookies. Each batch has 40 cookies with a mean of, say, 20g and all cookies weigh between 18-22g except for 1-3 that weigh 38g. How many batches would you need to produce where this outcome occurred before it’s “normal,” i.e., 1-3 extreme outliers can be expected every batch. Also, what if the machine made 3 batches with same mean and distribution of weight across 40 cookies but no outliers, and then the next X batches had 1-3 extreme outliers. What would the value of X need to be to suggest there’s likely a problem with the machine? Thanks so much!
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Jim Frost says
October 4, 2022 at 9:10 pm
Hi Bart,
There are different ways to answer this question.
In a real-world manufacturing setting, client requirements will set the spec limits for your cookies. Out of spec cookies are a problem. Typically, manufacturers will have some process related knowledge about the expected machine performance for being able to produce cookies that are in spec. In that way, they can compare actual performance to expected performance and determine whether the machine has a problem. Additionally, they can create control charts to determine whether the cookie machine is a stable process and perform capability analysis to get some standard capability values that they can compare to benchmark values.
That’s the approach from a real manufacturing perspective.
You can also look at it from a more distributional perspective. And you need to really define “outlier.” Some unusual values might be the regular part of the process and not real outliers. They’re expected. Like human heights follow a normal distribution and you’ll naturally see people who are unusually tall or short. In that case, it’s not indicative of a problem, just part the natural process. The same idea applies to your cookie machine. You’ll need to determine whether the unusual cookies are a natural part of the process’ distribution or indicative of a fault in the machine.
To make that type of determination, you’ll need to know the usual distribution of cookie weights for the machine from historical data. Then perform a new study and determine whether the machine is producing more unusual cookies than the usual distribution of cookie weights predicts.
Because I don’t know the previous capability of the machine, I can answer your question specifically, but that’s the general process.
In a real production process, the manufacture would use control charts to track their output and that would be the first way they know that something is out of whack. In general, control charts will do what you’re asking about. They’ll tell you if the proportion of defects (i.e., unusual values or outliers) is changing from sample to sample. The control limits on the chart indicate when a process is statistically out of control.
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peach says
February 11, 2022 at 7:13 pm
wow this helped me better understand an outlier thx
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6. Vyas says
February 1, 2022 at 9:18 pm
Hi Jim,
If we are measuring overall user experience using a CSAT with the Likert scale ranging from: (1) Very Unsatisfied (2) Unsatisfied (3) Neither Satisfied Nor Unsatisfied (4) Satisfied (5) Very Satisfied. And the dataset has only a few who scored (1) and (5). Can we say (1) and (2) are outliers in this self-evaluation? Or do we judge outliers from objective measurements (as per your example, measuring heights of a sample)? Kindly clarify. Thank you
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7. MikeA.says
November 12, 2021 at 1:42 pm
Great stuff. Thanks, Jim! A next-level question re: IQR: once we identify the outliers, and then examine why they were there in the first place, would it be advisable to remove them and run the numbers again? I’m thinking the reason for doing this would be to eliminate outliers that we know would never happen again– particularly if we adjust our processes to make SURE they don’t happen again! Also, I’m thinking that doing so would allow us to get to next-level refinements so that we can narrow our ranges going forward. What are your thoughts?
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Jim Frost says
November 13, 2021 at 11:44 pm
Hi Mike,
After you identify the potential outliers, if you decide that they really don’t belong in the dataset, which takes a lot of deliberation, then I’d certainly remove them and rerun your statistics. The trick is to really determine whether they belong in the dataset (they might be legitimate extreme values) or not. The link in this post to my other post about outliers discusses some of the ways to make that determination.
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Zohaib says
July 9, 2021 at 5:17 am
The DGP of multiple linear regression model is given
Y_i=0.3+2X_1i+1.5X_2i+ε_i
Where ε_i ~Norm(0,10)
Understand it DGP carefully and generate 500 observations of each variable in excel.
And prove that:
In case of normally distributed data, the value of SE (of estimators) are efficient, and t-statistics is valid, and parameters are not biased.
Generate outlier with value of 3000 (in Y_i) and show that how a one outlier violate all the distribution of the data in which the SE of parameters are not more efficient, t-statistics is not valid, and parameters become biased.
Note: Answers of above question should be given on official university answer sheet (which I have given you) but result of excel file should be paste in this world document, only word document for your excel results will be accepted.
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9. Jonas says
March 23, 2021 at 10:01 am
Hi Jim,
Great article!
Can you elaborate on why it is incorrect to use Grubs test several times to remove more than one outlier in a dataset? In a sufficiently large data set, why can’t you just run it several times?
I would think that one strategy to automate the procedure could be to take another few extra samples, then run Grubs test a few times.
(assuming we can afford to take maybe 30 samples or more )
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Jim Frost says
March 23, 2021 at 3:49 pm
Hi Jonas,
What happens if you repeat Grubs test is that it’ll tend to remove data points that are not outliers. The first outlier it finds is based on the entire distribution. Then, you remove an outlier and the distribution of the remaining data now has less variability. A point that was not an outlier might now appear to be an outlier because of the reduced variability. Statisticians base this recommendation to only use Grubbs test once per dataset due to its propensity for removing valid data points when you use the test multiple times.
I hope that helps!
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Joseph Lombardi says
February 23, 2021 at 4:20 pm
Hey, Jim.
I have a question re the Fitted Line Plot — Output as a function of Input. Visually, the one data point does not “fit the model.” If we look at the residuals, we should get a mean of zero. Can we not use the Standard Deviation of the residuals to calculate the Z-scores for THEM and then determine whether a datum or two can be omitted? Or is that subject to the same issues you mention above regarding the actual observed data?
If using the Z-scores of residuals is not a great idea, can we use percentiles instead? Can we just flag data with a residual of less then, say, the 1st percentile and greater than the 99th percentile? Again, I am referring to the residuals, here.
Cheers,
Joe
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Jim Frost says
February 25, 2021 at 4:23 pm
Hi Joe,
My overall point is to use all the methods carefully. Most of them have some drawback but I’m not saying to avoid them. In the end, it really comes down to your subject area knowledge and the investigation of candidate outliers. It’s always possible that an unusual value is part of the natural variation of the process rather than a problematic point!
The residuals you describe are known as standardize residuals and, yes, they do have a value equivalent to a Z-score. They are a good way to identify potential outliers. With normal residuals, you might see a value of X, but whether X is unusual depends on the data units, data variability, etc. Standardized residuals are good way to determine whether a residual is unusual given the properties of the data because it incorporates those factors. However, the same caveats for Z-values apply for standardized residuals. Just something to be aware of while assessing them. Additionally, remember that it’s normal for about 5% of the residuals to have standardized scores of +/-2. Again, use subject area knowledge and investigate particular data points.
Another useful type of residual are the deleted Studentized residuals. These are like the standardized residuals above but the calculations for the ith deleted studentized residual does not include the ith observation. That helps avoid the problem where the presence of an unusual residual actually causes its own standardized value to be lower because it’s inflating the residuals’ variability.
You could convert to percentiles as you describe. Most software I see use either +/- 2 or 3 standard deviations to identify candidate outliers.
In regression, there are multiple ways that an observation can be unusual. Residuals measure unusualness in the y-direction. But you can also have unusual observations in the X-direction. And points that individually affect the model’s coefficient estimates by a larger than usual degree. I can’t remember if you have my regression book, but I discuss those issues in it!
Take care!
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Hosam Salmansays
November 28, 2020 at 2:43 am
Dear Jim,
I hope all is great and well.
i have a master data sheet include few variables. In order to perform my regression, I need to make sure I get ride of the outliers. I understand that there are many ways to get the outliers out. I plan to use Box Plot method or Z Score method. The distributions of the data sometimes normally distributed, left skewed and right skewed. All over, non is consistent. The master data sheet will be resorted based on specific variables values. so I will create from the master data sheet few specific data sheets.
But the questions that need help are listed below;
How we deal with outliers when the master data sheet include various distributions. It is not consistent; some of them normally and the majority are skewed.
Should we apply one method to remove the outliers or we can apply more than one method, like these two methods. And when to be applied? Should this applied to the master data sheet or we still need to apply it after sorting the data as indicated above.
If we use the box plot to fix one column of variable, it will impact the other variables since it eliminate one complete row. That row may have other good test for other values.
Any advise or suggestions in general to deal with the outliers and at same time not impacting significantly the obtained data.
Thank you so much!
Hosam
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12. Bon says
August 3, 2020 at 1:23 pm
Hi Jim,
Is there a correct way to run a outlier analysis? What if there are 3 variate ( 12 variables) , is there any rulling about this? Thank you
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13. d says
August 1, 2020 at 10:16 pm
It best f you use several methods to find outliers. The true outliers will satisfy multiple methods.
You can compare the findings of the different methods and have confidence those data points can be treated as outliers when flagged by different methods independently.
It also helps to have a clear understand of what your dataset describes in reality, and what those outliers really represent, how they came to be in existence. If it’s likely they are errors, great, that’s more justification to ignore them.
If they are certainly correctly sampled you must consider what it means to remove them from your study, how their removal affects the integrity of your analysis.
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Jim Frost says
August 2, 2020 at 12:24 am
Hi D,
Thanks for writing! In terms of flagging observations for investigation, I’d agree that if multiple methods find the same values, there’s good reason to investigate them. However, flagging by multiple methods doesn’t necessarily increase the likelihood that removing those values is appropriate.
I definitely agree that understanding what your dataset describes and how the outliers came to be are both crucial tasks.
However, removal really depends on understanding how those values came to exist. That can get fairly complicated. I discuss those issues in my post about determining whether to remove outliers. Identifying the candidate is the easy part. You’re just looking for unusual values. Why they exist and what to do about them is where it can get complicated!
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KECHLER POLYCARPE says
May 26, 2020 at 9:47 pm
Sorry didn’t want to blind you brother. Thank you I’m studying it now. I sent you a message I had a question.
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Jim Frost says
May 26, 2020 at 10:11 pm
No worries! 🙂
I will reply to your other question soon.
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KECHLER POLYCARPE says
May 26, 2020 at 8:56 pm
HELLO JIM, IF YOU DON’T HAVE DATA AT START CAN YOU STILL CRAFT THE RESEARCH QUESTION THEN DESIGN A STUDY TO COLLECT DATA? IF SO HOW?
-Thank you
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Jim Frost says
May 26, 2020 at 9:30 pm
Hi Kechler, first, please don’t use ALL CAPS! It hurts my eyes!
Yes, you can craft your research question and study design before collecting data. In fact, that’s the preferred approach. If you start collecting data before you have a research question and design, it’s very likely that you won’t be collecting the necessary data to answer your question. I write about this in my post about conducting scientific studies with statistical analyses.
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HTITI Sarah says
April 15, 2020 at 6:49 am
Is it legitimate to detect outliers based on the Z-score for a large population ( 800 K observations) even if it’s not normal?
Or it’s more appropriate to use the IQR and then compute an Upper hinge and lower hinge ? or are there other methods to apply in this case ?
Thanks in advance
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17. SANJAYA KUMAR SUBBA says
April 13, 2020 at 12:59 am
Hello Sir Good morning,
i’m a research scholar and i’m comparing mean of two independent sample data sets ,(i.e stock price returns of to pricing method )now in order to test parametric test my data should be normally distributed,but when i test normality ,i found my data is not normal..as my guide is suggesting me to normalize my data using Z score and finding their areas under curve,but i;m not able to understand that,please help me sir to normalize data.
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18. Suruchi Sarvate says
January 28, 2020 at 1:03 am
Hi Jim,
I have a dataset with 11 columns and I have written a common function detect_outliers() to find outliers in the columns.
For first 6 columns, the function is working out but for rest of the 5 outliers , function returns empty list though the columns have outliers. U can see the code written below:
def detect_outliers(data):
outliers = []
threshold=3
mean = np.mean(data)
std = np.std(data)
for i in data:
z_score = (i-mean)/std
print(z_score)
if np.abs(z_score) > threshold:
outliers.append(i)
return outliers
As you can see, if I have taken the value of “threshold = 3”. For first six columns, the function is working out as z_score>3 for outliers.
But for rest of the columns, z_score for outliers is greater than 1 (z_score>1), so the threshold should be taken 1 for rest of the six coulmns.
Here I have 11 columns only in the dataset. But what if I have 1000 columns in my dataset. In that case,I can’t check the threshold for each and every column.
Please!!!! help me and reply at the earliest.
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Jim Frost says
January 28, 2020 at 11:23 pm
Hi Suruchi,
Why would you use a Z-score of 1 to detect outliers? I’m not sure why it’s not working but with such a low threshold you should have more detections. How many observations per column?
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Rutvij says
November 27, 2019 at 7:47 am
Hi Jim, Thanks for sharing details on outliers. I have one question, happy if you can advise me.
My questions is, I am building a MachineLearning model, I have traning dataset and testing dataset. I removed outliers from traning dataset and building ML model with good efficient level. Now, I did have large amount of outliers in testing dataset (which I have to submit as it is).
Now, in that my ML model is less efficient when I applied on unseen test dataset (with outliers).
Can you please advice me, how shall I achive more efficiency on test dataset. Plus, I don’t want to loose any observed values in the test dataset.
Thanks.
Rutvij
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20. Narasimha Patro says
October 29, 2019 at 2:01 am
How to treat outliers ??? Please help me
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Jim Frost says
October 29, 2019 at 11:35 am
Hi Narasimha,
Read my follow up post to this one: Guidelines for Removing and Handling Outliers.
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Denny Fernandez del Viso says
October 24, 2019 at 12:36 pm
I usually use a Q-Q plot to detect outliers – just a visualization of what you suggest as using the Z-score.
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Jim Frost says
October 24, 2019 at 3:26 pm
Hi Denny,
Thanks for the suggestion. Unusual Z-scores might stand out more in a plot than a list. Just be aware of the constraints on Z-scores in small samples and the fact that Z-scores themselves are sensitive to outliers.
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Brion Hurley says
October 10, 2019 at 1:37 pm
I haven’t seen this formula before related to Z-scores: (n−1) / √ n
Can you share more details about where this comes from? It’s not intuitive to me at first glance
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Jim Frost says
October 11, 2019 at 3:58 pm
Hi Brion,
I’ve added a reference to this post for this formula. The referenced article discusses this limitation in the context of finding outliers and it includes references to other sources where the limit was derived. In a nutshell, maximizing Z-scores depends on minimizing the standard deviation (or variance). As I showed earlier in this post, the outlier is far from the mean score. While it increases the mean, it drastically increases the standard deviation. The net result of both increases is that it limits the maximum Z-value. In small samples, this limitation is even greater and severely constrains the maximum absolute Z-scores.
In general, an outlier pulls the mean towards it and inflates the standard deviation. Both effects reduce it’s Z-score. Indeed, our outlier’s Z-score of ~3.6 is greater than 3, but just barely. The Z-score seems to indicate that the value is just across the boundary for being outlier. However, it’s truly a severe outlier when you observe how unusual it truly is. Both the boxplot and IQR method make this clear. And, simply observing the value compared to reasonable values, it very far beyond legitimately possible values for human height.
The article uses an example of a dataset with 5 values {0, 0, 0, 0, 1 million}. The Z-score for the value of 1 million is only 1.789! Not an outlier using Z-scores!
To quote the article, “The concept of a Z score as a measure of a value’s position within a data set in terms of standard deviations is intuitively appealing. Unfortunately, the behavior of Z is quite constrained for small data sets.”
To illustrate this constraint, I’m including the table below that lists the maximum absolute Z-scores by sample size. Note how absolute Z-scores can exceed 3 only when the sample size is 11 and greater.
I hope this helps!
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Fernando Augusto Deheza Zambrana says
October 10, 2019 at 8:12 am
Excellent work. Congratulations
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2021 AMC 12A Problems/Problem 14
Contents
1 Problem
2 Solution 1 (Properties of Logarithms)
3 Solution 2 (Properties of Logarithms)
4 Solution 3 (Estimations and Answer Choices)
5 Solution 4 (Properties of Logarithms)
6 Video Solution by Punxsutawney Phil
7 Video Solution by Hawk Math
8 Video Solution by OmegaLearn (Using Logarithmic Manipulations)
9 Video Solution by TheBeautyofMath (Using Magical Ability)
10 Video Solution by The Power of Logic
11 Video Solution (Logic and Simplification)
12 See also
Problem
What is the value of
Solution 1 (Properties of Logarithms)
We will apply the following logarithmic identity: which can be proven by the Change of Base Formula: Now, we simplify the expressions inside the summations: and Using these results, we evaluate the original expression: ~MRENTHUSIASM (Solution)
~JHawk0224 (Proposal)
Solution 2 (Properties of Logarithms)
First, we can get rid of the exponents using properties of logarithms: (Leaving the single in the exponent will come in handy later). Similarly, Then, evaluating the first few terms in each parentheses, we can find the simplified expanded forms of each sum using the additive property of logarithms: In we use the triangular numbers equation: Finally, multiplying the two logarithms together, we can use the chain rule property of logarithms to simplify: Thus, ~Joeya (Solution)
~MRENTHUSIASM (Reformatting)
Solution 3 (Estimations and Answer Choices)
In note that the addends are greater than for all
In note that the addends are greater than for all
We have the inequality which eliminates choices and We get the answer by either an educated guess or a continued approximation:
Observe that and Therefore, we obtain the following rough underestimation: From here, it should be safe to guess that the answer is
~MRENTHUSIASM
Solution 4 (Properties of Logarithms)
Using the identity simplify and . Now we have the product: With the reciprocal rule the logarithms cancel out leaving:
~PowerQualimit
Video Solution by Punxsutawney Phil
Video Solution by Hawk Math
Video Solution by OmegaLearn (Using Logarithmic Manipulations)
Video Solution by TheBeautyofMath (Using Magical Ability)
~IceMatrix
Video Solution by The Power of Logic
Video Solution (Logic and Simplification)
~Education, the Study of Everything
See also
2021 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 13Followed by
Problem 15
1•2•3•4•5•6•7•8•9•10•11•12•13•14•15•16•17•18•19•20•21•22•23•24•25
All AMC 12 Problems and Solutions
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.
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3726 | https://www.fisme.science.uu.nl/toepassingen/28768/documents/mathematics_b_day_2015.pdf | Mathematics B-day 2015 Friday November 13, 9:00h – 16:00 h Around the corner… 1 Exploration 1 – piano You have to move a heavy piano through a 1 meter wide corridor with a right-angled corner in it. The figure above shows the situation as seen from above. Can you manoeuvre the piano (1,54m long and 0,60m wide) around the corner? Examine this using, for example, a calculation, a drawing or an experiment. Exploration 2 – sofa As seen from above, a sofa is a rectangle that is 1,36m long and 0,78m wide. Can the sofa be manoeuvred around the corner? Explanation You have just seen two examples of rectangles, where you have probably already seen that sometimes you can, and sometimes you can't manoeuvre them around the corner, even when the rectangles have the same circumference. Today you will further examine moving objects through the corridor. The question is whether certain objects can be moved through a 1 meter wide corridor with a right-angled corner in it. In other words, in today's assignments you will be looking at a moving problem in two dimensions; you will study which two-dimensional mathematical shapes (line segment, rectangle, circle, ...) can be moved through the corridor. As it is a problem in two dimensions, the height of the objects is not considered. From now on, the width of the corridor is 1 (without a unit). You can assume that objects with width equal to 1 fit precisely through the straight part of the corridor. So, a square with side 1 and a circle with diameter 1 can be moved through the corridor. 2 Mathematics B-day - general information The structure of the day This Mathematics B-day assignment consists of explorations, assignments and your own research. Try to spend approximately half of the day on your own research. What to hand in? You will hand in a report by the end of the day. In this report you describe the results you found in the assignments and your own study. You don't have to include the explorations in your report. Write your report in such a way that it is clear and convincing. Of course, you will use relevant explanatory images as illustrations. Make sure you report is understandable for people who did not take part in the Mathematics B-day. That means that you must take care to introduce the assignments clearly, and that, where needed, you must refer back to what you found in previous assignments. In short: you write a clear and understandable report, supported by mathematical arguments. The quality of your report will definitely play a part in the assessment! It may be useful for your report to already start writing down the assignments and answers you found during the morning. Keep in mind that the whole report has to be handed in by 4 o'clock in the afternoon! Information online You will see that there is some information about this problem can be found online (for example, However, in today's assignment you will look at other shapes than this ‘sofa’. 3 Exploration 3 – half disk a. What is the maximum radius for a semi-circular disk that you can move through the corridor? b. Describe, as precise as you can, how to get the largest possible semi-circular disk through the corridor. For example: first put the semi-circular disk into position .., then move it over distance ..., then turn ... degrees around rotation point..., etc. Exploration 4 – circle A circle with a diameter of 1 can be moved through the corridor. Now, suppose that there is a square obstructing the outer corner (see figure). What can be the maximum size of the square, such that the circle with a diameter of 1 can still be moved through the corridor? Intermezzo Before answering more assignments about moving objects through the corridor, here are two assignments that may be of use to you later today. Assignment 1 – triangular geometry Given is an isosceles triangle ABC, where M is the midpoint of AC. Also, F is a point between B and C on the line BC and the line FM intersects the line AB in point E. Show that line segment EF is longer than line segment AC. Please note: even if you cannot prove it, you can still use the information that line segment EF is longer than line segment AC in later assignments. 4 Assignment 2 – motions You can move objects by sliding them (without rotating the object), or by rotating them around a centre (which we call R from now on). In the figure below an example of sliding and two examples of rotating are shown. For sliding, the following holds: AA' = BB', plus AA' and BB' are parallel. For rotating, the following holds: AR = A'R, BR = B'R and ∠ARA' = ∠BRB'. Please note: to perform a sliding motion in the corridor with the right-angled corner, all line segments (such as AA' and BB') must lie inside the corridor. For a rotation, all arcs (such as AA' en BB') must lie inside the corridor, but point R does not necessarily have to lie inside the corridor. In the figure below you can also go from the light figure to the dark one by means of a rotation. Show how you can determine the centre R of the rotation. 5 Back to moving objects through a corridor... Assignment 3 – Sticks In this assignment you will look into which sticks can be moved around the corner, and which ones cannot. Consider a stick to be a line segment of a certain length. a. Do some experiments (with transparent paper or a drawing) to try and discover which sticks can or cannot be moved around the corner. b. With the aid of assignment 1 you can find out which sticks certainly can't be moved around the corner. Which ones are those? What is the length of the longest stick that can be moved around the corner? c. Prove that you can in fact get the longest stick that you found in question b around the corner, and that it doesn't get stuck. Assignment 4 –strategies for the longest stick There are several ways to move the longest stick through the corridor. a. In assignment 2 you looked at rotations. You can move the longest stick around the corner with a single rotation around a point R. Where is the rotation point R of this rotation, such that the longest stick can be rotated around the corner in this way? Also give a starting position from which you can make the rotation. Hint: Consider the moment that both ends of the stick touch the outside walls and the stick also touches the inner corner of the wall. If this is a snapshot of the rotation, you can find the circle that is described by the ends of the stick. b. Make a sketch of the area covered by the stick in question a. c. You can also perform a motion where the ends of the stick always touch the outside wall of the corridor. Describe the path that the middle of the stick follows when it is moved through the corridor this way as precise as possible. d. Sketch the area covered by the stick in question c. Assignment 5 – largest rectangle In the first two explorations you already looked at rectangles. In this assignment you will look for the rectangle with the largest area that you can move through the corridor. a. What is the area of the largest rectangle, with the shortest side of the rectangle being 0.5, that can be moved through the corridor? b. Find the size of the rectangle with the largest area that you can move through the corridor by both sliding and rotating. The shortest side no longer has to be 0.5. c. Show that the rectangle that you found in question b can actually be moved through the corridor, and does not get stuck in the corner. 6 Assignment 6 – all rectangles You have already looked at the largest rectangle that can be moved through the corridor. But what about all other rectangles? The figure below shows a coordinate system. In it, five rectangles have been drawn such that they have one side on the x-axis and one on the y-axis. The top right corners of the rectangles are black if the rectangle can be moved through the corridor and white if it cannot. a. Where is the boundary between black and white points? b. Which equations describe the boundary between black and white points? Assignment 7 – U-shape You can attach two extra straight parts perpendicular to the ends of the longest stick, so that you get a U-shape. Find the maximum length of a U-shaped stick (length a + 2b in the figure) that can be moved through the corridor. Assignment 8 – L-shape Consider a stick with one right angle (an L-shape). What is the longest stick (length a+b in the figure) that can be moved through the corridor? 7 Your own study In the assignments so far you have already learned much about objects that can or cannot be moved through the corridor. Did you know that it is actually an unsolved problem what the largest object is that you can get through the corridor? In this section you will further study objects that can be moved through the corridor. You can choose what to study. For example, you may choose a shape that has not yet been considered in the assignments so far. Questions you could ask yourself can concern: - maximum length or area of the object, - different strategies to move the object through the corridor, - the area that the object covers while being moved. You can also continue with the research you have done so far. There are a few suggestions below. You can choose one, or you can be inspired by them: - Suppose that the longest stick is moved through the corridor as described in assignment 4a. You can add a lot of surface in the direction of the outer walls (see figure). Maybe even more if the stick isn't maximised! Study this.. - Suppose that the longest stick touches the outside walls, like in assignment 4c. You can add a lot of surface in the direction of the inner walls (see figure). Maybe even more if the stick isn't maximised! Study this. - Suppose that you have a U-shape, as in assignment 7. You can add some extra surface on the inside of the U (see figure). Now, if the long (back) side of the U is moved along the 8 outside walls of the corridor, how much surface can you add on the inside of the U? You could first consider the case where the long side is shorter than twice the width of the corridor. Can you also find - perhaps experimentally - the maximum shape if the long side is longer than twice the width of the corridor? - Find a two-dimensional object that can be moved around the corner with one single rotation (similar to assignment 4a). |
3727 | https://www.ck12.org/flexi/algebra-ii/sine-graph-and-cosine-graph/extend-the-domain-of-trigonometric-functions-using-the-unit-circle/ | Flexi answers - Extend the domain of trigonometric functions using the unit circle. | CK-12 Foundation
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Question
Extend the domain of trigonometric functions using the unit circle.
Flexi Says:
The unit circle is a circle with a radius of 1 centered at the origin (0,0) in the coordinate plane. It's often used to define the trigonometric functions sine, cosine, and tangent for all real numbers. The x-coordinate of a point on the unit circle represents the cosine of the angle formed by the positive x-axis and the line segment connecting the origin with that point. The y-coordinate represents the sine of that angle. For example, if we have an angle of π/4 radians (or 45 degrees), the point on the unit circle associated with this angle is (2/2,2/2). So, cos(π/4)=2/2 and sin(π/4)=2/2. By using the unit circle, we can extend the domain of the trigonometric functions to all real numbers. This is because we can associate every real number with an angle in radians. For example, an angle of 2 π radians corresponds to one full revolution around the circle, and so it corresponds to the same point as an angle of 0 radians. This allows us to define trigonometric functions even for negative angles or angles greater than 2 π radians.
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What is the domain, range and period of cosec(x)?What is the domain and range of trigonometric functions?Is the function f(x)=1+cos(x), for 0<=x<=pi, one-to-one?
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3728 | https://cvc.cervantes.es/aula/didactired/anteriores/diciembre_13/09122013_01.htm | CVC. DidactiRed.
Explicación
La narración en pasado alterna el pretérito indefinidopara contar hechos y hacer avanzar el relato, el pretérito imperfecto,para describir las circunstancias en las que se produjeron los hechos y el pretérito pluscuamperfectopara hablar de acciones pasadas y anteriores a otra acción pasada. Si las acciones relatadas tienen relación con el tiempo presente se utiliza el pretérito perfectoen lugar del pretérito indefinido.
Esa alternancia de tiempos se ve modificada en la lengua hablada. En la lengua hablada, al contar una anécdota, es muy frecuente que el hablante comience el relato en pasado y recurra al presente de indicativo para actualizar el relato y darle más dinamismo y verosimilitud. Este uso del presente se denomina presente narrativo. Este presente se alterna con otros tiempos del pasado de modo que se destaca entre ellos y proporciona mayor agilidad a la narración. Favorecen además el uso de este presente algunas expresiones que introducen una alteración repentina o imprevista en el relato de los acontecimientos. Esas expresiones son: de pronto, de repente, resulta que(registro coloquial), (y) en esto(solo en registro coloquial), va y dice, va y se pone(solo en registro coloquial), por poco y casi.
Íbamos dando un paseo y al cruzar un paso de cebra, por poco nos atropella una moto.
Fue una conferencia aburridísima… casi me duermo. Y estábamos en primera fila.
Estábamos a punto de irnos, y en esto, llega Ana y nos dice que han convocado una reunión urgente.
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3729 | https://pubchemlite.lcsb.uni.lu/e/compound/612 | PubChemLite - Lactic acid (C3H6O3)
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C10H14N2
DUOANANYKYXIQY-UHFFFAOYSA-N
atrazine
CID 612
Lactic acid
Structural Information
Molecular FormulaC 3 H 6 O 3SMILES CC(C(=O)O)O InChI InChI=1S/C3H6O3/c1-2(4)3(5)6/h2,4H,1H3,(H,5,6)InChIKeyJVTAAEKCZFNVCJ-UHFFFAOYSA-NCompound name 2-hydroxypropanoic acid Related CIDs
CID 612
CID 13144
CID 13145
CID 13146
CID 22197
CID 27653
CID 61503
CID 61512
CID 62137
CID 62358
CID 91435
CID 92422
CID 107689
CID 118182
CID 153528
CID 157644
CID 159769
CID 161164
CID 165341
CID 167293
CID 175901
CID 176209
CID 197973
CID 198046
CID 198085
CID 201470
CID 205924
CID 205936
CID 3007856
CID 3045418
CID 3048680
CID 3048865
CID 3048866
CID 3609400
CID 5282393
CID 5460161
CID 5460179
CID 6453109
CID 6536825
CID 11768525
CID 13890992
CID 14475308
CID 15774944
CID 15775272
CID 16213313
CID 16217546
CID 19789253
CID 23664767
CID 23666456
CID 23671663
CID 23671664
CID 23673452
CID 23687877
CID 24180812
CID 25022182
CID 25113465
CID 25199573
CID 44135619
CID 44144458
CID 46835179
CID 71310297
CID 71587589
CID 71587863
CID 73555295
CID 87103201
CID 91667946
CID 101591487
CID 131852013
CID 131856106
CID 134159358
CID 159465775
CID 165355906
2D Structure
11 Annotation Hits
79395 References
561806 Patents
90.03169 Da Monoisotopic Mass
-0.7 XlogP (predicted)
Profile
Predicted Collision Cross Section
| Adduct | m/z | Predicted CCS (Ų) |
---
| [M+H]+ | 91.038966 | 114.4 |
| [M+Na]+ | 113.02091 | 121.9 |
| [M-H]- | 89.024414 | 112.6 |
| [M+NH4]+ | 108.06551 | 136.6 |
| [M+K]+ | 128.99485 | 122.3 |
| [M+H-H2O]+ | 73.028950 | 110.9 |
| [M+HCOO]- | 135.02989 | 135.1 |
| [M+CH3COO]- | 149.04554 | 159.5 |
| [M+Na-2H]- | 111.00636 | 119.5 |
| [M]+ | 90.031141 | 113.1 |
| [M]- | 90.032239 | 113.1 |
m/z: mass to charge ratio of the adduct.
Predicted Collision Cross Section (CCS) values (Ų) per adduct calculated using CCSbase.
Literature stripe
Patent stripe
Copyright © Université du Luxembourg 2025. All rights reserved. |
3730 | https://eaglepubs.erau.edu/introductiontoaerospaceflightvehicles/chapter/conservation-of-momentum-momentum-equation/ | Skip to content
22 Momentum Equation
Introduction
The second physical principle used in deriving the governing equations that describe aerodynamic flows (or the flow of a fluid, in general) is the conservation of momentum, i.e., the application of Newton’s second law. This principle is one of the most important for solving aerodynamic motion because it allows solving problems involving the forces produced on or by flows, such as the lift and drag on a wing or the thrust produced by a jet or rocket engine.
From elementary physics, Newton’s second law is usually written as , where is the force on the system, is the mass of the system, and is the resulting linear acceleration of the system. However, the most general form of Newton’s second law is when it is written in terms of the time rate of change of momentum, i.e.,
(1)
where is the total linear momentum in this case; naturally, there are analogous equations for angular momentum. Remember that, in reference to “momentum,” the meaning is that it refers to the momentum of something, whether it is a mass, a fluid element, or something else. Also, note that the plural of momentum is “momenta,” not “momentums.”
Learning Objectives
Understand the principle of conservation of momentum when applied to a fluid.
Be able to set up the most general form of the momentum equation for a fluid, including all sources of forces.
Use the momentum and continuity equations to solve elementary flow problems.
Understand the development of the Euler and Navier-Stokes equations.
Momentum Equation – Integral Approach
The objective is to apply the conservation of linear momentum principle to a flow to find a mathematical expression for the forces produced on a fluid flow in terms of the familiar macroscopic flow field variables, such as density , velocity , and pressure, . Pressure comes into the problem of describing this behavior because any non-uniform pressure in a flow acting over an area can produce a force. Such non-uniform pressures are referred to as pressure gradients, as they cause an imbalance of forces on the flow, accelerating or decelerating it and thus altering its momentum.
Flow Model
Consider the previously used control volume approach, as shown in the figure below. Recall that the control volume is abbreviated as “C.V.” and the control surface as “C.S.” The volume is assumed to be fixed in space, with the fluid moving through it; this is a fluid moving in an Eulerian modeling approach. In general, the forces on the air, as it flows through the C.V., will manifest from two sources, namely:
Body forces: These include gravity, inertial accelerations (maybe even electromagnetic, electrostatic forces), etc., and may act on the fluid elements inside the C.V. The idea of a body force resulting from gravity was introduced when the hydrostatic equation was established, i.e., .
Surface forces: Pressure and viscous shear stress act on the area of the control surface, , leading to forces.
Pressure Forces
The pressure force acting on the elemental area is , where the negative sign indicates that the force is inward and in the opposite direction to the unit normal vector area, which always points outward by convention. Therefore, the total pressure force on the fluid inside the C.V. is the summation of the elemental forces over the entire C.S., i.e., the pressure force is given by
(2)
Body Forces
Now consider the body forces, where is defined as the net body force per unit mass exerted on the fluid inside . The body force on the elemental fluid volume is , and the total body force is the summation over the volume , i.e.,
(3)
An example of a body force is gravity. So in this case, the body force per unit mass is simply the acceleration under gravity, i.e., , so that in terms of the body force vector, then . By convention, the direction of is usually directed vertically upward, so the minus sign indicates that gravity is directed downward.
Viscous Forces
If the flow is viscous, and all fluids are viscous to a lesser or greater degree, then the viscous shear stresses that arise over the C.S. will exert a surface force. These forces depend on the fluid’s viscosity and the velocity gradients in the flow, making it potentially difficult to evaluate such forces. Therefore, the viscous effects can be recognized but not yet evaluated. The total viscous force on the C.S. as it passes through the C.V. will be given by
(4)
It is essential to understand that the actual quantitative evaluation of this integral may be complex and typically requires some effort, especially when turbulence is present. Nevertheless, its perpetual presence cannot be ignored.
Total Forces
Therefore, the total forces acting on the flow in the C.V. will be
(5)
which will constitute the net force that must appear on the left-hand side of the final momentum equation.
Momentum Terms
Now consider the right-hand side of the momentum equation, i.e., the term that must describe the fluid’s time rate of change of momentum. There will be two contributions to this, namely:
The net momentum flows out of the C.V. across the surface of area per unit of time.
The time rate of momentum change inside will only occur if there are any unsteady fluctuations in the flow properties.
Consider the first term. At any point inside the C.V. or on the C.S., the flow velocity is . The mass flow across the element is given by so the flow of momentum across will be
(6)
The net flow of momentum out of the control C.V. over the C.S. is the sum of all the elemental contributions, i.e.,
(7)
Now, consider the second term. The momentum of the fluid inside the element of volume will be , and so the total momentum contained inside the C.V. at a given time is
(8)
and its time rate of change (which can only result from unsteady flow fluctuations) will be
(9)
Final Form of the Momentum Equation
Combining all the above equations into the form gives an expression for momentum conservation in terms of the total forces and time rate of change of momentum of the flow as it sweeps through the C.V., i.e.,
(10)
which is another “star” governing equation. In other words, Eq. 10 states that Total Forces = Body Forces + Pressure Forces + Viscous Forces = Time rate of change of momentum inside from any unsteadiness in the flow + Net flow of momentum out of across the C.S. of area per unit time. This general equation applies to any flow, whether compressible or incompressible, viscous or inviscid, and so on.
Equation 10 is the momentum equation in its integral form, first established by Leonhard Euler. It will be apparent that it is a vector equation, so three scalar equations are rolled into one, i.e., one equation for each coordinate direction. If , then
(11)
What are the units of momentum?
When we speak of “momentum,” we mean the momentum of something. Momentum is a vector quantity, so it has both magnitude and direction. If denotes mass and denotes velocity, then the translational momentum is . Therefore, in terms of base dimensions
noting that the units inside the parentheses represent a force. Therefore, in the SI system, the units of the translational momentum of a fluid will be kg m s, which is also equivalent to a Newton-second, i.e., N s. In the USC system, the units of momentum will be slug ft s, which is also equivalent to a pound-second, i.e., lb s.
Simplifications to the Momentum Equation
As in the use of the continuity equation for practical problem solving, the apparent complexity of the general form of the momentum equation can be simplified by making justifiable assumptions; the advantage is that the solutions to the resulting sets of governing equations (mass, momentum, and energy) become commensurately easier. Remember that any assumptions must be justified, which may require information from an experiment or a judgment call, i.e., based on experience and expertise. Experience is acquired through repeated problem-solving, i.e., starting with exemplar problems in the classroom and then doing many more on one’s own.
Consider, for example, a steady flow. In this case, , so the reduced form of the momentum equation becomes
(12)
If the flow is inviscid, then . Therefore, the momentum equation simplifies to
(13)
If the flow is also incompressible, then is constant and can be factored out, i.e.,
(14)
If body forces can be neglected, then the pressure forces must balance the momentum flux, so that
(15)
Alternatively, if pressure forces are negligible or the momentum terms are much greater than the pressure terms, then the momentum equation becomes
(16)
Two-Dimensional Forms of the Momentum Equation
A two-dimensional simplification is often suitable for many practical applications, particularly in external flows or flows confined by walls. Volume integrals reduce to area integrals per unit depth, and surface integrals reduce to line integrals around the boundary of a planar control surface.
The component form of the momentum equation for a two-dimensional, steady, incompressible, inviscid flow becomes
(17)
These latter equations include both momentum flux and pressure forces. If pressure forces are negligible, the expressions reduce further to
(18)
For a control volume with uniform properties at the inlet and outlet and negligible contributions from the other surfaces, these simplify further to
(19)
where is the mass flow rate through the control volume, and , are the changes in velocity components.
One-Dimensional Forms of the Momentum Equation
In the case of a one-dimensional flow, where the velocity vector is aligned with the -axis and properties are uniform over the inlet and outlet cross sections, the momentum equation becomes
(20)
which may be written as
(21)
Further neglecting pressure forces, the momentum equation simplifies to
(22)
or in terms of the mass flow rate, then
(23)
where . These one-dimensional forms are commonly used in applications when the flow is nearly unidirectional and cross-sectional variations in the flow dimensions are small.
Learning the Exemplars
The extent to which the momentum equation can be simplified depends on the problem’s complexity and the strength of the justifications for each assumption. Learning to make these simplifications effectively comes from studying classic examples in fluid dynamics. A structured approach, starting with fundamental problems and progressively working up to more complex scenarios, provides a solid foundation for understanding.
Beyond textbook examples, applying fluid dynamics concepts to real-world problems and using simulation software can significantly enhance comprehension. Computational tools enable a deeper exploration of flow behavior, facilitating the visualization and analysis of various scenarios. Additionally, numerous online resources, including open-access courses and materials from educational institutions, provide valuable opportunities for further study and practice.
Check Your Understanding #1 – Force on a converging duct or nozzle
Reduce the general form of the momentum equation to a form suitable for application to the steady, compressible, one-dimensional flow of air through a converging duct (often called a contraction) of circular cross-section, i.e., a nozzle, as shown below.
Show solution/hide solution.
Begin by sketching the nozzle problem, illustrating the control surface and annotating both known and unknown inlet and outlet conditions. In this case, the net pressure force exerted by the nozzle on the fluid needs to be determined. The flow is assumed to be steady, so . The fluid is air, which has a relatively low density, and there is no change in elevation from one side of the nozzle to the other; therefore, it is possible to neglect all body forces, including the effects of gravity.
As previously derived, the general form of the momentum equation is
The reduced form for the assumptions made that the flow is steady without body forces leads to
The left-hand side can be written as
In this case, the force on the fluid, , which changes its momentum, needs to be determined; this is the unknown. Notice that this force can include both pressure forces and viscous forces. There is also an equal and opposite force on the nozzle, denoted as , which must ultimately be determined. Therefore, the momentum equation, in this case, is
The flow can be assumed to be one-dimensional, meaning that the flow properties change only in one dimension. This assumption simplifies the solution process while still being realistic. With the one-dimensional assumption, then
noting the signs of the term on the first pressure integral over and remembering that always points out of the control volume by convention. Also, the continuity of the flow means that
So, the final form of the momentum equation for this problem becomes
This force will be negative, i.e., to the left. The fact that this force is in the backward direction (opposite to the flow direction) is somewhat counterintuitive because momentum is being added to the fluid as it passes through the nozzle. However, when a person holds the nozzle, they must apply a forward force reaction to keep it in place and prevent it from moving backward. This reaction force is
which will be to the right. Notice also that this problem involves velocities and pressures. However, the Bernoulli equation(as discussed in the next chapter) cannot be applied to this problem because there are viscous losses as the fluid passes through the nozzle. However, the continuity equation applies to any fluid flow, whether viscous or inviscid, and thus applies in this case, as does the momentum equation.
Check Your Understanding #2 – Force on an engine test stand
A static thrust stand is designed for testing a jet engine. The following conditions are expected for a typical test: intake air velocity = 200 m/s, exhaust gas velocity = 500 m/s, intake and exit cross-sectional areas = 1 m, intake static pressure = -22.5 kPa (= 78.5 kPa absolute), intake static temperature = 268K, exhaust static pressure = 0 kPa (101 kPa absolute). Estimate the thrust reaction force to design the test stand to carry, if it must have a safety factor of 3.
Show solution/hide solution.
Assume that the inlet is identified as station 1 and the outlet as station 2. The density of the air at the inlet can be found using
From continuity considerations (assume 1-dimensional, steady flow), then
If the flow is additionally assumed to be inviscid and there are no gravitational effects, then , and the relevant form of the momentum equation is
The pressure integral on the left-hand side is
or
where is the force of the engine on the fluid. Assuming one-dimensional flow gives
noting the signs of the terms, and remembering that always points out of the control volume, or
noting that = . Finally, substituting the known values gives
(24)
For the conditions stated, the residual force on the fluid must be 83,736 N in the direction of fluid flow, and the reaction force on the test stand will be in the opposite direction. Because there is a factor of safety requirement of 3, the force to which the stand should be designed is about 251 kN.
Check Your Understanding #3 – Force on an angled plate
Water (= 1.94 slug ft) flows at a mass flow rate of = 6.0 slug sthrough a nozzle of circular cross-section with an entrance diameter in and an exit diameter of in. The water jet then hits an angled plate with a circular groove, where it is deflected while also increasing in its circular cross-section, as shown in the figure below.
Draw an appropriate control surface and volume with annotations to analyze this problem.
Determine the velocity of the water as it enters and exits the nozzle.
Determine the velocity of the water in the deflected jet.
Determine the magnitude and direction of the force on the plate caused by the water jet.
Show solution/hide solution.
The appropriate control surface and volume, along with annotations, are shown below for the analysis of this problem.
This part of the problem requires the use of the continuity equation in its one-dimensional form, i.e.,
The entrance area, , is
Therefore, the entrance velocity is
The exit jet area, , is
Therefore, the exit jet velocity is
The velocity of the water in the deflected jet can be found using continuity, i.e.
The deflected jet area is
Therefore, the deflected jet velocity is
The applicable form of the momentum equation is
We can ignore all pressure forces because the plate is surrounded by free air at ambient conditions, so there is no pressure force on the control volume. In scalar form, because we have forces in two directions, then
In the direction, the force on the fluid is
where is the change in the flow velocity in the direction. Therefore,
In the direction, the force on the fluid is
where is the change in the flow velocity in the direction. Therefore,
If the force on the fluid is
Then, the reaction force on the plate is
which is to the right and downward.
Angular Momentum
For most fluid mechanics problems, especially in incompressible flows, consideration of linear momentum is sufficient because rotational effects are either negligible or irrelevant to the problem’s primary focus. Angular momentum effects tend to be evident in cases with strong rotational characteristics or where torque calculations are required. For example, in turbomachinery, such as pumps, turbines, or compressors, the angular momentum equation is essential for analyzing the torque and rotational energy transfer within the control volume. Furthermore, specific fluid flow problems, such as vortices, involve the conservation of angular momentum as fluid particles follow circular or spiral paths. In these cases, the analysis can often use angular momentum principles to derive velocity profiles or understand the flow behavior.
The basic flow model used for angular momentum is shown in the figure below. To include angular momentum in the momentum equation, the angular momentum (moment of momentum), i.e., , over a control volume is defined by
(25)
where is the fluid density, is the position vector from some reference, and is the velocity vector. The net time rate of change of angular momentum from the control volume can be expressed as
(26)
The first term represents the local (time) rate of change of momentum within , and the second term accounts for the angular momentum flux across the control surface .
To balance the angular momentum, it is necessary to consider the external moments acting on the fluid inside the C.V. These moments include:
Moment of body forces. If is the body force per unit mass (e.g., gravity), the moment from the body forces is given by
(27)
2. Moment of surface forces. The pressure and viscous stresses on the surface contribute to the surface moment. Regarding pressure , the moment contribution is
(28)
where is the outward-directed unit normal vector area.
3. The moment caused by the stresses. The viscous contribution represents the integrated moment from viscous stresses, i.e., the integral of the moments caused by the shear stresses acting over the C.S.
4. A torque (which is also a moment) from mechanical sources. In many cases, the angular momentum of a fluid is caused by the action of a mechanical device driven through a shaft, such as a pump, propeller, or rotor. Therefore, in this case, represents this torque on the fluid.
With these definitions, the angular momentum balance for the fluid in a control volume becomes
(29)
Substituting the expressions for the moments gives
(30)
The complete angular momentum equation, combining the rate of change of angular momentum and the moments, is
(31)
This equation states that the net external moment acting on the control volume is balanced by the rate of change of angular momentum within and the flux of angular momentum across the surface .
Notice that the foundation of the angular momentum equation is the . This term represents the angular momentum per unit volume, also known as the moment of momentum density.
Linear Momentum Equation from the RTE
Recall that the Reynolds Transport Equation (RTE) can be expressed as
(32)
In the case of momentum, then , and for the system, any change in momentum is equivalent to a force, so
(33)
where is the net force acting on the fluid in the C.V., and is the substantial derivative. Recall that the net force depends on body forces, pressure forces, and viscous forces, i.e.,
(34)
Therefore, the RTE becomes
(35)
which is the linear momentum equation. Remember that the momentum equation is a vector equation.
For steady flow, the momentum equation becomes
(36)
For incompressible flow where is constant, the density term can be taken outside of the integral to get
(37)
Momentum Equation – Differential Form
The momentum equations in differential form are called the Euler equations. In 1757, Leonhard Euler published his foundational work on the dynamics of fluids, which included the differential form of the momentum equations now known as the Euler equations. The Euler equations describe the motion of an inviscid fluid, i.e., a fluid of limitingly diminishing viscosity. Euler’s work laid the groundwork for later developments in fluid mechanics, including the Navier-Stokes equations, which extend the Euler equations to include viscosity effects.
Classic Approach
To derive the Euler equations, the classic approach is to apply the principles of conservation of momentum to a small differential fluid element. Consider a small fluid element in a Cartesian coordinate system with dimensions , , and , as shown in the figure below. The volume of this fluid element is .
Pressure Forces
Consider the pressure forces in the direction, i.e., on the –face of area . Let be the change of pressure with respect to . The net pressure force in the direction is
(38)
The pressure gradient term, as a pressure force per unit mass, appeared previously in the derivation of the hydrostatic equation. Similarly, for the and directions, the net pressure forces are
(39)
Therefore, the net resultant pressure force on the entire fluid element is
(40)
Body Forces
Body forces per unit mass acting on the element can be described as
(41)
Gravity is an omnipresent body force, i.e., . It is often useful to isolate and take the term outside of other body force terms, i.e.,
(42)
Momentum Terms
According to Newton’s second law, this force is equal to the time rate of change of momentum of the fluid element, i.e.,
(43)
where is the substantial derivative. Therefore,
(44)
Cancelling out and rearranging gives
(45)
Similarly, for the and directions, then
(46)
Final Form
Combining the previous results gives the Euler equations for an inviscid fluid, i.e.,
(47)
In vector form, these equations can be written as
(48)
Expanding out the substantial derivatives, then in the (long-handed) scalar form, they are
(49)
Euler Equations from the RTE
The differential form of the continuity equation can also be derived from the Reynolds Transport Theorem (RTE). Using the divergence theorem, which converts a surface integral into a volume integral, the RTE for a fixed C.V. becomes
(50)
where in this case, . Expanding the time derivative term gives
(51)
Substituting this back into the original integral yields
(52)
This latter result indicates that the time rate of change of momentum depends on both the rate of change of density and the rate of change of velocity. Furthermore, because the volume is arbitrary, and the mass of a differential fluid element is , the force per unit mass is expressed as
(53)
The force per unit mass comprises pressure forces, body forces (e.g., gravitational or inertial), and viscous forces, leading to
(54)
which is the momentum equation.
Simplifications
The viscous force term, , is zero for an inviscid flow. Rewriting the equation without this term, then
(55)
Using the continuity equation, i.e.,
(56)
then the momentum equation simplifies to
(57)
which, in terms of the substantial derivative, is
(58)
as derived previously using the differential element approach.
For incompressible flow, the density is constant, so the continuity equation simplifies to
(59)
and the momentum equations are simplified to
(60)
which can be rewritten in terms of the substantial derivative as
(61)
For a steady flow, the time derivatives are zero. Then, the Euler equations are simplified to
(62)
In vector form, these equations can be written as
(63)
Expanding these substantial derivatives in the scalar form gives
(64)
Navier-Stokes Equations
The viscous force term, denoted , that appears in the most general forms of the momentum equation requires some work to evaluate. These terms arise because of the relative velocity of the flow in the form of velocity gradients. Consider again the small differential fluid element in Cartesian coordinates with dimensions , , and , as shown in the figure below. On each face, the direction of the shear stresses, denoted by , is shown.
The indices i and j represent the coordinate directions , , and , where i denotes the plane on which the stress acts and j denotes the direction of the stress deformation. Notice that one normal stress component acts perpendicular to each face, and two shear stress components act tangentially to the face. For example, on the face perpendicular to the -direction, there is a normal stress, , and the two shear stresses, and . The normal stresses, , , and , are often denoted as , and .
Viscous Forces
The net viscous force on the fluid in the direction is
(65)
thereby giving
(66)
The two other stresses in the direction are and , as shown in the figure below. Therefore, the total viscous force contribution in the direction is
(67)
The stresses , , and contribute to the force in the -direction, i.e., in aggregate they are
(68)
Finally, the three remaining viscous shear stresses contribute to the force in the direction, i.e.,
(69)
Collecting all of the contributions of the viscous stresses gives
(70)
Incorporating the stress terms into the momentum equations can be seen by first considering the -component, i.e.,
(71)
and cancelling out the term gives
(72)
Similarly, for the -momentum equation, then
(73)
and for the -momentum equation, then
(74)
Therefore, combining the scalar components gives in vector form (which applies to a viscous laminar flow without body forces), then
(75)
where the shear stress tensor is
With the body forces included, then
Evaluating the Shear Stresses
Newton’s law of viscosity states that the shear stress is proportional to the rate of deformation (strain rate) of the fluid. Therefore, for a fluid with constant viscosity , this relationship is written as
In a flow where the velocity components are in the -direction and in the -direction, the fluid’s deformation rate is determined by how the velocity changes in space. The strain rates for a fluid include terms that describe both normal (bulk) stresses and shear stress deformations. For the component of the shear stress, the strain rate is given by
where the term represents the rate at which the horizontal velocity changes with respect to the vertical coordinate , which contributes to shear. The term represents the rate at which the vertical velocity changes with respect to the horizontal coordinate , contributing to shear. The reason the strain rate has the sum of these two terms is because of the symmetry of the stress tensor, i.e., the stress (shear in the -plane) equals (shear in the -plane). Therefore, both velocity gradients and contribute to the shear stress. Therefore, the component is given by
For the -component, then
The momentum equation for the -component is given by
where are the components of the viscous stress tensor in the -direction. Substituting the expressions for , , and
Simplifying gives
This result can be written as
where is the Laplace operator. Similarly, for the -component, then
Therefore, the -component of the momentum equation can be written as
Finally, for the -component, then
Therefore, the -component of the momentum equation can be written as
Collecting all the viscous terms together, then
In this case, these are referred to as the Navier-Stokes (N-S) equations in their incompressible, laminar form. In vector form, the N-S equations are
With the body forces including gravity, the equation becomes
The N-S equations (no body forces) can also be written in the form
where is the kinematic viscosity. Finally, in their most elegant form, the incompressible N-S equations are written as
The Navier-Stokes equations are the most famous in fluid dynamics. These equations, a statement of Newton’s second law for a fluid, are fundamental to understanding fluid behavior. Despite their apparent simplicity, the N-S equations represent a set of coupled nonlinear partial differential equations, so solving them poses a significant challenge. The N-S equations remain the cornerstone for studying fluid dynamics.
Summary & Closure
The principle of conservation of momentum is a cornerstone in fluid dynamics, and its use is necessitated whenever forces act on a fluid. The underlying principle is Newton’s second law of motion and is mathematically represented through the formal derivation of the momentum equation. This fundamental equation accounts for various forces that influence fluid motion, including body forces, pressure forces, and viscous forces. The momentum and continuity equations form the basis for analyzing and solving fluid flow problems. Depending on the specific situation, the momentum equation can be simplified by making appropriate assumptions, such as steady, inviscid, or incompressible flow. However, these simplifications must be carefully justified based on the physical context of the flow. In scenarios where pressure and temperature play a significant role, it becomes imperative to incorporate energy conservation principles to describe fluid behavior fully. This comprehensive approach ensures that all relevant forces and interactions are considered, resulting in a more accurate and thorough understanding of fluid flow dynamics.
The challenges of solving the Navier-Stokes equations extend well beyond their mathematical form because engineers must apply them to real-world scenarios. The Navier-Stokes equations are frequently employed to model turbulent flows, but predicting turbulence is also inherently problematic. Engineers now employ computational fluid dynamics (CFD) methods to find approximate numerical solutions to the Navier-Stokes equations and gain insights into the development of turbulent flows around flight vehicles. The Euler equations are often used as an initial approximation in computational fluid dynamics (CFD) simulations.
5-Question Self-Assessment Quiz
For Further Thought or Discussion
Consider the flow through a reducing pipe with a 180bend. Show how to set up the equations to calculate the force of the fluid on the bend.
If the value of a force calculated using the momentum equation is negative, then what does this mean?
Consider some flow problems where viscous effects are dominant, and an inviscid assumption for modeling the flow would likely be invalid.
How might the momentum equation have to be modified to analyze turbulent flow?
How might non-Newtonian fluids be treated differently in the momentum equation than Newtonian fluids?
Additional Online Resources
To learn more about the conservation of momentum applied to flow problems, check out some of these additional online resources:
See an excellent, detailed explanation of the momentum equation in this video.
A video explaining how to find the force from a flow through a nozzle.
A (long!) video by an instructor explaining the conservation of momentum.
MIT OpenCourseWare: Fluid Mechanics. Free course materials, including lecture notes, assignments, and exams from MIT’s course on advanced fluid mechanics.
Coursera: Fluid Mechanics Courses. Universities worldwide offer various courses covering fundamental and advanced topics in fluid mechanics.
YouTube: NPTEL Lectures on Fluid Mechanics. Comprehensive lecture series from India’s National Programme on Technology Enhanced Learning (NPTEL), covering fluid mechanics in detail.
Wikipedia: Navier-Stokes Equations. An overview of the Navier-Stokes equations, their derivation, and applications in fluid dynamics.
HyperPhysics: Fluid Dynamics. A comprehensive resource for physics concepts, including fluid dynamics and the application of the conservation of momentum.
Journal of Fluid Mechanics. A leading journal that publishes high-quality research papers on all aspects of fluid mechanics.
Physics of Fluids. An influential journal offering articles on fluid dynamics, including experimental, theoretical, and computational studies.
SimScale: CFD Simulations. An online platform for computational fluid dynamics (CFD) simulations, providing practical applications of the conservation of momentum in fluid flow problems.
PhET Interactive Simulations: Fluid Mechanics. Interactive simulations for visualizing and understanding concepts in fluid mechanics.
Engaging with these resources can reinforce theoretical knowledge while building practical problem-solving skills. ↵
Finding closed-form solutions remains one of the Clay Mathematics Institute's Millennium Prizes, recognizing the challenges in their solution. ↵
License
Introduction to Aerospace Flight Vehicles Copyright © 2022–2025 by J. Gordon Leishman is licensed under a Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 International License, except where otherwise noted.
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3731 | https://arxiv.org/abs/1106.4325 | [1106.4325] On generalized Pólya urn models
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arXiv:1106.4325 (math)
[Submitted on 21 Jun 2011]
Title:On generalized Pólya urn models
Authors:May-Ru Chen, Markus Kuba
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Abstract:We study an urn model introduced in the paper of Chen and Wei, where at each discrete time step $m$ balls are drawn at random from the urn containing colors white and black. Balls are added to the urn according to the inspected colors, generalizing the well known Pólya-Eggenberger urn model, case m=1. We provide exact expressions for the expectation and the variance of the number of white balls after n draws, and determine the structure of higher moments. Moreover, we discuss extensions to more than two colors. Furthermore, we introduce and discuss a new urn model where the sampling of the m balls is carried out in a step-by-step fashion, and also introduce a generalized Friedman's urn model.
Comments:15 pages, no figures
Subjects:Probability (math.PR)
Cite as:arXiv:1106.4325 [math.PR]
(or arXiv:1106.4325v1 [math.PR] for this version)
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From: Markus Kuba [view email]
[v1] Tue, 21 Jun 2011 20:55:13 UTC (13 KB)
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3732 | https://www.cdc.gov/std/treatment-guidelines/urethritis-and-cervicitis.htm | Urethritis and Cervicitis - STI Treatment Guidelines
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Diseases Characterized by Urethritis and Cervicitis
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Urethritis
Nongonococcal Urethritis (NGU)
Cervicitis
Urethritis
Urethritis, as characterized by urethral inflammation, can result from either infectious or noninfectious conditions. Symptoms, if present, include dysuria, urethral pruritis, and mucoid, mucopurulent, or purulent discharge. Signs of urethral discharge on examination can also be present among persons without symptoms. Although N. gonorrhoeae and C. trachomatis are well established as clinically important infectious causes of urethritis, M. genitalium has been strongly associated with urethritis and, less commonly, prostatitis (691–697). If POC diagnostic tools (e.g., Gram, methylene blue [MB], or gentian violet [GV] stain microscopy) are unavailable, drug regimens effective against both gonorrhea and chlamydia should be administered. Further testing to determine the specific etiology is recommended for preventing complications, reinfection, and transmission because a specific diagnosis might improve treatment compliance, delivery of risk-reduction interventions, and partner services. Both chlamydia and gonorrhea are reportable to health departments. NAATs are preferred for detecting C. trachomatis and N. gonorrhoeae, and urine is the preferred specimen for males (553). NAAT-based tests for diagnosing T. vaginalis among men with urethritis have not been cleared by FDA; however, laboratories have performed the CLIA-compliant validation studies (698) needed to provide such testing.
Etiology
Multiple organisms can cause infectious urethritis. The presence of gram-negative intracellular diplococci (GNID) or purple intracellular diplococci (MB or GV) on urethral smear is indicative of presumed gonococcal infection, which is frequently accompanied by chlamydial infection. Nongonococcal urethritis (NGU), which is diagnosed when microscopy of urethral secretions indicate inflammation without GNID or MB or GV purple intracellular diplococci, is caused by C. trachomatis in 15%–40% of cases; however, prevalence varies by age group, with a lower proportion of disease occurring among older men (699). Documentation of chlamydial infection as NGU etiology is essential because of the need for partner referral for evaluation and treatment to prevent complications of chlamydia, especially for female partners. Complications of C. trachomatis–associated NGU among males include epididymitis, prostatitis, and reactive arthritis.
M. genitalium is associated with symptoms of urethritis and urethral inflammation and accounts for 15%–25% of NGU cases in the United States (691–693,696,697,700). Among men with symptoms of urethritis, M. genitalium was detected in 11% of those with urethritis in Australia (701), 12%–15% in the United Kingdom (702–704), 15% in South Africa (696), 19% in China (705), 21% in Korea, 22% in Japan (706), and 28.7% in the United States (range: 20.4%–38.8%) (697). Data are inconsistent regarding other Mycoplasma and Ureaplasma species as etiologic agents of urethritis (707). The majority of men with Ureaplasma infections do not have overt disease unless a high organism load is present.
T. vaginalis can cause urethritis among heterosexual men; however, the prevalence varies substantially by U.S. geographic region, age, and sexual behavior and within specific populations. Studies among men with and without overt urethritis in developed countries document relatively low rates of T. vaginalis in the Netherlands (0.5%) (708), Japan (1.3%) (706,709), the United States (2.4%) (710), and the United Kingdom (3.6%) (703). Studies in other countries have documented higher rates, such as in Croatia (8.2%) (711) and Zimbabwe (8.4%) (712), particularly among symptomatic patients.
Neisseria meningitidis can colonize mucosal surfaces and cause urethritis (713). Urogenital N. meningitidis rates and duration of carriage, prevalence of asymptomatic and symptomatic infection, and modes of transmission have not been systematically described; however, studies indicate that N. meningitidis can be transmitted through oral-penile contact (i.e., fellatio) (714–716). N. meningitidis has similar colony morphology appearance on culture and cannot be distinguished from N. gonorrhoeae on Gram stain. Identification of N. meningitidis as the etiologic agent with presumed gonococcal urethritis on the basis of Gram stain but negative NAAT for gonorrhea requires a confirmation by culture. Meningococcal urethritis is treated with the same antimicrobial regimens as gonococcal urethritis. Although evidence is limited regarding the risk for sexual transmission or recurrent infections with meningococcal urethritis, treatment of sex partners of patients with meningococcal urethritis with the same antimicrobial regimens as for exposure to gonococcal infection can be considered. No indication exists for treating persons with N. meningitidis identified in their oropharynx when not also associated with symptomatic urethritis.
In other instances, NGU can be caused by HSV, Epstein-Barr virus, and adenovirus (699) acquired by fellatio (i.e., oral-penile contact). In a retrospective review of 80 cases of HSV urethritis in Australia (717), the majority of infections were associated with HSV-1 with clinical findings of meatitis (62%), genital ulceration (37%), and dysuria (20%). Adenovirus can present with dysuria, meatal inflammation, and conjunctivitis (718). Enteric bacteria have been identified as an uncommon cause of NGU and might be associated with insertive anal intercourse (699).
Other bacterial pathogens have been implicated as potential causes of clinical urethritis, either in clustered case series or as sporadic cases such as Haemophilus influenzae and Haemophilus parainfluenzae (719–723). Haemophilus was identified in 12.6% of cases among 413 men (mostly MSM reporting insertive oral sex) (724), and high rates of azithromycin resistance (39.5%) were identified among Haemophilus urethritis patients (725). Individual case reports have linked NGU to multiple bacterial species, including Corynebacterium propinquum (726), Kurthia gibsonii (727), Corynebacterium glucuronolyticum (728,729), Corynebacterium striatrium (730), Aerococcus urinae (731), and Neisseria elongata (732). Diagnostic testing and treatment for less-common organisms are reserved for situations in which these infections are suspected (e.g., sexual partner with trichomoniasis, urethral lesions, or severe dysuria and meatitis) or when NGU is not responsive to recommended therapy.
Even in settings that provide comprehensive diagnostic testing, etiology can remain obscure in half of cases. Idiopathic NGU was reported in 772 (59%) of 1,295 first presentations of NGU among men seeking sexual health services in Australia (701). In a case-control study of 211 men with NGU symptoms in Denmark, no identifiable pathogen was identified in 24% of acute cases and 33% of chronic cases (733). NGU’s importance if not caused by a defined pathogen is uncertain; neither complications (e.g., urethral stricture or epididymitis) nor adverse outcomes among sex partners have been identified in these cases.
Associations between NGU and insertive anal and oral exposure have been reported (734), as have higher rates of BV-associated Leptotrichia or Sneathia species among heterosexual men with urethritis (735). These studies increase concern for possible undetected infectious rectal or vaginal pathogens, or alternatively, a transient reactive dysbiosis after exposure to a new microbiome or even a noninfectious reactive etiology (736).
Diagnostic Considerations
Clinicians should attempt to obtain objective evidence of urethral inflammation. If POC diagnostic tests (e.g., Gram stain or MB or GV microscopy) are unavailable, urethritis can be documented on the basis of any of the following signs or laboratory tests:
Mucoid, mucopurulent, or purulent discharge on examination.
Gram stain is a POC diagnostic test for evaluating urethritis that is highly sensitive and specific for documenting both urethritis and the presence or absence of gonococcal infection; MB or GV stain of urethral secretions is an alternative POC diagnostic test with performance characteristics similar to Gram stain; thus, the cutoff number for WBCs per oil immersion field should be the same (737).
Presumed gonococcal infection is established by documenting the presence of WBCs containing GNID in Gram stain or intracellular purple diplococci in MB or GV smears; men should be tested for C. trachomatis and N. gonorrhoeae by NAATs and presumptively treated and managed accordingly for gonococcal infection (see Gonococcal Infections).
If no intracellular gram-negative or purple diplococci are present, men should receive NAATs for C. trachomatis and N. gonorrhoeae and can be managed for NGU as recommended (see Nongonococcal Urethritis).
Gram stain of urethral secretions exist that demonstrate ≥2 WBCs per oil immersion field (738). The microscopy diagnostic cutoff might vary, depending on background prevalence (≥2 WBCs/high power field [HPF] in high-prevalence settings [STI clinics] or ≥5 WBCs/HPF in lower-prevalence settings).§§
Positive leukocyte esterase test on first-void urine or microscopic examination of sediment from a spun first-void urine demonstrating ≥10 WBCs/HPF.
Men evaluated in settings in which Gram stain or MB or GV smear is unavailable who meet at least one criterion for urethritis (i.e., urethral discharge, positive leukocyte esterase test on first void urine, or microscopic examination of first-void urine sediment with ≥10 WBCs/HPF) should be tested for C. trachomatis and N. gonorrhoeae by NAATs and treated with regimens effective against gonorrhea and chlamydia.
If symptoms are present but no evidence of urethral inflammation is present, NAATs for C. trachomatis and N. gonorrhoeae might identify infections (739). Persons with chlamydia or gonorrhea should receive recommended treatment, and sex partners should be referred for evaluation and treatment. If none of these clinical criteria are present, empiric treatment of men with symptoms of urethritis is recommended only for those at high risk for infection who are unlikely to return for a follow-up evaluation or test results. Such men should be treated with drug regimens effective against gonorrhea and chlamydia.
Nongonococcal Urethritis (NGU)
NGU is a nonspecific diagnosis that can have various infectious etiologies. C. trachomatis has been well established as an NGU etiology; however, prevalence varies across populations and accounts for <50% of overall cases (712,740–742). M. genitalium is estimated to account for 10%–25% of cases (696,697,701,703,704,706,733,743), and T. vaginalis for 1%–8% of cases depending on population and location (703,706,708,710,712). Other etiologies include different bacteria, such as Haemophilus species (724,725), N. meningitidis (713,716), HSV (706,717), and adenovirus (744). However, even when extensive testing is performed, no pathogens are identified in approximately half of cases (701,733).
Diagnostic Considerations
Clinical presentation can include urethral discharge, irritation, dysuria, or meatal pruritus (697,743,745). NGU is confirmed for symptomatic men when diagnostic evaluation of urethral secretions indicates inflammation, without evidence of diplococci by Gram, MB, or GV smear on microscopy (712,746,747). Visible discharge or secretions can be collected by a swab without inserting it into the urethra; if no visible secretions, the swab can be inserted into the urethral meatus and rotated, making contact with the urethral wall before removal. If microscopy is unavailable, urine testing for leukocyte esterase can be performed on first-void urine, and microscopic examination of sediment from a spun first-void urine demonstrating ≥10 WBCs/HPF has a high negative predictive value.
All men who have suspected or confirmed NGU should be tested for chlamydia and gonorrhea by using NAATs. A specific diagnosis can potentially reduce complications, reinfection, and transmission. M. genitalium testing should be performed for men who have persistent or recurrent symptoms after initial empiric treatment. Testing for T. vaginalis should be considered in areas or among populations with high prevalence, in cases where a partner is known to be infected, or for men who have persistent or recurrent symptoms after initial empiric treatment.
Treatment
Ideally, treatment should be pathogen based; however, diagnostic information might not be immediately available. Presumptive treatment should be initiated at NGU diagnosis. Doxycycline is highly effective for chlamydial urethral infections and is also effective for chlamydial infections of the rectum; it also has some activity against M. genitalium. In contrast, reports have increased of azithromycin treatment failures for chlamydial infection (748,749), and the incidence of macrolide resistance in M. genitalium also has been rapidly rising (697,702,705,750,751). Pharmacokinetic data indicate that changing azithromycin dosing from a single-dose strategy to a multiday strategy might protect against inducing resistance in M. genitalium infections (745,752) (see Mycoplasma genitalium).
Recommended Regimen for Nongonococcal Urethritis
Doxycycline 100 mg orally 2 times/day for 7 days
Alternative Regimens
Azithromycin 1 g, orally in a single dose
OR
Azithromycin 500 mg orally in a single dose; then 250 mg orally daily for 4 days
To maximize compliance with recommended therapies, medications should be dispensed on-site at the clinic, and, regardless of the number of doses involved in the regimen, the first dose should be directly observed. Erythromycin is no longer recommended for NGU because of its gastrointestinal side effects and dosing frequency. Levofloxacin is no longer recommended for NGU because of its inferior efficacy, especially for M. genitalium.
Management Considerations
To minimize transmission and reinfections, men treated for NGU should be instructed to abstain from sexual intercourse until they and their partners have been treated (i.e., until completion of a 7-day regimen and symptoms have resolved or for 7 days after single-dose therapy). Men with NGU should be tested for HIV and syphilis.
Follow-Up
Men should be provided their testing results obtained as part of the NGU evaluation. Those with a specific diagnosis of chlamydia, gonorrhea, or trichomoniasis should be offered partner services and instructed to return 3 months after treatment for repeat testing because of high rates of reinfection, regardless of whether their sex partners were treated (136,137,753,754) (see Chlamydial Infections; Gonococcal Infections; Trichomoniasis).
If symptoms persist or recur after therapy completion, men should be instructed to return for reevaluation and should be tested for M. genitalium and T. vaginalis. Symptoms alone, without documentation of signs or laboratory evidence of urethral inflammation, are insufficient basis for retreatment. Providers should be alert to the possible diagnosis of chronic prostatitis or chronic pelvic pain syndrome in men experiencing persistent perineal, penile, or pelvic pain or discomfort; voiding symptoms; pain during or after ejaculation; or new-onset premature ejaculation lasting for >3 months. Men with persistent pain should be referred to a urologist with expertise in pelvic pain disorders.
Management of Sex Partners
All sex partners of men with NGU within the preceding 60 days should be referred for evaluation and testing and presumptive treatment with a drug regimen effective against chlamydia. All partners should be evaluated and treated according to the management section for their respective pathogen; EPT could be an alternate approach if a partner is unable to access timely care. To avoid reinfection, sex partners should abstain from sexual intercourse until they and their partners are treated.
Persistent or Recurrent Nongonococcal Urethritis
The objective diagnosis of persistent or recurrent NGU should be made before considering additional antimicrobial therapy. Symptomatic recurrent or persistent urethritis might be caused by treatment failure or reinfection after successful treatment. Among men who have persistent symptoms after treatment without objective signs of urethral inflammation, the value of extending the duration of antimicrobials has not been demonstrated. Treatment failure for chlamydial urethritis has been estimated at 6%–12% (755). The most common cause of persistent or recurrent NGU is M. genitalium, especially after doxycycline therapy (756,757). Treatment failure for M. genitalium is harder to determine because certain men achieve clinical cure (i.e., resolution of symptoms) but can still have detectable M. genitalium in urethral specimens (758).
The initial step in recurrent urethritis is assessing compliance with treatment or potential reexposure to an untreated sex partner (697,743). If the patient did not comply with the treatment regimen or was reexposed to an untreated partner, retreatment with the initial regimen can be considered. If therapy was appropriately completed and no reexposure occurred, therapy is dependent on the initial treatment regimen. Ideally, diagnostic testing among men with recurrent or persistent symptoms, including those with gonorrhea, chlamydia, M. genitalium, and trichomoniasis, can be used to guide further management decisions.
T. vaginalis is also known to cause urethritis among men who have sex with women. In areas where T. vaginalis is prevalent, men who have sex with women with persistent or recurrent urethritis should be tested for T. vaginalis and presumptively treated with metronidazole 2 g orally in a single dose or tinidazole 2 g orally in a single dose; their partners should be referred for evaluation and treatment, if needed.
If T. vaginalis is unlikely (MSM with NGU or negative T. vaginalis NAAT), men with recurrent NGU should be tested for M. genitalium by using an FDA-cleared NAAT. Treatment for M. genitalium includes a two-stage approach, ideally using resistance-guided therapy. If M. genitalium resistance testing is available it should be performed, and the results should be used to guide therapy (see Mycoplasma genitalium). If M. genitalium resistance testing is not available, doxycycline 100 mg orally 2 times/day for 7 days followed by moxifloxacin 400 mg orally once daily for 7 days should be used. The rationale for this approach is that although not curative, doxycycline decreases the M. genitalium bacterial load, thereby increasing likelihood of moxifloxacin success (759). Higher doses of azithromycin have not been effective for M. genitalium after azithromycin treatment failures. Men with persistent or recurrent NGU after treatment for M. genitalium or T. vaginalis should be referred to an infectious disease or urology specialist.
Special Considerations
HIV Infection
NGU might facilitate HIV transmission (760). Persons with NGU and HIV infection should receive the same treatment regimen as those who do not have HIV.
Cervicitis
Two major diagnostic signs characterize cervicitis: 1) a purulent or mucopurulent endocervical exudate visible in the endocervical canal or on an endocervical swab specimen (commonly referred to as mucopurulent cervicitis), and 2) sustained endocervical bleeding easily induced by gentle passage of a cotton swab through the cervical os. Either or both signs might be present. Cervicitis frequently is asymptomatic; however, certain women might report an abnormal vaginal discharge and intermenstrual vaginal bleeding (e.g., especially after sexual intercourse). The criterion of using an increased number of WBCs on endocervical Gram stain in the diagnosis of cervicitis has not been standardized; it is not sensitive, has a low positive predictive value for C. trachomatis and N. gonorrhoeae infections, and is not available in most clinical settings (297,761). Leukorrhea, defined as >10 WBCs/HPF on microscopic examination of vaginal fluid, might be a sensitive indicator of cervical inflammation with a high negative predictive value (i.e., cervicitis is unlikely in the absence of leukorrhea) (762,763). Finally, although the presence of gram-negative intracellular diplococci on Gram stain of endocervical exudate might be specific for diagnosing gonococcal cervical infection when evaluated by an experienced laboratorian, it is not a sensitive indicator of infection (764).
Etiology
C. trachomatis or N. gonorrhoeae is the most common etiology of cervicitis defined by diagnostic testing. Trichomoniasis, genital herpes (especially primary HSV-2 infection), or M. genitalium (761,765–768) also have been associated with cervicitis. However, in many cases of cervicitis, no organism is isolated, especially among women at relatively low risk for recent acquisition of these STIs (e.g., women aged >30 years) (769). Limited data indicate that BV and frequent douching might cause cervicitis (770–772). The majority of persistent cases of cervicitis are not caused by reinfection with C. trachomatis or N. gonorrhoeae; other factors might be involved (e.g., persistent abnormality of vaginal flora, M. genitalium, douching or exposure to other types of chemical irritants, dysplasia, or idiopathic inflammation in the zone of ectopy). Available data do not indicate an association between group B streptococcus colonization and cervicitis (773,774). No specific evidence exists for a role for Ureaplasma parvum or Ureaplasma urealyticum in cervicitis (707,761,765,775,776).
Diagnostic Considerations
Because cervicitis might be a sign of upper genital tract infection (e.g., endometritis), women should be assessed for signs of PID and tested for C. trachomatis and N. gonorrhoeae with NAAT on vaginal, cervical, or urine samples (553) (see Chlamydial Infections; Gonococcal Infections). Women with cervicitis also should be evaluated for concomitant BV and trichomoniasis. Because sensitivity of microscopy for detecting T. vaginalis is relatively low (approximately 50%), symptomatic women with cervicitis and negative wet-mount microscopy for trichomonads should receive further testing (i.e., NAAT, culture, or other FDA-cleared diagnostic test) (see Trichomoniasis). Testing for M. genitalium with the FDA-cleared NAAT can be considered. Although HSV-2 infection has been associated with cervicitis, the utility of specific testing (i.e., PCR or culture) for HSV-2 is unknown. Testing for U. parvum, U. urealyticum, Mycoplasma hominis, or genital culture for group B streptococcus is not recommended.
Treatment
Multiple factors should affect the decision to provide presumptive therapy for cervicitis. Presumptive treatment with antimicrobials for C. trachomatis and N. gonorrhoeae should be provided for women at increased risk (e.g., those aged <25 years and women with a new sex partner, a sex partner with concurrent partners, or a sex partner who has an STI), if follow-up cannot be ensured, or if testing with NAAT is not possible. Trichomoniasis and BV should be treated if detected (see Bacterial Vaginosis; Trichomoniasis). For women at lower risk for STIs, deferring treatment until results of diagnostic tests are available is an option. If treatment is deferred and C. trachomatis and N. gonorrhoeae NAATs are negative, a follow-up visit to determine whether the cervicitis has resolved can be considered.
Recommended Regimen for Cervicitis
Doxycycline100 mg orally 2 times/day for 7 days
Consider concurrent treatment for gonococcal infection if the patient is at risk for gonorrhea or lives in a community where the prevalence of gonorrhea is high (see Gonococcal Infections).
Alternative Regimen
Azithromycin 1 g orally in a single dose
Other Management Considerations
To minimize transmission and reinfection, women treated for cervicitis should be instructed to abstain from sexual intercourse until they and their partners have been treated (i.e., until completion of a 7-day regimen or for 7 days after single-dose therapy) and symptoms have resolved. Women who receive a cervicitis diagnosis should be tested for syphilis and HIV in addition to other recommended diagnostic tests.
Follow-Up
Women receiving treatment should return to their provider for a follow-up visit to determine whether cervicitis has resolved. For women who are untreated, a follow-up visit gives providers an opportunity to communicate test results obtained as part of the cervicitis evaluation. Providers should treat on the basis of any positive test results and determine whether cervicitis has resolved. Women with a specific diagnosis of chlamydia, gonorrhea, or trichomoniasis should be offered partner services and instructed to return in 3 months after treatment for repeat testing because of high rates of reinfection, regardless of whether their sex partners were treated (753). If symptoms persist or recur, women should be instructed to return for reevaluation.
Management of Sex Partners
Management of sex partners of women treated for cervicitis should be tailored for the specific infection identified or suspected. All sex partners during the previous 60 days should be referred for evaluation, testing, and presumptive treatment if chlamydia, gonorrhea, or trichomoniasis was identified. EPT and other effective partner referral strategies are alternative approaches for treating male partners of women who have chlamydial or gonococcal infection (125–127) (see Partner Services). To avoid reinfection, sex partners should abstain from sexual intercourse until they and their partners are treated.
Persistent or Recurrent Cervicitis
Women with persistent or recurrent cervicitis despite antimicrobial therapy should be reevaluated for possible reexposure or treatment failure. If relapse or reinfection with a specific infection has been excluded, BV is not present, and sex partners have been evaluated and treated, management options for persistent cervicitis are undefined. In addition, the usefulness of repeated or prolonged administration of antimicrobial therapy for persistent symptomatic cervicitis remains unknown. The etiology of persistent cervicitis, including the potential role of M. genitalium (777), is unclear. M. genitalium might be considered for cases of cervicitis that persist after azithromycin or doxycycline therapy in which reexposure to an infected partner or medical nonadherence is unlikely. Among women with persistent cervicitis who were previously treated with doxycycline or azithromycin, testing for M. genitalium can be considered and treatment initiated on the basis of results of diagnostic testing (318) (see Mycoplasma genitalium). For women with persistent symptoms that are clearly attributable to cervicitis, referral to a gynecologic specialist can be considered for evaluation of noninfectious causes (e.g., cervical dysplasia or polyps) (778).
Special Considerations
HIV Infection
Women with cervicitis and HIV infection should receive the same treatment regimen as those who do not have HIV. Cervicitis can increase cervical HIV shedding, and treatment reduces HIV shedding from the cervix and thereby might reduce HIV transmission to susceptible sex partners (779–783).
Pregnancy
Diagnosis and treatment of cervicitis for pregnant women should follow treatment recommendations for chlamydia and gonorrhea (see Chlamydial Infections, Special Considerations, Pregnancy; Gonococcal Infections, Special Considerations, Pregnancy).
Contraceptive Management
According to U.S. Medical Eligibility Criteria for Contraceptive Use, 2016, leaving an IUD in place during treatment for cervicitis is advisable (58). However, current recommendations specify that an IUD should not be placed if active cervicitis is diagnosed (59).
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3733 | https://allen.in/jee/chemistry/barium | HomeJEE ChemistryBarium
Barium
Barium is a soft, silvery, alkaline earth metal with the atomic number 56, symbolised as Ba. Due to its high reactivity, barium is never found in nature as a free element. It belongs to Group 2 of the periodic table and is the fifth element in this group. Barium typically occurs in compounds such as barite (barium sulfate) and witherite (barium carbonate), often combined with sulfur, carbon, or oxygen.
1.0Introduction
Barium (Ba) is a chemical element with atomic number 56, located in Group 2 of the periodic table. It is a soft, silvery, alkaline earth metal known for its high reactivity, and as such, it is never found in its free form in nature.
Barium's story began in 1600 when Italian alchemist Vincenzo Casciarolo discovered unusual pebbles, later identified as barite (barium sulfate, BaSO₄), that would glow when heated. Barium was first recognised as a distinct element in 1774, although it wasn't isolated in its pure form until 1808 by Sir Humphry Davy through electrolysis. It is relatively light, with a density about half that of iron. Barium readily oxidises when exposed to air and reacts vigorously with water, producing barium hydroxide and releasing hydrogen gas. It also reacts with most non-metals, often forming toxic compounds.
2.0Physical properties of Barium
| Property | Details |
| Appearance | Silvery-white, shiny, malleable metal |
| Density | 3.6 grams per cubic centimetre |
| Melting Point | Approximately 725°C |
| Boiling Point | Approximately 1,640°C |
| Flame Color When Heated | Pale yellow-green |
| Atomic Number | 56 |
| Atomic Weight | 137.327 |
| Group | Group 2 (Alkaline Earth Metals) |
| Reactivity | Comparatively dense and reactive |
| Electrical Conductivity | Good electrical conductivity |
3.0Chemical Properties of Barium
Barium is highly reactive and readily oxidises when exposed to air. It reacts vigorously with acids, water, and organic solvents like carbon tetrachloride. For example, barium reacts with hydrochloric acid (HCl) to form barium chloride (BaCl₂) and hydrogen gas (H₂):
Ba + 2HCl → BaCl2 + H2
Precipitation Reactions: Barium ions (Ba²⁺) readily precipitate from solution when combined with certain anions, forming insoluble salts. For example:
With carbonate ions (CO₃²⁻), barium forms barium carbonate (BaCO₃):
Ba2+(aq) + CO32−(aq) ⇌ BaCO3(s)
With sulfate ions (SO₄²⁻), barium forms barium sulfate (BaSO₄).
With chromate ions (CrO₄²⁻), barium forms barium chromate (BaCrO₄):
Ba2+(aq)+CrO42−(aq) ⇌ BaCrO4(s)
With phosphate ions (PO₄³⁻), barium forms barium phosphate.
Cosmic Abundance: On a scale where silicon’s cosmic abundance is normalised to 10⁶ atoms, the estimated cosmic abundance of barium is relatively low at 3.7 atoms.
Oxidation State: In all known compounds, barium exhibits an oxidation state +2. This stable divalent state is due to the loss of two electrons from the outermost shell of barium atoms, forming Ba²⁺ ions. This characteristic makes it highly reactive and useful in various precipitation reactions to form solid compounds from solutions.
4.0Compounds of Barium
Barium Carbonate (BaCO₃)
Barium Carbonate (BaCO₃), also known as witherite, was discovered by William Withering in 1784. It is an inorganic white mineral with a molecular weight of 197.34 g/mol. It has a melting point of 811°C and a boiling point of 1360°C and reacts with calcium salts to produce calcium carbonate and barium sulfate. When combined with hydrochloric acid, it forms barium chloride, carbon dioxide, and water.
Uses of Barium Carbonate:
Widely used in the ceramics industry for products like plates and mugs.
Essential in making electrical ceramics such as capacitors and thermistors.
Raw material for magnetic components and optical glass production.
Used in brick manufacturing and as a flux and matting agent in various applications.
Barium Sulfate (BaSO₄)
Barium Sulfate (BaSO₄) is a white, crystalline, non-toxic compound with a molar mass of 233.43 g/mol. It has a high density of 4.49 g/mL, a melting point of 1580°C, and a boiling point of 1600°C. Insoluble in water and alcohol, it dissolves in concentrated acids. Due to its radio opacity and water insolubility, it is used for various medical and industrial applications.
Uses of Barium Sulfate:
Aids in diagnosing gastrointestinal diseases through GI imaging.
Used in oil well drilling fluids.
Acts as a filler in plastics to increase density.
Employed in oil paintings to alter consistency.
Serves as a radiopaque agent and is used in alloy manufacturing.
Measures soil pH.
Barium Chloride (BaCl₂)
Barium chloride is an ionic compound formed when barium, a Group 2 metal, donates its electrons to chlorine, a Group 17 halogen. Madam Curie obtained it in 1898 as a by-product of discovering radium.
Barium chloride is a white, hygroscopic solid with a molecular weight of 208.23 amu. It melts at 962°C, boils at 1560°C, and has a 3.856 g/cm³ density. It is highly soluble in water and methanol but insoluble in ethanol and ethyl acetate. Barium chloride is toxic, with a bitter, salty taste. Chemically, it forms a neutral aqueous solution, reacts with sulfate ions to produce barium sulfate, and reacts with oxalates and sodium hydroxide to form barium oxalate and barium hydroxide, respectively.
Uses of Barium Chloride:
In manufacturing pigments, rodenticides, and pharmaceuticals.
Purifies brine in caustic chlorine plants and is used in steel hardening and magnesium production.
Acts as a flux, water softener, and boiler compound.
Creates green flames in fireworks and is used in the leather and textile industries.
Used as a starting material for other barium salts and in water treatment plants.
In medicine, it stimulates the heart and muscles.
5.0Uses of Barium
Barium's applications are somewhat limited, but its compounds have diverse uses in various fields:
Vacuum Tubes: Barium, as a pure metal or alloyed with aluminium, removes unwanted gases (gettering) from vacuum tubes, including TV image tubes. Its low vapour pressure and reactivity with oxygen, nitrogen, carbon dioxide, and water make it ideal for this role. Additionally, barium can partially remove noble gases by absorbing them into its crystal lattice.
Alloy Production: Small amounts of barium are added to produce alloys, such as:
Steel and cast iron: Barium improves their properties.
Lead-tin soldering alloys: Barium enhances creep resistance.
Nickel alloys: Used in spark plugs.
Aluminum-silicon alloys (Silumin): Barium refines the alloy's structure.
Table of Contents
1.0Introduction
2.0Physical properties of Barium
3.0Chemical Properties of Barium
4.0Compounds of Barium
5.0Uses of Barium
Frequently Asked Questions
The most common compounds are barium sulfate (BaSO₄) and barium carbonate (BaCO₃). Barium sulfate is widely used in medical imaging, while barium carbonate is used in ceramics and as a rat poison.
Barium compounds, especially barium nitrate, are used in fireworks to produce a bright green flame.
Barium reacts vigorously with water, acids, and oxygen. It forms barium hydroxide when it reacts with water and barium chloride with hydrochloric acid. It also forms insoluble compounds when reacting with sulfate or carbonate ions.
Barium sulfate is used as a contrast agent in X-ray imaging of the digestive tract. It helps doctors visualise the oesophagus, stomach, and intestines during diagnostic procedures. |
3734 | https://www.geeksforgeeks.org/maths/division-property-of-equality/ | Division Property of Equality - GeeksforGeeks
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Division Property of Equality
Last Updated : 23 Jul, 2025
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Division Property of Equality is a fundamental concept in mathematics used to maintain balance in an equation. It states that if two numbers or expressions are equal, dividing both sides of the equation by the same non-zero number will keep them equal. This property is important when solving equations because it allows us to simplify and isolate variables. For example, if a = b, then a/c = b/c, as long as c ≠ 0.
In this article, we will discuss Division Property of Equality in detail.
Table of Content
What is Equality?
Properties of Equality
What is the Division Property of Equality?
Division Property of Equality Formula
Division Property of Equality: Practice Questions with Solutions
Practice Problems: Division Property of Equality
FAQs
What is Equality?
Equality refers to the condition of two values, expressions, or quantities being identical or equivalent in value, magnitude, or status. In mathematics, equality is represented by the symbol "=", indicating that the expressions on either side of the equals sign have the same value.
For example, in the equation: 5+3=85 + 3 = 85+3=8 This means that the sum of 5 and 3 is equal to 8.
Properties of Equality
Some of the common properties of equalities are:
Reflexive Property of Equality
Symmetric Property of Equality
Transitive Property of Equality
Substitution Property of Equality
Addition Property of Equality
Subtraction Property of Equality
Multiplication Property of Equality
Division Property of Equality
Distributive Property of Equality
In this article, we will discuss Division Property of Equality in detail.
What is the Division Property of Equality?
Division Property of Equality states that if two values or expressions are equal, then dividing both sides of the equation by the same non-zero number will not change the equality. In other words, if a = b, then dividing both a and b by the same non-zero number c will result in a/c = b/c, as long as c ≠ 0. This property is often used to solve equations where a variable needs to be isolated by eliminating a coefficient through division.
Division Property of Equality Formula
If a = b and c ≠ 0, then:
a/c = b/c
This property ensures that the equation remains balanced after division, just as it does with addition, subtraction, and multiplication.
Division Property of Equality: Practice Questions with Solutions
Question 1: Solve the equation 4x = 20.
Solution:
To isolate x, divide both sides by 4:
x = 20/4 = 5
So,x = 5.
Question 2: Solve the equation x/3 = 7.
Solution:
To solve for _x_, multiply both sides by 3:
x = 7⋅3 = 21
So,x = 21
Question 3: Solve the proportion 5/x = 15/9
Solution:
Cross-multiply to solve for x:
5 ⋅ 9 = 15 ⋅ x
⟹ 45 = 15x
Divide both sides by 15:
x = 45/15 = 3
So,x = 3
Question 4: Solve the equation(2y − 4)/5 = 6.
Solution:
First, multiply both sides by 5:
2y − 4 = 30
Then, add 4 to both sides:
2y = 34
Finally, divide both sides by 2:
y = 34/2 = 17
So,y = 17.
Question 5: If a/b = c/dand b = 4,d = 8, and c = 12, find_a_.
Solution:
Use the proportion formula:
a/4 = 12/8.
Cross-multiply to solve for a:
8a = 4 ⋅ 12
⟹ 8a = 48.
Divide both sides by 8:
a = 48/8 = 6
So,a = 6
Question 6: Solve the inequality 3x/4 > 6.
Solution:
First, multiply both sides by 4:
3x > 24
Then, divide both sides by 3:
x > 24/3 = 8
So,x > 8.
Practice Problems: Division Property of Equality
Problem 1:Solve 5 _x_ = 10 for _x_.
Problem 2:If a/6 = 4/5, find _a_.
Problem 3:Solve for z: 12z = −144
Problem 4:Solve for m: 6m = −42
Problem 5:Solve for n: 7n = 0
Problem 6:Solve for p: p/4 = 3
Problem 7:Solve for q: q/−5 = 7
Problem 8:Solve for r: r/9 = −11
Problem 9:Solve for s: 3s = 24
Problem 10: Solve for t: −4t = −28
Answer Key
x = 2
a = 4.8
z = −12
m = −7
n = 0
p = 12
q = −35
r = −99
s = 8
t = 7
Read More,
Properties of Equality
Inequalities
Multiplication Property of Equality
Subtraction Property of Equality
Transitive Property
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3735 | https://zhuanlan.zhihu.com/p/414504985 | 竹九节问题的另一种求解思路 - 知乎
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竹九节问题的另一种求解思路
首发于数学手记
切换模式
竹九节问题的另一种求解思路
飞入云端的梦
华北电力大学 电气工程硕士在读
收录于 · 数学手记
3 人赞同了该文章
《九章算术》中记载了这样一个问题：
今有竹九节，下三节容四升，上四节容三升。问中间二节欲均容各多少？
——《九章算术·均输》
传统的方法是使用等差数列进行计算：
设第 n n 节容积为 a n a_n ，则有如下方程
S 4=4 a 1+4×3 2 d=3 S_4=4a_1+\frac{4\times3}{2}d=3
S 9−S 6=9 a 1+9×8 2 d−6 a 1−6×5 2 d=4 S_9-S_6=9a_1+\frac{9\times8}{2}d-6a_1-\frac{6\times5}{2}d=4
解得 a 1=13 22,d=7 66 a_1=\frac{13}{22},d=\frac{7}{66}
因此第五节容积 a 5=a 1+4 d=67 66 a_5=a_1+4d=\frac{67}{66}
本人在整理数学史笔记的时候，想到了另一种方法，暂时没想好名字，姑且起名“杠杆平衡法”。
杠杆平衡法示意图
如图所示，前 4 4 节共 3 3 升，平均每节 3 4\frac{3}{4} 升；后 3 3 节共 4 4 升，平均每节 4 3\frac{4}{3} 升。
前 4 4 节重心距离支点（第 5 5 节） 2.5 2.5 节，后 3 3 节重心距离支点（第 5 5 节） 3 3 节。
显然两端力臂不相等，杠杆不平衡
解决方法之一是给一端的平均容量乘以对侧的力臂长。同时为保证单位不变，求和后除以两端重心总距离：
L 5=3 4×3+4 3×2.5 3+2.5=67 66 L_5=\frac{\color{blue}{\frac{3}{4}}\times\color{red}{3}+\color{red}{\frac{4}{3}}\times\color{blue}{2.5}}{\color{red}{3}+\color{blue}{2.5}}=\frac{67}{66}
结果与传统方法一致。
本方法的原理是，针对线性增加的序列，通过对差分进行加权平均，抵消不均匀的因素，从而得到正确答案。
发布于 2022-01-28 13:21
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火种
本质上是用a2.5和a8线性加权表示a5，an表示的离散点（n，an）落在同一条直线f（n）上，就可以用向量的线性加权公式了
2023-01-04
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mimimit
请问这个杠杆原理可以讲的再详细一点吗，没看懂为什么就算出来第五节容积了
2022-04-15
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3736 | https://theeffortfuleducator.com/2022/11/18/improving-multiple-choice-questioning/ | Skip to content
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Improving Multiple Choice Questioning
I am a big fan of the multiple choice question. I’ve written multiple articles discussing how to more efficiently and effectively utilize these questions in class. Probably one of my most seen tweets centers around the use of multiple choice questions:
With this article, though, I want to describe a method to more efficiently and effectively use multiple choice questioning as a tool for self-assessment and as an assessment of learning technique. It is incredibly simple and easy to implement with students in class and is also quite useful for students to utilize in their own, individual studies.
In fact, I’m using this method in class today as a form of retrieval practice.
Using multiple choice questioning traditionally involves providing students with a stem and, usually, 3-5 answer choices. Students choose which they choose to be correct and then move on to the next task. There’s nothing really wrong with this, but with some simple adjustments, students can be provided with the opportunity to think more deeply about the question and the possible answers.
So, here’s how this goes:
Instead of showing the question stem with all possible answers, I only show the stem. For instance:
Give the students 20 seconds or so to write down what answer they believe will be correct. For some question stems, like the one above, there may be multiple possible correct answers. If this is the case, students should write down as many possible correct answers as possible. Obviously, the more correct information they can recall at this point is a positive and indicates a quality level of understanding. This allows for the goal free effect. If they cannot come up with any possible correct answers, that should also communicate to students their level of understanding. Either their storage strength or retrieval strength isn’t strong enough.
Show the students the possible answers.
Instruct students to find the correct answer. Hopefully, it is one of the possible correct answers they wrote down. If that is the case, they can feel pretty confident in their knowledge (assuming it wasn’t a guess). If they didn’t have the correct possible answer written down prior to seeing the answer choices, but they can now identify the correct answer, that should also communicate to the students they can recognize but cannot recall the correct answer. Obviously, this is better than getting it completely wrong, but they may want to come back to this information for more review. Finally, if after revealing all of the possible answers, they still cannot successfully answer the question, this should communicate to the students that they have a poor understanding of the needed knowledge and this topic should probably be first on their list of to-be reviewed material.
There it is. Simple and easy. You can even manipulate an index card to have the question stem written on top of the card and then fold over the bottom to cover the possible answers so students can more easily utilize this on their own. And, of course, there are several other methods to manipulate the questions and answers…like instructing students, once seeing the possible answers, to change the question stem to make the other answers correct.
As important as this strategy is for assessing student knowledge, I believe the discussion that often accompanies it to be just as important for instructing students on how to study and what different levels of understanding ‘look like’ in action.
For instance, simply saying something like “If you can correctly recall the answer before the possible answer choices are made available for you, you can feel quite confident in your knowledge” may seem obvious to teachers, but students rarely think about their learning in terms of levels of understanding. They don’t realize if they can recall it without any retrieval cues, they can certainly recognize the correct answer when provided options.
Also, talking with students about the versatility of a strategy like this is important. There is no reason this cannot be used in other classes. It isn’t exclusive to middle school or high school or college learning. Proper learning is proper learning, no matter the level. Ask them, “How can you tailor this for other classes? How can you use this at home?”
Finally, an incredibly meaningful discussion to have with students is how this method of assessment is superior to simple rereading of notes or highlighting, which tends to be somewhat intuitively used by many students. While not as easy as just rereading notes, posing questions that require students to retrieve information is much more effective and efficient for remembering. And, again, this is valuable for learning in all classes at all levels.
So, how can you use this modification on the multiple choice question in your classroom? What would you change? How could you talk about this with your students? Let me know with a comment.
Feature image by Katerina Holmes on Pexels.com.
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7 thoughts on “Improving Multiple Choice Questioning”
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This is a really good tip – something I have never thought of trying but it makes absolute sense! Thank you.
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2. Pingback: Improving Multiple Choice Questioning — The Effortful Educator | Desde mi Salón
3. Brilliant post, Blake! Teachers in my retrieval practice sessions often wonder whether short answer or multiple choice retrieval practice questions are better, and the answer I initially learned from Pooja Agarwal was that ‘both are good’. Your post shows we can have the benefit of both at the same time! Thank you for this!
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3737 | https://en.wikibooks.org/wiki/Fool_Proof_Mathematics/CP1/Argand_diagrams | Fool Proof Mathematics/CP1/Argand diagrams - Wikibooks, open books for an open world
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1'"UNIQ--postMath-00000005-QINU"'Worked Examples:
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<Fool Proof Mathematics
Considering that a complex number is made of 2 parts, we can represent each complex number as coordinates, or vectors, on 2 planes: a real number, x-axis, and imaginary, y-axis, planes:
Using vector notation the above complex number can be represented as the vector (a b){\displaystyle a \choose b}. Therefore we can add and subtract complex numbers as vectors. Alternatively, we can represent each complex number by the angle it makes with the positive real axis and its' magnitude. Firstly, the angle is known as the argument: Argument Complex Analysis, the principal argument being in the region: −π<argz≤π{\displaystyle -\pi <\arg z\leq \pi }. The magnitude can be calculated as:|z|=a 2+b 2{\displaystyle |z|={\sqrt {a^{2}+b^{2}}}}. We can derive the modulus-argument form of a complex number given these facts and using basic trigonometric functions:
|z|=r,argz=θ z=a+b i cosθ=a r,sinθ=b r z=r(cosθ+i sinθ){\displaystyle {\begin{aligned}|z|&=r,\arg z=\theta \z&=a+bi\\cos \theta &={\dfrac {a}{r}},\sin \theta ={\dfrac {b}{r}}\z&=r(\cos \theta +i\sin \theta )\end{aligned}}}Several identities for complex numbers can be derived by using this form, which are displayed below. Due to the triviality of the derivations they have been omittedt:
|z 1 z 2|=|z 1||z 2|arg(z 1 z 2)=argz 1+argz 2|z 1 z 2|=|z 1||z 2|arg(z 1 z 2)=argz 1−argz 2{\displaystyle {\begin{aligned}|z_{1}z_{2}|&=|z_{1}||z_{2}|\\arg(z_{1}z_{2})&=\arg z_{1}+\arg z_{2}\|{\dfrac {z_{1}}{z_{2}}}|&={\dfrac {|z_{1}|}{|z_{2}|}}\\arg({\dfrac {z_{1}}{z_{2}}})&=\arg z_{1}-\arg z_{2}\end{aligned}}}Worked Examples:
[edit | edit source]
z 1=3(cos5 π 12+i sin5 π 12)z 2=4(cosπ 12+i sinπ 12){\displaystyle {\begin{aligned}z_{1}=3(\cos {\dfrac {5\pi }{12}}+i\sin {\dfrac {5\pi }{12}})\z_{2}=4(\cos {\dfrac {\pi }{12}}+i\sin {\dfrac {\pi }{12}})\\end{aligned}}}
Find |z 1 z 2|{\displaystyle |z_{1}z_{2}|}, arg(z 1 z 2){\displaystyle \arg(z_{1}z_{2})}
|z 1 z 2|=|z 1||z 2||z 1|=3,|z 2|=4∴|z 1 z 2|=12 arg(z 1 z 2)=argz 1+argz 2 argz 1=5 π 12,argz 2=π 12∴arg(z 1 z 2)=π 2{\displaystyle {\begin{aligned}|z_{1}z_{2}|&=|z_{1}||z_{2}|\|z_{1}|&=3,|z_{2}|=4\therefore |z_{1}z_{2}|=12\\arg(z_{1}z_{2})&=\arg z_{1}+\arg z_{2}\\arg z_{1}&={\dfrac {5\pi }{12}},\arg z_{2}={\dfrac {\pi }{12}}\therefore \arg(z_{1}z_{2})={\dfrac {\pi }{2}}\end{aligned}}}
Hence write z 1 z 2{\displaystyle z_{1}z_{2}}in modulus-argument form then Cartesian formː
z 1 z 2=12(cosπ 2+i sinπ 2)z 1 z 2=12(0+i)=12 i{\displaystyle {\begin{aligned}z_{1}z_{2}&=12(\cos {\dfrac {\pi }{2}}+i\sin {\dfrac {\pi }{2}})\z_{1}z_{2}&=12(0+i)=12i\end{aligned}}}
We can construct a triangle on the complex plane using 2 complex numbers where z 2−z 1{\textstyle z_{2}-z_{1}}makes the 3rd length.
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3738 | https://www.ebsco.com/research-starters/history/fleming-discovers-penicillin-molds | Research Starters Home
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Fleming Discovers Penicillin in Molds
The discovery of penicillin by Sir Alexander Fleming marked a significant turning point in the field of medicine during the early 20th century. Initially part of the burgeoning discipline of bacteriology, Fleming's research led him to observe that a mold, later identified as Penicillium notatum, was capable of destroying bacteria, specifically staphylococci, in a petri dish. This observation, made in 1928, laid the groundwork for the development of penicillin, the first widely used antibiotic, which would revolutionize treatment for bacterial infections. Despite its potential, Fleming's findings were largely overlooked until the late 1930s, when researchers Howard Florey and Ernst Boris Chain successfully isolated and tested penicillin, demonstrating its therapeutic power.
The clinical application of penicillin gained urgency during World War II, as the need for effective treatments for infected wounds became critical. Large-scale production techniques were developed, allowing for mass distribution of penicillin, which significantly reduced mortality rates from bacterial infections. This antibiotic not only transformed how common diseases like pneumonia and syphilis were treated but also enabled complex medical procedures, such as heart surgery and organ transplants, by mitigating the risk of infections. Today, penicillin is recognized as one of the greatest medical advancements of the 20th century, fundamentally changing the landscape of healthcare and the treatment of infectious diseases worldwide.
Published in: 2023
By: Neushul, Peter
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Fleming Discovers Penicillin in Molds
Date September, 1928
Alexander Fleming’s discovery of the antibiotic penicillin led to the development of a wonder drug that saved millions of lives.
Locale London, England
Key Figures
Alexander Fleming (1881-1955), English bacteriologist
Ernst Boris Chain (1906-1979), German English biochemist
Baron Florey (1898-1968), Australian pathologist
Almroth Edward Wright (1861-1947), British bacteriologist
Summary of Event
During the early twentieth century, scientists became increasingly interested in bacteriology, or the study of infectious disease. This field has since come to be called microbiology and includes the study of viruses, protozoa, fungi, and bacteria. Early bacteriologists were able to identify the sources of diseases such as pneumonia, syphilis, meningitis, gas gangrene, and tonsillitis. Prior to the discovery of antibiotics, exposure to bacteria such as streptococci, staphylococci, pneumococci, and tubercle bacilli resulted in serious and often fatal illness. Penicillin was the first of a series of twentieth century “wonder drugs” used to treat bacterial infections. This powerful antibiotic altered the lives of millions of patients who would otherwise have fallen prey to diseases caused by bacteria.
Sir Alexander Fleming began his scientific career in 1901, when he inherited a small legacy that enabled him to enter St. Mary’s Medical School in London. Fleming was a prizewinning student and a superb technician. He qualified as a doctor in 1906 and remained at St. Mary’s as a junior assistant to Sir Almroth Edward Wright, a prominent pathologist and well-known proponent of inoculation. In 1909, Fleming was one of the first to use Paul Ehrlich’s new arsenical compound, Salvarsan, for treatment of syphilis. He became renowned for his skilled administration of Salvarsan.
During World War I, Wright and Fleming joined the Royal Army Medical Corps and conducted wound research at a laboratory in Boulogne. Fleming was in charge of identifying the infecting bacteria by taking swabs from wounds before, during, and after surgery. His results showed that 90 percent of the samples contained Clostridium welchii, the anaerobic bacteria that cause gas gangrene. Although scientists could isolate the bacteria, they were uncertain as to the best method for combating diseases. Antiseptics were a known means for killing bacteria but were not always effective, especially when used in deep wounds. Wright and Fleming showed that white blood cells found in pus discharged from wounds had ingested bacteria. Fleming also demonstrated that contrary to popular opinion, when antiseptics were packed into a wound, bacteria survived in the crevasses. Antiseptics destroyed the body’s own defenses (white blood cells), allowing the remaining bacteria to create serious infection unimpeded. The horrors of bacterial infection during World War I had a lasting impact on Fleming, who decided to focus his postwar research on antibiotic substances. Fleming was convinced that the ideal antiseptic or bacteria-fighting agent should be highly active against microorganisms but harmless to the body’s own white blood cell defenses.
In 1921, Fleming observed the dissolving effect that a sample of his own nasal mucus had on bacteria growing in a petri dish. He isolated the antibiotic component of the mucus and named it “lysozyme.” Further research showed that lysozyme was also present in human blood serum, tears, saliva, pus, and milk. Fleming had discovered a universal biological protective mechanism that kills and dissolves most of the airborne bacteria that invade exposed areas of the body. He also found that lysozyme does not interfere with the body’s white blood cells. Surprisingly, the discovery of lysozyme, subsequently recognized as fundamentally important, received little attention from the scientific community. Ronald Hare, Fleming’s associate and biographer, attributes the neglect of lysozyme to “Fleming’s inability to express himself clearly and lucidly in either words or print.” The significance of lysozyme continued to be overlooked until researchers discovered its presence in white blood cells. Despite the neglect of lysozyme, Fleming continued to focus his research on antibiotics.
In September, 1928, Fleming noticed that a mold was growing in a petri dish containing strains of staphylococci and that bacteria surrounding the mold were being destroyed. It is likely that the source of the mold spores was the laboratory below Fleming’s, where mycologist C. J. La Touche was growing molds for research on allergies. Because of his interest in antibiotics, Fleming was conditioned to recognize immediately that an agent capable of dissolving staphylococci could be of great biological significance. He preserved the original culture plate and made a subculture of the mold in a tube of broth. Fleming’s mold was later identified as Penicillium notatum. Further experiments showed that the “mould juice” could be produced by several strains of Penicillium but not by other molds. The substance was nontoxic and did not interfere with the action of white blood cells.
Fleming described his findings in a paper titled “On the Antibacterial Action of Cultures of a Penicillium, with Special Reference to Their Use in the Isolation of B. Influenzae,” which appeared in the British Journal of Experimental Pathology in 1929. The unusual title refers to Fleming’s use of penicillin to isolate bacteria that were not vulnerable to penicillin. In his paper, Fleming described the mold extract and listed the sensitive bacteria. Most important, Fleming suggested that penicillin might be used in the treatment of infection. In addition to describing his experiments, Fleming also stated that the name “penicillin” would be used to refer to the mold broth filtrate. This article on penicillin came to be regarded as one of the most important medical papers ever written.
Fleming’s petri dish elicited little interest from colleagues at St. Mary’s, who were familiar with his previous work and assumed that this was an example of lysozyme being produced by a mold. Fleming knew this to be untrue, as lysozyme was incapable of destroying a pathogenic organism such as staphylococcus. Once again, Fleming’s limited ability as a writer and speaker left his audience content to shelve his latest discovery along with lysozyme. Even Ernst Boris Chain, who discovered Fleming’s paper during a literature search in 1936, “thought that Fleming had discovered a sort of mould lysozyme which, in contrast to egg white lysozyme, acted on a wide range of . . . pathogenic bacteria.”
During 1929, Fleming continued to investigate the antibiotic properties of penicillin, collecting data that clearly established the chemotherapeutic potential of penicillin. Fleming was unable to purify and concentrate penicillin adequately, and, hence, did not conduct clinical tests that could prove the effectiveness of the antibiotic in vivo. The significance of Fleming’s discovery was not recognized until 1940, when Baron Florey and Chain discovered the enormous therapeutic power of penicillin.
Significance
Fleming’s discovery of penicillin had no immediate impact on twentieth century medicine. By 1931, Fleming had discontinued work on the antibiotic and turned to the study of sulfa drugs. In 1940, Florey and Chain succeeded in concentrating and clinically testing penicillin, after which Fleming’s discovery gained enormous notoriety and he was showered with accolades and honors. In 1943, he was elected to fellowship in the Royal Society; in 1944, he was knighted; and in 1945, he received the Nobel Prize in Physiology or Medicine jointly with Florey and Chain.
Penicillin achieved particular notoriety because of World War II and the demand for an antibiotic that could halt diseases such as gas gangrene, which infected the wounds of numerous soldiers during World War I. With the help of Florey and Chain’s Oxford group, scientists at the U.S. Department of Agriculture’s Northern Regional Research Laboratory developed a highly efficient method for producing penicillin using fermentation. An excellent cornstarch medium was developed for both surface and submerged culture of penicillium molds. After an extended search, scientists also were able to isolate a more productive penicillium strain (Penicillium chrysogenum). By 1945, a strain was developed that produced five hundred times more penicillin than Fleming’s original mold.
During World War II, the U.S. Office of Scientific Research and Development’s Committee on Medical Research conducted large-scale clinical tests of penicillin on 10,838 patients. Their results provided doctors with effective methods and dosages for use of penicillin in treatment of many diseases.
American pharmaceutical companies were galvanized by the development of an efficient production technique and positive clinical tests. Corporations such as Merck, Pfizer, Squibb, and many others built large factories to mass-produce penicillin for use by the U.S. armed forces. The War Production Board increased production by allocating supplies and equipment to twenty-two U.S. chemical companies engaged in the production of the antibiotic. Penicillin prevented many of the horrendous casualties that Fleming witnessed during World War I.
Penicillin is regarded as among the greatest medical discoveries of the twentieth century. Almost every organ in the body is vulnerable to bacteria. Before penicillin, the only antimicrobial drugs available were quinine, arsenic, and sulfa drugs. Of these, only the sulfa drugs were useful for treatment of bacterial infection, but high toxicity precluded their use in many cases. With this limited arsenal, doctors were helpless as thousands died in epidemics caused by bacteria. Diseases such as pneumonia, meningitis, and syphilis are now treated with penicillin all over the world.
Penicillin and other antibiotics also had a broad impact on medicine as major procedures such as heart surgery, organ transplants, and management of severe burns became possible once the threat of bacterial infection was minimized. Fleming’s discovery brought about a revolution in medical treatment by offering an extremely effective solution to the enormous problem of infectious disease.
Bibliography
Epstein, Samuel, and Beryl Williams. Miracles from Microbes. New Brunswick, N.J.: Rutgers University Press, 1946. Presents the background of the development of several antibiotics, including penicillin and streptomycin.
Fleming, Alexander. “On the Antibacterial Action of Cultures of a Penicillium, with Special Reference to Their Use in the Isolation of B. Influenzae.” British Journal of Experimental Pathology 10 (1929): 226-236. Fleming’s first description of his discovery gives valuable insight into his experimental technique and understanding of the antibiotic potential of penicillin.
Hare, Ronald. The Birth of Penicillin, and the Disarming of Microbes. London: George Allen & Unwin, 1970. Provides a firsthand description of Fleming’s work by a scientist who worked at St. Mary’s at the time of Fleming’s discovery and was among those who witnessed Fleming’s early work with penicillin. Presents an interesting perspective on penicillin research before 1940.
Hobby, Gladys L. Penicillin: Meeting the Challenge. New Haven, Conn.: Yale University Press, 1985. Presents a good overall description of the roles played by Fleming, Florey, Chain, and numerous other scientists in the discovery, development, and eventual mass production of penicillin. Includes extensive footnotes.
Lax, Eric. The Mold in Dr. Florey’s Coat: The Story of the Penicillin Miracle. New York: Henry Holt, 2004. Relates the story of the discovery of penicillin and its development into a useful drug. Sheds light on the personalities of the scientists involved—Florey and Chain as well as Fleming. Includes bibliography and index.
Ludovici, L. J. Fleming: Discoverer of Penicillin. London: Andrew Dakers, 1952. Contemporary biography of Fleming reflects the fact that he was revered both by the author and by the general public.
Macfarlane, Gwyn. Alexander Fleming: The Man and the Myth. Cambridge, Mass.: Harvard University Press, 1984. Authoritative biography of Fleming draws on interviews as well as Fleming’s own notes to dispel somewhat the “Fleming myth” that abounds in earlier biographies.
Mateles, Richard I., ed. Penicillin: A Paradigm for Biotechnology. Chicago: Canadida Corporation, 1998. Volume reprints The History of Penicillin Production, a classic work first published in 1970, along with new chapters that address advances in penicillin research since that time. Also discusses the status of penicillin and its derivatives at the end of the twentieth century.
Maurois, André. The Life of Alexander Fleming. New York: E. P. Dutton, 1959. Authorized biography of Fleming includes two chapters on his penicillin research.
Ryan, Frank. The Forgotten Plague: How the Battle Against Tuberculosis Was Won—and Lost. Boston: Little, Brown, 1993. Written for nontechnical readers, a significant portion of this volume discusses the methodology behind the discovery and testing of many major antibiotics.
Sheehan, John C. The Enchanted Ring: The Untold Story of Penicillin. Cambridge, Mass.: MIT Press, 1982. Definitive history of the discovery, development, and marketing of penicillin, accessible to general readers.
Related Topics
Protozoa
Fungi
Bacteria
Alexander Fleming
Paul Ehrlich
World War I
White blood cell
Experiments (sociology)
Baron Florey
Royal Society
Nobel Prize
World War II
War Production Board |
3739 | https://www.youtube.com/watch?v=eR23nPNqf6A | Actually, you CAN divide by zero.
mCoding
245000 subscribers
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Description
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Posted: 3 Nov 2023
Yes, it's possible!
You've probably heard that you "can't" divide by zero, but why not? As it turns out, adding in the inverse of a number is a well-defined process in math, similar to how you can add in the solution to x^2 = -1. The result is a new number system. In this video, we find out what happens when you apply this process to add division by zero. The result is a pleasant mix of surprising and completely expected.
Notes:
Normal rules of algebra means a ring.
Topology is important too, but algebra alone is enough for 1/0.
Cup would be more correct symbol for union than plus, but this is YouTube :).
We did not need to start with the reals, adding 1/0 in any ring results in the zero ring.
― mCoding with James Murphy (
Normal algebra rules:
Localization:
Dyadic rationals:
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CHAPTERS
0:00 Intro
1:00 Localization
2:00 Zero inverse
Transcript:
Intro calculators computers and teachers throughout your
life have told you that you can't divide by zero but I'm here to tell you that you can it's true in the real numbers
division by zero isn't defined but neither is taking square root of -1 and yet mathematicians found a way to do it checking all the details might be a bit complex but ultimately the idea is just "throw in a number whose square is
-1 then follow the normal rules of algebra to figure out what this
new number system looks like" if you do this with the square root of -1
you end up with the familiar complex numbers they might have some weird properties compared to the reals but they're a perfectly
good system of numbers to work with can't we just do the same thing
with division by zero? can't we just throw in an inverse of
zero and see what happens? given that you've been told your whole
life that you can't divide by zero you might expect that for reasons
you don't fully understand you can't just throw in an inverse of a number the
same way you can throw in a solution to x^2 = -1 nope that's not the problem Localization the process of throwing in
the inverse of a number or set of numbers and seeing what
happens is called localization and it's no more difficult to do
than throwing in square root of -1 suppose you started with the integers there's no 1/2 in the
integers so let's throw it in in order to be able to multiply numbers
as usual we would also need to throw in all the multiples of 1/2 including negative ones and multiplying 1/2 with itself means we also
need 1/4 and all the multiples of a fourth and so on and so forth we need to add in anything that's
an integer divided by a power of two but then we're good to go we end up with what are called
the dyadic or binary rationals which is another weird but
perfectly valid number system if we also threw in 1/3, 1/4, 1/5, 1/6,
inverses of every positive whole number then we end up with all the
rational numbers--ratios of integers where the denominator is not zero this is a pretty standard way
to construct the rationals so what about zero inverse? Zero inverse well let's do it let's start with the real numbers
throw in an inverse of zero and just see what kind of number system pops out just like 1/2 is the unique
number such that 2 1/2 is 1 our inverse of zero is going to be the unique
number such that 0 the inverse of 0 is 1 so let's assume that's true then what? pretty quickly we see this implies that
1 equals 0 because anything 0 is 0 but wait--this is not a contradiction of course in the real
numbers 0 is different than 1 but in the real numbers x^2
can never be negative either and yet we're perfectly okay in the
complex numbers saying that i^2 is -1 just because we've discovered an unintuitive
property of this new number system that doesn't make it wrong let's just continue and see where it takes us take any number "x" in this system and note that
x is 1 x because 1 anything is that thing but because 1 = 0 this is also 0 x which
of course is 0 because 0 anything is 0 thus we've shown in our new number
system that every number equals 0 there's still no contradiction, no error in logic we've just invented what's called the Zero Ring where 0 is the only number 0 + 0 = 0, 0 - 0 = 0, 0 0 = 0, 0 / 0 = 0 and 1 is just another label for 0 feel free to check that this
number system satisfies all the normal rules of algebra it really does so it's not that you can't divide by zero it's just that if you allow it then all
numbers are zero and you get what you deserve |
3740 | https://flexbooks.ck12.org/cbook/ck-12-math-analysis-concepts/section/1.10/primary/lesson/linear-and-absolute-value-function-families-mat-aly/ | Skip to content
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1.10
Linear and Absolute Value Function Families
Written by:Raja Almukkahal | Larame Spence |
Fact-checked by:The CK-12 Editorial Team
Last Modified: Aug 01, 2025
On Tuesday, Mr. Varner's math class filed into the room, and gawked at the message on the whiteboard: "The first student to add together all of the numbers between 1 and 100 wins four free movie tickets to the theater next Friday!" Everyone grabbed a pencil and started adding: 1 + 2 + 3 + 4 + 5... No one was further than about 20 when Brian walked in, late as usual, looked at the white board for about 15 secs, and wrote: "5050" on the bottom. The surprised Mr. Varner handed Brian the tickets and told him to take his seat.
How did he come up with the answer so fast?
Linear and Absolute Value Function Families
In this Concept we will examine several families of functions. A family of functions is a set of functions whose equations have a similar form. The parent of the family is the equation in the family with the simplest form. For example, y = x2 is a parent to other functions, such as y = 2x2 - 5x + 3.
Linear Function Family
An equation is a member of the linear function family if it contains no powers of @$\begin{align}x\end{align}@$ greater than 1. For example, @$\begin{align}y = 2x\end{align}@$ and @$\begin{align}y = 2\end{align}@$ are linear equations, while @$\begin{align}y = x^2\end{align}@$ and @$\begin{align}y = \frac{1}{x}\end{align}@$ are non-linear.
Linear equations are called linear because their graphs form straight lines. As you may recall from your earlier studies of algebra, we can describe any line by its average rate of change, or slope, and its y-intercept. (In fact, it is the constant slope of a line that makes it a line!) These aspects of a line are easiest to identify if the equation of the line is written in slope-intercept form, or y = mx + b. The slope of the line is equal to the coefficient m, and the y-intercept of the line is the point (0, b).
Note that a line can be a member of a family such as the family of "linear functions", and also a member of a sub-family of linear functions with the same slope. The graphs of this subfamily will be a set of parallel lines. One particular subfamily of linear functions is the constant function subfamily. The line x = 5 is a constant function, as the function values are constant, or unchanging. The constant functions “sub-family” of linear functions is composed of functions whose graphs are horizontal lines.
Absolute Value Function Family
Let’s first consider the parent of the family: y = |x|. Because the absolute value of a number is that number’s distance from zero, all of the function values of an absolute value function will be non-negative. If x = 0, then y = |0| = 0. If x is positive, then the function value is equal to x. For example, the graph contains the points (1, 1), (2, 2), (3, 3), etc. However, when x is negative, the function value will be the opposite of the number. For example, the graph contains the points (-1, 1), (-2, 2), (-3, 3), etc. As you can see in the graph below, the absolute value function forms a “V” shape.
There are two important things to note about the graph of this kind of function. First, the absolute value graph has a vertex (a highest or lowest point) and a line of symmetry (a line that splits the function into equal and opposite 'halves'). For example, the graph of y = |x| has its vertex at (0, 0) and it is symmetric across the y-axis. Second, note that the graph is not curved, but composed of two straight portions. Every absolute value graph will take this shape, as long as the expression inside the absolute value is linear.
Piece-Wise Defined Functions
Consider again the function y = |x|. For positive x values, the graph resembles the identity function y = x. For negative x values, the graph resembles the function y = -x. We can express this relationship by defining the absolute value function in two pieces:
@$\begin{align}f(x) = \begin{cases}
-x, x < 0\
x, x \ge 0\
\end{cases}\end{align}@$
We can read this notation as: the function values are equal to -x if x is negative. The function values are equal to x if x is 0 or positive.
A piece-wise defined function does not have to represent a function that can already be written as a single equation, such as the absolute value function. For example, one “piece” may be from one function family, while another piece is from a different function family.
Examples
Example 1
Earlier, you were given a problem about Mr. Varner's math class giveaway.
He wrote the following message on the whiteboard: "The first student to add together all of the numbers between 1 and 100 wins four free movie tickets to the theater next Friday!" Everyone grabbed a pencil and started adding: 1 + 2 + 3 + 4 + 5... No one was further than about 20 when Brian walked in, looked at the white board for about 15 secs, and wrote: "5050" on the bottom. Mr. Varner handed Brian the tickets and told him to take his seat. How did he come up with the answer so fast?
Brian recognized that he didn't need to add each of the numbers individually, only the pairs: 100 + 0 = 100, so does 99 + 1, and 98 + 2, and so on. Since there are 50 pairs of 100, that adds up to 5000. The only number without a pair is 50, so it gets added to the total: 5050.
Function families represent this same sort of time-saver. By recognizing common bits of information and combining them in different ways, we can 'automate' some very complex-seeming processes.
Example 2
Identify the slope and the y-intercept of each line.
@$\begin{align}y=\frac{2}{3}x-1\end{align}@$
This line has slope (2/3) and the y-intercept is the point (0, -1).
@$\begin{align}y=5\end{align}@$
This is a horizontal line. The slope is 0, and the y-intercept is (0,5).
@$\begin{align}x=-2\end{align}@$
This is a vertical line. The slope is undefined, and the line does not cross the y-axis. (Note that this line is not a function!)
@$\begin{align}y=\frac{2}{3}x+3\end{align}@$
The slope of this line is 2/3, and the y-intercept is the point (0, 3).
Example 3
Graph the following: @$\begin{align}y=|2x-1|\end{align}@$ and @$\begin{align}y=|2x^2 -1|\end{align}@$.
The graph of @$\begin{align}y=|2x-1|\end{align}@$ makes a “V” shape, much like @$\begin{align}y=|x|\end{align}@$.
The function inside the absolute value, 2x+1, is linear, so the graph is composed of straight lines.
The graph of @$\begin{align}y=|2x^2-1|\end{align}@$ is curved, and it does not have a single vertex, but two “cusps.”
The function inside the absolute value is NOT linear, therefore the graph contains curves.
Example 4
Sketch a graph of the function
@$\begin{align}f(x) = \begin{cases}
x^2, x < 2\
x + 3, x \ge 2\
\end{cases}\end{align}@$
It is important to note that the pieces of a piece-wise defined function may or may not meet up. For example, in the graph of f(x) above, the function value is 4 at x = -2, but the piece of the graph that is defined by x + 3 is headed to the y value of 1. Therefore the two pieces do not meet.
Example 5
Find the x-intercepts of the function @$\begin{align}f(x) = 8|x - 7| - 64\end{align}@$.
To find the x-intercepts, set f(x) equal to 0, and solve for x:
@$\begin{align}0 = 8|x - 7| - 64\end{align}@$
@$\begin{align}64 = 8|x - 7|\end{align}@$
@$\begin{align}8 = |x - 7|\end{align}@$
@$\begin{align}8 = (x-7)\end{align}@$or@$\begin{align}8 = -(x-7)\end{align}@$
@$\begin{align}15 = x\end{align}@$ or @$\begin{align}-1 = x\end{align}@$
@$\begin{align}\therefore\end{align}@$ the x-intercepts are 15 and -1
Example 6
What is the graph of @$\begin{align}y = |x|\end{align}@$? How is that graph related to the graph of @$\begin{align}y = a|x-h| + k\end{align}@$? What happens to the graph of @$\begin{align}y = |x|\end{align}@$ when the equation changes to @$\begin{align}y = |x| - 5\end{align}@$?
The graph of @$\begin{align}y = |x|\end{align}@$ is shown below. You can either use a graphing tool, or plot points, noting that every positive x has a matching y, and every negative x matches with its positive equivalent as y. @$\begin{align}y = |x|\end{align}@$ is the simplest example of a graph in the absolute value function family, of which @$\begin{align}y = a|x-h| + k\end{align}@$ is the parent. Changes to @$\begin{align}a\end{align}@$, @$\begin{align}h\end{align}@$, and @$\begin{align}k\end{align}@$ shift the graph of @$\begin{align}y = |x|\end{align}@$ in different ways.
The graph of @$\begin{align}y = |x| - 5\end{align}@$ is below. It is clear that the -5 after the absolute value causes the graph to shift down 5 places.
Review
For questions 1-5, identify the family that each function belongs to.
@$\begin{align}y = |x - 7|\end{align}@$
@$\begin{align}y = 3x - 4\end{align}@$
@$\begin{align}f(x) = |x^2|\end{align}@$
@$\begin{align}|x| - 2 = y\end{align}@$
@$\begin{align}f(x) = x + \frac{3x}{2}\end{align}@$
Graph the following piecewise function by hand: @$\begin{align}f(x) = \begin{cases} x, x\geq 0\ -x, x < 0\ \end{cases}\end{align}@$
On your graphing calculator, graph the function @$\begin{align} f(x) = |x| -2\end{align}@$, and answer the following questions: a. What is the shape of the graph? b. Compare the graph to the graph in the problem above. What is the difference between the two graphs? c. What is the slope of the two lines that create the graph?
For each equation that follows, identify the coordinates of the vertex of the graph, without actually graphing.
@$\begin{align}f(x) = |6x|\end{align}@$
@$\begin{align}f(x) = |x - 6| + 8\end{align}@$
@$\begin{align}f(x) = |x +7| - 8\end{align}@$
@$\begin{align}f(x) = |x + 5|\end{align}@$
The graph of @$\begin{align}p(x) = |x|\end{align}@$ is shown below. If @$\begin{align}t(x) = -|x|-3\end{align}@$, how will the graph of @$\begin{align}t(x)\end{align}@$ be different from the graph of @$\begin{align}p(x)\end{align}@$?
Graph the absolute value equation, create your own table to justify values: @$\begin{align}f(x) = |x - 3|\end{align}@$
Graph the absolute value equation, create your own table to justify values: @$\begin{align}g(x) = |x+3|\end{align}@$
Identify the parent function for each set of linear functions. Graph each set of functions using a graphing calculator. Identify similarities and differences of each set.
a. @$\begin{align}f(x) = x -7\end{align}@$, b. @$\begin{align}f(x) = x - 2\end{align}@$, c. @$\begin{align}f(x) = x + 1\end{align}@$, d. @$\begin{align}f(x) = x + 5\end{align}@$,
e. @$\begin{align}f(x) = x + 10\end{align}@$
Parent Function:
Similarities:
Differences:
a. @$\begin{align}f(x) = \frac{2}{11}x\end{align}@$, b. @$\begin{align}f(x) = \frac{1}{2}x\end{align}@$, c. @$\begin{align}f(x) = \frac{2}{3}x\end{align}@$
Parent Function:
Similarities:
Differences:
a. @$\begin{align}f(x) = x -7\end{align}@$, b. @$\begin{align}f(x) = 2x\end{align}@$, c. @$\begin{align}f(x) = 4x\end{align}@$, d. @$\begin{align}f(x) = 2x + 5\end{align}@$, e. @$\begin{align}f(x) = 6x - 10\end{align}@$
Parent Function:
Similarities:
Differences:
Use the standard form of a linear equation: @$\begin{align}f(x) = ax + b\end{align}@$ and your investigations above to help you answer the following questions.
How does the a value affect the graph?
How does the b value affect the graph?
How are the domain values similar/different?
How are the range values similar/different?
Does the a and/or b value affect the domain?
Does the a and/or b value affect the range?
Review (Answers)
Click HERE to see the answer key or go to the Table of Contents and click on the Answer Key under the 'Other Versions' option.
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Sun, 23 Jun 2019 15:41:36 GMT
Core (Britt's page)
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Translation - Protein synthesis# The coupling of these two processes, and even mRNA degradation, is facilitated not only because transcription and translation happen in the same compartment but also because both of the processes happen in the same direction - synthesis of the RNA transcript happens in the 5' to 3' direction and translation reads the transcript in the 5' to 3' direction.
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3742 | https://www.pfizer.com/news/articles/what%E2%80%99s-y-chromosome-handed-down-father-son | Skip to main content
What’s on the Y Chromosome Handed Down From Father to Son?
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Among the many things parents hand down to their children are 23 pairs of chromosomes – those thread-like structures in the nucleus of every cell containing the genetic instructions for every person. We inherit a set of 23 chromosomes from our mothers and another set of 23 from our fathers.
One of those pairs are the chromosomes that determine the biological sex of a child – girls have an XX pair and boys have an XY pair, with very rare exceptions in certain disorders. Females always pass an X chromosome onto their offspring. If the father passes on an X chromosome, the baby will be genetically female, and if the father passes on a Y chromosome, the baby will be genetically male.
During that process of sexual reproduction, those two inherited chromosomes will “recombine” their genes, meaning that the chromosomes exchange genetic information with each other. Over the short term, this process of cross-talk means that the son or daughter has combinations of traits that aren’t necessarily identical to either parent. And over the long term, that genetic diversity helps to weed out traits that may be disadvantageous to a population.
In commemoration of Father’s Day, here’s a look at the Y chromosome and the role it plays in deciphering ancestry.
What’s on the Y Chromosome?
Though a short segment of the X and Y chromosomes are identical, gene sequencing has determined that more than 95 percent of the Y chromosome is unique to males – known as the male-specific region of the Y, or MSY. In fact, this region is so different from the X chromosome that the often-cited fact that humans are 99.9 percent genetically identical only applies when comparing people of the same sex.
Since only men have the Y chromosome, the genes on the MSY are thought to be involved in the determination of sex and development specific to males, including male fertility.
This part of the Y chromosome does not recombine sexually with the X chromosome during reproduction — meaning that when a father contributes a Y chromosome to the process of sexual reproduction, most of the genes on that chromosome don’t “mix” with the genes on the X. In genetic terms, they’re passed on asexually.
While genetic recombination allows for the expression of diverse traits in offspring, the mostly linear transmission of the Y chromosome isn’t necessarily a bad thing. In fact, that trait has allowed scientists to understand the history of male ancestry.
Using the Y Chromosome to Investigate Ancestry
The DNA in mitochondria – an organelle in the cell that produces energy – are used for genetic studies involving maternal lineage because the mitochondria has its own DNA distinct from the nuclear DNA, and we inherit mitochondrial DNA from our mothers only.
But since most of the Y chromosome is passed on without recombination, the DNA on that chromosome provides a genetic history of a man’s paternal ancestral line.
An example of this is a 2003 study that examined the genetic legacy of the Mongols, whose territory once spanned the largest contiguous land empire in human history.
The researchers studied the Y chromosomes of more than 2,100 men throughout a wide swath of Asia and found features that showed up in about 8 percent of the men in the region, from the Pacific rim to the Caspian Sea, even though it shows up in only 0.5 percent of men overall in the world. The variation in the lineage suggested that this feature on the Y chromosome originated in Mongolia about 1,000 years ago.
The researchers concluded that the rapid spread of that genetic feature on the Y chromosome could not have happened by chance but rather was likely the result of it being spread by the male descendants of Genghis Khan, the leader of the Mongol empire.
While it wasn’t possible to test Genghis Khan’s Y chromosome directly, researchers tested the Y chromosomes of the Hazaras of Pakistan, whose genealogy suggests they are direct male-line descendants of Genghis Khan. The study concluded that if the sample of more than 2,100 men was representative of the region, there would be about 16 million men with this Y chromosome feature – all presumably progeny of Genghis Khan.
Other studies have also identified other highly successful male lineages that started in China and Ireland.
And while not every father may be so genetically prosperous, what makes these studies possible is the thread-like substance that is the Y chromosome.
P
Pfizer Staff
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3743 | https://testbook.com/question-answer/when-sampling-is-done-without-replacement-then-sta--5feafbf1c605c99d67ea51bc | [Solved] When sampling is done without replacement then standard erro
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Variance and Standard Deviation
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When sampling is done without replacement then standard error of mean is:
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UGC Paper 2: Commerce_17th Oct 2020 Shift 1
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S E X¯=σ n
S E X¯=σ n N−n N−1
S E X¯=σ n N−1 N−n
S E X¯=σ n 1−n N
Answer (Detailed Solution Below)
Option 2 : S E X¯=σ n N−n N−1
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Standard Error:
The standard error of the mean is the standard deviation of the sampling distribution of the mean.
In other words, it is the standard deviation of a large number of sample means of the same sample size drawn from the same population.
The term standard error of the mean is commonly (though imprecisely) shortened to just standard error.
T he "standard error of the mean" refers to the standard deviation of the distribution of sample means taken from a population.
The smaller the standard error, the more representative the sample will be of the overall population.
In sampling without replacement, the formula for the standard deviation of all sample means for samples of size n must be modified by including a finite population correction. The formula becomes:where N is the population size.
Thus, option 2 is the correct answer.
When sampling with replacement the standard deviation of all sample means equals the standard deviation of the population divided by the square root of the sample size when sampling with replacement.
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More Variance and Standard Deviation Questions
Q1.The standard deviation of 100 observations is 10. If 5 is added to each observation and then divided by 20, then what will be the new standard deviation?
Q2.Which one of the following statements is correct?
Q3.Which one of the following is correct?
Q4.Let Σ i=1 9 x i 2=885 If M is the mean and σ is the standard deviation of x1, x2, x3.....x9 then what is the value of M2+σ2?
Q5.Let a, b∈ R. Let the mean and the variance of 6 observations –3, 4, 7, –6, a, b be 2 and 23, respectively. The mean deviation about the mean of these 6 observations is :
Q6.The standard deviation of the first 10 natural numbers is 3.028. What will be the standard deviation of the first 20 natural numbers?
Q7.If the sum of 10 observations is 12 and the sum of their squares is 18, then the standard deviation is:
Q8.The fourth central moment of a mesokurtic distribution is 243. Its standard deviation is:
Q9.The variance of 20 observations is 5. If each observation is multiplied by 3 then variance for obtained observations is:
Q10.What is the variance of first five positive integers?
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Q1.The mean of a data set comprising of 10 observations is 20. If one of the observations whose value is 20 is deleted and two new observations with values 14 and 15 are added to the data, then the mean of the new data set is
Q2.What is the arithmetic mean of 82,92,102,...,152?
Q3.The standard deviation of 100 observations is 10. If 5 is added to each observation and then divided by 20, then what will be the new standard deviation?
Q4.The arithmetic mean of 100 observations is 50. If 5 is subtracted from each observation and then divided by 20, then what is the new arithmetic mean?
Q5.Which one of the following statements is correct?
Q6.Which one of the following is correct?
Q7.The most appropriate graphical representation of the given frequency distribution is
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Q9.What is the height of the class?
Q10.What is the total number of students whose height is less than or equal to 165 cm?
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3744 | https://www.nsta.org/quantum-magazine-math-and-science?srsltid=AfmBOoqAjMcHC1i1OVwTuVHwlxnz24-mtA1Em6f45ucIPWcF_mdjd_Hl | Quantum: The Magazine of Math and Science
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Article Index
The articles are listed in alphabetical order by title. You can use your web browser’s “Find” function to search by author, title, description, department, or date (see the format used below).
Note: In addition to the articles listed below, each issue of Quantum contained math and physics problems, “brainteasers,” a solution section, a department called Gallery Q that tied works of art to topics in the magazine, a chess column (in the early years), a science crossword puzzle (from November/December 1992 on), and a bulletin board.
A
About the Triangle (it may be the simplest polygon, but it gave rise to an entire branch of mathematics), Mar/Apr00, p31 (Kaleidoscope)
Algebraic and Transcendental Numbers (2/3, e, pi, the square root of 2—things like that), N. Feldman, Jul/Aug00, p22 (Feature)
An Act of Divine Providence (Kepler excerpt), Yuly Danilov, May/Jun93, p41 (Anthology)
Adding Angles in Three Dimensions (taking a plane theorem into the realm of polyhedrons), A. Shirshov and A. Nikitin, May/Jun97, p46 (At the Blackboard)
The Advent of Radio (why radio was invented when it was), Pavel Bliokh, Nov/Dec96, p4 (Feature)
Adventures Among Pt-sets (math challenge), George Berzsenyi, Mar/Apr91, p55 (Contest)
The Adventures of Hans Pfaall and Fatty Pyecraft (questionable physics in stories by Poe and Wells), V. Nevgod, Jan90, p14 (Quantum Smiles)
Against the Current (evaluating fluid resistance), Alexander Mitrofanov, May/Jun96, p22 (Feature)
AHSME-AIME-USAMO-IMO (introduction to math competitions), Nov/Dec90, p52 (Happenings)
Airplanes in Ozone (effect of high-flying aircraft on stratospheric ozone), Albert Stasenko, May/Jun95, p20 (Feature)
Alexandrian Astronomy Today (the method found by Eratosthenes in the third century B.C. still works), Case Rijsdijk, Sep/Oct99, p35 (At the Blackboard)
All Bent Out of Shape (a look at many kinds of deformation), Sep/Oct95, p32 (Kaleidoscope)
All Sorts of Sorting (classification algorithms), P. Blekher and M. Kelbert, Jul/Aug97, p12 (Feature)
Always a New Face to Show (theorems about polyhedrons), May/Jun93, p32 (Kaleidoscope)
The Amazing Paraboloid (double refraction and energy redistribution), M. I. Feingold, Jul/Aug94, p40 (At the Blackboard)
The A-maze-ing Rubik’s Cube (a new variation), Vladimir Dubrovsky, Sep/Oct91, p64 (Toy Store)
The American Mathematics Correspondence School (I. M. Gelfand’s project for high school students), Nov/Dec93, p51 (Happenings)
The American Regions Mathematics League (summer competition), Mark Saul, May90, p56 (Happenings)
American Team Garners Six Gold Medals at 35th IMO (report on International Mathematical Olympiad), Nov/Dec94, p52 (Happenings)
Amusing Electrolysis (current thinking in chemistry), N. Paravyan, May/Jun98, p41 (In the Lab)
An Ant on a Tin Can (finding the shortest path from A to B), Igor Akulich, Sep/Oct97, p50 (At the Blackboard)
The Ancient Numbers Pi and Tau (approximating pi and using the golden ratio tau), Jan/Feb91, p28 (Kaleidoscope)
Animal Magnetism (physics challenge), Arthur Eisenkraft and Larry D. Kirkpatrick, May/Jun93, p28 (Physics Contest)
A. N. Kolmogorov (biographical sketch), Jan90, p38 (Innovators)
Anniversaries (satellites and science reform), Gerry Wheeler, Nov/Dec97, p2 (Front Matter)
The Annual Puzzle Party (report and samples), Anatoly Kalinin, Jul/Aug94, p56 (Toy Store)
Another Perpetual Motion Project? (a feasibility foray), A. Stasenko, Jan/Feb99, p39 (At the Blackboard)
The Anthropic Principle (humans and the universe), A. Kuzin, Jan/Feb99, p4 (Feature)
Anticipating Future Things (science education in 2044), Bill G. Aldridge, Jul/Aug94, p2 (Publisher’s Page)
“Are We Almost There, Captain?” (Columbus’s geographical problems), Glenn M. Edwards, Sep/Oct92, p52 (Looking Back)
Are You Relatively Sure? (relativity in its many forms), A. Leonovich, Sep/Oct96, p32 (Kaleidoscope)
Arithmetic Obstacles (analyzing the possibility of moving from one position to another), N. Vaguten, Jul/Aug99, p4 (Feature)
Arithmetic on Graph Paper (planar numbers, gnomons, Pythagorean triples, and triangular numbers), Semyon Gindikin, Mar/Apr95, p49 (At the Blackboard)
Around and Around She Goes (the motion of merry-go-rounds), Arthur Eisenkraft and Larry D. Kirkpatrick, Mar/Apr98, p30 (Physics Contest)
An Arresting Sight (the stroboscopic effect), V. Uteshev, Jul/Aug01, p30 (Now Showing)
As Easy as (a, b, c)? (Pythagorean triples), S. M. Voronin and A. G. Kulagin, Jan/Feb99, p34 (Feature)
The Ashen Light of the Moon (the how, when, and why of a faint lunar glow), Alexey Byalko, Sep/Oct96, p40 (In the Open Air)
The “Assayer” Weighs the Facts (Galileo excerpt), Yuly Danilov, Nov/Dec92, p43 (Anthology)
Atlantic Crossings (graphical method for motion problems), A. Rozental, Jul/Aug93, p46 (In Your Head)
Atmospherics (physics of the Earth’s atmosphere), A. V. Byalko, Mar/Apr91, p12 (Feature)
At Sixes and Sevens (math challenge), George Berzsenyi, May90, p35 (Contest)
Atwood’s Marvelous Machines (physics challenge), Arthur Eisenkraft and Larry D. Kirkpatrick, Jul/Aug93, p42 (Physics Contest)
Auxiliary Polynomials (solving equations with polynomials), L. D. Kurlyandchik and S. V. Fomin, Sep/Oct98, p42 (At the Blackboard)
Ax by Sea (actually, ax + by = c: approaches to Diophantine equations), Boris Kordemsky, Nov/Dec96, p22 (Feature)
B
Baby, It’s Cold Out There! (“cosmic cold” and thermal radiation), Albert Stasenko, Mar/Apr92, p12 (Feature)
Backtracking to Faraday’s Law (threshold voltage in electrolysis), Alexey Byalko, Jan/Feb94, p20 (Feature)
Bad Milk (a dynamic system gone sour), Dr. Mu, Sep/Oct97, p63 (Cowculations)
Ballpark Estimates (Fermi problems), David Halliday, May90, p30 (In Your Head)
Barn Again (a smooth move), Dr. Mu, Jul/Aug98, p62 (Cowculations)
Batteries and Bulbs (progressively more complicated circuits), Larry D. Kirkpatrick and Arthur Eisenkraft, Jul/Aug00, p32 (Physics Contest)
The Beetle and the Rubber Band (mind-stretching problem), Alexander A. Pukhov, Mar/Apr94, p42 (At the Blackboard)
Behind the Mirror (measuring the thickness of the reflecting layer), N. M. Rostovtsev, Jan/Feb96, p37 (In the Lab)
Behind the Scenes at the IMO (report and IMO questions), Vladimir Dubrovsky, Mar/Apr93, p53 (Happenings)
Bell Curve? What Bell Curve? (response to the May/June 1995 Publisher’s Page), Paul Horwitz, Jan/Feb96, p27 (Feedback)
Below Absolute Zero (who said it’s impossible?), Henry D. Schreiber, Jan/Feb97, p23 (Feature)
Be More Clever Than Chris! (Columbus’s egg trick), Yakov Perelman, Sep/Oct92, p54 (insert)
Bend This Sheet (developable surfaces), Dmitry Fuchs, Jan90, p16 (Feature)
Beyond the Reach of Ohm’s Law (interesting phenomena where the law doesn’t apply), Sergey Murzin, Mikhail Trunin, and Dmitry Shovkun, Nov/Dec94, p24 (Feature)
Billiard Math (reflections on simple optical reflection), Anatoly Savin, Nov/Dec96, p28 (Kaleidoscope)
The Birth of Low-temperature Physics (properties of helium near absolute zero), A. Buzdin and V. Tugushev, Jan/Feb01, p12 (Feature)
Bobbing for Knowledge (experiments with a hollow plastic ball), Pavel Kanayev, Mar/Apr95, p30 (In the Lab)
Bohr’s Quantum Leap (history of atomic theory), A. Korzhuyev, Jan/Feb99, p42 (Looking Back)
Boiling Liquid (how a bubble chamber works), A. Borovoi, Mar/Apr00, p54 (In the Lab)
Boing, Boing, Boing … (what happens after the second bounce, and the third, ...), Arthur Eisenkraft and Larry D. Kirkpatrick, Jul/Aug96, p30 (Physics Contest)
The Bombs Bursting in Air (a look at sample problems and their social significance), Arthur Eisenkraft and Larry D. Kirkpatrick, Sep/Oct96, p34 (Physics Contest)
Borsuk’s Problem (n-dimensionality meets combinatorics), Arkady Skopenkov, Sep/Oct96, p16 (Feature)
The Borsuk–Ulam Theory (horsing around with continuous functions on a circle), M. Krein and A. Nudelman, Jul/Aug00, p16 (Feature)
Botanical Geometry (triangular “flowers” and Torricelli circles), Sep/Oct90, p32 (Kaleidoscope)
Bottling Milk (so many bottle sizes!), Dr. Mu, Mar/Apr97, p63 (Cowculations)
The Bounding Main (physics of sea swells), Ivan Vorobyov, May/Jun94, p20 (Feature)
Boy-oh-buoyancy! (problems in fluid statics), Alexander Buzdin and Sergey Krotov, Sep/Oct90, p27 (Feature)
Braids and Knots (primer on knot theory), Alexey Sosinsky, Jan/Feb95, p10 (Feature)
Breakfast of Champions (acquiring a full set of baseball cards in boxes of cereal), Don Piele, Mar/Apr01, p55 (Informatics)
Breaking Up is Hard to Do (nuclear fission), Arthur Eisenkraft and Larry D. Kirkpatrick, Sep/Oct99, p30 (Physics Contest)
A Brewer and Two Doctors (origins of the law of conservation of energy), Gennady Myakishev, May/Jun96, p43 (Looking Back)
Bridging the Gap (between classical and quantum mechanics, teacher and student), Bill G. Aldridge, Nov/Dec95, p2 (Publisher’s Page)
A Brilliant Idea (poem), David Arns, Jul/Aug97, p33
Brocard Points (properties of points inside a triangle), V. Prasolov, Mar/Apr01, p22 (At the Blackboard)
Bubbles in Puddles (their size, shape, and longevity), Alexander Mitrofanov, Jul/Aug95, p4 (Feature)
A Burst of Green (mathematics of plant growth), Alexander Vedenov and Oleg Ivanov, May/Jun93, p10 (Feature)
Bushels of Pairs (graphical primer), Andrey N. Kolmogorov, Nov/Dec93, p4 (Feature)
But What Does It Mean? (the thinking behind the symbols), Bill G. Aldridge, Mar/Apr96, p2 (Publisher’s Page)
C
Calculating Pi (the contribution of Christiaan Huygens), Valery Vavilov, May/Jun92, p44 (Looking Back)
Calculus and Inequalities (three problems, one method), V. Ovsienko, Jan/Feb01, p38 (At the Blackboard)
Calendar Calculations (“Doomsday” rule), John Conway, Jan/Feb91, p46 (Mathematical Surprises)
“Can-do” Competitors in Canberra (report on the XXVI International Physics Olympiad), Larry D. Kirkpatrick, Nov/Dec95, p53 (Happenings)
Canopies and Bottom-flowing Streams (a spoonful of physics), Ivan Vorobyov, Jul/Aug95, p45 (In the Lab)
Cantor Cheese (recursive designs), Don Piele, Jan/Feb00, p53 (Informatics)
Can White Be Blacker Than Black? (black-body demonstration), V. V. Mayer, Sep/Oct92, p23 (In the Lab)
Can You Carry Water in a Sieve? (investigations of the surface layer), A. Dozorov, Jul/Aug00, p44 (In the Lab)
“Can You Hear Me?” (some thoughts on the history of human communication), Bill G. Aldridge, Jul/Aug96, p2 (Publisher’s Page)
Can You See the Magnetic Field? (using a TV as a detector), Alexander Mitrofanov, Jul/Aug97, p18 (Feature)
Can You Trace the Rays? (ray diagrams), A. Leonovich, May/Jun99, p28 (Kaleidoscope)
A Cardioid for a Mushroom Picker (the curvy path of a lost forager), S. Bogdanov, Jul/Aug99, p34 (At the Blackboard)
Card Party (a seating problem), Don Piele, Jul/Aug01, p50 (Informatics)
Carl Friedrich Gauss, Part I (a biographical sketch of a prince of mathematics), S. Gindinkin, Nov/Dec99, p14 (Feature)
Carl Friedrich Gauss, Part II, S. Gindinkin, Jan/Feb00, p10 (Feature)
The Case of the Mythical Beast (Holmes and the Helmholtz resonator), Roman Vinokur, Nov/Dec93, p10 (Feature)
Catch as Catch Can (the theory of gravitational capture), Y. Osipov, Jan/Feb92, p38 (Looking Back)
Catching Up on Rays and Waves (a rhapsody on wavelengths and the Stefan–Boltzmann law), Albert Stasenko, Jul/Aug00, p10 (Feature)
Cauchy and Induction (a simpler proof of his famous inequality), Y. Solovyov, Jan/Feb01, p37 (At the Blackboard)
Caught in the Web (interesting World Wide Web sites), Sep/Oct95, p53 (Happenings)
The Century of the Cycloid (historical patterns), S. G. Gindikin, Mar/Apr99, p36 (Looking Back)
A Chebyshev Polyplayground (recurrence relations applied to a famous set of formulas), N. Vasilyev and A. Zelevinsky, Sep/Oct99, p20 (Feature)
Chebyshev’s Problem (polynomials of least deviation from zero), S. Tabachnikov and S. Gashkov, Sep/Oct94, p12 (Feature)
The Chemical Elements (curiosities from the periodic table), Sheldon Lee Glashow, May90, p14 (Getting to Know …)
Chess Puzzles and Real Chess (what happens when the two worlds intersect), Yevgeny Gik, Sep/Oct96, p64 (Toy Store)
Chopping Up Pick’s Theorem (triangulation and polygonal partition), Nikolay Vasilyev, Jan/Feb94, p49 (At the Blackboard)
Chores (how long will they take on your theoretical farm?), Don Piele, Jul/Aug00, p55 (Informatics)
Circular Reasoning (inscribed angles), Mark Saul and Benji Fisher, Nov/Dec97, p34 (Gradus ad Parnassum)
A Circuitous Route (“relevance” in science education), Bill G. Aldridge, Jul/Aug93, p2 (Publisher’s Page)
Circuits and Symmetry (cutting down on algebra), Gary Haardeng-Pedersen, Jul/Aug95, p28 (At the Blackboard)
Circumcircles to the Rescue! (useful technique for solving certain problems), D. F. Izaak, Jan/Feb91, p32 (At the Blackboard)
The Clamshell Mirrors (physics challenge), Arthur Eisenkraft and Larry D. Kirkpatrick, Mar/Apr92, p48 (Physics Contest)
Clarity, Reality, and the Art of Photography (an examination of “depth of field”), Mark L. Biermann, Sep/Oct95, p26 (Feature)
Classic Writings from the History of Science (Plutarch’s “Concerning the Face Which Appears in the Orb of the Moon”), Yuly Danilov, Mar/Apr92, p42 (Anthology)
Click, click, click … (physics challenge), Arthur Eisenkraft and Larry Kirkpatrick, Sep/Oct90, p41 (Contest)
A Clock Wound for All Time (the Earth as a timepiece—can it measure its own age?), V. I. Kuznetsov, May/Jun97, p26 (Feature)
Cloud Formulations (a moist air mass went over the mountain), Arthur Eisenkraft and Larry D. Kirkpatrick, Jan/Feb95, p36 (Physics Contest)
Coalescing Droplets (surface tension and drops), A. Varlamov, May/Jun99, p26 (At the Blackboard)
Cold Boiling (just add water), S. Krotov and A. Chernoutsan, Jan/Feb99, p33 (In the Lab)
Colder Means Slower (the Arrhenius equation), Henry D. Schreiber, Jul/Aug97, p4 (Feature)
A Collapsible Saddle (model of a hyperbolic paraboloid), Vladimir Dubrovsky, Jan/Feb91, p56 (Toy Store)
Color Creation (partial “rainbows” in oil slicks), Arthur Eisenkraft and Larry D. Kirkpatrick, May/Jun97, p36 (Physics Contest)
Combinatorics-polynomials-probability (permutations and binomial coefficients), Nikolay Vasilyev and Victor Gutenmacher, Mar/Apr93, p18 (At the Blackboard)
Come, Bossy (rounding up the herd), Dr. Mu, May/Jun98, p63 (Cowculations)
Competitive Computing in Stockholm (1994 International Olympiad in Informatics), Donald T. Piele, Nov/Dec94, p53 (Happenings)
The Complete Quadrilateral (definition and peculiar properties), I. Sharygin, Jul/Aug97, p28 (Kaleidoscope)
Completing a Tetrahedron (a geometrical trick of the trade), I. F. Sharygin, Jul/Aug99, p46 (At the Blackboard)
Completing the Square (quadratic equations), Mark Saul and Titu Andreescu, Nov/Dec98, p35 (Gradus ad Parnassum)
The Conductor of a Set (an old problem revisited and feedback), George Berzsenyi, May/Jun96, p37 (Math Investigations)
Confessions of a Clock Lover (the cosmic consequences of switching hands), V. M. Babovic, Sep/Oct96, p44 (Horological Surprises)
Considerations of Continuity (wobbly chair and other problems), S. L. Tabachnikov, May90, p8 (Feature)
Constructing Quadratic Solutions (a novel use for compass and straightedge), A. A. Presman, Jan/Feb98, p42 (At the Blackboard)
Constructing Triangles from Three Given Parts (186 problems), George Berzsenyi, Jul/Aug94, p30 (Math Investigations)
Constructing Triangles from Three Located Points (20 out of 139 problems still need solving), George Berzsenyi, Sep/Oct94, p54 (Math Investigations)
Construction Program (regular polygons, Euler’s function, and Fermat numbers), Alexander Kirillov, Mar/Apr96, p10 (Feature)
Constructions with Compass Alone (Mohr–Mascheroni theorem), Dmitry Fuchs, May90, 47 (At the Blackboard)
Contact (number bit patterns), Dr. Mu, Nov/Dec98, p52 (Cowculations)
Contented Cows (finding all ways to sum digits in a number to zero), Dr. Mu, Jul/Aug99, p26 (Cowculations)
Continued Fractions (when close enough is good enough), Y. Nesterenko and E. Nikishin, Jan/Feb00, p22 (Feature)
Convection and Displacement Currents (nature of electricity), V. Dukov, Mar/Apr99, p4 (Feature)
A Conversation in a Streetcar (“lucky tickets” in Leningrad), A. Savin and L. Fink, Mar/Apr92, p23 (In Your Head)
Cooled by the Light (photonic refrigeration), I. Vorobyov, Sep/Oct93, p20 (Feature)
Cool Vibrations (fun with oscillations), Arthur Eisenkraft and Larry D. Kirkpatrick, Sep/Oct97, p46 (Physics Contest)
Core Dynamics (transformers explained), A. Dozorov, Mar/Apr99, p14 (Feature)
Counting Problems in Finite Groups (problems from Research Experiences for Undergraduates), George Berzsenyi, Jul/Aug97, p34 (Math Investigations)
Counting Random Paths (probability, symmetry, and random walk), Vladimir Dubrovsky, Jul/Aug93, p39 (Follow-up)
Creating Scientist-citizens (thoughts on “scientific literacy”), Bernard V. Khoury, Mar/Apr97, p2 (Front Matter)
The Creative Leap (Einstein’s science—everyone’s science), Gerry Wheeler, Jan/Feb97, p2 (Front Matter)
Criminal Geometry, or A Matter of Principle (Sherlock Holmes displays math prowess), D. V. Fomin, Sep/Oct91, p46 (Smiles)
Curiosity’s Natural Extension (feedback on National Science Education Standards editorial), Bill G. Aldridge, Jul/Aug95, p2 (Publisher’s Page)
Curved Reality (does Nature abhor a straight line?), Arthur Eisenkraft and Larry D. Kirkpatrick, Sep/Oct00, p30 (Contest Problem)
Cutting Facets (a simple problem with many hidden charms), Vladimir Dubrovsky, May/Jun96, p4 (Feature)
Cyberspace Exploration (cheap thrills and real science in the computer age), Bill G. Aldridge, Sep/Oct94, p2 (Publisher’s Page)
D
The Danger of Italian Restaurants (poem), David Arns, Sep/Oct98, p60 (Musings)
The Dark Power of Conventional Wisdom (Lobachevsky bicentenary), A. D. Alexandrov, Nov/Dec92, p4 (Feature)
The Death of a Star, Part I (poem), David Arns, Mar/Apr00, p53
The Death of a Star, Part II, David Arns, May/Jun00, p33
Delusion or Fraud? (dropping a needle to calculate pi), A. N. Zaydel, Sep/Oct90, p6 (Feature)
Democracy and Mathematics (voting paradoxes), Valery Pakhomov, Jan/Feb93, p4 (Feature)
Democratizing Expert Knowledge (climate change and science in society), Maurie J. Cohen, Jan/Feb98, p2 (Front Matter)
The Demoflush Figure (algebra where you least expect it), Linda P. Rosen, Jul/Aug97, p2 (Front Matter)
Depth of Knowledge (effects of air resistance), Arthur Eisenkraft and Larry Kirkpatrick, May/Jun98, p28 (Physics Contest)
Derivatives in Algebraic Problems (counting roots), Alexander Zvonkin, Nov/Dec93, p28 (At the Blackboard)
Desperately Seeking Susan on a Cylinder (a geometric approach to search and detection), A. Chkhartishvili and E. Shikin, Mar/Apr97, p10 (Feature)
Diamond Latticework (geometry of crystalline structures), R. V. Galiulin, Jan/Feb91, p6 (Feature)
Diamonds from a Jug (two tales with a brainteasing twist), Sergey Grabarchuk, Sep/Oct94, p63 (Toy Store)
Dielectrical Materialsm (the behavior of nonconducting objects in electric fields), Jan/Feb01, p28 (Kaleidoscope)
Diffraction in Laser Light (seeing diffraction patterns), D. Panenko, Mar/Apr99, p33 (In the Lab)
Disorder in the Court! (using energy “free of charge”), V. Fabricant, May90, p43 (Quantum Smiles)
Differing Differences (math challenge), George Berzsenyi, Nov/Dec91, p30 (Math Investigation)
Digitized Multiplication a la Steinhaus (math challenge), George Berzsenyi, Jul/Aug93, p27 (Math Investigations)
Dinosaurs in the Haystack (scientific method), Stephen Jay Gould, Sep/Oct92, p10 (Feature)
Direct Current Events (DC machines), I. Slobodetsky, Mar/Apr92, p52 (At the Blackboard)
The Discriminant at Work (a handy algebraic tool), Andrey Yegorov, Jan/Feb96, p34 (At the Blackboard)
Distinct Sums of Twosomes (pushing the lower bound), George Berzsenyi, Mar/Apr95, p39 (Math Investigations)
Divide and Conquer! (shortcut divisibility rules), Ruma Falk and Eyal Oshry, Mar/Apr99, p18 (Feature)
Divisibility Rules (problems in divisibility), Mark Saul and Titu Andreescu, Mar/Apr99, p43 (Gradus ad Parnassum)
Divisive Devices (Euclid’s algorithm, greatest common divisor, and fundamental theorem of arithmetic), V. N. Vaguten, Sep/Oct91, p36 (Feature)
Do As We Say … (diversity in Quantum), Bill G. Aldridge, Mar/Apr93 p2 (Publisher’s Page)
Dr. Matrix on the Wonders of 8 (observations of the “world’s greatest numerologist”), Martin Gardner, Jul/Aug95, p43 (Mathematical Surprises)
Does a Falling Pencil Levitate? (tabletop physics), Leaf Turner and Jane L. Pratt, Mar/Apr98, p22 (Feature)
Does Elementary Length Exist? (surprising implications of relativity and quantum mechanics), Andrey Sakharov, May/Jun97, p14 (Feature)
Doing It the Hard Way (multiple methodology), M. Tulchinsky, Sep/Oct92, p17 (Smiles)
Doppler Beats (sound frequency and relative motion), Larry D. Kirkpatrick and Arthur Eisenkraft, Jul/Aug98, p28 (Physics Contest)
Double, Double Toil and Trouble (boundary boiling of two liquids), A. Buzdin and V. Sorokin, May/Jun92, p52 (In the Lab)
Do You Get the Drift? (what the wind does to the snow), Lev Aslamazov, JanFeb93, p28 (sidebar)
Do You Have Potential? (the concept of potential), A. Leonovich, Nov/Dec97, p28 (Kaleidoscope)
Do You Know the Binding Energy? (a notion that unifies various types of physical interactions), May/Jun00, p28 (Kaleidoscope)
Do You Really Know Time? (it’s still a bit of a mystery), A. Leonovich, Sep/Oct99, p28 (Kaleidoscope)
Do You Really Know Vapors? (water behavior), A. Leonovich, Sep/Oct98, p32 (Kaleidoscope)
Do You Get the Drift? (behavior of blowing snow), Lev Aslamazov, Jan/Feb93, p28 (insert)
Do You Know Atoms and Their Nuclei? (broad outlines of an tiny, intricate world), A. Leonovich, Jan/Feb00, p28 (Kaleidoscope)
Do You Promise Not to Tell? (uses of constructive and destructive interference), Arthur Eisenkraft and Larry D. Kirkpatrick, Jan/Feb97, p30 (Physics Contest)
Dragon Curves (Chandler and Knuth’s famous design), Nikolay Vasilyev and Victor Gutenmacher, Sep/Oct95, p4 (Feature)
Dragon the Omnipresent (a proof of a remarkable property [see “Dragon Curves,” Sep/Oct95, and “Nesting Puzzles-Part II,” Mar/Apr96]), Vladimir Dubrovsky, Jul/Aug96, p34 (Follow-up)
Drops for the Crops (limits on the size of droplets), Yuly Bruk and Albert Stasenko, Mar/Apr94, p10 (Feature)
The Duke and His Chicken Incubator (seventeenth-century Florentine thermoscopes), Alexander Buzdin, Sep/Oct91, p51 (Looking Back)
Duracell Awards $100,000 to Young Inventors (results of Duracell/NSTA Scholarship Competition), May/Jun96, p53 (Happenings)
Dutch Treat (generating a sequence), Dr. Mu, Mar/Apr99, p55 (Cowculations)
E
East and West of Pythagoras by 30 degrees (math challenge), George Berzsenyi, Mar/Apr92, p51 (Math Investigations)
Educated Guesses (amusing Fermi problems), John A. Adam, Sep/Oct95, p20 (Feature)
Egyptian Fractions (an alternative method from the 17th century B.C.), George Berzsenyi, Nov/Dec94, p45 (Math Investigations)
Electrical and Mechanical Oscillations (current in an oscillating circuit), A. Kikoyin, Mar/Apr01, p48 (At the Blackboard)
Electric Currents on Coulomb Hills (the ups and downs of a circuit), E. Romishevsky, Jul/Aug99, p37 (At the Blackboard)
Electricity in the Air (surface charge density of the Earth), Arthur Eisenkraft and Larry D. Kirkpatrick, Nov/Dec93, p46 (Physics Contest)
Electric Multipoles (how a little order can weaken your potential), A. Dozorov, Sep/Oct99, p4 (Feature)
Electromagnetic Induction (intertwined lives of electricity and magnetism), Mar/Apr91, p32 (Kaleidoscope)
Elementary Functions (definitions from two perspectives), A. Veselov and S. Gindikin, Jul/Aug01, p22 (Feature)
The Elementary Particles (subatomic primer), Sheldon Lee Glashow, Sep/Oct90, p49 (Getting to Know …)
Elephant Ears (laws of scaling in the natural world), Arthur Eisenkraft and Larry D. Kirkpatrick, Nov/Dec97, p30 (Physics Contest)
Elevator Physics (free-falling balls), Arthur Eisenkraft and Larry Kirkpatrick, Mar/Apr99, p30 (Physics Contest)
11th Tournament of Towns (problems), Nov/Dec90, p51 (Happenings)
Embedding Triangles in Lattices (a classic problem from Math.Note at DEC), George Berzsenyi, Sep/Oct96, p38 (Math Investigations)
Endless Self-description (Hilgemeier’s “likeness sequence”), George Berzsenyi, Sep/Oct93, p17 (Math Investigations)
The Enigmatic Magnetic Force (the Lorentz force and the importance of accounting for small magnetic forces), E. Romishevsky, Jul/Aug00, p41 (At the Blackboard)
Enough Nerdiness (why the geek stereotype is so uncool), Dennis R. Harp and Harry Kloor, May/Jun98, p2 (Front Matter)
The Equalizer of a Triangle (a clever line that does double duty), George Berzsenyi, Mar/Apr97, p51 (Math Investigations)
Equation of the Gaseous State (handling twists with the ideal gas law), V. Belonuchkin, May/Jun00, p44 (At the Blackboard)
Equations Think For You (weeding out incorrect assumptions), V. Nakhshin, Jan90, p46 (At the Blackboard)
Ernst Abbe and “Carl Zeiss” (giants of optics), A. Vasilyev, Jul/Aug00, p46 (Looking Back)
Errorproof Coding (error detection and self-correction), Alexey Tolpygo, Mar/Apr93, p10 (Feature)
Errors in Geometric Proofs (searching for mistakes), S. L. Tabachnikov, Nov/Dec98, p37 (At the Blackboard)
Euclidean Complications (alternate geometries), I. Sabitov, Sep/Oct98, p20 (Feature)
Experiments of Frank and Hertz (putting Bohr’s quantum postulates to the test), A. Levashov, Mar/Apr00, p38 (Looking Back)
Exploring Every Angle (several approaches to the same problem), Boris Pritsker, Mar/Apr01, p38 (Problem Primer)
Exploring Remainders and Congruences (a set of exercises), A. Yegorov, May/Jun01, p32 (At the Blackboard)
Extra! Extra! Read All About It! (inductive incompetence), Vladimir Dubrovsky, Jul/Aug92, p43 (Smiles)
Extremists of Every Stripe (investing in Russia’s future), Bill G. Aldridge, Mar/Apr94, p2 (Publisher’s Page)
The Eye and the Sky (the art of seeing faint objects), V. Surdin, Jan/Feb00, p16 (Feature)
The Eyes Have It (workings of the human eye), Arthur Eisenkraft and Larry Kirkpatrick, May/Jun99, p30 (Physics Contest)
F
Fair and Squared! (quadratic equations in physics problems), Boris Korsunsky, May/Jun97, p53 (At the Blackboard)
Fantasy Chess (adding a rule or two), Yevgeny Gik, Sep/Oct90, p64 (Checkmate!)
Faraday’s Legacy (communicating a love of science), Laurence I. Gould, Nov/Dec98, p2 (Front Matter)
Farewell to JCMN (in memory of Basil Rennie), George Berzsenyi, May/Jun97, p40 (Math Investigations)
The Far from Dismal Science (sustainability and input-output economics), Dean Button, Faye Duchin, and Kurt Kreith, Sep/Oct97, p38 (Feature)
The Fast Game for Math Minds (the “Twenty-Four” challenge), Mar/Apr91, p52 (Happenings)
Feeding Rhythms and Algorithms (premier of computing column), Dr. Mu, Nov/Dec96, p37 (Cowculations)
The Fellowship of the Rings (vortices and turbulence), S. Shabanov and V. Shubin, Jul/Aug01, p37 (In the Lab)
Fermat’s Little Theorem (proving its value to mathematicians), V. Senderov and A. Spivak, May/Jun00, p14 (Feature)
Fertilizer with a Bang (investigating an explosive situation), B. Novozhilov, Sep/Oct00, p8 (Feature)
The Feuerbach Theorem (exploring the inscribed and escribed circles of triangles), V. Protasov, Nov/Dec99, p4 (Feature)
Fibonacci Strikes Again! (curious occurrences of a famous number sequence), Elliott Ostler and Neal Grandgenett, Jul/Aug92, p15 (Mathematical Surprises)
Field Pressure (the “pressure” of a static field), A. Chernoutsan, Sep/Oct00, p40 (At the Blackboard)
The Fifth International Olympiad in Informatics (problems and empanadas), Donald T. Piele, Mar/Apr94, p46 (Happenings)
Finding the Family Resemblence (an attempt to categorize number representation problems), George Berzsenyi, Jul/Aug96, p27 (Math Investigations)
Fire and Ice (report on the 1998 International Physics Olympiad), Sep/Oct98, p56 (Happenings)
The First Bicycle (each wheel consisted of two sticks), Albert Stasenko, Jan/Feb97, p44 (At the Blackboard)
The First Photon (the “vending machine” model), Arthur Eisenkraft and Larry D. Kirkpatrick, May/Jun95, p34 (Physics Contest)
Flexible in the Face of Adversity (topological transformations), A. P. Veselev, Sep/Oct90, p12 (Feature)
Flexible Polyhedral Surfaces (bending the rules), V. A. Alexandrov, Sep/Oct98, p4 (Feature)
Flexland Revisited (new forms of “flexlife”), Alexander Panov and Anatoly Kalinin, Jul/Aug93, p64 (Toy Store)
Flights of Fancy? (the upper limits of arrow shooting), V. Drozdov, Mar/Apr01, p40 (In the Open Air)
A Flight to the Sun (the challenges of sending a probe to the nearest star), Alexey Byalko, Nov/Dec96, p16 (Feature)
Fluids and Fault Lines (why large earthquakes are rather rare), G. Golytsyn, Jan/Feb00, p4 (Feature)
Fluids and Gases on the Move (a look at fluid mechanics), L. Leonovich, Jan/Feb96, p28 (Kaleidoscope)
Flux and Fixity (quantifying the energy stored in a magnetic field), V. Novikov, Jan/Feb01, p6 (Feature)
Fly Zapper (kill ’em and count ’em), Dr. Mu, Nov/Dec98, p62 (Cowculations)
Focusing Fields (finding the magnetic field that will focus charged particles), Arthur Eisenkraft and Larry D. Kirkpatrick, Jan/Feb96, p32 (Physics Contest)
Focusing on the Fleet (Archimedean victory at sea), Sergey Semenchinsky, Sep/Oct93, p28 (In the Lab)
Foiled by the Coanda Effect (an alternative way of explaining lift), Jef Raskin, Sep/Oct94, p4 (Feature)
Follow the Bouncing Buckyball (fullerenes and other carbonic architecture), Sergey Tikhodeyev, May/Jun94, p8 (Feature)
The Force Behind the Tides (understanding the attraction of the Moon), V. E. Belonuchkin, May/Jun98, p10 (Feature)
Forcing the Issue (Newtonian mechanics), Mar/Apr92, p32 (Kaleidoscope)
Forked Roads and Forked Tongues (a logical lie detector), P. Blekher, Nov/Dec97, p10 (Feature)
Formulas for Sin nx and Cos nx (handy mnemonic devices), Dmitry Fuchs, May/Jun93, p48 (At the Blackboard)
For the Love of Her Subject (interview with Marina Ratner), Julia Angwin, Jul/Aug94, p44 (Profile)
The Fourth State of Matter (plasma physics), Alexander Kingsep, Sep/Oct93, p4 (Feature)
The Friction and Pressure of Skating (glaciers and Carnot theorem), Alexey Chernoutsan, Jul/Aug94, p25 (At the Blackboard)
Friction, Fear, Friends, and Falling (mountaineering physics), John Wylie, Jul/Aug92, p4 (Feature)
Friezing Our Way into Summer (zigzag frieze patterns), John Conway, May90, p50 (Mathematical Surprises)
From a Roman Myth to the Isoperimetric Problem (searching for the greatest area given equal perimeters), I. F. Sharygin, Jan/Feb97, p34 (At the Blackboard)
From a Snowy Swiss Summit to the Apex of Geometry (biographical sketch of Jacob Steiner), I. M. Yaglom, Nov/Dec93, p35 (Looking Back)
From Cherokee Math to Tubby Genes (educational content on the World Wide Web), Tim Weber, May/Jun97, p2 (Front Matter)
From Mouse to Elephant (cell size and other zoological constants), Anatoly Mineyev, Mar/Apr96, p18 (Feature)
From the Edge of the Universe to Tartarus (Hesiod meets modern physics), Albert Stasenko, Mar/Apr96, p4 (Feature)
From the Pages of History (talking dolls, singing goblets, a fountain that spurts on command), A. Varlamov, Mar/Apr01, p34 (Looking Back)
From the Prehistory of Radio (Faraday, Maxwell, Hertz, and Popov), S. M. Rytov, May90, p39 (Looking Back)
The Fruits of Kepler’s Struggle (discovering the laws of orbital motion), B. E. Belonuchkin, Jan/Feb92, p18 (Feature)
Fuel Economy on the Moon (harnessing the Moon’s gravity), A. Stasenko, Jan/Feb00, p38 (At the Blackboard)
Functional Equations and Groups (and how to solve them), Y. S. Brodsky and A. K. Slipenko, Nov/Dec98, p14 (Feature)
The Fundamental Particles (combining quarks), Larry D. Kirkpatrick and Arthur Eisenkraft, Mar/Apr01, p30 (Physics Contest)
Fun with Liquid Nitrogen (latent heat of vaporization), Arthur Eisenkraft and Larry D. Kirkpatrick, Mar/Apr94, p38 (Physics Contest)
Further Adventures in Flexland (two-way hinges and flexchains), Alexey Panov, May/Jun92, p64 (Toy Store)
G
The Gambler, the Aesthete, and St. Pete (probabilities and payoffs), Leon Taylor, Jan/Feb98, p20 (Feature)
The Game of Battleships (achieving naval superiority on a paper sea), Yevgeny Gik, Nov/Dec96, p56 (Toy Store)
The Game of Bop (mathematical wordplay), Sheldon Lee Glashow, Sep/Oct92, p27 (In Your Head)
Generalizing Monty’s Dilemma (whether to stick with a choice or switch), John P. Georges and Timothy V. Craine, Mar/Apr95, p16 (Feature)
Generating Functions (problem-solving methods), S. M. Voronin and A. G. Kulagin, May/Jun99, p8 (Feature)
Getting It Together with “Polyominoes” (approach to tiling problems based on group theory), Dmitry V. Fomin, Nov/Dec91, p20 (Feature)
Genealogical Threes (using Euclid’s theorem to generate Pythagorean triples), A. A. Panov, Nov/Dec90, p36 (Looking Back)
Geometric Summation (infinite algebraic tilings), M. Apresyan, May/Jun94, p30 (In Your Head)
Geometric Surprises (a collection of elegant oddities), A. Savin, Jul/Aug00, p28 (Kaleidoscope)
Geometry in the Pagoda (classic problems of the great Japanese geometers), George Berzsenyi, Jan/Feb95, p48 (Math Investigations)
The Geometry of Population Genetics (color blindness and the Hardy-Weinberg law), I. M. Yaglom, May90, p24 (Feature)
Geometry of Sliding Vectors (modeling forces acting on rigid bodies having a definite size and shape), Y. Solovyov and A. Sosinsky, Mar/Apr00, p18 (Feature)
Georg Cantor (an anniversary review of his achievements), Vladimir Tikhomirov, Nov/Dec95, p48 (Looking Back)
The Giants (on whose shoulders Newton stood), Vladimir Belonuchkin, Jul/Aug95, p38 (Looking Back)
Gingerbread Man (creating computer graphics), Dr. Mu, Jan/Feb98, p55 (Cowculations)
Giving Astronomy Its Due (the role of astronomy in human history), A. Mikhailov, Jul/Aug01, p46 (At the Blackboard)
Glancing at the Thermometer … (computing the coefficient of thermal expansion), M. I. Kaganov, Jan/Feb93, p26 (At the Blackboard)
Gliding Home (propelling a glider long distances), Albert Stasenko, Mar/Apr99, p21 (At the Blackboard)
Glittering Performances (XXII International Physics Olympiad), Arthur Eisenkraft and Larry D. Kirkpatrick, Nov/Dec91, p53 (Happenings)
Global Change (commentary on events in the former Soviet Union), Bill G. Aldridge, Mar/Apr92, p2 (Publisher’s Page)
Going Around in Circles (math challenge), George Berzsenyi, Sep/Oct91, p35 (Math Investigations)
Going to Extremes (using the “extremity rule”), A. L. Rosenthal, Nov/Dec90, p8 (Feature)
The Golden Ratio in Baseball (Fibonacci in sport statistics), Dave Trautman, Mar/Apr96, p30 (Mathematical Surprises)
Go “Mod” with Your Equations (remainders and congruences), Andrey Yegorov, May/Jun92, p24 (Feature)
The Good Old Pythagorean Theorem (proofs and generalizations), V. N. Beryozin, Jan/Feb94, p24 (Feature)
A Good Question (active thought versus passive absorption), Bill G. Aldridge, Sep/Oct90, p3 (Publisher’s Page)
A Good Theory (or two—from Newton and Bohr), Arthur Eisenkraft and Larry D. Kirkpatrick, Jan/Feb01, p30 (Physics Contest)
Grand Illusions (apparent violations of light’s speed limit), A. D. Chernin, Jan/Feb92, p24 (At the Blackboard)
Graphs and Grafs (a little graph theory and practice), Anatoly Savin, Nov/Dec95, p32 (Kaleidoscope)
Gravitational Redshift (determining a star’s characteristics from photonic redshift), Arthur Eisenkraft and Larry D. Kirkpatrick, Nov/Dec95, p34 (Physics Contest)
The Great Art (controversial origins of “Cardano’s formula”), Semyon Gindikin, May/Jun95, p40 (Looking Back)
The Great Law (Newton and gravitational attraction), V. Kuznetsov, Sep/Oct99, p38 (Looking Back)
The Greek Alphabet (a physicist’s guide), Sheldon Lee Glashow, Mar/Apr92, p40 (Getting to Know …)
The Green Flash (an unusual spectacle at the close of day), Lev Tarasov, Jan/Feb97, p38 (In the Open Air)
A Gripping Story (how to calculate static friction), Alexey Chernoutsan, Mar/Apr96, p40 (At the Blackboard)
Group Velocity (a wider application of wave motion equations), Helio Waldman, Nov/Dec00, p47 (At the Blackboard)
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Halving It All (curiosities of planar bisection), Mark E. Kidwell and Mark D. Meyerson, Mar/Apr92, p6 (Feature)
Halving Some More (segments of constant area), Dmitry Fuchs and Sergey Tabachnikov, Mar/Apr92, p26 (Feature)
Hands-on (or -off?) Science (thermal sensitivity), Alexey Byalko, Nov/Dec97, p4 (Feature)
Hands-on Topology (experiments with the Möbius strip), Boris Kordemsky, Nov/Dec95, p64 (Toy Store)
Happy Birthday, Uncle Paul! (Erdös turns eighty-one), George Berzsenyi, May/Jun94, p28 (Math Investigations)
Happy New Year! (publisher resolves to learn Russian), Bill G. Aldridge, Jan/Feb91, p5 (Publisher’s Page)
Hard-core Heavenly Bodies (ionic crystal, Young’s modulus, and planetary mass), Yuly Bruk and Albert Stasenko, Jul/Aug93, p34 (Feature)
Head over Heels (mechanics of an odd top), Sergey Krivoshlykov, May/Jun95, p62 (Toy Store)
Health and Long Life (travel notes: mad cows and the Brontës), Bill G. Aldridge, May/Jun96, p2 (Publisher’s Page)
Heart Waves (behavior of electrical waves in the heart), A. S. Mikhailov, Nov/Dec91, p12 (Feature)
Heating Water from the Top (moving boundaries and waves under water), V. Pentegov, Nov/Dec99, p41 (In the Lab)
High-Speed Conservation (physics at near-light speeds), A. Korzhuyev, Sep/Oct98, p38 (At the Blackboard)
High-speed Hazards (a radical method of combatting the effects of very large accelerations), I. Vorobyov, May/Jun00, p24 (Feature)
Hindsight (when to hold ‘em and when to fold ‘em), Dr. Mu, Nov/Dec97, p55 (Cowculations)
The History of a Fall (what happens to a drip as it drops), Leonid Guryashkin and Albert Stasenko, Mar/Apr95, p10 (Feature)
Hit or Miss (Perelman problems), Nov/Dec92, p32 (Kaleidoscope)
Holding Up Under Pressure (modeling bridges), Alexander Borovoy, Jan90, p30 (In the Lab)
Holes in Graphs (functions that are both continuous and discontinuous), Michael H. Brill and Michael Stueben, Sep/Oct91, p12 (Feature)
Hollow Molecules (belated insert to “Follow the Bouncing Buckyball”), David E. H. Jones, Mar/Apr95, p53 (Addendum)
Homemade Pendulums (describing their motion), G. L. Kotkin, Mar/Apr98, p38 (In the Lab)
Home on the Range (functional primer), Andrey N. Kolmogorov, Sep/Oct93, p10 (Feature)
Homogeneous Equations (more equation solving), L. Ryzhkov and Y. Ionin, May/Jun98, p43 (At the Blackboard)
The Horrors of Resonance (are you in for a rough landing?), A. Stasenko, Mar/Apr98, p45 (At the Blackboard)
Horseflies and Flying Horses (matters of scale in the animal world), A. Zherdev, May/Jun94, p32 (Kaleidoscope)
A Horse is a Horse (of Course, of Course) (shenanigans with fractions), A. S. Yarsky, May90, p43 (Quantum Smiles)
How About a Date? (physics challenge), Arthur Eisenkraft and Larry D. Kirkpatrick, Mar/Apr93, p30 (Physics Contest)
How Big Am I, Really? (poem), David Arns, Jul/Aug98, p55 (Musings)
How Do We Breathe? (physics in alveoli), K. Y. Bogdanov, May90, p4 (Feature)
How Enlightened Are You? (straight answers to crooked questions about light), Alexander Leonovich, May/Jun96, p32 (Kaleidoscope)
How Long Does a Comet Live? (an attempt at an estimate), S. Varlamov, May/Jun01, p4 (Feature)
How Many Bubbles are in Your Bubbly? (the physics of gas dissolved in luquids), A. Stasenko, Nov/Dec00, p44 (In the Lab)
How Many Divisors Does a Number Have? (a classic problem with many interconnections), Boris Kotlyar, Mar/Apr96, p24 (Feature)
How’s Your Astronomy? (collection of heavenly facts and questions), May/Jun95, p32 (Kaleidoscope)
How the Ball Bounces (physics challenge), Arthur Eisenkraft and Larry D. Kirkpatrick, Mar/Apr91, p54 (Contest)
How to Escape the Rain (to run or to walk?), I. F. Akulich, May/Jun98, p38 (In the Open Air)
Hula Hoop (circular animation), Dr. Mu, Jan/Feb99, p54 (Cowculations)
Hurling at the Abyss (oscillating too-short bridges), A. Stasenko, Nov/Dec98, p43 (At the Blackboard)
Hydroparadoxes (when fluids forsake model behavior), S. Betyaev, Jul/Aug98, p20 (Feature)
Hyperbolic Tension (measuring the coefficient of surface tension), I. I. Vorobyov, Jan/Feb98, p30 (In the Lab)
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I Can See Clearly Now (poem), David Arns, Nov/Dec98, p8
An Ideal Gas Gets Real (and relativity visits electromagnetic induction), Albert Stasenko and Alexey Chernoutsan, Sep/Oct93, p42 (At the Blackboard)
Image Charge (electrostatic investigation), Larry D. Kirkpatrick and Arthur Eisenkraft, Jul/Aug99, p30 (Physics Contest)
The Importance of Studying the Physics of Sound Insulation (a detective story), Roman Y. Vinokur, Nov/Dec95, p18 (Feature)
Important Components of Learning Components (a different approach to vectors), Boris Korsunsky, Jan/Feb95, p45 (Sticking Points)
Incandescent Bulbs (illuminating thermal expansion), D. C. Agrawal and V. J. Menon, Jan/Feb98, p35 (At the Blackboard)
An Incident on the Train (air pressure in a tunnel), Carlo Camerlingo and Andrey Varlamov, Nov/Dec90, p42 (At the Blackboard)
Inequalities Become Equalities (a baker’s dozen problems with a common ingredient), A. Egorov, Mar/Apr00, p42 (At the Blackboard)
The Inevitability of Black Holes (Schwarzschild radius, principle of equivalence), William A. Hiscock, Mar/Apr93, p26 (Feature)
Infinite Descent (a method with wide applicability), Lev Kurlyandchik and Grigory Rozenblume, Jul/Aug96, p10 (Feature)
In Focus (optics and your eyes), A. Dozorov, Sep/Oct98, p48 (At the Blackboard)
In Foucault’s Footsteps (a simple experiment on the Coriolis force), M. Emelyanov, A. Zharkov, V. Zagainov, and V. Matochkin, Nov/Dec96, p26 (In the Lab)
In Memoriam: Paul Erdös 1913–1996) (an appreciation of the great problem master), George Berzsenyi, Nov/Dec96, p40 (Math Investigations)
The Ins and Outs of Circles (inscribed and circumscribed circles), I. F. Sharygin, Nov/Dec97, p38 (At the Blackboard)
Inscribe, Subtend, Circumscribe (variations on a geometric theme), Vladimir Uroyev and Mikhail Shabunin, Nov/Dec96, p10 (Feature)
In Search of a Definition of Surface Area (working through a paradoxical result), Vladimir Dubrovsky, Mar/Apr91, p6 (Feature)
In Search of Perfection (numbers equal to the sum of their own divisors), I. Depman, May/Jun01, p8 (Feature)
Interacting Bodies (all about collisions), A. Leonovich, Jan/Feb99, p28 (Kaleidoscope)
Internal Energy and Heat (why Q is in the reference tables, not ΔU), Alexey Chernoutsan, Jul/Aug97, p38 (Fundamentals)
In the Curved Space of Relativistic Velocities (link between relativity and hyperbolic geometry), Vladimir Dubrovsky, Mar/Apr93, p34 (Feature)
Interstellar Bubbles (a phase in the life cycle of stars), S. Silich, Nov/Dec97, p14 (Feature)
In the Planetary Net (the potential in gravitational fields), V. Mozhayev, Jan/Feb98, p4 (Feature)
Inversion (useful transformation), Vladimir Dubrovsky, Sep/Oct92, p40 (Getting to Know …)
Invincible Mephisto! (computer chess), Y. Gik, Jan90, p56 (Checkmate!)
An Invitation to the Bathhouse (physics in the Russian banya), I. I. Mazin, Sep/Oct90, p20 (Feature)
IOI 2000 (report from the International Olympiad of Informatics in Beijing), Don Piele, Jan/Feb01, p55 (Informatics)
Irrationality and Irreducibility (how are they connected?), V. A. Oleynikov, May/Jun97, p22 (Feature)
Irregular Regular Polygons (a math problem found in a dictionary), Eric D. Carlson and Sheldon L. Glashow, Jul/Aug95, p48 (At the Blackboard)
Is Bingo Fair? (parlor probability), Mark Krosky, May/Jun98, p4 (Feature)
Is This What Fermat Did? (fast factorization), B. A. Kordemsky, Sep/Oct91, p17 (At the Blackboard)
It All Depends on Your Attitude (getting oriented in outer space), Bernice Kastner, Jan/Feb92, p12 (Feature)
It’s All Greek to Me! (symbols in math and science), Bill G. Aldridge, Jan/Feb95, p2 (Publisher’s Page)
It’s Beautiful—But Is It Science? (waves in a Viking painting), Albert Stasenko, Jan90, p8 (Feature)
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Jesse James Discovers the Heat Equation (using spreadsheets for diffusion processes), Kurt Kreith, May/Jun95, p26 (Feature)
Jewels in the Crown (mathematical induction), Mark Saul, Jul/Aug92, p10 (Feature)
Jingle Bell? (bell-ringing in a vacuum), N. Paravyan, Nov/Dec97, p27 (In the Lab)
Jules Verne’s Cryptogram (cracking a code to save a life), G. A. Gurevich, Sep/Oct90, p44 (Looking Back)
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Karate Chop (physics of tameshiwari), A. Biryukov, May/Jun99, p14 (Feature)
Keeping Cool and Staying Put (heat pumps and rope tension), Alexander Buzdin, May/Jun93, p17 (At the Blackboard)
Keeping Track of Points (trajectories, tracks, and displacements), Sep/Oct93, p32 (Kaleidoscope)
Kith and Kin (friendly numbers and twin primes), Jan90, p28 (Kaleidoscope)
Knots, Links, and Their Polynomials (Reidemeister moves, Conway polynomial, and other aspects of knot theory), Alexey Sosinsky, Jul/Aug95, p8 (Feature)
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Landau’s License Plate Game (math prowess of a great physicist), M. I. Kaganov, Mar/Apr93, p47 (In Your Head)
Langton’s Ant (a spinoff of Conway’s Game of Life), Don Piele, Mar/Apr00, p63 (Informatics)
Laser Levitation (lifting with light), Arthur Eisenkraft and Larry D. Kirkpatrick, May/Jun94, p38 (Physics Contest)
Laser Pointer (the underlying physics), S.Obukhov, Nov/Dec00, p14 (Feature)
The Last Problem of the Cube (“God’s algorithm” for Rubik’s immortal cube), Vladimir Dubrovsky, Mar/Apr95, p61 (Toy Store)
Late Light from Mercury (gravitational refraction), Yakov Smorodinsky, Nov/Dec93, p40 (In the Lab)
Latin Rectangles (exercise in combinatorics), V. Shevelyov, Mar/Apr91, p18 (Feature)
Latin Triangles (a puzzle and a model of Schwarz’s boot), D. Bernshtein, Mar/Apr91, p64 (Toy Store)
Lattices and Brillouin Zones (polygonal patterns), A. B. Goncharov, Nov/Dec98, p4 (Feature)
Launch into International Space Year! (guide to ISY activities), Jan/Feb92, p53 (Happenings)
Lazy-day Antidotes (light summertime problems to quicken the mind), Jul/Aug95, p32 (Kaleidoscope)
The Leaky Pendulum (physics challenge), Arthur Eisenkraft and Larry D. Kirkpatrick, Nov/Dec91, p28 (Physics Contest)
Learning About (Not By) Osmosis (discovery and applications), Alexander Borovoy, Nov/Dec91, p48 (In the Lab)
Learning from a Virus (applying system dynamics to the spread of an illness), Matthias Ruth, Sep/Oct97, p28 (Feature)
The Legacy of al-Khwarizmi (the origins of algebra), Z. D. Usmanov and I. Hodjiev, Jul/Aug98, p26 (Looking Back)
The Legacy of Norbert Wiener (Part I: childhood, boyhood, youth), Nov/Dec94, p47 (Innovators)
The Legacy of Norbert Wiener (Part II: Brownian motion and beyond), Jan/Feb95, p41 (Innovators)
The Legacy of Norbert Wiener (Part III: from feedback to cybernetics), Mar/Apr95, p42 (Innovators)
Less Heat and More Light (properties of the “ideal black body”), Y. Amstislavsky, Nov/Dec95, p4 (Feature)
Let’s Not Be Dense About It! (facts, questions, and problems about density), A. A. Leonovich, May/Jun97, p32 (Kaleidoscope)
Letters from the Editors (notes by the editors in chief), Jan90, p6
Lewis Carroll’s Sleepless Nights (two “pillow problems” in probability), Martin Gardner, Mar/Apr95, p40 (Mathematical Surprises)
Liberté, Égalité, Géométrie (Gaspard Monge—father of descriptive geometry), V. Lishevsky, Nov/Dec00, p20 (Feature)
Life on an Accelerating Skateboard (toward an improved definition of “weight”), Albert A. Bartlett, Sep/Oct95, p49 (Follow-up)
Light at the End of the Tunnel (invariants and monovariants), Dmitry Fomin and Lev Kurlyandchik, Mar/Apr94, p16 (Feature)
Light in a Dark Room (history of the camera obscura), V. Surdin and M. Kartashev, Jul/Aug99, p40 (Looking Back)
Lightning in a Crystal (story of the LED), Yury R. Nosov, Nov/Dec90, p12 (Feature)
Light Pressure (are sunny days more burdensome?), S. V. Gryslov, May/Jun98, p36 (Looking Back)
The Limits to Growth Revisited (a primer on exponential growth, overshoot, and dynamic modeling), Kurt Kreith, Sep/Oct97, p4 (Feature)
The Little House on the Tundra (keeping the foundation from melting the ground), A. Tokarev, Jul/Aug00, p38 (At the Blackboard)
A Little Lens Talk (“paper” and “real” lenses), Alexander Zilberman, May/Jun94, p35 (At the Blackboard)
Local Fields Forever (looking at gravity and acceleration), Arthur Eisenkraft and Larry D. Kirkpatrick, Jan/Feb98, p32 (Physics Contest)
The Long and Short of It (ruminations on the notion of “length”), Anatoly Savin, Mar/Apr96, p32 Kaleidoscope)
The Long Road to Longitude (how we finally became “coordinated”), A. A. Mikhailov, Mar/Apr97, p42 (Looking Back)
Look, Ma—No Calculus! (a spreadsheet approach to population dynamics), Kurt Kreith, Nov/Dec94, p15 (Feature)
The Lorentz/FitzGerald Diet (poem), David Arns, Jan/Feb99, p41
Lost in a Forest (Bellman’s problem: how to get out in the shortest time?), George Berzsenyi, Nov/Dec95, p41 (Math Investigations)
Love and Hate in the Molecular World (the “emotions” of dipoles), Albert Stasenko, Nov/Dec94, p10 (Feature)
Lunar Ironies (it is made of cheese!), M. A. Koretz and Z. L. Ponizovsky, Jul/Aug93, p24 (Smiles)
Lunar Launch Pad (could a volcano have given birth to a satellite of the Earth of Sun?), A. Stasenko, Mar/Apr01, p44 (Forces of Nature)
Lunar Miscalculation (how to get stranded in the pitch-dark mountains), Bill G. Aldridge, Nov/Dec96, p2 (Publisher’s Page)
The Lunes of Hippocrates (an early attempt to square the circle), V. N. Berezin, Jan/Feb98, p39 (Looking Back)
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A Magical Musical Formula (soundless guitar tuning), P. Mikheyev, Jan/Feb95, p30 (In the Lab)
The Magic of 3 x 3 (specifically, a magic square of squares), Martin Gardner, Jan/Feb96, p24 (Mathematical Surprises)
Magnetic Fieldwork (measuring magnetic fields), D. Tselykh, Sep/Oct98, p46 (In the Lab)
Magnetic Levitation Comes of Age (superconductivity applied), Thomas D. Rossing and John R. Hull, Mar/Apr95, p22 (Feature)
Magnetic Monopoly (in search of the magnetic monopole), John Wylie, May/Jun95, p4 (Feature)
Magnetic Personality (Hans Christian ײsted), V. Kartsev, May/Jun99, p42 (Looking Back)
Magnetic Vee (a constant current I in a wire shaped like a V), Larry D. Kirkpatrick and Arthur Eisenkraft, Mar/Apr00, p34 (Physics Contest)
Magnets, Charges, and Planets (the search for connections among forces), Albert Stasenko, May/Jun97, p42 (At the Blackboard)
A Magnificant Obsession (perfect numbers), Michael H. Brill and Michael Stueben, Jan/Feb93, p18 (Feature)
Make Yourself Useful, Diana (the Moon as a radio telescope antenna), P. V. Bliokh, Mar/Apr92, p34 (Feature)
Making the Crooked Straight (linearizing mechanism for the steam engine), Yury Solovyov, Nov/Dec90, p20 (Feature)
The Many Faces of Ice (the physics of frozen water), A. Zaretsky, Jul/Aug01, p6 (Feature)
Many Happy Returns (the tricky business of returning from space), Albert Stasenko, Jul/Aug01, p16 (Feature)
Many Ways to Multiply (a survey of techniques), Anatoly Savin, Mar/Apr01, p28 (Kaleidoscope)
The Mapmaker’s Tale (Four Color Theorem goes awry), Sheldon Lee Glashow, May/Jun93, p46 (Smiles)
Marching Orders (finite group primer), Alexey Sosinsky, Nov/Dec91, p6 (Feature)
The Markov Equation (an elegant solution to a Diophantine equation), M. Krein, Jan/Feb00, p42 (At the Blackboard)
Mars or Bust! (problems related to exploring the Red Planet), Arthur Eisenkraft and Larry D. Kirkpatrick, Mar/Apr97, p34 (Physics Contest)
Martin Gardner’s “Royal Problem” (generalization of a chessboard problem), Jesse Chan, Peter Laffin, and Da Li, Sep/Oct93, p45 (Follow-up)
A Mathematical Handbook with No Figures (silliness with a purpose), Yuly Danilov, May/Jun94, p42 (Quantum Smiles)
Mathematical Hopscotch (discontinuous Q&A), Jul/Aug94, p28 (Kaleidoscope)
The Mathematician, the Physicist, and the Engineer (science jokes on the Internet), May/Jun96, p48 (Quantum Smiles)
Mathematics: 1900–1950 (an overview), V. Tikhomirov, Mar/Apr00, p4 (Feature)
Mathematics in Living Organisms (calculating cats), M. Berkenblit and E. Glagoleva, Nov/Dec92, p34 (Feature)
Mathematics in Perpetual Motion (imaginary elliptical engine), Anatoly Savin, Jul/Aug94, p4 (Feature)
Math Relay Races (relay problems from the trenches), Don Barry, May/Jun98, p26 (At the Blackboard)
Matter and Gravity (material points and extended objects), Sep/Oct00, p28 (Kaleidoscope)
Matter and Magnetism (a quick tour), Mau/Jun01, p28 (Kaleidoscope)
Maximizing the Greatest (revisiting a GCD problem), George Berzsenyi, May/Jun95, p39 (Math Investigations)
Meandering down to the Sea (natural curvature of riverbeds), Lev Aslamazov, Jul/Aug92, p34 (In the Lab)
The Mean Value of a Function (stretching an arithmetic concept), Yury Ionin and Alexander Plotkin, Nov/Dec95, p26 (Feature)
The Medians (multiple proofs of a well-known theorem), Vladimir Dubrovsky, Nov/Dec94, p32 (Kaleidoscope)
Meeting No Resistance (high-temperature superconductivity), Alexander Buzdin and Andrey Varlamov, Sep/Oct91, p6 (Feature)
A Meeting of Minds (US-Soviet science teachers conference), Bill G. Aldridge, Sep/Oct91, p4 (Publisher’s Page)
Merry-go-round Kinematics (a dynamic game of cherry tossing), Albert Stasenko, Sep/Oct96, p48 (At the Blackboard)
Message from Afar (poem), David Arns, May/Jun99, p48 (Musings) [reprinted Nov/Dec99, p9]
Mighty Ether Has Struck Out (poem), David Arns, May/Jun97, p30
Milk Routes (the best whey into town), Dr. Mu, Mar/Apr98, p55 (Cowculations)
Minimal Surfaces (wire contours and soap films), A. Fomenko, May/Jun00, p4 (Feature)
Mirror Full of Water (wet optics), Arthur Eisenkraft and Larry D. Kirkpatrick, Jul/Aug94, p32 (Physics Contest)
Miss or Hit (more Perelman problems), Mar/Apr93, p32 (Kaleidoscope)
Modeling a Tornado (cyclone in a jar), V. Mayer, May/Jun00, p42 (In the Lab)
Models of Efficiency (problems and facts about work, power, and efficiency), Sep/Oct94, p32 (Kaleidoscope)
The Modest Experimentalist, Henry Cavendish (scientist who didn’t publish results), S. Filonovich, Jan/Feb91, p41 (Looking Back)
Molecular Interactions Up Close (fundamental forces), G. Myakishev, May/Jun00, p8 (Feature)
Molecular Intrigue (how small are molecules?), A. Leonovich, Jan/Feb98, p28 (Kaleidoscope)
A Moon of Steel (loony research), M. A. Koretz and Z. L. Ponizovsky, Jul/Aug93, p24 (Smiles)
The Moscow Correspondence School in Quantum (sample problems from a school without walls), I. M. Gelfand, Mar/Apr91, p42 (Math by Mail)
The “Most Inertial” Reference Frame (the universe’s relict radiation), Gennady Myakishev, Mar/Apr95, p48 (In Your Head)
The Most Mysterious Shape of All (a spiral primer), Mar/Apr95, p32 (Kaleidoscope)
The Most Profit with the Least Effort (Chebyshev’s “The Drawing of Geographic Maps”), Yuly Danilov, Sep/Oct94, p35 (Anthology)
Mushrooms and X-ray Astronomy (natural collimator), Alexander Mitrofanov, Jul/Aug94, p10 (Feature)
Musical Chairs (a variant, and a question), Don Piele, May/Jun01, p54 (Informatics)
Moving Matter (using a pendulum to measure speed), Arthur Eisenkraft and Larry D. Kirkpatrick, May/Jun96, p34 (Physics Contest)
The Multidimensional Cube (an introduction to multidimensional space), Vladimir Dubrovsky, Sep/Oct96, p4 (Feature)
The Music of Physicists (amusing anecdotes about Einstein, Bunsen, Planck, and Rutherford), Sep/Oct90, p54 (Quantum Smiles)
The Mystery of Figure No. 51 (descendant of tangram), Alexey Panov, Sep/Oct92, p63 (Toy Store)
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The Name Game of the Elements (confusion and politics in chemistry), Henry D. Schreiber, Sep/Oct96, p24 (Feature)
Nascent Non-Euclidean Geometry (revisiting a geometry classic), N. I. Lobachevsky, May/Jun99, p20 (Feature)
The Natural Logarithm (derivation of an unnatural-looking number), Bill G. Aldridge, Nov/Dec90, p26 (Getting to Know …)
The Nature of an Ideal Gas (implications of the model), A. Leonovich, May/Jun98, p32 (Kaleidoscope)
The Nature of Light (the Compton effect), Arthur Eisenkraft and Larry D. Kirkpatrick, Nov/Dec96, p30 (Physics Contest)
Nature’s Fireworks (inner workings of the auroras), A. K. Kikoyin, Jan/Feb92, p50 (Feature)
The Near and Far of It (limitations of optical instruments), A. Stasenko, Mar/Apr01, p24 (Feature)
Nesting Puzzles (part I: The Tower of Hanoi and Panex), Vladimir Dubrovsky, Jan/Feb96, p53 (Toy Store)
Nesting Puzzles (part II: Chinese rings and the return of the dragon), Vladimir Dubrovsky, Mar/Apr96, p61 (Toy Store)
Neutrinos and Supernovas (physics challenge), Arthur Eisenkraft and Larry D. Kirkpatrick, Nov/Dec90, p35 (Contest)
Neutrons Seek the Murderer (neutron activation analysis), A. S. Shteinberg, May/Jun92, p20 (In the Lab)
The New Earth (physics of a hollow Earth), A. Stasenko, Jul/Aug99, p16 (Feature)
Nine Solutions to One Problem (classic triangle problem), Constantine Knop, May/Jun94, p46 (At the Blackboard)
Nonreπeating, Πatternless, and Πerπetually Aππproximated (aspects of pi), Nov/Dec00, p28 (Kaleidoscope)
Nonstandardly Continued Fractions (infinite processes with simple answers), George Berzsenyi, Jan/Feb96, p39 (Math Investigations)
Not All is Revealed (the Uncertainty Principle and other forms of indeterminacy), Albert Stasenko, Nov/Dec96, p42 (At the Blackboard)
Not a Silver Bullet—A Golden Opportunity (how technology can improve science education), Stanley Litow, Jan/Feb01, p3 (Front Matter)
Notes of a Traveler (education in the US and USSR), Bill G. Aldridge, Nov/Dec90, p2 (Publisher’s Page)
The Notion of Vicinity (math challenge), George Berzsenyi, Nov/Dec92, p18 (Math Investigations)
Nudging Our Way to a Proof (using the method of small perturbations), Galina Balk, Mark Balk, and Vladimir Boltyansky, Mar/Apr95, p4 (Feature)
Number Cells (numerical destinations), Thomas Hagspihl, Nov/Dec97, p41 (At the Blackboard)
Number Show (a handful of numerical tricks), Ivan Depman and Naum Vilenkin, Mar/Apr96, p46 (In Your Head)
Numbers in Our Genes (quantification in molecular biology), Bill G. Aldridge, May/Jun94, p2 (Publisher’s Page)
Number Systems (Babylonian, Roman, Mayan, and beyond), Isaak Yaglom, Jul/Aug95, p22 (Feature)
Numeral Roamings (exploring nontraditional mathematical operations), A. Egorov and A. Kotova, Mar/Apr98, p16 (Feature)
Numerical Data in Geometry Problems (new angles to problem solving), S. V. Ovchinnikov and I. F. Sharygin, May/Jun99, p37 (At the Blackboard)
O
Obtaining Symmetric Inequalities (Muirhead’s Theorem), S. Dvoryaninov and E. Yasinovyi, Nov/Dec99, p44 (At the Blackboard)
The Oceanic Phone Booth (large-scale waveguides), Andrey Varlamov and Alexey Malyarovsky, May/Jun93, p36 (Feature)
Of Amoebas and Men (amoeba in a dinner jacket), Alexey Sosinsky, Jan90, p44 (Looking Back)
Of Combs and Coulombs (a smorgasbord of electrical questions and facts), A. Leonovich, Jan/Feb97, p28 (Kaleidoscope)
Off into Space (jumping out of the plane), Vladimir Dubrovsky and Igor Sharygin, Jan/Feb92, p44 (Feature)
Of Microscopes, E-mail, and Word of Mouth (questions of survival), Bill G. Aldridge, May/Jun93, p2 (Publisher’s Page)
An Old Algorithm (taking square roots), Y. Solovyov, Mar/Apr00, p51 (At the Blackboard)
An Old Fact and Some New Ones (shape-numbers and number-shapes), John Conway, Sep/Oct90, p24 (Mathematical Surprises)
Olympiad Honors (report on the 40th International Mathematical Olympiad), Jan/Feb00, p41 (Happenings)
Olympian Effort (reminiscences of Moscow competitions), V. Tikhomirov, Mar/Apr00, p32 (At the Blackboard)
Olympic Recap from England (XXXI International Physics Olympiad), Mary Mogge, Nov/Dec00, p26 (Happenings)
The Omnipresent and Omnipotent Neutrino (brief history, current research), Chris Waltham, Jul/Aug93, p10 (Feature)
One Problem After Another (chain questions), B. M. Bolotovsky, Jan90, p13 (Quantum Smiles)
One’s Best Approach (summing up reciprocals), O. T. Izhboldin and L. D. Kurlyandchik, Mar/Apr99, p24 (At the Blackboard)
Ones Up Front in Powers of Two (Fractional Parts Theorem), Vladimir Boltyansky, Nov/Dec93, p16 (Feature)
One, Two, Many (“primitive” counting method of scientists), May/Jun92, p32 (Kaleidoscope)
On Kaleidoscopes (a look at them in all their dimensions), E. B. Vinberg, May/Jun97, p4 (Feature)
On Quasiperiodic Sequences (an unexpected use of graph paper), A. Levitov, A. Sidorov, and A. Stoyanovsky, Sep/Oct00, p34 (At the Blackboard)
On the Edge (compassless constructions), Igor Sharygin, Mar/Apr98, p28 (Kaleidoscope)
On the Nature of Space Magnetism (the cosmic “hydromagnetic dynamo”), Alexander Ruzmaykin, Sep/Oct95, p12 (Feature)
On the Quantum Nature of Heat (finding direction in chaos), V. Mityugov, Nov/Dec99, p10 (Feature)
Optics for a Stargazer (can one see stars at noon from the bottom of a well?), Vladimir Surdin, Sep/Oct94, p18 (Feature)
The Orbit of Triangles (attractors and “butterflies”), George Berzsenyi, Mar/Apr96, p43 (Math Investigations)
The Orchard Problem (planting trees but maintaining a view), Vladimir Jankovic, Jan/Feb96, p16 (Feature)
Ordered Sets (ordered triplets, some generalizations, and interesting inequalities), L. Pinter and I. Khegedysh, Jul/Aug98, p43 (At the Blackboard)
Ornamental Groups (Escher and symmetry groups), Vladimir Dubrovsky, Nov/Dec91, p32 (Kaleidoscope)
Osmosis the Magnificent (powerful, yes; perpetual …?), Norayr Paravyan, Jul/Aug96, p39 (In the Lab)
The Other Half of What You See (more on derivatives in algebraic problems), Vladimir Dubrovsky, Nov/Dec93, p44 (Follow-up)
Our Old Magnetic-Enigmatic Friend (follow-up to “The Enigmatic Magnetic Force”), E. Romishevsky, Nov/Dec00, p38 (At the Blackboard)
Out of Flexland (“gasping starfish” and more), Vladimir Dubrovsky, Jul/Aug92, p63 (Toy Store)
Out Standing in the Field (divvying up the purse in a shortened season), Dr. Mu, Jul/Aug97, p55 (Cowculations)
Out to Pasture (finding the digital product root of 123456789), Dr. Mu, Jul/Aug99, p54 (Cowculations)
Overshooting the Limits (reappraising Malthus with computer simulations), Bob Eberlein, Sep/Oct97, p14 (Feature)
P
The Painter’s Paradox (covering an infinite surface), A. A. Panov, Mar/Apr91, p10 (Quantum Smiles)
Painting the Digital World (surface areas of pixels and voxels), Michael H. Brill, Mar/Apr99, p10 (Feature)
Panting Dogs, Aromatic Blooms, and Tea in a Saucer (tales of evaporation in the natural world), Andrey Korzhuyev, Nov/Dec94, p30 (In the Open Air)
A Partial History of Fractions (unit fractions, sexagesimal fractions, decimal fractions, binary fractions …), N. Vilenkin, Sep/Oct00, p26 (Nomenclatorium)
A Party of Wise Guys (14th Annual Puzzle Party), Anatoly Kalinin, Jul/Aug95, p63 (Toy Store)
Patterns of Predictability (symmetry, anisotropy, and Ohm’s law), S. N. Lykov and D. A. Parshin, Nov/Dec91, p36 (Feature)
Peering into Potential Wells (a common aspect of three disparate objects), K. Kikoin, May/Jun01, p12 (Feature)
Penrose Patterns and Quasi-crystals (tiling and a high-tech alloy), V. Koryepin, Jan/Feb94, p12 (Feature)
Perfect Shuffle (an algorithm for a deck of six cards), Don Piele, Nov/Dec00, p55 (Informatics)
Periodic Binary Sequences (generating 0’s and 1’s), George Berzsenyi, Nov/Dec93, p50 (Math Investigations)
Periodic Functions in Hiding (math challenge), George Berzsenyi, Sep/Oct92, p39 (Math Investigations)
A Permutator’s Bag of Tricks (solutions to rolling-block puzzles) Vladimir Dubrovsky, Jan/Feb94, p62 (Toy Store)
The Pharaoh’s Golden Staircase (dynamic programming and Bellman’s formula), M. Reytman, Mar/Apr94, p4 (Feature)
Phlogiston and the Magnetic Field (outgrown concepts), Stephanie Eatman, Fraser Muir, and Hugh Hickman, Mar/Apr94, p35 (Looking Back)
Photosynthesism (artificial barriers between disciplines), Bill G. Aldridge, Jul/Aug92, p2 (Publisher’s Page)
Physical Optics and Two Camels (two-beam interference and the limits of far-sightedness), A. Stasenko, Sep/Oct99, p44 (At the Blackboard)
Physics Fights Frauds (scientific sleuthing), I. Lalayants and A. Milovanova, Jan/Feb93, p10 (Feature)
Physics for Fools (hare-brained experiments for crackpots), V. F. Yakovlev, Nov/Dec90, p17 (Quantum Smiles)
Physics in the Kitchen (simple experiments with boiling water), I. I. Mazin, Sep/Oct97, p54 (In the Lab)
Physics in the News (calculus and the laws of scaling), Albert A. Bartlett, May/Jun96, p50 (At the Blackboard)
Physics Limericks (finished and unfinished rhymes), Robert Resnick, Sep/Oct90, p52 (Quantum Smiles)
The Physics of Chemical Reactions (molecular kinetics), O. Karpukhin, Nov/Dec00, p4 (Feature)
The Physics of Walking (oscillations and parametric resonance), I. Urusovsky, Sep/Oct00, p20 (Feature)
A Physics Soufflé (having enough information, or too much), Arthur Eisenkraft and Larry D. Kirkpatrick, Jul/Aug97, p30 (Physics Contest)
Physics Without Fancy Tools (summertime scientific observations), A. Dozorov, Jul/Aug01, p28 (Kaleidoscope)
A Pigeonhole for Every Pigeon (math challenge), George Berzsenyi, Sep/Oct90, p40 (Contest)
Pigeons in Every Pigeonhole (application of the Dirichlet principle), Alexander Soifer and Edward Lozansky, Jan90, p24 (Feature)
Ping-Pong in the Sink (Bernoullian behavior), Alexey Byalko, Jul/Aug93, p48 (In the Lab)
Pins and Spin (a bowling problem with a twist), Arthur Eisenkraft and Larry D. Kirkpatrick, Jul/Aug95, p34 (Physics Contest)
A Pivotal Approach (applying rotation in problem solving), Boris Pritsker, May/Jun96, p44 (At the Blackboard)
The Pizza Theorem—Part I (equality of off-center slices), George Berzsenyi, Jan/Feb94, p29 (Math Investigations)
The Pizza Theorem—Part II (including the Calzone Theorem), George Berzsenyi, Mar/Apr94, p29 (Math Investigations)
Planar Graphs (can you make the connections?), A. Y. Olshansky, Jan/Feb98, p10 (Feature)
A Planetary Air Brake (viscous drag and the slowing of the Earth), D. C. Agrawal and V. J. Menon, Mar/Apr97, p40 (At the Blackboard)
Planetary Building Blocks (blueprints for creating terra firma), V. Mescheryakov, Jul/Aug98, p4 (Feature)
Playing with the Ordinary (exploring everyday phenomena), Sep/Oct92, p32 (Kaleidoscope)
Play It Again … (inducing strange repetitions), John Conway, Nov/Dec90, p30 (Mathematical Surprises)
The Play of Light (results of a “slight” change in the rules), Dmitry Tarasov and Lev Tarasov, May/Jun96, p10 (Feature)
The Pointed Meeting of a Triangle’s Altitudes (various ways of proving a well-known theorem), I. F. Sharygin, Jul/Aug99, p28 (Kaleidoscope)
Points of Interest (unique locations within a triangle), I. F. Sharygin, Mar/Apr98, p34 (At the Blackboard)
A Polarizer in the Shadows (life and physics of Etienne Malus), Andrey Andreyev, Jan/Feb94, p44 (Looking Back)
A Portrait of Poisson (one of the founders of modern mathematical physics), B. Geller and Y. Bruk, Mar/Apr91, p21 (Innovators)
Portrait of Three Puzzle Graces (Rubiklike games and group theory), Vladimir Dubrovsky, Nov/Dec91, p63 (Toy Store)
The Power of Dimensional Thinking (problem-solving method), Yuly Bruk and Albert Stasenko, May/Jun92, p34 (Feature)
The Power of Likeness (strengths and weaknesses of analogy), S. R. Filonovich, Sep/Oct91, p22 (Feature)
The Power of the Sun and You (surprises of scale), V. Lange and T. Lange, Jul/Aug96, p16 (Feature)
A Prelude to the Study of Physics (models and their role in science), Robert J. Sciamanda, Nov/Dec96, p45 (Fundamentals)
The Price of Resistance (“kitchen experiments” on how the medium “pushes back”), S. Betyayev, Sep/Oct00, p38 (In the Lab)
Prime Time (prime number infinitude), G. A. Galperin, Jan/Feb99, p10 (Feature)
A Princess of Mathematics (excerpt from autobiography of Sofya Kovalevskaya), Yuly Danilov, Jan/Feb94, p37 (Anthology)
Principles of Vortex Theory (inside the hydronamics of Helmholtz), N. Zhukovsky, Mar/Apr00, p26 (Feature)
The Problem Book of Anania of Shirak (ancient Armenian mathematics), Yuly Danilov, Mar/Apr93, p42 (Looking Back)
The Problem Book of History (mathematical approach to the past), Yuly Danilov, Sep/Oct93, p47 (Looking Back)
The Problem of Eight Points (intersecting lines), N. B. Vasiliev, Jan/Feb99, p25 (At the Blackboard)
Problem Racing (formulating math problems out of everyday experiences), Gary Sherman, Mar/Apr91, p45 (In Your Head)
Problems Beget Problems (follow-up on previously published problems), George Berzsenyi, Sep/Oct95, p40 (Math Investigations)
Problems Teach Us How to Think (as Euler said, “My pencil is sometimes more clever than my head”), V. Proizvolov, Jan/Feb01, p42 (Problem Primer)
Programming Challenges (problems from the 1994 IOI), Jan/Feb95, p49 (Happenings)
Ptolemy’s Trigonometry (proving and using his theorem), V. Zatakavai, jan/Feb01, p40 (Looking Back)
Q
The Quadratic (something old, something new), Vladimir Boltyansky, Sep/Oct95, p45 (At the Blackboard)
The Quadratic Trinomial (combining algebraic and geometric reasoning), A. Bolibruch, V. Uroev, and M. Shabunin, May/Jun00, p36 (At the Blackboard)
Quantum in Outer Space and the Inner Space of Art (International Space Year and Kvant art), Bill G. Aldridge, May90, p3 (Publisher’s Page)
The Quantum Nature of Light (visible proof of quanta), D. Sviridov and R. Sviridova, Nov/Dec98, p28 (Looking Back)
Quaternions (simple operations with complex numbers), A. Mishchenko and Y. Solovyov, Sep/Oct00, p4 (Feature)
Queens on a Cylinder (cylindrical and toroidal chess [see “Torangles and Torboards,” Mar/Apr94]), Alexey Tolpygo, May/Jun96, p38 (Follow-up)
Questioning Answers (in every ending is a beginning), Barry Mazur, Jan/Feb97, p4 (Feature)
A Question of Complexity (and the need to simplify in solving physics problems), Arthur Eisenkraft and Larry D. Kirkpatrick, Nov/Dec99, p32 (Physics Contest)
R
Raising the Boats or Lowering the Water (misuse of the National Science Education Standards), Bill G. Aldridge, May/Jun95, p2 (Publisher’s Page)
Ramanujan the Phenomenon (India’s inspired mathematician), S. G. Gindikin, Mar/Apr98, p4 (Feature)
Randomly Seeking Cipollino (introduction to random walk), S. Sobolev, Jul/Aug93, p20 (Feature)
Reaching Back (extending a helping hand), Bill G. Aldridge, Nov/Dec91, p5 (Publisher’s Page)
Rearranging Sums (math challenge), George Berzsenyi, Jan/Feb91, p18 (Contest)
Reflection and Refraction (a look at optics), Sep/Oct91, p32 (Kaleidoscope)
Relativistic Conservation Laws (special relativity is no excuse not to obey conservation laws), Larry D. Kirkpatrick and Arthur Eisenkraft, Nov/Dec00, p30 (Physics Contest)
Relativity of Motion (frames of reference), A. I. Chernoutsan, Mar/Apr99, p44 (At the Blackboard)
Remarkable Geometric Formulas (algebraic relations), I. F. Sharygin, Mar/Apr99, p28 (Kaleidoscope)
Remarkable Limits (generated by classical means), M. Crane and A. Nudelman, Jul/Aug97, p34 (At the Blackboard)
Repartitioning the World (population and the powers of two), V. Arnold, Jan/Feb00, p34 (Digit Demographics)
Resistance in the Multidimensional Cube (a physical application of a math concept), F. Nedemeyer and Y. Smorodinsky, Sep/Oct96, p12 (Feature)
Restricted Distances (math challenge), George Berzsenyi, Jan/Feb92, p31 (Math Investigations)
Returning to a Former State (Rubik’s Cube and periodicity), A. Savin, Nov/Dec99, p28 (Kaleidoscope)
Revisiting Napoleon’s Theorem (via the internet), George Berzsenyi, Jul/Aug95, p37 (Math Investigations)
Revisiting the N-cluster Problem (a classic problem from Math.Note at DEC), George Berzsenyi, Jan/Feb97, p47 (Math Investigations)
A Revolution Absorbed (how non-Euclidean geometry entered the mainstream), E. B. Vinberg, Jan/Feb97, p18 (Feature)
Revolutionary Teaching (the Ecole Polytechnique in Paris), Yuri Solovyov, Mar/Apr98, p26 (Looking Back)
The Riddle of the Etruscans (gold spheres on jewelry), A. S. Alexandrov, Sep/Oct91, p42 (In the Lab)
A Ride on Sierpinski’s Carpet (fractals in the mind and in nature), I. M. Sokolov, May/Jun92, p6 (Feature)
Rigidity of Convex Polyhedrons (solid solutions), N. P. Dolbilin, Sep/Oct98, p8 (Feature)
Ripples on a Cosmic Sea (graviational waves), Shane S. Larson, Mar/Apr01, p4 (Feature)
Rising Star (a problem of wave interference), Arthur Eisenkraft and Larry D. Kirkpatrick, Sep/Oct94, p44 (Physics Contest)
Rivers, Typhoons, and Molecules (all are affected by the Coriolis force), Albert Stasenko, Jul/Aug98, p38 (At the Blackboard)
Rock ’n’ No Roll (rocking cliffs), A. Mitrofanov, Mar/Apr01, p18 (Feature)
The Rolling Cubes (solutions and records), Vladimir Dubrovsky, May/Jun94, p62 (Toy Store)
Rolling Wheels (design considerations facing the engineer), Arthur Eisenkraft and Larry D. Kirkpatrick, May/Jun00, p30 (Physics Contest)
Rook versus Knight (twists in a common endgame), Yevgeny Gik, Nov/Dec90, p64 (Checkmate!)
A Rotating Capacitor (electromagnetic fields and motion), A. Stasenko, May/Jun99, p34 (At the Blackboard)
Row, Row, Row Your Boat (physics challenge), Arthur Eisenkraft and Larry D. Kirkpatrick, Jan/Feb93, p42 (Physics Contest)
A Royal Problem (marital tension on a chessboard), Martin Gardner and Andy Liu, Jul/Aug93, p30 (Checkmate!)
Rubik Art (monumental designs built from the classic cube), May/Jun97, p31 (Toy Store)
Russian Bazaar (economic hard times), Bill G. Aldridge, May/Jun92, p2 (Publisher’s Page)
S
Sally Ride (biographical sketch), Jan90, p39 (Innovators)
Satellite Aerodynamic Paradox (orbital irregularities), A. Mitrofanov, Jan/Feb99, p18 (Feature)
The Satellite Paradox (acceleration upon entering atmosphere), Y. G. Pavlenko, Mar/Apr93, p50 (At the Blackboard)
Savoring Science (piquancy of primary sources), Bill G. Aldridge, Nov/Dec93, p2 (Publisher’s Page)
The School Bus and the Mud Puddles (inclusion-exclusion theory), Thomas P. Dence, Jan/Feb95, p24 (Feature)
Science and Fanaticism (reflections on public policy), Bill G. Aldridge, Sep/Oct92, p2 (Publisher’s Page)
The Science of Pole Vaulting (materials and techniques), Peter Blanchonette and Mark Stewart, May/Jun01, p48 (Physical Education)
The Science of the Jump-Shot (basketball kinematics), Roman Vinokur, Jan/Feb93, p46 (At the Blackboard)
Science vs. the UFO (solving a tranformational puzzle), Will Oakley, Jan/Feb92, p84 (Toy Store)
Science with Charm (communicating the simplicity of physics), Bernard V. Khoury, Mar/Apr98, p2 (Front Matter)
sciLINKS: The World’s a Click Away (techy textbooks), Gerald F. Wheeler, Sep/Oct98, p2 (Front Matter)
Scores and SNO in Sudbury (report on the 1997 International Physics Olympiad), Nov/Dec97, p44 (Happenings)
Sea Sounds (underwater refraction of sound waves), Arthur Eisenkraft and Larry D. Kirkpatrick, Mar/Apr96, p34 (Physics Contest)
Sea Waves (describing wave motion), L. A. Ostrovsky, Nov/Dec98, p20 (Feature)
The Secret of the Venerable Cooper (Johannes Kepler and mysterious barrels), M. B. Balk, May90, p36 (Looking Back)
Seeing is Believing (visual proofs of the Pythagorean theorem), Daniel J. Davidson and Louis H. Kauffman, Jul/Aug97, p24 (Feature)
Selecting the Best Alternative (mathematical programming and problems of management), V. Gutenmakher and Zh. Rabbot, Nov/Dec99, p36 (At the Blackboard)
Self-propelled Sprinkler Systems (an attempt at applying Segner’s wheel), A. Stasenko, May/Jun01, p40 (In the Open Air)
Self-similar Mosaics (when the whole is the sum of its parts), N. Dolbilin, Jul/Aug00, p4 (Feature)
Shady Computations (a paradox at the boundary of dark and light), Chauncey W. Bowers, Nov/Dec96, p34 (At the Blackboard)
Shake, Rattle, and Roll (physics challenge), Arthur Eisenkraft and Larry D. Kirkpatrick, May/Jun92, p40 (Physics Contest)
Shall We Light a Fire in the Fireplace? (an equation that seems to say: “Don’t bother”), Victor Lange, Jan/Feb96, p40 (At the Blackboard)
Shape Numbers (exploring a Fermat hypothesis), A. Savin, Sep/Oct00, p14 (Feature)
Shapes and Sizes (math challenge), George Berzsenyi, Nov/Dec 90, p34 (Contest)
Sharing a Point (a handy method for a common geometric challenge), I. Sharygin, Jul/Aug00, p35 (At the Blackboard)
Shortest Networks (Jacob Steiner’s famous problem), E. Abakumov, O. Izhboldin, L. Kurlyandchik, and N. Netsvetayev, May/Jun93, p4 (Feature)
Shortest Path (Edsger Kijkstra and his algorithm), Don Piele, May/Jun00, p54 (Informatics)
Short Takes (jokes, cartoons), Mar/Apr91, p11 (Quantum Smiles)
The Short, Turbulent Life of Evariste Galois (a revolutionary in politics and math), Y. P. Solovyov, Nov/Dec91, p42 (Looking Back)
Shouting into the Wind (quantifying how sounds fade on windy days), G. Kotkin, Nov/Dec00, p40 (In the Open Air)
Signals, Graphs, and Kings on a Torus (ensuring error-free communication), A. Futer, Nov/Dec95, p12 (Feature)
A Simple Capacity for Heat (specific heat and molecular motion), Valeryan Edelman, Nov/Dec93, p22 (Feature)
The Simplicity of Mathematics (complications of life, Stone Age math), Jan/Feb91, p48 (Quantum Smiles)
The Sines and Cosines You Do and Don’t Know (survey with linguistic digressions), Nov/Dec93, p32 (Kaleidoscope)
Sink or Swim (whales and buoyancy), N. Rodina, May/Jun00, p34 (In the Open Air)
Sir Isaac Newton (poem), David Arns, p14, Mar/Apr98
Six Challenging Dissection Tasks (and the birth of “high-phi”), Martin Gardner, May/Jun94, p26 (Mathematical Surprises)
Sky (poem), David Arns, Mar/Apr99, p54
Slinking Around (springy physics), Diar Chokin, Nov/Dec92, p64 (Toy Store)
Slipping Silage (how to calculate the amount of stolen hay), Dr. Mu, May/Jun97, p63 (Cowculations)
Smale’s Horseshoe (a venture in symbolic dynamics), Yuly Ilyashenko and Anna Kotova, May/Jun95, p12 (Feature)
Smoky Mountain (why the air is warmer on the leeward side), Ivan Vorobyov, Nov/Dec95, p38 (At the Blackboard)
A Snail That Moves Like Light (physics challenge), Arthur Eisenkraft and Larry D. Kirkpatrick, Sep/Oct91, p28 (Physics Contest)
Solar Calculator (accurate thinking about precision), Bill G. Aldridge, Sep/Oct96, p2 (Publisher’s Page)
Solving for the Slalom (understand the forces and it’s all downhill from there), A. Abrikosov, Nov/Dec99, p20 (Feature)
Some Mathematical Magic (“magic squares” and a magic tesseract), John Conway, Mar/Apr91, p28 (Mathematical Surprises)
Some Things Never Change (problem solving with invariants), Yury Ionin and Lev Kurlyandchik, Sep/Oct93, p34 (Feature)
Songs That Shatter and Winds That Howl (sound thinking), Jan/Feb94, p32 (Kaleidoscope)
Sound Power (intense acoustic waves), O. V. Rudenko and V. O. Cherkezyan, Sep/Oct98, p26 (Feature)
Sources, Sinks, and Gaussian Spheres (physics challenge), Arthur Eisenkraft and Larry D. Kirkpatrick, Jul/Aug92, p24 (Physics Contest)
So What’s the Joke? (the damage done by a computer virus), Bill G. Aldridge, Jan/Feb96, p2 (Publisher’s Page)
So What’s the Point? (replacing algebra with geometry in vector analysis), Gary Haardeng-Pedersen, Mar/Apr96, p48 (At the Blackboard)
So, What’s Wrong? (debunking problematic solutions), I. F. Sharygin, Jul/Aug98, p34 (Feature)
Space Physics: A Voyage of Adventure (near-Earth phenomena), M. Frank Ireton, Sue Cox Kauffman, Ron Morse, and Mark Pesses, Nov/Dec92, p40 (Poster)
Spinning Gold from Straw (how two secrets can add up to one certainty), S. Artyomov, Y. Gimatov, and V. Fyodorov, Jul/Aug96, p20 (Feature)
Spinning in a Jet Stream (Bernoulli, Magnus, and a vacuum cleaner), Stanislav Kuzmin, Sep/Oct94, p49 (In the Lab)
Split Image (behavior of light in a broken lens), Arthur Eisenkraft and Larry D. Kirkpatrick, Sep/Oct95, p36 (Physics Contest)
Sportin’ Life (physics of free throws and field goals), Arthur Eisenkraft and Larry Kirkpatrick, Jan/Feb99, p30 (Physics Contest)
Square or not Square? (recognizing which numbers can’t be perfect squares), Mark Saul and Titu Andreescu, Jul/Aug99, p49 (Gradus ad Parnassum)
Squaring the Hyperbola (a different approach to logarithms and exponents), Andrey Yegorov, Mar/Apr97, p26 (Feature)
Squeaky Doors, Squealing Tires, and Singing Violins (dry friction), I. Slobodetsky, Nov/Dec92, p46 (In the Lab)
A Star is Born (gravity backs a stellar production), V. Surdin, Mar/Apr00, p12 (Feature)
The Steiner–Lehmus Theorem (addressing angle bisectors), I. F. Sharygin, Nov/Dec98, p26 (At the Blackboard)
Stirring Up Bubbles (vapor cones and vortices in a boiling liquid), T. Polyakova, V. Zablotsky, and O. Tsyganenko, Mar/Apr97, p52 (In the Lab)
The Stomachion (Archimedean game), Yuly Danilov, Jan/Feb93, p64 (Toy Store)
Stop on Red, Go on Green … (what to do on yellow?), Arthur Eisenkraft and Larry D. Kirkpatrick, Jan/Feb94, p34 (Physics Contest)
The Story of a Dewdrop (surface shape and phase equilibrium), A. A. Abrikosov, Sep/Oct92, p34 (Feature)
A Strange Emperor and a Strange General (psychology and numerical avalanches), Igor Akulich, May/Jun94, p16 (Feature)
Stretching Exercise (solutions to challenging geometry problems), Donald Barry, Jul/Aug01, p12 (Feature)
Strips on a Board (close packing in two dimensions), Boris Kotlyar, Nov/Dec94, p63 (Toy Store)
Strolling to Chebyshev’s Theorem (problems in honor of the Chebyshev centennial), Victor Ufnarovsky, Nov/Dec94, p4 (Feature)
Student Inventors Show Their Stuff (Duracell/NSTA Scholarship Competition winners), May/Jun95, p52 (Happenings)
Suds Studies (soap films and bubbles), P. Kanaev, Jul/Aug98, p47 (In the Lab)
Suggestive Tilings (new material, old topics revisited), Vladimir Dubrovsky, Jul/Aug94, p36 (Follow-up)
A Summer Festival of Puzzlers (twelve problems from twelve countries), Jul/Aug93, p32 (Kaleidoscope)
Summer Study in New York and Tartu, Maryland and Moscow (Science and Mathematics International Institutes), May90, p54 (Happenings)
Summertime, and the Choosin’ Ain’t Easy (ice cream counting problem), Kurt Kreith, Jul/Aug92, p28 (At the Blackboard)
Summing Up (curiosities of single-digit addition), Mark Lucianovic, Jul/Aug92, p51 (Student Corner)
The Sum of Minima and the Minima of Sums (a general method for proving many well-known inequalities), R. Alekseyev and L. Kurlyandchik, Jan/Feb01, p34 (At the Blackboard)
Superconducting Magnet (how its superconducting switch works), Arthur Eisenkraft and Larry D. Kirkpatrick, Nov/Dec94, p36 (Physics Contest)
The Superfluidity of Helium II (loss of viscosity at a low temperature), Alexander Andreyev, Jan90, p34 (Feature)
Superheated by Equations (mathematics of heat exchange), Dmitry Fomin, Jul/Aug93, p4 (Feature)
Superprime Beef (superprimes and repusprimes), Dr. Mu, Jan/Feb97, p55 (Cowculations)
The Superproblem of Space Flight (origins of Tsiolkovsky formula), Albert Stasenko, Jul/Aug94, p20 (Feature)
Surfing the Electromagnetic Spectrum (an array of questions and facts), Jan/Feb95, p32 (Kaleidoscope)
Surprises of Conversion (proving the converse of theorems), I. Kushnir, Mar/Apr96, p38 (Sticking Points)
Surprises of the Cubic Formula (an equation of little use and much significance), Dmitry Fuchs and Irene Klumova, May/Jun96, p16 (Feature)
Suspending Belief (calculating a bridge’s curve), Y. S. Petrov, Jul/Aug93, p28 (At the Blackboard)
Swinging from Star to Star (accelerating a spacecraft into the cosmos), Vladimir Surdin, Mar/Apr97, p4 (Feature)
Swinging Techniques (parametric resonance), Alexey Chernoutsan, May/Jun93, p64 (Toy Store)
Swords into Plowshares (Russian wingships and California fires), Bill G. Aldridge, Jan/Feb94, p2 (Publisher’s Page)
Symmetry in Algebra (getting started with group theory), Mark Saul and Titu Andreescu, Mar/Apr98, p43 (Gradus ad Parnassum)
Symmetry, Part II (polynomial equations and their roots), Mark Saul and Titu Andreescu, May/Jun98, p34 (Gradus ad Parnassum)
Symmetry in Algebra, Part III (using the factor theorem), Mark Saul and Titu Andreescu, Jul/Aug98, p41 (Gradus ad Parnassum)
The Symmetry of Chance (introduction to geometric probability), Nikolay Vasilyev, May/Jun93, p22 (Feature)
Symmetry on the Chessboard (accidental and intentional symmetry), Yevgeny Gik, May90, p64 (Checkmate!)
T
Tackling Twisted Hoops (invariants and untangling challenges), S. Matveyev, Nov/Dec00, p8 (Feature)
Tactile Microscopes (sensing techniques), A. Volodin, Jan/Feb93, p36 (Feature)
Taking Advantage (hard times for Russian science), Bill G. Aldridge, Jan/Feb93, p2 (Publisher’s Page)
Taking a Flying Leap (Hooke’s law on a South Seas island), A. A. Dozorov, Sep/Oct90, p10 (At the Blackboard)
Taking on Triangles (recounting a solution, with fruitful “side trips”), A. Kanel and A. Kovaldzhi, Mar/Apr01, p10 (Feature)
Taking the Earth’s Temperature (how hot is the Earth’s core?), Alexey Byalko, Jan/Feb95, p4 (Feature)
A Tale of One City (Tournament of Towns report), Andy Liu, May/Jun94, p50 (Happenings)
The Talking Wave of the Future (fiber optics), Yury Nosov, Nov/Dec92, p12 (Feature)
A Talk with Professor I. M. Gelfand (reminiscences of a mathematical boyhood), recorded by V. S. Retakh and A. B. Sosinsky, Jan/Feb91, p20 (Feature)
Tartu in the Summer of ’91 (math program for American and Soviet students), Mark Saul, Mar/Apr92, p56 (Happenings)
A Tell-tale Trail and a Chemical Clock (two experiments with alternating current), N. Paravyan, Sep/Oct95, p42 (In the Lab)
Temperature, Heat, and Thermometers (overview of temperature and its measurement), A. Kikoyin, May90, p16 (Feature)
Thanks for Your Support! (end-of-year ruminations), Bill G. Aldridge, Mar/Apr91,p3 (Publisher’s Page)
The Theorem of Menelaus (the secant line), B. Orach, May/Jun01, p44 (At the Blackboard)
The Thermodynamic Universe (does time have a beginning and an end?), I. D. Novikov, Mar/Apr98, p10 (Feature)
Think Fast! (order-of-magnitude estimates in physics), G. V. Meledin, Mar/Apr91, p36 (Feature)
Think Twice, Code Once (cutting a tree trunk into boards), Dr. Mu, May/Jun99, p55 (Cowculations)
This Just In … (exchange of scientific views in the daily press), Jan/Feb91, p48 (Quantum Smiles)
Thoroughly Modern Diophantus (the arithmetic of elliptic curver), Y. Solovyov, Sep/Oct99, p10 (Feature)
The Three Chords Theorem (new version of the first part of Ptolemy’s theorem), Shikong Le and Lioukan Chen, Jul/Aug01, p48 (At the Blackboard)
Three Golds and Two Silvers in Italy (report on the XXX International Physics Olympiad), Mary Mogge and Leaf Turner, Nov/Dec99, p52 (Happenings)
Three Metaphysical Tales (profound thoughts of lines, light, and planets), A. Filonov, Mar/Apr94, p28 (Quantum Smiles)
Three Paths to Mt. Fermat-Euler (primes and squares), Vladimir Tikhomirov, May/Jun94, p4 (Feature)
Three Physicists and One Log (which physicist bears the brunt?), Roman Vinokur, Mar/Apr97, p48 (At the Blackboard)
Thrills by Design (physics in the amusement park), Arthur Eisenkraft and Larry D. Kirkpatrick, Sep/Oct93, p38 (Physics Contest)
Through a Glass Brightly (remarkable properties of green glass), B. Fabrikant, Sep/Oct90, p34 (In the Lab)
Through the Decimal Point (quadratics and 10-adic numbers), A. B. Zhiglevich and N. N. Petrov, Jul/Aug94, p16 (Feature)
Throwing the Book at Them (critique of textbooks), Bill G. Aldridge, Nov/Dec94, p2 (Publisher’s Page)
Tied into Knot Theory (the basics of mathematical knots), O. Viro, May/Jun98, p16 (Feature)
Time to Move On … (celebrating Quantum’s 12 years), Arthur Eisenkraft, Jul/Aug01, p3 (Front Matter)
The Tip of the Iceberg (physics challenge), Arthur Eisenkraft and Larry D. Kirkpatrick, Sep/Oct92, p24 (Physics Contest)
To Calculate or Guess—You Decide! (the virtues of guessing), I. Akulich, Mar/Apr91, p47 (In Your Head)
To Err Is Human (correction is the key), Bill G. Aldridge, Nov/Dec92, p2 (Publisher’s Page)
To Flexland with Mr. Flexman (two flexible toys), Alexey Panov, Mar/Apr92, p64 (Toy Store)
Tomahawk Throwing Made Easy (physics of getting the hatchet to stick), V. A. Davydov, Nov/Dec90, p4 (Feature)
A Topless Roller Coaster (physics challenge), Arthur Eisenkraft and Larry D. Kirkpatrick, Nov/Dec92, p28 (Physics Contest)
Topology and the Lay of the Land (mathematical topography), Mikhail Shubin, Sep/Oct92, p4 (Feature)
Topsy-turvy Pyramids (rolling-block puzzles), Vladimir Dubrovsky, Sep/Oct93, p63 (Toy Store)
Torangles and Torboards (toroidal constructions), Vladimir Dubrovsky, Mar/Apr94, p63 (Toy Store)
The Torch is Passed (introducing NSTA’s new Executive Director), Bill G. Aldridge, Sep/Oct95, p2 (Publisher’s Page)
Tori, Tori, Tori! (bagels and beyond), Mar/Apr94, p32 (Kaleidoscope)
Toroidal Currency (money with a twist), Martin Gardner, Sep/Oct94, p52 (Mathematical Surprises)
The Tournament of Towns (international math competition), Nikolay Konstantinov, Jan90, p50 (Happenings)
The Toy that Drove the Universe (critique of the “anthropic principle”), Jef Raskin, Nov/Dec99, p49
Trees Worthy of Paul Bunyan (physics and tree growth), Anatoly Mineyev, Jan/Feb94, p4 (Feature)
Triad and True (puzzles based on invariants), Vladimir Dubrovsky, Jan/Feb95, p62 (Toy Store)
Triangles of Differences (math challenge), George Berzsenyi, May/Jun92, p30 (Math Investigations)
Triangles of Sums (math challenge), George Berzsenyi, Jul/Aug92, p53 (Math Investigations)
Triangles with the Right Stuff (a special case of right triangles), L. D. Kurlyandchik, Jul/Aug98, p32 (Kaleidoscope)
Triangular Surgery (problems in which polygons are sliced into triangles), O. Izhboldin and L. Kurlyandchik, Nov/Dec00, p34 (At the Blackboard)
Tricky Rearrangements (more rolling-block puzzles), Vladimir Dubrovsky, Nov/Dec93, p63 (Toy Store)
A Trio of Topics (center of mass, electricity in metals, time travel) A. I. Chernoutsan and Andrey Varlamov, Sep/Oct92, p47 (At the Blackboard)
True on the Face of It (clockwork refutation of Zeno’s paradox), Gordon Moyer, Jul/Aug95, p16 (Feature)
Tunnel Trouble (dropping an apple in a tunnel through the Earth), Arthur Eisenkraft and Larry D. Kirkpatrick, Jan/Feb00, p32 (Physics Contest)
Turning Algebraic Identities into Geometric Inequalities (a use for complex numbers), Zalman Skopets, Sep/Oct94, p41 (At the Blackboard)
Turning the Incredible into the Obvious (non-Euclidean geometry), Vladimir Boltyansky, Sep/Oct92, p18 (Feature)
Turning the Tides (understanding the attraction of the Moon), V. E. Belonuchkin, May/Jun98, p10 (Feature)
2-adic Numbers (introduction to Hensel distances), B. Becker, S. Vostokov, and Y. Ionin, Jul/Aug99, p21 (Feature)
Two Physics Tricks (reluctant water becomes a fountain), V. Mayer and E. Mamayeva, Mar/Apr91, p35 (In the Lab)
Tycho, Lord of Uraniborg (a portrait of the great astronomer Tycho Brahe), J. D. Haines, Sep/Oct00, p25 (Looking Back)
U
Unidentical Twins (using conjugate numbers to tame irrationalitites), V. N. Vaguten, Nov/Dec97, p20 (Feature)
Uninscribable Polyhedrons? (Proving the Steinitz theorem), E. Andreev, May/Jun01, p18 (Feature)
The Universe Discovered (from contemplation to calculation), Yury Solovyov, May/Jun92, p12 (Feature)
A Universe of Questions (what we know about the universe), Yakov Zeldovich, Jan/Feb92, p6 (Feature)
The Unlimited Appeal of The Limits to Growth (it sparked the debate on “sustainable” economies), Tim Weber, Sep/Oct97, p2 (Front Matter)
An Unsinkable Disk (hands-on hydraulics), A. Luzin, Sep/Oct99, p42 (In the Lab)
Up the Down Incline (gravity defied? or obeyed unusually?), Alexander Mitrofanov, Mar/Apr96, p44 (In the Lab)
Up, Up, and Away (hot air rising), Arthur Eisenkraft and Larry Kirkpatrick, Sep/Oct98, p34 (Physics Contest)
The USA Computing Olympiad (report), Donald T. Piele, May/Jun93, p51 (Happenings)
The USA Mathematical Talent Search (competition without time pressure), George Berzsenyi, Sep/Oct90, p56 (Happenings)
Using Cents to Sense Surface Tension (experiments with pennies and fluids), Mary E. Stokes and Henry D. Schreiber, Mar/Apr01, p42 (In the Lab)
US Physics Team Places Third in Beijing (XXV International Physics Olympiad), Nov/Dec94, p50 (Happenings)
US Team Places Second at IMO (report), Cecil Rousseau and Daniel Ullman, Jan/Feb93, p51 (Happenings)
US Wins Gold at the International Physics Olympiad (report), Larry D. Kirkpatrick, Nov/Dec92, p51 (Happenings)
V
Vacuum (making something out of nothing), A. Semenov, Jul/Aug99, p12 (Feature)
Van der Waals and his Equation (making an ideal gas real), B. Yavelov, Nov/Dec97, p36 (Looking Back)
Van der Waerden’s Pathological Function (examining a “miserable sore”), B. Martynov, Jul/Aug98, p12 (Feature)
Van Rooman’s Challenge (solving a baffling equation), Yury Solovyov, Jan90, p42 (Looking Back)
Variations on a Theme (the Arithmetic Mean-Geometric Mean inequality), Mark Saul and Titu Andreescu, Jan/Feb98, p37 (Gradus ad Parnassum)
Vavilov’s Paradox (apparent violation of energy conservation law), V. A. Fabrikant, Jul/Aug92, p49 (At the Blackboard)
A Venusian Mystery (the riddle of her rotation), Vladimir Surdin, Jul/Aug96, p4 (Feature)
The View from the Masthead (new subtitle, departure, clarification), Bill G. Aldridge, Sep/Oct93, p2 (Publisher’s Page)
The View through a Bamboo Screen (birth of the modulation collimator), Minoru Oda, Jan/Feb92, p34 (Feature)
Vikings and Voltmeters (report on the 1996 International Physics Olympiad), Dwight E. Neuenschwander, Sep/Oct96, p52 (Happenings)
A Viscous River Runs Through It (the engine-saving properties of motor oil), Henry D. Schreiber, Nov/Dec95, p42 (In the Lab)
Visionary Science (atmospheric anomalies), V. Novoseltsev, May/Jun98, p21 (Feature)
Volta, Oersted, and Faraday (titans of electricity), A. Vasilyev, Jul/Aug01, p34 (Looking Back)
Volumes without Integrals (the Cavalieri principle), I. F. Sharygin, Mar/Apr97, p32 (Kaleidoscope)
W
Wacky Pyramids (cubic trisection), Yakov Smorodinsky, Mar/Apr93, p64 (Toy Store)
Wake Up! (brainteasers for vacationers), Anatoly Savin, Jul/Aug92, p32 (Kaleidoscope)
Walker in a Winter Wonderland (musings inspired by The Flying Circus of Physics), Alexander Borovoy, May90, p52 (In the Lab)
Walking on Water (physics of unusual modes of locomotion), K. Bogdanov, Jan/Feb91, p36 (Feature)
A Walk on the Sword’s Edge (literally—is it possible?), V. Meshcheryakov, Jan/Feb96, p10 (Feature)
Warp Speed (traveling faster than light), Arthur Eisenkraft and Larry Kirkpatrick, Nov/Dec98, p32 (Physics Contest)
The “Water Worm” (the Archimedean screw), M. Golovey, Jan/Feb97, p40 (In the Lab)
A Watery View and Waterloo (waves, mirages, and the sounds of battle), A. Stasenko, Mar/Apr00, p48 (In the Open Air)
Wave Interference (light diffractin and interference patterns), L. Bakanina, Jul/Aug01, p42 (At the Blackboard)
The Wave Mechanics of Erwin Schrödinger (an aspect of quantum theory), A. Vasilyev, May/Jun01, p36 (Looking Back)
Wave on a Car Tire (limitations to speed), L. Grodko, Nov/Dec98, p10 (Feature)
Waves Beneath the Waves (ocean acoustics), L. Brekhovskikh and V. Kurtepov, Jan/Feb98, p16 (Feature)
Wave Watching (investigation of a fundamental phenomenon), L. Aslamazov and I. Kikoyin, Jan/Feb91, p12 (Feature)
Weighing an Astronaut (weight-watching while weightless), Arthur Eisenkraft and Larry D. Kirkpatrick, Mar/Apr95, p36 (Physics Contest)
Weightlessness in a Car? (road-trip physics), Sergei Pikin, Jul/Aug98, p31 (At the Blackboard)
Weightlessness in a Magic Box (some assembly required), A. Dozorov, May/Jun99, p41 (In the Lab)
Welcome to International Space Year! (introduction to special ISY issue of Quantum), L. A. Fisk, Jan/Feb92, p2 (Guest Page)
Welcome to Quantum! (birth of Quantum), Bill G. Aldridge, Jan90, p5 (Publisher’s Page)
What a Commotion! (molecular motion), May90, p32 (Kaleidoscope)
What Did the Conductor Say? (mathematical induction), Mikhail Gerver, Jul/Aug92, p38 (Feature)
What Goes Up … (physics challenge), Arthur Eisenkraft and Larry D. Kirkpatrick, Jan/Feb92, p32 (Physics Contest)
What Happens at the Boundary (surface tension, surface films, and other interesting phenomena), A. Borovoy and Y. Klimov, Jan/Feb01, p46 (In the Lab)
What Harmony Means (exploration of harmonic mean), Vladimir Dubrovsky and Anatoly Savin, Jan/Feb93, p32 (Kaleidoscope)
What I Learned in Quantum Land (poem), David Arns, Jan/Feb98, p52 (Musings)
What is Elegance? (what mathematicians say), Julia Angwin, Jan/Feb95, p34 (Ruminations)
What is Thought? (and where does it happen?), V. Meshcheryakov, May/Jun01, p22 (Feature)
What Little Stars Do (physics of twinkling), Pavel Bliokh, Mar/Apr94, p22 (Feature)
What’s New in the Solar System? (applying old laws of orbital motion), Nov/Dec90, p32 (Kaleidoscope)
What’s That You See? (misperception of light), B. M. Bolotovsky, Mar/Apr93, p4 (Feature)
What’s the “Best” Answer? (it’s not just a matter of getting the right answer), Boris Kordemsky, Jul/Aug96, p28 (Kaleidoscope)
What the Seesaw Taught (physics challenge), Arthur Eisenkraft and Larry D. Kirkpatrick, Jan/Feb91, p19 (Contest)
What You Add is What You Take (going backward to go forward), Andrey Yegorov, Nov/Dec94, p40 (At the Blackboard)
When a Body Meets a Body (Giant Impact theory of the Moon’s formation), A. G. W. Cameron, Jan/Feb95, p16 (Feature)
When Days Are Months (physics challenge), Arthur Eisenkraft and Larry D. Kirkpatrick, May90, p34 (Contest)
When Things Fall Apart (exercises in stability), Boris Korsunsky, May/Jun99, p4 (Feature)
When Trojans and Greeks Collide (the challenge of multibody systems), I. Vorobyov, Sep/Oct99, p16 (Feature)
Where Do Problems Come From? (the art of problem composition), I. Sharygin, Jan/Feb01, p18 (Feature)
Where is Last Year's Winter? (a problem of heat exchange), A. Stasenko, May/Jun01, p30 (In the Lab)
While the Water Evaporates … (time enough to think about the rate of the process), Mikhail Anfimov and Alexey Chernoutsan, Jul/Aug96, p25 (At the Blackboard)
Whirlwinds over the Runway (vortices generated by large jet planes), Albert Stasenko, Jul/Aug97, p42 (At the Blackboard)
Whistling in Space (electromagnetic signals from outer space), Pavel Bliokh, Mar/Apr97, p18 (Feature)
Who Needs a Lofty Tower? (the Earth’s rotation adds a twist to certain experiments), A. Stasenko, May/Jun00, p39 (At the Blackboard)
Who Owns Roman Numerals? (the history and practice of I’s, V’s, X’s …), Steven Schwartzman, Jan/Feb96, p4 (Feature)
Why Are the Cheese Holes Round? (transmission of pressure), Sergey Krotov, Nov/Dec90, p46 (In Your Head)
Why Doesn’t the Sack Slide? (impulsive sliding friction), Alexey Chernoutsan, May/Jun97, p50 (In the Lab)
Why Don’t Planes Fly with Cats and Dogs? (flight dynamics), S. K. Betyaev, Sep/Oct98, p14 (Feature)
Why Is a Burnt Match Bent? (playing with fire), V. Milman, Nov/Dec98, p40 (In the Lab)
Why Is the Sky Blue? (the physics behind the sky’s colors), Alexander Buzdin and Sergei Krotov, Mar/Apr98, p47 (In the Open Air)
Why Study Mathematics? (putting practicality in its place), Vladimir Arnold, Sep/Oct94, p24 (Feature)
Why Won’t Weeble Wobbly Go to Bed? (the physics of a “light-headed” toy), L. Borovinsky, May/Jun96, p64 (Toy Store)
The Wind in the Quicksilver (backward ion flow in mercury amalgams), Ivan Vorobyov, Jan/Feb96, p20 (Feature)
Winning Strategies (solutions to previous Kaleidoscope), Vladimir Dubrovsky, Sep/Oct95, p61 (Toy Store)
Wobbling Nuclear Drops (macrolaws in microworlds), Yuly Bruk, Maxim Zelnikov, and Albert Stasenko, Jan/Feb97, p12 (Feature)
The Wolf, the Baron, and Isaac Newton (action and reaction and more), V. A. Fabrikant, Nov/Dec91, p24 (Smiles)
The Wonderland of Poincaria (Lobachevsky bicentenary), Simon Gindikin, Nov/Dec92, p20 (Feature)
Word and Image (hints on how to read Quantum), Bill G. Aldridge, Mar/Apr95, p2 (Publisher’s Page)
The World3 Model (a graphic representation of a system dynamics model), Sep/Oct97, p32 (Kaleidoscope)
The World According to Malthus and Volterra (mathematical theory of the struggle for existence), Constantine Bogdanov, Jul/Aug92, p18 (Feature)
World-class Physics in Colonial Williamsburg (IPO report), Larry D. Kirkpatrick, Sep/Oct93, p51 (Happenings)
The World in a Bubble (sustainability in closed ecological systems), Joshua L. Tosteson, Sep/Oct97, p20 (Feature)
The World Puzzle Championship (report and sample puzzles), Vladimir Dubrovsky, Jul/Aug96, p55 (Toy Store)
The Worm Problem of Leo Moser—Part I (math challenge), George Berzsenyi, Jan/Feb93, p41 (Math Investigations)
The Worm Problem of Leo Moser—Part II (math challenge), George Berzsenyi, Mar/Apr93, p16 (Math Investigations)
The Worm Problem of Leo Moser—Part III (math challenge), George Berzsenyi, May/Jun93, p21 (Math Investigations)
A Wrinkle in Reality (excerpt from Lobachevsky’s New Elements of Geometry), Yuly Danilov, Jul/Aug92, p44 (Anthology)
The WRITE Stuff (how the USA 2000 Informatics Team was formed), Don Piele, Sep/Oct00, p53 (Informatics)
Y
Young US Mathematicians Excel in Bombay (report on the 1996 International Mathematical Olympiad), Sep/Oct96, p55 (Happenings)
About Quantum
About Quantum: The Magazine of Math and Science
Published from 1990 to 2001 by the National Science Teachers Association (NSTA), Quantum was a lively, handsomely illustrated bimonthly magazine of math and science (primarily physics). In addition to its feature articles, Quantum's departments included At the Blackboard (the beauty and usefulness of equations), In the Lab (hands-on science), Kaleidoscope (a collection of snippets designed to consolidate your grasp of a given topic), How Do You Figure? (challenging problems in physics and math), Brainteasers (fun problems requiring a minimum of math background), Looking Back (biographical and historical pieces), and Gallery Q (an exploration of links between art and science).
Quantum actively engaged the reader, posing questions and pursuing ideas as if they're brand new (even if they're a thousand years old). Most Quantum articles included problems for the reader to work through, and each issue contained an answer section. Some articles were elegant expositions of sophisticated concepts, and some gave an unexpected twist to a well-known idea or phenomenon. Others showed that there is no such thing as a silly question. For instance, one article asked: "Why are the holes in Swiss cheese round?" Quantum in effect said: smirk at your own risk—there's knowledge to be had in such "naive" questions!
Who published Quantum?
Quantum was published by the National Science Teachers Association (NSTA) and Springer-Verlag New York, Inc. NSTA was responsible for the editorial side of the magazine, and Springer printed and distributed it. NSTA worked closely with Quantum Bureau of the Russian Academy of Sciences in selecting and preparing material, and Quantum's advisory board consisted of members of the American Association of Physics Teachers (AAPT) and the National Council of Teachers of Mathematics (NCTM).
Where did Quantum come from?
It came from Kvant, Quantum's Russian-language parent magazine. Any given issue of Quantum was not, however, simply a translation of the corresponding issue of Kvant. NSTA used translations of selected articles from Kvant and supplemented them with original material generated in the US. After months of negotiations, the first issue of Quantum appeared in January 1990.
Kvant is a journal of math and physics founded in 1970 by two prominent Soviet scientists: the mathematician A. N. Kolmogorov and the physicist I. K. Kikoyin. (Kvant means "quantum" in Russian.) It was an outgrowth of a Russian educational tradition that encourages top-notch scientists to extend their teaching to high school students and to write challenging but accessible material for them. Kvant helped form an entire generation of Soviet scientists, many of whom wrote for Kvant / Quantum.
Kvant is still in production, and its archive is online.
What's in Quantum?
You'll find articles that make you think and articles that make you wonder. You'll find brainteasers and mathematical amusements. You'll find articles that teach you tricks of the trade, articles that broaden your sense of the scientific endeavor, and articles that bring classical notions to life. You'll find challenging problems in physics and math, and you'll find answers, hints, and solutions in the back of each issue. And you'll find full-color, thought-provoking artwork by award-winning Russian and American artists. Nominally, Quantum's target audience was high school and college students and their teachers. But actually, Quantum was aimed at the student in all of us.
Quantum awards
Quantum books
National Science Teaching Association
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405 E Laburnum Avenue Ste 3
Richmond, VA 23222
(T) 703.524.3646 (F) 703.243.7177 |
3745 | https://fiveable.me/key-terms/combinatorics/ncr | NCr - (Combinatorics) - Vocab, Definition, Explanations | Fiveable | Fiveable
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Combinatorics
NCr
🧮combinatorics review
key term - NCr
Citation:
MLA
Definition
nCr, or 'n choose r', represents the number of ways to choose r elements from a set of n elements without considering the order of selection. This mathematical concept is crucial for calculating combinations, especially in situations where the arrangement of selected items doesn't matter, as is often the case in probability and combinatorial problems.
5 Must Know Facts For Your Next Test
The formula for nCr is given by $$nCr = \frac{n!}{r!(n - r)!}$$, where n! denotes the factorial of n.
nCr is only defined for values where 0 ≤ r ≤ n; if r > n or r < 0, nCr equals 0.
The values of nCr are symmetric, meaning that $$nCr = nC(n - r)$$, allowing for easier calculations depending on the value of r.
When r equals 0 or n, nCr equals 1, representing the single way to select none or all elements from the set.
In practical applications, nCr is commonly used in probability problems and scenarios like lottery selections where order does not matter.
Review Questions
How does the concept of nCr relate to real-life situations where combinations are applicable?
nCr can be applied to many real-life scenarios such as forming committees or selecting teams. For example, if you need to choose 3 members from a group of 10 for a committee, using nCr helps you determine how many different groups can be formed without worrying about the order. This illustrates how important nCr is in organizing groups or selections in everyday decisions.
Discuss how the factorial function is utilized in calculating nCr and why it is significant in this context.
The factorial function is essential for calculating nCr as it provides a way to determine the total arrangements possible for selecting items. In the formula $$nCr = \frac{n!}{r!(n - r)!}$$, both the numerator and denominator involve factorials. This shows how many ways we can arrange all items and then adjusts for overcounting by removing arrangements of selected and unselected items. The factorial thus helps break down combinations into manageable calculations.
Evaluate the impact of understanding nCr on solving complex problems involving probability and statistics.
Grasping the concept of nCr greatly enhances one's ability to tackle complex probability problems and statistical analysis. For instance, when evaluating outcomes in a game involving random selections, knowing how to apply nCr allows for accurate calculations of winning combinations or potential outcomes. This understanding leads to better decision-making based on calculated probabilities and aids in analyzing data sets effectively.
Related terms
Factorial:
The product of all positive integers up to a specified number n, denoted as n!, which is used in calculating combinations and permutations.
Permutation:
An arrangement of objects in a specific order. Unlike combinations, permutations consider the sequence of selected items.
Binomial Theorem:
A formula that describes the algebraic expansion of powers of a binomial expression, which is closely related to combinations and provides a way to calculate coefficients using nCr.
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3746 | https://www.maths.tcd.ie/pub/Maths/Courseware/374/Primality.pdf | Chapter 1 The Fundamental Theorem of Arithmetic 1.1 Prime numbers If a, b ∈Z we say that a divides b (or is a divisor of b) and we write a | b, if b = ac for some c ∈Z.
Thus −2 | 0 but 0 ∤2.
Definition 1.1 The number p ∈N is said to be prime if p has just 2 divisors in N, namely 1 and itself.
Note that our definition excludes 0 (which has an infinity of divisors in N) and 1 (which has just one).
Writing out the prime numbers in increasing order, we obtain the sequence of primes 2, 3, 5, 7, 11, 13, 17, 19, . . .
which has fascinated mathematicians since the ancient Greeks, and which is the main object of our study.
Definition 1.2 We denote the nth prime by pn.
Thus p5 = 11, p100 = 541.
It is convenient to introduce a kind of inverse function to pn.
Definition 1.3 If x ∈R we denote by π(x) the number of primes ≤x: π(x) = ∥{p ≤x : p prime}∥.
1–1 374 1–2 Thus π(1.3) = 0, π(3.7) = 2.
Evidently π(x) is monotone increasing, but discontinuous with jumps at each prime x = p.
Theorem 1.1 (Euclid’s First Theorem) The number of primes is infinite.
Proof ▶Suppose there were only a finite number of primes, say p1, p2, . . . , pn.
Let N = p1p2 · · · pn + 1.
Evidently none of the primes p1, . . . , pn divides N.
Lemma 1.1 Every natural number n > 1 has at least one prime divisor.
Proof of Lemma ▷The smallest divisor d > 1 of n must be prime. For otherwise d would have a divisor e with 1 < e < d; and e would be a divisor of n smaller than d.
◁ By the lemma, N has a prime factor p, which differs from p1, . . . , pn.
◀ Our argument not only shows that there are an infinity of primes; it shows that pn < 22n; a very feeble bound, but our own. To see this, we argue by induction. Our proof shows that pn+1 ≤p1p2 · · · pn + 1.
But now, by our inductive hypothesis, p1 < 221, p2 < 222, . . . , pn < 22n.
It follows that pn+1 ≤221+22+···+2n But 21 + 22 + · · · + 2n = 2n+1 −1 < 2n+1.
Hence pn+1 < 22n+1.
It follows by induction that pn < 22n, for all n ≥1, the result being trivial for n = 1.
374 1–3 This is not a very strong result, as we said. It shows, for example, that the 5th prime, in fact 11, is < 225 = 232 = 4294967296.
In general, any bound for pn gives a bound for π(x) in the opposite direction, and vice versa; for pn ≤x ⇐ ⇒π(x) ≥n.
In the present case, for example, we deduce that π(22y) ≥[y] > y −1 and so, setting x = 22y, π(x) ≥log2 log2 x −1 > log log x −1.
for x > 1. (We follow the usual convention that if no base is given then log x denotes the logarithm of x to base e.) The Prime Number Theorem (which we shall make no attempt to prove) asserts that pn ∼n log n, or, equivalently, π(x) ∼ x log x.
This states, roughly speaking, that the probability of n being prime is about 1/ log n. Note that this includes even numbers; the probability of an odd number n being prime is about 2/ log n. Thus roughly 1 in 6 odd numbers around 106 are prime; while roughly 1 in 12 around 1012 are prime.
(The Prime Number Theorem is the central result of analytic number theory since its proof involves complex function theory. Our concerns, by contrast, lie within algebraic number theory.) There are several alternative proofs of Euclid’s Theorem. We shall give one below. But first we must establish the Fundamental Theorem of Arithmetic (the Unique Factorisation Theorem) which gives prime numbers their central rˆ ole in number theory; and for that we need Euclid’s Algorithm.
1.2 Euclid’s Algorithm Proposition 1.1 Suppose m, n ∈N, m ̸= 0. Then there exist unique q.r ∈N such that n = qm + r, 0 ≤r < m.
Proof ▶For uniqueness, suppose n = qm + r = q′m + r′, 374 1–4 where r < r′, say. Then (q′ −q)m = r′ −r.
The number of the right is < m, while the number on the left has absolute value ≥m, unless q′ = q, and so also r′ = r.
We prove existence by induction on n. The result is trivial if n < m, with q = 0, r = n. Suppose n ≥m. By our inductive hypothesis, since n −m < n, n −m = q′m + r, where 0 ≤r < m. But then n = qm + r, with q = q′ + 1.
◀ Remark: One might ask why we feel the need to justify division with remainder (as above), while accepting, for example, proof by induction. This is not an easy question to answer.
Kronecker said, “God gave the integers. The rest is Man’s.” Virtually all number theorists agree with Kronecker in practice, even if they do not accept his theology. In other words, they believe that the integers exist, and have certain obvious properties.
Certainly, if pressed, one might go back to Peano’s Axioms, which are a stan-dard formalisation of the natural numbers. (These axioms include, incidentally, proof by induction.) Certainly any properties of the integers that we assume could easily be derived from Peano’s Axioms.
However, as I heard an eminent mathematician (Louis Mordell) once say, “If you deduced from Peano’s Axioms that 1+1 = 3, which would you consider most likely, that Peano’s Axioms were wrong, or that you were mistaken in believing that 1 + 1 = 2?” Proposition 1.2 Suppose m, n ∈N. Then there exists a unique number d ∈N such that d | m, d | n, and furthermore, if e ∈N then e | m, e | n = ⇒e | d.
Definition 1.4 We call this number d the greatest common divisor of m and n, and we write d = gcd(m, n).
Proof ▶Euclid’s Algorithm is a simple technique for determining the greatest common divisor gcd(m, n) of two natural numbers m, n ∈N. It proves inci-dentally — as the Proposition asserts — that any two numbers do indeed have a greatest common divisor (or highest common factor).
374 1–5 First we divide the larger, say n, by the smaller. Let the quotient be q1 and let the remainder (all we are really interested in) be r1: n = mq1 + r1.
Now divide m by r1 (which must be less than m): m = r1q2 + r2.
We continue in this way until the remainder becomes 0: n = mq1 + r1, m = r1q2 + r2, r1 = r2q3 + r3, . . .
rt−1 = rt−2qt−1 + rt, rt = rt−1qt.
The remainder must vanish after at most m steps, for each remainder is strictly smaller than the previous one: m > r1 > r2 > · · · Now we claim that the last non-zero remainder, d = rt say, has the required property: d = gcd(m, n) = rt.
In the first place, working up from the bottom, d = rt | rt−1, d | rt and d | rt−1 = ⇒d | rt−2, d | rt−1 and d | rt−2 = ⇒d | rt−3, . . .
d | r3 and d | r2 = ⇒d | r1, d | r2 and d | r1 = ⇒d | m, d | r1 and d | m = ⇒d | n.
Thus d | m, n; so d is certainly a divisor of m and n.
On the other hand, suppose e is a divisor of m and n: e | m, n.
374 1–6 Then, working downwards, we find successively that e | m and e | n = ⇒e | r1, e | r1 and e | m = ⇒e | r2, e | r2 and e | r1 = ⇒e | r3, . . .
e | rt−2 and e | rt−1 = ⇒e | rt.
Thus e | rt = d.
We conclude that our last non-zero remainder rt is number we are looking for: gcd(m, n) = rt.
◀ It is easy to overlook the power and subtlety of the Euclidean Algorithm. The algorithm also gives us the following result.
Theorem 1.2 Suppose m, n ∈N. Let gcd(m, n) = d.
Then there exist integers x, y ∈Z such that mx + ny = d.
Proof ▶The Proposition asserts that d can be expressed as a linear combination (with integer coefficients) of m and n. We shall prove the result by working backwards from the end of the algorithm, showing successively that d is a linear combination of rs and rs+1, and so, since rs+1 is a linear combination of rs−1 and rs, d is also a linear combination of rs−1 and rs.
To start with, d = rt.
From the previous line in the Algorithm, rt−2 = qtrt−1 + rt.
Thus d = rt = rt−2 −qtrt−1.
But now, from the previous line, rt−3 = qt−1rt−2 + rt−1.
Thus rt−1 = rt −3 −qt−1rt−2.
374 1–7 Hence d = rt−2 −qtrt −1 = rt−2 −qt(rt−3 −qt−1rt−2) = −qtrt−3 + (1 + qtqt−1)rt−2.
Continuing in this way, suppose we have shown that d = asrs + bsrs+1.
Since rs−1 = qs+1rs + rs+1, it follows that d = asrs + bs(rs−1 −qs+1rs) = bsrs−1 + (as −bsqs+1)rs.
Thus d = as−1rs−1 + bs−1rs, with as−1 = bs, bs−1 = as −bsqs+1.
Finally, at the top of the algorithm, d = a0r0 + b0r1 = a0r0 + b0(m −q1r0) = b0m + (a0 −b0q1)r0 = b0m + (a0 −b0q1)(n −q0m) = (b0 −a0q0 + b0q0q1)m + (a0 −b0q0)n, which is of the required form.
◀ Example: Suppose m = 39, n = 99. Following Euclid’s Algorithm, 99 = 2 · 39 + 21, 39 = 1 · 21 + 18, 21 = 1 · 18 + 3, 18 = 6 · 3.
Thus gcd(39, 99) = 3.
Also 3 = 21 −18 = 21 −(39 −21) = −39 + 2 · 21 = −39 + 2(99 −2 · 39) = 2 · 99 −5 · 39.
374 1–8 Thus the Diophantine equation 99x + 39y = 3 has the solution x = 2, y = −5.
(By a Diophantine equation we simply mean a polynomial equation to which we are seeking integer solutions.) This solution is not unique; we could, for example, add 39 to x and subtract 99 from y. We can find the general solution by subtracting the particular solution we have just found to give a homogeneous linear equation. Thus if x′, y′ ∈Z also satisfies the equation then X = x′ −x, Y = y′ −y satisfies the homogeneous equation 99X + 39Y = 0, ie 33X + 13Y = 0, the general solution to which is X = 13t, Y = −33t for t ∈Z. The general solution to this diophantine equation is therefore x = 2 + 13t, y = −5 −33t (t ∈Z).
It is clear that the Euclidean Algorithm gives a complete solution to the general linear diophantine equation ax + by = c.
This equation has no solution unless gcd(a, b) | c, in which case it has an infinity of solutions. For if (x, y) is a solution to the equation ax + by = d, and c = dc′ then (c′x, c′y) satisfies ax + by = c, and we can find the general solution as before.
374 1–9 Corollary 1.1 Suppose m, n ∈Z. Then the equation mx + ny = 1 has a solution x, y ∈Z if and only if gcd(m, n) = 1.
It is worth noting that we can improve the efficiency of Euclid’s Algorithm by allowing negative remainders. For then we can divide with remainder ≤m/2 in absolute value, ie n = qm + r, with −m/2 ≤r < m/2. The Algorithm proceeds as before; but now we have m ≥|r0/2| ≥|r1/22| ≥. . . , so the Algorithm concludes after at most log2 m steps.
This shows that the algorithm is in class P, ie it can be completed in polyno-mial (in fact linear) time in terms of the lengths of the input numbers m, n — the length of n, ie the number of bits required to express n in binary form, being [log2 n] + 1.
Algorithms in class P (or polynomial time algorithms) are considered easy or tractable, while problems which cannot be solved in polynomial time are consid-ered hard or intractable. RSA encryption — the standard techniqhe for encrypting confidential information — rests on the belief — and it should be emphasized that this is a belief and not a proof — that factorisation of a large number is intractable.
Example: Taking m = 39, n = 99, as before, the Algorithm now goes 99 = 3 · 39 −18, 39 = 2 · 18 + 3, 18 = 6 · 3, giving (of course) gcd(39, 99) = 3, as before.
1.3 Ideals We used the Euclidean Algorithm above to show that if gcd(a, b) = 1 then there we can find u, v ∈Z such that au + bv = 1.
There is a much quicker way of proving that such u, v exist, without explicitly computing them.
Recall that an ideal in a commutative ring A is a non-empty subset a ⊂A such that 374 1–10 1. a, b ∈a = ⇒a + b ∈a; 2. a ∈a, c ∈A = ⇒ac ∈a.
As an example, the multiples of an element a ∈A form an ideal ⟨a⟩= {ac : c ∈A}.
Such an ideal is said to be principal.
Proposition 1.3 Every ideal a ⊂Z is principal.
Proof ▶If a = 0 (by convention we denote the ideal {0} by 0) the result is trivial: a = ⟨0⟩. We may suppose therefor that a ̸= 0.
Then a must contain integers n > 0 (since −n ∈a = ⇒n ∈a). Let d be the least such integer. Then a = ⟨d⟩.
For suppose a ∈a. Dividing a by d, a = qd + r, where 0 ≤r < d.
But r = a + (−q)d ∈a.
Hence r = 0; for otherwise r would contradict the minimality of d. Thus a = qd, ie every element a ∈a is a multiple of d.
◀ Now suppose a, b ∈Z. Consider the set of integers I = {au + bv : u, v ∈Z}.
It is readily verified that I is an ideal.
According to the Proposition above, this ideal is principal, say I = ⟨d⟩.
But now a ∈I = ⇒d | a, b ∈I = ⇒d | b.
On the other hand, e | a, e | b = ⇒e | au + bv = ⇒e | d.
374 1–11 It follows that d = gcd(a, b); and we have shown that the diophantine equation au + bv = d always has a solution.
In particular, if gcd(a, b) = 1 we can u, v ∈Z such that au + bv = 1.
This proof is much shorter than the one using the Euclidean Algorithm; but it suffers from the disadvantage that it provides no way of computing d = gcd(a, b), and no way of solving the equation au + bv = d.
In effect, we have taken d as the least of an infinite set of positive integers, using the fact that the natural numbers N are well-ordered, ie every subset S ⊂N has a least element.
1.4 The Fundamental Theorem of Arithmetic Proposition 1.4 (Euclid’s Lemma) Suppose p ∈N is a prime number; and sup-pose a, b ∈Z. Then p | ab = ⇒p | a or p | b.
Proof ▶Suppose p | ab, p ∤a. We must show that p | b. Evidently gcd(p, a) = 1.
Hence, by Corollary 1.1, there exist x, y ∈Z such that px + ay = 1.
Multiplying this equation by b, pxb + aby = b.
But p | pxb and p | aby (since p | ab). Hence p | b.
◀ 374 1–12 Theorem 1.3 Suppose n ∈N, n > 0. Then n is expressible as a product of prime numbers, n = p1p2 · · · pr, and this expression is unique up to order.
Remark: We follow the convention that an empty product has value 1, just as an empty sum has value 0. Thus the theorem holds for n = 1 as the product of no primes.
Proof ▶We prove existence by induction on n, the result begin trivial (by the remark above) when n = 1. We know that n has at least one prime factor p, by Lemma 1.1, say n = pm.
Since m = n/p < n, we may apply our inductive hypothesis to m, m = q1q2 · · · qs.
Hence n = pq1q2 · · · qs.
Now suppose n = p1p2 · · · pr = m = q1q2 · · · qs.
Since p1 | n, it follows by repeated application of Euclid’s Lemma that p1 | qj for some j. But then it follows from the definition of a prime number that p1 = qj.
Again, we argue by induction on n. Since n/p1 = p2 · · · pr = q1 · · · ˆ qj · · · qs (where the ‘hat’ indicates that the factor is omitted), and since n/p1 < n, we deduce that the factors p2, . . . , pr are the same as q1, . . . , ˆ qj, . . . , qs, in some order.
Hence r = s, and the primes p1, · · · , pr and q1, . . . , qs are the same in some order.
◀ We can base another proof of Euclid’s Theorem (that there exist an infinity of primes) on the fact that if there were only a finite number of primes there would not be enough products to “go round”.
Thus suppose there were just m primes p1, . . . , pm.
374 1–13 Let N ∈N. By the Fundamental Theorem, each n ≤N would be expressible in the form n = pe1 1 · · · pem m .
(Actually, we are only using the existence part of the Fundamental Theorem; we do not need the uniqueness part.) For each i (1 ≤i ≤m), pei i | n = ⇒pei i ≤n = ⇒pei i ≤N = ⇒2ei ≤N = ⇒ei ≤log2 N.
Thus there are at most log2 N +1 choices for each exponent ei, and so the number of numbers n ≤N expressible in this form is ≤(log2 N + 1)m.
So our hypothesis implies that (log2 N + 1)m ≥N for all N.
But in fact, to the contrary, X > (log2 X + 1)m = log X log 2 + 1 !m for all sufficiently large X. To see this, set X = ex. We have to show that ex > x log 2 + 1 !m .
Since x log 2 + 1 < 2x if x ≥3, it is sufficient to show that ex > (2x)m for sufficiently large x. But ex > xm+1 (m + 1)!
if x > 0, since the expression on the right is one of the terms in the power-series expansion of ex. Thus the inequality holds if xm+1 (m + 1)! > (2x)m, 374 1–14 ie if x > 2m(m + 1)!.
We have shown therefore that m primes are insufficient to express all n ≤N if N ≥e2m(m+1)!.
Thus our hypothesis is untenable; and Euclid’s theorem is proved.
Our proof gives the bound pn ≤e2m(m+1)!.
which is even worse than the bound we derived from Euclid’s proof. (For it is easy to see by induction that (m + 1)! > em for m ≥2. Thus our bound is worse than een, compared with 22n by Euclid’s method.) We can improve the bound considerably by taking out the square factor in n.
Thus each number n ∈N (n > 0) is uniquely expressible in the form n = d2p1 . . . pr, where the primes p1, . . . , pr are distinct. In particular, if there are only m primes then each n is expressible in the form n = d2pe1 1 · · · pem m , where now each exponent ei is either 0 or 1.
Consider the numbers n ≤N. Since d ≤√n ≤ √ N, the number of numbers of the above form is ≤ √ N2m.
Thus we shall reach a contradiction when √ N2m ≥N, ie N ≤22m.
This gives us the bound pn ≤22n, better than 22n, but still a long way from the truth.
374 1–15 1.5 The Fundamental Theorem, recast We suppose throughout this section that A is an integral domain. (Recall that an integral domain is a commutative ring with 1 having no zero divisors, ie if a, b ∈A then ab = 0 = ⇒a = 0 or b = 0.) We want to examine whether or not the Fundamental Theorem holds in A — we shall find that it holds in some commutative rings and not in others. But to make sense of the question we need to re-cast our definition of a prime.
Looking back at Z, we see that we could have defined primality in two ways (excluding p = 1 in both cases): 1. p is prime if it has no proper factors, ie p = ab = ⇒a = 1 or b = 1.
2. p is prime if p | ab = ⇒p | a or p | b.
The two definitions are of course equivalent in the ring Z. However, in a general ring the second definition is stronger: that is, an element satisfying it must satisfy the first definition, but the converse is not necessarily true. We shall take the second definition as our starting-point.
But first we must deal with one other point. In defining primality in Z we actually restricted ourselves to the semi-ring N, defined by the order in Z: N = {n ∈Z : n ≥0}.
However, a general ring A has no natural order, and no such semi-ring, so we must consider all elements a ∈A.
In the case of Z this would mean considering −p as a prime on the same footing as p. But now, for the Fundamental Theorem to make sense, we would have to regard the primes ±p as essentially the same.
The solution in the general ring is that to regard two primes as equivalent if each is a multiple of the other, the two multiples necessarily being units.
Definition 1.5 An element ϵ ∈A is said to be a unit if it is invertible, ie if there is an element η ∈A such that ϵη = 1.
We denote the set of units in A by A×.
For example, Z× = {±1}.
Proposition 1.5 The units in A form a multiplicative group A×.
374 1–16 Proof ▶This is immediate. Multiplication is associative, from the definition of a ring; and η = ϵ−1 is a unit, since it has inverse ϵ.
◀ Now we can define primality.
Definition 1.6 Suppose a ∈A is not a unit, and a ̸= 0. Then 1. a is said to be irreducible if a = bc = ⇒b or c is a unit.
2. a is said to be prime if a | bc = ⇒a | b or p | b.
Proposition 1.6 If a ∈A is prime then it is irreducible.
Proof ▶Suppose a = bc.
Then a | b or a | c.
We may suppose without loss of generality that a | b. Then a | b, b | a = ⇒a = bϵ, where ϵ is a unit; and a = bc = bϵ = ⇒c = ϵ.
◀ Definition 1.7 The elements a, b ∈A are said to be equivalent, written a ∼b, if b = ϵa for some unit ϵ.
In effect, the group of units A× acts on A and two elements are equivalent if each is a transform of the other under this action.
Now we can re-state the Fundamental Theorem in terms which make sense in any integral domain.
Definition 1.8 The integral domain A is said to be a unique factorisation domain if each non-unit a ∈A, a ̸= 0 is expressible in the form a = p1 · · · pr, where p1, . . . , pr are prime, and if this expression is unique up to order and equiv-alence of primes.
374 1–17 In other words, if a = q1 · · · qs is another expression of the same form, then r = s and we can find a permutation π of {1, 2, . . . , r} and units ϵ1, ϵ2, . . . , ϵr such that qi = ϵipπ(i) for i = 1, 2, . . . , r.
Thus a unique factorisation domain (UFD) is an integral domain in which the Fundamental Theorem of Arithmetic is valid.
1.6 Principal ideals domains Definition 1.9 The integral domain A is said to be a principal ideal domain if every ideal a ∈A is principal, ie a = ⟨a⟩= {ac : c ∈A} for some a ∈A.
Example: By Proposition 1.3, Z is a principal ideal domain.
Our proof of the Fundamental Theorem can be divided into two steps — this is clearer in the alternative version outlined in Section 1.3 — first we showed that that Z is a principal ideal domain, and then we deduced from this that Z is a unique factorisation domain.
As our next result shows this argument is generally available; it is the tech-nique we shall apply to show that the Fundamental Theorem holds in a variety of integral domains.
Proposition 1.7 A principal ideal domain is a unique factorisation domain.
Proof ▶Suppose A is a principal ideal domain.
Lemma 1.2 A non-unit a ∈A, a ̸= 0 is prime if and only if it is irreducible, ie a = bc = ⇒a is a unit or b is a unit.
Proof of Lemma ▷By Proposition 1.6, a prime is always irreducible.
Conversely, if a = p1 · · · pr is irreducible then evidently r = 1, and a is prime.
◁ 374 1–18 Now suppose a is neither a unit nor 0; and suppose that a is not expressible as a product of primes. Then a is reducible, by the Lemma above: say a = a1b1, where a1, b1 are non-units. One at least of a1, b1 is not expressible as a product of primes; we may assume without loss of generality that this is true of a1.
It follows by the same argument that a1 = a2b2, where a2, b2 are non-units, and a2 is not expressible as a product of primes.
Continuing in this way, a = a1b1, a1 = a2b2, a2 = a3b3, . . . .
Now consider the ideal a = ⟨a1, a2, a3, . . . ⟩.
By hypothesis this ideal is principal, say a = ⟨d⟩.
Since d ∈a, d ∈⟨a1, . . . , ar⟩= ⟨ar⟩ for some r. But then ar+1 ∈⟨d⟩= ⟨ar⟩.
Thus ar | ar+1, ar+1 | ar = ⇒ar = ar+1ϵ = ⇒br+1 = ϵ, where ϵ is a unit, contrary to construction.
Thus the assumption that a is not expressible as a product of primes is unten-able; a = p1 · · · pr.
To prove uniqueness, we argue by induction on r, where r the smallest number such that a is expressible as a product of r primes.
Suppose a = p1 · · · pr = q1 · · · qs.
Then p1 | q1 · · · qs = ⇒p1 | qj for some j. Since qj is irreducible, by Proposition 1.6, it follows that qj = p1ϵ, where ϵ is a unit.
374 1–19 We may suppose, after re-ordering the q’s that j = 1. Thus p1 ∼q1.
If r = 1 then a = p1 = ϵp1q2 · · · qs = ⇒1 = ϵq2 · · · qs.
If s > 1 this implies that q2, . . . , qs are all units, which is absurd. Hence s = 1, and we are done.
If r > 1 then q1 = ϵp1 = ⇒p2p3 · · · pr = (ϵq2)q3 · · · qs (absorbing the unit ϵ into q2). The result now follows by our inductive hypothesis.
◀ 1.7 Polynomial rings If A is a commutative ring (with 1) then we denote by A[x] the ring of polynomials p(x) = anxn + · · · + a0 (a0, . . . , an ∈A).
Note that these polynomials should be regarded as formal expressions rather than maps p : A →A; for if A is finite two different polynomials may well define the same map.
We identify ainA with the constant polynomial f(x) = a. Thus A ⊂A[x].
Proposition 1.8 If A is an integral domain then so is A[x].
Proof ▶Suppose f(x) = amxm + · · · + a0, g(x) = bnxn + · · · + b0, where am ̸= 0, bn ̸= 0. Then f(x)g(x) = (ambn)xm+n + · · · + a0b0; and the leading coefficient ambn ̸= 0.
◀ Proposition 1.9 The units in A[x] are just the units of A: (A[x])× = A×.
374 1–20 Proof ▶It is clear that a ∈A is a unit (ie invertible) in A[x] if and only if it is a unit in A.
On the other hand, no non-constant polynomial F(x) ∈A[x] can be invertible, since deg F(x)G(x) ≥deg F(x) if G(x) ̸= 0.
◀ If A is a field then we can divide one polynomial by another, obtaining a remainder with lower degree than the divisor. Thus degree plays the rˆ ole in k[x] played by size in Z.
Proposition 1.10 Suppose k is a field; and suppose f(x), g(x) ∈k[x], with g(x) ̸= 0. Then there exist unique polynomials q(x), r(x) ∈k[x] such that f(x) = g(x)q(x) + r(x), where deg r(x) < deg g(x).
Proof ▶We prove the existence of q(x), r(x) by induction on deg f(x).
Suppose f(x) = amxm + · · · + a0, g(x) = bnxn + · · · + b0, where am ̸= 0, bn ̸= 0.
If m < n then we can take q(x) = 0, r(x) = f(x). We may suppose therefore that m ≥n. In that case, let f1(x) = f(x) −(am/bn)xm−n.
Then deg f1(x) < deg f(x).
Hence, by the inductive hypothesis, f1(x) = g(x)q1(x) + r(x), where deg r(x) < deg g(x); and then f(x) = g(x)q(x) + r(x), with q(x) = (am/bn)xm−n + q1(x).
For uniqueness, suppose f(x) = g(x)q1(x) + r1(x) = g(x)q2(x) + r2(x).
374 1–21 On subtraction, g(x)q(x) = r(x), where q(x) = q2(x) −q1(x), r(x) = r1(x) −r2(x).
But now, if q(x) ̸= 0, deg(g(x)q(x)) ≥deg g(x), deg r(x) < deg g(x).
This is a contradiction. Hence q(x) = 0, ie q1(x) = q2(), r1(x) = r2().
◀ Proposition 1.11 If k is a field then k[x] is a principal ideal domain.
Proof ▶As with Z we can prove this result in two ways: constructively, using the Euclidean Algorithm; or non-constructively, using ideals. This time we take the second approach.
Suppose a ⊂k[x] is an ideal. If a = 0 the result is trivial; so we may assume that a ̸= 0.
Let d(x) ∈a be a polynomial in a of minimal degree. Then a = ⟨d(x)⟩.
For suppose f(x) ∈a. Divide f(x) by d(x): f(x) = d(x)q(x) + r(x), where deg r(x) < deg d(x). Then r(x) = f(x) −d(x)q(x) ∈a since f(x), d(x) ∈a. Hence, by the minimality of deg d(x), r(x) = 0, ie f(x) = d(x)q(x).
◀ By Proposition 1.7 this gives the result we really want.
374 1–22 Corollary 1.2 If k is a field then k[x] is a unique factorisation domain.
Every non-zero polynomial f(x) ∈k[x] is equivalent to a unique monic poly-nomial, namely that obtained by dividing by its leading term. Thus each prime, or irreducible, polynomial p(x) ∈k[x] has a unique monic representative; and we can restate the above Corollary in a simpler form.
Corollary 1.3 Each monic polynomial f(x) = xn + an−1xn−1 + · · · + a0 can be uniquely expressed (up to order) as a product of irreducible monic polyno-mials: f(x) = p1(x) · · · pr(x).
1.8 Postscript We end this Chapter with a result that we don’t really need, but which we have come so close to it would be a pity to omit.
Suppose A is an integral domain. Let K be the field of fractions of A. (Recall that K consists of the formal expressions a b, with a, b ∈A, b ̸= 0; where we set a b = c d if ad = bc.
The map a 7→a 1 : A →K is injective, allowing us to identify A with a subring of K.) The canonical injection A ⊂K evidently extends to an injection A[x] ⊂K[x].
Thus we can regard f(x) ∈A[x] as a polynomial over K.
Proposition 1.12 If A is a unique factorisation domain then so is A[x].
Proof ▶First we must determine the primes in A[x].
Lemma 1.3 The element p ∈A is prime in A[x] if and only if it is prime in A.
374 1–23 Proof of Lemma ▷It is evident that p prime in A[x] = ⇒p prime in A.
Conversely, suppose p is prime in A; We must show that if F(x), G(x) ∈A[x] then p | F(x)G(x) = ⇒p | F(x) or p | G(x).
In other words, p ∤F(x), p ∤G(x) = ⇒p ∤F(x)G(x).
Suppose F(x) = amxm + · · · + a0, G(x) = bnxn + · · · + b0; and suppose p ∤F(x), p ∤G(x).
Let ar, bs be the highest coefficients of f(x), g(x) not divisible by p. Then the coefficient of xr+s in f(x)g(x) is a0br+s + a1br+s−1 + · · · + arbs + · · · + ar+sb0 ≡arbs mod p, since all the terms except arbs are divisible by p. Hence p | arbs = ⇒p mod ar or p mod bs, contrary to hypothesis. In other words, p ∤F(x)G(x).
◁ Lemma 1.4 Suppose f(x) ∈K[x]. Then f(x) is expressible in the form f(x) = αF(x), where α ∈K and F(x) = anxn + · · · + a0 ∈A[x] with gcd(a0, . . . , an) = 1; and the expression is unique up to multiplication by a unit, ie if f(x) = αF(x) = βG(x), where G(x) has the same property then G(x) = ϵF(x), α = ϵβ for some unit ϵ ∈A.
374 1–24 Proof of Lemma ▷Suppose f(x) = αnxn + · · · + α0.
Let αi = ai bi , where ai, bi ∈A; and let b = Y bi.
Then bf(x) = bnxn + · · · + b0 ∈A[x].
Now let d = gcd(b0, . . . , bn).
Then f(x) = (b/d)(cnxn + · · · + c0) is of the required form, since gcd(c0, . . . , cn) = 1.
To prove uniqueness, suppose f(x) = αF(x) = βG(x).
Then G(x) = γF(x), where γ = α/β.
In a unique factorisation domain A we can express any γ ∈K in the form γ = a b, with gcd(a, b) = 1, since we can divide a and b by any common factor.
Thus aF(x) = bG(x).
Let p be a prime factor of b. Then p | aF(x) = ⇒p | F(x), contrary to our hypothesis on the coefficients of F(x). Thus b has no prime factors, ie b is a unit; and similarly a is a unit, and so γ is a unit.
◁ Lemma 1.5 A non-constant polynomial F(x) = anxn + · · · + a0 ∈A[x] is prime in A[x] if and only if 374 1–25 1. F(x) is prime (ie irreducible) in K(x); and 2. gcd(a0, . . . , an) = 1.
Proof of Lemma ▷Suppose F(x) is prime in A[x]. Then certainly gcd(a0, . . . , an) = 1, otherwise F(x) would be reducible.
Suppose F(x) factors in K[x]; say F(x) = g(x)h(x).
By Proposition 1.4, g(x) = αG(x), h(x) = βH(x), where G(x), H(x) have no factors in A. Thus F(x) = γG(x)H(x), where γ ∈K. Let γ = a/b, where a, b ∈A and gcd(a, b) = 1. Then bF(x) = aG(x)H(x).
Suppose p is a prime factor of b. Then p | G(x) or p | H(x), neither of which is tenable. Hence b has no prime factors, ie b is a unit. But now F(x) = ab−1G(x)H(x); and so F(x) factors in A[x].
Conversely, suppose F(x) has the two given properties. We have to show that F(x) is prime in A[x].
Suppose F(x) | G(x)H(x) in A[x].
If F(x) is constant then F(x) = a ∼1 by the second property, so F(x) | G(x) and F(x) | H(x).
We may suppose therefore that deg F(x) ≥1. Since K[x] is a unique factori-sation domain (Corollary to Proposition 1.11), F(x) | G(x) or F(x) | H(x) 374 1–26 in K[x]. We may suppose without loss of generality that F(x) | G(x) in K[x], say G(x) = F(x)h(x), where h(x) ∈K[x].
By Lemma 1.4 we can express h(x) in the form h(x) = αH(x), where the coefficients of H(x) are factor-free. Writing α = a b, with gcd(a, b) = 1, we have bG(x) = aF(x)H(x).
Suppose p is a prime factor of b. Then p | a or p | F(x) or p | H(x), none of which is tenable. Hence b has no prime factors, ie b is a unit. Thus F(x) | G(x) in A[x].
◁ Now suppose F(x) = anxn + · · · a0 ∈A[x] is not a unit in A[x].
If F(x) is constant, say F(x) = a, then the factorisation of a into primes in A is a factorisation into primes in A[x], by Lemma 1.3. Thus we may assume that deg F(x) ≥1.
Since K[x] is a unique factorisation domain (Corollary to Proposition 1.11), F(x) can be factorised in K[x]: F(x) = anp1(x) · · · ps(x), where p1(x), . . . , ps(x) are irreducible monic polynomials in K[x].
By Lem-mas 1.4 and 1.5 each pi(x) is expressible in the form pi(x) = αiPi(x), where Pi(x) is prime in A[x].
Thus F(x) = αP1(x) · · · Pr(x), 374 1–27 where α = anα1 · · · αr ∈K.
Let α = a b, where gcd(a, b) = 1. Then bF(x) = aP1(x) · · · Pr(x).
Let p be a prime factor of b. Then p | Pi(x) for some i, contrary to the definition of Pi(x). Hence b has no prime factors, ie b is a unit.
If a is a unit then we can absorb ϵ = a/b into P1(x): F(x) = Q(x)P2(x) · · · Pr(x), where Q(x) = (a/b)P1(x).
If a is not a unit then ab−1 = p1 · · · ps, where p1, . . . , ps are prime in A (and so in A[x] by Lemma 1.3); and F(x) = p1 · · · psP1(x) · · · Pr(x), as required.
Finally, to prove uniqueness, we may suppose that deg F(x) ≥1, since the result is immediate if F(x) = a is constant.
Suppose F(x) = p1 · · · psP1(x) · · · Pr(x) = q1 · · · qs′Q1(x) · · · Qr′(x).
Each Pi(x), Qj(x) is prime in K[x] by Lemma 1.5. Since K[x] is a unique factorisation domain (Corollary to Proposition 1.11) it follows that r = r′ and that after re-ordering, Qi(x) = αPi(x), where α ∈K×. Let α = a/b with gcd(a, b) = 1. Then aPi(x) = bQi(x).
If p is a prime factor of b then p | bQi(x) = ⇒p | Qi(x), 374 1–28 contrary to the definition of Qi(x). Thus b has no prime factors, and is therefore a unit. Similarly a is a unit. Hence Qi(x) = ϵiPi(x), where ϵi ∈A is a unit.
Setting ϵ = Y i ϵi, we have p1 · · · ps = ϵq1 · · · qs′.
Since A is a unique factorisation domain, s = s′ and after re-ordering, qj = ηjpj, where ηj ∈A is a unit.
We conclude that the prime factors of F(x) are unique up to order and equiv-alence (multiplication by units), ie A[x] is a unique factorisation domain.
◀ Example: There is unique factorisation in Z[x], since Z is a principal ideal domain by Proposition 1.3 and so a unique factorisation domain by Proposition 1.7.
Note that Z[x] is not a principal ideal domain, since eg the ideal a = ⟨2, x⟩, consisting of all polynomials F(x) = anxn + · · · + a0 with a0 even, is not principals: a ̸= ⟨G(x)⟩.
For if it were, its generator G(x) would have to be constant, since a contains non-zero constants, and deg G(x)H(x) ≥deg G(x) if H(x) ̸= 0. But if G(x) = d then a ∩Z = ⟨2⟩= ⇒d = ±2, ie a consists of all polynomials with even coefficients. Since x ∈a is not of this form we conclude that a is not principal.
Chapter 2 Number fields 2.1 Algebraic numbers Definition 2.1 A number α ∈C is said to be algebraic if it satisfies a polynomial equation f(x) = xn + a1xn−1 + · · · + an = 0 with rational coefficients ai ∈Q.
For example, √ 2 and i/2 are algebraic.
A complex number is said to be transcendental if it is not algebraic. Both e and π are transcendental. It is in general extremely difficult to prove a number transcendental, and there are many open problems in this area, eg it is not known if πe is transcendental.
Proposition 2.1 The algebraic numbers form a field ¯ Q ⊂C.
Proof ▶If α satisfies the equation f(x) = 0 then −α satisfies f(−x) = 0, while 1/α satisfies xnf(1/x) = 0 (where n is the degree of f(x)). It follows that −α and 1/α are both algebraic. Thus it is sufficient to show that if α, β are algebraic then so are α + β, αβ.
Suppose α satisfies the equation f(x) ≡xm + a1xm−1 + · · · + am = 0, and β the equation g(x) ≡xn + b1xn−1 + · · · + bn = 0.
Consider the vector space V = ⟨αiβj : 0 ≤i < m, 0 ≤j < n⟩ over Q spanned by the mn elements αiβj. Evidently α + β, αβ ∈V.
2–1 374 2–2 But if θ ∈V then the mn + 1 elements 1, θ, θ2, . . . , θmn are necessarily linearly dependent (over Q), since dim V ≤mn. In other words θ satisfies a polynomial equation of degree ≤mn. Thus each element θ ∈V is algebraic. In particular α + β and αβ are algebraic.
◀ 2.2 Minimal polynomials and conjugates Recall that a polynomial p(x) is said to be monic if its leading coefficient — the coefficient of the highest power of x — is 1: p(x) = xn + a1xn−1 + · · · + an.
Proposition 2.2 Each algebraic number α ∈¯ Q satisfies a unique monic polyno-mial m(x) of minimal degree.
Proof ▶Suppose α satisfies two monic polynomials m1(x), m2(x) of minimal degree d. Then α also satisfies the polynomial p(x) = m1(x) −m2(x) of degree < d; and if p(x) ̸= 0 then we can make it monic by dividing by its leading coefficient. This would contradict the minimality of m1(x). Hence m1(x) = m2(x).
◀ Definition 2.2 The monic polynomial m(x) satisfied by α ∈¯ Q is called the min-imal polynomial of α. The degree of the algebraic number α is the degree of its minimal polynomial m(x).
Proposition 2.3 The minimal polynomial m(x) of α ∈¯ Q is irreducible.
Proof ▶Suppose to the contrary m(x) = f(x)g(x) where f(x), g(x) are of lower degrees than m(x). But then α must be a root of one of f(x), g(x).
◀ Definition 2.3 Two algebraic numbers α, β are said to be conjugate if they have the same minimal polynomial.
374 2–3 Proposition 2.4 An algebraic number of degree d has just d conjugates.
Proof ▶If the minimal poynomial of α is m(x) = xd + a1xd−1 + · · · + ad, then by definition the conjugates of α are the d roots α1 = α, α2, . . . , αd of m(x): m(x) = (x −α1)(x −α2) · · · (x −αd).
These conjugates are distinct, since an irreducible polynomial m(x) over Q is necessarily separable, ie it cannot have a repeated root. For if α were a repeated root of m(x), ie (x −α)2 | m(x) then (x −α) | m′(x), and so (x −α) | d(x) = gcd(m(x), m′(x)).
But d(x) | m(x) and 1 ≤deg(d(x)) ≤d −1, contradicting the irreducibility of m(x).
◀ 2.3 Algebraic number fields Proposition 2.5 Every subfield K ⊂C contains the rationals Q: Q ⊂K ⊂C.
Proof ▶By definition, 1 ∈K. Hence n = 1 + · · · + 1 ∈K for each integer n > 0.
By definition, K is an additive subgroup of C. Hence −1 ∈K; and so −n = (−1)n ∈K for each integer n > 0. Thus Z ⊂K.
Finally, since K is a field, each rational number r = n d ∈K where n, d ∈Z with d ̸= 0.
◀ We can consider any subfield K ⊂C as a vector space over Q.
374 2–4 Definition 2.4 An number field (or more precisely, an algebraic number field) is a subfield K ⊂C which is of finite dimension as a vector space over Q. If dimQ = d then K is said to be a number field of degree d.
Proposition 2.6 There is a smallest number field K containing the algebraic numbers α1, . . . , αr.
Proof ▶Every intersection (finite or infinite) of subfields of C is a subfield of C; so there is a smallest subfield K containing the given algebraic numbers, namely the intersection of all subfields containing these numbers. We have to show that this field is a number field, ie of finite dimension over Q.
Lemma 2.1 Suppose K ⊂C is a finite-dimensional vector space over Q. Then K is a number field if and only if it is closed under multiplication.
Proof of Lemma ▷If K is a number field then it is certainly closed under multi-plication.
Conversely, if this is so then K is closed under addition and multiplication; so we only have to show that it is closed under division by non-zero elements.
Suppose α ∈V, α ̸= 0. Consider the map x 7→αx : V →V.
This is a linear map over Q; and it is injective since αx = 0 = ⇒x = 0.
Since V is finite-dimensional it follows that the map is surjective; in particular, αx = 1 for some x ∈V , ie α is invertible. Hence V is a field.
◁ Now suppose αi is of degree di (ie satisfies a polynomial equation of degree di over Q). Consider the vector space (over Q) V = ⟨αi1 1 · · · αir r : 0 ≤i1 < d1, · · · , 0 ≤ir < dr⟩.
It is readily verified that αiV ⊂V, and so V V ⊂V, ie V is closed under multiplication.
It follows that V is a field; and since any field containing α1, . . . , αr must contain these products, V is the smallest field containing α1, . . . , αr. Moreover V is a number field since dimQ V ≤d1 · · · dr.
◀ 374 2–5 Definition 2.5 We denote the smallest field containing α1, . . . , αr ∈C by Q(α1, . . . , αr).
Proposition 2.7 If α is an algebraic number of degree d then each element γ ∈ Q(α) is uniquely expressible in the form a0 + a1α + · · · + ad−1αd−1 (a0, a1, . . . , ad−1 ∈Q).
Proof ▶It follows as in the proof of Proposition 2.6 that these elements do con-stitute the field Q(α). And if two of the elements were equal then α would satisfy an equation of degree < d, which could be made monic by dividing by the leading coefficient.
◀ A number field of the form K = Q(α), ie generated by a single algebraic number α, is said to be simple. Our next result shows that, surprisingly, every number field is simple. The proof is more subtle than might appear at first sight.
Proposition 2.8 Every number field K can be generated by a single algebraic number: K = Q(α).
Proof ▶It is evident that K = Q(α1, . . . , αr); for if we successively adjoin algebraic numbers αi+1 ∈K \ Q(α1, . . . , αr) then dim Q(α1) < dim Q(α1, α2) dim Q(α1, α2, α3) < and so K must be attained after at most dimQ K adjunctions.
Thus it is suffient to prove the result when r = 2, ie to show that, for any two algebraic numbers α, β, Q(α, β) = Q(γ).
Let p(x) be the minimal polynomial of α, and q(x) the minimal polynomial of β. Suppose α1 = α, . . . , αm are the conjugates of α and β1 = β, . . . , βn the conjugates of β. Let γ = α + aβ, where a ∈Q is chosen so that the mn numbers αi + aβj are all distinct. This is certainly possible, since αi + aβj = αi′ + aβj′ ⇐ ⇒a = αi′ −αi βj −βj′ .
374 2–6 Thus a has to avoid at most mn(mn −1)/2 values.
Since α = γ −aβ, and p(α) = 0, β satisfies the equation p(γ −ax) = 0.
This is a polynomial equation over the field k = Q(γ).
But β also satisfies the equation q(x) = 0.
It follows that β satisfies the equation d(x) = gcd(p(γ −ax), q(x)) = 0.
Now (x −β) | d(x) since β is a root of both polynomials. Also, since d(x) | q(x) = (x −β1) · · · (x −βn), d(x) must be the product of certain of the factors (x −βj). Suppose (x −βj) is one such factor. Then βj is a root of p(γ −ax), ie p(γ −aβj) = 0.
Thus γ −aβj = αi for some i. Hence γ = αi + aβj.
But this implies that i = 1, j = 1, since we chose a so that the elements αi + aβj were all distinct.
Thus d(x) = (x −β).
But if u(x), v(x) ∈k[x] then we can compute gcd(u(x), v(x)) by the eu-clidean algorithm without leaving the field k, ie u(x), v(x) ∈k[x] = ⇒gcd(u(x), v(x)) ∈k[x].
374 2–7 In particular, in our case x −β ∈k = Q(γ).
But this means that β ∈Q(γ); and so also α = γ −aβ ∈Q(γ).
Thus α, β ∈Q(γ) = ⇒Q(α, β) ⊂Q(γ) ⊂Q(α, β).
Hence Q(α, β) = Q(γ).
◀ 2.4 Algebraic integers Definition 2.6 A number α ∈C is said to be an algebraic integer if it satisfies a polynomial equation f(x) = xn + a1xn−1 + · · · + an = 0 with integral coefficients ai ∈Z. We denote the set of algebraic integers by ¯ Z.
Proposition 2.9 The algebraic integers form a ring ¯ Z with Z ⊂¯ Z ⊂¯ Q.
Proof ▶Evidently Z ⊂¯ Z, since n ∈Z satisfies the equation x −n = 0.
We have to show that α, β ∈¯ Z = ⇒α + β, αβ ∈¯ Z.
Lemma 2.2 The number α ∈C is an algebraic integer if and only if there exists a finitely-generated (but non-zero) additive subgroup S ⊂C such that αS ⊂S.
374 2–8 Proof of Lemma ▷Suppose α ∈¯ Z; and suppose the minimal polynomial of α is m(x) = xd + a1xd−1 + · · · + ad, where a1, . . . , ad ∈Z. Let S be the abelian group generated by 1, α, . . . , αd−1: S = ⟨1, α, . . . , αd−1⟩.
Then it is readily verified that αS ⊂S.
Conversely, suppose S is such a subgroup.
◁ If α is a root of the monic polynomial f(x) then −α is a root of the monic polynomial f(−x). It follows that if α is an algebraic integer then so is −α. Thus it is sufficient to show that if α, β are algebraic integers then so are α + β, αβ.
Suppose α satisfies the equation f(x) ≡xm + a1xm−1 + · · · + am = 0 (a1, . . . , am ∈Z), and β the equation g(x) ≡xn + b1xn−1 + · · · + bn = 0 (b1, . . . , bn ∈Z).
Consider the abelian group (or Z-module) M = ⟨αiβj : 0 ≤i < m, 0 ≤j < n⟩ generated by the mn elements αiβj. Evidently α + β, αβ ∈V.
As a finitely-generated torsion-free abelian group, M is isomorphic to Zd for some d. Moreover M is noetherian, ie every increasing sequence of subgroups of M is stationary: if S1 ⊂S2 ⊂S3 · · · ⊂M then for some N, SN = SN+1 = SN+2 = · · · .
Suppose θ ∈M. Consider the increasing sequence of subgroups ⟨1⟩⊂⟨1, θ⟩⊂⟨1, θ, θ2⟩⊂· · · .
This sequence must become stationary; that is to say, for some N θN ∈⟨1, θ, . . . , θN−1⟩.
In other words, θ satisfies an equation of the form θN = a1θN−1 + a2θN−2 + · · · .
Thus every θ ∈M is an algebraic integer. In particular α+β and αβ are algebraic integers.
◀ 374 2–9 Proposition 2.10 A rational number c ∈Q is an algebraic integer if and only if it is a rational integer: ¯ Z ∩Q = Z.
Proof ▶Suppose c = m/n, where gcd(m, n) = 1; and suppose c satisfies the equation xd + a1xd−1 + · · · + ad = 0 (ai ∈Z).
Then md + a1md−1n + · · · + adnd = 0.
Since n divides every term after the first, it follows that n | md. But that is incompatible with gcd(m, n) = 1, unless n = 1, ie c ∈Z.
◀ Proposition 2.11 Every algebraic number α is expressible in the form α = β n, where β is an algebraic integer, and n ∈Z.
Proof ▶Let the minimal polynomial of α be m(x) = xd + a1xd−1 + · · · + ad, where a1, . . . , ad ∈Q. Let the lcm of the denominators of the ai be n. Then bi = nai ∈Z (1 ≤i ≤d).
Now α satisfies the equation nxd + b1xd−1 + · · · + bd = 0.
It follows that β = nα satisfies the equation xd + b1xd−1 + (nb2)xd−2 + · · · + (nd−1bd = 0.
Thus β is an integer, as required.
◀ The following result goes in the opposite direction.
Proposition 2.12 Suppose α is an algebraic integer. Then we can find an alge-braic integer β ̸= 0 such that αβ ∈Z.
374 2–10 Proof ▶Let the minimal polynomial of α be m(x) = xd + a1xd−1 + · · · + ad, where a1, . . . , ad ∈Z. Recall that the conjugates of α, α1 = α, . . . , αd are the roots of the minimal equation.
Each of these conjugates is an algebraic integer, since its minimal equation m(x) has integer coefficients. Hence β = α2 · · · αd is an algebraic integer; and αβ = α1α2 · · · αd = ±ad ∈Z.
◀ 2.5 Units Definition 2.7 A number α ∈C is said to be a unit if both α and 1/α are alge-braic integers.
Any root of unity, ie any number satisfying xn = 1 for some n, is a unit.
But these are not the only units; for example, √ 2 −1 is a unit.
The units form a multiplicative subgroup of ¯ Q×.
2.6 The Integral Basis Theorem Proposition 2.13 Suppose A is a number ring. Then we can find γ1, . . . , γd ∈A such that each α ∈A is uniquely expressible in the form α = c1γ1 + cdγd with c1, . . . , cd ∈Z.
In other words, as an additive group A ∼ = Zd.
We may say that γ1, . . . , γd is a Z-basis for A.
Proof ▶Suppose A is the ring of integers in the number field K. By Proposi-tion ??, K = Q(α).
374 2–11 By Proposition 2.12, α = β m, where β ∈¯ Z, m ∈Z. Since Q(β) = Q(α), we may suppose that α is an integer.
Let m(x) = xd + a1xd−1 + · · · + ad be the minimal polynomial of α; and let α1 = α, . . . , αd be the roots of this polynomial, ie the conjugates of α.
Note that these conjugates satisfy exactly the same set of polynomials over Q; for p(α) = 0 ⇐ ⇒m(x) | p(x) ⇐ ⇒p(αi) = 0.
Now suppose β ∈A. Then β = b0 + b1α + · · · bd−1αd−1, where b0, . . . , bd−1 ∈Q, say β = f(α) with f(x) ∈Q[x].
Let βi = b0 + b1αi + · · · bd−1αd−1 i for i = 1, . . . , d.
Each βi satisfies the same set of polynomials over Q as β. for p(β) = 0 ⇐ ⇒p(f(α)) = 0 ⇐ ⇒p(f(αi)) = 0 ⇐ ⇒p(βi) = 0.
In particular, each βi has the same minimal polynomial as β, and so each βi is an integer.
We may regard the formulae for the βi as linear equations for the coefficients b0, . . . , bd−1: b0 + α1b1 + · · · αd−1bd−1 = β1, . . .
b0 + αdb1 + · · · αd−1 d bd−1 = βd.
We can write this as a matrix equation D b0 .
.
.
bd−1 = β1 .
.
.
βd 374 2–12 where D is the matrix D = 1 α1 . . .
αd−1 1 .
.
.
. . .
. . .
.
.
.
1 αd . . .
αd−1 d .
By a familiar argument, det 1 x1 . . .
xd−1 1 .
.
.
. . .
. . .
.
.
.
1 xd . . .
xd−1 d = Y i<j (xi −xj).
(The determinant vanishes whenever xi = xj since then two rows are equal.
Hence (xi −xj) is a factor for each pair i, j; from which the result follows on comparing degrees and leading coefficients.) Thus det D = Y i 0. Thus each bi is expressible in the form bi = γi n , where γi ∈¯ Z ∩Q = Z.
In other words, each β ∈A is expressible in the form β = coδ0 + · · · + cd−1δd−1, where δi = αi n and ci ∈Z (0 ≤i < d).
The elements coδ0 + · · · + cd−1δd−1 (ci ∈Z) form a finitely-generated and torsion-free abelian group C, of rank d; and A is a subgroup of C of finite index. We need the following standard result from the theory of finitely-generated abelian groups.
374 2–13 Lemma 2.3 If S ⊂Zd is a subgroup of finite index then S ∼ = Zd Proof of Lemma ▷We have to construct a Z-index for S. We argue by induction on d.
Choose an element e = (e1, . . . , ed) ∈S with least positive last coordinate ed. Suppose s = (s1, . . . , sd) ∈S.
Then sd = qe, or we could find an element of S with smaller last coordinate. Thus s −qe = (t1, . . . , td−1, 0).
Hence S = Ze ⊕T, where T = S ∩Zd−1 (identifying Zd−1 with the subgroup of Zd formed by the d-tuples with last coor-dinate 0).
The result follows on applying the inductive hypothesis to T.
◁ The Proposition follows on applying the Lemma to A ⊂C ∼ = Zd.
◀ 2.7 Unique factorisation in number rings As we saw in Chapter 1, a principal ideal domain is a unique factorisation domain.
The converse is not true; there is unique factorisation in Z[x], but the ideal ⟨2, x⟩ is not principal. Our main aim in this Section is to show that the converse does hold for number rings A: A principal ideal domain ⇐ ⇒A unique factorisation domain.
We suppose throughout the Section that A is a number ring, ie the ring of integers in a number field K.
374 2–14 Proposition 2.14 Suppose a ⊂A is a non-zero ideal. Then the quotient-ring A/a is finite.
Proof ▶Take α ∈a, α ̸= 0. By Proposition 1.8, we can find β ∈A, β ̸= 0 such that a = αβ ∈Z.
We may suppose that a > 0. Then ⟨a⟩⊂⟨α⟩⊂a.
Thus α ≡β mod a = ⇒α ≡β mod a.
By Proposition ??, A has an integral basis γ1, . . . , γd, ie each α ∈A is (uniquely) expressible in the form α = c1γ1 + cdγd with c1, . . . , cd ∈Z. It follows that α is congruent moda to one of the numbers r1γ1 + rdγd (0 ≤ri < a).
Thus ∥A/⟨a⟩∥= ad.
Hence ∥A/a∥≤ad.
◀ Proposition 2.15 The number ring A is a unique factorisation domain if and only if it is a principal ideal domain.
Proof ▶We know from Chapter 1 that A principal ideal domain = ⇒A unique factorisation domain.
We have to proce the converse.
Let us suppose therefore that the number ring A is a unique factorisation do-main.
374 2–15 Lemma 2.4 Suppose α = ϵπe1 1 · · · πer r , β = ϵ′πf1 1 · · · πfr r .
Let δ = πmin(e1,f1) 1 · · · πmin(er,fr) r .
Then δ = gcd(α, β) in the sense that δ | α, δ | β and δ′ | α, δ | β = ⇒δ′ | δ.
Proof of Lemma ▷This follows at once from unique factorisation.
◁ Lemma 2.5 If β1 ≡β2 mod α then gcd(α, β1) = gcd(α, β2).
Proof of Lemma ▷It is readily verified that if β1 = β2 + αγ then δ | α, β1 ⇐ ⇒δ | α, β2.
◁ We say that α, β are coprime if gcd(α, β) = 1.
It follows from the Lemma that we may speak of a congruence class ¯ β mod α being coprime to α.
Lemma 2.6 The congruence classes modα coprime to α form a multiplicative group (A/⟨α⟩)× .
Proof of Lemma ▷We have gcd(α, β1β2) = 1 ⇐ ⇒gcd(α, β1) = 1, gcd(α, β2) = 1.
Thus (A/⟨α⟩)× is closed under multiplication; and if β is coprime to α then the map ¯ γ 7→¯ β¯ γ : (A/⟨α⟩)× →(A/⟨α⟩)× is injective, and so surjective since A/⟨α⟩is finite. Hence (A/⟨α⟩)× is a group.
◁ 374 2–16 Lemma 2.7 Suppose gcd(α, β) = δ.
Then we can find u, v ∈A such that αu + βv = δ.
Proof of Lemma ▷We may suppose, on dividing by δ, that gcd(α, β) = 1, and so ¯ β ∈(A/⟨α⟩)× .
Since this group is finite, ¯ βn = 1 for some n > 0. In other words, βn ≡1 mod α, ie βn = 1 + αγ, ie αu + βv = 1 with u = −γ, v = βn−1.
◁ We can extend the definition of gcd to any set (finite or infinite) of numbers αi ∈A (i ∈I).
and by repeated application of the last Lemma we can find βi (all but a finite number equal to 0) such that X i∈I αiβi = gcd i∈I (αi).
Applying this to the ideal a, let δ = gcd α∈a(α).
Then δ = X αiβi ∈a; and so a = ⟨δ⟩.
◀ Chapter 3 Quadratic Number Fields 3.1 The fields Q(√m) Definition 3.1 A quadratic field is a number field of degree 2.
Recall that this means the field k has dimension 2 as a vector space over Q: dimQ k = 2.
Definition 3.2 The integer m ∈Z is said to be square-free if m = r2s = ⇒r = ±1.
Thus ±1, ±2, ±3, ±5, ±6, ±7, ±10, ±11, ±13, . . .
are square-free.
Proposition 3.1 Each quadratic field is of the form Q(√m) for a unique square-free integer m ̸= 1.
Recall that Q(√m) consists of the numbers x + y√m (x, y ∈Q).
Proof ▶Suppose k is a quadratic field. Let α ∈k \ Q. Then α2, α, 1 are linearly dependent over Q, since dimQ k = 2. In other words, α satisfies a quadratic equation a0α2 + a1α + a2 = 0 with a0, a1, a2 ∈Q. We may assume that a0, a1, a2 ∈Z. Then α = −a1 + q a2 1 −4a0a2 2a0 3–1 374 3–2 Thus q a2 1 −4a0a2 = 2a0α + a1 ∈k.
Let a2 1 −4a0a2 = r2m where m is square-free. Then √m = 1 r q a2 1 −4a0a2 ∈k.
Thus Q ⊂Q(√m) ⊂k.
Since dimQ k = 2, k = Q(√m).
To see that different square-free integers m1, m2 give rise to different quadratic fields, suppose m1 ∈Q(√m2), say m1 = x + y√m2 (x, y ∈Q) Squaring, x1 = x2 + m2y2 + 2xy√m2.
Thus either x = 0 or y = 0 or √m2 ∈Q, all of which are absurd.
◀ When we speak of the quadratic field Q(√m) it is understood that m is a square-free integer ̸= 1.
Definition 3.3 The quadratic field Q(√m) is said to be real if m > 0, and imag-inary if m < 0.
This is a natural definition since it means that Q(√m) is real if and only if Q(√m) ⊂R.
374 3–3 3.2 Conjugates and norms Proposition 3.2 The map x + y√m 7→x −y√m is an automorphism of Q(√m); and it is the only such automorphism apart from the identity map.
Proof ▶The map clearly preserves addition. It also preserves multiplication, since (x + y√m)(u + v√m = (xu + yvm) + (xv + yu)√m, and so (x −y√m)(u −v√m = (xu + yvm) −(xv + yu)√m.
Since the map is evidently bijective, it is an automorphism.
Conversely, if θ is an automorphism of Q(√m) then θ preserves the elements of Q; in fact if α ∈Q(√m) then θ(α) = α ⇐ ⇒α ∈Q.
Thus θ(√m)2 = θ(m) = m = ⇒θ(√m) = ±√m, giving the identity automorphism and the automorphism above.
◀ Definition 3.4 If α = x + y√m (x, y ∈Q) then we write ¯ α = x −y√m (x, y ∈Q) and we call ¯ α the conjugate of α.
Note that if Q(√m) is imaginary (ie m < 0) then the conjugate ¯ α coincides with the usual complex conjugate.
Definition 3.5 We define the norm N(α) of α ∈Q(√m) by N(α) = α¯ α.
Thus if α = x + y√m (x, y ∈Q) then N(α) = (x + y√m)(x −y√m) = x2 −my2.
Proposition 3.3 1. N(α) ∈Q; 374 3–4 2. N(()α = 0 ⇐ ⇒α = 0; 3. N(αβ) = N(α)N(β); 4. If a ∈Q then N(a) = a2; 5. If m < 0 then N(α) ≥0.
Proof ▶All is clear except perhaps the third part, where N(αβ) = (αβ)(αβ) = (αβ)(¯ α¯ β) = (α¯ α)(β ¯ β) = N(α)N(β).
◀ 3.3 Integers Proposition 3.4 Suppose k = Q(√m), where m ̸= 1 is square-free.
1. If m ̸≡1 mod 4 then the integers in k are the numbers a + b√m, where a, b ∈Z.
2. If m ≡1 mod 4 then the integers in k are the numbers a 2 + b 2 √m, where a, b ∈Z and a ≡b mod 2, ie a, b are either both even or both odd.
Proof ▶Suppose α = a + b√m ( b ∈Q) is an integer. Recall that an algebraic number α is an integer if and only if its minimal polynomial has integer coefficients. If y = 0 the minimal polynomial of α is x −a. Thus α = a is in integer if and only if a ∈Z (as we know of course since ¯ Z ∩Q = Z).
If y ̸= 0 then the minimal polynomial of α is (x −a)2 −mb2 = x2 −2ax + (a2 −mb2).
374 3–5 Thus α is an integer if and only if 2a ∈Z and a2 −mb2 ∈Z.
Suppose 2a = A, ie a = A 2 .
Then 4a2 ∈Z, a2 −mb2 ∈Z = ⇒4mb2 ∈Z = ⇒4b2 ∈Z = ⇒2b ∈Z since m is square-free. Thus b = B 2 , where B ∈Z.
Now a2 −mb2 = A2 −mB2 4 ∈Z, ie A2 −mB2 ≡0 mod 4.
If A is even then 2 | A = ⇒4 | A2 = ⇒4 | mB2 = ⇒2 | B2 = ⇒2 | B; and similarly 2 | B = ⇒4 | B2 = ⇒4 | A2 = ⇒2 | A.
Thus A, B are either both even, in which case a, b ∈Z, or both odd, in which case A2, B2 ≡1 mod 4, so that 1 −m ≡0 mod 4, ie m ≡1 mod 4.
Conversely if m ≡1 mod 4 then A, B odd = ⇒A2 −mB2 ≡0 mod 4 = ⇒a2 −mb2 ∈Z.
◀ It is sometimes convenient to express the result in the following form.
374 3–6 Corollary 3.1 Let ω = √m if m ̸≡1 mod 4, 1+√m 2 if m ≡1 mod 4.
Then the integers in Q(√m) form the ring Z[ω].
Examples: 1. The integers in the gaussian field Q(i) are the gaussian integers a + bi (a, b ∈Z) 2. The integers in Q( √ 2) are the numbers a + b √ 2 (a, b ∈Z).
3. The integers in Q(√−3) are the numbers a + bω (a, b ∈Z) where ω = 1 + √−3 2 .
Proposition 3.5 If α ∈Q(√m) is an integer then N(α) ∈Z.
Proof ▶If α is an integer then so is its conjugate ¯ α (since α, ¯ α satisfy the same polynomial equations over Q). Hence N(α) ∈¯ Z ∩Q = Z.
◀ 3.4 Units Proposition 3.6 An integer ϵ ∈Q(√m) is a unit if and only if N(ϵ) = ±1.
Proof ▶Suppose ϵ is a unit, say ϵη = 1.
374 3–7 Then N(ϵ)N(η) = N(1) = 1.
Hence N(ϵ) = ±1.
Conversely, suppose N(ϵ) = ±1, ie ϵ¯ ϵ = ±1.
Then ϵ−1 = ±¯ ϵ is an integer, ie ϵ is a unit.
◀ Proposition 3.7 An imaginary quadratic number field contains only a finite num-ber of units.
1. The units in Q(i) are ±1, ±i; 2. The units in Q(√−3) are ±1, ±ω, ±ω2, where ω = (1 + √−3)/2.
3. In all other cases the imaginary quadratic number field Q(√m) (where m < 0) has just two units, ±1.
Proof ▶We know of course that ±1 are always units.
Suppose ϵ = a + b√m is a unit. Then N)ϵ) = a2 + (−m)b2 = 1 by Proposition 3.6. In particular (−m)b2 ≤1.
If m ≡3 mod 4 then a, b ∈Z; and so b = 0 unless m = −1 in which case b = ±1 is a solution, giving a = 0, ie ϵ = ±i.
If m ≡1 mod 4 then b may be a half-integer, ie b = B/2, and (−m)b2 = (−m)B2/4 > 1 if B ̸= 0, unless m = −3 and B = ±1, in which case A = ±1. Thus we get four additional units in Q(√−3), namely ±ω, ±ω2.
◀ 374 3–8 Proposition 3.8 Every real quadratic number field Q(√m) (where m > 0) con-tains an infinity of units. More precisely, there is a unique unit η > 1 such that the units are the numbers ±ηn (n ∈Z) Proof ▶The following exercise in the pigeon-hole principle is due to Kronecker.
Lemma 3.1 Suppose α ∈R. There are an infinity of integers m, n with m > 0 such that |mα −n| < 1 n.
Proof of Lemma ▷Let {x} denote the fractional part of x ∈R. Thus {x} = x −[x], where [x] is the integer part of x.
Suppose N is a positive integer. Let us divide [0, 1) into N equal parts: [0, 1/N), [1/N, 2/N), . . . , [(N −1)/N, 1).
Consider how the N + 1 fractional parts {0}, {α}, {2α}, . . . , {Nα} fall into these N divisions.
Two of the fractional parts — say {rα} and {sα}, where r < s — must fall into the same division. But then |{sα} −{rα}| < 1/N, ie |(sα −[sα]) −(rα −[rα])| < N.
Let m = s −r, n = [sα] −[rα].
Then |mα −n| < 1/N ≤1/m.
◁ Lemma 3.2 There are an infinity of a, b ∈Z such that |a2 −b2m| < 2√m + 1.
374 3–9 Proof of Lemma ▷We apply Kronecker’s Lemma above with α = √m. There are an infinity of integers a, b > 0 such that |a −b√m| < 1/b.
But then a < b√m + 1, and so a + b√m < 2b√m + 1 Hence |a2 −b2m| = (a + b√m)|a −b√m| < (2b√m + 1)/b ≤2√m + 1.
◁ It follows from this lemma that there are an infinity of integer solutions of a2 −b2m = d for some d < 2√m + 1.
But then there must be an infinity of these solutions (a, b) with the same re-mainders modd.
Lemma 3.3 Suppose α1 = a1 + b1 √m, α2 = a2 + b2 √m, where a2 1 −b2 1 = d = a2 2 −b2 2 and a1 ≡a2 mod d, b1 ≡b2 mod d.
Then α1 α2 is an algebraic integer.
Proof of Lemma ▷Suppose a2 = a1 + mr, b2 = b1 + ms.
Then α2 = α1 + dβ, 374 3–10 where β = r + s√m.
Hence α1 α2 = α1 ¯ α2 α2 ¯ α2 = α1 ¯ α2 d = α1( ¯ α1 + d¯ β) d = α1 ¯ α1 d + ¯ β = d d + β = 1 + β, which is an integer.
◁ Now suppose (a1, b1), (a2, b2) are two such solutions. Then ϵ = α1 α2 is an integer, and N(ϵ) = N(α1) N(α2) = d d = 1.
Hence ϵ is a unit, by Proposition 3.6.
Since there are an infinity of integers α satisfying these conditions, we obtain an infinity of units if we fix α1 and let α2 vary. In particular there must be a unit ϵ ̸= ±1.
Just one of the four units ±ϵ, ±ϵ−1 must lie in the range (1, ∞). (The others are distributes one each in the ranges (−∞, −1), (−1, 0) and (0, 1).) Suppose then that ϵ = a + b√m > 1.
Then |ϵ−1| < 1, and so ¯ ϵ = ±ϵ−1 ∈(−1, 1), ie −1 < a −b√m < 1.
374 3–11 Adding these two inequalities, 0 < 2a, ie a > 0.
On the other hand, ϵ > ¯ ϵ = ⇒b > 0.
It follows that there can only be a finite number of units in any range 1 < ϵ ≤c.
In particular, if ϵ > 1 is a unit, then there is a smallest unit η in the range 1 < η ≤ϵ.
Evidently η is the least unit in the range 1 < η.
Now suppose ϵ is a unit ̸= ±1. As we observed, one of the four units ±ϵ, ±ϵ−1 must lie in the range (1, ∞). We can take this in place of ϵ, ie we may assume that ϵ > 1.
Since ηn →∞, ηr ≤ϵ < ηr+1 for some r ≥1. Hence 1 ≤ϵη−r < η.
Since η is the smallest unit > 1, this implies that ϵη−1 = 1, ie ϵ = ηr.
◀ 374 3–12 3.5 Unique factorisation Suppose A is an integral domain. Recall that if A is a principal ideal domain, ie each ideal A ⊂A can be generated by a single element a, a = ⟨a⟩, then A is a unique factorisation domain, ie each a ∈A is uniquely expressible — up to order, and equivalence of primes — in the form a = ϵπe1 1 · · · πer r , where ϵ is a unit, and π1, . . . , πr are inequivalent primes.
We also showed that if A is the ring of integers in an algebraic number field k then the converse is also true, ie A principal ideal domain ⇐ ⇒A unique factorisation domain .
Proposition 3.9 The ring of integers Z[ω] in the quadratic field Q(√m is a prin-cipal ideal domain (and so a unique factorisation domain) if m = −11, −7, −3, −2, −1, 2, 3, 5, 13.
Proof ▶We take |N(α)| as a measure of the size of α ∈Z[ω].
Lemma 3.4 Suppose α, β ∈Z[ω[, with β ̸= 0. Then there exist γ, ρ ∈Z[ω] such that α = βγ + ρ with |N(ρ)| < |N(β)|.
In other words, we can divide α by β, and get a remainder ρ smaller than β.
Proof of Lemma ▷Let α β = x + y√m where x, y ∈Q.
Suppose first that m ̸≡1 mod 4. We can find integers a, b such that |x −a|, |y −b| ≤1 2.
Let γ = a + b√m.
374 3–13 Then γ ∈Z[ω]; and α β −γ = (x −a) + (y −b)√m.
Thus N(α β −γ) = (x −a)2 −m(y −b)2.
If now m < 0 then 0 ≤N(α β −γ) ≤1 + m 4 , yielding |N(α β −γ)| < 1 if m = −2 or −1; while if m > 0 then −m 4 ≤N(α β −γ) ≤1 4, yielding |N(α β −γ)| < 1 if m = 2 or 3.
On the other hand, if m ≡1 mod 4 then we can choose a, b to be integers or half-integers. Thus we can choose b so that N(y −b) ≤1 4; and then we can choose a so that N(x −a) ≤1 2.
(Note that a must be an integer or half-integer according as b is an integer or half-integer; so we can only choose a to within an integer.) If m < 0 this gives 0 ≤N(α β −γ) ≤4 + m 16 , yielding |N(α β −γ)| < 1 if m = −11, −7 or −3; while if m > 0 then −m 16 ≤N(α β −γ) ≤1 4, 374 3–14 yielding |N(α β −γ)| < 1 if m = 5 or 13.
Thus in all the cases listed we can find γ ∈Z[ω] such that |N(α β −γ)| < 1 Multiplying by β, |N(α −βγ)| < |N(β)|, which gives the required result on setting ρ = α −βγ, ie α = βγ + ρ.
◁ Now suppose a ̸= 0 is an ideal in Z[ω]. Let α ∈a (α ̸= 0) be an element minimising |N(α)|. (Such an element certainly exists, since |N(α)| is a positive integer.) Now suppose β ∈a. By the lemma we can find γ, ρ ∈Z[ω] such that β = αγ + ρ with |N(ρ)| < |N(α)|.
But ρ = β −αγ ∈a.
Thus by the minimality of |N(α)|, N(α) = 0 = ⇒ρ = 0 = ⇒β = αγ = ⇒β ∈⟨α⟩.
Hence a = ⟨α⟩.
◀ Remarks: 1. We do not claim that these are the only cases in which Q(√m) — or rather the ring of integers in this field — is a unique factorisation domain. There are certainly other m for which it is known to hold; and in fact is not known if the number of such m is finite or infinite. But the result is easily estab-lished for the m listed above.
374 3–15 2. On the other hand, unique factorisation fails in many quadratic fields. For example, if m = −5 then 6 = 2 · 3 = (1 + √ −5)(1 − √ −5) Now 2 is irreducible in Z[ √ 5], since a2 + 5b2 = 2 has no solution in integers. Thus if there were unique factorisation then 2 | 1 + √ −5 or 2 | 1 − √ −5, both of which are absurd.
As an example of a real quadratic field in which unique factorisation fails, consider m = 10. We have 6 = 2 · 3 = (4 + √ 10)(4 − √ 10) The prime 2 is again irreducible; for a2 −10b2 = ±2 has no solution in integers, since neither ±2 is a quadratic residue mod 10. (The quadratic residues mod10 are 0, ±1, ±4, 5.) Thus if there were unique factorisation we would have 2 | 4 + √ 10 or 2 | 4 − √ 10, both of which are absurd.
3.6 The splitting of rational primes Throughout n this section we shall assume that the integers Z[ω] in Q(√m) form a principal ideal domain (and so a unique factorisation domain).
Proposition 3.10 Let p ∈N be a rational prime. Then p either remains a prime in Z[ω], or else p = ±π¯ π, where π is a prime in Z[ω]. In other words, p has either one or two prime factors; and if it has two then these are conjugage.
Proof ▶Suppose p = ϵπ1 · · · πr.
374 3–16 Then N(π1) · · · N(πr) = N(p) = p2.
Since N(πi) is an integer ̸= 1, it follows that either r = 1, ie p remains a prime, or else r = 2 with N(π1) = ±p, N(π2) = ±p.
In this case, writing π for π1, p = ±N(π) = ±π¯ π.
◀ We say that p splits in Q(√m) in the latter case, ie if p divides into two prime factors in Z[ω]. We say that p ramifies if these two prime factors are equal, ie if p = ϵπ2, Corollary 3.2 The rational prime p ∈N splits if and only if there is an integer α ∈Z[ω] with N(α) = ±p.
Proposition 3.11 Suppose p ∈N is an odd prime with p ∤m. Then p splits in Q(√m) if and only if m is a quadratic residue modp, ie if and only if x2 ≡m mod p for some x ∈Z.
Proof ▶Suppose x2 ≡m mod p.
Then (x −√m)(x + √m) = pq for some q ∈Z.
If now p is prime in Z[ω] (where it is assumed, we recall, that there is unique factorisation). Then p | x −√m or p | x + √m, both of which are absurd, since for example p | x −√m = ⇒x −√m = p(a + b√m) = ⇒pb = −1, where b is (at worst) a half-integer.
◀ It remains to consider two cases, p | m and p = 2.
Proposition 3.12 If the rational prime p | m then p ramifies in Q(√m).
374 3–17 Proof ▶We have (√m)2 = m = pq, for some q ∈Z. If p remains prime then p | √m = ⇒N(p) | N(√m) = ⇒p2 | −m, which is impossible, since m is square-free.
Hence p = ±π¯ π, and √m = πα for some α ∈Z[ω]. Note that α cannot contain ¯ π as a factor, since this would imply that p = ±π¯ π | √m, which as we have seen is impossible.
Taking conjugates −√m = ¯ π¯ α.
Thus ¯ π | √m.
Since the factorisation of √m is (by assumption) unique, ¯ π ∼π, ie p ramifies.
◀ Proposition 3.13 The rational prime 2 remains prime in Z[ω] if and only if m ≡5 mod 8.
Moreover, 2 ramifies unless m ≡1 mod 4.
Proof ▶We have dealt with the case where 2 | m, so we may assume that m is odd.
Suppose first that m ≡3 mod 4.
In this case (1 −√m)(1 + √m) = 1 −m = 2q.
If 2 does not split then 2 | 1 −√m or 2 | 1 + √m, 374 3–18 both of which are absurd.
Thus 2 = ±π¯ π, where π = a + b√m (a, b ∈Z), say. But then ¯ π = a −b√m = π + 2b√m.
Since π | 2 is follows that π | ¯ π; and similarly ¯ π | π.
Thus ¯ π = ϵπ, where ϵ is a unit; and so 2 ramifies.
Now suppose m ≡1 mod 4.
Suppose 2 splits, say a2 −mb2 = ±2, where a, b are integers or half-integers. If a, b ∈Z then a2 −mb2 ≡0, ±1 mod 4, since a2, b2 ≡0 or 1 mod 4.
Thus a, b must be half-integers, say a = A/2, b = B/2, where A, B are odd integers. In this case, A2 −mB2 = ±8.
Hence A2 −mB2 ≡0 mod 8 But A2 ≡B2 ≡1 mod 8, and so A2 −mB2 ≡1 −m mod 8.
Thus the equation is insoluble if m ≡5 mod 8, ie 2 remains prime in this case.
Finally, if m ≡1 mod 8 374 3–19 then 1 −√m 2 · 1 + √m 2 = 1 −m 4 = 2q.
If 2 does not split then 2 | 1 −√m 2 or 2 | 1 + √m 2 , both of which are absurd.
Suppose 2 = ±π¯ π, where π = A + B√m 2 , with A, B odd; and ¯ π = A −B√m 2 = π −B√m.
Thus π | ¯ π = ⇒π | B√m = ⇒N(π) | N(B√m) = ⇒±2 | B2m, which is impossible since B, m are both odd. Hence 2 is unramified in this case.
◀ 3.7 Quadratic residues Definition 3.6 Suppose p is an odd rational prime; and suppose a ∈Z. Then the Legendre symbol is defined by a p !
= 0 if p | a 1 if p ∤a and a is a quadratic residue modp −1 if a is a quadratic non-residue modp Proposition 3.14 Suppose p is an odd rational prime; and suppose a, b ∈Z.
Then a p ! b p !
= ab p !
.
374 3–20 Proof ▶The resul is trivial if p | a or p | b; so we may suppose that p ∤a, b.
Consider the group-homomorphism θ : (Z/p)× →(Z/p)× : ¯ x 7→¯ x2.
Since ker θ = {±1} it follows from the First Isomorphism Theorem that |im θ| = p −1 2 , and so (Z/p)×/ im θ ∼ = C2 = {±1}.
The result follows, since im θ = {¯ a ∈(Z/p)× : a p !
= 1}.
◀ Proposition 3.15 Suppose p is an odd rational prime; and suppose a ∈Z. Then a(p−1)/2 ≡ a p !
mod p.
Proof ▶The resul is trivial if p | a; so we may suppose that p ∤a.
By Lagrange’s Theorem (or Fermat’s Little Theorem) ap−1 ≡1 mod p.
Thus a(p−1)/22 ≡1 mod p; and so a(p−1)/2 ≡±1 mod p.
Suppose a is a quadratic residue, say a ≡b2 mod p.
Then a p−1 2 ≡bp−1 ≡1 mod p.
Thus a p !
= 1 = ⇒a p−1 2 ≡1 mod p.
374 3–21 As we saw in the proof of Proposition 3.14, exactly half, ie p−1 2 of the numbers 1, 2, . . . , p −1 are quadratic residues. On the other hand, the equation x p−1 2 −1 = 0 over the field Fp = Z/(p) has at most p−1 2 roots. It follows that a p !
= 1 ⇐ ⇒a p−1 2 ≡1 mod p; and so a p !
≡a p−1 2 mod p; ◀ Corollary 3.3 If p ∈N is an odd rational prime then −1 p !
= 1 if p ≡1 mod 4, −1 if p ≡3 mod 4.
Proof ▶By the Proposition, −1 p !
≡(−1) p−1 2 mod p.
If p ≡1 mod 4, say p = 4m + 1, then p −1 2 = 2m; while if p ≡3 mod 4, say p = 4m + 3, 374 3–22 then p −1 2 = 2m + 1.
◀ It is sometimes convenient to take the remainder r ≡a mod p in the range −p 2 < r < p 2.
We may say that a has negative remainder modp if −p 2 < r < 0.
Thus 13 has negative remainder mod7, since 13 ≡−1 mod 7.
Proposition 3.16 Suppose p ∈N is an odd rational prime; and suppose p ∤a.
Then a p !
= (−1)µ, where µ is the number of numbers among 1, 2a, . . . , p −1 2 a with negative remainders.
Suppose, for example, p = 11, a = 7. Then 7 ≡−4, 14 ≡3, 21 ≡−1, 28 ≡−5, 35 ≡2 mod 11.
Thus µ = 3.
Proof ▶Suppose 1 ≤r ≤p −1 2 .
Then just one of the numbers a, 2a, . . . p −1 2 a has remainder ±r.
For suppose ia ≡r mod p, ja ≡−r mod p.
374 3–23 Then (i + j)a ≡0 mod p = ⇒p | i + j which is impossible since 1 ≤i + j ≤p −1.
It follows (by the Pigeon-Hole Principle) that just one of the congruences ia ≡±r mod p (1 ≤i ≤p −1 2 ) is soluble for each r.
Multiplying together these congruences, a · 2a · · · p −1 2 a ≡(−1)µ1 · 2 · · · p −1 2 mod p, ie a p−1 2 1 · 2 · · · p −1 2 ≡(−1)µ1 · 2 · · · p −1 2 mod p, and so a p−1 2 ≡(−1)µ mod p.
Since a p !
≡a p−1 2 mod p by Proposition 3.15, we conclude that a p !
≡(−1)µ mod p.
◀ Proposition 3.17 If p ∈N is an odd rational prime then 2 p !
= 1 if p ≡±1 mod 8, −1 if p ≡±3 mod 8.
Proof ▶Consider the numbers 2, 4, . . . , p −1.
The number 2i will have negative remainder if p 2 < 2i < p, 374 3–24 ie p 4 < i < p 2.
Thus the µ in Proposition 3.16 is given by µ = p 2 − p 4 .
We consider p mod 8. If p ≡1 mod 8, say p = 8m + 1, then p 2 = 4m, p 4 = 2m, and so µ = 2m.
If p ≡3 mod 8, say p = 8m + 3, then p 2 = 4m + 1, p 4 = 2m, and so µ = 2m + 1.
If p ≡5 mod 8, 374 3–25 say p = 8m + 5, then p 2 = 4m + 2, p 4 = 2m + 1, and so µ = 2m + 1.
If p ≡7 mod 8, say p = 8m + 7, then p 2 = 4m + 3, p 4 = 2m + 1, and so µ = 2m + 2.
◀ Corollary 3.4 If p ∈N is an odd rational prime then −2 p !
= 1 if p ≡1 or 3 mod 8, −1 if p ≡5 or 7 mod 8.
Proof ▶This follows from the Proposition and the Corollary to Proposition 3.15, since −2 p !
= −1 p ! 2 p !
, by Proposition 3.14.
◀ Proposition 3.18 If p ∈N is an odd rational prime then 3 p !
= 1 if p ≡±1 mod 12, −1 if p ≡±5 mod 12.
374 3–26 Proof ▶If 0 < i < p 2 then 0 < 3i < 3p 2 .
Thus 3i has negative remainder if p 2 < 3i < p, ie p 6 < i < p 3.
Thus µ = p 3 − p 6 .
If p ≡1 mod 6, say p = 6m + 1, then p 3 = 2m, p 6 = m, and so µ = m.
If p ≡5 mod 6, say p = 6m + 5, then p 3 = 2m + 1, p 6 = m, and so µ = m + 1.
The result follows.
◀ 374 3–27 Corollary 3.5 If p ∈N is an odd rational prime then −3 p !
= 1 if p ≡1 mod 6, −1 if p ≡5 mod 6.
Proof ▶This follows from the Proposition and the Corollary to Proposition 3.15, since −3 p !
= −1 p ! 3 p !
, by Proposition 3.14.
◀ Proposition 3.19 If p ∈N is an odd rational prime then 5 p !
= 1 if p ≡±1 mod 10, −1 if p ≡±3 mod 10.
Proof ▶If 0 < i < p 2 then 0 < 5i < 5p 2 .
Thus 5i has negative remainder if p 2 < 5i < p or 3p 2 < i < 2p, ie p 10 < i < p 5 or 3p 10 < i < 2p 5 .
Thus µ = p 5 − p 10 + 2p 5 − 3p 10 .
If p ≡1 mod 12, say p = 10m + 1, then p 5 = 2m, p 10 = m, 2p 5 = 4m, 3p 10 = 3m, and so µ = 2m.
The other cases are left to the reader.
◀ 374 3–28 3.8 Gauss’ Law of Quadratic Reciprocity Proposition 3.16 provides an algorithm for computing the Legendre symbol, as illustrated in Propositions 3.17–3.19, perfectly adequate for our purposes. How-ever, Euler discovered and Gauss proved a remarkable result which makes com-putation of the symbol childishly simple. This result — The Law of Quadratic Reciprocity — has been called the most beautiful result in Number Theory, so it would be a pity not to mention it, even though — as we said — we do not really need it.
Proposition 3.20 Suppose p, q ∈N are two distinct odd rational primes. Then q p ! p q !
= −1 if p ≡q ≡3 mod 4, 1 otherwise.
Another way of putting this is to say that q p ! p q !
= (−1) p−1 2 q−1 2 .
Proof ▶Let S = {1, 2, . . . , p −1 2 }, T = {1, 2, . . . , q −1 2 }.
We shall choose remainders modp from the set {−p 2 < i < p 2} = −S ∪{0} ∪S, and remainders modq from the set {−q 2 < i < q 2} = −T ∪{0} ∪T.
By Gauss’ Lemma (Proposition 3.16), q p !
= (−1)µ, p q !
= (−1)ν, where µ = ∥{i ∈S : qi mod p ∈−S}∥, ν = ∥{i ∈T : pi mod q ∈−T}∥.
By ‘qi mod p ∈−S’ we mean that there exists a j (necessarily unique) such that qi −pj ∈−S.
374 3–29 But now we observe that, in this last formula, 0 < i < p 2 = ⇒0 < j < q 2.
The basic idea of the proof is to associate to each such contribution to µ the ‘point’ (i, j) ∈S × T. Thus µ = ∥{(i, j) ∈S × T : −p 2 < qi −pj < 0}∥; and similarly ν = ∥{(i, j) ∈S × T : 0 < qi −pj < q 2}∥, where we have reversed the order of the inequality on the right so that both for-mulae are expressed in terms of (qi −pj).
Let us write [R] for the number of integer points in the region R ⊂R2. Then µ = [R1], ν = [R2], where R1 = {(x, y) ∈R : −p 2 < qx−py < 0}, R2 = {(x, y) ∈R : 0 < qx−py < q 2}, and R denotes the rectangle R = {(x, y) : 0 < x < p 2, 0 < y < p 2}.
The line qx −py = 0 is a diagonal of the rectangle R, and R1, R2 are strips above and below the diago-nal (Fig 3.8).
This leaves two triangular regions in R, R3 = {(x, y) ∈R : qx −py < −p 2}, R4 = {(x, y) ∈R : qx −py > q 2}.
We shall show that, surprisingly perhaps, reflection in a central point sends the integer points in these two regions into each other, so that [R3] = [R4].
Since R = R1 ∪R2 ∪R3 ∪R4, it will follow that [R1] + [R2] + [R3] + [R4] = [R] = p −1 2 q −1 2 , 374 3–30 R3 R4 R1 R2 Figure 3.1: p = 11, q = 7 ie µ + ν + [R3] + [R4] = p −1 2 q −1 2 .
But if now [R3] = [R4] then it will follow that µ + ν ≡p −1 2 q −1 2 mod 2, which is exactly what we have to prove.
It remains to define our central reflection. Note that reflection in the centre (p 4, q 4) of the rectangle R will not serve, since this does not send integer points into integer points. For that, we must reflect in a point whose coordinates are integers or half-integers.
We choose this point by “shrinking” the rectangle R to a rectangle bounded by integer points, ie the rectangle R′ = {1 ≤x ≤p −1 2 , 1 ≤y ≤q −1 2 }.
Now we take P to be the centre of this rectangle, ie P = (p + 1 4 , q + 1 4 ).
The reflection is then given by (x, y) 7→(X, Y ) = (p + 1 − x, q + 1 −y).
374 3–31 It is clear that reflection in P will send the integer points of R into themselves.
But it is not clear that it will send the integer points in R3 into those in R4, and vice versa. To see that, let us shrink these triangles as we shrank the rectangle. If x, y ∈Z then qx −py < −p 2 = ⇒qx −py ≤−p + 1 2 ; and similarly qx −py > q 2 = ⇒qx −py ≥q + 1 2 .
Now reflection in P does send the two lines qx −py = −p + 1 2 , qx −py = q + 1 2 into each other; for qX −pY = q(p + 1 −x) −p(q + 1 −y) = (q −p) −(qx −py), and so qx −py = −p + 1 2 ⇐ ⇒qX −pY = (q −p) + p + 1 2 = q + 1 2 .
We conclude that [R3] = [R4].
Hence [R] = [R1] + [R2] + [R3] + [R4] ≡µ + ν mod 2, and so µ + ν ≡[R] = p −1 2 q −1 2 .
◀ Example: Take p = 37, q = 47. Then 37 47 !
= 47 37 !
since 37 ≡1 mod 4 = 10 37 !
= 2 37 ! 5 37 !
= − 5 37 !
since 37 ≡−3 mod 8 = − 37 5 !
since 5 ≡1 mod 4 = − 2 5 !
= −(−1) = 1.
374 3–32 Thus 37 is a quadratic residue mod47.
We could have avoided using the result for 2 p !
: 10 37 !
= −27 37 !
= −1 37 ! 3 37 !3 = (−1)18 37 3 !
= 1 3 !
= 1.
3.9 Some quadratic fields We end by applying the results we have established to a small number of quadratic fields.
3.9.1 The gaussian field Q(i) Proposition 3.21 1. The integers in Q(i) are the gaussian integers a + bi (a, b ∈Z) 2. The units in Z[i] are the numbers ±1, ±i.
3. The ring of integers Z[i] is a principal ideal domain (and so a unique fac-torisation domain).
4. The prime 2 ramifies in Z[i]: 2 = −i(1 + i)2.
The odd prime p splits in Z[i] if and only if p ≡1 mod 4, in which case it splits into two conjugate but inequivalent primes: p = ±π¯ π.
374 3–33 Proof ▶This follows from Propositions 3.4, 3.7, 3.9, 3.11–3.13, and the Corollary to Proposition 3.15.
◀ Factorisation in the gaussian field Q(i) gives interesting information on the expression of a number as a sum of two squares.
Proposition 3.22 An integer n > 0 is expressible as a sum of two squares, n = a2 + b2 (a, b ∈Z) if and only if each prime p ≡3 mod 4 occurs to an even power in n.
Proof ▶Suppose n = a2 + b2 = (a + bi)(a −bi).
Let a + bi = ϵπe1 1 · · · πer r .
Taking norms, n = N(a + bi) = N(π1)e1 · · · N(πr)er.
Suppose p ≡3 mod 4.
Then p remains prime in Z[i], by Proposition 3.21.
Suppose pe ∥a + ib, ie pe | a + ib but pe+1 ∤a + ib.
Then pe ∥a −ib, since a + ib = peα = ⇒a −ib = pe¯ α, on taking conjugates. Hence p2e ∥n = (a + ib)(a −ib), ie p appears in n with even exponent.
We have shown, incidentally, that if p ≡3 mod 4 then p2e ∥n = a2 + b2 = ⇒pe | a, pe | b.
374 3–34 In other words, each expression of n as a sum of two squares n = a2 + b2 is of the form n = (pea′)2 + (peb′)2, where n p2e = a′2 + b′2.
We have shown that each prime p ≡3 mod 4 must occur with even exponent in n. Conversely, suppose that this is so.
Each prime p ≡1 mod 4 splits in Z[i], by Proposition 3.21, say p = πpπp.
Also, 2 ramifies in Z[i]: 2 = −i(1 + i)2.
Now suppose n = 2e23e35e5 · · · , where e3, e7, e11, e19, . . . are all even, say p ≡3 mod 4 = ⇒ep = 2fp.
Let α = α2α3α5 · · · , where αp = (1 + i)e2 if p = 2, πep p if p ≡1 mod 4, pfp if p ≡3 mod 4.
Then N(αp) = pep in all cases, and so N(α) = Y p N(αp) = Y p pep = n.
Thus if α = a + bi then n = a2 + b2.
◀ 374 3–35 It’s worth noting that this argument actually gives the number of ways of ex-pressing n as a sum of two squares, ie the number of solutions of n = a2 + b2 (a, b ∈Z).
For the number of solutions is the number of integers α ∈Z[i] such that n = N(()α) = α¯ α.
Observe that when p ≡1 mod 3 in the argument above we could equally well have taken αp = πr¯ πs for any r, s ≥0 with r + s = ep.
There are just ep + 1 ways of choosing αp in this way.
It follows from unique factorisation that the choice of the αp for p ≡1 mod 4 determines α up to a unit, ie the general solution is α = ϵ(1 + i)e2 Y p≡1 mod 4 αp Y p≡3 mod 4 pfp.
Since there are four units, ±1, ±i, we conclude that the number of ways of ex-pressing n as a sum of two sqares is 4 Y p≡1 mod 4 (ep + 1).
Note that in this calculation, each solution n = a2 + b2 with 0 < a < b gives rise to 8 solutions: n = (±a)2 + (±b)2, n = (±b)2 + (±a)2.
To these must be added solutions with a = 0 or with a = b. The former occurs only if n = m2, giving 4 additional solutions: n = 02 + (±m)2 = (±m)2 + 02; while the latter occurs only if n = 2m2, again giving 4 additional solutions: n = (±m)2 + (±m)2.
374 3–36 We conclude that the number of solutions with a, b ≥0 is 1 2 Q p≡1 mod 4(ep + 1) if n ̸= m2, 2m2 1 2 Q p≡1 mod 4(ep + 1) + 1 if n = m2 or 2m2.
This is of course assuming that p ≡3 mod 4 = ⇒2 | ep, without which there are no solutions.
In particular, each prime p ≡1 mod 4 is uniquely expressible as a sum of two squares n = a2 + b2 (0 < a < b), eg 53 = 22 + 72.
As another example, 108 = 2233 cannot be expressed as a sum of two squares, since e3 = 3 is odd.
3.9.2 The field Q( √ 3) Proposition 3.23 1. The integers in Q( √ 3) are the numbers a + b √ 3 (a, b ∈Z) 2. The units in Z[ √ 3] are the numbers ±ηn (n ∈Z), where η = 2 + √ 3.
3. The ring of integers Z[ √ 3] is a principal ideal domain (and so a unique factorisation domain).
4. The primes 2 and 3 ramify in Z[ √ 3]: 2 = η−1(1 + √ 3)2, 3 = ( √ 3)2.
The odd prime p ̸= 3 splits in Z[ √ 3] if and only if p ≡±1 mod 12, in which case it splits into two conjugate but inequivalent primes: p = ±π¯ π.
Proof ▶This follows from Propositions 3.4, 3.8, 3.9, 3.11–3.13, and Proposi-tion 3.18.
◀ 374 3–37 3.9.3 The field Q( √ 5) Proposition 3.24 1. The integers in Q( √ 5) are the numbers a + bω (a, b ∈Z), where ω = 1 + √ 5 2 .
2. The units in Z[ √ 5] are the numbers ±ωn (n ∈Z).
3. The ring of integers Z[ω] is a principal ideal domain (and so a unique fac-torisation domain).
4. The prime 5 ramifies in Z[ω]: 5 = ( √ 5)2.
The prime p ̸= 5 splits in Z[ω] if and only if p ≡±1 mod 10, in which case it splits into two conjugate but inequivalent primes: p = ±π¯ π.
Proof ▶This follows from Propositions 3.4, 3.8, 3.9, 3.11–3.13, and Proposi-tion 3.19.
◀ Chapter 4 Mersenne and Fermat numbers 4.1 Mersenne numbers Proposition 4.1 If n = am −1 (a, m > 1) is prime then 1. a = 2; 2. m is prime.
Proof ▶In the first place, (a −1) | (am −1); so if a > 2 then n is certainly not prime.
Suppose m = rs, where r, s > 1. Evidently (x −1) | (xs −1) in Z[x]; explicitly xs −1 = (x −1)(xs−1 + xs−2 + xs−3 + · · · + 1).
Subsitituting x = ar, (ar −1) | (ars −1) = am −1.
Thus if am −1 is prime then m has no proper factors, ie m is prime.
◀ Definition 4.1 The numbers Mp = 2p −1, where p is prime, are called Mersenne numbers.
4–1 374 4–2 The numbers M2 = 3, M3 = 7, M5 = 31, M7 = 127 are all prime. However, M11 = 2047 = 23 · 89.
(It should be emphasized that Mersenne never claimed the Mersenne numbers were all prime. He listed the numbers Mp for p ≤257, indicating which were prime, in his view. His list contained several errors.) The following heuristic argument suggests that there are probably an infinity of Mersenne primes. (Webster’s Dictionary defines ‘heuristic’ as: providing aid or direction in the solution of a problem but otherwise unjustified or incapable of justification.) By the Prime Number Theorem, the probability that a large number n is prime is ≈ 1 log n.
In this estimate we are including even numbers. Thus the probability that an odd number n is prime is ≈ 2 log n.
Thus the probability that Mp is prime is ≈ 2 p log 2.
So the expected number of Mersenne primes is ≈ 2 log 2 X 1 pn where pn is the nth prime.
But — again by the Prime Number Theorem — pn ≈n log n.
Thus the expected number of Mersenne primes is ≈ 2 log 2 X 1 n log n = ∞, since X 1 n log n diverges, eg by comparison with Z X 1 x log x = log log X + C.
374 4–3 4.1.1 The Lucas-Lehmer test Mersenne numbers are important because there is a simple test, announced by Lucas and proved rigorously by Lehmer, for determining whether or not Mp is prime. (There are many necessary tests for primality, eg if p is prime then 2p ≡2 mod p.
What is rare is to find a necessary and sufficient test for the primality of numbers in a given class, and one which is moreover relatively easy to implement.) For this reason, all recent “record” primes have been Mersenne primes.
We shall give two slightly different versions of the Lucas-Lehmer test. The first is only valid if p ≡3 mod 4, while the second applies to all Mersenne num-bers. The two tests are very similar, and equally easy to implement. We are giving the first only because the proof of its validity is rather simpler. So it should be viewed as an introduction to the second, and true, Lucas-Lehmer test.
Both proofs are based on arithmetic in quadratic fields: the first in Q( √ 5), and the second in Q( √ 3); and both are based on the following result.
Proposition 4.2 Suppose α is an integer in the field Q(√m); and suppose P is an odd prime with P ∤m. Then αP ≡ α if P m !
= 1, ¯ α if P m !
= −1.
Proof ▶Suppose α = a + b√m, where a, b are integers if m ̸≡1 mod 4, and half-integers if m ≡1 mod 4.
In fact these cases do not really differ; for 2 is invertible modP, so we may consider a as an integer modP if 2a ∈Z. Thus αP ≡aP + P 1 !
aP−1b√m + P 2 !
aP−2bm + · · · + bPm P −1 2 √m mod P.
Now P | P r !
if 1 ≤r ≤P −1. Hence αP ≡aP + bPm P −1 2 √m mod P By Fermat’s Little Theorem, aP ≡a mod P, bP ≡b mod P.
374 4–4 Also m P −1 2 ≡ m P !
mod P, by Proposition 3.15. Thus αP ≡a + b P m !√m mod P, ie m P !
= 1 = ⇒αP ≡α mod P, m P !
= −1 = ⇒αP ≡¯ α mod P.
◀ Corollary 4.1 For all integers α in Q(√m, αP 2 ≡α mod P.
We may regard this as the analogue of Fermat’s Little Theorem aP ≡a mod P for quadratic fields.
There is another way of establishing this result, which we shall sketch briefly.
It depends on considering the ring A = Z[ω]/(P).
formed by the remainders α mod P of integers α in Q(√m).
There are P 2 elements in this ring, since each α ∈Z[ω] is congruent modP to just one of the numbers a + b√m where a, b ∈Z and 0 ≤a, b < P.
There are no nilpotent elements in the ring A if P ∤m; for if α = a + b√m then P | α2 = ⇒P | 2ab, P | a2 + b2m = ⇒P | a, b.
374 4–5 Thus α2 ≡0 mod P = ⇒α ≡0 mod P, from which it follows that, if n > 0, αn ≡0 mod P = ⇒α ≡0 mod P, A ring without non-zero nilpotent elements is said to be semi-simple. It is not hard to show that a finite semi-simple commutative ring is a direct sum of fields.
Now there is just one field (up to isomorphism) containing pe elements for each prime power pe, namely the galois field GF(pe). It follows that either 1. Z[ω]/(P) ∼ = GF(P 2); or 2. Z[ω]/(P) ∼ = GF(P) ⊕GF(P).
The non-zero elements in GF(pe) form a multiplicative group GF(pe)× with pe −1 elements. It follows from Legendre’s Theorem that a ̸= 0 = ⇒ape−1 = 1 in GF(pe). Hence ape = a for all a ∈GF(pe).
Thus in the first case, αP 2 ≡α for all α ∈Z[ω]/(P); while in the second case we even have αP ≡α for all α ∈Z[ω]/(P), since this holds in each of the constituent fields.
In the first case we can go further. The galois field GF(pe) is of characteristic p, ie pa = a + · · · a = 0, for all ainGF(pe). Also, the map a 7→ap is an automorphism of GF(pe). (This follows by essentially the same argument that we used above to show that αP ≡α or ¯ α above.) In particular, the map α 7→αP mod P is an automorphism of our field Z[ω]/(P).
374 4–6 On the other hand, the map α 7→¯ α is also an automorphism of Z[ω]/(P), since P | α = ⇒P | ¯ α.
Moreover, this is the only automorphism of Z[ω]/(P) apart from the identity map, since any automorphism must send √m mod P 7→±√m mod P.
The automorphism α 7→αP mod P is not the identity map, since the equation xP −x = 0 has at mos P solutions in the field Z[ω]/(P). We conclude that αP ≡¯ α mod P.
If Z[ω] is a principal ideal domain the second case arises if and only if P splits, which by Proposition 3.14 occurs when m P !
= 1.
Explicitly, if P = π1π2, then Z[ω]/(P) ∼ = Z[ω]/(π1) ⊕Z[ω]/(π2) ∼ = GF(P) ⊕GF(P).
Proposition 4.3 Suppose p ≡3 mod 4. Let the sequence rn be defined by r1 = 3, rn+1 = r2 n −2.
Then Mp is prime if and only if Mp | rp−1.
Proof ▶We work in the field Q( √ 5). By Proposition 3.4, the integers in this field are the numbers a + bω (a, b ∈Z) where ω = 1 + √ 5 2 .
By Proposition 3.9, there is unique factorisation in the ring of integers Z[ω].
374 4–7 Lemma 4.1 If rn is the sequence defined in the Proposition then rn = ω2n + ω−2n for each n ≥1.
Proof of Lemma ▷Let us set sn = ω2n + ω−2n for n ≥0. Then s2 n = ω2n + ω−2n2 = ω2n+1 + 2 + ω−2n+1 = sn+1 + 2, ie sn+1 = s2 n −2.
Also s0 = ω + ω−1 = ω −¯ ω = √ 5, and so s1 = s2 0 −2 = 3.
We conclude that rn = sn = ω2n + ω−2n for all n ≥1.
◁ Let us suppose first that Mp is prime. Let us write P = Mp.
Lemma 4.2 We have 5 P !
= −1.
Proof of Lemma ▷Since 24 ≡1 mod 5 it follows that 2p ≡23 mod 5 ≡3 mod 5.
374 4–8 Hence P = 2p −1 ≡2 mod 5; and so, by Proposition 3.19, 5 P !
= −1.
◁ It follows from this Lemma and Proposition 4.2 that αP ≡¯ α mod P for all α ∈Z[ω]. In particular, ωP ≡¯ ω mod P.
Hence ωP+1 ≡ω¯ ω mod P ≡N(ω) mod P ≡−1 mod P.
In other words, ω2p ≡−1 mod P.
Thus ω2p + 1 ≡0 mod P.
Dividing by ω2p−1, ω2p−1 + ω−2p−1 ≡0 mod P, ie rp−1 ≡0 mod P.
Conversely, suppose P is a prime factor of Mp. Then Mp | rp−1 = ⇒rp−1 ≡0 mod P = ⇒ω2p−1 + ω−2p−1 ≡0 mod P = ⇒ω2p + 1 ≡0 mod P = ⇒ω2p ≡−1 mod P.
But this implies that the order of ω mod P is 2p+1. For ω2p+1 = (ω2p)2 ≡1 mod P, so if the order of ω mod P is d then d | 2p+1 = ⇒d = 2e 374 4–9 for some e ≤p + 1; and if e ≤p then ω2p ≡1 mod P.
On the other hand, by the Corollary to Proposition 4.2, ωP 2 ≡ω mod P = ⇒ωP 2−1 ≡1 mod P.
Hence 2p+1 | P 2 −1 = (P + 1)(P −1).
Now gcd(P + 1, P −1) = 2.
It follows that 2p | P + 1 or 2p | P −1.
The latter is impossible since 2p > Mp ≥P > P −1; while 2p | P + 1 = ⇒2p ≤P + 1 = ⇒Mp = 2p −1 ≤P = ⇒P = Mp.
◀ Now for the ‘true’ Lucas-Lehmer test. As we shall see, the proof is a little harder, which is why we gave the earlier version.
Proposition 4.4 Let the sequence rn be defined by r1 = 4, rn+1 = r2 n −2.
Then Mp is prime if and only if Mp | rp−1.
Proof ▶We work in the field Q( √ 3). By Proposition 3.4, the integers in this field are the numbers a + b √ 3 (a, b ∈Z).
By Proposition 3.9, there is unique factorisation in the ring of integers Z[ √ 3].
We set η = 1 + √ 3, ϵ = 2 + √ 3.
Lemma 4.3 The units in Z[ √ 3] are the numbers ±ϵn (n ∈N).
374 4–10 Proof of Lemma ▷It is sufficient, by Proposition 3.8, to show that ϵ is the smallest unit > 1. And from the proof of that Proposition, we need only consider units of the form a + b √ 3 with a, b ≥0.
Thus the only possible units in the range (1, ϵ) are √ 3 and 1+ √ 3 = η, neither of which is in fact a unit, since N( √ 3) = −3, N(η) = −2, whereas a unit must have norm ±1, by Proposition 3.6.
◁ Lemma 4.4 If rn is the sequence defined in the Proposition then rn = ϵ2n−1 + ϵ−2n−1 for each n ≥1.
Proof of Lemma ▷Let us set sn = ϵ2n−1 + ϵ−2n−1 for n ≥1. Then s2 n = ϵ2n−1 + ϵ−2n−12 = ϵ2n + 2 + ϵ−2n = sn+1 + 2, ie sn+1 = s2 n −2.
Also s1 = ϵ + ϵ−1 = ϵ + ¯ ϵ = 4.
We conclude that rn = sn = ϵ2n−1 + ϵ−2n−1 for all n ≥1.
◁ Suppose first that P = Mp is prime.
Lemma 4.5 We have 3 P !
= −1.
374 4–11 Proof of Lemma ▷We have Mp = 2p −1 ≡(−1)p −1 mod 3 ≡−1 −1 mod 3 ≡1 mod 3; while Mp ≡−1 mod 4.
By the Chinese Remainder Theorem there is just one remainder mod12 with these remainders mod3 and mod4; and that is 7 ≡−5 mod 12. For any odd prime p, Mp ≡7 mod 12 Hence 3 P !
= −1.
by Proposition 3.18, ◁ It follows from this Lemma and Proposition 4.2 that αP ≡¯ α mod P for all α ∈Z[ √ 3]. In particular, ϵP ≡¯ ϵ mod P.
Hence ϵP+1 ≡ϵ¯ ϵ mod P ≡N(ϵ) mod P ≡1 mod P.
In other words, ϵ2p ≡1 mod P.
It follows that ϵ2p−1 ≡± mod P.
We want to show that in fact ϵ2p−1 ≡−1 mod P.
This is where things get a little trickier than in the first version of the Lucas-Lehmer test. In effect, we need a number with negative norm. To this end we introduce η = 1 + √ 3.
Lemma 4.6 1. N(η) = −2.
374 4–12 2. η2 = 2ϵ.
Proof of Lemma ▷This is a matter of simple verification: N(η) = 1 −3 = −2, while η2 = (1 + √ 3)2 = 4 + 2 √ 3 = 2ϵ.
◁ By Proposition /refMersenneLemma, ηP ≡¯ η mod P, and so ηP+1 ≡η¯ η −2 mod P, ie η2p ≡−2 mod P.
By the Lemma, this can be written (2ϵ)2p−1 ≡−2 mod P, ie 22p−1ϵ2p−1 ≡−2 mod P, But by Proposition 3.14, 2 P −1 2 = 22p−1−1 ≡ 2 P !
mod P ≡1 mod P, by Proposition 3.17, since P = 2p −1 ≡−1 mod 8.
Thus 22p−1 ≡2 mod P 374 4–13 and so 2ϵ2p−1 ≡−2 mod P.
Hence ϵ2p−1 ≡−1 mod P.
Thus ϵ2p−1 + 1 ≡0 mod P.
Dividing by ϵ2p−2, ϵ2p−2 + ϵ−2p−2 ≡0 mod P, ie rp−1 ≡0 mod P.
Conversely, suppose P is a prime factor of Mp. Then Mp | rp−1 = ⇒rp−1 ≡0 mod P = ⇒ϵ2p−2 + ϵ−2p−2 ≡0 mod P = ⇒ϵ2p−1 + 1 ≡0 mod P = ⇒ϵ2p−1 ≡−1 mod P.
But (by the argument we used in the proof of the first Lucas-Lehmer test) this implies that the order of ϵ mod P is 2p.
On the other hand, by the Corollary to Proposition 4.2, ϵP 2 ≡ϵ mod P = ⇒ϵP 2−1 ≡1 mod P.
Hence 2p | P 2 −1 = (P + 1)(P −1).
Now gcd(P + 1, P −1) = 2.
It follows that 2p−1 | P + 1 or 2p−1 | P −1.
In either case, 2p−1 ≤P + 1 = ⇒P ≥2p−1 −1 = Mp −1 2 = ⇒P ≥Mp 3 = ⇒Mp P < 3.
Since Mp is odd, this implies that P = Mp, ie Mp is prime.
◀ 374 4–14 4.1.2 Perfect numbers Mersenne numbers are also of interest because of their intimate connection with perfect numbers.
Definition 4.2 For n ∈N, n > 0 we denote the number of divisors of n by d(n), and the sum of these divisors by σ(n).
Example: Since 12 has divisors 1, 2, 3, 4, 6, 12, d(12) = 6, σ(12) = 28.
Definition 4.3 The number n ∈N is said to be perfect if σ(n) = 2n, ie if n is the sum of its proper divisors.
Example: The number 6 is perfect, since 6 = 1 + 2 + 3.
Proposition 4.5 If Mp = 2p −1 is a Mersenne prime then 2p−1(2p −1) is perfect.
Conversely, every even perfect number is of this form.
Proof ▶In number theory, a function f(n) defined on {n ∈N : n > 0} is said to be multiplicative if gcd(m, n) = 1 = ⇒f(mn) = f(m)f(n).
If the function f(n) is multiplicative, and n = pe1 1 · · · per r then f(n) = f(pe1 1 ) · · · f(per r ).
Thus the function f(n) is completely determined by its value f(pe) for prime powers.
374 4–15 Lemma 4.7 The functions d(n) and σ(n) are both multiplicative.
Proof of Lemma ▷Suppose gcd(m, n) = 1; and suppose d | mn.
Then d is uniquely expressible in the form d = d1d2 (d1 | m, d2 | n).
In fact d1 = gcd(d, m), d2 = gcd(d, n).
It follows that d(mn) = d(m)d(n); and σ(mn) = X d|mn d = X d1|m d1 X d2|n d2 = σ(m)σ(n).
◁ Now suppose n = 2p−1Mp where Mp is prime. Since Mp is odd, gcd(2p−1, Mp) = 1.
Hence σ(n) = σ(2p−1)σ(Mp).
If P is prime then evidently σ(P) = 1 + P.
On the other hand, σ(P e) = 1 + P + P 2 + · · · + P e = P e+1 −1 P −1 .
In particular, σ(2e) = 2e+1 −1.
Thus σ(2p−1) = 2p −1 = Mp, 374 4–16 while σ(Mp) = Mp + 1 = 2p.
We conclude that σ(n) = 2pMp = 2n.
Conversely, suppose n is an even perfect number. We can write n (uniquely) in the form n = 2em where m is odd. Since 2e and m are coprime, σ(n) = σ(2e)σ(m) = (2e+1 −1)σ(m).
On the other hand, if n is perfect then σ(n) = 2n = 2e+1m.
Thus 2e+1 −1 2e+1 = m σ(m).
The numerator and denominator on the left are coprime. Hence m = d(2e+1 −1), σ(m) = d2e+1, for some d ∈N.
If d > 1 then m has at least the factors 1, d, m. Thus σ(m) ≥1 + d + m = 1 + d2e+1, contradicting the value for σ(m) we derived earlier.
It follows that d = 1. But then σ(m) = 2e+1 = m + 1.
Thus the only factors of m are 1 and m, ie m = 2e+1 −1 = Me+1 is prime. Setting e + 1 = p, we conclude that n = 2p−1Mp, where Mp is prime.
◀ It is an unsolved problem whether or not there are any odd perfect numbers.
The first 4 even perfect numbers are 21M2 = 6, 22M3 = 28, 24M5 = 496, 26M7 = 8128.
(In fact these are the first 4 perfect numbers, since it is known that any odd perfect number must have at least 300 digits!) 374 4–17 4.2 Fermat numbers Proposition 4.6 If n = am + 1 (a, m > 1) is prime then 1. a2 is even; 2. m = 2e.
Proof ▶If a is odd then n is even and > 2, and so not prime.
Suppose m has an odd factor, say m = rs, where r is odd. Since xr + 1 = 0 when x = −1, it follows by the Remainder Theorem that (x + 1) | (xr + 1).
Explicitly, xr + 1 = (x + 1)(xr−1 −xr−2 + · · · −x + 1).
Substituting x = ys, (ys + 1) | (ym + 1) in Z[x]. Setting y = a, (as + 1) | (ars + 1) = (am + 1).
In particular, am + 1 is not prime.
Thus if am + 1 is prime then m cannot have any odd factors. In other words, m = 2e.
◀ Definition 4.4 The numbers Fn = 22n + 1 (n = 0, 1, 2, . . . ) are called Fermat numbers.
Fermat hypothesized — he didn’t claim to have a proof — that all the numbers F0, F1, F2, . . .
are prime. In fact this is true for F0 = 3, F1 = 5, F2 = 17, F3 = 257, F4 = 65537.
374 4–18 However, Euler showed in 1747 that F5 = 232 + 1 = 4294967297 is composite. In fact, no Fermat prime beyond F4 has been found.
The heuristic argument we used above to suggest that the number of Mersenne primes is probably infinite now suggests that the number of Fermat primes is probably finite.
For by the Prime Number Theorem, the probability of Fn being prime is ≈2/ log Fn ≈2 · 2−n.
Thus the expected number of Fermat primes is 2 ≈ X 2−n = 4 < ∞.
This argument assumes that the Fermat numbers are “independent”, as far as primality is concerned. It might be argued that our next result shows that this is not so. However, the Fermat numbers are so sparse that this does not really affect our heuristic argument.
Proposition 4.7 The Fermat numbers are coprime, ie gcd(Fm, Fn) = 1 if m ̸= n.
Proof ▶Suppose gcd(Fm, Fn) > 1.
Then we can find a prime p (which must be odd) such that p | Fm, p | Fn.
Now the numbers {1, 2, . . . , p −1} form a group (Z/p)× under multiplication modp. Since p | Fm, 22m ≡−1 mod p.
It follows that the order of 2 mod p (ie the order of 2 in (Z/p)×) is exactly 2m+1.
For certainly 22m+1 = (22m)2 ≡1 mod p; and so the order of 2 divides 2m+1, ie it is 2e for some e ≤m + 1. But if e ≤m then 22m ≡1 mod p, whereas we just saw that the left hand side was ≡−1 mod p. We conclude that the order must be 2m+1.
374 4–19 But by the same token, the order is also 2n+1. This is a contradiction, unless m = n.
◀ We can use this result to give a second proof of Euclid’s Theorem that there are an infinity of primes.
Proof ▶Each Fermat number Fn has at least one prime divisor, say qn. But by the last Proposition, the primes q0, q1, q2, . . .
are all distinct.
◀ We end with a kind of pale imitation of the Lucas-Lehmer test, but now applied to Fermat numbers.
Proposition 4.8 The Fermat number Fn = 22n + 1 is prime if and only if 3 Fn−1 2 ≡−1 mod Fn.
Proof ▶Suppose P = Fn is prime.
Lemma 4.8 We have Fn ≡5 mod 12.
Proof of Lemma ▷Evidently Fn ≡1 mod 4; while Fn ≡(−1)2n + 1 mod 3 ≡2 mod 3.
By the Chinese Remainder Theorem these two congruences determine Fn mod 12; and observation shows that Fn ≡5 mod 12.
◁ It follows from this Lemma, and Proposition 3.18, that 3 P !
= −1.
Hence 3 P −1 2 ≡−1 mod P 374 4–20 by Proposition 3.14.
Conversely, suppose 3 Fn−1 2 ≡−1 mod Fn; and suppose P is a prime factor of Fn. Then 3 Fn−1 2 ≡−1 mod P, ie 322n−1 ≡−1 mod P.
It follows (as in the proof of the Lucas-Lehmer theorems) that the order of 3 mod P is 22n.
But by Fermat’s Little Theorem, 3P−1 ≡1 mod P.
Hence 22n | P −1, ie Fn −1 | P −1.
Since P | Fn this implies that Fn = P, ie Fn is prime.
◀ This test is more-or-less useless, even for quite small n, since it will take an inordinate time to compute the power, even working modulo Fn. However, it does give a short proof — which we leave to the reader — that F5 is composite.
It may be worth noting why this test is simpler than its Mersenne analogue.
In the case of Mersenne primes P = Mp we had to introduce quadratic fields because the analogue of Fermat’s Little Theorem, αP 2−1 ≡1 mod P, then allowed us to find elements of order P +1 = 2p. In the case of Fermat primes P = Fn Fermat’s Little Theorem aP−1 = a22n ≡1 mod P suffices. |
3747 | https://www.ahpo.net/assets/2013_prof_251_rcophth_standards_for_the_retrieval_of_human_ocular_tissue_1.pdf | Eye retrieval © The Royal College of Ophthalmologists 2013 All rights reserved For permission to reproduce any of the content contained herein please contact beth.barnes@rcophth.ac.uk 2013/PROF/251 Author: Francisco Figueiredo Review Date: July 2016 Page 1 of 20 STANDARDS FOR THE RETRIEVAL OF HUMAN OCULAR TISSUE USED IN TRANSPLANTATION, RESEARCH AND TRAINING Revised July 2013 CONTENTS Section Page 1. Purpose 3 1.1. Who the Standards are for 1.2. What the Standards cover 1.3. Using the Standards 1.4. Review of the Standards 2. Eye donation and the supply of ocular tissue in the UK 4 3. Consent 5 3.1. Consent for transplantation 3.2. Coroner/Procurator Fiscal 3.3. Information for relatives 3.4. Consent for research and training 3.5. Unsuitability for transplantation and disposal of tissue 3.6. Consent for blood sample and testing 3.7. Consent for seeking further information 3.8. Record of consent 4. Donor Age 6 4.1. Upper age 4.2. Lower age 5. Post-mortem time 6 6. Medical assessment of donors 6 6.1. Providing donor information 6.2. Sources of information about donors 6.3. Information required 7. Eye retrieval 8 7.1. Eye retrievers 7.2. NHSBT Human Tissue Transport box 7.3. Retrieval site risk assessment 7.4. Donor identification 7.5. Physical examination of donor 7.6. Blood sample 7.7. Enucleation 7.8. Restoring the donor’s appearance 7.9. Packaging, labelling and transport to an eye bank 8. Responsibilities 11 8.1. Donor hospitals/retrieval centres 8.2. NHSBT 8.3. Eye banks Eye retrieval © The Royal College of Ophthalmologists 2013 All rights reserved For permission to reproduce any of the content contained herein please contact beth.barnes@rcophth.ac.uk 2013/PROF/251 Author: Francisco Figueiredo Review Date: July 2016 Page 2 of 20 8.4. RCOphth Ocular Tissue Transplantation Standards Group 8.5. NHSBT Ocular Tissue Advisory Group 8.6. Joint UK Blood Services/NIBSC Professional Advisory Committee (JPAC) 8.7. JPAC Standing Advisory Committee on Tissues and Cell Therapy Products (SAC-TCTP) 8.8. DH Advisory Committee on Blood, Tissues and Organs (SaBTO) 9. References and bibliography 15 10. Abbreviations 15 11. Contact details for advice and further information 16 Annex 1. Contraindications to ocular tissue transplantation 16 Annex 2. Enucleation protocol 19 Eye retrieval © The Royal College of Ophthalmologists 2013 All rights reserved For permission to reproduce any of the content contained herein please contact beth.barnes@rcophth.ac.uk 2013/PROF/251 Author: Francisco Figueiredo Review Date: July 2016 Page 3 of 20 1. Purpose More than 3500 patients in the United Kingdom benefit each year from ocular tissue transplants. While the overwhelming majority of these procedures are corneal transplants, sclera and ocular stem cells are also transplanted. The successful outcome of these transplants depends not only on surgical and clinical expertise but on the quality of the tissue and, critically, on the steps taken to minimize the risk of disease transmission from donor to recipient. Human tissue is also required for research into the causes and treatment of ocular disease and for surgical training. The purpose of these Standards for the Retrieval of Human Ocular Tissue used in Transplantation, Research and Training (“the Standards”) is to provide professional guidance for individuals and organizations involved in eye donation. 1.1. Who the Standards are for • These Standards are intended to assist medical and other NHS staff who may be involved in eye donation by setting out the standards and responsibilities that must be met in order for donated ocular tissue to be used primarily for the treatment of patients but also for research and training. 1.2. What the Standards cover • Consent or, in Scotland, authorisation for the removal and use of ocular tissue from deceased donors. Throughout this document, use of the term “consent” will be taken to include both consent and authorisation • The information needed to determine the suitability of a donor according to current government regulations and professional guidance. • Eye retrieval, including the collection of a blood sample and restoring a donor’s cosmetic appearance following enucleation. • The responsibilities of individuals and organizations involved in eye retrieval. 1.3. Using the Standards • Eye banks are licensed by the Human Tissue Authority (“the HTA”) and are obliged under the Human Tissue (Quality and Safety for Human Application) Regulations 2007 (“the Quality and Safety Regulations”), which transposed the EU Tissues and Cells Directive and its accompanying Commission Directives into UK law, to have written third party agreements (“TPA”) with donor centres retrieving eyes on their behalves. These Standards provide the rationale and professional guidance for such TPAs. • These Standards provide the basis for training in eye retrieval and a resource for staff undertaking eye retrieval. 1.4. Review of the Standards • These Standards will be reviewed at least annually by the Ocular Tissue Transplantation Standards Group (“OTTSG”), a sub-committee of the Professional Standards Committee of the Royal College of Ophthalmologists (“the RCOphth”) and updated accordingly on the RCOphth website (www.rcophth.ac.uk). The OTTSG sub-committee may be contacted through the RCOphth (see Section 11 for contact details) for advice and further guidance. Eye retrieval © The Royal College of Ophthalmologists 2013 All rights reserved For permission to reproduce any of the content contained herein please contact beth.barnes@rcophth.ac.uk 2013/PROF/251 Author: Francisco Figueiredo Review Date: July 2016 Page 4 of 20 • A competency framework for tissue donation and retrieval is currently under review by NHS Blood and Transplant (“NHSBT”) Tissue Services. Module 9 of this framework deals specifically with competency to undertake eye retrieval. 2. Eye donation and the supply of ocular tissue in the UK Eye donation involves a number of well-defined steps, including: • Consent • Medical and behavioural assessment of potential donor • Eye retrieval by enucleation (in situ excision of corneoscleral discs is also practised in some countries but is not covered in these Standards) • Processing and storage of ocular tissue Lawful consent for eye donation must be obtained and the potential donor’s medical background investigated. Specialist Nurses – Organ Donation (“SNOD”), Tissue Donor Co-ordinators, the NHSBT Tissue Services National Referral Centre, and other specially trained NHS staff are responsible for obtaining consent and for providing information about the medical history of eye donors. The latter is obtained through discussion with donor families, from GPs and other relevant sources. The information is recorded on standardized NHSBT Consent, Patient Assessment and GP Questionnaire forms. Trained non-medical staff provide invaluable support for ophthalmologists in helping to ensure that eyes are retrieved wherever possible. NHSBT funds a number of Eye Retrieval Schemes in hospitals around the country to provide dedicated support for eye donation. Training in eye retrieval is provided for nurses, mortuary staff, and other hospital staff by the eye banks and through these Eye Retrieval Schemes. Surgeons who undertake ocular tissue transplantation must also be fully aware of the high standards required in all aspects of eye donation and should provide opportunities for medical staff in training to learn about and participate in eye retrieval. Queries on policy for retrieval should be directed to the relevant regional representative on the NHSBT Ocular Tissue Advisory Group (“OTAG”).The process of donation involves substantial commitment and effort by a number of staff in the donor hospital, including SNODs and tissue coordinators, the nursing and medical staff who cared for the donor, patient affairs officers, and mortuary staff. These individuals are taking on extra tasks in order to help patients who need ocular tissue transplants: if eyes are not retrieved when offered, their cooperation may easily be lost. If there are difficulties in attending donors, others who are involved should always be kept informed. Following eye retrieval, the eyes are sent to an eye bank for processing, storage of corneas and sclerae, and subsequent distribution to hospitals. The eye banks are responsible through their respective HTA Licences for the quality and safety of ocular tissue used in transplantation. There are currently four eye banks in the UK, namely the Corneal Transplant Service (“the CTS”) eye banks in Bristol and Manchester, and the eye banks at the Queen Victoria Hospital, East Grinstead, and Moorfields Eye Hospital, London. There are also laboratories providing cells for the treatment of ocular surface disease. The great majority of eyes are retrieved on behalf of the CTS Eye Banks. The CTS Eye Banks are contracted to NHSBT to process, store and supply ocular tissue for patients throughout the UK. Eyes sent to the CTS eye banks are retrieved in accordance with Third Party Agreements (“TPA”) between the eye banks and the employing authority of the individuals carrying out the eye retrievals. Although these Standards assume in some sections that eyes from donors are contributed to the CTS Eye Banks, it is recognized that this is not always the case; but the same principles and standards must still be applied. Eye retrieval © The Royal College of Ophthalmologists 2013 All rights reserved For permission to reproduce any of the content contained herein please contact beth.barnes@rcophth.ac.uk 2013/PROF/251 Author: Francisco Figueiredo Review Date: July 2016 Page 5 of 20 3. Consent 3.1. Consent for transplantation The Human Tissue Act 2004 and the Human Tissue (Scotland) Act 2006 (“the HT Acts”) require, respectively, specific consent or authorisation for the donation and storage of tissue for transplantation and other specified purposes, including research, education or training, quality assurance and clinical audit. If a person has expressed a wish to be an eye donor or not to donate the eyes, for example through the National Organ Donor Register or in a will, that consent or refusal to donate is paramount and should not be overridden by relatives except in exceptional circumstances. In the absence of prior consent given by a potential donor, consent may be given by a person in a qualifying relationship as defined in the HT Acts. There are some differences between the two HT Acts and it is important that consent is obtained according to the relevant legislation. 3.2. Coroner/Procurator Fiscal If the death has been referred to a Coroner/Procurator Fiscal, permission from the Coroner/Procurator Fiscal must be obtained in addition to the consent mentioned in Section 3.1 before proceeding with eye retrieval. If in doubt, always check first with the Coroner/Procurator Fiscal before proceeding with eye retrieval. 3.3. Information for relatives Relatives must be given sufficient and accurate information on which to base their decision. 3.4. Consent for research and training While the primary purpose of eye donation will almost always be transplantation, there is also a significant need for ocular tissue both for research into human eye diseases and for surgical training. The HT Acts require separate consent for these purposes and relatives should always be asked about these additional uses of tissue. 3.5. Unsuitability for transplantation and disposal of tissue Relatives should be informed that not every cornea will be suitable for transplantation, but that suitability cannot be determined before the eyes have been collected. Corneas and other parts of the eye that are unsuitable for transplantation may nevertheless be suitable for research or education/training. The HT Acts require separate consent for these purposes and relatives should always be asked about these additional uses of tissue. If the tissue, however, is not going to be used, relatives should be informed that the tissue will be disposed of in a lawful manner according to the HT Acts. 3.6. Consent for a blood sample and testing Consent should also be obtained for a sample of the donor’s blood to be taken for the testing of viral and other microbiological markers of transmissible disease. Relatives should be told that they will be informed of any positive results that may have implications for their own health. 3.7. Consent for seeking further information Relatives should also be asked for their permission to seek further information about a donor’s medical history and behavioural background from the donor’s medical records, GP and other relevant healthcare professionals. Eye retrieval © The Royal College of Ophthalmologists 2013 All rights reserved For permission to reproduce any of the content contained herein please contact beth.barnes@rcophth.ac.uk 2013/PROF/251 Author: Francisco Figueiredo Review Date: July 2016 Page 6 of 20 3.8. Record of consent It is strongly recommended as good practice that consent is recorded by a specially trained and experienced healthcare professional, such as a SNOD or Tissue Co-ordinator, using the NHSBT Consent/Authorisation forms and Management Process Document or their equivalent. This is a requirement of the TPA governing the retrieval of eyes to be sent to the CTS Eye Banks. If consent is taken over the telephone, the relatives should still be asked all of the questions on the NHSBT form and their answers recorded on the form by the healthcare professional. The names of the person giving consent and of the healthcare professional obtaining consent must always be recorded correctly and legibly. 4. Donor Age 4.1. Upper age Provided that the corneal endothelium is carefully examined by microscopy before transplantation to exclude those corneas with low endothelial cell densities, endothelial damage, or other abnormalities, there is currently no need to set an upper age limit for eye donation. However, an upper age limit may be imposed for operational reasons as a means to manage the availability of tissue at times of reduced surgical demand. 4.2. Lower age The lower age limit is less certain. Generally, there will be very little demand for corneas from donors under three-years old; however, tissue from such young donors may be important for other uses such as research and training. 5. Post-mortem time Enucleation should be carried out as soon as possible after a donor’s death, but post mortem times up to 24 hours are acceptable. It is a statutory requirement of the Quality and Safety Regulations that the blood sample must be taken within 24 hours of death. 6. Medical assessment of donors There is an overriding responsibility to transplant recipients to assure as far as possible the quality and safety of donated ocular tissue. 6.1. Providing donor information 6.1.1. It is the responsibility of the donor centre typically through SNODs, the NHSBT National Referral Centre or local Tissue Donor Coordinators to obtain most of the information required to determine the suitability of the donor. 6.1.2. The medical and behavioural history of potential donors must be investigated rigorously taking into account current government and professional guidance, and the outcome of these inquiries should be fully documented. It is strongly recommended that the NHSBT Patient Assessment form is used to record the family interview and that the NHSBT GP form is used to obtain information from the donor’s GP. It is a requirement of the CTS Eye Bank TPA that these forms or their equivalent are used when eyes are to be sent to the CTS Eye Banks. Eye retrieval © The Royal College of Ophthalmologists 2013 All rights reserved For permission to reproduce any of the content contained herein please contact beth.barnes@rcophth.ac.uk 2013/PROF/251 Author: Francisco Figueiredo Review Date: July 2016 Page 7 of 20 6.1.3. The main Medical Contraindications to Donation and Transplantation of Ocular Tissue are listed in Annex 1. The Quality and Safety Regulations set out the minimum standards for donor selection. Further government and professional guidance are provided by the Department of Health Advisory Committee on the Safety of Blood, Tissues and Organs (“SaBTO”) and the Joint UK Blood Services/NIBSC Professional Advisory Committee (“JPAC”). Annex 1 is kept under regular review but is by no means exhaustive. Specific advice may be sought from the eye banks and from the UK Blood Services Tissue Donor Selection Guidelines – Deceased Donors (www.transfusionguidelines.org.uk). The JPAC Standing Advisory Committee in Tissues and Cell Therapy Products (“SAC-TCTP”) advises JPAC on selection criteria for tissue donors. 6.1.4. In some instances, based on the information available at the time, it will be clear that a local decision not to proceed with the donation should be made. 6.1.5. If there is no immediate reason to exclude the donor and the eyes are to be sent to an eye bank, all required information must be provided in order for the eye bank to be able to determine the suitability of the donor. 6.1.6. If the eyes are to be sent to a CTS eye bank, the NHSBT Ocular Tissue Donor Information form must be completed as fully as possible by the person retrieving the eyes. 6.2. Sources of information about donors These include: • Family member/partner or others close to the donor – this individual need not be the person giving consent • GP • Hospital medical records • Consultant/Senior Nursing Staff with clinical responsibility for the deceased • Post mortem examination request form 6.3. Information required 6.3.1. With reference to the list of main medical contraindications (Annex 1), information should be sought about the following: • Immediate cause of death • Infusions of blood and fluids. Where infusions have been administered, and a pre-infusion blood sample is not available, complete details of all fluids administered in the 48 hours previous to death and the donor’s weight must be recorded in order to be able to estimate the extent of plasma dilution. Plasma dilution of 50% or more may invalidate the serological tests for markers of transmissible disease but poses less risk where nucleic acid technology (“NAT”) testing is used. This guidance is currently under review by the Joint United Kingdom Blood Transfusion Services and National Institute of Biological Standards and Control Professional Advisory Committee (JPAC) and may change with NAT testing. • HIV, hepatitis, HTLV or syphilis infection, known or suspected, or behavioural activity that would put the donor at risk of acquiring these infections • Other infectious disease • Previous surgery or medical treatment, including organ or tissue transplants and past history of transfusion Eye retrieval © The Royal College of Ophthalmologists 2013 All rights reserved For permission to reproduce any of the content contained herein please contact beth.barnes@rcophth.ac.uk 2013/PROF/251 Author: Francisco Figueiredo Review Date: July 2016 Page 8 of 20 • Diseases of unknown aetiology and CNS disorders, including CJD/vCJD and the risk factors for CJD/vCJD • Malignancies • Eye disease or ocular surgery 6.3.2. The most relevant life partner of the donor or, where there is none, a close family member should be interviewed. The person asked to give consent to the donation under the HT Acts may not be the most relevant person to provide information about the donor’s medical and behavioural history. The name, contact details, and relationship to the donor of the person(s) interviewed to provide medical and behavioural history should be recorded. The family and relevant life partner must be informed that a sample of the deceased’s blood will be tested for HIV, hepatitis B, hepatitis C, HTLV and syphilis. The family and/or relevant life partner should be asked not only about the deceased’s past medical history, but about any behavioural activity that would place the deceased at increased risk of HIV, HBV, HCV, or HTLV (see Annex 1). 6.3.3. The deceased’s GP should be contacted as a potentially important source of information. If it is not possible to do this before the eye retrieval takes place, either the donor centre or the eye bank will subsequently contact the GP. If the donor centre is to contact the GP, this must be made clear to avoid GPs being contacted twice and written evidence from the GP should subsequently be passed to the eye bank using the NHSBT GP form . 6.3.4. If the donor died in hospital, medical records, if available, should be checked and/or the deceased’s medical history discussed with the Consultant that had clinical responsibility for the patient. 6.3.5. If a post mortem examination of the donor is pending, the reason for the autopsy request must be ascertained to check that there is not a suspected medical contraindication (e.g., a neurological condition). 6.3.6. If a sample of the donor’s blood is tested locally, the mandatory tests for HIV, HBsAg, HBc, HTLV and syphilis must be carried out only by an accredited test laboratory. However, it is good practice and expected that the blood sample will be sent to the eye bank together with retrieved eye(s). 6.3.7. The person responsible for investigating the potential donor’s medical and behavioural history must confirm that these Standards have been applied by completing fully the required documentation. 7. Eye retrieval A competency framework for eye donation and retrieval is currently under review by NHSBT. 7.1. Eye retrievers 7.1.1. Eye retrieval must be carried out by a person who is competent in enucleation. 7.1.2. If the retrieval of eyes is to be performed by someone who was not responsible for obtaining consent and investigating the medical and behavioural background of the donor, the enucleation must not take place until the eye retriever is personally satisfied that consent has been obtained, that all relevant sources of medical information have been checked, and that where information is awaited, there should be no immediate reason to believe that the retrieval should not take place, Eye retrieval © The Royal College of Ophthalmologists 2013 All rights reserved For permission to reproduce any of the content contained herein please contact beth.barnes@rcophth.ac.uk 2013/PROF/251 Author: Francisco Figueiredo Review Date: July 2016 Page 9 of 20 especially where there may be an infectious risk to the person retrieving the eyes. To this end, the individual referring the donor to the eye retriever must provide the name and position of the person who obtained consent and confirm that the NHSBT Consent form and NHSBT Patient Assessment forms have been completed and that there is no reason not to proceed with the eye retrieval. The eye retriever must record this information on the NHSBT Ocular Tissue Donor Information form to confirm that the eye retriever has been informed that lawful consent has been obtained. However, it is good practice for the eye retriever to have, if possible, a copy of the consent form, especially as some mortuaries may not allow the removal of tissue without having written confirmation of consent. 7.2. NHSBT Human Tissue Transport box 7.2.1. The NHSBT Human Tissue Transport box (available through NHSBT) contains: • a set of sterile, single-use instruments with a paper wrapper for use as a drape • 2 x EDTA blood sample tubes • alcohol swabs for cleaning the skin around the eyes and the eye lids • sterile saline for irrigating eyes • sterile pots, 25 G needles, eye stands, cotton balls and saline for creating moist chambers • eye caps and cotton balls for restoring the donor’s appearance • documentation comprising an enucleation protocol, the list of medical contraindications, the NHSBT Ocular Tissue Donor Information and Retrieval Site Risk Assessment forms. 7.2.2. Additional required items not included in the transport box: • 1 kg of melting ice is needed to keep the contents of the transport box between 0°C and 8°C during transportation to the eye bank • 10-ml syringe and 19 G needle for taking the blood sample • Prep gloves, sterile gloves and appropriate protective clothing 7.3. Retrieval site risk assessment 7.3.1. It is a requirement of the Quality and Safety Regulations that a risk assessment is carried out to ensure that the retrieval site is suitable and appropriate for the removal of tissue from a deceased donor. This focuses principally on the need to be able to treat a donor with dignity and respect as well as quality and safety issues. 7.3.2. The risk assessment should be documented and must be carried out prior to eye retrieval. To this end an NHSBT Tissue Retrieval Site Risk Assessment form is provided with the enucleation kit in the NHSBT Human Tissue Transport Box. This risk assessment must be carried out for every eye retrieval as circumstances may change even within the same premises. 7.4. Donor identification 7.4.1. Correct identification of the donor is critical to avoid illegally removing tissue from a cadaver without consent and without any investigation of medical or behavioural history. 7.4.2. It is strongly recommended as good practice for identification of the donor to be confirmed by the eye retriever and another person, such as mortuary staff, hospital Eye retrieval © The Royal College of Ophthalmologists 2013 All rights reserved For permission to reproduce any of the content contained herein please contact beth.barnes@rcophth.ac.uk 2013/PROF/251 Author: Francisco Figueiredo Review Date: July 2016 Page 10 of 20 site manager, nursing or medical staff. If the donor is not in a hospital or hospice, confirmation of identification will have to rely on, for example, care home staff, funeral directors, and the donor’s relatives. 7.4.3. In hospitals and hospices, the donor should be identified by the wrist or ankle tag using name, date of birth, hospital number and any other available identifiers. The means of identification and persons confirming identification should be recorded. 7.5. Physical examination of the donor 7.5.1. It is a requirement of the Quality and Safety Regulations that a physical examination of the donor is undertaken. 7.5.2. It is appreciated that eye retrievers are likely to be working alone and the extent of such a physical examination will be limited, especially as some eye donors are dressed. However, the eye retriever should perform a physical examination sufficient to satisfy the retriever that no physical impediment to eye retrieval is apparent, often by examining those parts of a donor’s body that are readily accessible. The areas examined and findings such as tattoos, piercings and scars should be recorded on a body map such as that provided in the NHSBT Ocular Tissue Donor Information form. 7.6. Blood sample 7.6.1. A sample of the donor’s blood, at least 5 ml, must be sent to the eye bank with the donor’s eyes. The blood sample must be sufficient for serological and NAT testing, both of which are required. If, exceptionally, the blood sample is to be tested locally, the testing laboratory must be accredited and the serological and NAT testing carried out using test kits of acceptable sensitivity and specificity. 7.6.2. If an ante-mortem blood sample taken not more than 7 days before death is not available, a blood sample should be taken from the deceased as soon after death as possible and not more than 24 hours after death. This is a statutory requirement of the Quality and Safety Regulations. The quality of the sample is critical to the reliability of the serological tests for markers of transmissible disease. The blood should be taken from a site away from infusion lines where there is a likelihood of sample dilution. The preferred sites are the brachiocephalic, subclavian or femoral veins. 7.6.3. The blood sample should be placed in EDTA blood sample tube(s). 7.6.4. The sample tube(s) must be clearly labelled with the date, donor’s name, date of birth, and one other identifier (e.g., hospital name). 7.6.5. The syringe and needle must be disposed of immediately and safely after use. 7.7. Enucleation 7.7.1. A standard enucleation protocol, such as that provided in the NHSBT Human Tissue Transport Box, should be followed (see Annex 2). 7.7.2. A set of sterile, single-use instruments must be used. The instruments must be disposed of immediately and safely after use. Eye retrieval © The Royal College of Ophthalmologists 2013 All rights reserved For permission to reproduce any of the content contained herein please contact beth.barnes@rcophth.ac.uk 2013/PROF/251 Author: Francisco Figueiredo Review Date: July 2016 Page 11 of 20 7.7.3. The moist chambers must be labelled clearly with the date, donor’s name, date of birth and one other identifier (e.g., hospital name), indicating left or right eye. 7.8. Restoring the donor’s appearance 7.8.1. The final cosmetic result is of critical importance both out of respect for the donor and because family or friends may wish to view the body. The orbits should be packed with cotton wool and the lids closed over plastic eye caps to restore the original profile of the lids. Any bleeding from the sockets or bruising around the orbits following enucleation should be recorded on the body map (see section 7.5). 7.9. Packaging, labelling and transport to an eye bank 7.9.1. Labelling • It is essential that the moist chambers and the blood sample tube are clearly and correctly labelled with the date, donor’s name, date of birth and at least one other identifier (e.g., hospital name). It should be borne in mind that eye banks may receive in any one day the eyes from several donors. Absent or incomplete labelling may result in the eyes being discarded owing to uncertainty about donor identification. 7.9.2. Packaging • The eyes must be packed securely, for example in an NHSBT Human Tissue Transport Box, with the blood sample, and required documentation. For eyes being sent to a CTS Eye Bank, the documentation comprises the NHSBT Retrieval Site Risk Assessment form, an NHSBT Ocular Tissue Donor Information form completed to the best of the eye retriever’s knowledge, and any other information that may be available at the time, such as a consent form, a medical history check list, or an NHSBT GP form. • The box must be packed according to the instructions provided, including at least 1 kg of ice to ensure correct maintenance of temperature during transport. 7.9.3. Transport to a CTS Eye Bank • The box should be closed using the supplied tamper-evident security tag. • The eye retriever should contact NHSBT when the eyes are ready for collection, providing specific details of the location and reporting the security tag number. NHSBT will specify the eye bank address, which should then be clearly written on the label provided and attached to the side of the box. The eyes must be kept at a secure location until they are collected. 8. Responsibilities 8.1. Donor Hospitals/Retrieval Centres In practice, the following may be undertaken by staff from different hospitals; e.g., the initial approach to a bereaved family is likely to be from medical or nursing staff in the Donor Hospital; SNODs or Tissue Donor Coordinators may obtain consent and undertake investigation of medical/behavioural background, and a member of staff from an eye bank or from another hospital (Retrieval Centre) may collect the blood sample and retrieve the eyes. Eye retrieval © The Royal College of Ophthalmologists 2013 All rights reserved For permission to reproduce any of the content contained herein please contact beth.barnes@rcophth.ac.uk 2013/PROF/251 Author: Francisco Figueiredo Review Date: July 2016 Page 12 of 20 • To provide the donor’s family and/or most relevant life partner with accurate and relevant information • To ensure that lawful consent has been obtained for the removal, storage and use of ocular tissue. • To ensure that all available sources of medical and behavioural history of the donor have been checked and recorded, including the donor’s relatives and/or most relevant life partner • To obtain a blood sample from the donor and to retrieve the eyes according to these Standards • If the eyes are to be sent to a CTS eye bank: - To comply with the CTS Eye Bank Third Party Agreement signed by the eye retriever’s employing NHS Trust/Board or other NHS organization. The CTS Eye Banks cannot accept eyes from Donor Centres that have not signed the TPA. - To report all relevant donor information to NHSBT - To specify sources of information (e.g., GP, pending post-mortem report) that have not been checked and to agree who will be responsible for subsequently obtaining the information - To inform NHSBT immediately of any relevant donor information that is obtained after the donor has been referred to NHSBT - To complete the NHSBT Ocular Tissue Donor Information form and Retrieval Site Risk Assessment form, and to confirm that this Standard has been followed - To provide copies of test results if the donor’s blood sample is tested locally for the mandatory markers of transmissible disease. - To provide copies of the Consent form, the Patient Assessment form and, if available, the GP questionnaire. - To notify NHSBT that the eye retrieval has been completed and to specify a location for collection of the eyes • To send a letter of thanks to the donor family 8.2. NHSBT • To accept donor referrals and to record donor information provided by the donor centre • To check that all relevant information has been obtained or will be obtained • To agree who will be responsible for obtaining outstanding information • To arrange transport of the eyes through its contracted suppliers to an eye bank • To provide feedback to the donor centre about use of corneas and/or sclerae from local donors Eye retrieval © The Royal College of Ophthalmologists 2013 All rights reserved For permission to reproduce any of the content contained herein please contact beth.barnes@rcophth.ac.uk 2013/PROF/251 Author: Francisco Figueiredo Review Date: July 2016 Page 13 of 20 • To notify the donor centre of positive results from tests for markers of transmissible disease that may have health implications for the donor’s family and/or most relevant life partner in accordance with the NHSBT Donor Care Policy. 8.3. Eye Banks • To receive the eyes and to process, store and distribute ocular tissue • To contact GPs and/or pathologists for information not available at the time of referral • To determine the suitability of the donor on the basis of the information provided by the donor centre and the results of the mandatory tests for markers of transmissible disease • To determine the suitability of tissue for transplantation • To answer specific technical and/or medical queries from donor centres about eye donation and to provide general information about eye and tissue donation for healthcare professionals and the lay public. 8.4. RCOphth Ocular Tissue Transplantation Standards Group (OTTSG) • To review these Standards in line with statutory requirements, and government and professional recommendations, guidance and good practice. To develop and provide standards for the retrieval, storage and transplantation of the cornea, sclera and all other ocular and non-ocular tissues into the human eye, including collection of data on the outcome of such procedures, and the distribution of such tissue for research or training. To provide advice and support to consultant colleagues and medical staff in training involved with eye donation, retrieval and transplantation. 8.5. NHSBT Ocular Tissue Advisory Group (OTAG) • To consider operational aspects of transplantation including eye retrieval, allocation and data analysis and to monitor activity and outcome. To recommend, as necessary, changes to the nationally agreed protocols, (i.e., some protocols and practices are remit of HTA, others are College or NHSBT,) to recognise clinical governance issues and ensure, as far as possible, that national standards of good practice are in place with regard to waiting list criteria and organ allocation that provide equity of access to transplantation. Together with OTTSG to advise NHSBT on the standards for ocular tissue transplantation, eye donation and retrieval. To liaise as necessary with the RCOphth and other organizations in the development of national standards. 8.6. Joint UK Blood Services/NIBSC Professional Advisory Committee (JPAC) • To provide professional advice and guidance on tissue donor selection criteria based on recommendations from SAC-TCTP and SaBTO. 8.7. JPAC Standing Advisory Committee on Tissues and Cell Therapy Products (SAC-TCTP) • To keep under review the UK Blood Services Tissue Donor Selection Guidelines and to refer recommendations to JPAC for decision concerning donor selection criteria. 8.8. DH Advisory Committee on the Safety of Blood, Tissues and Organs (SaBTO) Eye retrieval © The Royal College of Ophthalmologists 2013 All rights reserved For permission to reproduce any of the content contained herein please contact beth.barnes@rcophth.ac.uk 2013/PROF/251 Author: Francisco Figueiredo Review Date: July 2016 Page 14 of 20 • To provide government advice on donor selection and testing. These Standards were reviewed by the RCOphth Ocular Tissue Transplantation Standards Group, May 2013 Eye retrieval © The Royal College of Ophthalmologists 2013 All rights reserved For permission to reproduce any of the content contained herein please contact beth.barnes@rcophth.ac.uk 2013/PROF/251 Author: Francisco Figueiredo Review Date: July 2016 Page 15 of 20 9. References and bibliography 9.1. Department of Health (including agencies and legal documents) Human Tissue Authority (www.hta.gov.uk) • Human Tissue Act 2004 • Human Tissue (Scotland) Act 2006 • Human Tissue (Quality and Safety for Human Application) Regulations 2007 • EU Tissues and Cells Directive (2004/23/EC) • EU Commission Directive (2006/17/EC) • EU Commission Directive (2006/86/EC) • Human Tissue Authority Directions 003/2010 • Human Tissue Authority Codes of Practice Data Protection Act 1998 (revised) Variant Creutzfeldt-Jakob Disease (vCJD): Minimising the Risk of Transmission. Health Services Circular 1999/178, NHS Executive, 13 August 1999 (www.doh.gov.uk/coinh.htm) UK Blood Services Tissue Donor Selection Guidelines – Deceased Donors (www.transfusionguidelines.org.uk) Advisory Committee on the Safety of Blood, Tissues and Organs (SaBTO) Guidance on the Microbiological Safety of Human Organs, Tissues and Cells used in Transplantation (www.dh.gov.uk) 9.2. Professional organisations The Royal College of Ophthalmologists European Eye Bank Association (www.eeba.net) • EEBA Standards British Association for Tissue Banking (www.batb.org.uk) British Transplantation Society 10. Abbreviations CTS Corneal Transplant Service HTA Human Tissue Authority HT Acts Human Tissue Act 2004 and Human Tissue (Scotland) Act 2006 JPAC Joint UK Blood Services/NIBSC Professional Advisory Group NHSBT NHS Blood and Transplant NIBSC National Institute for Biological Standards and Control ODT Organ Donation and Transplantation (Directorate of NHSBT) OTAG NHSBT Ocular Tissue Advisory Group OTTSG RCOphth Ocular Tissue Transplantation Standards Group Eye retrieval © The Royal College of Ophthalmologists 2013 All rights reserved For permission to reproduce any of the content contained herein please contact beth.barnes@rcophth.ac.uk 2013/PROF/251 Author: Francisco Figueiredo Review Date: July 2016 Page 16 of 20 RCOphth Royal College of Ophthalmologists SAC-TCTP JPAC Standing Advisory Committee on Tissues and Cell Therapy Products SNOD Specialist Nurse – Organ Donation 11. Contact details for advice or further information Human Tissue Authority (www.hta.gov.uk) 151 Buckingham Palace Road Victoria London SW1W 9SZ Tel: 020 7269 1900 Email: enquiries@hta.gov.uk Standards and Advisory Groups Ocular Tissue Transplantation Standards Group – contact through RCOphth (Professional Standards) Prof Francisco C Figueiredo, MD, PhD, FRCOphth (Chair) Ocular Tissue Advisory Group – contact through NHSBT Prof Stephen B Kaye, MD, FRCOphth (Chair) The Royal College of Ophthalmologists (www.rcophth.ac.uk) 17 Cornwall Terrace, London, NW1 4QW Tel: 020 7935 0702 NHSBT Organ Donation and Transplantation (www.nhsbt.nhs.uk) NHS Blood & Transplant Fox Den Road Stoke Gifford Bristol BS34 8RR Tel: 0117 975 7575 Eye banks (HTA licensed) CTS Bristol Eye Bank Bristol Eye Hospital Lower Maudlin Street Bristol BS1 2LX Tel: 0117 342 4438 Fax: 0117 904 6624 Director: Professor John Armitage (w.j.armitage@bristol.ac.uk) Medical Advisor: Mr Derek Tole MD, FRCOphth (derek.tole@uhbristol.nhs.uk) CTS Manchester Eye Bank Manchester Royal Eye Hospital Oxford Road Manchester M13 9WH Tel: 0161 276 5623 Fax: 0161 276 5610 Manager: Isaac Zambrano PhD (isaac.zambrano@cmft.nhs.uk) Medical Advisor: Mrs Fiona Carley FRCOphth (fiona.carley@cmft.nhs.uk) East Grinstead Eye Bank The Queen Victoria Hospital Holtye Road East Grinstead Sussex RH19 3DZ Tel: 01342 410 210 Fax: 01342 414106 Eye retrieval © The Royal College of Ophthalmologists 2013 All rights reserved For permission to reproduce any of the content contained herein please contact beth.barnes@rcophth.ac.uk 2013/PROF/251 Author: Francisco Figueiredo Review Date: July 2016 Page 17 of 20 Medical Director: Mr Damian Lake FRCOphth Moorfields Eye Bank Moorfields Eye Hospital City Road London EC1V 2PD Tel: 0207 253 1199 Fax: 0207 253 4696 Medical Director: Mr Stephen Tuft MD, FRCOphth ANNEX 1. Contraindications to ocular tissue transplantation These are the main exclusion criteria, but the list is not exhaustive and further advice may be required (www.transfusionguidelines.org.uk). Check www.rcophth.ac.uk for updates. 1. INFECTIONS 1.1 acquired immunodeficiency syndrome (HIV/AIDS) 1.2 viral hepatitis (A, B, or C) 1.3 HTLV 1.4 seropositivity: anti-HIV, HBsAg, anti-HBc (if anti-HBs <100 IU/L), anti-HCV, anti-HTLV, syphilis 1.5 behaviour leading to risk of contracting HIV, hepatitis or HTLV1 1.6 tattoos and body piercing within the 4 months before death2 1.7 acupuncture within the 4 months before death3 1.8 imprisonment within the 12 months before death4 1.9 bleeding disorders treated with blood-derived coagulation concentrates5 1.10 viral encephalitis or encephalitis of unknown origin, viral meningitis6 1.11 rabies 1.12 congenital rubella 1.13 tuberculosis 1.14 Reyes syndrome 1.15 progressive multifocal leukoencephalopathy 1.16 septicaemia6 2. PREVIOUS SURGERY/MEDICAL TREATMENT 2.1 immunosuppression7 2.2 receipt of an organ transplant7,8 2.3 receipt of dura mater or brain/spinal surgery before August 19928 2.4 receipt of human pituitary hormones8 2.5 receipt of a cornea, sclera or other human tissue allograft8 3. Unknown AETIOLOGY AND CNS DISORDERS 3.1 death from unknown cause9 3.2 Creutzfeldt-Jakob disease10 and central nervous system diseases of unknown aetiology (e.g., Alzheimer’s disease, other dementias, Parkinson’s disease, multiple sclerosis, motor neurone disease) 4. MALIGNANCIES Eye retrieval © The Royal College of Ophthalmologists 2013 All rights reserved For permission to reproduce any of the content contained herein please contact beth.barnes@rcophth.ac.uk 2013/PROF/251 Author: Francisco Figueiredo Review Date: July 2016 Page 18 of 20 4.1 leukaemia, lymphoma, myeloma, polycythaemia (unless confirmed as secondary polycythaemia), sideroblastic anaemia and myelodysplastic syndrome 5. EYE DISEASES 5.1 active ocular inflammation/uveitis 5.2 any congenital or acquired disorders of the eye, or previous ocular surgery (including corneal laser surgery), that would preclude successful graft outcome 5.3 retinoblastoma 5.4 malignant tumours of the anterior segment Notes: 1 The Department of Health Blood & Tissue Safety Entry excludes donors that: • thought they needed a test for HIV/AIDS, HTLV or hepatitis. • are HIV positive • are HTLV positive • are a hepatitis B carrier • are a hepatitis C carrier • are a man who has ever had oral or anal sex with another man, even if a condom or other protective was used • have ever received money or drugs for sex • have ever injected, or been injected with, drugs, even a long time ago or only once. This includes bodybuilding drugs. Donation may be possible if a doctor prescribed the drugs. • have had sex within the last 12 months (even if a condom or other protective was used) with - a partner who is or thinks they may be: HIV or HTLV positive a hepatitis B carrier a hepatitis C carrier. - (if a woman): a man who has ever had oral or anal sex with another man, even if they used a condom or other protective - a partner who has ever received money or drugs for sex - a partner who has ever injected, or been injected with, drugs, even a long time ago or only once. This includes bodybuilding drugs. This may not apply if a doctor prescribed the drugs. – a partner who has or may have been sexually active in parts of the world where HIV/AIDS is very common. This includes most countries in Africa. 2Instruments used for tattoos and body piercing have transmitted infections. Allowing a deferral period of 4 months helps to ensure that infections tested for will be detected but this assumes that a validated NAT test for hepatitis C is negative. In the absence of the NAT test, the deferral period will be 6 months. 3Acupuncture needles have transmitted infections. Allowing a deferral period of 4 months helps to ensure that infections tested for will be detected but this assumes that a validated NAT test for hepatitis C is negative. In the absence of the NAT test, the deferral period will be 6 months. The acupuncture must not have been performed for a condition that is itself a reason for excluding the donor. The acupuncture must have been performed by a suitably qualified healthcare professional on NHS premises. Acupuncture performed outside the NHS but by a qualified individual registered with a statutory body (GMC, NMC, GDC, HPC), is acceptable. 4A deceased person cannot answer questions about ‘at risk behaviour’ that may have occurred while in prison and relatives are unlikely to know. Being held in a police cell for less than 96 hours may not exclude the donor. 5Treatment with blood-derived coagulation concentrates are very likely at risk of having transfusion acquired infections. Sexual partners of persons that have received blood-derived coagulation concentrates should not donate tissue if less than 6 months since last sexual contact. Eye retrieval © The Royal College of Ophthalmologists 2013 All rights reserved For permission to reproduce any of the content contained herein please contact beth.barnes@rcophth.ac.uk 2013/PROF/251 Author: Francisco Figueiredo Review Date: July 2016 Page 19 of 20 6Viraemia and viral meningitis are absolute contraindications. Bacterial forms of septicaemia or meningitis may be acceptable at the discretion of the eye bank Medical Director but only when the corneas are to be stored by organ culture. 7Immunosuppression invalidates the serological tests for markers of infectious disease such as HIV. Use of NAT testing may allow use of the tissue. 8Increased risk of CJD/vCJD transmission. 9Death from unknown cause is not a contraindication provided a post-mortem examination is pending and the result will be known before the tissue is transplanted. 10Covers individuals with CJD (sporadic, familial or iatrogenic) or variant CJD, and individuals identified as being at risk of CJD/vCJD. ANNEX 2. Enucleation protocol Only single-use instruments are to be used. This protocol assumes use of an enucleation kit as provided in the NHSBT Human Tissue Transport box, which is available through NHSBT (see s.11 Contact Details). It is advisable to take the blood sample before proceeding with the enucleation (see s.7.6) 1. Wearing prep gloves and protective clothing, open eyelids and irrigate eyes with a sterile saline to remove debris, mucus and foreign matter. Clean face around the eyes, over eyelids, bridge of nose and eyebrows using alcohol wipes. Care should be taken not to touch the cornea with the alcohol wipes during this procedure. Remove and discard the prep gloves. 2. Open the wrapper of a single-use instrument pack to create a sterile field. Put on sterile gloves and place the eye sheet included in the instrument pack over the donor’s head and neck. 3. Insert lid speculum and perform a peritomy close to, but not up to, the limbus using fine-toothed forceps and scissors. It is important to leave a frill of conjunctiva attached at the limbus to protect the epithelial stem cell niche. Tenon’s capsule is pushed back by entering each of the four quadrants with the scissors and performing a blunt dissection. 4. Isolate the lateral rectus muscle with a muscle hook, insert artery forceps between hook and sclera and clamp muscle. Cut the muscle between the hook and the clamp. Using the muscle hook, isolate and divide each of the remaining rectus muscles in turn, cutting with the scissors between the muscle hook and the sclera. The oblique muscles may then be lifted with the muscle hook and cut, although this is not obligatory. Care must be taken not to rub the cornea against the speculum or instruments. 5. Gently lift the eye with the artery forceps. Insert the enucleation scissors from the medial side, and, keeping the scissors almost vertical, locate the optic nerve by moving the scissors gently from side to side. Still keeping the scissors almost vertical, cut the optic nerve while maintaining gently upward pressure on the eye with the artery forceps. This should ensure that a stump of optic nerve at least 5 mm long remains attached to the eye. 6. Once the optic nerve has been severed, gently raise the eye from the orbit, excising residual tissue with the enucleation scissors. 7. Carefully transfer the eye to a plastic eye stand, passing the stump of the optic nerve through the hole in the base of the stand and secure the eye on the stand (e.g., by placing a sterile 25G hypodermic needle through the protruding optic nerve). Place the eye stand and eye (cornea uppermost) on top of a cotton wool ball (or gauze swab) moistened with saline in a sterile pot (moist chamber). The eye must not be immersed in liquid in the moist chamber and should not be covered with gauze or cotton wool. Eye retrieval © The Royal College of Ophthalmologists 2013 All rights reserved For permission to reproduce any of the content contained herein please contact beth.barnes@rcophth.ac.uk 2013/PROF/251 Author: Francisco Figueiredo Review Date: July 2016 Page 20 of 20 8. Remove the speculum and repeat the procedure with the other eye. 9. Pack orbits with cotton wool and, with the aid of eye caps, restore the original appearance of the donor. 10. Clearly label the moist chambers with the date, donor’s name, date of birth, hospital name and whether LEFT or RIGHT eye. 11. For eyes being sent to one of the CTS Eye Banks, complete an NHSBT Ocular Tissue Donor Information form and Retrieval Site Risk Assessment form MUST BE COMPLETED. Pack the moist chambers and the blood sample into an NHSBT Human Tissue Transport box. Fill a plastic bag with melting ice (at least 1 kg) and place bag into the box. Place completed NHSBT Ocular Tissue Donor Information form and Retrieval Site Risk Assessment into the box. Contact the Duty Office at NHSBT (Tel: 0117 975 7580) to report donor details and to arrange transport for the eyes. The box must be sealed using the safety numbered tag provided in the retrieval pack. |
3748 | https://math.answers.com/math-and-arithmetic/What_is_the_formula_for_calculating_the_length_of_the_space_diagonal_of_a_right_rectangular_prism_on_the_basis_of_the_prism%27s_length_l_width_w_and_height_h | What is the formula for calculating the length of the space diagonal of a right rectangular prism on the basis of the prism's length l width w and height h? - Answers
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What is the formula for calculating the length of the space diagonal of a right rectangular prism on the basis of the prism's length l width w and height h?
Anonymous ∙ 10 y ago
Updated: 7/29/2025
The formula for calculating the length of the space diagonal ( d ) of a right rectangular prism is given by the equation ( d = \sqrt{l^2 + w^2 + h^2} ), where ( l ) is the length, ( w ) is the width, and ( h ) is the height of the prism. This formula derives from the Pythagorean theorem, applied in three dimensions.
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### What is the formula for calculating the length of the space diagonal of a right rectangular prism on the basis of the prisms length width and height? It is the extension of Pythagoras's theorem to 3-d. d2 = l2 + w2 + h2
### What is the formula for calculating the volume of a rectanglular prism? The formula for calculating the volume of a rectangular prism is: Length x width x height Example, if length=2cm, width=4cm, and height is 2cm your answer would be: 2 x 4 x 2 = 16 cm3
### What is the formula used to find the length of a rectangular prism? The answer depends on what information you are given: (volume, breadth and height), (surface are, breadth and height), (principle diagonal, breadth and height), (mass, density, breadth and height) or some other set.
### What is the formula of calculating area of a rectangular solid? First of a;; it is NOT a rectangular solid, but a CUBOID. Thformuloa for calculating volume is baase x perpendicular width x perpendicular height. Algebraically V = l X w X h
### What is the formular for calculating the volume of a rectangular pyramid? The formula for calculating the volume of a rectangular pyramid is ( V = \frac{1}{3} \times \text{Base Area} \times \text{Height} ). Here, the Base Area is the area of the rectangular base, calculated as ( \text{length} \times \text{width} ). Thus, the complete formula can be expressed as ( V = \frac{1}{3} \times (\text{length} \times \text{width}) \times \text{height} ).
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3749 | https://www.canr.msu.edu/news/using_the_right_planting_density_is_critical_for_optimum_yield_and_revenue | MSU Extension
Using the right planting density is critical for optimum yield and revenue for vegetable crops
Mathieu Ngouajiongouajio@msu.edu, Michigan State University Extension, Department of Horticulture -
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Since both low and high crop densities reduce yield and total revenue, it is important to calibrate the planter for optimum density.
With the return of warm weather, the planting of warm season vegetables like cucumber and squash will be at full swing over the next couple of days. During these busy times, it is important to remember that the planting density used may have significant impacts not only on yield, but also on net revenue. Planting densities that are either too low or too high may result in economic losses.
Impact of low planting density
When the planting density is too low, each individual plant may perform at its maximum capacity, but there are not enough plants as a whole to reach the optimum yield. Therefore, total yield of the crop becomes a limiting factor. Also, when there is too much space left between plants, weed growth is promoted, which could increase weeding costs.
Impact of a high planting density
If the planting density is too high, plants may compete against each other, known as intra-specific competition. Under those conditions, the performance of individual plants becomes a limiting factor for maximum crop yield.
Using the right density
As the planting density increases, the total crop yield increases and reaches a maximum, at which point further increase in planting density results in reduced yield. It is, therefore, important to use appropriate planting density, keeping in mind the fact that the density that provides the highest yield and the density that provides the highest revenue may be different. Growers should shoot for both optimum yield and revenue when balancing their planting density.
Tips for achieving the right planting density
Calibrate your seeder each time of use, especially if you change seed batches or the seeder has been sitting around for a long time or has been moved.
Remember that for some crops seed size may vary from batch to batch and smaller seed could increase your planting density (unless you are using a planter with high accuracy in seed singulation).
If you change your row spacing, remember to recalibrate the seeder for the target plant density. For example, reducing row spacing from 30 inches to 24 inches without recalibrating the seeder will increase the planting density by 25 percent.
Maintain a uniform planting depth for uniform emergence.
Using planting density to maximize economic value of the crop: The case of pickling cucumber
Profitability of pickling cucumber (as is the case for many other crops) is not just a function of total fruit weight, but is also dependent on seed cost and fruit selling price. Therefore, seed cost should be included in the analyses of studies designed to identify optimum pickling cucumber densities. With an arbitrary 5 percent margin of error, a study conducted under our growing conditions showed that optimum economic value is obtained with densities between 72,000 and 120,000 plants per acre (Figure 1). Optimum density for highest economic value varies depending on seed cost and cucumber selling price. The higher the seed cost, the lower the optimum density. Also, the lower the selling price, the lower the optimum density. Other factors that should be taken into account include cultivars, growing conditions and timing of harvest.
Figure 1. Economic value of pickling cucumber as affected by planting density.
Please note this is not the net profit to the grower because the cost of the seed is the only input used in the calculation. In our economic analysis, we used an average cost of $1.65 per thousand seeds and a selling price of $3.25 per bushel for grade 2 and 3 (marketable grades for machine-harvested cucumbers). Economic value was obtained by multiplying yield by selling price and subtracting seed cost.
Dr .Ngouajio’s work is funded in part by MSU’s AgBioResearch.
This article was published by Michigan State University Extension. For more information, visit To have a digest of information delivered straight to your email inbox, visit To contact an expert in your area, visit or call 888-MSUE4MI (888-678-3464).
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Maths Cramers Rule
What is Cramer’s Rule?
Cramer’s rule is a method that uses determinants to solve systems of equations that have the same number of equations as variables. It is named after the Swiss mathematician Gabriel Cramer, who first published it in 1750.
How does Cramer’s Rule work?
Cramer’s rule works by calculating the determinant of the coefficient matrix and the determinants of the numerator matrices. The determinant of a matrix is a single numerical value that can be calculated from a square matrix.
For a system of equations with $n$ equations and $n$ variables, Cramer’s rule can be used to solve for the value of each variable. The formula for Cramer’s rule is:
$$x_i = \dfrac{\det(A_i)}{\det(A)}$$
$x_i$ is the value of the $i$-th variable
$A$ is the coefficient matrix of the system of equations
$A_i$ is the matrix formed by replacing the $i$-th column of $A$ with the column of constants
Example
To illustrate how Cramer’s rule works, consider the following system of equations:
$$2x + 3y = 5$$
$$4x - y = 3$$
The coefficient matrix for this system is:
$$A = \begin{bmatrix} 2 & 3 \ 4 & -1 \end{bmatrix}$$
The determinant of $A$ is:
$$\det(A) = (2)(-1) - (3)(4) = -14$$
The determinant of the numerator matrix for $x$ is:
$$\det(A_x) = \begin{vmatrix} 5 & 3 \ 3 & -1 \end{vmatrix} = -14$$
The determinant of the numerator matrix for $y$ is:
$$\det(A_y) = \begin{vmatrix} 2 & 5 \ 4 & 3 \end{vmatrix} = -14$$
Therefore, the solutions to the system of equations are:
$$x = \dfrac{\det(A_x)}{\det(A)} = \dfrac{-14}{-14} = 1$$
$$y = \dfrac{\det(A_y)}{\det(A)} = \dfrac{-14}{-14} = 1$$
Advantages and Disadvantages of Cramer’s Rule
Cramer’s rule is a simple and straightforward method for solving systems of equations. However, it can be computationally inefficient for large systems of equations. Additionally, Cramer’s rule can only be used to solve systems of equations that have the same number of equations as variables.
Conclusion
Cramer’s rule is a useful tool for solving systems of equations. However, it is important to be aware of its limitations. For large systems of equations, it is often more efficient to use other methods, such as Gaussian elimination or matrix inversion.
Cramer’s Rule For 3×3 Matrix
Cramer’s rule is a method that uses determinants to solve systems of equations that have the same number of equations as variables. It is applicable to systems of linear equations with coefficients that are real numbers.
Understanding Cramer’s Rule
Cramer’s rule provides a formula to find the solution to each variable in a system of equations. The formula involves calculating the determinant of the coefficient matrix and the determinants of the numerator matrices.
Formula for Cramer’s Rule
For a system of three linear equations with three variables, Cramer’s rule is given by the following formulas:
$$x = \dfrac{\Delta_x}{\Delta}$$
$$y = \dfrac{\Delta_y}{\Delta}$$
$$z = \dfrac{\Delta_z}{\Delta}$$
where:
$x, y, z$ are the variables to be solved
$\Delta$ is the determinant of the coefficient matrix
$\Delta_x, \Delta_y, \Delta_z$ are the determinants of the numerator matrices
Calculating the Determinants
To apply Cramer’s rule, you need to calculate the following determinants:
$\Delta$: the determinant of the coefficient matrix
$\Delta_x$: the determinant of the numerator matrix obtained by replacing the first column of the coefficient matrix with the constants on the right-hand side of the equations
$\Delta_y$: the determinant of the numerator matrix obtained by replacing the second column of the coefficient matrix with the constants on the right-hand side of the equations
$\Delta_z$: the determinant of the numerator matrix obtained by replacing the third column of the coefficient matrix with the constants on the right-hand side of the equations
Example of Cramer’s Rule
Consider the following system of linear equations:
$$2x + 3y + 4z = 5$$
$$-x + 2y + 3z = 6$$
$$3x - y + 2z = 7$$
To solve this system using Cramer’s rule, we first calculate the determinants:
$$\Delta = \begin{vmatrix} 2 & 3 & 4 \ -1 & 2 & 3 \ 3 & -1 & 2 \end{vmatrix} = 27$$
$$\Delta_x = \begin{vmatrix} 5 & 3 & 4 \ 6 & 2 & 3 \ 7 & -1 & 2 \end{vmatrix} = -18$$
$$\Delta_y = \begin{vmatrix} 2 & 5 & 4 \ -1 & 6 & 3 \ 3 & 7 & 2 \end{vmatrix} = -63$$
$$\Delta_z = \begin{vmatrix} 2 & 3 & 5 \ -1 & 2 & 6 \ 3 & -1 & 7 \end{vmatrix} = 90$$
Now, we can solve for $x, y, z$:
$$x = \dfrac{\Delta_x}{\Delta} = \dfrac{-18}{27} = \dfrac{-2}{3}$$
$$y = \dfrac{\Delta_y}{\Delta} = \dfrac{-63}{27} = \dfrac{-7}{3}$$
$$z = \dfrac{\Delta_z}{\Delta} = \dfrac{90}{27} = \dfrac{10}{3}$$
Therefore, the solution to the system of equations is $x = \dfrac{-2}{3}, y = \dfrac{-7}{3}, z = \dfrac{10}{3}$.
Conclusion
Cramer’s rule provides a systematic method for solving systems of linear equations using determinants. While it is a useful tool, it is important to note that Cramer’s rule may not be the most efficient method for solving large systems of equations. For larger systems, other methods such as matrix inversion or Gaussian elimination may be more efficient.
Conditions for Cramer’s Rule
Cramer’s rule is a method that uses determinants to solve systems of equations that have the same number of equations as variables. However, Cramer’s rule can only be used if the coefficient matrix of the system is a square matrix and non-singular.
Conditions for Cramer’s Rule
For a system of linear equations to be solved using Cramer’s rule, the following conditions must be met:
The system must have the same number of equations as variables.
The coefficient matrix of the system must be a square matrix.
The coefficient matrix of the system must be non-singular.
Non-singular Matrix
A non-singular matrix is a square matrix that has a non-zero determinant. In other words, a non-singular matrix is a matrix that is invertible.
Example of a Non-singular Matrix
The following matrix is a non-singular matrix:
$$\begin{bmatrix} 1 & 2 \ 3 & 4 \end{bmatrix}$$
The determinant of this matrix is 1(4) - 2(3) = -2. Since the determinant is not zero, the matrix is non-singular.
Example of a Singular Matrix
The following matrix is a singular matrix:
$$\begin{bmatrix} 1 & 2 \ 2 & 4 \end{bmatrix}$$
The determinant of this matrix is 1(4) - 2(2) = 0. Since the determinant is zero, the matrix is singular.
Cramer’s rule is a useful tool for solving systems of linear equations, but it can only be used if the coefficient matrix of the system is a square matrix and non-singular.
Important Points on Cramer’s Rule
Cramer’s rule is a method that uses determinants to solve systems of equations that have the same number of equations as variables. It is named after the Swiss mathematician Gabriel Cramer, who first published it in 1750.
Cramer’s rule can be used to solve systems of equations that are in the form:
$$a_1x + b_1y = c_1$$
$$a_2x + b_2y = c_2$$
where $a_1, a_2, b_1, b_2, c_1,$ and $c_2$ are constants.
To solve a system of equations using Cramer’s rule, we first need to calculate the determinant of the coefficient matrix:
$$D = \begin{vmatrix} a_1 & b_1 \ a_2 & b_2 \end{vmatrix}$$
If $D = 0$, then the system of equations has no solution or infinitely many solutions. If $D \neq 0$, then the system of equations has a unique solution, which can be found using the following formulas:
$$x = \dfrac{\begin{vmatrix} c_1 & b_1 \ c_2 & b_2 \end{vmatrix}}{D}$$
$$y = \dfrac{\begin{vmatrix} a_1 & c_1 \ a_2 & c_2 \end{vmatrix}}{D}$$
Here are some important points to remember about Cramer’s rule:
Cramer’s rule can only be used to solve systems of equations that have the same number of equations as variables.
If the determinant of the coefficient matrix is $0$, then the system of equations has no solution or infinitely many solutions.
If the determinant of the coefficient matrix is not $0$, then the system of equations has a unique solution, which can be found using the formulas given above.
Cramer’s rule is a useful tool for solving systems of equations, but it can be computationally expensive for large systems of equations.
There are other methods for solving systems of equations, such as Gaussian elimination and matrix inversion, that may be more efficient for large systems of equations.
Understanding Determinant Properties.
Determinants are mathematical objects that are used to represent the area or volume of a geometric shape. They can also be used to represent the transformation of a vector under a linear transformation. Determinants have a number of important properties that make them useful in a variety of mathematical applications.
Properties of Determinants
The following are some of the most important properties of determinants:
Determinants are multilinear. This means that the determinant of a matrix is a linear function of each row and column of the matrix.
Determinants are alternating. This means that the determinant of a matrix changes sign if any two rows or columns are interchanged.
The determinant of a triangular matrix is the product of its diagonal elements.
The determinant of a matrix is zero if any two rows or columns are identical.
The determinant of a matrix is equal to the product of its eigenvalues.
Applications of Determinants
Determinants have a number of important applications in mathematics, including:
Finding the area or volume of a geometric shape.
Determining whether a matrix is invertible.
Solving systems of linear equations.
Finding the eigenvalues and eigenvectors of a matrix.
Determinants are a powerful mathematical tool that has a variety of applications. By understanding the properties of determinants, you can use them to solve a variety of mathematical problems.
Cramers Rule FAQs
Cramer’s rule is a method that is used to solve systems of equations that have the same number of equations as variables. It is a useful tool for solving systems of linear equations, but it can also be used to solve systems of nonlinear equations.
Here are some frequently asked questions about Cramer’s rule:
What is Cramer’s rule?
Cramer’s rule is a method for solving systems of linear equations that have the same number of equations as variables. It is based on the idea of determinants, which are numbers that can be calculated from a matrix.
How does Cramer’s rule work?
Cramer’s rule works by calculating the determinant of the coefficient matrix of the system of equations. This determinant is then used to calculate the values of the variables in the system of equations.
When can Cramer’s rule be used?
Cramer’s rule can be used to solve any system of linear equations that has the same number of equations as variables. However, it is most commonly used to solve systems of equations that have a small number of variables (two or three).
What are the advantages of Cramer’s rule?
Cramer’s rule is a relatively simple method to use, and it can be used to solve systems of equations that have a small number of variables. It is also a very accurate method, and it can be used to solve systems of equations that have coefficients that are not integers.
What are the disadvantages of Cramer’s rule?
Cramer’s rule can be difficult to use when the system of equations has a large number of variables. It is also not a very efficient method, and it can be computationally expensive to solve systems of equations that have a large number of variables.
Are there any alternatives to Cramer’s rule?
There are a number of alternatives to Cramer’s rule, including:
Gaussian elimination
LU decomposition
QR decomposition
Cholesky decomposition
These methods are all more efficient than Cramer’s rule, and they can be used to solve systems of equations that have a large number of variables.
Conclusion
Cramer’s rule is a useful tool for solving systems of linear equations that have the same number of equations as variables. It is a relatively simple method to use, and it can be used to solve systems of equations that have coefficients that are not integers. However, it is not a very efficient method, and it can be difficult to use when the system of equations has a large number of variables.
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3751 | https://api.pageplace.de/preview/DT0400.9781420010541_A25103714/preview-9781420010541_A25103714.pdf | Series Editor KENNETH H. ROSEN DISCRETE MATHEMATICS AND ITS APPLICATIONS Handbook of Combinatorial Designs Second Edition Juergen Bierbrauer, Introduction to Coding Theory Kun-Mao Chao and Bang Ye Wu, Spanning Trees and Optimization Problems Charalambos A. Charalambides, Enumerative Combinatorics Henri Cohen, Gerhard Frey, et al., Handbook of Elliptic and Hyperelliptic Curve Cryptography Charles J. Colbourn and Jeffrey H. Dinitz, Handbook of Combinatorial Designs, Second Edition Steven Furino, Ying Miao, and Jianxing Yin, Frames and Resolvable Designs: Uses, Constructions, and Existence Randy Goldberg and Lance Riek, A Practical Handbook of Speech Coders Jacob E. Goodman and Joseph O’Rourke, Handbook of Discrete and Computational Geometry, Second Edition Jonathan L. Gross and Jay Yellen, Graph Theory and Its Applications, Second Edition Jonathan L. Gross and Jay Yellen, Handbook of Graph Theory Darrel R. Hankerson, Greg A. Harris, and Peter D. Johnson, Introduction to Information Theory and Data Compression, Second Edition Daryl D. Harms, Miroslav Kraetzl, Charles J. Colbourn, and John S. Devitt, Network Reliability: Experiments with a Symbolic Algebra Environment Leslie Hogben, Handbook of Linear Algebra Derek F. Holt with Bettina Eick and Eamonn A. O’Brien, Handbook of Computational Group Theory David M. Jackson and Terry I. Visentin, An Atlas of Smaller Maps in Orientable and Nonorientable Surfaces Richard E. Klima, Neil P. Sigmon, and Ernest L. Stitzinger, Applications of Abstract Algebra with Maple™ and MATLAB®, Second Edition Patrick Knupp and Kambiz Salari, Verification of Computer Codes in Computational Science and Engineering William Kocay and Donald L. Kreher, Graphs, Algorithms, and Optimization Donald L. Kreher and Douglas R. Stinson, Combinatorial Algorithms: Generation Enumeration and Search Series Editor Kenneth H. Rosen, Ph.D.
and DISCRETE MATHEMATICS ITS APPLICATIONS Continued Titles Charles C. Lindner and Christopher A. Rodgers, Design Theory Hang T. Lau, A Java Library of Graph Algorithms and Optimization Alfred J. Menezes, Paul C. van Oorschot, and Scott A. Vanstone, Handbook of Applied Cryptography Richard A. Mollin, Algebraic Number Theory Richard A. Mollin, Codes: The Guide to Secrecy from Ancient to Modern Times Richard A. Mollin, Fundamental Number Theory with Applications Richard A. Mollin, An Introduction to Cryptography, Second Edition Richard A. Mollin, Quadratics Richard A. Mollin, RSA and Public-Key Cryptography Carlos J. Moreno and Samuel S. Wagstaff, Jr., Sums of Squares of Integers Dingyi Pei, Authentication Codes and Combinatorial Designs Kenneth H. Rosen, Handbook of Discrete and Combinatorial Mathematics Douglas R. Shier and K.T. Wallenius, Applied Mathematical Modeling: A Multidisciplinary Approach Jörn Steuding, Diophantine Analysis Douglas R. Stinson, Cryptography: Theory and Practice, Third Edition Roberto Togneri and Christopher J. deSilva, Fundamentals of Information Theory and Coding Design Lawrence C. Washington, Elliptic Curves: Number Theory and Cryptography Series Editor KENNETH H. ROSEN DISCRETE MATHEMATICS AND ITS APPLICATIONS Boca Raton London New York Chapman & Hall/CRC is an imprint of the Taylor & Francis Group, an informa business Handbook of Combinatorial Designs Second Edition Edited by Charles J. Colbourn Jeffrey H. Dinitz CRC Press Taylor & Francis Group 6000 Broken Sound Parkway NW, Suite 300 Boca Raton, FL 33487‑2742 © 2007 by Taylor & Francis Group, LLC CRC Press is an imprint of Taylor & Francis Group, an Informa business No claim to original U.S. Government works Printed in the United States of America on acid‑free paper 10 9 8 7 6 5 4 3 2 1 International Standard Book Number‑10: 1‑58488‑506‑8 (Hardcover) International Standard Book Number‑13: 978‑1‑58488‑506‑1 (Hardcover) This book contains information obtained from authentic and highly regarded sources. Reprinted material is quoted with permission, and sources are indicated. A wide variety of references are listed. Reasonable efforts have been made to publish reliable data and information, but the author and the publisher cannot assume responsibility for the validity of all materials or for the consequences of their use. No part of this book may be reprinted, reproduced, transmitted, or utilized in any form by any electronic, mechanical, or other means, now known or hereafter invented, including photocopying, microfilming, and recording, or in any informa‑ tion storage or retrieval system, without written permission from the publishers.
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Library of Congress Cataloging‑in‑Publication Data Handbook of combinatorial designs / edited by Charles J. Colbourn, Jeffrey H. Dinitz. ‑‑ 2nd ed.
p. cm. ‑‑ (Discrete mathematics and its applications) ISBN 1‑58488‑506‑8 1. Combinatorial designs and configurations. I. Colbourn, Charles J. II. Dinitz, Jeffrey H., 1952‑ III. Title. IV. Series.
QA166.25.H36 2007 511’.6‑‑dc22 2006050484 Visit the Taylor & Francis Web site at and the CRC Press Web site at Preface The Purpose of this Book The CRC Handbook of Combinatorial Designs (Second Edition) is a reference book of combinatorial designs, including constructions of designs, existence results, proper-ties of designs, and many applications of designs. The handbook is not a textbook in-troduction to combinatorial design theory, but rather a comprehensive, easy-to-access reference for facts about designs.
Organization of the Handbook The presentation is organized into seven main parts, each divided into chapters.
The first part contains a brief introduction and history. The next four parts center around four main classes of combinatorial designs: balanced incomplete block de-signs (II), orthogonal arrays and latin squares (III), pairwise balanced designs (IV), and Hadamard and orthogonal designs (V). Part VI then contains sixty-five shorter chapters (placed in alphabetical order) devoted to other classes of designs—naturally, many of these classes are closely connected with designs in Parts II–V. Part VII gives mathematical and computational background of particular use in design theory.
How to Use this Book This handbook is not meant to be used in a sequential way, but rather for direct access to information about relevant topics. Each chapter makes references to other chapters for related information; in particular, the See Also listing at the end of each chapter provides useful links.
An extensive index and the table of contents provide first entry points for most material. For readers new to the area, the first chapter, Opening the Door, can assist in finding a suitable starting point.
This handbook can be used on its own as a reference book or in conjunction with numerous excellent available texts on the subject.
Although material at the research frontier is often included, the handbook emphasizes basic, useful information.
Combinatorial design theory has extensive interactions with numerous areas of mathematics: group theory, graph theory, number theory and finite fields, finite ge-ometry, and linear algebra. It also has applications ranging from routine to complex in numerous disciplines: experimental design, coding theory, cryptography, and com-puter science, among others. The handbook treats all of these, providing information for design theorists to understand the applications, and for users of design theory to obtain pointers in to the rest of the handbook, and hence to the mathematics of combinatorial designs.
Notes for the User A home page for the CRC Handbook of Combinatorial Designs is being maintained.
It is located at Notes, corrections, and updates will be disseminated at that site.
Fixed numbering conventions are used. This book has seven main parts numbered using roman numerals I–VII. Each part is divided into chapters, which in turn have sections; all are numbered. So, for example, the third section in the fourth chapter of the second part is §II.4.3. This is the manner to which it is referred when referenced from any other chapter. When referenced from within Chapter II.4, it is referenced Preface simply as §4.3. Occasionally, sections contain subsections. However, to avoid four level numbers, such subsections are not numbered; rather, the beginning of a subsec-tion is given by the symbol ‘▷’ preceding the title of the subsection. Numbering of Theorems, Definitions, Remarks, Corollaries, Tables, Algorithms, and the like is con-secutive within each chapter. So, for example, in §II.4.2, Theorem II.4.8 (displayed with “4.8” exdented) is followed by Definition II.4.9, which is followed by Remark II.4.10. As before, the roman numeral is omitted if both the referenced theorem and the reference to it are in the same chapter. Definitions are displayed in gray boxes.
Forward and reverse citations are given. There is a single bibliography after Part VII. However, each chapter provides a list of the reference numbers for items cited therein. In addition, the last entry of each item in the bibliography (in angle brackets) enumerates the page numbers on which the item is cited. This is often useful to uncover further relationships among the topics treated.
Original references are not, in general, given. Instead, each chapter references surveys and textbooks in which the information is readily accessible. These sources should be consulted for more complete lists of references. One exception is Chapter I.2, Design Theory: Antiquity to 1950, which provides original references. Each chapter covers the basic material, but sufficient pointers to find more specialized information are typically given.
The state of the subject is continually advancing. Each chapter presents, as far as possible, information that is current at the time of writing. Inevitably, improvements will occur! Readers aware of improvements on, or corrections to, the material pre-sented here are invited to contact the author(s) of the chapter(s) involved and also the editors-in-chief.
Differences Between the Editions The second edition exhibits a substantial increase in size over the first; the book is over 30% larger, with 235 additional pages! While the basic structure has been retained, a new introductory part has been added to provide a general overview and a historical perspective. Material on Hadamard and related designs has been collected edition has been incorporated into the remaining parts, in the belief that the effort to distinguish theory from application is no longer appropriate. Finally, the nearly 90 separate bibliographies of the first edition have been amalgamated to form a single bibliography for the Handbook, to provide yet another means to find the right “starting point” on a search for desired information. This bibliography contains roughly 2200 references and in itself provides an excellent resource.
Often material that is closely related has been combined into a single chapter; in the process, virtually all of the content of the first edition is preserved, but the reader may find small changes in its location. The material is completely updated, and generally enlarged, in the second.
The major tables from the first edition (BIBDs, MOLS, PBDs, Hadamard matrices) have been definitive sources for the current knowledge, and all are updated in the second edition. We do not believe that designs whose heyday seems to have passed deserve less attention than the topics of most active current research. The basics of yesteryear continue to be important. Nevertheless, because the discipline continues to evolve, many new topics have been incorporated.
Particular emphasis is on topics that bridge with many areas of application, as well as with other areas of mathematics.
In the second edition, the new chapters are: Opening the Door (I.1), Design The-ory: Antiquity to 1950 (I.2), Triple Systems (II.2), Group Divisible Designs (IV.4), Optical Orthogonal Codes (V.9), Bent Functions (VI.4), Block-Transitive Designs (VI.5), Covering Arrays (VI.10), Deletion-Correcting Codes (VI.14), Graph Embed-together into a new part (V). On the other hand, the Applications part of the first Preface dings and Designs (VI.25), Grooming (VI.27), Infinite Designs (VI.30), Low Density Parity Check Codes (VI.33), Magic Squares (VI.34), Nested Designs (VI.36), Perfect Hash Families (VI.43), Permutation Codes and Arrays (VI.44), Permutation Polyno-mials (VI.45), Pooling Designs (VI.46), Quasi-3 Designs (VI.47), Supersimple Designs (VI.57), Tur´ an Systems (VI.61), Divisible Semiplanes (VII.3), Linear Algebra and De-signs (VII.7), Designs and Matroids (VII.10), and Directed Strongly Regular Graphs (VII.12). We are pleased to have been able to add much new material to the Handbook.
A Note of Appreciation We must begin by acknowledging the excellent work done by the 110 contributors.
Each kindly agreed to write a chapter on a specific topic, their job made difficult by the style and length restrictions placed upon them. Many had written hundreds of pages on their topic, yet were asked to summarize in a few pages. They also were asked to do this within the fairly rigid stylistic framework that is relatively consistent throughout the book. The editors were merciless in their attempt to uphold this style, and offer a special thanks to the copyeditor, Carole Gustafson, for assistance. Many chapters were edited extensively. We thank the contributors for their cooperation.
Many chapters have profited from comments from a wide variety of readers, in addition to the contributors and editors.
A compilation of this size can only be accomplished with the help of a small army of people. The concept developed and realized in the first edition from 1996 was the cumulative effort of approximately one hundred people.
Many have extensively updated their earlier chapters, and some generously agreed to contribute new material. Many chapters in the second edition employ a conceptual design and/or material first written by others. We thank again authors from both editions for worrying less about whose name is on the byline for the material than the fact that it is a good reflection on our discipline. The concept of the Handbook owes much to those who compiled extensive tables prior to the first edition, such as the BIBD tables of Rudi Mathon and Alex Rosa, and the MOLS table of Andries Brouwer. For the first edition, Ron Mullin coordinated a substantial effort to make the first tables of PBDs, and Don Kreher generated extensive tables of t-designs; in the process, both developed new and useful ways to present these tables, updated in the second edition. We thank Ron and Don again for their great contributions.
As in the first edition, we again thank Debbie Street and Vladimir Tonchev for their extensive help in organizing the material in statistical designs and coding the-ory, respectively. For the second edition, we especially thank Hadi Kharaghani for organizing (and writing much of) the material in Part V. Special thanks too to Leo Storme for the complete rewrite and reorganization of the chapter on finite geometry.
For going above and beyond with their help on the second edition, we also thank Julian Abel, Dan Archdeacon, Frank Bennett, Ian Blake, Warwick de Launey, Peter Dukes, Gennian Ge, Malcolm Greig, Jonathan Jedwab, Teresa Jin, Esther Lamken, Bill Martin, Gary McGuire, Patric ¨ Osterg˚ ard, David Pike, Donald Preece, Alex Rosa, Joe Rushanan, Doug Stinson, Anne Street, Tran van Trung, Ian Wanless, and Zhu Lie. Thanks also to the excellent logistical support from the publisher, particularly to Nora Konopka and Wayne Yuhasz for the first edition, and to Helena Redshaw, Judith Simon, and Bob Stern for the second.
Finally, we thank Sue Dinitz and Violet Syrotiuk for their love and support.
Putting together this extensive volume has taken countless hours; we thank them for their patience during it all.
Charles J. Colbourn Jeffrey H. Dinitz To Violet, Sarah, and Susie, with love.
CJC To my father, Simon Dinitz, who has inspired me in all that I have done.
Happy 80th birthday!
JHD Editors-in-Chief Charles J. Colbourn Computer Science and Engineering Arizona State University U.S.A.
Jeffrey H. Dinitz Mathematics and Statistics University of Vermont U.S.A Contributors R. Julian R. Abel University of New South Wales Australia Lars D. Andersen University of Aalborg Denmark Ian Anderson University of Glasgow U.K.
Lynn Margaret Batten Deakin University Australia Lucien B´ en´ eteau INSA France Frank E. Bennett Mount Saint Vincent University Canada Jean-Claude Bermond INRIA Sophia-Antipolis France Anton Betten Colorado State University U.S.A.
J¨ urgen Bierbrauer Michigan Technological University U.S.A.
Ian F. Blake University of Toronto Canada Andries E. Brouwer Technical University of Eindhoven Netherlands Darryn Bryant University of Queensland Australia Marco Buratti Universita di Perugia Italy Peter J. Cameron University of London U.K.
Contributors Yeow Meng Chee Card View Pte Ltd.
Singapore Leo G. Chouinard II University of Nebraska U.S.A.
Wensong Chu Arizona State University U.S.A.
Charles J. Colbourn Arizona State University U.S.A.
David Coudert INRIA Sophia-Antipolis France Robert Craigen University of Manitoba Canada Anne Delandtsheer Universit´ e Libre de Bruxelles Belgium Warwick de Launey Center for Communications Research U.S.A.
Michel M. Deza ´ Ecole Normale Sup´ erieure France Jeffrey H. Dinitz University of Vermont U.S.A.
Peter J. Dukes University of Victoria Canada Saad El-Zanati Illinois State University U.S.A.
Anthony B. Evans Wright State University U.S.A.
Norman J. Finizio University of Rhode Island U.S.A.
Dalibor Froncek University of Minnesota Duluth U.S.A.
Gennian Ge Zhejiang University P.R. China Peter B. Gibbons University of Auckland New Zealand Chistopher D. Godsil University of Waterloo Canada Solomon W. Golomb University of Southern California U.S.A.
K. Gopalakrishnan East Carolina University U.S.A.
Daniel M. Gordon Center for Communications Research U.S.A.
Ken Gray The University of Queensland Australia Malcolm Greig Greig Consulting Canada Terry S. Griggs The Open University U.K.
Hans-Dietrich O. F. Gronau Universit¨ at Rostock Germany Harald Gropp Universit¨ at Heidelberg Germany Contributors A. S. Hedayat University of Illinois at Chicago U.S.A.
Tor Helleseth University of Bergen Norway Sylvia A. Hobart University of Wyoming U.S.A.
Spencer P. Hurd The Citadel U.S.A.
Frank K. Hwang National Chiao Tung University Taiwan Yury J. Ionin Central Michigan University U.S.A.
Robert Jajcay Indiana State University U.S.A.
Dieter Jungnickel Universit¨ at Augsburg Germany Hadi Kharaghani University of Lethbridge Canada Gholamreza B. Khosrovshahi IPM Iran Christos Koukouvinos National Technical University of Athens Greece Earl S. Kramer University of Nebraska–Lincoln U.S.A.
Donald L. Kreher Michigan Technological University U.S.A.
Joseph M. Kudrle University of Vermont U.S.A.
Esther R. Lamken San Francisco U.S.A.
Reinhard Laue Universit¨ at Bayreuth Germany Charles F. Laywine Brock University Canada Vladimir I. Levenshtein Keldysh Inst. Applied Mathematics Russia P. C. Li University of Manitoba Canada Charles C. Lindner Auburn University U.S.A.
Spyros S. Magliveras Florida Atlantic University U.S.A.
Alireza Mahmoodi York University Canada William J. Martin Worcester Polytechnic Institute U.S.A.
Rudolf Mathon University of Toronto Canada Gary McGuire National University of Ireland Ireland Sarah B. Menard University of Vermont U.S.A.
Contributors Eric Mendelsohn University of Toronto Canada Ying Miao University of Tsukuba Japan J. P. Morgan Virginia Tech U.S.A.
Gary L. Mullen The Pennsylvania State University U.S.A.
Ronald C. Mullin Florida Atlantic University U.S.A.
Akihiro Munemasa Tohoku University Japan William Orrick Indiana University U.S.A.
Patric R. J. ¨ Osterg˚ ard Helsinki University of Technology Finland Stanley E. Payne University of Colorado at Denver U.S.A.
Alexander Pott Universit¨ at Magdeberg Germany Donald A. Preece University of London U.K.
Chris Rodger Auburn University U.S.A.
Alexander Rosa McMaster University Canada Gordon F. Royle University of Western Australia Australia Mikl´ os Ruszink´ o Hungarian Academy of Sciences Hungary Dinesh G. Sarvate College of Charleston U.S.A.
Nabil Shalaby Memorial University of Newfoundland Canada James B. Shearer IBM U.S.A.
Mohan S. Shrikhande Central Michigan University U.S.A.
Jozef ˇ Sir´ aˇ n University of Auckland New Zealand Ken W. Smith Central Michigan University U.S.A.
Hong-Yeop Song Yonsei University Korea Sung Y. Song Iowa State University U. S. A.
Edward Spence University of Glasgow U.K.
Douglas R. Stinson University of Waterloo Canada Leo Storme Ghent University Belgium Contributors Anne Penfold Street The University of Queensland Australia Deborah J. Street University of Technology, Sydney Australia Herbert Taylor University of Southern California U.S.A.
Joseph A. Thas Ghent University Belgium Vladimir D. Tonchev Michigan Technological University U.S.A.
David C. Torney Los Alamos National Laboratory U.S.A.
G. H. John van Rees University of Manitoba Canada Tran van Trung University of Duisburg–Essen Germany Robert A. Walker II Arizona State University U.S.A.
Walter D. Wallis Southern Illinois University U.S.A.
Ian M. Wanless Monash University Australia Bridget S. Webb The Open University U.K.
Ruizhong Wei Lakehead University Canada Hugh Williams University of Calgary Canada Richard M. Wilson California Institute of Technology U.S.A.
Jianxing Yin Suzhou University China L. Zhu Suzhou University China Contents I Introduction 1 Opening the Door Charles J. Colbourn 3 2 Design Theory: Antiquity to 1950 Ian Anderson, Charles J. Colbourn, Jeffrey H. Dinitz, Terry S. Griggs 11 II Block Designs 1 2-(v, k, λ) Designs of Small Order Rudolf Mathon, Alexander Rosa 25 2 Triple Systems Charles J. Colbourn 58 3 BIBDs with Small Block Size R. Julian R. Abel, Malcolm Greig 72 4 t-Designs with t ≥3 Gholamreza B. Khosrovshahi, Reinhard Laue 79 5 Steiner Systems Charles J. Colbourn, Rudolf Mathon 102 6 Symmetric Designs Yury J. Ionin, Tran van Trung 110 7 Resolvable and Near-Resolvable Designs R. Julian R. Abel, Gennian Ge, Jianxing Yin 124 III Latin Squares 1 Latin Squares Charles J. Colbourn, Jeffrey H. Dinitz, Ian M. Wanless 135 2 Quasigroups Frank E. Bennett, Charles C. Lindner 152 Contents 3 Mutually Orthogonal Latin Squares (MOLS) R. Julian R. Abel, Charles J. Colbourn, Jeffrey H. Dinitz 160 4 Incomplete MOLS R. Julian R. Abel, Charles J. Colbourn, Jeffrey H. Dinitz 193 5 Self-Orthogonal Latin Squares (SOLS) Norman J. Finizio, L. Zhu 211 6 Orthogonal Arrays of Index More Than One Malcolm Greig, Charles J. Colbourn 219 7 Orthogonal Arrays of Strength More Than Two Charles J. Colbourn 224 IV Pairwise Balanced Designs 1 PBDs and GDDs: The Basics Ronald C. Mullin, Hans-Dietrich O. F. Gronau 231 2 PBDs: Recursive Constructions Malcolm Greig, Ronald C. Mullin 236 3 PBD-Closure R. Julian R. Abel, Frank E. Bennett, Malcolm Greig 247 4 Group Divisible Designs Gennian Ge 255 5 PBDs, Frames, and Resolvability Gennian Ge, Ying Miao 261 6 Pairwise Balanced Designs as Linear Spaces Anton Betten 266 V Hadamard Matrices and Related Designs 1 Hadamard Matrices and Hadamard Designs Robert Craigen, Hadi Kharaghani 273 2 Orthogonal Designs Robert Craigen, Hadi Kharaghani 280 3 D-Optimal Matrices Hadi Kharaghani, William Orrick 296 4 Bhaskar Rao Designs Warwick de Launey 299 5 Generalized Hadamard Matrices Warwick de Launey 301 Contents 6 Balanced Generalized Weighing Matrices and Conference Matrices Yury J. Ionin, Hadi Kharaghani 306 7 Sequence Correlation Tor Helleseth 313 8 Complementary, Base, and Turyn Sequences Hadi Kharaghani, Christos Koukouvinos 317 9 Optical Orthogonal Codes Tor Helleseth 321 VI Other Combinatorial Designs 1 Association Schemes Christopher D. Godsil, Sung Y. Song 325 2 Balanced Ternary Designs Spencer P. Hurd, Dinesh G. Sarvate 330 3 Balanced Tournament Designs Esther R. Lamken 333 4 Bent Functions Douglas R. Stinson 337 5 Block-Transitive Designs Anne Delandtsheer 339 6 Complete Mappings and Sequencings of Finite Groups Anthony B. Evans 345 7 Configurations Harald Gropp 353 8 Correlation-immune and Resilient Functions K. Gopalakrishnan, Douglas R. Stinson 355 9 Costas Arrays Herbert Taylor, Jeffrey H. Dinitz 357 10 Covering Arrays Charles J. Colbourn 361 11 Coverings Daniel M. Gordon, Douglas R. Stinson 365 12 Cycle Decompositions Darryn Bryant, Chris Rodger 373 13 Defining Sets Ken Gray, Anne Penfold Street 382 Contents 14 Deletion-correcting Codes Vladimir I. Levenshtein 385 15 Derandomization K. Gopalakrishnan, Douglas R. Stinson 389 16 Difference Families R. Julian R. Abel, Marco Buratti 392 17 Difference Matrices Charles J. Colbourn 411 18 Difference Sets Dieter Jungnickel, Alexander Pott, Ken W. Smith 419 19 Difference Triangle Sets James B. Shearer 436 20 Directed Designs Frank E. Bennett, Alireza Mahmoodi 441 21 Factorial Designs Deborah J. Street 445 22 Frequency Squares and Hypercubes Charles F. Laywine, Gary L. Mullen 465 23 Generalized Quadrangles Stanley E. Payne 472 24 Graph Decompositions Darryn Bryant, Saad El-Zanati 477 25 Graph Embeddings and Designs Jozef ˇ Sir´ aˇ n 486 26 Graphical Designs Yeow Meng Chee, Donald L. Kreher 490 27 Grooming Jean-Claude Bermond, David Coudert 494 28 Hall Triple Systems Lucien B´ en´ eteau 496 29 Howell Designs Jeffrey H. Dinitz 499 30 Infinite Designs Peter J. Cameron, Bridget S. Webb 504 31 Linear Spaces: Geometric Aspects Lynn Margaret Batten 506 Contents 32 Lotto Designs (Ben) P. C. Li, G. H. John van Rees 512 33 Low Density Parity Check Codes Ian F. Blake 519 34 Magic Squares Joseph M. Kudrle, Sarah B. Menard 524 35 Mendelsohn Designs Eric Mendelsohn 528 36 Nested Designs J. P. Morgan 535 37 Optimality and Efficiency: Comparing Block Designs Deborah J. Street 540 38 Ordered Designs, Perpendicular Arrays, and Permutation Sets J¨ urgen Bierbrauer 543 39 Orthogonal Main Effect Plans Deborah J. Street 547 40 Packings Douglas R. Stinson, Ruizhong Wei, Jianxing Yin 550 41 Partial Geometries Joseph A. Thas 557 42 Partially Balanced Incomplete Block Designs Deborah J. Street, Anne Penfold Street 562 43 Perfect Hash Families Robert A. Walker II, Charles J. Colbourn 566 44 Permutation Codes and Arrays Peter J. Dukes 568 45 Permutation Polynomials Gary L. Mullen 572 46 Pooling Designs David C. Torney 574 47 Quasi-3 Designs Gary McGuire 576 48 Quasi-Symmetric Designs Mohan S. Shrikhande 578 49 (r, λ)-designs G. H. John van Rees 582 Contents 50 Room Squares Jeffrey H. Dinitz 584 51 Scheduling a Tournament J. H. Dinitz, Dalibor Froncek, Esther R. Lamken, Walter D. Wallis 591 52 Secrecy and Authentication Codes K. Gopalakrishnan, Douglas R. Stinson 606 53 Skolem and Langford Sequences Nabil Shalaby 612 54 Spherical Designs Akihiro Munemasa 617 55 Starters Jeffrey H. Dinitz 622 56 Superimposed Codes and Combinatorial Group Testing Charles J. Colbourn, Frank K. Hwang 629 57 Supersimple Designs Hans-Dietrich O. F. Gronau 633 58 Threshold and Ramp Schemes K. Gopalakrishnan, Douglas R. Stinson 635 59 (t,m,s)-Nets William J. Martin 639 60 Trades A. S. Hedayat, Gholamreza B. Khosrovshahi 644 61 Tur´ an Systems Mikl´ os Ruszink´ o 649 62 Tuscan Squares Wensong Chu, Solomon W. Golomb, Hong-Yeop Song 652 63 t-Wise Balanced Designs Earl S. Kramer, Donald L. Kreher 657 64 Whist Tournaments Ian Anderson, Norman J. Finizio 663 65 Youden Squares and Generalized Youden Designs Donald A. Preece, Charles J. Colbourn 668 VII Related Mathematics 1 Codes Vladimir D. Tonchev 677 Contents 2 Finite Geometry Leo Storme 702 3 Divisible Semiplanes Rudolf Mathon 729 4 Graphs and Multigraphs Gordon F. Royle 731 5 Factorizations of Graphs Lars D. Andersen 740 6 Computational Methods in Design Theory Peter B. Gibbons, Patric R. J. ¨ Osterg˚ ard 755 7 Linear Algebra and Designs Peter J. Dukes, Richard M. Wilson 783 8 Number Theory and Finite Fields Jeffrey H. Dinitz, Hugh C. Williams 791 9 Finite Groups and Designs Leo G. Chouinard II, Robert Jajcay, Spyros S. Magliveras 819 10 Designs and Matroids Peter J. Cameron, Michel M. Deza 847 11 Strongly Regular Graphs Andries E. Brouwer 852 12 Directed Strongly Regular Graphs Andries E. Brouwer, Sylvia A. Hobart 868 13 Two-Graphs Edward Spence 875 Bibliography and Index Bibliography 883 Index 967 Part I Introduction I.1 Opening the Door 3 It seems to me that whatever else is beautiful apart from absolute beauty is beautiful because it partakes of that absolute beauty, and for no other reason. Do you accept this kind of causality? Yes, I do.
Well, now, that is as far as my mind goes; I cannot understand these other ingenious theories of causation.
If someone tells me that the reason why a given object is beautiful is that it has a gorgeous color or shape or any other such attribute, I disregard all these other explanations—I find them all confusing—and I cling simply and straightforwardly and no doubt foolishly to the explanation that the one thing that makes the object beautiful is the presence in it or association with it, in whatever way the relation comes about, of absolute beauty. I do not go so far as to insist upon the precise details—only upon the fact that it is by beauty that beautiful things are beautiful. This, I feel, is the safest answer for me or anyone else to give, and I believe that while I hold fast to this I cannot fall; it is safe for me or for anyone else to answer that it is by beauty that beautiful things are beautiful. Don’t you agree?
(Plato, Phaedo, 100c-e) 1 Opening the Door Charles J. Colbourn 1.1 Example The Fano plane.
Seven points Three points on a line Every two points define a line Seven lines Three lines through a point Every two lines meet at a point 1.2 Remarks Opening the door into the world of combinatorial designs, and stepping through that door, is familiar to some and mysterious to many. What lies on the other side? And where is the map to guide us?
There is territory both charted and uncharted; the guide map is at best known only in part.
Meandering through this world without fixed direction can be both enjoyable and instructive; yet most have a goal, a question whose answer is sought.
This introduction explores, with the simplest examples, the landmarks of the world of combinatorial designs. While not prescribing a path, it provides some direction.
Perhaps the most vexing problem is to recognize the object sought when it is encountered; the disguises are often ingenious, but the many faces of one simple object often reveal its importance. Example 1.1 is recast in many different ways, to acquaint the reader with correspondences that are useful in recognizing an object when its representation changes. This exercise provides a first view of the large body of material in the rest of the handbook, to which pointers are provided for the reader to follow.
1.1 Balance ▷ Sets and Blocks 1.3 Example The Fano plane as a set system.
{0,1,2}, {0,3,4}, {0,5,6}, {1,3,5}, {1,4,6}, {2,3,6}, {2,4,5} Using elements {0, 1, 2, 3,4,5, 6}, label the points in Example 1.1. For each line, form a set that contains the elements corre-sponding to the points on that line. The result is a collection of sets, a set system.
4 Opening the Door I.1 1.4 Remark The set system of Example 1.3 exhibits in many ways a desideratum that is central to the understanding of combinatorial designs, the desire for balance. Some of the ways are considered here.
• All sets have the same size k = 3. The set system is k-uniform.
• All elements occur in precisely r = 3 of the sets. The set system is r-regular or equireplicate with replication r.
• Every pair of distinct elements occurs as a subset of exactly λ = 1 of the sets.
The set system is 2-balanced with index λ = 1.
• Every pair of distinct sets intersects in exactly µ = 1 elements. Equivalently, the symmetric difference of every two distinct sets contains exactly four elements; the set system is 4-equidistant.
Balance is a key to understanding combinatorial designs, not just a serendipitous feature of one small example.
1.5 Consider a general finite set system. It has a (finite) ground set V of size v whose members are elements (or points or varieties), equipped with a (finite) collection B of b subsets of V whose members are blocks (or lines or treatments). Then • K = {|B| : B ∈B} is the set of block sizes; • R = {|{B : x ∈B ∈B}| : x ∈V } is the set of replication numbers, • for t ≥0 integer, Λt = {|{B : T ⊆B ∈B}| : T ⊆V, |T| = t} is the set of t-indices. Indeed, Λ0 = {b}, and Λ1 = R.
• M = {|B ∩B′| : B, B′ ∈B, B ̸= B′}, the set of intersection numbers.
1.6 Remark Restricting any or all of K, R, M, or the {Λt}s imposes structure on the set system; restrictions to a single element balance that structure in some regard.
Table 1.7 considers the effect of restricting one or more of these sets to be singletons, providing pointers into the remainder of the handbook. Indeed, set systems so defined are among the most central objects in all of design theory and are studied extensively.
The language with which they are discussed in this chapter varies from set systems to hypergraphs, matrices, geometries of points and lines, graph decompositions, codes, and the like; the balance of the underlying set system permeates all.
1.7 Table Some of the set systems that result when at least one of the balance sets K, R, M, or the {Λt} is a singleton.
K R Λ2 Λt M Other Name Section {k} {r} {λ2} {λt} t-design II.4 {k} {r} {λ2} {λt} λt = 1 Steiner system II.5 {k} {r} {λ} balanced incomplete block de-sign (BIBD) II.1, II.3 {k} {r} {λ} {µ} k = r symmetric design (SBIBD) II.6 {k} {r} {λ} {µ} λ = µ = 1 projective plane II.6, VII.2.1 {r} {λ} (r, λ)-design VI.49 {λ} pairwise balanced design (PBD) IV.1 {λ} λ = 1 linear space IV.1, IV.6 {λt} t-wise balanced design VI.63 1.8 Remarks A natural relaxation is obtained by specifying an upper or a lower bound on values arising in the set, or by specifying a small number of permitted values; the short form {x} is adopted to permit all values greater than or equal to x, and {x} to indicate all values less than or equal to x. Then certain generalizations have been well studied; see Table 1.9.
I.1.1 Balance 5 1.9 Table Some set systems that result when K is a singleton and M or {Λt} is bounded.
K R Λ2 Λt M Other Name Section {k} {λt} t-covering VI.11 {k} {µ} constant weight code VII.1 {k} {λt} t-packing VI.40 ▷ Algebras and Arrays 1.10 Example A quasigroup and a latin square.
⊗0 1 2 3 4 5 6 0 0 2 1 4 3 6 5 1 2 1 0 5 6 3 4 2 1 0 2 6 5 4 3 3 4 5 6 3 0 1 2 4 3 6 5 0 4 2 1 5 6 3 4 1 2 5 0 6 5 4 3 2 1 0 6 Use the blocks of Example 1.3 to define the binary operation ⊗at left, as follows. When {x, y, z} is a block, define x ⊗y = z; then define x ⊗x = x.
Delete the headline and sideline, reorder rows and columns, and rename symbols to get the array on the right.
1 7 2 6 5 4 3 4 5 6 3 7 1 2 3 6 5 7 4 2 1 5 4 3 2 1 7 6 2 1 7 5 6 3 4 7 2 1 4 3 6 5 6 3 4 1 2 5 7 1.11 Remarks Example 1.10 introduces another kind of balance. The operation ⊗defined has a simple property: the equation a⊗b = c always has a unique solution for the third variable given the other two. This is reminiscent of groups, yet the algebra defined has no identity element and hence no notion of inverses. It weakens the properties of finite groups, and is a quasigroup. As an n×n array, it is equivalent to say that every symbol occurs in every row once, and also in every column once; the array is a latin square.
Part III focuses on latin squares and their generalizations; see particularly §III.1 for latin squares, §III.2 for quasigroups, and §VI.22 for generalizations to balanced squares in which every cell contains a fixed number of elements greater than one.
1.12 Example For the square at left in Example 1.10, form a set {x, y+7, z+14} whenever x ⊗y = z. The resulting sets are: {0, 7, 14}, {0, 8, 16}, {0, 9, 15}, {0, 10, 18}, {0, 11, 17}, {0, 12, 20}, {0, 13, 19}, {1, 7, 16}, {1, 8, 15}, {1, 9, 14}, {1, 10, 19}, {1, 11, 20}, {1, 12, 17}, {1, 13, 18}, {2, 7, 15}, {2, 8, 14}, {2, 9, 16}, {2, 10, 20}, {2, 11, 19}, {2, 12, 18}, {2, 13, 17}, {3, 7, 18}, {3, 8, 19}, {3, 9, 20}, {3, 10, 17}, {3, 11, 14}, {3, 12, 15}, {3, 13, 16}, {4, 7, 17}, {4, 8, 20}, {4, 9, 19}, {4, 10, 14}, {4, 11, 18}, {4, 12, 16}, {4, 13, 15}, {5, 7, 20}, {5, 8, 17}, {5, 9, 18}, {5, 10, 15}, {5, 11, 16}, {5, 12, 19}, {5, 13, 14}, {6, 7, 19}, {6, 8, 18}, {6, 9, 17}, {6, 10, 16}, {6, 11, 15}, {6, 12, 14}, {6, 13, 20}.
1.13 Remarks Example 1.12 gives a representation of the quasigroup in Example 1.10 once again as blocks. In this set system with 49 blocks on 21 points, every point occurs in 7 blocks, every block has three points; yet some pairs occur once while others occur not at all, in the blocks. Indeed pairs inside the groups {0, 1, 2, 3,4,5,6}, {7, 8, 9, 10, 11, 12, 13}, and {14, 15, 16, 17,18, 19, 20} are precisely those that do not appear. This is a group divisible design (see §IV.1). Balance with respect to occurrence of pairs could be recovered by adjoining the three groups as blocks; the result exhibits pair balance but loses equality of block sizes because blocks of size 3 and 7 are present; this is a pairwise balanced design (see §IV.1). Instead, placing a copy of the 7 point design on each of the groups balances both pair occurrences and block sizes, and, indeed, provides a larger balanced incomplete block design, this one on 21 points having 70 (= 49 + 3 × 7) blocks.
Returning to the group divisible design, yet another balance property is found; every block meets every group in exactly one point. This defines a transversal design; see §III.3.
6 Opening the Door I.1 1.14 Example Two mutually orthogonal latin squares of side 7.
+ 0 1 2 3 4 5 6 0 0 1 2 3 4 5 6 1 1 2 3 4 5 6 0 2 2 3 4 5 6 0 1 3 3 4 5 6 0 1 2 4 4 5 6 0 1 2 3 5 5 6 0 1 2 3 4 6 6 0 1 2 3 4 5 ⊙0 1 2 3 4 5 6 0 0 2 4 6 1 3 5 1 1 3 5 0 2 4 6 2 2 4 6 1 3 5 0 3 3 5 0 2 4 6 1 4 4 6 1 3 5 0 2 5 5 0 2 4 6 1 3 6 6 1 3 5 0 2 4 +, ⊙0 1 2 3 4 5 6 0 0,0 1,2 2,4 3,6 4,1 5,3 6,5 1 1,1 2,3 3,5 4,0 5,2 6,4 0,6 2 2,2 3,4 4,6 5,1 6,3 0,5 1,0 3 3,3 4,5 5,0 6,2 0,4 1,6 2,1 4 4,4 5,6 6,1 0,3 1,5 2,0 3,2 5 5,5 6,0 0,2 1,4 2,6 3,1 4,3 6 6,6 0,1 1,3 2,5 3,0 4,2 5,4 1.15 Remark The operation ⊗in Example 1.10 is by no means the only way to define a quasigroup on 7 symbols. In Example 1.14, two more are shown, with operations + and ⊙. A natural question is whether they are different from each other in an essential manner, or different from ⊗. The definition of + has the same columns as that of ⊙, but in a different order. While different as algebras, they are nonetheless quite similar.
They are introduced here to note yet another kind of balance. When superimposing two n × n latin squares, one could expect every ordered pair of symbols to arise any number of times between 0 and n; but the average number of times is precisely one.
This example demonstrates that perfect balance can be achieved: every pair arises exactly once in the superposition. As latin squares, these are mutually orthogonal.
Section III.3 explores these, in their various disguises as quasigroups, as squares, and as transversal designs.
1.16 Example For the square at left in Example 1.10, form an array whose columns are precisely the triples (x, y, z)T for which x ⊗y = z.
0000000111111122222223333333444444455555556666666 0123456012345601234560123456012345601234560123456 0214365210563410265434563012365042163412505432106 1.17 Remark Example 1.16 gives yet another view of the same structure. In the 3 × 49 array, each of the 7 symbols occurs 7 times in each row. However, much more is true.
Choosing any two rows, there are 49 = 7·7 pairs of entries possible, and 49 occurring.
As one might expect, the case of interest is that of balance, when every pair occurs the same number of times (here, once). This is an orthogonal array; see §III.6 and §III.7.
While playing a prominent role in combinatorial design theory, they play a larger role yet in experimental design in statistics; see §VI.21.
▷ Matrices, Vectors, and Codes 1.18 Example Begin with the set system representation of the Fano plane in Example 1.3 and construct the 7 × 7 matrix of 0s and 1s on the right by placing a 1 in cell (a, b) if the element a is in set b; place a 0 otherwise.
1.19 Remark The matrix in Example 1.18 is the incidence ma-trix of the Fano plane. Generally, the rows are indexed by the points of the design and the columns by the blocks. Im-portant properties of designs can be determined by matrix properties of these incidence matrices; see §VII.7.
1 1 1 0 0 0 0 1 0 0 1 1 0 0 1 0 0 0 0 1 1 0 1 0 1 0 1 0 0 1 0 0 1 0 1 0 0 1 1 0 0 1 0 0 1 0 1 1 0 1.20 Example Apply the permutation (0)(1)(2 5 3)(4)(6) to the symbols in Example 1.3.
Form a 7 × 7 matrix with columns indexed by blocks and rows indexed by symbols.
I.1.1 Balance 7 −−−−−−−− −−−+ −+ ++ −+ −−+ −++ −+ + −−+ −+ −+ + + −−+− −−+ + + −−+ −+ −+ + + −− −−+ −+ + +− Place a −in the cell of row x and column y when symbol x appears in block y, and a + otherwise. Finally add an initial row and an initial column consisting entirely of −to get the array at the left. Invert + and −to get the array on the right.
+ + + + + + ++ + + + −+ −−− + −+ + −+ −− + −−+ + −+− + −−−+ + −+ + + −−−+ +− + −+ −−−++ + + −+ −−−+ 1.21 Remark Of course, the balance observed previously repeats in the matrix on the left in Example 1.20. In this presentation, however, the balance is seen in a new light. Taking + to be +1 and −to be −1, it is possible for two rows to have inner product between −8 and +8; yet here every pair of distinct rows have inner product exactly equal to 0 (and indeed, so do any two distinct columns). This property over {±1} defines a Hadamard matrix; see §V.1. Such matrices have the remarkable property that they maximize the matrix determinant over all n×n {±1}-matrices; see §V.3. Generalizing to arithmetic over algebras other than ({±1}, ·) yields classes of orthogonal matrices and designs (§V.2).
The matrix on the right shares the same properties, being the negative of the one on the left. But a surprise is in store. Juxtapose the two horizontally, and delete the two constant columns to form an 8 × 14 matrix. Every two distinct columns of the result have either 0 or 4 identical entries. For each column form a set that contains the indices of the + entries in that column; this yields 14 blocks each of size 4, on a set of 8 elements. Every element occurs in 7 blocks, every 2-subset of elements in 6 blocks, and every 3-subset of elements in exactly one block. This is a Steiner system and a 3-design (a Hadamard 3-design).
1.22 Example In Example 1.20, delete the constant row and constant column, and treat the remaining 7 rows as {±1}-vectors. These vectors are: [--+-+++] [+--+-++] [++--+-+] [+++--+-] [-+++--+] [+-+++--] [-+-+++-] 1.23 Remark In communicating between a sender and a receiver, synchronization is a ma-jor issue, as demonstrated by the message ...EATEATEATEATEATEATEATEA..., which might say EAT, ATE, or TEA depending on your timing. Knowing the message, how-ever, enables one to determine the timing. Suppose then that a sender repeatedly sent [+-+++--]. If the receiver is “in phase,” it multiplies the expected vector with the received one, and computes 7. If “out of phase,” it instead computes −1 no matter what the nonzero phase shift is. This autocorrelation can be used to synchronize, and the balance plays a practical purpose; see §V.9 and §V.7.
1.24 Example An error-correcting code.
Replace + by 1 and −by 0 in the two Hadamard ma-trices of Example 1.20, and treat the 16 rows of the resulting matrices as binary vectors of length 8.
1.25 Remark The sixteen binary vectors of Example 1.24 have the property that every two distinct vectors differ in exactly 4 or in all 8 positions. A sender and receiver can agree that in order to communicate sixteen information symbols, they are to use these 16 “codewords.” If a single bit error is made in transmission, there is exactly one codeword that could have been intended; if two bit errors are made, the receiver knows that an error occurred but does not know which two bit errors account for the incorrect received word. This is a 1-error-correcting code and a 2-error-detecting code; see §VII.1.
8 Opening the Door I.1 ▷ Graphs 1.26 Example An embedding of the Fano plane on the torus.
0 0 0 0 1 2 4 6 3 5 Draw the complete graph on 7 vertices (K7) on the torus (identify the left and right borders, and the top and bot-tom borders, to form the torus—you may need to pur-chase a second Handbook to do this).
The number of vertices is 7 and the number of edges is 21, so the num-ber of faces (by Euler’s formula) is 14.
1.27 Remark The triangular faces in the drawing of Example 1.26 have the property that they can be colored with two colors, so that every edge belongs to one face of each color. Consider then the faces of one color; every edge (2-subset of vertices) appears in exactly one of the chosen faces (3-subset of vertices). Seven vertices, seven faces, three vertices on every face, three faces at every vertex, and for each edge exactly one face that contains it. See §VI.25 for embeddings of graphs in surfaces that correspond to designs.
1.28 Example In the graph shown, label points on the left by seven different symbols, and each point on the right by the 3-subset of symbols for its adjacent vertices. The seven 3-subsets thereby cho-sen form the blocks of a balanced incomplete block design, and this incidence graph shows the element-block incidence structure.
1.29 Remark Example 1.28 gives a purely graph-theoretic representation of the running example; see §VII.4. Again a surprise is in store. Consider Example 1.26 further.
Surface duality implies that, by interchanging the roles of faces and vertices in the drawing, the embedding of a dual graph is obtained. What is the structure of the dual graph in this case? It is shown in Example 1.28.
1.30 Example The graph in Example 1.28 has a hamiltonian cycle. Re-draw the graph with the hamiltonian cycle on the exterior. This is the Heawood graph, the smallest 3-regular graph whose shortest cycle has length six (also called a cage); see §VII.4.
1.31 Example A decomposition of the complete graph into triangles.
1.32 Remark In another vernacular, Example 1.26 shows that K7 can be decomposed into edge-disjoint copies of K3 (the triangle). The toroidal embedding is not needed to see this, however. Example 1.31 explicitly shows seven K3s that edge-partition K7.
Another line of investigation opens: When can one edge-partition a graph G into copies of a graph H? Combinatorial design theory concentrates on cases when G is a complete graph. When H is also complete, another view of balanced incomplete block designs is obtained. But when H is a cycle, a different generalization is encountered, cycle systems; see §VI.12. Further, when H is arbitrary, a graph design is obtained; see §VI.24.
Imagine that the set of all possible edges is to be partitioned into subgraphs, so that no subgraph has too many edges; in addition, every vertex in a subgraph that has one or more edges at it incurs a cost.
The goal is to minimize cost.
This is achieved by choosing subgraphs that are complete whenever possible, and hence block designs provide solutions.
Exactly such a problem arises in “grooming” traffic in optical networks; see §VI.27.
I.1.1 Balance 9 1.33 Example Start with blocks {0, 1, 3}, {0, 4, 5}, {0, 2, 6}, {1, 2, 4}, {1, 5, 6}, {2, 3, 5}, and {3, 4, 6}. Arrange the points 0, . . ., 6 on the perimeter of a circle consecutively, and place an additional point, ∞, at the center of the circle. Now form seven graphs numbered also 0, . . ., 6. In the ith, place the edge {∞, i}, and for all x ∈{0, . . ., 6}\{i} place the edge {x, y} if and only if {i, x, y} is a block.
1.34 Remark In Example 1.33, each of the seven graphs is a perfect matching or 1-factor on the eight points. The seven 1-factors share no edges; they partition the edges of K8 into 1-factors. This is a 1-factorization; see §VII.5. Playing a round-robin tournament with 8 players over 7 time-p erio ds can b e accomplished using the matches shown here; see §VI I.51.
1.35 Example An orthogonal 1-factorization and a Room square.
∞0 26 45 13 ∞1 03 56 24 ∞2 14 06 35 46 ∞3 25 01 05 ∞4 36 12 23 16 ∞5 04 15 34 02 ∞6 Above is shown another 1-factorization, obtained from Example 1.33 by reflecting the first picture through the vertical axis.
Remarkably it has the property that every 1-factor of this factorization shares at most one edge with that of Example 1.33, such one factorizations are orthogonal. Form a 7 × 7 square (at left) with rows indexed by the factors of Example 1.33 and columns by the factors in its re-flection. In each cell, place the edge (if any) that the factors share.
1.36 Remark Example 1.35 gives a square in which every cell is either empty or contains an unordered pair (edge); every unordered pair occurs in exactly one cell; and every row and every column contains every symbol exactly once. This is a Room square; see §VI.50. Generalizing to factorizations of regular graphs other than complete ones leads to Howell designs; see §VI.29.
▷ Finite Geometry 1.37 Example In the picture at right, label the three corner points by binary vectors (0,0,1), (0,1,0), and (1,0,0). When (a, b, c) and (x, y, z) are points, so also is (a ⊕x, b ⊕y, c ⊕z) = (a, b, c) ⊕(x, y, z); the operation ⊕is addition modulo 2. Then (1,1,0), (1,0,1), (0,1,1), and (1,1,1) are also points—but (0,0,0) is not. Lines are defined by all sets of three distinct points whose sum under ⊕is (0,0,0).
1.38 Remark This is a geometry of points and lines that is finite; as with plane geometry, any two points define a line, and two lines meet in at most one point. Unlike usual geometry in the plane, however, there are no two lines parallel; every two lines meet in exactly one point. This is a finite projective geometry, one of a number of finite geometries (§VII.2). Geometries in which lines can have different sizes generalize this notion to linear spaces; see §VI.31.
10 Opening the Door I.1 1.2 Symmetry 1.39 Remark That the one simple structure in Example 1.1 can demonstrate balance in so many different settings is remarkable, and its vast generalizations through these different representations underpin many topics in combinatorial design theory. But the example leads to much more. It exhibits symmetry as well as balance. As a first step, note that permuting the three coordinate positions in Example 1.37 relabels the points. Yet the picture remains unchanged! This describes a sixfold symmetry.
1.40 Example In Z7 form the 3-set {1, 2, 4} consisting of the three nonzero squares (the quadratic residues). Form seven sets by adding each element of Z7 in turn to this base block. These seven sets give the set system representation of the Fano plane.
1.41 Remark Example 1.40 gives another representation of the recurring example, as trans-lates of the quadratic residues modulo 7; this opens the door into number theory (§VII.8). More than that, there is a sevenfold symmetry by translation through Z7.
Example 1.31 depicted this symmetry already, but in Example 1.40 a more compact representation is given.
1.42 Example The automorphism group of the Fano plane.
(0 1 2 3 4 5 6) (0 3) (1) (2 6 4 5) (0 3 1) (2) (4 5 6) (0) (1 3) (2) (6) (4 5) The four permutations on the left generate a permu-tation group Γ of order 168. Under the action of Γ, the orbit of {0, 1, 3} contains 7 triples; that of {0, 1, 5} contains seven (different) triples; and that of {0, 1, 2} contains the remaining 21 triples.
1.43 Remark Example 1.42 constructs the recurring example via the action of a group; indeed the orbit of {0, 1, 3} under Γ provides the seven triples. As there are no other group actions that preserve this orbit, Γ is the full automorphism group of the design.
See §VII.9 for relevant descriptions of groups. In this representation, many symmetries are provided, not all of which are easily seen geometrically. One particular symmetry of interest is that the seven points lie in a single orbit (point transitivity); a second is that all blocks lie in the same orbit (block transitivity). See §VII.5.
1.44 Example Begin again with the 3-set {1, 2, 4} in Z7 from Example 1.40. Calculate differences modulo 7 among these three elements to obtain ±1, ±2, and ±3, all nonzero differences in Z7.
1.45 Remark In Example 1.42, not only is there a transitive action on the seven points, the action is a single 7-cycle.
In Example 1.44, this is seen by the occurrence of every nonzero difference among the p oints once. Such cyclic designs have received sets) and §VI.16 for cyclic blo ck designs (difference families); also see §V.5 and §VI.17 for generalizations of orthogonal arrays defined by group actions.
1.3 Parallelism 1.46 Example On {0, 1, 2, 3,4, 5, 6}× {0, 1, 2}, form the 70 sets B = {{(i, 0), (i, 1), (i, 2)} : i = 0, . . ., 6} ∪{{(i + 1, j), (i + 2, j), (i + 4, j)} : i = 0, . . ., 6, j = 0, 1, 2} ∪{{(i + 3, a), (i + 5, b), (i + 6, c)} : i = 0, . . ., 6, {a, b, c} = {0, 1, 2}}.
Arithmetic in the first component is done modulo 7, and in the second modulo 3.
1.47 Remark Example 1.46 contains three copies of Example 1.31 within it; they lie on the points {0, 1, 2, 3, 4,5,6} × {j} for j ∈{0, 1, 2}. Every two of the 21 points occur particular attention. See §I I.18 for cyclic and transitive symmetric designs (difference I.1.4 Closing 11 together in exactly one of the 70 sets. This is a recursive construction of a larger design from a smaller one. Now rewrite Example 1.46.
1.48 Example For i ∈{0, 1, 2, 3, 4,5,6} let Pi = {{(i, 0), (i, 1), (i, 2)}, {(i + 1, 0), (i + 2, 0), (i + 4, 0)}, {(i + 3, 0), (i + 5, 1), (i + 6, 2)}, {(i + 1, 1), (i + 2, 1), (i + 4, 1)}, {(i + 3, 1), (i + 5, 2), (i + 6, 0)}, {(i + 1, 2), (i + 2, 2), (i + 4, 2)}, {(i + 3, 2), (i + 5, 0), (i + 6, 1)}}.
For j ∈{0, 1, 2} let Qj = {{(i + 3, j), (i + 5, j + 2), (i + 6, j + 1)} : i = 0, . . ., 6}. The 70 sets in B are partitioned into ten classes (P0, P1, . . ., P6, Q1, Q2, Q3) of seven sets each in this manner.
1.49 Remark In Example 1.48, consider one of the classes in the partition. It contains seven sets each of size 3, chosen from a set system on 21 points. Asking again for “balance,” one would hope that every point occurs in exactly one set of the class. And indeed it does! Each class contains every point exactly once; it is a parallel class or a resolution class. The whole design is partitioned into resolution classes; it is resolvable. The partition into classes is a resolution. Geometrically, one might think of the sets as lines, and a resolution class as a spanning set of parallel lines. Then resolvability is interpreted as a “parallelism.” Resolvability is a key concept in combinatorial design theory. See §II.7 and §IV.5. Indeed resolvability also arises extensively in tournament scheduling (§VI.3,§VI.51).
1.4 Closing 1.50 Remark This short introduction has afforded only a cursory glance at the many ways in which such simple objects are generated, classified, enumerated, generalized, and applied. The main themes of combinatorial design theory, balance and symmetry, recur throughout the handbook. Nevertheless, there is more to learn from the initial tour. Often the representation chosen dictates the questions asked; one seemingly simple object appears in a myriad of disguises. To explore this world further, one piece of advice is in order.
Be mindful of the ease with which any combinatorial design can arise in an equivalent formulation, with different vernacular and different lines of research. The intricate web of connections among designs may not be a blessing to those entering the area, but it is nonetheless a strength.
2 Design Theory: Antiquity to 1950 Ian Anderson Charles J. Colbourn Jeffrey H. Dinitz Terry S. Griggs Those who do not read and understand history are doomed to repeat it. (Harry Truman) At the time of writing, every passing year sees the addition of more than five hundred new published papers on combinatorial designs and likely thousands more employing the results and techniques of combinatorial design theory. The roots of modern design theory are diverse and often unexpected. Here a brief history is given.
It is intended to provide first steps in tracing the evolution of ideas in the field.
12 Design Theory: Antiquity to 1950 I.2 The literature on latin squares goes back at least 300 years to the monograph Koo-Soo-Ryak by Choi Seok-Jeong (1646–1715); he uses orthogonal latin squares of order 9 to construct a magic square and notes that he cannot find orthogonal latin squares of order 10.
It is unlikely that this was the first appearance of latin squares.
Ahrens remarks that latin square amulets go back to medieval Islam (c1200), and a magic square of al-Buni, c 1200, indicates knowledge of two 4 × 4 orthogonal latin squares. In 1723, a new edition of Ozanam’s four-volume treatise presented a card puzzle that is equivalent to finding two orthogonal latin squares of order 4. Then in 1776, Euler presented a paper (De Quadratis Magicis) to the Academy of Sciences in St. Petersburg in which he again constructed magic squares of orders 3, 4, and 5 from orthogonal latin squares. He posed the question for order 6, now known as Euler’s 36 Officers Problem.
Euler was unable to find a solution and wrote a more extensive paper in 1779/1782. He conjectured that no solution exists for order 6. Indeed he conjectured further that there exist orthogonal latin squares of all orders n except when n ≡2 (mod 4): et je n’ai pas h´ esit´ e d’en conclure qu’on ne saurait produire aucun quarr´ e complet de 36 cases, et que la mˆ eme impossibilit´ e s’´ etende aux cas de n = 10, n = 14 et en g´ en´ eral a tous les nombres impairement pairs.
No progress was to be made on this problem for over a century, although it was not neglected by mathematicians of the day. Indeed, Gauss and Schumacher (see ) corresponded in 1842 about work of Clausen establishing the impossibility when n = 6 and conjecturing (independently of Euler) the impossibility when n ≡2 (mod 4), which work was apparently not published. In 1900, by an exhaustive search, Tarry showed that none exists for n = 6, and in 1934 the statisticians Fisher and Yates gave a simpler proof.
Petersen in 1901 and Wernicke in 1910 published fallacious proofs of Euler’s conjecture for n ≡2 (mod 4). Then in 1922 MacNeish published another, after pointing out the error in Wernicke’s approach! In 1942 Mann described the construction of orthogonal latin squares using orthomorphisms. The falsity of the general Euler conjecture was finally estab-lished in 1958 when Bose and Shrikhande constructed two orthogonal squares of order 22; and then in 1960 Bose, Shrikhande, and Parker established that two orthogonal latin squares of order n exist for all n ≡2 (mod 4), other than 2 and 6.
The study of block designs can be traced back to 1835, when Pl¨ ucker , in a study of algebraic curves, encountered a Steiner triple system of order 9, and claimed that an STS(m) could exist only when m ≡3 (mod 6). In 1839 he correctly revised this condition to m ≡1, 3 (mod 6) . Pl¨ ucker’s discovery that every nonsingular cubic has 9 points of inflection, defining 12 lines each of 3 points, and with each pair of points on a line, illustrates the early connection between designs and geometry.
Another discovery of Pl¨ ucker concerned the double tangents of fourth order curves, namely, a design of 819 4-element subsets of a 28-element set such that every 3-element subset is in exactly one of the 4-element subsets, a 3-(28, 4, 1) design. Pl¨ ucker raised the question of constructing in general a t-(m, t + 1, 1) design.
In England, a prize question of Woolhouse in the Lady’s and Gentleman’s Diary of 1844 asked: determine the number of combinations that can be made out of n symbols, p symbols in each; with this limitation, that no combination of q symbols, which may appear in any one of them shall be repeated in any other.
In 1847 Kirkman dealt with the existence of such a system in the case p = 3 and q = 2, constructing STS(m) for all m ≡1, 3 (mod 6). The next case to be solved, p = 4 and q = 3, was done more than a century later by Hanani in 1960. The reason the triad systems (as Kirkman called them) are not now called Kirkman triple I.2 Design Theory: Antiquity to 1950 13 systems but rather Steiner triple systems lies in the fact that Steiner , apparently unaware of Kirkman’s work, asked about their existence in 1853; subsequently Reiss (again) settled their existence, and later Witt named the systems after Steiner. A particular irony is that Steiner had actually asked a related but different question, which was taken up by Bussey in 1914 . The name Kirkman has instead come to be attached to resolvable Steiner systems, on account of Kirkman’s famous 15 schoolgirls problem : fifteen young ladies in a school walk out three abreast for seven days in succession; it is required to arrange them daily, so that no two shall walk twice abreast.
The first solution to the schoolgirls problem to appear in print was by Cayley in 1850 followed in the same year by another by Kirkman himself. Kirkman’s so-lution involved an ingenious combination of an STS(7) and a (noncyclic) Room square of side 7. In 1863, Cayley pointed out that his solution could also be presented in this way. Cayley’s and Kirkman’s solutions are nonisomorphic as Kirkman triple systems but isomorphic as Steiner triple systems, being the two different resolutions into parallel classes of the point-line design of the projective geometry PG(3, 2). In 1860 Peirce found all three solutions of the 15 schoolgirls problem having an automorphism of order 7, and in 1897 Davis published an elegant geometric solu-tion based on representing the girls as the 8 corners, the 6 midpoints of the faces, and the centre of a cube. In 1912, Eckenstein published a bibliography of Kirkman’s schoolgirl problem, consisting of 48 papers. The 7 nonisomorphic solutions of the 15 schoolgirls problem were enumerated by Mulder and Cole . The first pub-lished solution of the existence of a Kirkman triple system of order m, KTS(m), for all m ≡3 (mod 6) was by Ray-Chaudhuri and Wilson in 1971. The problem had in fact already been solved by the Chinese mathematician Lu Jiaxi at least eight years previously, but the solution had remained unpublished because of the political upheavals of the time. Lu also solved the corresponding problem for blocks of size 4; these results were eventually published in his collected works in 1990.
Prior claim to the authorship of the 15 schoolgirls problem was made by Sylvester in 1861, hotly refuted by both Kirkman and Woolhouse . But Sylvester raised further questions about such systems, for example: can all 455 triples from a 15-element set be arranged into 13 disjoint KTS(15)s, thereby allowing a walk for each of the 13 weeks of a school term, without any three girls walking together twice?
The problem was finally solved in 1974 by Denniston although both Kirkman and Sylvester had solutions for the corresponding problem with 9 instead of 15 (indeed they had the first example of a large set of designs). Surely neither Sylvester nor Denniston would have anticipated that Denniston’s solution would form the basis in 2005 of a musical score, Kirkman’s Ladies , about which the composer states: The music seems to come from some point where precise organization meets near chaos.
The problem of the existence of large sets of KTS(n) in the general case of n ≡3 (mod 6) is still unsolved. Sylvester also asked about the existence of a large set of 13 disjoint STS(15)s. In this case, the general problem of large sets of STS(n) is solved for all n ≡1, 3 (mod 6), n ̸= 7, mainly by work of Lu [1482, 1483] but also Teirlinck . Cayley had shown that the maximum number of disjoint STS(7)s on the same set is 2. In 1917, Bays determined that there are precisely 2 nonisomorphic large sets of STS(9).
Kirkman’s solution of the 15 schoolgirls problem was extended brilliantly in 1852– 3 by Anstice [103, 104]; he generalised the case where n = 15 to any integer of the form n = 2p + 1, where p is prime, p ≡1 (mod 6). For the first time making use of primitive roots in the construction of designs (a method basic to Bose’s 1939 paper) and introducing the method of difference families, Anstice constructed an infinite class 14 Design Theory: Antiquity to 1950 I.2 of cyclic Room squares, cyclic Steiner triple systems, and 2-rotational Kirkman triple systems. His starter-adder approach to the construction of Room squares includes the Mullin–Nemeth starters as a special case.
In 1857 Kirkman used difference sets (in the equivalent formulation of perfect partitions) to construct cyclic finite projective planes of orders 2, 3, 4, 5, and 8, introducing at the same time the concept of a multiplier of a difference set. He also wrote that he thought that a solution for 6 is “improbable,” for 7 is “very likely” (because 7 is prime) and that he did not see why one should not be discovered for 9. He did not comment on order 10. A few years earlier, Kirkman had shown how to construct affine planes of prime order, and, hence, essentially by adding points at infinity, finite projective planes of all prime orders (although he did not use the geometric terminology). This paper also showed how to construct certain families of pairwise balanced designs. Such designs were to prove invaluable in later combinatorial constructions; indeed, in the same paper Kirkman himself used them to solve the schoolgirls problem for 5 · 3m+1 symbols. Kirkman’s work on combinatorial designs is indeed seminal. To quote Biggs : Kirkman has established an incontestable claim to be regarded as the founding father of the theory of designs. Among his contemporaries, only Sylvester attempted anything com-parable, and his papers on Tactic seem to be more concerned with advancing his claims to have discovered the subject than with advancing the subject itself. Not until the Tactical Memoranda of E. H. Moore in 1896 is there another contribution to rival Kirkman’s.
In 1867, Sylvester investigated a tiling problem and constructed what he called anallagmatic pavements, which are equivalent to Hadamard matrices. He showed how to construct such of order 2n from a solution of order n; twenty-six years later Hadamard showed that Hadamard matrices give the largest possible determi-nant for a matrix whose entries are bounded by 1. Hadamard showed that the order of such a matrix had to be 1, 2, or a multiple of 4, and he constructed matrices of orders 12 and 20. In 1898 Scarpis showed how to construct Hadamard matrices of order 2k · p(p + 1) whenever p is a prime for which a Hadamard matrix of order p + 1 exists. In 1933 Paley used squares in Fq to construct Hadamard matrices of order q + 1 when q ≡3 (mod 4) and of order 2(q + 1) when q ≡1 (mod 4). He also showed that if m = 2n, then the 2m (±1)-sequences of length m can be parti-tioned in such a way as to form the rows of 2m−n Hadamard matrices. In a footnote Paley acknowledged that some of his results had been announced by Gilman in the 1931 Bulletin of the AMS, but without proof.
In the same journal issue in 1933, Todd pointed out the connection between Hadamard matrices and Hadamard designs; designs with these parameters were relevant to work of Coxeter (also in the same journal issue) on regular compound polytopes! At this time the matrices were not yet called Hadamard matrices; Paley called them U-matrices. Williamson first called them Hadamard matrices in 1944. The existence of Hadamard matrices for all admissible orders remains open.
Cayley introduced the word tactic for the general area of designs. The word config-uration, used nowadays more generally, was given a specific meaning in 1876 by Reye in terms of geometrical structures. Essentially, a configuration was defined to be a system of v points and b lines with k points on each line and r lines through each point, with at most one line through any two points (and hence any two lines intersecting in at most one point). If v = b (and therefore k = r), the configuration is symmetric and denoted by vk; thus, for example, the Fano plane is 73, the Pappus con-figuration is 93, and the Desargues configuration is 103. In fact, the first appearance of the Petersen graph is not in Petersen’s paper of 1891 but in 1886 by Kempe as the graph of the Desargues configuration. In 1881, Kantor showed that there exist one 83, three 93, and ten 103 configurations. Six years later, Martinetti I.2 Design Theory: Antiquity to 1950 15 showed there were thirty-one 113 configurations. In the following years, Martinetti constructed symmetric 2-configurations (two points are connected by at most 2 lines), with parameters (74)2, (84)2, and (94)2 as well as the biplanes (115)2 and (166)2. The biplane (115)2, also a Hadamard design 2-(11, 5, 2), had previously been constructed by Kirkman in 1862. Symmetric configurations v4 were first studied in 1913 by Merlin . In 1942, Levi unified the treatment of configurations with questions in algebra, geometry, and design theory.
Interest in counting configurations is reflected in the enumerative work on Steiner systems. It was known early on that there are unique STS(7) and STS(9). In 1897, Zulauf showed that the known STS(13)s fall into two isomorphism classes.
Using the connection between STS(13)s and symmetric configurations 103, in 1899 De Pasquale determined that only two isomorphism classes are possible. The same result was also obtained two years later by Brunel. The enumeration of nonisomorphic STS(15)s was first done by Cole, Cummings, and White. In 1913, both Cole and White published papers introducing ideas for enumeration and a year later Cummings classified all 23 STS(15)s which have a subsystem of order 7. Then in 1919, White, Cole, and Cummings in a remarkable memoir, succeeded in determining that there are precisely 80 nonisomorphic STS(15)s. In 1940, Fisher , unaware of this catalogue, also generated STS(15)s; he used trades on Pasch configurations to find 79 of the 80 systems. The veracity of White, Cole, and Cummings’ work was confirmed in 1955 by Hall and Swift in one of the first cases in which digital computers were used to catalogue combinatorial designs.
The number of nonisomorphic STS(19)s was published only in 2004 , perhaps understandably as it is 11,084,874,829.
Enumeration of latin squares has a more complex history (see ); early results are by Euler in 1782, Cayley and Frolov in 1890, Tarry in 1900, MacMahon in 1915, Sch¨ onhardt in 1930, Fisher and Yates in 1934, and Sade and Saxena in 1951.
By the end of the 19th century there was no formal body of work called graph theory, but some graph theoretic results were in existence in the language of config-urations or designs. One-factorizations of K2n are resolvable 2-(2n, 2, 1) designs, and the “classical” method of construction appeared in Lucas’ 1883 book ; the con-struction was credited to Walecki and can be described in terms of chords of a circle or in terms of a difference family now known as a patterned starter. Remarkably, in 1847 Kirkman already gave a construction based on a lexicographic method (in fact the greedy algorithm), which gives a schedule isomorphic to that of Lucas– Walecki. Lucas’ book also describes the construction of round-dances (or hamiltonian decompositions of the complete graph) by means of what we now call a terrace. Enu-meration results for nonisomorphic 1-factorizations of K2n span nearly a century. In 1896, Moore reported that there are 6 distinct 1-factorizations of K6 that fall into a single isomorphism class. The 6 nonisomorphic 1-factorizations of K8 appear in Dickson and Safford . Gelling showed that the number of nonisomorphic 1-factorizations of K10 is 396, and Dinitz, Garnick, and McKay showed that there are 526,915,620 nonisomorphic 1-factorizations of K12.
In 1891, Netto , unaware of Kirkman’s work, gave four constructions for Steiner triple systems: (i) an STS(2n+1) from an STS(n), (already used by Kirkman), (ii) an STS(mn) from an STS(m) and an STS(n), (iii) an STS(p) where p is a prime of the form 6m+1, and (iv) an STS(3p) where p is of the form 6m+5. The construction given by Netto for (iii) involves primitive roots, but is not the construction usually associated with his name. These constructions enabled Netto to construct STS(n) for all admissible n < 100 except for n ∈{25, 85}. In 1893 Moore completed Netto’s proof, giving a construction of an STS(w + u(v −w)) from an STS(u) and an STS(v) with an STS(w) subsystem. In he also proved that for all admissible 16 Design Theory: Antiquity to 1950 I.2 v > 13, there exist at least two nonisomorphic STS(v). In 1895 and 1899, Moore [1625, 1627] determined the automorphism groups of the unique STS(7) and 3-(8, 4, 1) designs and studied the intersection properties of the 30 realizations on the same base set of both systems.
Undoubtedly the most far-reaching paper by Moore is his “Tactical Memoranda I–III” . This was published in 1896 in the American Journal of Mathematics (founded in 1878 by Sylvester during his seven-year stay in the U.S. as professor at Johns Hopkins University). It contains many jewels hidden in 40 pages of difficult and often bewildering terminology and notation. After a brief introductory section (memorandum I), memorandum II introduces a plethora of different types of designs, including with order and/or with repeated block elements, the first occurrence of both of these concepts. Here we find the construction using finite fields of a complete set of q −1 mutually orthogonal latin squares (MOLS) of order q whenever q is a prime power (a result rediscovered much later by Bose in 1938 and by Stevens in 1939 independently). One also finds the theorem, later rediscovered by MacNeish , that if there exist t MOLS of order m and of order n, then there exist t MOLS of order mn. Further, it contains the construction of many 1-rotational designs including Kirkman triple systems and work on orthogonal arrays. Memorandum III contains the first general results on whist tournaments and triple-whist tournaments.
Whist tournaments Wh(4n) had been presented by Mitchell in 1891 for several small values of n; Moore showed how to construct several infinite families of such designs. These are resolvable 2-(4n, 4, 3) designs with extra properties, and are the first examples of nested designs to appear in the literature. Further in this section, Moore points out the relationship between whist tournaments Wh(4n) and resolvable 2-(4n, 4, 1) designs and gives a cyclic construction of such designs in the case where 4n = 3p + 1, where p is a prime of the form 4m + 1.
At this time much work was being done on finite geometries and tactical configurations. For example, Fano described a number of finite geometries in 1892, and in 1906 Veblen and Bussey used finite projective geometries to construct many designs, in particular finite projective planes of all prime power orders.
Triple systems were also studied by Heffter whose approach to the subject stemmed from his study of triangulations of complete graphs on orientable surfaces, relating these to twofold triple systems. In 1891 he gave a particularly simple construction for twofold triple systems of order 12s + 7 . Although Kirkman, nearly 40 years earlier, had pointed to the existence of an indecomposable 2-(7, 3, 3) design without repeated blocks, this seems to be the first known infinite class of designs with λ > 1.
In 1896, Heffter introduced his famous first difference problem, in relation to the construction of cyclic STS(6s + 1), and a year later both the first and second difference problems appeared .
Heffter’s difference problems were eventually solved in 1939 by Peltesohn , and later a particularly elegant approach by Skolem [1924, 1925] was given in 1957– 58. Skolem’s interest in Steiner triple systems was longstanding. In 1927 he wrote notes for the second edition of Netto’s book , whose first edition appeared in 1901; there he constructed STS(v) for all products of primes of the form 6n + 1 and showed that if v = 6s + 1 then there are at least 2s such STS(v). Then in , he introduced the idea of a pure Skolem sequence of order n and proved that these exist if and only if n ≡0, 1 (mod 4). In he extended this idea to that of a hooked Skolem sequence, the existence of which for all admissible n, along with that of pure Skolem sequences, would constitute a complete solution to Heffter’s first difference problem. O’Keefe proved that a hooked Skolem sequence exists if and only if n ≡2, 3 (mod 4).
In 1923–34, Bays [170, 171] undertook a systematic study of group actions on I.2 Design Theory: Antiquity to 1950 17 cyclic STS, and also in on SQS; this culminated in a remarkable theorem, the Bays–Lambossy theorem demonstrating that isomorphism of cyclic structures of prime orders is the same as multiplier equivalence.
The period between the two world wars saw a major step forward in the study of statistical experimental design. Already by 1788, de Palluel had used a 4 × 4 latin square to define an experimental layout for wintering sheep (see for a modern treatment of his work).
In 1924 Knut Vik described the use of certain latin squares in field experiments. But in 1926 , Fisher, at Rothamsted Experimental Station, indicated how orthogonal latin squares could be used in the construction of experiments. This led to detailed work in listing and enumerating all latin squares of small orders; in particular, in 1934 Fisher and Yates (Fisher’s successor at Rothamsted when he left to take up a chair in London) enumerated all 6 × 6 latin squares and established that no two were orthogonal.
Then, in a 1935 paper delivered to the Royal Statistical Society, Yates drew attention to the importance of balanced block designs for statistical design. In Yates called them symmetrical incomplete randomized blocks, the present terminology of balanced incomplete block designs being introduced by Bose in 1939; but the present use of v, b, r, k, and λ goes back to Yates (except that Yates originally used t for treatment instead of v for variety).
It was also in 1935 that Yates gave the first formal treatment of the factorial designs proposed by Fisher in 1926. In 1938, Fisher and Yates published their important book Statistical Tables for Biological, Agricultural and Medical Research which, along with much statistical data, presented what was known about latin squares of small order, including complete sets of orthogonal latin squares of orders 3, 4, 5, 7, 8, and 9, as well as tables of balanced incomplete block designs with replication number r up to 10. Some gaps would be filled by Bose the following year, and some gaps would be ruled out by Fisher’s inequality.
The years 1938–39 saw the publication of several important papers.
Peltesohn gave a complete solution to Heffter’s difference problems and hence established the existence of cyclic Steiner triple systems for all admissible orders with the exception of v = 9. Singer used hyperplanes in PG(2, q) to construct cyclic difference sets, thereby establishing the existence of cyclic finite projective planes (a few of which had been found by Kirkman) for all prime power orders. Singer’s cyclic difference sets gave the first important family of difference sets apart from those implicitly in the work of Paley, although many examples of difference families had been used by Anstice and Netto. In 1942 Bose gave an affine analogue of Singer’s theorem. Here he gave the first examples of relative difference sets and suggested the notion of a group divisible design (GDD). Chowla constructed difference sets using biquadratic residues in 1944. In 1947, Hall published an important paper on cyclic projective planes.
Difference sets were studied in depth after World War II; the important multiplier theorem of Hall and Ryser dates from 1951.
In 1938, there was also the remarkable paper by Witt on Steiner systems, detailing the existence of several Steiner systems with t = 3, including various families obtained geometrically and the Steiner systems S(3, 4, 26) and S(3, 4, 34) constructed by Fitting in 1914. But of greater interest, the paper also includes the systems S(5, 8, 24) and S(5, 6, 12) with the Mathieu groups M24 and M12, respectively, as their automorphism groups. Witt gives a construction of these systems as a single orbit of a starter block under the action of the groups PSL2(23) and PSL2(11), respectively, and proofs of the uniqueness of the designs as well as of their derived subsystems. He also gives a proof of the uniqueness of the system S(3, 5, 17) and of the nonexistence of a system S(4, 6, 18).
The results on the Mathieu systems had already appeared in 1931 by Carmichael [437, 438]. However in 1868 in the Educational Times, Lea both proposed the 18 Design Theory: Antiquity to 1950 I.2 problem and gave a solution to constructing the system S(4, 5, 11). The next year another solution appeared, this time by none other than Kirkman , who went on to try to construct a system S(4, 5, 15). The fact that this system does not exist was not shown until 1972 by Mendelsohn and Hung . Because it is relatively easy to construct the system S(5, 6, 12) from the S(4, 5, 11), it is perhaps a surprise that Kirkman did not do so.
Instead, this was left to Barrau in 1908 who, using purely combinatorial methods, proved the existence and uniqueness of both the systems S(4, 5, 11) and S(5, 6, 12).
The work of Fisher and Yates created growing interest in designs among statisti-cians, and in 1939 Bose , in Calcutta, published a major paper on the subject.
To quote Colbourn and Rosa : Sixty years on, the paper of Bose forms a watershed. Before then, while combinatorial design theory arose with some frequency in other researches, it was an area without a basic theme, and without substantial application. After the paper of Bose, the themes of the area that recur today were clarified, and the importance of constructing designs went far beyond either recreational interest, or as a secondary tactical problem in a larger algebraic or geometric study.
Bose presents the study of balanced incomplete block designs as a coherent theory, using finite fields, finite projective geometries, and difference methods to create many families of designs that contain many old and many new examples. There is much material on designs with λ > 1. The use of difference families, both pure and mixed, carried on the work of Anstice of which he was unaware.
Bose’s paper appeared in volume 9 of the Annals of Eugenics. It is remarkable what combinatorial design theory appeared in that volume of 1939. Stevens discusses the existence of complete sets of orthogonal latin squares; Savur gives (yet) another construction of STSs; and Norton discusses 7×7 latin and graeco-latin squares. Stevens and Norton had presented their work at the 1938 meeting of the British Association in Cambridge, where Youden introduced the experimental designs subsequently known as Youden squares.
These arrays are equivalent to latin rectangles whose columns are the blocks of a symmetric balanced design. That this representation is always possible is a consequence of Philip Hall’s 1935 theorem on systems of distinct representatives (or, equivalently, K¨ onig’s theorem), widely known as the Marriage Problem, which has had many applications in the study of designs.
In 1938, Bose proved that the existence of a complete set of n −1 MOLS of order n is equivalent to the existence of a finite projective plane of order n. Then in 1938, Bose and Nair introduced the ideas of association schemes and partially balanced incomplete block designs. In 1942, Bose gave nontrivial examples of balanced incomplete block designs with repeated blocks. The term association scheme was introduced by Bose and Shimamoto in 1952. Bose moved to the US in 1949, and later with Shrikhande and Parker established the falsity of the Euler conjecture, as mentioned above. This work showed clearly the usefulness of pairwise balanced designs.
While Bose’s 1939 paper exploits the algebra of finite fields, the connection with algebra explored at the time was much broader. In 1877, Cayley [444, 445] discussed the structure of the multiplication table of a group (its Cayley table). Schr¨ oder, from 1890 through 1905, published a two-thousand page treatise on algebras with a binary operation, and extensively discussed quasigroups with various restrictions.
Frolov and Sch¨ onhardt exploited the connection between latin squares and quasigroups. However, only in the 1930s and 1940s did the area take form. In 1935, Moufang studied a specific class of quasigroups, introducing in the process the Moufang loops.
In 1937, Ore and Hausmann first used the term quasigroup in I.2 Design Theory: Antiquity to 1950 19 the sense that we do today, which left a need for a term that conveyed the additional structure of the algebras studied by Moufang. The distinction between quasigroups and loops was made explicit by Albert in 1943 [73, 74], and a general theory of quasigroups was laid by Bruck in 1944 [354, 355]. The field of universal algebra builds on these threads. The important work of Baer [133, 134] in 1939 adopted a geometric view of quasigroups as nets, and the essential connections between orthogonal latin squares and nets ensured that universal algebra and combinatorial designs were on converging paths.
Experimental design continued to generate much of the impetus for the design theory that we know today. Fisher’s inequality that b ≥v in a block design dates from 1940. Orthogonal arrays of strength two were introduced by Plackett and Burman in 1943, and for general strength by Plackett and Rao in 1946. Weighing designs were introduced in 1944 by Hotelling and in 1945 by Kishen . Also in 1945, Finney introduced fractional replication of factorial arrangements.
The applications in experimental design were, as it turned out, indicative of many more applications to come. In 1950, Belevitch considered a problem in conference telephony, and in the process introduced conference matrices.
The connection to digital communication is much more extensive.
In 1948, the new subject of information and coding theory intervened as if from nowhere. However, as MacWilliams wrote in 1968: Our history, of course, begins with Shannon. However, in the pre-Shannon era there was a fairly abundant growth of primeval flora and fauna, resulting in deposits of valuable fuel which we are now beginning to discover.
The fuels are perhaps not the expected ones. In 1949, Shannon and Weaver cite earlier work in statistical physics on entropy, and in the complexity of biological sys-tems, as influential sources. However it came about, Shannon in 1948 created a new field, information theory, in a single stroke. In this seminal paper, Shannon gives a perfect error-correcting code provided to him by Hamming. As well explained in , Hamming had made substantial advances in a theory of error-correcting codes but his discovery was to be tied up in patent disclosure for nearly three years. Mean-while Golay saw Hamming’s example in Shannon’s paper, and in 1949 he published perfect error-correcting codes for p symbols, p a prime . In 1950, the patent issue resolved, Hamming published a much fuller exposition. In 1954, Reed developed linear codes. It may appear that this work evolved in isolation, but quickly connections with experimental design and finite geometry set much of experimental design theory, combinatorial design theory, and coding theory on paths that share much to the current day.
Shannon’s paper on information theory in 1948 launched coding theory into a burst of incredible growth. But Shannon’s next work , in 1949, was to be ap-preciated only much later. He developed a mathematical theory of secrecy systems, a precursor to deep connections between design theory and cryptography that have been extensively developed in the past 20 years. Indeed, many applications in com-munications, cryptography, and networking that continue to drive the field today were all anticipated in the literature of the middle of the last century.
The important theorems of Bruck–Ryser , Chowla–Ryser , and Bose– Connor on the existence of symmetric designs and configurations appeared in 1949, 1950, and 1952, respectively. Of course, many topics in the Handbook were not developed until after 1950, where this history ends. However, it may be appropriate to record the first occurrence of some of these.
In 1943, Bhattacharya showed that 2-(v, 3, 2) designs exist for all v ≡0, 1 20 Design Theory: Antiquity to 1950 I.2 (mod 3), and in 1961, Hanani proved that the admissible conditions for the existence of 2-(v, 3, λ) designs are sufficient for all λ. Necessary and sufficient condi-tions for the existence of directed triple systems were given by Hung and Mendelsohn in 1973 and for Mendelsohn triple systems by Mendelsohn in 1969. Also in , Hanani determined the necessary and sufficient conditions for the existence of BIBDs with block size 4 and any value of λ. In 1972, Hanani did the same for BIBDs with block size 5 with the exception of the nonexistent 2-(15, 5, 2) design, a fact previously discovered by Nandi in 1945. Three years later, Hanani published a survey paper over 100 pages long, containing all of his previous existence results for block sizes 3, 4 and 5; there he extended the results to block size 6 and any λ ≥2, showing sufficiency of the basic necessary conditions for existence except for the nonexistent 2-(21, 6, 2) design. The existence spectrum for block size 6 and λ = 1 remains open.
Notwithstanding the earlier examples, particularly by Anstice, Room squares are so-named after their introduction by Room in 1955.
In 1968, Stanton and Mullin reintroduced the starter-adder method of construction, and the exis-tence problem for Room squares of side 2n + 1 for all n ̸= 3 was finally completed in 1973 . Room squares of sides up to 29 were known by Howell in the 1890s in the context of bridge tournaments where use was also made of a generalization of Room squares, now known as Howell designs. The mathematical study of Howell designs dates from Hung and Mendelsohn in 1974. The first paper on self-orthogonal latin squares (SOLS) was by Stein in 1957. Spouse avoiding mixed doubles round robin tournaments (SAMDRR) were introduced by Brayton, Coppersmith, and Hoffman in 1974 and were shown to be closely related to SOLS. Balanced tour-nament designs were introduced by Haselgrove and Leech in 1977. Early results on lotto designs were given by Hanani, Ornstein, and S´ os in 1964. Generalized polygons were introduced by Tits in 1959 and partial geometries by Bose in 1963. Golay sequences were introduced by Golay in 1961 . Block’s theorem that under the action of an automorphism group, the number of block orbits is at least as large as the number of point orbits, dates from 1967.
In 1973, Wilson proved that given t, k, and v, there exist t-(v, k, λ) designs whenever the admissibility conditions for the design are satisfied and provided λ is sufficiently large. But the designs may have repeated blocks. In 1987, Teirlinck proved that t-designs without repeated blocks exist for all t.
The landmark contributions of Euler in 1782, Kirkman in 1847, Moore in 1896, and Bose in 1939 are central to the development of combinatorial design theory; each in its own way charted a new course for the field. With the comfort of a long historical perspective, these landmarks are easily seen; we leave it to others to write the history of the last five decades with the same benefit of historical perspective. Among the results of more modern times, however, the contributions of Wilson [2145, 2146, 2151] in demonstrating that the elementary necessary conditions for the existence of pairwise balanced designs (and block designs) are asymptotically sufficient stands out. Indeed, the techniques developed by Wilson have placed pairwise balanced designs and their close relatives among the cornerstones of modern design theory.
Many threads are interwoven to form the fabric of the area of combinatorial de-signs explored in this handbook. Combinatorial design theory continues to develop as a subject that is at once pure mathematics, applicable mathematics, and applied mathematics.
I.2.1 A Timeline 21 2.1 A Timeline A very brief list of biographical data about the main contributors mentioned is given; only name, year and place of birth, and year and place of death, are given.
Name Years Birthplace Place of Death Ozanam, Jacques 1640–1717 Bouligneux Paris Choi Seok-Jeong 1646–1715 Korea Korea Euler, Leonhard 1707–1783 Basel St Petersburg Crett´ e de Palluel, Francois 1741–1798 Drancy-les-Noues Dugny, France Steiner, Jakob 1796–1863 Utzenstorf Bern Pl¨ ucker, Julius 1801–1868 Elberfeld Bonn Kirkman, Thomas Penyngton 1806–1895 Bolton Bowden, UK Peirce, Benjamin 1809–1880 Salem MA Cambridge MA Woolhouse, Wesley Stoker Barker 1809–1893 North Shields London Anstice, Robert Richard 1813–1853 Madeley Wigginton, UK Sylvester, James Joseph 1814–1897 London London Cayley, Arthur 1821–1895 Richmond Cambridge Reye, Theodor 1838–1919 Ritzeb¨ uttell W¨ urzburg Petersen, Julius Peter Christian 1839–1910 Sorø Copenhagen Schr¨ oder, Ernst 1841–1902 Mannheim Karlsruhe Lucas, Francois Eduard Anatole 1842–1891 Amiens Paris Tarry, Gaston 1843–1913 Villefranche Le Havre Netto, Eugen 1848–1919 Halle Giessen Mitchell, John Templeton 1854–1914 Glasgow Chicago, IL Martinetti, Vittorio 1859–1936 Scorzalo Milan Eckenstein, Oscar Johannes Ludwig 1859–1921 London Aylesbury Howell, Edwin Cull 1860–1907 Nantucket MA Washington DC Scarpis, Umberto 1861–1921 Padova Bologna Cole, Frank Nelson 1861–1926 Ashland MA New York NY White, Henry Seely 1861–1943 Cazenovia NY Poughkeepsie NY Heffter, Lothar 1862–1962 K¨ oslin Freiburg Moore, Eliakim Hastings 1862–1932 Marietta OH Chicago IL Hadamard, Jacques 1865–1963 Versailles Paris Wernicke, August Ludwig Paul 1866–?
Leipzig St. Louis MO Cummings, Louise Duffield 1870–1947 Hamilton ON ?
Fano, Gino 1871–1952 Mantua Verona Barrau, Johan Anthony 1873–1953 Oisterwijk Utrecht Carmichael, Robert Daniel 1879–1967 Goodwater AL Urbana IL?
Veblen, Oswald 1880–1960 Decorah IA Brooklin ME Bays, Severin 1885–1972 La Joux, Fribourg Fribourg Skolem, Thoralf Albert 1887–1963 Sandsvaer Oslo Fisher, Ronald Aylmer 1890–1962 London Adelaide Youden, William John 1900–1971 Townsville, Aus.
Washington DC Bose, Raj Chandra 1901–1987 Hoshangabad, India Fort Collins CO Yates, Frank 1902–1994 Manchester Harpenden Golay, Marcel J E 1902–1989 Neuchˆ atel Lausanne Baer, Reinhold 1902–1979 Berlin Zurich Hall, Philip 1904–1982 London Cambridge Moufang, Ruth 1905–1977 Darmstadt Frankfurt Mann, Henry Berthold 1905–2000 Vienna Tucson AZ 22 Design Theory: Antiquity to 1950 I.2 Name Years Birthplace Place of Death Paley, Raymond Edward Alan Christopher 1907–1933 Bournemouth BanffAB Chowla, Sarvadaman 1907–1995 London Laramie WY Todd, John Arthur 1908–1994 Liverpool Croydon Hall, Marshall 1910–1990 St. Louis MO London Witt, Ernst 1911–1991 Alsen Hamburg Hanani, Haim 1912–1991 Slupca, Poland Haifa Bruck, Richard Hubert 1914–1991 Pembroke ON Madison WI Hamming, Richard Wesley 1915–1998 Chicago IL Monterey CA Shannon, Claude Elwood 1916–2001 Gaylord MI Medford MA Mendelsohn, Nathan Saul 1917–2006 New York NY Toronto ON Ryser, Herbert John 1923–1985 Milwaukee WI Pasadena CA Parker, Ernest Tilden 1926–1991 Oakland MI Urbana IL Leech, John 1926–1992 Weybridge, UK Firth of Clyde Lu, Jiaxi 1935–1983 Shanghai Baotou, China See Also §VI.34 History of magic squares.
[153, 154] Kirkman triple systems.
A history for triple systems.
Biographies of Cayley, Kirkman, Sylvester from 1916.
The history of loops and quasigroups.
Orthogonal latin squares, early history.
References Cited: [68, 73, 74, 103, 104, 133, 134, 153, 154, 161, 169, 170, 171, 172, 181, 249, 269, 283, 300, 301, 302, 303, 304, 305, 306, 307, 308, 309, 322, 354, 355, 356, 407, 437, 438, 442, 443,444, 445, 446, 501, 503, 578, 593, 618, 634, 659, 690, 703, 715, 769, 798, 807, 817, 818, 819, 820, 821,822, 834, 862, 887, 923, 924, 1003, 1012, 1019, 1020, 1031, 1035, 1036, 1039, 1042, 1046, 1066, 1076, 1077, 1078,1135,1154,1155,1212,1250,1268,1277,1300,1301,1302,1303,1304,1305,1306,1307,1382, 1418,1442,1482,1483,1484,1488,1493,1498,1499,1502,1520,1528,1573,1586,1588,1593,1618, 1624,1625,1626,1627,1640,1644,1653,1656,1664,1670,1671,1686,1695,1701,1712,1713,1726, 1727,1731,1734,1747,1748,1754,1755,1775,1781,1786,1799,1800,1817,1836,1843,1844,1846, 1853,1857,1891,1892,1893,1920,1924,1925,1953,1955,1956,1961,1985,1996,1997,2004,2012, 2015,2030,2034,2036,2081,2097,2100,2129,2135,2142,2145,2146,2147,2149,2151,2159,2167, 2168,2179,2180,2181,2195,2219] Part II Block Designs II.1 2-(v, k, λ) Designs of Small Order 25 1 2-(v, k, λ) Designs of Small Order Rudolf Mathon Alexander Rosa 1.1 Definition and Basics 1.1 A balanced incomplete block design (BIBD) is a pair (V, B) where V is a v-set and B is a collection of b k-subsets of V (blocks) such that each element of V is contained in exactly r blocks and any 2-subset of V is contained in exactly λ blocks. The numbers v, b, r, k, and λ are parameters of the BIBD.
1.2 Proposition Trivial necessary conditions for the existence of a BIBD(v, b, r, k, λ) are (1) vr = bk, and (2) r(k −1) = λ(v −1). Parameter sets that satisfy (1) and (2) are admissible.
1.3 A BIBD (X, D) is a subdesign of a BIBD (V, B) if X ⊆V and D ⊆B. The subdesign is proper if X ⊂V .
1.4 Proposition If a (v, k, λ) design has a proper (w, k, λ) subdesign, then w ≤v−1 k−1.
1.5 Remark The three parameters v, k, and λ determine the remaining two as r = λ(v−1) k−1 and b = vr k . Hence one often writes (v, k, λ) design to denote a BIBD(v, b, r, k, λ). The notation 2-(v, k, λ) design is also used, because BIBDs are t-designs with t = 2. See §II.4. When λ = 1, the notation S(2, k, v) is also employed in the literature, because such BIBDs are Steiner systems. See §II.5. The notations Sλ(2, k, v) or (v, k, λ) BIBD for a (v, k, λ) design are also in common use.
1.6 A BIBD (V, B) with parameters v, b, r, k, λ is complete or full if it is simple and contains v k blocks.
decomposable if B can be partitioned into two nonempty collections B1 and B2 so that (V, Bi) is a (v, k, λi) design for i = 1, 2.
derived (from a symmetric design) (X, D) if for some D ∈D, the collection of blocks {D′ ∩D : D′ ∈D, D ̸= D′} = B.
Hadamard if v = 4n −1, k = 2n −1, and λ = n −1 for some integer n ≥2. See §V.1.
m-multiple if v, b m, r m, k, λ m are the parameters of a BIBD.
nontrivial if 3 ≤k < v.
quasi-symmetric if every two distinct blocks intersect in either µ1 or µ2 elements; the block intersection graph is strongly regular.
See §VI.48 and §VI.11.
residual (of a symmetric design) (X, D) if for some D ∈D, the collection of blocks {D′ \ D : D′ ∈D, D ̸= D′} = B.
resolvable (an RBIBD) if there exists a partition R of its set of blocks B into parallel classes, each of which in turn partitions the set V ; R is a resolution. See §II.7 simple if it has no repeated blocks.
symmetric if v = b, or equivalently k = r. See §II.6.
26 2-(v, k, λ) Designs of Small Order II.1 1.7 The incidence matrix of a BIBD (V, B) with parameters v, b, r, k, λ is a v × b matrix A = (aij), in which aij = 1 when the ith element of V occurs in the jth block of B, and aij = 0 otherwise.
1.8 Theorem If A is the incidence matrix of a (v, k, λ)-design, then AAT = (r −λ)I + λJ and JA = k ˆ J, where I is a v × v identity matrix, J is a v × v all ones matrix, and ˆ J is a v × b all ones matrix. Moreover, any matrix A satisfying these conditions also satisfies λ(v −1) = r(k −1) and bk = vr; when k < v, it is the incidence matrix of a (v, k, λ) design. (See §VII.7.3) 1.9 Theorem (Fisher’s inequality) If a BIBD(v, b, r, k, λ) exists with 2 ≤k < v, then b ≥v.
1.10 Proposition An additional trivial necessary condition for the existence of an RBIBD is (3) k|v. A nontrivial condition is that b ≥v + r −1, a result that extends the inequality in Theorem 1.9. See §II.7.3.
1.11 Two BIBDs (V1, B1), (V2, B2) are isomorphic if there exists a bijection α : V1 →V2 such that B1α = B2. Isomorphism of resolutions of BIBDs is defined similarly. An automorphism is an isomorphism of a design with itself. The set of all automorphisms of a design forms a group, the (full) automorphism group. An automorphism group of the design is any subgroup of the full automorphism group.
1.12 Remark If (V, B) is a BIBD(v, b, r, k, λ) with automorphism group G, the action of G partitions B into classes (orbits). A set of orbit representatives is a set of starter blocks or base blocks. Together with the group G, a set of base blocks can be used to define a design.
1.13 A BIBD (V, B) with parameters v, b, r, k, λ and automorphism group G is cyclic if G contains a cycle of length v.
regular if G contains a subgroup G′ of order v that acts transitively on the elements.
k-rotational if some automorphism has one fixed point and k cycles each of length (v −1)/k.
transitive if for any two elements x and y, there is an automorphism mapping x to y.
1.14 The complement of a design (V, B) is (V, B), where B = {V \B : B ∈B}.
1.15 Proposition The complement of a design with parameters (v, b, r, k, λ) is a design with parameters (v, b, b −r, v −k, b −2r + λ).
1.16 Remarks 1. In view of Proposition 1.15, one usually considers designs with v ≥2k and obtains the rest by taking the complement.
2. Complement has also been used, when (V, B) is a simple 2-(v, b, r, k, λ) design, to denote the 2-(v, v k −b, v−1 k−1 −r, k, v−2 k−2 −λ) design obtained by taking all k-subsets not in B as blocks. The term supplement is used here for this second kind of complementation.
1.2 Small Examples 1.17 Remark To conserve space, designs are displayed in a k×b array in which each column contains the elements (taken from the decimal digits and roman letters) forming a II.1.2 Small Examples 27 block.
For a discussion of how to find the automorphism group of each of these designs, see Remarks VII.6.107.
1.18 Example The unique (6, 3, 2) design and the unique (7, 3, 1) design (see also Example II.6.4).
0000011122 0001123 1123423433 1242534 2345554545 3654656 1.19 Table The four nonisomorphic (7, 3, 2) designs.
1: 00000011112222 2: 00000011112222 11335533443344 11335533443344 22446655666655 22446656565656 3: 00000011112222 4: 00000011112223 11334533453344 11234523453344 22456646565656 23456664565656 1.20 Table The 10 nonisomorphic (7, 3, 3) designs.
1: 000000000111111222222 2: 000000000111111222222 111333555333444333444 111333555333444333444 222444666555666666555 222444666556566566556 3: 000000000111111222222 4: 000000000111111222222 111333455333445333444 111333455333445333444 222445666456566566556 222445666466556556566 5: 000000000111111222222 6: 000000000111111222223 111333445333445333445 111233455233445333444 222456566456566456566 223445666546566566565 7: 000000000111111222223 8: 000000000111111222223 111233455233445333444 111233445233455333444 223445666645566566556 223455666644566566556 9: 000000000111111222223 10: 000000000111111222233 111233445233445333454 111223445223345334544 223456566546566456665 234356566465656456656 1.21 Table The four nonisomorphic (8, 4, 3) designs.
1: 00000001111222 2: 00000001111222 11123342334334 11123342334334 22554666455455 22554666455455 34675777677766 34675777767676 3: 00000001111222 4: 00000001111224 11123342334334 11122332233335 22554566465455 24645454545466 34677677577676 35767767667577 1.22 Example The unique (9, 3, 1) design.
000011122236 134534534547 268787676858 28 2-(v, k, λ) Designs of Small Order II.1 1.23 Table The 36 nonisomorphic (9, 3, 2) designs.
1: 000000001111112222223344 2: 000000001111112222223344 113355773344663344556655 113355773344663344556655 224466885577888866777788 224466885578787866787878 3: 000000001111112222223344 4: 000000001111112222223344 113355673344673344556655 113355773344663344555656 224467885568787867687878 224466885758786768788877 5: 000000001111112222223344 6: 000000001111112222223344 113355773344663344555656 113355673344673344555656 224466885758786867787887 224467885657887868678787 7: 000000001111112222223344 8: 000000001111112222223344 113355673344673344555656 113355673344673344555656 224467885658787867687887 224467885658787867688778 9: 000000001111112222223344 10: 000000001111112222223344 113355673344663344555756 113355773344563344565656 224467885858776767886878 224466885768786857877887 11: 000000001111112222223344 12: 000000001111112222223344 113355673344573344565656 113355673344573344565656 224467885678687856788787 224467885678687856877887 13: 000000001111112222223344 14: 000000001111112222223344 113355673344573344565656 113355673344573344565656 224467885867686758788787 224467885867686758877887 15: 000000001111112222223344 16: 000000001111112222223344 113346673345573344555656 113345673345673344556655 224557884676887868677887 224567884586787867687878 17: 000000001111112222223344 18: 000000001111112222223344 113345673345673344555656 113345673345663344555756 224567884856786778687887 224567884857786778686887 19: 000000001111112222223344 20: 000000001111112222223344 113345673345663344555657 113345673345563344566755 224567884857787867686788 224567884876785678877868 21: 000000001111112222223344 22: 000000001111112222223344 113345673345563344576655 113345673345563344575656 224567884876785867687788 224567884786876857687887 23: 000000001111112222223344 24: 000000001111112222233333 113345673345563344565756 112445672445674455644556 224567884876786758876887 233567883567887868778687 25: 000000001111112222233333 26: 000000001111112222233334 112445672445674455644556 112445672345673455644565 233567883568787867878687 233567884567888768778678 27: 000000001111112222233334 28: 000000001111112222233334 112445672345673455644565 112445672345673455644565 233567884568787868778687 233567884568788767878678 29: 000000001111112222233334 30: 000000001111112222233334 112445672345673455644565 112445672345673455644565 233567884576886878778876 233567884576888678778678 31: 000000001111112222233334 32: 000000001111112222233334 112445672345673455644565 112445672345673455644565 233567884586787768868877 233567884586788768767788 33: 000000001111112222233344 34: 000000001111112222233344 112355672344673345645655 112355672344673345645556 234467885568787868776878 234467885658786778887867 35: 000000001111112222233344 36: 000000001111112222233344 112345672345673345645655 112345672345673345645655 234567885468787876886778 234567885468787886776878 II.1.2 Small Examples 29 1.24 Table The 11 nonisomorphic (9, 4, 3) designs.
1: 000000001111122223 2: 000000001111122223 111233562334433444 111233562334433444 225544777556666555 225544777565656565 346678888787878786 346678888788787786 3: 000000001111122223 4: 000000001111122223 111233462334533444 111233462334533444 225545777456666555 225545777456666555 346768888788778687 346768888877878687 5: 000000001111122223 6: 000000001111122223 111233462334533444 111233462334533444 225545677457666555 225545677457666555 346778888688778687 346788788678878687 7: 000000001111122223 8: 000000001111122223 111233462334533444 111233462334533444 225545777465656565 225545676465757565 346768888788787687 346778888877868876 9: 000000001111122223 10: 000000001111122223 111233462334533444 111233452334533444 225545676465757565 225546677467656565 346788788778868786 346787888578888677 11: 000000001111122234 111223352233533446 246454674745656557 357867886887887768 1.25 Table The three nonisomorphic (10, 4, 2) designs.
1: 000000111122233 2: 000000111122233 3: 000000111122233 112356234534544 112356234534544 112345234534545 244778767658656 244778767658656 246867867647667 356899899899787 356899998889797 357989979858998 1.26 Table The unique (11, 5, 2) design and the unique (13, 4, 1) design.
00000111223 0000111223345 11234236354 1246257364789 24567457465 385a46b57689a 35889898677 9c7ba8cb9cabc 769aaaa99a8 1.27 Table The two nonisomorphic (13, 3, 1) designs.
1: 00000011111222223334445556 2: 00000011111222223334445556 13579b3469a3467867868a7897 13579b3469a3467867868a7897 2468ac578bc95acbbacc9bbac9 2468ac578bc95abcbcac9babc9 30 2-(v, k, λ) Designs of Small Order II.1 1.28 Table The 80 nonisomorphic (15, 3, 1) designs.
1: 00000001111112222223333444455556666 2: 00000001111112222223333444455556666 13579bd3478bc3478bc789a789a789a789a 13579bd3478bc3478bc789a789a789a789a 2468ace569ade65a9edbcdecbeddebcedcb 2468ace569ade65a9edbcdecbededcbdebc 3: 00000001111112222223333444455556666 4: 00000001111112222223333444455556666 13579bd3478bc3478bc789a789a789a789a 13579bd3478bc3478bc789a789a789a789a 2468ace569ade65a9edbcdedebcedcbcbed 2468ace569ade65a9edbcdedbecedcbcebd 5: 00000001111112222223333444455556666 6: 00000001111112222223333444455556666 13579bd3478bc3478bc789a789a789a789a 13579bd3478bc3478bc789a789a789a789a 2468ace569ade65a9edbceddecbcbdeedbc 2468ace569ade65a9edbceddecbcdbeebdc 7: 00000001111112222223333444455556666 8: 00000001111112222223333444455556666 13579bd3478bc3478bc789a789a789a789a 13579bd3478bc34789c78ab789a789a789a 2468ace569ade65a9edbdececbdcedbdbce 2468ace569ade65abedc9deedcbdebcbcde 9: 00000001111112222223333444455556666 10: 00000001111112222223333444455556666 13579bd3478bc34789c78ab789a789a789a 13579bd3478bc34789c78ab789a789a789a 2468ace569ade65abedc9deedbcdecbbcde 2468ace569ade65abedc9deedcbdcbebedc 11: 00000001111112222223333444455556666 12: 00000001111112222223333444455556666 13579bd3478bc34789c789a789a78ab789a 13579bd3478bc34789c789a789a789a78ab 2468ace569ade65abedcdbeecdbd9cebecd 2468ace569ade65abedcdbeecdbbecdd9ce 13: 00000001111112222223333444455556666 14: 00000001111112222223333444455556666 13579bd3478bc34789c789a789a78ab789a 13579bd3478bc34789c789a78ab789a789a 2468ace569ade65abedcebdedcbd9cebcde 2468ace569ade65abedcebdd9ceedcbbcde 15: 00000001111112222223333444455556666 16: 00000001111112222223333444455556666 13579bd3478bc34789a78ac789a789b789a 13579bd3478bc34789a789a78bc789a789a 2468ace569ade65bcdee9bddbecadcecebd 2468ace569ade65bcdeedcba9eddebccbed 17: 00000001111112222223333444455556666 18: 00000001111112222223333444455556666 13579bd3478bc34789a789a789c78ab789a 13579bd3478bc34789a78ac789b789a789a 2468ace569ade65bcdeedcbabedd9cecebd 2468ace569ade65bcede9bdadcedebccbde 19: 00000001111112222223333444455556666 20: 00000001111112222223333444455556666 13579bd3478bc34789a789a78ab789c789a 13579bd3478bc34789a78ac789b789a789a 2468ace569ade65bdceebdcc9deaebddceb 2468ace569ade65bdece9bdacdedbcecebd 21: 00000001111112222223333444455556666 22: 00000001111112222223333444455556666 13579bd3478ac34789b789a789a78ab789c 13579bd3478ac34789b789a789c789a78ab 2468ace569bde65acedbdcecedbd9ceeabd 2468ace569bde65acedbdceeabdcedbd9ce 23: 00000001111112222223333444455566667 24: 00000001111112222223333444455566667 13579bd3478bc34589c789a58ab789789aa 13579bd3478bc34589c789a589a78978aba 2468ace569ade67abedcdbed9cebececdbd 2468ace569ade67abedcdbebecdecdd9ceb 25: 00000001111112222223333444455566667 26: 00000001111112222223333444455566667 13579bd3478bc34589c789a589a78978aba 13579bd3478bc3458ac789a589a789789ab 2468ace569ade67abedcedbecbdbdcd9cee 2468ace569ade67b9edcbededbcacddecbe 27: 00000001111112222223333444455566667 28: 00000001111112222223333444455566667 13579bd3478bc34589a78ab589c789789aa 13579bd3478bc34589a78ab589c789789aa 2468ace569ade67bcded9ceaebdcdeebcdb 2468ace569ade67bcedd9ceaebdedccbdeb 29: 00000001111112222223333444455566667 30: 00000001111112222223333444455566667 13579bd3478bc34589a789a58bc789789aa 13579bd3478bc34589a78ab589c789789aa 2468ace569ade67bcedebdca9eddeccdbeb 2468ace569ade67bdced9ceabedcedecbdb 31: 00000001111112222223333444455566667 32: 00000001111112222223333444455566667 13579bd3478bc34589a789a58bc789789aa 13579bd3478bc34589a78ab589c789789aa 2468ace569ade67bdcedbeca9edcedecbdb 2468ace569ade67bdcec9deaebddceebdcb 33: 00000001111112222223333444455566667 34: 00000001111112222223333444455566667 13579bd3478bc34589a78ab589c789789aa 13579bd3478bc34589a789a589c78978baa 2468ace569ade67bdecc9deaebdecddbceb 2468ace569ade67bdecdbceaebdecdc9edb 35: 00000001111112222223333444455566678 36: 00000001111112222223333444455566678 13579bd3478bc34569a689a579a78a789bc 13579bd3478bc34569a689a579a78a789bc 2468ace569ade78bcdedbecedcbc9daebed 2468ace569ade78bceddbceecdbd9caebed II.1.2 Small Examples 31 37: 00000001111112222223333444455566678 38: 00000001111112222223333444455566678 13579bd3478bc34569a68ab579c78978aa9 13579bd3478bc34569a68ab579a78978ac9 2468ace569ade78bced9dceaebdceddbebc 2468ace569ade78bdce9cdecedbadebecdb 39: 00000001111112222223333444455566678 40: 00000001111112222223333444455566678 13579bd3478bc34569a68ab579c78978aa9 13579bd3478bc34569a689a579a78a789bc 2468ace569ade78bdce9dceabedecdcebdb 2468ace569ade78bdcecbedecdbd9caebed 41: 00000001111112222223333444455566678 42: 00000001111112222223333444455566678 13579bd3478bc34569a689c57ab78a789a9 13579bd3478bc34569a68ab579c78978aa9 2468ace569ade78bdceabed9dceecdcebbd 2468ace569ade78bdec9cdeaebddeccbebd 43: 00000001111112222223333444455566678 44: 00000001111112222223333444455566678 13579bd3478bc34569a68ac579b78978aa9 13579bd3478bc34569a689a579a78a789bc 2468ace569ade78bdec9ebdacdeedcbcedb 2468ace569ade78bdeccbdeedcbc9daebed 45: 00000001111112222223333444455566678 46: 00000001111112222223333444455566678 13579bd3478bc34569a68ac579a78978ab9 13579bd3478bc34569a68ab579c78978aa9 2468ace569ade78becd9bededbcacdcdbee 2468ace569ade78becd9dceabedceddcbeb 47: 00000001111112222223333444455566678 48: 00000001111112222223333444455566678 13579bd3478bc34569a68ab579c78978aa9 13579bd3478bc34568a69ab578c78979aa9 2468ace569ade78bedc9cdeabeddeccdbeb 2468ace569ade79becd8dceabedcdedcbeb 49: 00000001111112222223333444455566679 50: 00000001111112222223333444455566678 13579bd3478bc34568a689a578a78978abc 13579bd3478bc34568a69ac578a78979ab9 2468ace569ade79becddebccdbeadec9bed 2468ace569ade79becd8bededbcadccdbee 51: 00000001111112222223333444455566678 52: 00000001111112222223333444455566678 13579bd3478bc34568968ab579c79a78aa9 13579bd3478bc345689689a579c79a78aab 2468ace569ade7abecd9dce8ebddcecbdbe 2468ace569ade7abecdcdeb8ebddceb9dce 53: 00000001111112222223333444455566678 54: 00000001111112222223333444455566679 13579bd3478bc345689689a579c79a78aab 13579bd3478bc345689689a578c78a78aab 2468ace569ade7abecdbdec8ebddcec9dbe 2468ace569ade7abecdbecd9ebdcded9cbe 55: 00000001111112222223333444455566678 56: 00000001111112222223333444455566678 13579bd3478bc34568969ab578c78a79aa9 13579bd3478bc345689689a579c79a78aab 2468ace569ade7abecd8cde9ebdcdedbcbe 2468ace569ade7abedcdceb8ebdcdeb9cde 57: 00000001111112222223333444455566678 58: 00000001111112222223333444455566678 13579bd3478bc34568968ab579c79a78aa9 13579bd3478bc345689689a579c79a78aab 2468ace569ade7abedc9cde8ebdcdedbcbe 2468ace569ade7abedcbced8ebdcded9cbe 59: 0000000111111222222333344445556667a 60: 0000000111111222222333344445556667a 13579bd3478bc345689689a578a789789cb 13579bd3478bc34568968a9578a789789cb 2468ace569ade7ebacdcbeddb9caecedbde 2468ace569ade7ebacddecbcb9dadeebcde 61: 00000001111112222223333444455556666 62: 00000001111112222223333444455566667 13579bd3478ac34789b789c789a78ab789a 13579bd3478ac34589b78ab589a789789ca 2468ace569bde65aecdbaedecdbd9cecdbe 2468ace569bde67adcec9edbedcdceeabdb 63: 00000001111112222223333444455566667 64: 00000001111112222223333444455566667 13579bd3478ac34589b78ab589a789789ca 13579bd3478ac34589a789a589b78b789ca 2468ace569bde67aecdd9cebcdecdeeabdb 2468ace569bde67cdbeecdbaecdd9ebaedc 65: 00000001111112222223333444455566678 66: 00000001111112222223333444455566678 13579bd3478ac34569b68ac579a789789ba 13579bd3478ac34568968ac579a79b789ba 2468ace569bde78dacee9bdbedcacecdbde 2468ace569bde7bdacee9bd8edcacecdbde 67: 00000001111112222223333444455566679 68: 00000001111112222223333444455566678 13579bd3478ac34568a68ac578b789789ab 13579bd3478ac34569b68ac579a789789ba 2468ace569bde79cbdee9bdaecdbeddacce 2468ace569bde78adcee9bdbedccdeacbde 69: 00000001111112222223333444455566679 70: 0000000111111222222333344445556667a 13579bd3478ac3456a868ac578b789789ab 13579bd3478ac34568968ab5789789789cb 2468ace569bde79bcede9bdaecddecbadce 2468ace569bde7dbaece9cdceabadebcdde 71: 0000000111111222222333344445556667a 72: 0000000111111222222333344445556667a 13579bd3478ac34568968ab5789789789cb 13579bd3478ac34568968ab5789789789cb 2468ace569bde7cabdee9cddeabbecacdde 2468ace569bde7dcbaee9cdaecbbedadcde 32 2-(v, k, λ) Designs of Small Order II.1 73: 00000001111112222223333444455566679 74: 0000000111111222222333344445556667a 13579bd3478ac34568a689a578b78c789ab 13579bd3478ac345698689a578b789789cb 2468ace569bde7c9bdeecdbae9dbeddacce 2468ace569bde7abdecedbcce9daedbacde 75: 0000000111111222222333344445556667a 76: 00000001111112222223333444455566678 13579bd3478ac345689689a578b789789cb 13579bd3478ac34569b68ab578978979aac 2468ace569bde7cdbaeedbcae9dbecacdde 2468ace569bde7da8cee9cdcbaedebcdbed 77: 00000001111112222223333444455566678 78: 00000001111112222223333444455566678 13579bd3478ac34569a68ab578979b789ac 13579bd3478ac34569b689c578b78a79aa9 2468ace569bde7d8cebe9cdacebdcebaded 2468ace569bde7ac8edeabd9cdedebbdcec 79: 00000001111112222223333444455566678 80: 00000001111112222223333444455566678 13579bd3478ac34568b689c57ab79a789a9 13579bd3469ac34578b678a58ab78979c9a 2468ace569bde79ecadaebd8dcecdbbdeec 2468ace578bde96aecdbcded9cebecaeddb 1.29 Table Properties of the 80 STS(15)s ((15,3,1) BIBDs). |G| is the order of the au-tomorphism group. CI is the chromatic index, the minimum number of colors with which the blocks can be colored so that no two intersecting blocks receive the same color; when CI = 7, the design is resolvable. PC is the number of parallel classes.
Sub is the number of (7,3,1)-subdesigns, and Pa is the number of Pasch configurations (four triples on six points).
|G| CI PC Sub Pa # |G| CI PC Sub Pa # |G| CI PC Sub Pa 1 20160 7 56 15 105 2 192 8 24 7 73 3 96 9 8 3 57 4 8 9 8 3 49 5 32 8 16 3 49 6 24 8 12 3 37 7 288 7 32 3 33 8 4 9 4 1 37 9 2 9 2 1 31 10 2 9 6 1 31 11 2 9 6 1 23 12 3 9 1 1 32 13 8 9 4 1 33 14 12 9 0 1 37 15 4 8 8 1 25 16 168 9 0 1 49 17 24 8 12 1 25 18 4 9 4 1 25 19 12 7 16 1 17 20 3 9 1 1 20 21 3 9 1 1 20 22 3 8 4 1 17 23 1 9 1 0 18 24 1 9 0 0 19 25 1 9 1 0 20 26 1 9 0 0 23 27 1 9 3 0 14 28 1 9 2 0 15 29 3 9 0 0 19 30 2 9 3 0 14 31 4 9 5 0 18 32 1 9 2 0 13 33 1 9 1 0 12 34 1 9 1 0 12 35 3 9 0 0 13 36 4 9 1 0 10 37 12 9 5 0 6 38 1 9 4 0 9 39 1 9 1 0 12 40 1 9 0 0 13 41 1 9 1 0 12 42 2 9 5 0 8 43 6 9 3 0 10 44 2 9 1 0 8 45 1 9 2 0 9 46 1 9 2 0 7 47 1 9 1 0 10 48 1 9 1 0 8 49 1 9 2 0 7 50 1 8 7 0 6 51 1 9 2 0 9 52 1 9 0 0 9 53 1 9 1 0 10 54 1 9 2 0 11 55 1 9 2 0 9 56 1 9 1 0 8 57 1 8 4 0 5 58 1 9 3 0 8 59 3 9 0 0 13 60 1 9 6 0 7 61 21 7 7 1 14 62 3 9 0 0 7 63 3 8 6 0 7 64 3 9 3 0 10 65 1 9 2 0 7 66 1 9 3 0 6 67 1 8 4 0 5 68 1 9 1 0 6 69 1 8 4 0 5 70 1 9 2 0 9 71 1 9 2 0 5 72 1 9 4 0 5 73 4 9 9 0 6 74 4 9 3 0 8 75 3 8 6 0 7 76 5 9 1 0 10 77 3 9 1 0 2 78 4 9 9 0 6 79 36 8 17 0 6 80 60 9 11 0 0 II.1.2 Small Examples 33 1.30 Table The five nonisomorphic (15, 7, 3) designs.
1: 000000011112222 2: 000000011112222 3: 000000011112222 111335533443344 111335533443344 111335533443344 222446655666655 222446655666655 222446655666655 37b797978787878 37b797978787878 37b797978787878 48c8a8a9a9aa9a9 48c8a8a9a9aa9a9 48c8a8a9aa99aa9 59dbddbbccbbccb 59dbddbbccbcbbc 59dbddbbccbcbbc 6aeceecdeededde 6aeceecdeeddeed 6aeceecdedeeded 4: 000000011112222 5: 000000011112222 111335533443344 111335533443344 222446655666655 222446656565656 37b797879787878 37b797977888877 48c8a9a8aa99aa9 48c8a8a9aa99aa9 59dbdbcdbcbcbbc 59dbddbbccbcbbc 6aecedeecdeeded 6aeceecdedeeded 1.31 Example The unique (16, 4, 1) design.
00000111122223333456 147ad456945684567897 258be7b8cc79a98abbac 369cfadefefbddcfefed 1.32 Table The three nonisomorphic (16, 6, 2) designs.
1: 0000001111222334 2: 0000001111222334 3: 0000001111222334 1123452345345455 1123452345345455 1123452345345455 2667896789877666 2667896789877666 2667896789877666 37aabcdbaa998987 37aabcdbaa998987 37aabcdbaa998987 48bddeeccbabccba 48bddeeccbacbbca 48bddeeccbcabbac 59ceffffedfeddef 59ceffffedfdeedf 59ceffffeddfeefd 1.33 Table The six nonisomorphic (19, 9, 4) designs.
1: 0000000001111122223 2: 0000000001111122223 1111233562334433444 1111233562334433444 2225544777556666555 2225544777556666555 3346678888787878786 3346678888787878786 499acab99a9b9aa9b99 499acab99a9ba99ab99 5aebdcdabddcbcbccaa 5aebdcdabcdcbcbdcaa 6bfegefcdefecdfedbd 6bfegefdcffeddeedbc 7cgfhghfehgghfgfege 7cgfhghefhgggehfegf 8dhiiiihgiihighifih 8dhiiiihgiihhiigfih 3: 0000000001111122223 4: 0000000001111122223 1111233562334433444 1111233562334433444 2225544777565656565 2225544777565656565 3346678888788787786 3346678888788787786 499acab99a99abba999 499acab99a9ab99ba99 5aebdcdabbcdcccdbaa 5aebdcdabdbccccdbaa 6bfegefdcgfeddeedcb 6bfegefdcegfddeecbd 7cgfhghfehhgefffefg 7cgfhghfehhgeffffge 8dhiiiighiiihgghiih 8dhiiiighiihgiighih 34 2-(v, k, λ) Designs of Small Order II.1 5: 0000000001111122223 6: 0000000001111122223 1111233462334533444 1111233452334533444 2225545777456666555 2225546677467656565 3346768888788778687 3346787888578888677 499aca9a9abb999ab99 499aaab99bc9a99ab99 5aebdcdbbcdccabdcaa 5aebcddbcddcbabccaa 6bfeegfdcffeddeedbc 6bfegefcdefecdfeddb 7cgfghhefhggeghffge 7cgfhgghehhgfggfefe 8dhiiiihgiihihighif 8dhiiihifiiighihgih 1.34 Table The 18 nonisomorphic (25, 4, 1) designs. Their respective automorphism group orders are: 504,63,9,9,9,150,21,6,3,3,3,3,3,3,3,3,1,1.
1: 00000000111111122222223333344445555666778899aabbil 134567ce34578cd34568de468bh679f78ag79b9aabcddecejm 298dfbhkea6g9kf7c9afkg5cgfihdgifchi8ejjcjdfhgfghkn iaolgmjnmbohnljonblhmjjdlknmeklnekmkinlimimonooloo 2: 0000000011111112222222333334444555566667778889abil 13457bce34589cd3456ade489eh6acf7bdg79ab9ab9abdecjm 2689gdfka76fekg798fckh5cfgidghichfi8chjjdfgjehfgkn iloahnjmbmohlnjobngmljjdknmeklnekmlkinmnilmliooooo 3: 0000000011111112222222333334444555566667778889abil 13457bde34589ce3456acd489eh6acf7bdg79ab9ab9abcdejm 2689gcfka76fdkg798fehk5cgfidhgicfhi8hcjjfdejgfghkn iloahmjnbmohnljobngljmjdkmneknleklmkminlniimlooooo 4: 0000000011111112222222333334444555566667778889abil 13457aef3458bcf34569dg489dh69ef7acg79bc9acabdcdejm 2689bcjha769djg7b8aejh5cbfkdagkebhk8gieihdifefghkn ilodgknmemohklnocnfkmljgminhnilflimkjnmljnmjlooooo 5: 00000000111111122222223333344445555666778899abcfil 134567cd34578de34568ce468ag67bh789f79aab9bacdedgjm 298abhgkba69fhk79bgakf5cdfiedgicehi8djejjcbfghehkn ieolfmjnmcognjlondlhmjjhlknmfklngkmkinilmiolmnoooo 6: 000000001111111222222333334444555566667778899abcde 123468cj23479af3458bg456ah57bi78bd89ce9adabacghiff 5ad97fek6be8gdl7ch9em89fcn6gdkfickgjdlhmeekblhijjg oignbhmlkjhcinmlfjdonmeikoajlonlgmomhnkoijnfolmnok 7: 000000001111111222222333333444445555566666777889al 124789bh2589ace39abde47abcf8bcde79cdf78adg89b9cabm 365egifj46fhjgi5gikhf6ihjegjikfggkjehfhekiadcbdcdn dcaolmnk7bolmnk8olmnj9nolmknolmhmnolilmnojkjheifgo 8: 000000001111111222222333333444455566667777889aacee 12459bdf24569cg3458bh4568bi59bh9ac78ab89cgaddbedfl 3786cihjd78haek96kfgja7fgclgkdiimfi9djebkjcjffhmgn oanegklmifbmljnecolmnjdkhnmlmeojnhnoglmhloiknokoio 9: 000000001111111222222233333444455556677889abcdefgh 13456789345678a34567894679c67ad68be9aab9bdecijkijk 2fbcdgiadg9jcfbaehficb5b8eg89ch7adfchdfgefghmlllmn onklehjmmlikehnjnmgkdloilhkmjfinkgjolomnokijnnmooo 10: 000000001111111222222233333444455566577889abccdfgh 13456789345678a345678b467ad679e69c9b89aabfghdeeijk 2jadebchekbdc9f9ciaedg5f8bg8gbf7afcihjdiemllikjlmn onmkgfliilnmhgjljmhnfkokmchnidhlegojjkokonnmnmlooo II.1.3 Parameter Tables 35 11: 0000000011111112222222333334444555566778899acdefgh 1345678b3456789345678a467bc679d68ae9daebcabbkijijk 2iace9fgdjbgcah9cibhdf5a8fg8bfh79gfcfdgehedclmllmn ojmdhnklekniflmlekmjgnokmhnniglljhmojokoiikjmnnooo 12: 0000000011111112222222333334444555566778899accdfgh 1345678a345678b34567894679e67ac68bd9fagbfabbdeeijk 2j9ebcmickal9dibdielaj5a8df8beg79chcgdhehgfhjiklmn olhgdfnkhmfnegjgfnhmckomkininjljlkmokoiojnmlnmlooo 13: 0000000011111112222222333334444555566677899accdfgh 134578ae34568bc345679d467ad78be689c7898ababbdeeijk 2f6b9cjg9g7dakh8ahbeif5lbigl9ifamjffhcgdegfhikjlmn omkdhlnienimfljjclgnmkonckhmdjhenkgjiokoolnmnmlooo 14: 0000000011111112222222333334444555566677899accdfgh 1345789c34568ad34567be467ad78be689c7898ababbdeeijk 2h6beajg9f7bckh8agd9if5lbifl9jgamiffhcgdegfhjiklmn omidnflkenjglmikclmhnjoncjgmdkhenkhkjoioonmlnmlooo 15: 0000000011111112222222333334444555566677889abcdeil 1345689b345679a34578ab467ad789e689c7bc9daeghffghjm 2cg7afdi8dhfbjcf6eg9kc59ebhacbfdbag8ehcfdgijkijkkn lnkjohemknigomejinhomdmligoljhoklfonjmkmimnnnllloo 16: 0000000011111112222222333334444555566677889abcdeil 134568ab345679b345789a467ae789c689d7be9cadfghfghjm 2cf7g9dk8dgafich6ebfcj59dbgaebhcbaf8dfegchkijijkkn lnjihoemjnkohmeiknogdmmlkfoligojlhonimjmkmnnnllloo 17: 0000000011111112222222333333344445555666677778889a 147adgjm4569chi4569bdf45689ae9bcf9ach89abbcdeabicd 258behkn78ebfkl87caekgdb7cfglgkehefgjfildfighegjdj 369cfiloadgjmnonlohimjimjhoknmojlkinoknmhnkomolmln 18: 0000000011111112222222333333344445555666677778889a 147adgjm4569cef4569bcd45689ae9bdf9abe89acbcdiabcfd 258behkn78bhkil87jafghhc7jbigcekggfihelhdheglgkfij 369cfiloadgjmnoikoemnlmlfndokolnjmnjomnkinjomlohkm 1.3 Parameter Tables 1.35 Table Admissible parameter sets of nontrivial BIBDs with r ≤41 and k ≤v/2. Earlier listings of BIBDs by Hall , Takeuchi , and Kageyama and papers by Hanani and Wilson are frequently referenced. Multiples of known designs are included; although their existence is trivially implied, information concerning their number and resolvability usually is not.
The admissible parameter sets of nontrivial BIBDs satisfying r ≤41, 3 ≤k ≤v/2 and conditions (1) and (2) of Proposition 1.2 are ordered lexicographically by r, k, and λ (in this order).
The column “Nd” contains the number Nd(v, b, r, k, λ) of pairwise nonisomorphic BIBD(v, b, r, k, λ) or the best known lower bound for this number. The column “Nr” contains a dash (-) if condition (3) of Proposition 1.10 is not satisfied.
Otherwise it contains the number Nr of pairwise nonisomorphic resolutions of BIBD(v, b, r, k, λ)’s or the best known lower bound. The number of nonisomorphic RBIBDs is not necessarily Nr. Indeed, there are seven nonisomorphic resolutions of BIBD(15,35,7,3,1)s but only four nonisomorphic RBIBD(15,35,7,3,1)s (see Example II.2.76). The symbol ? indicates that the existence of the corresponding BIBD (RBIBD, respectively) is in doubt. The meanings of the “Comments” are: 36 2-(v, k, λ) Designs of Small Order II.1 m#x m-multiple of an existing BIBD #x m#x m-multiple of #x that does not exist or whose existence is un-decided R#x (D#x) residual (derived) design of #x that exists R#x (D#x) residual (derived) design of #x that does not exist or whose existence is undecided #x+#y union of two designs on the same set of elements #x↓#y design #x is a design (V, B) with parameters (v, b, r, k, λ); design #y is a design (V, D) with parameters (v, b′, b −r, k + 1, r −λ); add a new point ∞to each block of B to obtain b B; then (V, b B∪D) is a design with parameters (v + 1, b + b′, b, k + 1, r).
PG (AG) projective (affine) geometry (see §VII.2) ×1 BIBD does not exist by Bruck–Ryser–Chowla (BRC) Theorem (see §II.6.2) ×2 BIBD is a residual of a BIBD that does not exist by the BRC theorem, and λ = 1 or 2 ×3 RBIBD does not exist by Bose’s condition (see Theorem II.7.28).
HD RBIBD(4t, 8t −2, 4t −1, 2t, 2t −1) exists from a symmetric (Hadamard) BIBD(4t −1, 4t −1, 2t −1, 2t −1, t −1); see §V.1.
Typically no references are given under “Ref” for multiple, derived, or residual designs of known BIBDs. A trivial formula giving Nd(v, mb, mr, k, mλ) ≥n + 1 is often used provided Nd(v, b, r, k, λ) ≥n, m ≥2, n ≥1 (similarly for Nr). The column “Where?” gives a pointer to an explicit construction.
No v b r k λ Nd Nr Comments, Ref Where?
1 7 7 3 3 1 1 - PG(2,2) II.6.4 2 9 12 4 3 1 1 1 R#3,AG(2,3) 1.22 3 13 13 4 4 1 1 - PG(2,3) 1.26 4 6 10 5 3 2 1 0 R#7,×3 1.18 5 16 20 5 4 1 1 1 R#6,AG(2,4) 1.31 6 21 21 5 5 1 1 - PG(2,4) VI.18.73 7 11 11 5 5 2 1 -1.26 8 13 26 6 3 1 2 - 1.27 9 7 14 6 3 2 4 - 2#1,D#20 1.19 10 10 15 6 4 2 3 - R#13 1.25 11 25 30 6 5 1 1 1 R#12,AG(2,5) 12 31 31 6 6 1 1 - PG(2,5) VI.18.73 13 16 16 6 6 2 3 - 1.32 14 15 35 7 3 1 80 7 PG(3,2) [1241, 1544] 1.28 15 8 14 7 4 3 4 1 R#20,AG2(3,2) [1241, 898] 1.21 16 15 21 7 5 2 0 0 R#19,×2 17 36 42 7 6 1 0 0 R#18,×2,AG(2,6) 18 43 43 7 7 1 0 - ×1,PG(2,6) 19 22 22 7 7 2 0 - ×1 20 15 15 7 7 3 5 - PG2(3, 2) 1.30 21 9 24 8 3 2 36 9 2#2,D#40 1.23 22 25 50 8 4 1 18 - [1339, 1945] 1.34 23 13 26 8 4 2 2461 - 2#3 24 9 18 8 4 3 11 - D#41 1.24 25 21 28 8 6 2 0 - R#28,×2 26 49 56 8 7 1 1 1 R#27,AG(2,7) 27 57 57 8 8 1 1 - PG(2,7) VI.18.73 28 29 29 8 8 2 0 - ×1 29 19 57 9 3 1 11084874829 - VI.16.12 30 10 30 9 3 2 960 - D#54 [537, 856, 1174] VI.16.81 II.1.3 Parameter Tables 37 No v b r k λ Nd Nr Comments, Ref Where?
31 7 21 9 3 3 10 - 3#1 1.20 32 28 63 9 4 1 ≥4747 ≥7 [1346, 1533, 1548] III.1.8 33 10 18 9 5 4 21 0 R#41,×3 VI.16.85 34 46 69 9 6 1 0 - 35 16 24 9 6 3 18920 - R#40 II.6.30 36 28 36 9 7 2 8 0 R#39,×3 37 64 72 9 8 1 1 1 R#38,AG(2,8) 38 73 73 9 9 1 1 - PG(2,8) VI.18.73 39 37 37 9 9 2 4 - II.6.47 40 25 25 9 9 3 78 - II.6.47 41 19 19 9 9 4 6 - 1.33 42 21 70 10 3 1 ≥62336617 ≥63745 [518, 1263] VI.16.12 43 6 20 10 3 4 4 1 2#4 [1241, 976] 44 16 40 10 4 2 ≥2.2 · 106 339592 2#5 [696, 1267] 45 41 82 10 5 1 ≥15 - VI.16.16 46 21 42 10 5 2 ≥22998 - 2#6 47 11 22 10 5 4 4393 - 2#7,D#63 48 51 85 10 6 1 ?
-49 21 30 10 7 3 3809 0 R#54,×3 [1016, 1241, 1946] 50 36 45 10 8 2 0 - R#53,×2 51 81 90 10 9 1 7 7 R#52,AG(2,9) [679, 1373] 52 91 91 10 10 1 4 - PG(2,9) [679, 1373] VI.18.73 53 46 46 10 10 2 0 - ×1 54 31 31 10 10 3 151 - II.6.47 55 12 44 11 3 2 242995846 74700 D#84 VI.16.81 56 12 33 11 4 3 ≥17172470 5 D#85 [1632, 1938] VI.16.83 57 45 99 11 5 1 ≥16 ? VI.16.31 58 12 22 11 6 5 11603 1 R#63,HD [1016, 1241, 1743] 59 45 55 11 9 2 ≥16 0 R#62,×3 [1241, 692] 60 100 110 11 10 1 0 0 R#61,AG(2,10) 61 111 111 11 11 1 0 - PG(2,10) 62 56 56 11 11 2 ≥5 - II.6.47 63 23 23 11 11 5 1106 - [1016, 1172, 1943] VI.18.73 64 25 100 12 3 1 ≥1014 - VI.16.12 65 13 52 12 3 2 ≥1897386 - 2#8,D#96 66 9 36 12 3 3 22521 426 3#2 [1543, 1707] 67 7 28 12 3 4 35 - 4#1 68 37 111 12 4 1 ≥51402 - [587, 1349] VI.16.14 69 19 57 12 4 2 ≥423 - VI.16.15 70 13 39 12 4 3 ≥3702 - 3#3,D#97 [1548, 1940] 71 10 30 12 4 4 13769944 - 2#10 72 25 60 12 5 2 ≥118884 ≥748 2#11 [1225, 2062] 73 61 122 12 6 1 ?
-74 31 62 12 6 2 ≥72 - 2#12 75 21 42 12 6 3 ≥236 - VI.16.18 76 16 32 12 6 4 ≥111 - 2#13 77 13 26 12 6 5 19072802 - D#98 [1016, 1267] 78 22 33 12 8 4 0 - R#85 79 33 44 12 9 3 ≥3375 - R#84 [1551, 1999] 80 55 66 12 10 2 0 - R#83,×2 81 121 132 12 11 1 ≥1 ≥1 R#82,AG(2,11) 82 133 133 12 12 1 ≥1 - PG(2,11) VI.18.73 83 67 67 12 12 2 0 - ×1 84 45 45 12 12 3 ≥3752 - VI.18.73 85 34 34 12 12 4 0 - ×1 38 2-(v, k, λ) Designs of Small Order II.1 No v b r k λ Nd Nr Comments, Ref Where?
86 27 117 13 3 1 ≥1011 ≥1.4 · 1013 AG(3,3) [1546, 2148] VI.16.12 87 40 130 13 4 1 ≥106 ≥2 PG(3,3) VI.16.14 88 66 143 13 6 1 ≥1 ? II.3.32 89 14 26 13 7 6 15111019 0 R#98,×3 [1267, 1999] 90 27 39 13 9 4 ≥2.45 · 108 68 R#97,AG2(3, 3) [1241, 1380, 1381] 91 40 52 13 10 3 ?
0 R#96,×3 92 66 78 13 11 2 ≥2 0 R#95,×3 [1241, 114] 93 144 156 13 12 1 ?
? R#94,AG(2,12) 94 157 157 13 13 1 ?
- PG(2,12) 95 79 79 13 13 2 ≥2 - II.6.47 96 53 53 13 13 3 0 - ×1 97 40 40 13 13 4 ≥1108800 - PG2(3, 3) VI.18.73 98 27 27 13 13 6 208310 - VI.18.73 99 15 70 14 3 2 ≥685521 ≥36 2#14,D#140 [734, 1548] 100 22 77 14 4 2 ≥7921 - VI.16.15 101 8 28 14 4 6 2310 4 2#15 102 15 42 14 5 4 ≥896 0 2#16,D#141 [1265, 2031] VI.16.85 103 36 84 14 6 2 ≥5 ≥2 2#17 [1228, 2130] VI.16.86 104 15 35 14 6 5 ≥117 - D#142 [1016, 1533] VI.16.86 105 85 170 14 7 1 ?
-106 43 86 14 7 2 ≥4 - 2#18 VI.16.30 107 29 58 14 7 3 ≥1 -VI.16.30 108 22 44 14 7 4 ≥3393 - 2#19 VI.16.30 109 15 30 14 7 6 ≥57810 - 2#20,D#143 110 78 91 14 12 2 0 - R#113,×2 111 169 182 14 13 1 ≥1 ≥1 R#112,AG(2,13) 112 183 183 14 14 1 ≥1 - PG(2,13) VI.18.73 113 92 92 14 14 2 0 - ×1 114 31 155 15 3 1 ≥6 · 1016 - VI.16.12 115 16 80 15 3 2 ≥1013 - D#169 VI.16.13 116 11 55 15 3 3 ≥436800 - VI.16.24 117 7 35 15 3 5 109 - 5#1 118 6 30 15 3 6 6 0 3#4 [976, 1548] 119 16 60 15 4 3 ≥6 · 105 ≥6 · 105 3#5,D#170 120 61 183 15 5 1 ≥10 - VI.16.16 121 31 93 15 5 2 ≥1 -VI.16.17 122 21 63 15 5 3 ≥109 - 3#6 123 16 48 15 5 4 ≥294 - D#171 [352, 734, 1016] 124 13 39 15 5 5 ≥76 - [1016, 2059, 734] VI.16.17 125 11 33 15 5 6 ≥127 - 3#7 [324, 1224] 126 76 190 15 6 1 ≥1 - II.3.32 127 26 65 15 6 3 ≥1 - VI.16.89 128 16 40 15 6 5 ≥25 - D#172 [1016, 1533, 734] 129 91 195 15 7 1 ≥2 ? VI.16.70 130 16 30 15 8 7 ≥9 · 107 5 R#143,AG3(4, 2),HD [1271, 1319] 131 21 35 15 9 6 ≥104 - R#142 [1016, 404] 132 136 204 15 10 1 ?
-133 46 69 15 10 3 ?
-134 28 42 15 10 5 ≥3 - R#141 135 56 70 15 12 3 ≥4 - R#140 136 91 105 15 13 2 0 0 R#139,×2 137 196 210 15 14 1 0 0 R#138,×2,AG(2,14) 138 211 211 15 15 1 0 - ×1,PG(2,14) 139 106 106 15 15 2 0 - ×1 140 71 71 15 15 3 ≥72 - [1005, 1832] II.6.47 II.1.3 Parameter Tables 39 No v b r k λ Nd Nr Comments, Ref Where?
141 43 43 15 15 5 0 - ×1 142 36 36 15 15 6 ≥25634 - VI.18.73 143 31 31 15 15 7 ≥22478260 - PG3(4, 2) VI.18.80 144 33 176 16 3 1 ≥1013 ≥4494390 [1548, 562] VI.16.12 145 9 48 16 3 4 16585031 149041 4#2 146 49 196 16 4 1 ≥769 - [396, 978, 587] VI.16.14 147 25 100 16 4 2 ≥17 - 2#22 148 17 68 16 4 3 ≥542 - D#185 [1999, 734] 149 13 52 16 4 4 ≥2462 - 4#3 150 9 36 16 4 6 270474142 - 2#24 151 65 208 16 5 1 ≥2 ≥1 [514, 587] VI.16.16 152 81 216 16 6 1 ?
-153 21 56 16 6 4 ≥1 - 2#25 II.3.32 154 49 112 16 7 2 ≥1 ≥1 2#26 155 113 226 16 8 1 ?
-156 57 114 16 8 2 ≥1362 - 2#27 157 29 58 16 8 4 ≥2 - 2#28 VI.16.30 158 17 34 16 8 7 ≥28 - D#186 [734, 1999, 2060] 159 145 232 16 10 1 ?
-160 25 40 16 10 6 ≥43 - R#172 [1236, 1938] 161 33 48 16 11 5 ≥19 0 R#171,×3 [1241, 352] 162 177 236 16 12 1 ?
-163 45 60 16 12 4 ≥1 - R#170 164 65 80 16 13 3 ?
0 R#169,×3 165 105 120 16 14 2 ?
- R#168 166 225 240 16 15 1 ?
? R#167,AG(2,15) 167 241 241 16 16 1 ?
- PG(2,15) 168 121 121 16 16 2 ?
-169 81 81 16 16 3 ?
-170 61 61 16 16 4 ≥6 - II.6.47 171 49 49 16 16 5 ≥12146 - II.6.47 172 41 41 16 16 6 ≥115307 - II.6.47 173 18 102 17 3 2 ≥4 · 1014 ≥173 D#217 [734, 1040, 1548] VI.16.81 174 52 221 17 4 1 ≥206 ≥30 [392, 587, 1377] VI.16.14 175 35 119 17 5 2 ≥1 ≥1 [1999, 1] VI.16.17 176 18 51 17 6 5 ≥582 ≥2 D#218 [1999, 1243, 734] VI.16.86 177 35 85 17 7 3 ≥2 ? [1, 1044, 1548] 178 120 255 17 8 1 ≥94 ≥1 [1729, 1877] 179 18 34 17 9 8 ≥103 0 R#186,×3 [404, 1241, 1999] 180 52 68 17 13 4 ≥6 0 R#185,×3 [1241, 2065] 181 120 136 17 15 2 0 0 R#184,×2 182 256 272 17 16 1 ≥189 ≥189 R#183,AG(2,16) [1207, 1208] 183 273 273 17 17 1 ≥22 - PG(2,16) [660, 1208, 1207] VI.18.73 184 137 137 17 17 2 0 - ×1 185 69 69 17 17 4 ≥4 - II.6.47 186 35 35 17 17 8 ≥108131 - VI.18.73 187 37 222 18 3 1 ≥1010 - [1463, 1999] VI.16.12 188 19 114 18 3 2 ≥2 · 109 - 2#29,D#231 189 13 78 18 3 3 ≥3 · 109 - 3#8 190 10 60 18 3 4 ≥961 - 2#30 191 7 42 18 3 6 418 - 6#1 [1743, 1938] 192 28 126 18 4 2 ≥139 ≥8 2#32 193 10 45 18 4 6 ≥14819 - 3#10 194 25 90 18 5 3 ≥1017 ≥1017 3#11 195 10 36 18 5 8 ≥135922 5 2#33 [1271, 1545] 40 2-(v, k, λ) Designs of Small Order II.1 No v b r k λ Nd Nr Comments, Ref Where?
196 91 273 18 6 1 ≥4 - [534, 1607] VI.16.18 197 46 138 18 6 2 ≥1 - 2#34 VI.16.18 198 31 93 18 6 3 ≥1022 - 3#12 199 19 57 18 6 5 ≥1535 - D#232 II.7.46 200 16 48 18 6 6 ≥108 - 3#13,2#35 201 28 72 18 7 4 ≥392 ? 2#36 202 64 144 18 8 2 ≥121 ≥121 2#37 203 145 290 18 9 1 ?
-204 73 146 18 9 2 ≥3500 - 2#38 205 49 98 18 9 3 ≥1 - VI.16.30 206 37 74 18 9 4 ≥852 - 2#39 207 25 50 18 9 6 ≥79 - 2#40 208 19 38 18 9 8 ≥108 - 2#41,D#233 209 55 99 18 10 3 ?
-210 100 150 18 12 2 ?
-211 34 51 18 12 6 ≥2 - R#218 212 85 102 18 15 3 ?
- R#217 213 136 153 18 16 2 ?
- R#216 214 289 306 18 17 1 ≥1 ≥1 R#215,AG(2,17) 215 307 307 18 18 1 ≥1 - PG(2,17) VI.18.73 216 154 154 18 18 2 ?
-217 103 103 18 18 3 0 - ×1 218 52 52 18 18 6 0 - ×1 219 39 247 19 3 1 ≥1044 ≥1626684 [1463, 562] VI.16.12 220 20 95 19 4 3 ≥10040 ≥204 D#270 [1999, 623, 734] VI.16.83 221 20 76 19 5 4 ≥10067 ≥14 D#271 [734, 1042] VI.16.85 222 96 304 19 6 1 ≥1 ? II.3.32 223 153 323 19 9 1 ?
?
224 20 38 19 10 9 ≥1016 3 R#233,HD 225 39 57 19 13 6 ?
0 R#232,×3 226 96 114 19 16 3 ?
0 R#231,×3 227 153 171 19 17 2 0 0 R#230,×2 228 324 342 19 18 1 ?
? R#229,AG(2,18) 229 343 343 19 19 1 ?
- PG(2,18) 230 172 172 19 19 2 0 - ×1 231 115 115 19 19 3 ?
-232 58 58 19 19 6 0 - ×1 233 39 39 19 19 9 ≥5.87 · 1014 - V.1.28 234 21 140 20 3 2 ≥5 · 1014 ≥79 2#42,D#307 235 9 60 20 3 5 5862121434 203047732 5#2 236 6 40 20 3 8 13 1 4#4 237 61 305 20 4 1 ≥18132 - [1999, 587] VI.16.61 238 31 155 20 4 2 ≥43 - [1999, 734] VI.16.15 239 21 105 20 4 3 ≥26320 - D#308 [1999, 734] VI.16.15 240 16 80 20 4 4 ≥6 · 105 ≥6 · 105 4#5 241 13 65 20 4 5 ≥103 - 5#3 242 11 55 20 4 6 ≥348 - VI.16.15 243 81 324 20 5 1 ≥1 - VI.16.16 244 41 164 20 5 2 ≥6 - 2#45 245 21 84 20 5 4 ≥109 - 4#6,D#309 II.7.46 246 17 68 20 5 5 ≥7260 - [514, 734] VI.16.17 247 11 44 20 5 8 ≥4394 - 4#7 248 51 170 20 6 2 ≥446 - 2#48 VI.16.91 249 21 70 20 6 5 ≥1 - D#310 VI.16.18 250 21 60 20 7 6 ≥3810 ≥1 2#49,D#311 II.1.3 Parameter Tables 41 No v b r k λ Nd Nr Comments, Ref Where?
251 36 90 20 8 4 ≥2 - 2#50 [1, 2071] 252 81 180 20 9 2 ≥1169 ≥1169 2#51 253 181 362 20 10 1 ?
-254 91 182 20 10 2 ≥46790 - 2#52 255 61 122 20 10 3 ≥1 - VI.16.30 256 46 92 20 10 4 ≥1 - 2#53 VI.16.30 257 37 74 20 10 5 ≥1 - VI.16.30 258 31 62 20 10 6 ≥152 - 2#54 259 21 42 20 10 9 ≥4 - D#312 [1999, 248] 260 111 185 20 12 2 ?
-261 45 75 20 12 5 ?
-262 141 188 20 15 2 ?
-263 57 76 20 15 5 ?
- R#271 264 36 48 20 15 8 ≥1 - 265 76 95 20 16 4 ≥1 - R#270 [1999, 514] 266 171 190 20 18 2 ?
- R#269 267 361 380 20 19 1 ≥1 ≥1 R#268,AG(2,19) 268 381 381 20 20 1 ≥1 - PG(2,19) VI.18.73 269 191 191 20 20 2 ?
-270 96 96 20 20 4 ≥2 - VI.18.73 271 77 77 20 20 5 0 - ×1 272 43 301 21 3 1 ≥5 · 1064 - [1042, 1463] VI.16.12 273 22 154 21 3 2 ≥3 · 109 - D#336 VI.16.81 274 15 105 21 3 3 ≥1015 ≥1011 3#14 275 8 56 21 3 6 3077244 - VI.16.82 276 7 49 21 3 7 1508 - 7#1 [1743, 1938] 277 64 336 21 4 1 ≥1.4 · 1031 ≥2.5 · 1037 VI.16.14 278 8 42 21 4 9 8360901 10 3#15 279 85 357 21 5 1 ≥3.2 · 1038 ≥1 PG(3,4) [1546, 587] VI.16.70 280 15 63 21 5 6 ≥2211 ≥149 3#16 VI.16.17 281 106 371 21 6 1 ≥1 - II.3.32 282 36 126 21 6 3 ≥1 ≥1 3#17 [1042, 1] VI.16.86 283 22 77 21 6 5 ≥3 - D#337,#6↓#153 VI.16.18 284 16 56 21 6 7 ≥1 - #13+#128 285 127 381 21 7 1 ?
-286 64 192 21 7 2 ≥1 - II.3.32 287 43 129 21 7 3 ≥1 - 3#18 VI.16.30 288 22 66 21 7 6 ≥1 - 3#19,D#338 II.7.47 289 19 57 21 7 7 ≥1 - II.7.46 290 15 45 21 7 9 ≥108 - 3#20 291 57 133 21 9 3 ≥1 - 292 190 399 21 10 1 ?
?
293 22 42 21 11 10 ≥2 0 R#312,×3 [1241, 1999, 1013] 294 232 406 21 12 1 ?
-295 274 411 21 14 1 ?
-296 92 138 21 14 3 ?
-297 40 60 21 14 7 ?
- R#311 298 295 413 21 15 1 ?
-299 50 70 21 15 6 ≥1 - R#310 300 64 84 21 16 5 ≥10810800 ≥157 R#309,AG2(3,4) [1374, 1223] 301 85 105 21 17 4 ?
0 R#308,×3 302 120 140 21 18 3 ?
- R#307 303 190 210 21 19 2 ?
0 R#306,×3 304 400 420 21 20 1 ?
? R#305,AG(2,20) 305 421 421 21 21 1 ?
- PG(2,20) 42 2-(v, k, λ) Designs of Small Order II.1 No v b r k λ Nd Nr Comments, Ref Where?
306 211 211 21 21 2 ?
-307 141 141 21 21 3 0 - ×1 308 106 106 21 21 4 0 - ×1 309 85 85 21 21 5 ≥213964 - PG2(3,4) VI.18.73 310 71 71 21 21 6 ≥2 - II.6.47 311 61 61 21 21 7 0 - ×1 312 43 43 21 21 10 ≥82 - [1013, 2066] VI.18.73 313 45 330 22 3 1 ≥6 · 1076 ≥84 [1042, 1463, 1548] VI.16.12 314 12 88 22 3 4 ≥20476 ≥3 2#55 315 34 187 22 4 2 ≥1 - VI.16.15 316 12 66 22 4 6 ≥1.7 · 106 ≥1 2#56 317 45 198 22 5 2 ≥17 ? 2#57 318 111 407 22 6 1 ≥1 - II.3.32 319 12 44 22 6 10 ≥11604 545 2#58 [324, 1271] 320 133 418 22 7 1 ?
?
321 45 110 22 9 4 ≥1353 ? 2#59 322 100 220 22 10 2 ≥1 ? 2#60 IV.2.67 323 221 442 22 11 1 ?
-324 111 222 22 11 2 ?
- 2#61 325 56 112 22 11 4 ≥2696 - 2#62 326 45 90 22 11 5 ≥1 - VI.16.30 327 23 46 22 11 10 ≥1103 - 2#63,D#351 328 287 451 22 14 1 ?
-329 45 66 22 15 7 ?
? R#338 330 56 77 22 16 6 ≥3 - R#337 331 133 154 22 19 3 ?
0 R#336,×3 332 210 231 22 20 2 0 - R#335,×2 333 441 462 22 21 1 0 0 R#334,×2,AG(2,21) 334 463 463 22 22 1 0 - ×1,PG(2,21) 335 232 232 22 22 2 0 - ×1 336 155 155 22 22 3 ?
-337 78 78 22 22 6 ≥3 - II.6.47 338 67 67 22 22 7 0 - ×1 339 24 184 23 3 2 ≥3 · 109 ≥1 D#404 [1040, 1548] VI.16.81 340 24 138 23 4 3 ≥1 ≥1 D#405 [144, 1042] VI.16.83 341 24 92 23 6 5 ≥1 ≥1 D#406 [848, 1042] VI.16.86 342 70 230 23 7 2 ≥1 ? VI.16.88 343 24 69 23 8 7 ≥1 ? D#407 VI.16.88 344 70 161 23 10 3 ?
?
345 231 483 23 11 1 ?
?
346 24 46 23 12 11 ≥1027 130 R#351,HD [1271, 1375, 1999] 347 231 253 23 21 2 0 0 R#350,×2 348 484 506 23 22 1 0 0 R#349,×2,AG(2,22) 349 507 507 23 23 1 0 - ×1,PG(2,22) 350 254 254 23 23 2 0 - ×1 351 47 47 23 23 11 ≥55 - VI.18.73 352 49 392 24 3 1 ≥6 · 1014 - [1042, 1463] VI.16.12 353 25 200 24 3 2 ≥1014 - 2#64,D#438 354 17 136 24 3 3 ≥4968 - VI.16.24 355 13 104 24 3 4 ≥108 - 4#8 356 9 72 24 3 6 ≥107 ≥105 6#2 357 7 56 24 3 8 5413 - 8#1 [1743, 1938] 358 73 438 24 4 1 ≥107 - VI.16.61 359 37 222 24 4 2 ≥4 - 2#68 360 25 150 24 4 3 ≥1022 - 3#22,D#439 II.1.3 Parameter Tables 43 No v b r k λ Nd Nr Comments, Ref Where?
361 19 114 24 4 4 ≥424 - 2#69 362 13 78 24 4 6 ≥108 - 6#3 363 10 60 24 4 8 ≥1759614 - 4#10 364 9 54 24 4 9 ≥106 - 3#24 365 25 120 24 5 4 ≥1017 ≥1017 4#11,D#440 366 121 484 24 6 1 ≥1 - [1042, 1607] VI.16.31 367 61 244 24 6 2 ≥1 - 2#73 VI.16.18 368 41 164 24 6 3 ≥1 -VI.16.18 369 31 124 24 6 4 ≥1022 - 4#12 370 25 100 24 6 5 ≥1 - D#441 371 21 84 24 6 6 ≥1 - 3#25,2#75 372 16 64 24 6 8 ≥108 - 4#13 373 13 52 24 6 10 ≥2572157 - 2#77 374 49 168 24 7 3 ≥1052 ≥1052 3#26 375 169 507 24 8 1 ?
-376 85 255 24 8 2 ≥1 - VI.16.30 377 57 171 24 8 3 ≥1063 - 3#27 378 43 129 24 8 4 ≥1 - VI.16.30 379 29 87 24 8 6 ≥1 - 3#28 VI.16.30 380 25 75 24 8 7 ≥1 - D#442 VI.16.31 381 22 66 24 8 8 ≥1 - 2#78 VI.16.30 382 33 88 24 9 6 ≥3376 - 2#79 383 55 132 24 10 4 ?
- 2#80 384 25 60 24 10 9 ≥3 - D#443 [607, 2144] 385 121 264 24 11 2 ≥365 ≥365 2#81 386 265 530 24 12 1 ?
-387 133 266 24 12 2 ≥9979200 - 2#82 388 89 178 24 12 3 ?
-389 67 134 24 12 4 ≥1 - 2#83 VI.16.30 390 45 90 24 12 6 ≥3753 - 2#84 391 34 68 24 12 8 ≥1 - 2#85 VI.16.30 392 25 50 24 12 11 ≥10 - D#444 [607, 2144] 393 105 180 24 14 3 ?
-394 85 136 24 15 4 ?
-395 46 69 24 16 8 ≥1 - R#407 396 69 92 24 18 6 ?
- R#406 397 115 138 24 20 4 ?
- R#405 398 161 184 24 21 3 ?
- R#404 399 49 56 24 21 10 ≥1 - VI.16.32 400 253 276 24 22 2 0 - R#403,×2 401 529 552 24 23 1 ≥1 ≥1 R#402,AG(2,23) 402 553 553 24 24 1 ≥1 - PG(2,23) VI.18.73 403 277 277 24 24 2 0 - ×1 404 185 185 24 24 3 0 - ×1 405 139 139 24 24 4 ?
-406 93 93 24 24 6 0 - ×1 407 70 70 24 24 8 ≥28 - [1185, 620] II.6.47 408 51 425 25 3 1 ≥6 · 1053 ≥9419 [1463, 1976] VI.16.12 409 6 50 25 3 10 19 0 5#4 [1174, 1548] 410 76 475 25 4 1 ≥169574 ≥1 [396, 587, 1047] VI.16.14 411 16 100 25 4 5 ≥106 ≥106 5#5 412 101 505 25 5 1 ≥3 - VI.16.62 413 51 255 25 5 2 ≥1 - VI.16.17 414 26 130 25 5 4 ≥1 - D#471 II.7.46 415 21 105 25 5 5 ≥109 - 5#6 44 2-(v, k, λ) Designs of Small Order II.1 No v b r k λ Nd Nr Comments, Ref Where?
416 11 55 25 5 10 ≥3337 - 5#7 417 126 525 25 6 1 ≥2 ≥1 [679, 1042, 1578] VI.16.92 418 176 550 25 8 1 ?
?
419 226 565 25 10 1 ?
-420 76 190 25 10 3 ?
-421 46 115 25 10 5 ≥1 - 422 26 65 25 10 9 ≥19 - D#472 [1042, 1556] 423 276 575 25 12 1 ?
?
424 26 50 25 13 12 ≥1 0 R#444,×3 [1241, 1999] 425 351 585 25 15 1 ?
-426 51 85 25 15 7 ?
-427 36 60 25 15 10 ≥1 - R#443 428 51 75 25 17 8 ?
0 R#442,×3 429 76 100 25 19 6 ≥1 0 R#441,×3 430 476 595 25 20 1 ?
-431 96 120 25 20 5 ≥1 - R#440 432 126 150 25 21 4 ?
0 R#439,×3 433 176 200 25 22 3 ?
0 R#438,×3 434 276 300 25 23 2 ?
0 R#437,×3 435 576 600 25 24 1 ?
? R#436,AG(2,24) 436 601 601 25 25 1 ?
- PG(2,24) 437 301 301 25 25 2 ?
-438 201 201 25 25 3 ?
-439 151 151 25 25 4 0 - ×1 440 121 121 25 25 5 ≥1 - II.6.47 441 101 101 25 25 6 ≥1 -VI.18.73 442 76 76 25 25 8 0 - ×1 443 61 61 25 25 10 ≥24 - II.6.47 444 51 51 25 25 12 ≥1 - V.1.39 445 27 234 26 3 2 ≥1011 ≥910 2#86,D#511 446 40 260 26 4 2 ≥106 ≥1 2#87 447 14 91 26 4 6 ≥4 - [1042, 1588] II.5.29 448 105 546 26 5 1 ≥1 ≥1 IV.2.2 449 66 286 26 6 2 ≥1 ≥1 2#88 450 27 117 26 6 5 ≥1 - D#512 VI.16.86 451 14 52 26 7 12 ≥1363846 1363486 2#89 452 92 299 26 8 2 ≥1 - 453 27 78 26 9 8 ≥8072 ≥13 2#90,D#513 454 235 611 26 10 1 ?
-455 40 104 26 10 6 ≥1 ? 2#91 456 66 156 26 11 4 ≥494 ? 2#92 457 144 312 26 12 2 ≥1 ? 2#93 IV.2.67 458 313 626 26 13 1 ?
-459 157 314 26 13 2 ?
- 2#94 460 105 210 26 13 3 ?
-461 79 158 26 13 4 ≥940 - 2#95 462 53 106 26 13 6 ≥1 - 2#96 VI.16.30 463 40 80 26 13 8 ≥390 - 2#97 464 27 54 26 13 12 ≥208311 - 2#98,D#514 465 40 65 26 16 10 ≥1 - R#472 466 105 130 26 21 5 ?
0 R#471,×3 467 300 325 26 24 2 0 - R#470,×2 468 625 650 26 25 1 ≥33 ≥33 R#469,AG(2,25) 469 651 651 26 26 1 ≥17 - PG(2,25) VI.18.73 470 326 326 26 26 2 0 - ×1 II.1.3 Parameter Tables 45 No v b r k λ Nd Nr Comments, Ref Where?
471 131 131 26 26 5 ?
-472 66 66 26 26 10 ≥588 - [2067, 1720] II.6.47 473 55 495 27 3 1 ≥6 · 1076 - VI.16.12 474 28 252 27 3 2 ≥1055 - D#564 VI.16.13 475 19 171 27 3 3 ≥1017 - 3#29 476 10 90 27 3 6 ≥1012 - 3#30 477 7 63 27 3 9 17785 - 9#1 [1743, 1938] 478 28 189 27 4 3 ≥1032 ≥1026 3#32,D#565 479 55 297 27 5 2 ≥1 ≥1 [1042, 848] IV.2.67 480 10 54 27 5 12 ≥108 0 3#33 481 136 612 27 6 1 ≥1 - II.3.32 482 46 207 27 6 3 ≥1028 - 3#34 [1042, 416] VI.16.90 483 28 126 27 6 5 ≥2 - D#566 VI.16.18 484 16 72 27 6 9 ≥18921 - 3#35 485 28 108 27 7 6 ≥5047 ≥1 3#36,D#567 [848, 1224] VI.16.87 486 64 216 27 8 3 ≥1077 ≥1077 3#37 487 217 651 27 9 1 ?
-488 109 327 27 9 2 ≥1 - VI.16.57 489 73 219 27 9 3 ≥1090 - 3#38 490 55 165 27 9 4 ≥1 - VI.16.30 491 37 111 27 9 6 ≥1037 - 3#39 492 28 84 27 9 8 ≥3 - D#568 [1548, 2130] II.7.48 493 25 75 27 9 9 ≥1028 - 3#40 494 19 57 27 9 12 ≥1016 - 3#41 495 55 135 27 11 5 ?
?
496 100 225 27 12 3 ?
-497 28 63 27 12 11 ≥246 - D#569 [1236, 1379, 710] 498 325 675 27 13 1 ?
?
499 28 54 27 14 13 ≥9 · 1021 7570 R#514,HD [1271, 1319, 2039] 500 190 342 27 15 2 ?
-501 55 99 27 15 7 ?
-502 460 690 27 18 1 ?
-503 154 231 27 18 3 ?
-504 52 78 27 18 9 ≥2 - R#513 505 91 117 27 21 6 ?
- R#512 506 208 234 27 24 3 ?
- R#511 507 325 351 27 25 2 ?
0 R#510,×3 508 676 702 27 26 1 ?
? R#509,AG(2,26) 509 703 703 27 27 1 ?
- PG(2,26) 510 352 352 27 27 2 ?
-511 235 235 27 27 3 0 - ×1 512 118 118 27 27 6 0 - ×1 513 79 79 27 27 9 ≥1463 - II.6.47 514 55 55 27 27 13 ≥1 - [1999, 514] V.1.28 515 57 532 28 3 1 ≥1090 ≥1 VI.16.12 516 15 140 28 3 4 ≥1015 ≥1011 4#14 517 9 84 28 3 7 ≥330 ≥9 7#2 518 85 595 28 4 1 ≥1015 - [1042, 332] VI.16.14 519 43 301 28 4 2 ≥1 -VI.16.15 520 29 203 28 4 3 ≥1 - D#586 II.7.46 521 22 154 28 4 4 ≥7922 - 2#100 522 15 105 28 4 6 ≥31300 - VI.16.15 523 13 91 28 4 7 ≥108 - 7#3 524 8 56 28 4 12 ≥2310 31 4#15 [1174, 1548] 525 15 84 28 5 8 ≥104 ≥1 4#16,2#102 46 2-(v, k, λ) Designs of Small Order II.1 No v b r k λ Nd Nr Comments, Ref Where?
526 141 658 28 6 1 ≥1 - 527 36 168 28 6 4 ≥5 ≥3 4#17,2#103 528 21 98 28 6 7 ≥1 - #75+#153 529 15 70 28 6 10 ≥118 - 2#104 530 169 676 28 7 1 ≥1 - [1044, 1548] VI.16.65 531 85 340 28 7 2 ≥2 - 2#105 [1, 959] VI.16.30 532 57 228 28 7 3 ≥1 - VI.16.30 533 43 172 28 7 4 ≥4 - 4#18,2#106 534 29 116 28 7 6 ≥1 - 2#107,D#587 535 25 100 28 7 7 ≥1 - VI.16.31 536 22 88 28 7 8 ≥35 - 4#19,2#108 537 15 60 28 7 12 ≥108 - 4#20 538 50 175 28 8 4 ≥1 - VI.16.89 539 225 700 28 9 1 ?
?
540 85 238 28 10 3 ?
-541 309 721 28 12 1 ?
-542 78 182 28 12 4 ?
- 2#110 543 45 105 28 12 7 ≥1 - #84+#163 544 169 364 28 13 2 ≥765 ≥765 2#111 545 365 730 28 14 1 ?
-546 183 366 28 14 2 ≥109 - 2#112 547 92 184 28 14 4 ?
- 2#113 548 53 106 28 14 7 ≥1 - VI.16.30 549 29 58 28 14 13 ≥1 - D#588 550 36 63 28 16 12 ≥8784 - R#569 [1236, 710] 551 477 742 28 18 1 ?
-552 57 84 28 19 9 ?
0 R#568,×3 553 561 748 28 21 1 ?
-554 141 188 28 21 4 ?
-555 81 108 28 21 7 ≥1 - R#567 556 57 76 28 21 10 ?
-557 99 126 28 22 6 ?
- R#566 558 162 189 28 24 4 ?
- R#565 559 225 252 28 25 3 ?
0 R#564,×3 560 351 378 28 26 2 0 - R#563,×2 561 729 756 28 27 1 ≥7 ≥7 R#562,AG(2,27) 562 757 757 28 28 1 ≥3 - PG(2,27) VI.18.73 563 379 379 28 28 2 0 - ×1 564 253 253 28 28 3 ?
-565 190 190 28 28 4 0 - ×1 566 127 127 28 28 6 ?
-567 109 109 28 28 7 ≥1 -VI.18.73 568 85 85 28 28 9 ?
-569 64 64 28 28 12 ≥8784 - VI.18.73 570 30 290 29 3 2 ≥2 · 1051 ≥1 D#655 [1040, 1548] VI.16.81 571 88 638 29 4 1 ≥2 ≥1 [1042, 332, 1047] IV.2.2 572 30 174 29 5 4 ≥1 ≥4 D#656 [1042, 623] VI.16.85 573 30 145 29 6 5 ≥1 ≥1 D#657 [1042, 33] VI.16.86 574 175 725 29 7 1 ≥1 ? VI.16.31 575 117 377 29 9 2 ≥1 ? VI.16.70 576 30 87 29 10 9 ≥1 ? D#658 VI.16.88 577 117 261 29 13 3 ?
?
578 378 783 29 14 1 ?
?
579 30 58 29 15 14 ≥1 0 R#588,×3 [1241, 2144] II.1.3 Parameter Tables 47 No v b r k λ Nd Nr Comments, Ref Where?
580 88 116 29 22 7 ?
0 R#587,×3 581 175 203 29 25 4 ?
0 R#586,×3 582 378 406 29 27 2 ?
0 R#585,×3 583 784 812 29 28 1 ?
? R#584,AG(2,28) 584 813 813 29 29 1 ?
- PG(2,28) 585 407 407 29 29 2 ?
-586 204 204 29 29 4 ?
-587 117 117 29 29 7 0 - ×1 588 59 59 29 29 14 ≥1 -VI.18.73 589 61 610 30 3 1 ≥2 · 1024 - [1042, 1463] VI.16.12 590 31 310 30 3 2 ≥6 · 1016 - 2#114,D#677 591 21 210 30 3 3 ≥1024 ≥1021 3#42 592 16 160 30 3 4 ≥1013 - 2#115 593 13 130 30 3 5 ≥108 - 5#8 594 11 110 30 3 6 ≥436801 - 2#116 595 7 70 30 3 10 54613 - 10#1 596 6 60 30 3 12 34 1 6#4,3#43 [1174, 1548] 597 46 345 30 4 2 ≥1 - VI.16.15 598 16 120 30 4 6 ≥1015 ≥1015 6#5 599 10 75 30 4 10 ≥29638 - 5#10 600 121 726 30 5 1 ≥1 - VI.16.62 601 61 366 30 5 2 ≥11 - 2#120 602 41 246 30 5 3 ≥1046 - 3#45 603 31 186 30 5 4 ≥1 - 2#121,D#678 604 25 150 30 5 5 ≥1017 ≥1017 5#11 605 21 126 30 5 6 ≥1024 - 6#6 606 16 96 30 5 8 ≥12 - 2#123 607 13 78 30 5 10 ≥31 - 2#124 608 11 66 30 5 12 ≥106 - 6#7 609 151 755 30 6 1 ≥1 - [1042, 2144] VI.16.54 610 76 380 30 6 2 ≥1 - 2#126 611 51 255 30 6 3 ≥1 - 3#48 VI.16.18 612 31 155 30 6 5 ≥1023 - 5#12,D#679 613 26 130 30 6 6 ≥1 - 2#127 614 16 80 30 6 10 ≥108 - 5#13 615 91 390 30 7 2 ≥3 ? 2#129 616 21 90 30 7 9 ≥1018 ≥1 3#49 617 36 135 30 8 6 ≥3 - 3#50 [1, 1497, 2130] 618 16 60 30 8 14 ≥9 · 107 ≥6 2#130 619 81 270 30 9 3 ≥10108 ≥10108 3#51 620 21 70 30 9 12 ≥104 - 2#131 621 271 813 30 10 1 ?
-622 136 408 30 10 2 ?
- 2#132 623 91 273 30 10 3 ≥10125 - 3#52 624 55 165 30 10 5 ≥2 - VI.16.30 625 46 138 30 10 6 ≥1 - 2#133,3#53 626 31 93 30 10 9 ≥152 - 3#54,D#680 627 28 84 30 10 10 ≥5 - 2#134 [1548, 2130] II.7.48 628 166 415 30 12 2 ?
-629 56 140 30 12 6 ≥5 - 2#135 630 34 85 30 12 10 ≥1 - VI.16.30 631 91 210 30 13 4 ?
? 2#136 632 196 420 30 14 2 ?
? 2#137 633 421 842 30 15 1 ?
-634 211 422 30 15 2 ?
- 2#138 635 141 282 30 15 3 ?
-48 2-(v, k, λ) Designs of Small Order II.1 No v b r k λ Nd Nr Comments, Ref Where?
636 106 212 30 15 4 ?
- 2#139 637 85 170 30 15 5 ?
-638 71 142 30 15 6 ≥9 - 2#140 639 61 122 30 15 7 ≥1 - VI.16.30 640 43 86 30 15 10 ≥1 - 2#141 VI.16.30 641 36 72 30 15 12 ≥25635 - 2#142 642 31 62 30 15 14 ≥106 - 2#143,D#681 643 171 285 30 18 3 ?
-644 286 429 30 20 2 ?
-645 96 144 30 20 6 ?
-646 58 87 30 20 10 ?
- R#658 647 301 430 30 21 2 ?
-648 116 145 30 24 6 ?
- R#657 649 145 174 30 25 5 ≥1 - R#656 650 261 290 30 27 3 ?
- R#655 651 406 435 30 28 2 0 - R#654,×2 652 841 870 30 29 1 ≥1 ≥1 R#653,AG(2,29) 653 871 871 30 30 1 ≥1 - PG(2,29) VI.18.73 654 436 436 30 30 2 0 - ×1 655 291 291 30 30 3 ?
-656 175 175 30 30 5 ≥2 - [331, 1839] VI.18.73 657 146 146 30 30 6 0 - ×1 658 88 88 30 30 10 0 - ×1 659 63 651 31 3 1 ≥1042 ≥82160 PG(5,2) [1463, 1548] VI.16.12 660 32 248 31 4 3 ≥1 ≥1 D#729 [1042, 144] VI.16.83 661 125 775 31 5 1 ≥1.9 · 1097 ≥5.2 · 10109 AG(3,5) IV.2.2 662 156 806 31 6 1 ≥1.6 · 10116 ≥1 PG(3,5) IV.2.2 663 63 279 31 7 3 ≥1 ≥1 [1042, 1] VI.16.87 664 32 124 31 8 7 ≥1 ≥1 D#730 [1042, 1714] VI.16.87 665 63 217 31 9 4 ≥1 ≥1 VI.16.87 666 280 868 31 10 1 ?
?
667 435 899 31 15 1 ?
?
668 32 62 31 16 15 ≥1028 ≥1 R#681,AG4(5,2),HD 669 63 93 31 21 10 ≥1017 0 R#680 [1900, 2058] 670 125 155 31 25 6 ≥1012 ≥1012 R#679,AG2(3, 5) 671 156 186 31 26 5 ?
0 R#678,×3 672 280 310 31 28 3 ?
0 R#677,×3 673 435 465 31 29 2 0 0 R#676,×2 674 900 930 31 30 1 0 0 R#675,×2,AG(2,30) 675 931 931 31 31 1 0 - ×1,PG(2,30) 676 466 466 31 31 2 0 - ×1 677 311 311 31 31 3 ?
-678 187 187 31 31 5 0 - ×1 679 156 156 31 31 6 ≥1017 - PG2(3, 5) VI.18.73 680 94 94 31 31 10 0 - ×1 681 63 63 31 31 15 ≥1017 - PG4(5, 2) VI.18.73 682 33 352 32 3 2 ≥1013 ≥4.4 · 106 2#144,D#775 683 9 96 32 3 8 ≥107 ≥105 8#2 684 97 776 32 4 1 ≥5985 - [332, 396] VI.16.61 685 49 392 32 4 2 ≥770 - 2#146 686 33 264 32 4 3 ≥1 - D#776 II.7.46 687 25 200 32 4 4 ≥1022 - 4#22 688 17 136 32 4 6 ≥1 - 2#148 689 13 104 32 4 8 ≥108 - 8#3 690 9 72 32 4 12 ≥108 - 4#24 II.1.3 Parameter Tables 49 No v b r k λ Nd Nr Comments, Ref Where?
691 65 416 32 5 2 ≥3 ≥1 2#151 692 81 432 32 6 2 ≥1 - 2#152 IV.2.7 693 33 176 32 6 5 ≥1 - D#777 VI.16.18 694 21 112 32 6 8 ≥1 - 4#25,2#153 695 49 224 32 7 4 ≥1052 ≥1052 4#26 696 225 900 32 8 1 ?
-697 113 452 32 8 2 ≥1 - 2#155 VI.16.64 698 57 228 32 8 4 ≥1063 - 4#27 699 33 132 32 8 7 ≥1 - D#778 II.7.46 700 29 116 32 8 8 ≥3 - 4#28,2#157 701 17 68 32 8 14 ≥12 - 2#158 702 145 464 32 10 2 ?
- 2#159 703 25 80 32 10 12 ≥44 - 2#160 704 33 96 32 11 10 ≥20 ? 2#161,D#779 705 177 472 32 12 2 ?
- 2#162 706 45 120 32 12 8 ≥1 - 2#163 707 33 88 32 12 11 ≥1 - D#780 708 65 160 32 13 6 ?
? 2#164 709 105 240 32 14 4 ?
- 2#165 710 225 480 32 15 2 ?
? 2#166 711 481 962 32 16 1 ?
-712 241 482 32 16 2 ?
- 2#167 713 161 322 32 16 3 ?
-714 121 242 32 16 4 ≥1 - 2#168 VI.16.31 715 97 194 32 16 5 ?
-716 81 162 32 16 6 ?
- 2#169 717 61 122 32 16 8 ≥1 - 2#170 718 49 98 32 16 10 ≥45 - 2#171 719 41 82 32 16 12 ≥115308 - 2#172 720 33 66 32 16 15 ≥1 - D#781 721 305 488 32 20 2 ?
-722 369 492 32 24 2 ?
-723 93 124 32 24 8 ?
- R#730 724 217 248 32 28 4 ?
- R#729 725 465 496 32 30 2 0 - R#728,×2 726 961 992 32 31 1 ≥1 ≥1 R#727,AG(2,31) 727 993 993 32 32 1 ≥1 - PG(2,31) VI.18.82 728 497 497 32 32 2 0 - ×1 729 249 249 32 32 4 ?
-730 125 125 32 32 8 0 - ×1 731 67 737 33 3 1 ≥1035 - [1042, 1463] VI.16.12 732 34 374 33 3 2 ≥1041 - D#811 VI.16.81 733 23 253 33 3 3 ≥2 · 1014 - VI.16.24 734 12 132 33 3 6 ≥108 ≥1 3#55 735 7 77 33 3 11 155118 - 11#1 736 100 825 33 4 1 ≥5985 ≥1 [332, 396, 1047] IV.2.2 737 12 99 33 4 9 ≥1010 ≥1 3#56 738 45 297 33 5 3 ≥1054 ≥1 3#57 [2, 1548] 739 166 913 33 6 1 ?
-740 56 308 33 6 3 ≥1 - IV.2.7 741 34 187 33 6 5 ≥1 - D#812 VI.16.86 742 16 88 33 6 11 ≥1 - #13+#484 743 12 66 33 6 15 ≥11604 ≥12 3#58 744 232 957 33 8 1 ≥1 ≥1 745 45 165 33 9 6 ≥35805 ? 3#59 50 2-(v, k, λ) Designs of Small Order II.1 No v b r k λ Nd Nr Comments, Ref Where?
746 100 330 33 10 3 ≥1 ? 3#60 IV.2.67 747 331 993 33 11 1 ?
-748 166 498 33 11 2 ?
-749 111 333 33 11 3 ≥1 - 3#61 IV.2.67 750 67 201 33 11 5 ≥1 - VI.16.30 751 56 168 33 11 6 ≥1071 - 3#62 752 34 102 33 11 10 ≥1 - D#813 VI.16.88 753 31 93 33 11 11 ≥1 - II.7.46 754 23 69 33 11 15 ≥1103 - 3#63 755 364 1001 33 12 1 ?
-756 155 341 33 15 3 ?
-757 496 1023 33 16 1 ≥1 ≥1 758 34 66 33 17 16 ≥1 0 R#781,×3 [1241, 2110] 759 133 209 33 21 5 ?
-760 56 88 33 21 12 ?
- R#780 761 694 1041 33 22 1 ?
-762 232 348 33 22 3 ?
-763 100 150 33 22 7 ?
-764 78 117 33 22 9 ?
-765 64 96 33 22 11 ?
- R#779 766 760 1045 33 24 1 ?
-767 100 132 33 25 8 ≥1 0 R#778,×3 [1241, 1013] 768 144 176 33 27 6 ?
- R#777 769 232 264 33 29 4 ?
0 R#776,×3 770 320 352 33 30 3 ?
- R#775 771 496 528 33 31 2 ?
0 R#774,×3 772 1024 1056 33 32 1 ≥11 ≥11 R#773,AG(2,32) 773 1057 1057 33 33 1 ≥6 - PG(2,32) VI.18.28 774 529 529 33 33 2 ?
-775 353 353 33 33 3 0 - ×1 776 265 265 33 33 4 ?
-777 177 177 33 33 6 ?
-778 133 133 33 33 8 ≥1 - VI.18.73 779 97 97 33 33 11 ?
-780 89 89 33 33 12 0 - ×1 781 67 67 33 33 16 ≥1 - VI.18.73 782 69 782 34 3 1 ≥4 · 1041 ≥1 [1042, 1047, 1463] VI.16.12 783 18 204 34 3 4 ≥4 · 1014 ≥1 2#173 784 52 442 34 4 2 ≥207 ≥1 2#174 785 18 153 34 4 6 ≥1 - VI.16.84 786 35 238 34 5 4 ≥2 ≥2 2#175,D#869 787 171 969 34 6 1 ≥1 - II.3.32 788 18 102 34 6 10 ≥4 ≥3 2#176 789 35 170 34 7 6 ≥3 ≥1 2#177,D#870 VI.16.87 790 120 510 34 8 2 ≥1 ≥1 2#178 791 18 68 34 9 16 ≥103 ≥1 2#179 792 35 119 34 10 9 ≥1 - D#871 793 341 1054 34 11 1 ?
?
794 52 136 34 13 8 ≥1 ? 2#180 795 35 85 34 14 13 ≥1 - D#872 796 120 272 34 15 4 ?
? 2#181 797 256 544 34 16 2 ≥477603 ≥477603 2#182 798 545 1090 34 17 1 ?
-799 273 546 34 17 2 ≥1012 - 2#183 800 137 274 34 17 4 ≥1 - 2#184 II.1.3 Parameter Tables 51 No v b r k λ Nd Nr Comments, Ref Where?
801 69 138 34 17 8 ≥1 - 2#185 802 35 70 34 17 16 ≥1854 - 2#186,D#873 803 715 1105 34 22 1 ?
-804 69 102 34 23 11 ?
0 R#813,×3 805 154 187 34 28 6 ?
- R#812 806 341 374 34 31 3 ?
0 R#811,×3 807 528 561 34 32 2 0 - R#810,×2 808 1089 1122 34 33 1 0 0 R#809,×2,AG(2,33) 809 1123 1123 34 34 1 0 - ×1 810 562 562 34 34 2 0 - ×1 811 375 375 34 34 3 ?
-812 188 188 34 34 6 0 - ×1 813 103 103 34 34 11 ?
-814 36 420 35 3 2 ≥2 · 1050 ≥1 D#961 [1040, 1548] VI.16.81 815 15 175 35 3 5 ≥1015 ≥1011 5#14 816 6 70 35 3 14 48 0 7#4 [1174, 1548] 817 36 315 35 4 3 ≥1 ≥1 D#962 [1042, 144] VI.16.83 818 16 140 35 4 7 ≥1015 ≥1015 7#5 819 8 70 35 4 15 ≥2224 82 5#15 [697, 1271] 820 141 987 35 5 1 ≥1 - VI.16.16 821 71 497 35 5 2 ≥1 - VI.16.17 822 36 252 35 5 4 ≥2 - D#963 [1042, 2130] VI.16.85 823 29 203 35 5 5 ≥2 - [1042, 380] II.7.46 824 21 147 35 5 7 ≥1024 - 7#6 825 15 105 35 5 10 ≥1 ≥1 5#16,#102+#280 826 11 77 35 5 14 ≥106 - 7#7 827 36 210 35 6 5 ≥1 ≥1 #103+#282,D#964 [1042, 144] 828 211 1055 35 7 1 ?
-829 106 530 35 7 2 ≥1 - 830 71 355 35 7 3 ≥1 - VI.16.30 831 43 215 35 7 5 ≥1 - 5#18,#106+#287 832 36 180 35 7 6 ≥1 - D#965 II.7.46 833 31 155 35 7 7 ≥5 - PG2(4, 2) II.7.46 834 22 110 35 7 10 ≥1 - 5#19,#108+#288 835 16 80 35 7 14 ≥1 - #131↓#259 836 15 75 35 7 15 ≥109 - 5#20 837 36 140 35 9 8 ≥4 ≥6 D#966 [1042, 1629] VI.16.87 838 316 1106 35 10 1 ?
-839 106 371 35 10 3 ?
-840 64 224 35 10 5 ≥1 - II.3.32 841 46 161 35 10 7 ?
-842 36 126 35 10 9 ≥2 - D#967 [1042, 2130] 843 22 77 35 10 15 ≥1 - #104↓#290 844 176 560 35 11 2 ?
?
845 36 105 35 12 11 ≥1 ≥1 D#968 [1042, 1566] VI.16.88 846 456 1140 35 14 1 ?
-847 92 230 35 14 5 ?
-848 66 165 35 14 7 ?
-849 36 90 35 14 13 ≥1 - D#969 850 246 574 35 15 2 ?
-851 99 231 35 15 5 ?
-852 36 84 35 15 14 ≥1 - D#970, #142+#264 853 176 385 35 16 3 ?
?
854 561 1155 35 17 1 ?
?
855 36 70 35 18 17 ≥91 ≥91 R#873,HD [404, 2110] 52 2-(v, k, λ) Designs of Small Order II.1 No v b r k λ Nd Nr Comments, Ref Where?
856 96 168 35 20 7 ?
-857 351 585 35 21 2 ?
-858 141 235 35 21 5 ?
-859 51 85 35 21 14 ?
- R#872 860 85 119 35 25 10 ?
- R#871 861 316 395 35 28 3 ?
-862 136 170 35 28 7 ?
- R#870 863 64 80 35 28 15 ?
-864 204 238 35 30 5 ?
- R#869 865 561 595 35 33 2 0 0 R#868,×2 866 1156 1190 35 34 1 ?
? R#867,AG(2,34) 867 1191 1191 35 35 1 ?
- PG(2,34) 868 596 596 35 35 2 0 - ×1 869 239 239 35 35 5 ?
-870 171 171 35 35 7 ?
-871 120 120 35 35 10 ?
-872 86 86 35 35 14 0 - ×1 873 71 71 35 35 17 ≥9 - [2110, 619] VI.18.73 874 73 876 36 3 1 ≥1034 - [1042, 1463] VI.16.12 875 37 444 36 3 2 ≥1010 - 2#187,D#991 876 25 300 36 3 3 ≥1025 - 3#64 877 19 228 36 3 4 ≥1017 - 4#29 878 13 156 36 3 6 ≥1017 - 6#8 879 10 120 36 3 8 ≥1012 - 4#30 880 9 108 36 3 9 ≥1014 ≥105 9#2 881 7 84 36 3 12 412991 - 12#1 882 109 981 36 4 1 ≥1.6 · 1013 - VI.16.61 883 55 495 36 4 2 ≥1 - IV.2.7 884 37 333 36 4 3 ≥1040 - 3#68,D#992 885 28 252 36 4 4 ≥1032 ≥1026 4#32 886 19 171 36 4 6 ≥1020 - 3#69 887 13 117 36 4 9 ≥108 - 9#3 888 10 90 36 4 12 ≥109 - 6#10 889 145 1044 36 5 1 ≥1 ≥1 [1042, 41] VI.16.70 890 25 180 36 5 6 ≥1038 ≥1038 6#11 891 10 72 36 5 16 ≥108 27121734 4#33,2#195 [1548, 1633] 892 181 1086 36 6 1 ≥1 - [1042, 2144] VI.16.54 893 91 546 36 6 2 ≥5 - 2#196 894 61 366 36 6 3 ≥1 - 3#73 VI.16.18 895 46 276 36 6 4 ≥1 - 4#34,2#197 896 37 222 36 6 5 ≥2 - D#993 897 31 186 36 6 6 ≥1051 - 6#12 898 21 126 36 6 9 ≥1 - 3#75 899 19 114 36 6 10 ≥1 - 2#199 900 16 96 36 6 12 ≥1019 - 6#13 901 13 78 36 6 15 ≥1011 - 3#77 902 217 1116 36 7 1 ≥1 ? VI.16.30 903 28 144 36 7 8 ≥5432 ≥1 4#36 [1, 1224] 904 64 288 36 8 4 ≥1077 ≥1077 4#37 905 22 99 36 8 12 ≥1 - 3#78 VI.16.30 906 289 1156 36 9 1 ?
-907 145 580 36 9 2 ≥1 - 2#203 908 97 388 36 9 3 ≥1 - VI.16.30 909 73 292 36 9 4 ≥1090 - 4#38 910 49 196 36 9 6 ≥5 - 2#205 II.1.3 Parameter Tables 53 No v b r k λ Nd Nr Comments, Ref Where?
911 37 148 36 9 8 ≥1037 - 4#39,D#994 912 33 132 36 9 9 ≥1039 - 3#79 II.7.46 913 25 100 36 9 12 ≥1028 - 4#40 914 19 76 36 9 16 ≥1016 - 4#41 915 325 1170 36 10 1 ?
-916 55 198 36 10 6 ?
- 3#80,2#209 917 121 396 36 11 3 ≥10188 ≥10188 3#81 918 397 1191 36 12 1 ?
-919 199 597 36 12 2 ?
-920 133 399 36 12 3 ≥10208 - 3#82 921 100 300 36 12 4 ?
- 2#210 922 67 201 36 12 6 ≥1 - 3#83 VI.16.30 923 45 135 36 12 9 ≥1057 - 3#84 924 37 111 36 12 11 ≥1 - D#995 II.7.46 925 34 102 36 12 12 ≥2 - 3#85,2#211 VI.16.88 926 469 1206 36 14 1 ?
-927 505 1212 36 15 1 ?
-928 85 204 36 15 6 ?
- 2#212 929 136 306 36 16 4 ?
- 2#213 930 289 612 36 17 2 ≥3481 ≥3481 2#214 931 613 1226 36 18 1 ?
-932 307 614 36 18 2 ≥8 · 1013 - 2#215 933 205 410 36 18 3 ?
-934 154 308 36 18 4 ?
- 2#216 935 103 206 36 18 6 ?
- 2#217 936 69 138 36 18 9 ?
-937 52 104 36 18 12 ?
- 2#218 938 37 74 36 18 17 ≥1 - D#996 939 685 1233 36 20 1 ?
-940 115 207 36 20 6 ?
-941 721 1236 36 21 1 ?
-942 91 156 36 21 8 ?
-943 49 84 36 21 15 ?
- R#970 944 253 414 36 22 3 ?
-945 55 90 36 22 14 ?
- R#969 946 208 312 36 24 4 ?
-947 70 105 36 24 12 ?
- R#968 948 91 126 36 26 10 ?
- R#967 949 105 140 36 27 9 ?
- R#966 950 973 1251 36 28 1 ?
-951 145 180 36 29 7 ?
0 R#965,×3 952 1045 1254 36 30 1 ?
-953 175 210 36 30 6 ?
- R#964 954 217 252 36 31 5 ?
0 R#963,×3 955 280 315 36 32 4 ?
- R#962 956 385 420 36 33 3 ?
- R#961 957 595 630 36 34 2 ?
- R#960 958 1225 1260 36 35 1 ?
? R#959,AG(2,35) 959 1261 1261 36 36 1 ?
- PG(2,35) 960 631 631 36 36 2 ?
-961 421 421 36 36 3 ?
-962 316 316 36 36 4 0 - ×1 963 253 253 36 36 5 ?
-964 211 211 36 36 6 0 - ×1 965 181 181 36 36 7 ?
-54 2-(v, k, λ) Designs of Small Order II.1 No v b r k λ Nd Nr Comments, Ref Where?
966 141 141 36 36 9 ?
-967 127 127 36 36 10 ?
-968 106 106 36 36 12 0 - ×1 969 91 91 36 36 14 0 - ×1 970 85 85 36 36 15 ?
-971 75 925 37 3 1 ≥10196 ≥1 [1013, 1042, 1174] VI.16.12 972 112 1036 37 4 1 ≥1.69 · 1013 ≥210 [1042, 396] IV.2.2 973 75 555 37 5 2 ≥1 ≥1 [1042, 41] VI.16.31 974 186 1147 37 6 1 ≥1 ≥1 [1042, 958] IV.2.2 975 112 592 37 7 2 ≥1 ? 976 297 1221 37 9 1 ?
?
977 408 1258 37 12 1 ?
?
978 75 185 37 15 7 ≥1 ≥1 979 112 259 37 16 5 ?
?
980 630 1295 37 18 1 ?
?
981 38 74 37 19 18 ≥1 0 R#996,×3 [1241, 2110] 982 75 111 37 25 12 ?
0 R#995,×3 983 112 148 37 28 9 ?
? R#994 984 186 222 37 31 6 ≥1 0 R#993,×3 [1241, 1773] 985 297 333 37 33 4 ?
0 R#992,×3 986 408 444 37 34 3 ?
0 R#991,×3 987 630 666 37 35 2 0 0 R#990,×2 988 1296 1332 37 36 1 ?
? R#989,AG(2,36) 989 1333 1333 37 37 1 ?
- PG(2,36) 990 667 667 37 37 2 0 - ×1 991 445 445 37 37 3 0 - ×1 992 334 334 37 37 4 0 - ×1 993 223 223 37 37 6 ≥1 - 994 149 149 37 37 9 ?
-995 112 112 37 37 12 ?
-996 75 75 37 37 18 ≥1 - V.1.39 997 39 494 38 3 2 ≥1044 ≥89 2#219,D#1071 998 58 551 38 4 2 ≥1 - IV.2.7 999 20 190 38 4 6 ≥1 ≥4 2#220 1000 20 152 38 5 8 ≥1 ≥3 2#221 1001 96 608 38 6 2 ≥1 ≥1 2#222 1002 39 247 38 6 5 ≥1 - D#1072 VI.16.18 1003 77 418 38 7 3 ≥2 ? [1042, 1] VI.16.70 1004 20 95 38 8 14 ≥1 - VI.16.88 1005 153 646 38 9 2 ≥1 ? 2#223 [958, 1548] VI.16.31 1006 115 437 38 10 3 ?
-1007 20 76 38 10 18 ≥1016 ≥4 2#224 1008 77 266 38 11 5 ?
?
1009 210 665 38 12 2 ?
-1010 39 114 38 13 12 ≥1 ? 2#225,D#1073 VI.16.88 1011 96 228 38 16 6 ?
? 2#226 1012 153 342 38 17 4 ?
? 2#227 1013 324 684 38 18 2 ≥1 ? 2#228 IV.2.67 1014 685 1370 38 19 1 ?
-1015 343 686 38 19 2 ?
- 2#229 1016 229 458 38 19 3 ?
-1017 172 344 38 19 4 ?
- 2#230 1018 115 230 38 19 6 ?
- 2#231 1019 77 154 38 19 9 ?
-1020 58 116 38 19 12 ?
- 2#232 II.1.3 Parameter Tables 55 No v b r k λ Nd Nr Comments, Ref Where?
1021 39 78 38 19 18 ≥5.87 · 1014 - 2#233,D#1074 1022 666 703 38 36 2 ?
- R#1025 1023 1369 1406 38 37 1 ≥1 ≥1 R#1024,AG(2,37) 1024 1407 1407 38 38 1 ≥1 - PG(2,37) VI.18.82 1025 704 704 38 38 2 ?
-1026 79 1027 39 3 1 ≥1056 - VI.16.12 1027 40 520 39 3 2 ≥6 · 1024 - D#1163 VI.16.13 1028 27 351 39 3 3 ≥1028 ≥1030 3#86 1029 14 182 39 3 6 ≥2 · 1034 - VI.16.82 1030 7 91 39 3 13 1033129 - 13#1 1031 40 390 39 4 3 ≥1033 ≥1033 3#87,D#1164 1032 40 312 39 5 4 ≥1 ≥1 D#1165 [1, 1042] VI.16.85 1033 196 1274 39 6 1 ≥1 - II.3.32 1034 66 429 39 6 3 ≥1 ≥1 3#88 1035 40 260 39 6 5 ≥1 - D#1166 VI.16.86 1036 16 104 39 6 13 ≥1 - #13+#742 1037 14 91 39 6 15 ≥1 - VI.16.86 1038 14 78 39 7 18 ≥1011 0 3#89 1039 40 195 39 8 7 ≥1 ≥1 D#1167 [1, 1042] VI.16.87 1040 105 455 39 9 3 ≥1 - II.3.32 1041 27 117 39 9 12 ≥1017 ≥1017 3#90 1042 40 156 39 10 9 ≥1 ≥3 3#91,D#1168 [1042, 1629] VI.16.87 1043 66 234 39 11 6 ≥21584 ? 3#92 1044 144 468 39 12 3 ≥1 ? 3#93 IV.2.67 1045 40 130 39 12 11 ≥1 - D#1169 VI.16.88 1046 469 1407 39 13 1 ?
-1047 235 705 39 13 2 ?
-1048 157 471 39 13 3 ≥1 - 3#94 IV.2.67 1049 118 354 39 13 4 ?
-1050 79 237 39 13 6 ≥10125 - 3#95 1051 53 159 39 13 9 ≥1 - 3#96 1052 40 120 39 13 12 ≥1033 - 3#97,D#1170 1053 37 111 39 13 13 ≥1 - II.7.46 1054 27 81 39 13 18 ≥208311 - 3#98 1055 40 104 39 15 14 ≥1 - D#1171 1056 222 481 39 18 3 ?
-1057 703 1443 39 19 1 ?
?
1058 40 78 39 20 19 ≥1 ≥1 R#1074,HD 1059 196 364 39 21 4 ?
-1060 976 1464 39 26 1 ?
-1061 326 489 39 26 3 ?
-1062 196 294 39 26 5 ?
-1063 76 114 39 26 13 ?
- R#1073 1064 66 99 39 26 15 ?
-1065 209 247 39 33 6 ?
- R#1072 1066 456 494 39 36 3 ?
- R#1071 1067 703 741 39 37 2 0 0 R#1070,×2 1068 1444 1482 39 38 1 0 0 R#1069,×2,AG(2,38) 1069 1483 1483 39 39 1 0 - ×1,PG(2,38) 1070 742 742 39 39 2 0 - ×1 1071 495 495 39 39 3 ?
-1072 248 248 39 39 6 0 - ×1 1073 115 115 39 39 13 0 - ×1 1074 79 79 39 39 19 ≥2091 - VI.18.73 1075 81 1080 40 3 1 ≥1048 ≥107 AG(4,3) [1463, 1548] VI.16.12 56 2-(v, k, λ) Designs of Small Order II.1 No v b r k λ Nd Nr Comments, Ref Where?
1076 21 280 40 3 4 ≥1024 ≥1021 4#42 1077 9 120 40 3 10 ≥108 ≥105 10#2 1078 6 80 40 3 16 76 1 8#4,4#43 1079 121 1210 40 4 1 ≥1013 - [332, 375] VI.16.61 1080 61 610 40 4 2 ≥104 - 2#237 1081 41 410 40 4 3 ≥1 - D#1192 II.7.46 1082 31 310 40 4 4 ≥1 - 2#238 1083 25 250 40 4 5 ≥1023 - 5#22 1084 21 210 40 4 6 ≥1 - 2#239 1085 16 160 40 4 8 ≥1015 ≥1015 8#5 1086 13 130 40 4 10 ≥1014 - 10#3 1087 11 110 40 4 12 ≥1 - 2#242 1088 9 90 40 4 15 ≥106 - 5#24 1089 161 1288 40 5 1 ≥1 - VI.16.16 1090 81 648 40 5 2 ≥1 - 2#243 1091 41 328 40 5 4 ≥1046 - 4#45,D#1193 1092 33 264 40 5 5 ≥1 - VI.16.17 1093 21 168 40 5 8 ≥1024 - 8#6 1094 17 136 40 5 10 ≥1 - 2#246 1095 11 88 40 5 16 ≥107 - 8#7 1096 201 1340 40 6 1 ≥1 - II.3.32 1097 51 340 40 6 4 ≥1 - 4#48,2#248 1098 21 140 40 6 10 ≥1 - 5#25,2#249 1099 49 280 40 7 5 ≥1053 ≥1053 5#26 1100 21 120 40 7 12 ≥1018 ≥1 4#49,2#250 1101 281 1405 40 8 1 ?
-1102 141 705 40 8 2 ≥1 - 1103 71 355 40 8 4 ≥1 - VI.16.30 1104 57 285 40 8 5 ≥1063 - 5#27 1105 41 205 40 8 7 ≥1 - D#1194 II.7.46 1106 36 180 40 8 8 ≥3 - 4#50,2#251 1107 29 145 40 8 10 ≥1 - 5#28,#157+#379 1108 21 105 40 8 14 ≥1 - VI.16.30 1109 81 360 40 9 4 ≥10108 ≥10108 4#51 1110 361 1444 40 10 1 ?
-1111 181 724 40 10 2 ≥2 - 2#253 [1, 959] VI.16.30 1112 121 484 40 10 3 ≥1 - VI.16.31 1113 91 364 40 10 4 ≥10125 - 4#52 1114 73 292 40 10 5 ≥1 - VI.16.30 1115 61 244 40 10 6 ≥10 - 2#255 1116 46 184 40 10 8 ≥1 - 4#53,2#256 1117 41 164 40 10 9 ≥1 - D#1195 II.7.46 1118 37 148 40 10 10 ≥1 - 2#257 1119 31 124 40 10 12 ≥152 - 4#54 1120 25 100 40 10 15 ≥1 - #160+#384 1121 21 84 40 10 18 ≥5 - 2#259 1122 441 1470 40 12 1 ?
-1123 111 370 40 12 4 ?
- 2#260 1124 45 150 40 12 10 ≥1 - 2#261,#84+#543 1125 481 1480 40 13 1 ?
?
1126 105 300 40 14 5 ?
-1127 561 1496 40 15 1 ?
-1128 141 376 40 15 4 ?
- 2#262 1129 81 216 40 15 7 ≥1 - 1130 57 152 40 15 10 ?
- 2#263 II.1.3 Parameter Tables 57 No v b r k λ Nd Nr Comments, Ref Where?
1131 36 96 40 15 16 ≥1 - 2#264 1132 76 190 40 16 8 ≥1 - 2#265 1133 171 380 40 18 4 ?
- 2#266 1134 361 760 40 19 2 ≥7417 ≥7417 2#267 1135 761 1522 40 20 1 ?
-1136 381 762 40 20 2 ≥3 · 1016 - 2#268 1137 191 382 40 20 4 ?
- 2#269 1138 153 306 40 20 5 ?
-1139 96 192 40 20 8 ≥3 - 2#270 1140 77 154 40 20 10 ?
- 2#271 1141 41 82 40 20 19 ≥1 - D#1196 1142 121 220 40 22 7 ?
-1143 921 1535 40 24 1 ?
-1144 231 385 40 24 4 ?
-1145 93 155 40 24 10 ?
-1146 65 104 40 25 15 ?
- R#1171 1147 1001 1540 40 26 1 ?
-1148 81 120 40 27 13 ≥1013 ≥1013 R#1170,AG3(4, 3) 1149 217 310 40 28 5 ?
-1150 91 130 40 28 12 ?
- R#1169 1151 1161 1548 40 30 1 ?
-1152 291 388 40 30 4 ?
-1153 117 156 40 30 10 ?
- R#1168 1154 156 195 40 32 8 ?
- R#1167 1155 221 260 40 34 6 ?
- R#1166 1156 273 312 40 35 5 ?
- R#1165 1157 351 390 40 36 4 ?
- R#1164 1158 481 520 40 37 3 ?
0 R#1163,×3 1159 741 780 40 38 2 0 - R#1162,×2 1160 1521 1560 40 39 1 ?
? R#1161,AG(2,39) 1161 1561 1561 40 40 1 ?
- PG(2,39) 1162 781 781 40 40 2 0 - ×1 1163 521 521 40 40 3 ?
-1164 391 391 40 40 4 ?
-1165 313 313 40 40 5 0 - ×1 1166 261 261 40 40 6 0 - ×1 1167 196 196 40 40 8 0 - ×1 1168 157 157 40 40 10 0 - ×1 1169 131 131 40 40 12 ?
-1170 121 121 40 40 13 ≥1029 - PG3(4,3) VI.18.73 1171 105 105 40 40 15 ≥4 - 1172 42 574 41 3 2 ≥6 · 1024 ≥1 D [1040, 1548] VI.16.81 1173 124 1271 41 4 1 ≥2 ≥1 [332, 1042, 1047] III.2.9 1174 165 1353 41 5 1 ≥15 ≥1 [396, 1042] III.2.9 1175 42 287 41 6 5 ≥2 ≥2 D [1, 2, 1042] VI.16.86 1176 42 246 41 7 6 ≥1 ≥1 D [1, 1042] VI.16.87 1177 288 1476 41 8 1 ≥1 ≥1 II.7.50 1178 370 1517 41 10 1 ?
?
1179 247 779 41 13 2 ?
?
1180 42 123 41 14 13 ≥1 ? D 1181 247 533 41 19 3 ?
?
1182 780 1599 41 20 1 ?
?
1183 42 82 41 21 20 ≥1 0 R#1196,×3 1184 124 164 41 31 10 ?
0 R#1195,×3 1185 165 205 41 33 8 ?
0 R#1194,×3 58 Triple Systems II.2 No v b r k λ Nd Nr Comments, Ref Where?
1186 288 328 41 36 5 ?
0 R#1193,×3 1187 370 410 41 37 4 ?
0 R#1192,×3 1188 780 820 41 39 2 ?
0 R#1191,×3 1189 1600 1640 41 40 1 ?
? R#1190,AG(2,40) 1190 1641 1641 41 41 1 ?
- PG(2,40) 1191 821 821 41 41 2 ?
-1192 411 411 41 41 4 ?
-1193 329 329 41 41 5 ?
-1194 206 206 41 41 8 0 - ×1 1195 165 165 41 41 10 ?
-1196 83 83 41 41 20 ≥1 - VI.18.73 See Also §II.2 BIBDs with block size 3.
§II.3 Existence results on designs with “small” block size.
§II.7 Resolvable designs.
§II.6 Symmetric designs.
§II.5 Steiner systems.
§IV.2 PBD constructions make BIBDs.
A general introduction to combinatorics and in particular to de-sign theory.
Contains exhaustive catalogues of BIBDs and their resolutions in electronic form.
References Cited: [1, 2, 4, 18, 19, 21, 22, 33, 41, 43, 46, 114, 124, 127, 144, 180, 246, 248, 272, 323, 324, 331, 332, 352, 375, 380, 382, 392, 393, 396, 404, 416, 514, 518, 528, 534, 537, 562, 587, 607,619, 620, 623, 660, 679, 688, 692, 693, 694, 696, 697, 710, 734, 790, 827, 848, 856, 898, 957, 958, 959,976, 978, 1005,1013,1016,1034,1040,1042,1044,1047,1053,1087,1138,1151,1172,1174,1183,1184,1185, 1188,1207,1208,1223,1224,1225,1228,1236,1239,1241,1243,1251,1262,1263,1264,1265,1267, 1268,1271,1319,1321,1339,1346,1347,1349,1372,1373,1374,1375,1377,1378,1379,1380,1381, 1398,1463,1496,1497,1533,1537,1543,1544,1545,1546,1547,1548,1549,1551,1556,1566,1578, 1588,1605,1606,1607,1609,1610,1612,1628,1629,1632,1633,1665,1704,1705,1707,1714,1720, 1729,1743,1773,1832,1833,1839,1877,1900,1903,1912,1938,1940,1941,1943,1944,1945,1946, 1976,1999,2031,2039,2043,2045,2058,2059,2060,2062,2063,2064,2065,2066,2067,2071,2090, 2110,2130,2144,2148] 2 Triple Systems Charles J. Colbourn 2.1 Definitions and Examples 2.1 A triple system (TS(v, λ)) (V, B) is a set V of v elements together with a collection B of 3-subsets (blocks or triples) of V with the property that every 2-subset of V occurs in exactly λ blocks B ∈B. The size of V is the order of the TS. It is a Steiner triple system, or STS(v), when λ = 1.
2.2 Examples A TS(6, 2) is given in Example 1.18. A Steiner triple system of order 7 is given in Example I.1.3.
II.2.2 Direct Constructions 59 2.3 Remark A Steiner triple system of order v can exist only when v −1 is even because every element occurs with v −1 others, and in each block in which it occurs it appears with two other elements. Moreover, every block contains three pairs and hence v 2 must be a multiple of 3. Thus, it is necessary that v ≡1, 3 (mod 6). This condition was shown to be sufficient in 1847.
2.4 Theorem A Steiner triple system of order v exists if and only if v ≡1, 3 (mod 6).
2.5 Theorem A TS(v, λ) exists if and only if v ̸= 2 and λ ≡0 (mod gcd(v −2, 6)).
2.2 Direct Constructions 2.6 Let Wn be an (n + 1)-dimensional vector space over F2. A punctured subspace of Wn is obtained from a subspace by removing the zero vector. The n-dimensional projective space PG(n, 2) is the set of all punctured subspaces of Wn, ordered by inclusion. Punctured subspaces of dimension one are points, and those of dimension two are lines.
2.7 Theorem (Projective triple system) The points and lines of PG(n, 2) form the elements and triples of an STS(2n+1 −1).
2.8 Construction A projective triple system of order 2k−1. Let the points be the nonzero vectors of length k over F2 and {a, b, c} be a block if a + b + c = 0.
2.9 For Vn an n-dimensional vector space over F3, the n-dimensional affine space AG(n, 3) is the set of all subspaces of Vn and their cosets, ordered by set inclu-sion. Points are vectors, and lines are translates of 1-dimensional subspaces.
2.10 Theorem (Affine triple system) The points and lines of AG(n, 3) form the elements and triples of an STS(3n).
2.11 Theorem Let p be a prime, n ≥1, and pn ≡1 (mod 6). Let Fpn be a finite field on a set X of size pn = 6t + 1 with 0 as its zero element, and ω a primitive root of unity. Then {{ωi + j, ω2t+i + j, ω4t+i + j} : 0 ≤i < t, j ∈X} (with computations in Fpn) is the set of blocks of an STS(pn) on X.
2.12 Theorem (Netto triple system) Let p be a prime, n ≥1, and pn ≡7 (mod 12). Let Fpn be a finite field on a set X of size pn = 6t + 1 = 12s + 7 with 0 as its zero element and ω a primitive root of unity. Then {{ω2i + j, ω2t+2i + j, ω4t+2i + j} : 0 ≤i < t, j ∈X} forms the blocks of an STS(pn) on X.
2.13 Remark Constructions VI.53.17 and VI.53.19 give a direct method for producing STSs using Skolem sequences. Algorithm VII.6.71 gives a fast computational method for constructing STSs.
2.3 Recursive Constructions 2.14 Construction (Moore) If there exist an STS(u) and an STS(v) that contains an STS(w) subsystem (see §2.5), then there exists an STS(u(v −w) + w). Because every STS(v) contains a subsystem of orders 0 and 1, if there exist an STS(u) and an STS(v), then there exist an STS(uv) and an STS(u(v −1) + 1).
2.15 Construction An STS(2v + 1) from an STS(v). Let (V, B) be an STS(v) on element set V . Let X = (V × {0, 1}) ∪{∞}. For every block {x, y, z} ∈B, form blocks {(x, α), (y, β), (z, γ)} for α, β, γ ∈{0, 1} with α + β + γ ≡0 (mod 2). Adjoin the blocks {∞, (x, 0), (x, 1)} for x ∈V .
60 Triple Systems II.2 2.16 Remark Construction 2.15 applied to Example I.1.40 produces an STS(15).
This STS(15) contains within it an STS(7) on the elements {0, . . ., 6}× {0}, a subdesign of the STS.
2.17 Construction An STS(3v) from an STS(v) (Bose Construction). Let (V, ⊕) be an idempotent, commutative quasigroup. On V × Z3, let D = {(x, i), (y, i), (x ⊕y, (i + 1) mod 3)} : x, y ∈V, x ̸= y, i ∈Z3}, and C = {{(x, 0), (x, 1), (x, 2)} : x ∈V }.
Then (V × Z3, C ∪D) is an STS(3v).
2.4 Isomorphism and Automorphism 2.18 Two set systems (V, B) and (W, D) are isomorphic if there is a bijection (isomor-phism) φ from V to W so that the number of times B appears as a block in B is the same as the number of times φ(B) = {φ(x) : x ∈B} appears as a block in D. An isomorphism from a set system to itself is an automorphism.
2.19 Remark Up to isomorphism, there are a unique STS(7); a unique STS(9) (Exam-ple 1.22); two STS(13)s (Table 1.26); 80 STS(15)s (Table 1.28); and 11,084,874,829 STS(19)s (, also §VII.6). There are a unique TS(6, 2), 4 TS(7, 2)s (Table 1.19), 10 TS(7, 3)s (Table 1.20), and 36 TS(9, 2)s (Table 1.23). See §II.1 for further enumer-ation results.
2.20 Theorem (see ) The number of STS(v)s is vv2( 1 6 +o(1)) as v →∞.
2.21 Theorem Determining isomorphism of triple systems (of arbitrary index) is poly-nomial time equivalent to determining isomorphism of graphs.
2.22 Theorem Isomorphism of STS(v)s can be determined in vlog v+O(1) time.
2.23 Remark The vlog v+O(1) method can be parallelized: Isomorphism of STS(v)s can be determined in O((logv)2) time with O(vlog v+2) processors .
2.24 A TS(v, λ) is cyclic if it admits an automorphism of order v, and k-rotational if it admits an automorphism containing one fixed point and k cycles each of length v−1 k .
2.25 Theorem (see ) There is a cyclic triple system of order v and index λ if and only if: 1. λ ≡1, 5 (mod 6) and v ≡1, 3 (mod 6), (v, λ) ̸= (9, 1); 2. λ ≡2, 10 (mod 12) and v ≡0, 1, 3, 4,7, 9 (mod 12), (v, λ) ̸= (9, 2); 3. λ ≡3 (mod 6) and v ≡1 (mod 2); 4. λ ≡4, 8 (mod 12) and v ≡0, 1 (mod 3); 5. λ ≡6 (mod 12) and v ≡0, 1, 3 (mod 4); or 6. λ ≡0 (mod 12) and v ≥3.
2.26 Two cyclic TS(v, λ)s are (multiplier) equivalent if, when they share the cyclic auto-morphism i 7→(i + 1) mod v, there is an isomorphism from one to the other of the form i 7→µ · i mod v.
2.27 Theorem (Bays–Lambossy, see ) If B1 and B2 are cyclic collections of subsets of the same set V for |V | a prime, then the two collections are isomorphic if and only if they are multiplier equivalent.
II.2.4 Isomorphism and Automorphism 61 2.28 Remark Theorem 2.27 holds more generally than for prime orders, but does not extend to all orders. Here are three difference families generating three isomorphic but inequivalent cyclic TS(16, 2)s: 1.
{{0, 1, 2}, {0, 2, 9}, {0, 3, 6}, {0, 4, 8}, {0, 5, 10}}; 2.
{{0, 1, 3}, {0, 1, 6}, {0, 2, 7}, {0, 3, 10}, {0, 4, 8}}; 3.
{{0, 1, 7}, {0, 1, 10}, {0, 2, 5}, {0, 2, 13}, {0, 4, 8}}.
This design has a full automorphism group of order 128. Indeed there are isomorphic but inequivalent STS(v)s as well; the smallest example has order v = 49 = 72 .
2.29 Table The number of (multiplier) inequivalent cyclic TS(v, λ)s for small v and λ. When x/y is given, the first number is the count for simple systems, the second number for all.
v\λ 1 2 3 4 5 6 7 5 1/1 0/1 6 0 1/2 0 7 1/1 1/2 1/3 1/4 1/5 0/7 0/8 8 1/12 9 0 0 1/4 1/3 0/4 0/24 1/27 10 0 9/21 0 11 6/8 6/209 12 0/4 62/180 62/3561 13 1/1 7/9 23/47 71/253 117/1224 117/5285 71/20347 14 0 15 2/2 0/9 237/421 902/2981 0/15686 6577/375372 ?/?
16 67/71 6916/21742 66418/2460490 2.30 Table The number of (multiplier) inequivalent cyclic STS(v)s for 19 ≤v ≤57.
Order # Order # Order # Order # 19 4 21 7 25 12 27 8 31 80 33 84 37 820 39 798 43 9508 45 11616 49 157340 51 139828 55 3027456 57 2353310 2.31 Theorem A cyclic STS(n · u) containing a cyclic sub-STS(u) exists if and only if n · u ≡1, 3 (mod 6), n · u ̸= 9, u ≡1, 3 (mod 6), u ̸= 9, and if u ≡3 (mod 6), then n ̸= 3.
2.32 Theorem (see ) A 1-rotational TS(v, λ) exists if and only if λ = 1 and v ≡3, 9 (mod 24) λ ≡1, 5 (mod 6), λ > 1 and v ≡1, 3 (mod 6) λ ≡2, 4 (mod 6) and v ≡0, 1 (mod 3) λ ≡3 (mod 6) and v ≡1 (mod 2) λ ≡0 (mod 6) and v ̸= 2 2.33 Theorem Let v, k be positive integers such that 1 ≤k ≤(v −1)/2. Then a k-rotational STS(v) exists if and only if all of v ≡1, 3 (mod 6); v ≡3 (mod 6) if k = 1; v ≡1 (mod k); and v ̸≡7, 13, 15, 21 (mod 24) if (v −1)/k is even.
2.34 Theorem Almost all Steiner triple systems have no automorphism other than the trivial one.
62 Triple Systems II.2 2.35 Table ( with corrections) The type of an automorphism is the lengths of the cycles of points induced by its action (written in exponential notation omitting the fixed points). In the table on the left, the numbers of STS(19)s that admit one or more basic automorphisms, those of type am for a ≥2 prime, are given. The systems are partitioned by the order of the automorphism group (Ord) and the subset of basic automorphisms represented; to distinguish subsets arising for the same group order, a letter indicates the class (Cl) of the partition. In the table on the right, classes are refined to further indicate the remaining automorphisms. Systems whose group is generated by a single basic automorphism are omitted in the second table. Any type of automorphism not explicitly mentioned does not arise as the type of an automorphism of an STS(19).
Basic Automorphisms Ord Cl191 29 36 28 26 34 #STSs 432 ⋆ ⋆ ⋆ ⋆ 1 171 ⋆ ⋆ 1 144 ⋆ ⋆ ⋆ 1 108 ⋆ ⋆ ⋆ ⋆ 1 96 ⋆ ⋆ ⋆ 1 57 ⋆ ⋆ 2 54 ⋆ ⋆ ⋆ 2 32 ⋆ ⋆ 3 24 ⋆ ⋆ ⋆ 11 19 ⋆ 1 18 a ⋆ ⋆ 1 b ⋆ ⋆ ⋆ 2 c ⋆ ⋆ ⋆ 6 d ⋆ ⋆ 2 16 ⋆ ⋆ 13 12 a ⋆ ⋆ ⋆ 8 b ⋆ ⋆ 7 c ⋆ ⋆ 12 d ⋆ ⋆ ⋆ 10 9 ⋆ 19 8 a ⋆ ⋆ 84 b ⋆ 17 6 a ⋆ ⋆ 14 b ⋆ ⋆ 14 c ⋆ ⋆ 116 d ⋆ ⋆ 10 e ⋆ ⋆ 28 4 a ⋆ ⋆ 839 b ⋆ 662 c ⋆ 620 3 a ⋆ 12664 b ⋆ 64 2 a ⋆ 169 b ⋆ 78961 c ⋆ 70392 totals 4 184128858064572150 124 164758 Nonbasic Automorphisms Cl 92 63 3262 2144 2182 82 44 2262 2243 # 432 ⋆ ⋆ ⋆ ⋆ 1 171 ⋆ 1 144 ⋆ ⋆ ⋆ ⋆ 1 108 ⋆ ⋆ 1 96 ⋆ ⋆ 1 57 2 54 ⋆ 2 32 ⋆ ⋆ 3 24 ⋆ 11 18a ⋆ 1 18b ⋆ 2 18c ⋆ 6 18d ⋆ 2 16 ⋆ ⋆ 1 ⋆ 1 ⋆ ⋆ ⋆ 5 ⋆ 6 12a ⋆ 8 12b 7 12c 12 12d ⋆ 10 9 ⋆ 9 10 8a ⋆ 2 82 8b ⋆ ⋆ 5 ⋆ ⋆ 10 ⋆ ⋆ 2 6a ⋆ 14 6b 14 6c ⋆ 104 12 6d ⋆ 10 6e 28 4a 839 4b ⋆ 498 ⋆ 153 11 4c ⋆ 48 572 totals1015 137 518 16 4 185 24 48 II.2.5 Subsystems, Holes, and Embedding 63 2.5 Subsystems, Holes, and Embedding 2.36 A partial triple system PTS(v, λ) is a set V of v elements and a collection B of triples, so that each unordered pair of elements occurs in at most λ triples of B. Its leave is the multigraph on vertex set V in which the edge {x, y} appears λ −s times when there are precisely s triples of B containing {x, y}. An incomplete triple system of order v and index λ, and with a hole of size w, is a PTS(v, λ) (V, B) with the property that for some W ⊆V , |W| = w, 1. if x, y ∈W, no triple of B contains {x, y}; and 2. if x ∈V \ W and y ∈V , then exactly λ triples of B contain {x, y}.
Such a partial system is denoted ITS(v, w; λ). When an ITS(v, w; λ) (V, B) exists having a hole on W ⊆V , if there is a TS(w, λ) (W, D), the system (V, B ∪D) is a TS(v, λ). In this system, the TS(w, λ) is a subsystem.
2.37 Theorem (Doyen–Wilson Theorem) Let v, w ≡1, 3 (mod 6) and v ≥2w + 1.
Then there exists an STS(v) containing an STS(w) as a subdesign.
2.38 Theorem An ITS(v, w; λ) exists if and only if 1. w = 0 and λ ≡0 (mod gcd(v −2, 6)), 2. v = w, or 3. 0 < w < v and (a) v ≥2w + 1, (b) λ v 2 − w 2 ≡0 (mod 3) , and (c) λ(v −1) ≡λ(v −w) ≡0 (mod 2).
2.39 Theorem A subsystem-free STS(v) exists for all v ≡1, 3 (mod 6).
2.40 Let (V, B) be a PTS(v, λ) and (W, D) a TS(w, µ) for which V ⊆W and B is a submultiset of D. Then (W, D) is an enclosing of (V, B). When λ = µ, the enclosing is an embedding. When v = w, the enclosing is an immersion. When both equalities hold, the enclosing is a completion. An enclosing is faithful if the only triples of D wholly contained in V are those in B.
2.41 Theorem Any partial Steiner triple system of order v can be embedded in a Steiner triple system of order w if w ≡1, 3 (mod 6) and w ≥2v + 1.
2.42 Conjecture Every PTS(v, 1) has an enclosing in a TS(w, µ) in which (w −v) + (µ −1) ≤ 3 if v ≡4 (mod 6), v ≥16, 2 otherwise.
2.43 Remark Even the determination of when a TS(v, λ) can be enclosed in a TS(w, µ) faithfully is not solved. Two partial results are in Theorems 2.44 and 2.45.
2.44 Theorem A TS(v, λ) has a faithful enclosing in a TS(w, µ) whenever µ ≥λ, µ ≡0 (mod gcd(w −2, 6)), and w ≥2v + 1.
2.45 Theorem A TS(v, λ) has a faithful enclosing in a TS(w, 2λ) if and only if w ≥ (3v −1)/2.
64 Triple Systems II.2 2.6 Leaves, Excesses, and Neighborhoods 2.46 Theorem The maximum number of triples in a PTS(v, λ) is ⌊v 3⌊λ(v−1) 2 ⌋⌋−1 if v ≡2(6) and λ ≡4(6) or v ≡5(6) and λ ≡1(3) ⌊v 3⌊λ(v−1) 2 ⌋⌋ otherwise 2.47 Theorem Let G be a graph on v ≡1 (mod 2) vertices with every vertex of degree 0 or 2, having 0 (mod 3) edges if v ≡1, 3 (mod 6) or 1 (mod 3) edges if v ≡5 (mod 6). Then G is the leave of a PTS(v, 1) unless v = 7 and G = 2{C3} ∪K1, or v = 9 and G = C4 ∪C5.
2.48 Conjecture Let G be a graph with maximum degree k on v ≥4k + 2 vertices and m edges in which the degree of every vertex is congruent to v −1 modulo 2, and for which v 2 −m ≡0 (mod 3). Then G is the leave of a PTS(v, 1).
2.49 Let B be a collection of 3-subsets on element set V of cardinality v. For every 2-subset {x, y} of V , let λxy be the number of triples in B in which {x, y} occurs.
If λxy ≥λ for all {x, y} ⊂V , (V, B) is a covering CT(v, λ) whose excess is the multigraph (V, E) in which for every 2-subset {x, y} the edge {x, y} appears λxy −λ times in E. The covering is minimal if every triple in B contains a pair {w, z} for which λwz = λ, and minimum if B is the smallest among all CT(v, λ)s.
2.50 Theorem The number of triples in a minimum CT(v, λ) is µ(v, λ) = ⌈v 3⌈λ(v−1) 2 ⌉⌉+ 1 if v ≡2(6) and λ ≡2(6) or v ≡5(6) and λ ≡2(3) ⌈v 3⌈λ(v−1) 2 ⌉⌉ otherwise 2.51 Theorem Every multigraph with vertex degrees from {0, 2} on v ≡1 (mod 2) vertices, having e edges with e ≡0 (mod 3) when v ≡1, 3 (mod 6) or e ≡2 (mod 3) when v ≡5 (mod 6), is an excess of a CT(v, 1).
2.52 The neighborhood of an element x in a TS(v, λ) is the λ-regular multigraph on v −1 vertices whose edges are the pairs occurring in triples with x.
2.53 Theorem [571, 524] Let G be an n-vertex λ-regular simple graph, other than two disjoint triangles. Then if λ ≡0 (mod gcd(n −1, 6)), G is a λ-neighborhood. In-deed for n ≥8, n ≡0, 2 (mod 3), every λ-regular n-vertex multigraph with λ ≡0 (mod gcd(n −1, 6)) is a λ-neighborhood.
2.54 A double star of triples based on a pair {x, y} of a v-set V is a partial triple system on V so that every triple contains at least one of {x, y}, and every element z ∈V {x, y} appears in exactly one triple with x and exactly one triple with y. Equivalently, a double star is a simultaneous specification of the neighborhoods of x and y, a double neighborhood.
2.55 Theorem Every double star on n ≡1, 3 (mod 6) elements can be completed to a Steiner triple system of order n.
2.7 Intersection, Support, and Large Sets 2.56 Let (V, B) and (V, D) be two STS(v)s. Their intersection size is |B ∩D|. They are disjoint when their intersection size is zero. A set of v −2 STS(v)s {(V, Bi) : i = 1, . . ., v −2} is a large set if every two systems in the set are disjoint.
II.2.7 Intersection, Support, and Large Sets 65 2.57 Theorem Let (V, B) and (V, D) be two STS(v)s. Then there exists a permuta-tion π on V such that (V, B) and (π(V ), π(D)) are disjoint.
2.58 Theorem Let v ≡1, 3 (mod 6), v ≥15, and 0 ≤d ≤ v 2 with d ̸∈{ v 2 − 5, v 2 −3, v 2 −2, v 2 −1}. Then there exist two STS(v)s whose intersection size is d.
2.59 Let (V, B) be a TS(v, λ). The support size is the number of distinct triples in B.
2.60 Theorem [561, 565, 578] Let mv = ⌈v(v−1) 6 ⌉, sv = ⌈v(v+2) 6 ⌉, and Mv,λ = max λv(v−1) 6 , v 3 .
Define PS(v, λ) = ∅if λ ̸≡0 (mod gcd(v−2, 6)). Otherwise when v−2 ̸= λ, PS(v, λ) is: v (mod 12) PS(v, λ) 0, 4 sv, sv + 2, sv + 3, . . ., Mv,λ 1, 3, 7, 9 mv, mv + 4, mv + 6, mv + 7, . . ., Mv,λ 2 sv + 7, . . ., Mv,λ 5, 11 mv + 6, mv + 9, mv + 10, . . ., Mv,λ 6, 10 sv + 1, sv + 2, . . ., Mv,λ 8 sv + 6, sv + 8, sv + 9, . . ., Mv,λ When λ = v −2, PS(v, λ) is defined similarly, but with the omission of {Mv,λ − 5, Mv,λ−3, Mv,λ −2, Mv,λ−1}. Then for v ̸∈{6, 7, 8, 9, 10, 12, 14}, and λ ≥1, there is a TS(v, λ) with support size s if and only if s ∈PS(v, λ) except possibly when v ≡8 (mod 12) and s = sv + 7.
2.61 For a TS(v, λ), let ci be the number of distinct blocks that are each i-times repeated in the system. The fine structure of the TS is (c1, c2, . . ., cλ).
2.62 Theorem [569, 570] Let v ≥17. Then (3s −2t + δ, t, ⌊v(v −1)/6⌋−s) is the fine structure of a TS(v, 3) if and only if • v ≡1, 3 (mod 6); δ = 0; 0 ≤t ≤s ≤v(v −1)/6; s ̸∈{1, 2, 3, 5}; and (t, s) ̸∈ {(1, 4), (2, 4), (3, 4), (1, 6), (2, 6), (3, 6), (5, 6), (2, 7), (5, 7), (1, 8), (3, 8), (5, 8)} or • v ≡5 (mod 6); δ = 1; 0 ≤t < s ≤⌊v(v −1)/6⌋; s ̸∈{1, 2, 4, 5}; and (t, s) ̸∈ {(1, 3), (2, 3), (0, 6), (1, 6), (2, 6), (3, 6), (0, 7), (1, 7), (2, 7), (3, 7), (5, 7), (0, 8), (1, 8), (2, 8), (6, 8), (7, 8), (1, 9)}.
2.63 Theorem [1482, 1483, 2015] There exists a large set of STS(v)s if and only if v ≡1, 3 (mod 6) and v ̸= 7.
2.64 Theorem When v ≡0, 4 (mod 6), there are v−2 2 disjoint TS(v, 2)s and when v ≡2 (mod 6) there are v−2 6 disjoint TS(v, 2)s . When v ≡5 (mod 6), there are v−2 3 disjoint TS(v, 3)s . From these large sets and Theorem 2.63, it follows that a simple TS(v, λ) exists if and only if λ ≤v −2 and λ ≡0 (mod gcd(v −2, 6)).
2.65 An overlarge set of STS(v)s is a partition of all triples on v + 1 points into v + 1 disjoint STS(v)s.
2.66 Theorem The v + 1 derived designs of a Steiner quadruple system of order v + 1 form an overlarge set of STS(v)s; hence, these exist for all v ≡1, 3 (mod 6).
2.67 Two STS(v)s are almost disjoint if they share exactly one triple. A large set of mutually almost disjoint (MAD) STS(v)s is a set of k STS(v)s that are pairwise almost disjoint, and for which every triple on v points appears in at least one of the STSs.
66 Triple Systems II.2 2.68 Theorem (see ) There exists a large set of k MAD STS(v)s only if k ∈{v, v +1} when v > 7. A large set of v MAD STS(v)s exists if and only if v ≡1, 3 (mod 6).
2.69 Conjecture A large set of v + 1 MAD STS(v)s exists if and only if v ≡1, 3 (mod 6) and v ≥13. (Solutions are known for v = 13 and v = 15.) 2.70 Remark The existence of large sets of Kirkman triple systems of order v when v ≡ 3 (mod 6), known as Sylvester’s Problem, remains open.
See for a survey.
Sylvester posed the problem when v = 15 and settled existence when v is a power of three.
2.71 Example A large set of KTS(15)s. Consider the KTS(15) in which the rows form the parallel classes: 0,1,9 2,4,12 5,10,11 7,8,α 3,6,β 0,2,7 3,4,8 5,6,12 9,11,α 1,10,β 0,3,11 1,7,12 6,8,10 2,5,α 4,9,β 0,4,6 1,8,11 2,9,10 3,12,α 5,7,β 0,5,8 1,2,3 6,7,9 4,10,α 11,12,β 0,10,12 3,5,9 4,7,11 1,6,α 2,8,β 1,4,5 2,6,11 3,7,10 8,9,12 0,α,β Under the automorphism fixing α and β and mapping i 7→(i + 1) mod 13, thirteen disjoint KTS(15)s are produced; this is a large set.
2.72 Two STS(v)s (V, A) and (V, B) are orthogonal if A ∩B = ∅, and whenever {{a, b, w}, {x, y, w}} ⊂A and {{a, b, s}, {x, y, t}} ⊂B, one has s ̸= t.
2.73 Theorem If v ≡1, 3 (mod 6), v ≥7, and v ̸= 9, then there exist two orthogonal STS(v)s.
2.74 Theorem There exist three mutually orthogonal STS(v)s whenever v ≡1, 3 (mod 6), v ≥19, except possibly when v = 6n + 3 for n ∈{3, 17, 27, 29, 31, 34, 35, 37, 38, 40, . . ., 51, 57, 58, 59}.
2.8 Resolvability 2.75 A set of blocks is a partial parallel class (PPC) if no two blocks in the set share an element. A PPC is an almost parallel class if it contains v−1 3 blocks; when it contains v 3 blocks, it is a parallel class or resolution class. A partition of all blocks of a TS(v, λ) into parallel classes is a resolution and the STS is resolvable. An STS(v) together with a resolution of its blocks is a Kirkman triple system, KTS(v).
2.76 Example In 1850, the Reverend Thomas P. Kirkman posed the question: Kirkman’s schoolgirl problem: Fifteen young ladies in a school walk out three abreast for seven days in succession: it is required to arrange them daily, so that no two walk twice abreast.
This is equivalent to finding a KTS(15), a resolution of some Steiner triple system of order 15. Of the eighty nonisomorphic STS(15)s (Table II.1.28), exactly four are resolvable.
However, there are seven nonisomorphic resolved systems, seven KTS(15)s. These are given here, using the numbering of the underlying STSs and displaying blocks as columns.
II.2.8 Resolvability 67 The Seven Solutions of the Kirkman Schoolgirl Problem # Monday Tuesday Wednesday Thursday Friday Saturday Sunday 1a adefg abcdf abcdg abcfg abcde abcef abceg bjhik hemkj jmehi dlikh fhlik lidjh ndhij cnmol ignol kofln enjmo gjomn mkgon ofklm 1b adefg abcdf abcdg abcfg abcde abceg abcef bjhik hemkj jmehi dilhj flhij ldikh nhdik cnmol ignol kofln ekonm gnkmo mfjno ojglm 7a adefg abceg abcdf abcfg abcde abcef abcdg bjikh hdlkj jemhi dlihk fmhij lidhj nheki comln ifonm kgnlo enjmo goknl mkgon ojfml 7b adefg abceg abcdf abcfg abcde abcdg abcef bjikh hdlkj jemhi dmhji flikh lheik nidjh comln ifonm kgnlo eoknl gnjmo mjfno okglm 15a abcfg abcde abceg abcdf adefg abcdg abcef dihmj fhikl ldkih nejih bhkji hmejk jldhi eklno gjnmo mfojn ogmlk conlm iofnl kngmo 15b abcfg abcde abceg abcdf adefg abcef abcdg dihmj fhikl ldkih nejih bjhik hmdkj jlehi eklno gjnmo mfojn ogmlk cnmol iognl knfom 61 adefg abcfg abcde abcdf abcdg abcef abceg bijlh dijhk fhimk heljk jmehi lkdhi ndhij cknom elmno gjonl ignom kofln mngoj ofkml 2.77 Theorem [1484, 1781] There exists a KTS(v) if and only if v ≡3 (mod 6).
2.78 Theorem Let v, w ≡3 (mod 6) and v ≥3w. Then there exists a KTS(v) containing a sub-KTS(w) (the resolution of the KTS(v) extends that of the KTS(w)).
2.79 A Hanani triple system is an STS(v) with a partition of its blocks into (v −1)/2 maximum parallel classes, and a single partial parallel class with (v −1)/6 blocks.
2.80 Theorem A Hanani triple system of order v exists if and only if v ≡1 (mod 6) and v ̸∈{7, 13}.
2.81 A block-coloring of a TS(v, λ) is an assignment of colors to the blocks so that no two blocks of the same color intersect (equivalently, a partition of the blocks into partial parallel classes).
2.82 Conjecture (Special case of the Erd˝ os–Faber–Lov´ asz Conjecture) Every STS(v) admits a block-coloring with at most v colors.
2.83 Theorem (see ) There is a triple system TS(v, λ) that admits a block-coloring in l λv(v−1)/6 ⌊v/3⌋ m colors, in which at most one color class is not a partial parallel class with ⌊v 3⌋blocks, whenever λ ≡0 (mod gcd(v −2, 6)) and v ≥3, except when v = 6 and λ ≡2 (mod 4), (v, λ) = (13, 1), or v = 7 and λ ∈{1, 3, 5}.
2.84 Remark Not every Steiner triple system has even a single parallel class. The minimum size of a maximum cardinality partial parallel class in an STS(v) remains open.
2.85 Theorem Every STS(v) contains a partial parallel class containing at least (v/3)− cv1/2(ln v)3/2 blocks, for c an absolute constant.
2.86 Conjecture (see ) For all v ≡1, 3 (mod 6), v ≥15, there exists an STS(v) whose largest partial parallel class has fewer than ⌊v/3⌋blocks.
68 Triple Systems II.2 2.9 Configurations in Steiner Triple Systems 2.87 A (k, ℓ)-configuration in an STS(v) is simply a set of ℓlines whose union contains precisely k points. It is constant if every STS(v) contains the same number of copies of the configuration, variable otherwise.
2.88 Remark Each configuration with at most three lines is constant. The number of 1-line configurations is just the number of triples, v(v −1)/6. There are two 2-line configurations. A pair of disjoint triples occurs v(v −1)(v −3)/72 times, while a pair of intersecting triples appears v(v −1)(v −3)/8 times. The five 3-line configurations Bi, i = 1, . . ., 5 are B1 B2 B3 B4 B5 Set nv = v(v −1)(v −3). The number bi of times that Bi occurs satisfies b1 = nv(v −7)(v2 −19v + 96)/1296; b2 = nv(v −7)(v −9)/48; b3 = nv(v −4)/48; b4 = nv(v −7)/8; b5 = nv/6 = 4a2/3.
2.89 Remark There are 16 4-line configurations, denoted Ci for i = 1, . . ., 16: C1 C2 C3 C4 C5 C6 C7 C8 C9 C10 C11 C12 C13 C14 C15 C16 Let ci be the number of occurrences of Ci. Five are constant: c4 = nv(v −7)(v −9)(v −11)/288; c7 = nv(v −5)(v −7)/384; c8 = nv(v −7)(v −9)/16; c11 = nv(v −7)/4; and c15 = nv/6.
2.90 Theorem The number of occurrences of each variable 4-line configuration is determined by the number of occurrences of any single one of them. Indeed, c1 = nv(v −9)(v −10)(v −13)(v2 −22v + 141)/31104 + c16 c2 = nv(v −9)(v −10)(v2 −22v + 129)/576 −6c16 c3 = nv(v −9)2(v −11)/128 + 3c16 c5 = nv(v −9)(v2 −20v + 103)/48 + 12c16 c6 = nv(v −9)(v −10)/36 −4c16 c9 = nv(v −9)2/8 −12c16 c10 = nv(v −8)/8 + 3c16 c12 = nv(v −9)/4 + 12c16 c13 = nv(v2 −18v + 85)/48 −4c16 c14 = nv/4 −6c16 2.91 Theorem [949, 1472] For every v ≡1, 3 (mod 6), v ̸∈{7, 13}, there is an anti-Pasch Steiner triple system, an STS(v) in which no four triples are isomorphic to the Pasch configuration C16.
2.92 Theorem A mitre configuration with 7 points and 5 lines is pictured on the right. There is an anti-mitre STS(v) (one that con-tains no configuration isomorphic to the mitre configuration) if and only if v ≡1, 3 (mod 6) and v ̸= 9.
II.2.10 Coloring and Independent Sets 69 2.93 Theorem Let P(v) be the maximum number of Pasch configurations in a Steiner triple system on v points, and let M(v) = v(v −1)(v −3)/24. Then P(v) ≤M(v), with equality if and only if v = 2d+1 −1 (the STS is projective).
Moreover, for v ̸= 2d+1 −1, limsupv→∞P(v)/M(v) = 1.
2.94 Theorem The number of mitres in an STS(v) is at most v(v−1)(v−3)(v−7)/24, and equality is attained if and only if v is a power of 3 and the STS(v) is a Hall triple system (see §VI.28).
2.95 For ℓ≥2, an STS(v) is ℓ-sparse if there is no set of σ + 2 points that underlie σ triples for 1 < σ ≤ℓ.
2.96 Remarks In 1976, Erd˝ os conjectured that an ℓ-sparse STS(v) exists for every integer ℓ≥2. Every STS(v) is 3-sparse. An STS is 4-sparse exactly when it is anti-Pasch; see Theorem 2.91. It is 5-sparse when it is both anti-Pasch and anti-mitre. A 5-sparse STS(v) is known to exist when v ≡3 (mod 6) and v ≥21, and for many other orders . However a complete characterization is not known. Only finitely many 6-sparse STSs are known; see . No ℓ-sparse STS(v) is known for any ℓ≥7.
2.97 An STS(v) (V, B) is G-uniform if for every triple {x, y, z}, the graph on vertex set V \ {x, y, z} with edges {{a, b} : {a, b, w} ∈B, a, b ∈V \ {x, y, z}, w ∈{x, y}} is isomorphic to G. When G consists of a single (v −3)-cycle, the STS is perfect.
2.98 Remark The projective, affine, Hall, and Netto triple systems are all uniform. However perfect systems are known for only 14 orders: 7, 9, 25, 33, 79, 139, 367, 811, 1531, 25771, 50923, 61339, 69991, and 135859 .
2.10 Coloring and Independent Sets 2.99 In a set system (V, B), a set W ⊆V is independent if there is no B ∈B with B ⊆W.
2.100 Theorem The size of a largest independent set in a TS(v, λ) is (v + 1)/2 if v ≡3 (mod 4) or v ≡1 (mod 4) and λ ≡0 (mod 2), v 2 if v ≡0 (mod 2), (v −1)/2 if v ≡1 (mod 4) and λ ≡1 (mod 2) 2.101 Theorem The size of the smallest maximum independent set in an STS(v) is bounded below by c1 √ v ln v and above by c2 √ v lnv, for absolute constants c1 and c2.
2.102 A set system S = (V, B) is (weakly) m-colorable if V can be partitioned into m sets, each of which is independent. It is (weakly) m-chromatic if it is weakly m-colorable but not weakly (m −1)-colorable.
2.103 Theorem For all m ≥3, there exists vm so that for every v > vm with v ≡1, 3 (mod 6), there exists a weakly m-chromatic STS(v).
2.104 Theorem (see ) There is a 3-chromatic STS(v) whenever v ≡1, 3 (mod 6).
2.105 Theorem There is a 4-chromatic STS(v) whenever v ≡1, 3 (mod 6) and v ≥21.
There is no 4-chromatic STS(v) for v ≤15, and the case v = 19 is in doubt.
70 Triple Systems II.2 2.11 Decomposability 2.106 Remark Theorem 2.64 constructs triple systems that are decomposable into Steiner triple systems.
More generally, a TS(v, λ) is decomposable if its block set is the (multiset) union of the block sets of a TS(v, λ1) and a TS(v, λ2) with 1 ≤λ1 ≤ λ2 < λ. It is indecomposable otherwise. For fixed v, there exist only finitely many indecomposable TS(v, λ)s .
2.107 Theorem Whenever λ ≡0 (mod gcd(v −2, 6)) and v ≥24λ −5, an indecom-posable, simple TS(v, λ) exists.
2.12 Nested and Derived Triple Systems 2.108 A TS(v, λ) (V, B) is nested in (a 2-(v, 4, λ) design) (V, D) when there is a mapping ψ : B 7→V for which the collection {B ∪{ψ(B)} : B ∈B} is equal to D.
2.109 Theorem [535, 1965] Whenever there exists a TS(v, λ), there exists a nested TS(v, λ).
2.110 Remarks It is not known whether there is any TS(v, λ) that cannot be nested. Indeed, for v ≡1 (mod 6) and v ≤61, every STS(v) admitting a cyclic automorphism can be nested so that the resulting 2-(v, 4, 2) design shares the same cyclic automorphism.
2.111 Conjecture (Nov´ ak ) For v ≡1 (mod 6), every STS(v) admitting a cyclic auto-morphism can be nested so that the resulting 2-(v, 4, 2) design shares the same cyclic automorphism.
2.112 Remark (see ) Triple systems can also be extended to designs of higher strength t, and can be produced as derived designs of 3-(v, 4, λ) designs. Every STS(v) is the derived design of some 3-(v, 4, 1) design when v ≤15. However, it is open whether every STS(v) is derived for larger values of v.
2.13 Directed and Mendelsohn Triple Systems 2.113 The transitive tuple (x, y, z) contains the three ordered pairs (x, y), (x, z), and (y, z).
A directed triple system DTS(v, λ) is a v-set V and a collection of transitive tuples (blocks) each having distinct elements of V for which every ordered pair is contained in exactly λ blocks. The underlying TS(v, 2λ) is obtained by omitting the ordering.
2.114 Theorem Every TS(v, 2λ) underlies a DTS(v, λ).
2.115 The cyclic tuple ⟨x, y, z⟩contains the three ordered pairs (x, y), (y, z), and (z, x).
A Mendelsohn triple system MTS(v, λ) is a v-set V and a collection of cyclic tuples (blocks) each having distinct elements of V for which every ordered pair is contained in exactly λ blocks.
2.116 Theorem 1. An MTS(v, λ) exists if and only if λv(v −1) ≡0 (mod 3) and (v, λ) ̸= (6, 1).
2. Not every TS(v, 2λ) underlies an MTS(v, λ); indeed, it is NP-complete to decide.
II.2.13 Directed and Mendelsohn Triple Systems 71 See Also §I.1 Connections of STS(v)s to other designs.
§I.2 Triple systems are often the first class of designs studied in the history of designs.
§II.1 Existence and enumeration results.
§II.4 Triple systems are obtained by derivation on designs of higher strength t.
§II.5 Steiner quadruple systems are extensions of STSs.
§II.7 Kirkman triple systems are special types of resolvable designs.
§III.2 Steiner triple systems are equivalent to Steiner quasigroups.
§IV.3 Generalizations to pairwise balanced designs.
§IV.4 Existence results for GDDs with block size three.
§VI.2 Balanced ternary designs with block size three.
§VI.3 Triple systems arise in balanced tournament scheduling.
§VI.5 “Large” automorphism groups of triple systems.
§VI.7 More on configurations.
§VI.11 CTs are coverings with block size three.
§VI.12 Triple systems are cycle decompositions with cycle length three.
§VI.13 Defining sets of triple systems.
§VI.16 Cyclic triple systems are difference families with block size three.
§VI.20 Directed triple systems.
§VI.24 Triple systems are graph decompositions into triangles.
§VI.25 Embeddings of triple systems on surfaces.
§VI.28 Hall triple systems are STSs in which every three points not on a triple generate an sub-STS(9).
§VI.30 Triple systems on an infinite number of points.
§VI.33.1 KTSs are used to produce low density parity check codes.
§VI.35 Mendelsohn triple systems.
§VI.36 Nested triple systems are a class of nested designs.
§VI.40 PTSs are packings with block size three.
§VI.53 Skolem sequences are used to construct triple systems.
§VI.56 STSs are used to produce cover-free families.
§VI.51 Triple systems arise in tournament scheduling.
§VI.60 Trades are used extensively to prove theorems on support and intersection.
§VII.1 Point codes and block codes of STSs.
§VII.5 STSs are equivalent to Steiner 1-factorizations of graphs.
§VII.6 Many algorithmic techniques and complexity results have been established for triple systems.
§VII.7 Connections with linear algebra.
A comprehensive introduction to triple systems.
References Cited: [78,130,267,268,361,442,523,524,535,541,549,551,552,556,561,565,569,570, 571,573,574,578,581,589,638,690,712,745,747,784,785,786,787,826,945,949,955,970,1004, 1268,1300,1423,1462,1464,1472,1482,1483,1484,1586,1603,1668,1670,1687,1737,1739,1741, 1781,1792,1957,1965,2008,2009,2011,2015,2095,2163,2164,2204] 72 BIBDs with Small Block Size II.3 3 BIBDs with Small Block Size R. Julian R. Abel Malcolm Greig 3.1 Existence Results for BIBDs for Specific k and λ 3.1 Remarks The existence question for BIBDs with small k is: Given k and λ, for what values of v are Proposition II.1.2 (integrality conditions) and Theorem II.1.9 (Fisher’s inequality) sufficient? This problem has been addressed for all λ when k ≤9. For k ≥10, much less is known, although for some such k, there are known bounds on v for existence of near-resolvable BIBD(v, k, k −1)s (and hence also BIBD(v, k, k + 1)s with v ≡1 (mod k); see §II.7.5).
The notation B(k, λ) is used for the set of v values for which a BIBD(v, k, λ) is known. Theorem IV.3.7 establishes that, for every fixed k and λ, there is a constant C(k, λ), so that if v > C(k, λ) and v satisfies λ(v−1) ≡0 (mod k−1) and λv(v−1) ≡0 (mod k(k −1)), then there exists a BIBD(v, k, λ). Table 3.3 summarizes the known results for k ≤9.
3.2 Proposition For any v, k, if λmin is the minimum λ for which the integrality conditions are satisfied, then λmin divides k(k −1) (if k is odd) or k(k −1)/2 (if k is even).
3.3 Table A summary of known results.
Possible Possible Largest k, λ Reference Values Exceptions Exceptions Possible of v Exception 3,1 1,3 mod 6 none none 3,2 0,1 mod 3 none none 3,3 1 mod 2 none none 3,6 all none none 4,1 1,4 mod 12 none none 4,2 1 mod 3 none none 4,3 0,1 mod 4 none none 4,6 all none none 5,1 1,5 mod 20 none none 5,2 1,5 mod 10 15 none 5,4 0,1 mod 5 none none 5,5 1 mod 4 none none 5,10 1 mod 2 none none 5,20 all none none 6,1 [41, 46, 1138] 1,6 mod 15 16,21,36,46 Table 3.4 801 6,2 1,6 mod 15 21 none 6,3 1 mod 5 none none 6,5 0,1 mod 3 none none 6,15 all none none II.3.1 Existence Results for BIBDs for Specific k and λ 73 Possible Possible Largest k, λ Reference Values Exceptions Exceptions Possible of v Exception 7,1 [3, 4, 961, 1184] 1,7 mod 42 43 Tables 3.5, 3.6 2605 7,2 [4, 42] 1,7 mod 21 22 Table 3.7 994 7,3 [42, 1044, 2194, 2214] 1,7 mod 14 none none 7,6 0,1 mod 7 none none 7,7 1 mod 6 none none 7,14 1 mod 3 none none 7,21 [42, 1044] 1 mod 2 none none 7,42 all none none 8,1 [3, 961] 1,8 mod 56 none Tables 3.8, 3.9 3753 8,2 1,8 mod 28 29,36 Table 3.10 589 8,4 [19, 1372] 1 mod 7 22 none 8,7 [19, 2182] 0,1 mod 8 none none 8,14 0,1 mod 4 none none 8,28 0,1 mod 2 none none 8,56 all none none 9,1 [3, 961] 1,9 mod 72 none Tables 3.11, 3.12 16497 9,2 1,9 mod 36 none Table 3.13 1845 9,3 1,9 mod 24 none Table 3.13 385 9,4 1,9 mod 18 none Table 3.13 783 9,6 1,9 mod 12 none Table 3.13 213 9,8 0,1 mod 9 none none 9,9 1 mod 8 none none 9,12 1,3 mod 6 none none 9,18 1 mod 4 none none 9,24 0,1 mod 3 none none 9,36 1 mod 2 none none 9,72 all none none 3.4 Table Values of v for which existence of a BIBD(v, 6, 1) remains undecided.
51 61 81 166 226 231 256 261 286 316 321 346 351 376 406 411 436 441 471 501 561 591 616 646 651 676 771 796 801 3.5 Table Values of t for which existence of a BIBD(42t + 1, 7, 1) remains undecided.
2 3 5 6 12 14 17 19 22 27 33 37 39 42 47 59 62 3.6 Table Values of t for which existence of a BIBD(42t + 7, 7, 1) remains undecided.
3 19 34 39 3.7 Table Values of v for which existence of a BIBD(v, 7, 2) remains undecided.
274 358 574 694 988 994 3.8 Table Values of t for which existence of a BIBD(56t + 1, 8, 1) remains undecided.
2 3 4 5 6 7 14 19 20 21 22 24 25 26 27 28 31 32 34 35 39 40 46 52 59 61 62 67 3.9 Table Values of t for which existence of a BIBD(56t + 8, 8, 1) remains undecided.
3 11 13 20 22 23 25 26 27 28 74 BIBDs with Small Block Size II.3 3.10 Table Values of v, λ with λ > 1 for which existence of a BIBD(v, 8, λ) remains undecided. There is no BIBD for (v, k, λ) = (22, 8, 4), (29, 8, 2), or (36, 8, 2).
v λ v λ v λ v λ v λ v λ 365 2 393 2,5 477 2 484 2 533 2 540 2 589 2 785 3,5 1121 3,5 1128 3,5 1177 3,5 3.11 Table Values of t for which existence of a BIBD(72t + 1, 9, 1) remains undecided.
2 3 4 5 7 11 12 15 20 21 22 24 27 31 32 34 37 38 40 42 43 45 47 50 52 53 56 60 61 62 67 68 75 76 84 92 94 96 102 132 174 191 194 196 201 204 209 3.12 Table Values of t for which existence of a BIBD(72t + 9, 9, 1) remains undecided.
2 3 4 5 12 13 14 18 22 23 25 26 27 28 31 33 34 38 40 41 43 46 47 52 59 61 62 67 68 76 85 93 94 102 103 139 148 174 183 192 202 203 209 229 3.13 Table Values (v, k, λ) with λ > 1 for which existence of a BIBD(v, 9, λ) remains undecided.
(177,9,3) (189,9,2) (213,9,6) (253,9,2) (315,9,4) (345,9,3) (385,9,3) (459,9,4) (505,9,2) (765,9,2) (783,9,4) (837,9,2) (1197,9,2) (1837,9,2) (1845,9,2) 3.14 Table Parameter sets (v, k, λ) with λ(v −1) ≡0 (mod (k −1)), λv(v −1) ≡0 (mod k(k −1)), v ≤43 and k ≤v/2 for which no BIBD exists. These nonexistence results are generally due to Fisher’s inequality or to the known necessary conditions for symmetric and quasi-residual designs. The nonexistence of a (22, 8, 4) design was shown by computer search .
Fisher: (16,6,1) (21,6,1) (25,10,3) (34,12,2) (36,15,2) (36,15,4) Symmetric: (22,7,2) (29,8,2) (34,12,4) (43,7,1) (43,15,5) Quasi residual: (15,5,2) (21,6,2) (36,6,1) (22,8,4) (36,8,2) 3.15 Remark The only BIBD(v, k, λ)s with v ≤43 and k ≤v/2 whose existence remains unsolved have the parameters: (39,13,6), (40,14,7), and (42,10,3).
3.16 Remark There exists a k-GDD of type kt(k−1)+1 (or equivalently, a BIBD(k(k −1)t + k, k, 1) containing a parallel class) if any one of the following conditions is satisfied: 1. k ≤5.
2. k = 6 and t ̸= 1, 2, 7, 9, 11, 39, 50, or 51.
3. k = 7, t is not in Table 3.6, and t ̸= 24.
4. k = 8 and t is not in Table 3.9.
5. k = 9, t is not in Table 3.12, and t ̸= 16, 20, 48, 60, 92, 104, 147, 166, 187, 191, or 205.
3.17 Remark k-GDDs of type (k(k −1))t can sometimes be useful for BIBD constructions.
For k = 3, 4, and 5, k-GDDs of type (k(k−1))t exist for all t ≥k. For k = 6, not much is known, as most recursive constructions require the existence of k-GDDs of types (k(k −1))x and (k(k −1))x+1 for some small value of x; at present, none is known for k = 6. Tables 3.18 through 3.21 summarize known results for k ∈{6, 7, 8, 9} and t < 100.
3.18 Table Values of t < 100 for which existence of a 6-GDD of type 30t is known.
6 16 21 26 31 36 41 51 61 66 71 76 78 81 86 90 91 96 II.3.2 Recursive Constructions 75 3.19 Table Values of t < 100 for which existence of a 7-GDD of type 42t is known.
7 8 9 15 17 28 33 36 41 49 50 56 57 63 64 65 70 71 72 73 77 78 80 81 85 88 91 92 96 97 99 3.20 Table Values of t < 100 for which existence of a 8-GDD of type 56t is known.
8 9 10 15 16 17 29 30 33 36 37 41 43 44 49 50 51 53 54 55 57 58 63 64 65 71 72 73 74 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 3.21 Table Values of t < 100 for which existence of a 9-GDD of type 72t is known.
9 10 17 19 49 54 55 57 58 64 65 66 73 74 80 81 89 90 91 3.2 Recursive Constructions 3.22 Theorem A BIBD(v, k, λ) exists if any one of the following conditions holds: 1. v = w+ r/λ, where there exist a BIBD(r/λ, k, λ) and an RBIBD(w, k −1, λ) whose blocks form r parallel classes.
2. v = (w −1)m + 1, λ = λ1λ2, where a (k, λ1)-GDD of type (k −1)(w−1)/(k−1), a BIBD((k −1)m + 1, k, λ1λ2), and TDλ2(k, m) exist.
3. v = v1(v2 −α) + α, λ = λ1λ2 where α is 0 or 1, a TDλ2(k, v2 −α) exists, and BIBD(v1, k, λ1) and BIBD(v2, k, λ) both exist.
3.23 Remark Most recursive BIBD constructions start with a GDD (usually a truncated transversal design), use Wilson’s Fundamental GDD construction (Theorem IV.2.5, using Remark IV.2.6) to obtain a larger GDD, and finally fill in the groups of the larger GDD. Some examples follow.
3.24 Example Suppose TD(10, t) exists and 0 ≤u ≤t.
Then a {9, 10}-GDD of type t9u1 exists, and because 9-GDD of type 8x exists for x ∈{9, 10}, a 9-GDD of type (8t)9(8u)1 also exists. Therefore, if α = 1 or 9 and {8t + α, 8u + α} ⊂B(9, 1), then 8(9t + u) + α ∈B(9, 1).
3.25 Example Suppose TD(9, t) exists and 0 ≤u ≤t. Then a {8, 9}-GDD of type t8u1 exists, and because a 7-GDD of type 42x exists for x ∈{8, 9}, a 7-GDD of type (42t)8(42u)1 also exists. Therefore, if α = 1 or 7 and {42t + α, 42u + α} ⊂B(7, 1), then 42(8t + u) + α ∈B(7, 1).
3.26 Example Suppose TD(8, t) exists and 0 ≤u ≤t. Then a {7, 8}-GDD of type t7u1 exists, and because a (7, 3)-GDD of type 2x exists for x ∈{7, 8}, a (7, 3)-GDD of type (2t)7(2u)1 also exists. Therefore, if {2t + 1, 2u + 1} ⊂B(7, 3), then 2(7t + u) + 1 ∈ B(7, 3).
3.27 Example Suppose TD(9, t) exists and 0 ≤u, w ≤t. Then a {7, 8, 9}-GDD of type t7u1w1 exists, and because (7, 21)-GDDs of type 2x exist for x ∈{7, 8, 9}, a (7, 21)-GDD of type (2t)7(2u)1(2w)1 also exists. Therefore, if {2t + 1, 2u + 1, 2w + 1} ⊂ B(7, 21), then 2(7t + u + w) + 1 ∈B(7, 21).
3.28 Example Suppose there exists an RBIBD(w, k, 1) whose blocks form r parallel classes.
Then if x ≤r, adding each of x infinite points to a different parallel class gives a {k, k + 1}-GDD of type 1wx1. If k, k −1 are prime powers, then k-GDDs of type (k −1)t exist for t = k, k + 1; hence, a k-GDD of type (k −1)w((k −1)x)1 exists.
Therefore, if (k −1)x + 1 ∈B(k, 1), then (k −1)(w + x) + 1 ∈B(k, 1). |
3752 | https://brighterly.com/math/digit/ | What is a Digit in Math ⭐ Definition, Types, Examples, Facts
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Digit in Math – Definition with Examples
Digit in Math – Definition with Examples
Jo-ann Caballes
Updated on January 2, 2024
Table of Contents
Welcome to another enlightening post from Brighterly, your reliable companion in making the world of numbers simpler, enjoyable, and comprehensive for children. Today, we dive into one of the most fundamental aspects of mathematics — the digit. Despite being the smallest unit in mathematics, digits form the basis of all numeric systems and play a crucial role in arithmetic and beyond. What is a digit in math? What role does it play? How does its position in a number impact the number’s value? We aim to answer these questions and more in this all-inclusive guide. At Brighterly, we believe that a strong foundation in understanding digits and their properties can make children’s journey through the broader aspects of mathematics a breeze.
What are Digits in Math?
When we think about mathematics, we often picture numbers, operations, and complex formulas. However, the most basic building block of all numbers in math is the humble digit. But what exactly is a digit? In the world of arithmetic, a digit is any one of the ten symbols: 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9. These simple yet powerful symbols lay the foundation for all number systems, which are essential for our everyday math calculations. Whether we’re counting apples, measuring distances, or solving complex equations, digits are the tools we employ in our mathematical arsenal. It’s not just about the numbers; understanding the essence of digits can help enhance our comprehension and appreciation of math as a whole.
Definition of a Digit in Mathematics
In mathematics, a digit is defined as a single character used in numerals. As previously mentioned, we have ten digits in the base-10 or decimal number system that we use daily. These are 0-9. These numbers form the basis of all our mathematical operations, whether it’s simple addition or complex calculus. Each digit has a unique value, and when they come together to form larger numbers, their position also determines the overall number’s value. For instance, in the number 123, 1, 2, and 3 are all digits, but 1 represents hundred, 2 represents twenty, and 3 represents three. Hence, digits and their position within a number play a critical role in mathematics.
The Role of Digits in Mathematics
From simple arithmetic operations to complex equations, digits form the foundation of all mathematical activities. Each digit carries a unique value that contributes to the overall value of a number. When performing operations like addition, subtraction, multiplication, or division, we use these digits and their respective values. More importantly, in larger numbers, the position of a digit determines its place value, contributing significantly to the number’s total value. For instance, in the number 3456, the digit 3 is in the thousands place, 4 in the hundreds, 5 in the tens, and 6 in the ones place. The concept of place value is essential in the world of mathematics and is often one of the first concepts taught in primary school math education.
Properties of Digits
Each digit, from 0 through 9, possesses unique properties that make them an integral part of the number system. For example, zero (0) is the only digit that can be placed at the end of a number to increase its value tenfold. Meanwhile, one (1) is the identity for multiplication; any number multiplied by one stays the same. Each digit also has its set of prime and composite numbers. For instance, 2 and 3 are the only prime digits, while 4, 6, 8, and 9 are composite. Understanding these properties of digits aids in the comprehension of larger, more complex numbers and mathematical operations.
The Importance of Each Digit in a Number
Every digit plays a critical role in determining a number’s value. Take the number 789, for example. Here, 7 is in the hundreds place, 8 in the tens place, and 9 in the ones place. Without each digit in its proper place, the number would have a completely different value. For example, 987 is not the same as 789. This principle is called place value and is fundamental to our understanding of the number system. Recognizing the value of each digit in a number is vital for various mathematical operations, including addition, subtraction, multiplication, and division.
Difference Between Digits and Numbers
One common area of confusion in mathematics is the difference between digits and numbers. While they might seem synonymous, there is a significant distinction. A digit is a single symbol used to make numbers. There are only ten digits: 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9. On the other hand, a number can be made up of one or more digits. For instance, 7 is a number made up of a single digit, while 77 is a number made up of two digits. Understanding this distinction is key to mastering many aspects of mathematics.
Multi-Digit Multiplication Worksheets PDF
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Multi Digit Multiplication Worksheet PDF
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Multi Digit Multiplication Worksheet
At Brighterly, we believe that practice is the key to mastery. That’s why we invite you to explore our digit worksheets, where you can find an array of additional practice questions, complete with answers.
Understanding Place Value of Digits
Place value is a fundamental concept in mathematics, built around the idea that the value of a digit depends on its position in the number. In the decimal number system, the rightmost place value is ‘ones’, followed by ‘tens’, ‘hundreds’, ‘thousands’, and so on. Each position holds ten times the value of the position to its right. For example, in the number 4567, the digit 7 is in the ‘ones’ place, 6 is in the ‘tens’ place, 5 is in the ‘hundreds’ place, and 4 is in the ‘thousands’ place. Understanding place value is crucial for all operations in mathematics, especially when carrying or borrowing in addition and subtraction.
Writing Numbers Using Digits
The beauty of digits is that we can use them to represent any number, no matter how large or small. By combining different digits in varying orders and positions, we can create all the numbers we use in daily life. For instance, the number ‘twenty-three’ is written using two digits: 2 and 3. The 2, being in the tens place, represents twenty, and the 3 represents three. So when we write numbers using digits, we need to understand the place value of each digit. This fundamental skill forms the basis of all numerical literacy.
Practice Problems on Digits in Mathematics
To better grasp the concept of digits, it’s important to practice with some exercises. Let’s consider a few examples:
Identify the place value of 3 in the number 12345.
Write the number seventy-nine using digits.
What is the difference between the digit 2 and the number 22?
These exercises can help children understand the concept of digits, how they form numbers, and the role of place value. As they gain confidence, they can progress to more complex problems, enhancing their mathematical skills.
Multiple Digit Division Worksheets
Multiply 3 Digits By 1 Digit Worksheet
Conclusion
Every learning journey comes to an end, but each end is a new beginning. In our journey today at Brighterly, we’ve delved into the world of digits, exploring their definition, role, properties, and significance in our numeric system. We learned how each digit plays a pivotal role in the formation of numbers and how changing their position can impact the overall number’s value. We distinguished between digits and numbers and understood the immense power of place value.
Yet, as we conclude, we should remember that this is just a stepping stone into the vast ocean of mathematics. The real magic of digits unfolds as we climb higher on the ladder of mathematical concepts, opening doors to more complex calculations and number theories. At Brighterly, we are committed to making each step of this journey as enlightening and engaging as possible. After all, our goal is to light the spark of curiosity that fuels the quest for knowledge in every child.
Frequently Asked Questions on Digits in Mathematics
What is a digit?
A digit is a single character or symbol used in numerals. In the decimal number system that we use daily, there are only ten digits: 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9. These digits are the basic units used to represent all numbers, and each digit carries a unique value that contributes to the overall value of the number it’s part of.
What is the difference between a digit and a number?
A digit is a single symbol used in representing numbers. There are only ten digits (0-9). On the other hand, a number is a mathematical object used to count, measure, and label. It can be composed of one or more digits. For instance, 3 is a digit that also represents the number 3. However, 33 is a number composed of two 3 digits.
What is place value?
Place value is a system in mathematics that assigns value to a digit based on its position within a number. In the decimal number system, the rightmost position is ‘ones’, followed by ‘tens’, ‘hundreds’, ‘thousands’, and so on as we move to the left. Each position is ten times the value of the position to its right. So in the number 456, the digit 4 is in the hundreds place, 5 is in the tens place, and 6 is in the ones place. Place value is a fundamental concept that underpins our understanding and operations of numbers.
Sources of Information
1. Wikipedia – Digit
2. National Council of Teachers of Mathematics
3. U.S. Department of Education
Jo-ann Caballes
14 articles
As a seasoned educator with a Bachelor’s in Secondary Education and over three years of experience, I specialize in making mathematics accessible to students of all backgrounds through Brighterly. My expertise extends beyond teaching; I blog about innovative educational strategies and have a keen interest in child psychology and curriculum development. My approach is shaped by a belief in practical, real-life application of math, making learning both impactful and enjoyable.
Table of Contents
What are Digits in Math?
Definition of a Digit in Mathematics
The Role of Digits in Mathematics
Properties of Digits
The Importance of Each Digit in a Number
Difference Between Digits and Numbers
Multi-Digit Multiplication Worksheets PDF
Multi Digit Multiplication Worksheet PDF
Understanding Place Value of Digits
Writing Numbers Using Digits
Practice Problems on Digits in Mathematics
Conclusion
Frequently Asked Questions on Digits in Mathematics
What is a digit?
What is the difference between a digit and a number?
What is place value?
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3753 | https://www.engineeringtoolbox.com/archimedes-principle-d_1845.html | Engineering ToolBox - Resources, Tools and Basic Information for Engineering and Design of Technical Applications!
Archimedes' Law
Forces acting on bodies submerged in fluids.
Archimedes' principle states that:
"If a solid body floats or is submerged in a liquid - the liquid exerts an upward thrust force - a buoyant force - on the body equal to the gravitational force on the liquid displaced by the body."
The buoyant force can be expressed as
FB = W
= V γ
= V ρ g (1)
FB = buoyant force acting on submerged or floating body (N, lbf)
W = weight (gravity force) of displaced liquid (N, lbf)
V = volume of body below surface of liquid (m3, ft3)
γ = ρ g = specific weight of fluid (weight per unit volume) (N/m3, lbf/ft3)
ρ = density of fluid (kg/m3, slugs/ft3)
g = acceleration of gravity (9.81 m/s2, 32.174 ft/s2)
Example - Density of a Body that floats in Water
A floating body is 95% submerged in water with density 1000 kg/m3.
For a floating body the buoyant force is equal to the weight of the water displaced by the body.
FB = W
or
Vb ρb g = Vw ρw g
where
Vb = volume body (m3)
ρb = density of body (kg/m3)
Vw = volume of water (m3)
ρw = density of water (kg/m3)
The equation can be transformed to
ρb = Vw ρw / Vb
Since 95% of the body is submerged
0.95 Vb = Vw
and the density of the body can be calculated as
ρb = 0.95 Vb (1000 kg/m3) / Vb
= 950 kg/m3
Example - Buoyant force acting on a Brick submerged in Water
A standard brick with actual size 3 5/8 x 2 1/4 x 8 (inches) is submerged in water with density 1.940 slugs /ft3 . The volume of the brick can be calculated
Vbrick = (3 5/8 in) (2 1/4 in) (8 in)
= 65.25 in3
The buoyant force acting on the brick is equal to the weight of the water displaced by the brick and can be calculated as
FB = (( 65.25 in3) / (1728 in/ft3)) ( 1.940 slugs/ft3) ( 32.174 ft/s2)
= 2.36 lbf
The weight or the gravity force acting on the brick - common red brick has specific gravity 1.75 - can be calculated to
WB = (2.36 lbf) 1.75
= 4.12 lbf
The resulting force acting on the brick can be calculated as
W(WB - FB) = (4.12 lbf) - (2.36 lbf)
= 1.76 lbf
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3754 | https://ftransp.files.wordpress.com/2013/05/ejercicios-conduccion-unidimensional-incropera.pdf | 1' j 1 ' ' ! : ¡ 134 Capítulo 3 • Conducción unidimensional de estado estable nes para las resistencias de conducción que pertenecen a cada una de las tres geome-trías comunes. También debe estar familiarizado eón el hecho de saber en qué fornia la ecuación de calor y la ley de Fourier sirven para obtener las distribuciones de tempe-ratura y los flujos correspondientes. Las implicaciones de una fuente de energía inter-namente distribuida también deben entenderse con claridad. Finalmente, debe apreciar el importante papel que las -superficies. extendidas juegan en el diseño de los sistemas térmicos y debe tener la facilidad de efectuar diseños y ejecutar cálculos para tales su-perficies. Bibliografia l. Fried, E., "Thermal Conduction Contribution to Heat Transfer at Contacts", en R. P. Tye, ed., Thermal Con-ductivity, vol. 2, Academic Press, Lo11dres, 1969. 2. Bid, J. C., y V. W. Antonetti, "A Small Scale Thermal Contact Resistance of Aluminium Against Silicon", en C. L. Tien, V. P. Carey y J. K. Ferre!, eds., Heat Trans-fer-1986, vol. 2, Hemisphere, Nueva York, 1986, pp. 659-664 .
. 3. 4. Snaith, B., P. W. O'Callaghan y S. D. Probert, "Intersti-tial Materials for Controlling Thermal Conductances across Pressed Metallic Contacts", Appl. Energy, 16, 175, 1984. Yovanovich, M. M., "Theory and Application of Cons-tricÚon and Spreading Resistance Concepts for Micro-electronic Thermal Management", presentado en el In-ternational Symposium on Cooling Technology for Elec tronic Equipment, Honolulú, 1987. 5. Peterson, G. P. y L. S. Fletcher, "Therrnal Contact Re-sistance of Silicon Chip Bonding Materials", Procee-Problemas Pared plana 3.1 Considere la pared plana de la figura 3 .1, que separa los fluidos caliente y frío a temperaturas T "', 1 y T "'· 2, respectivamente. Con e1 uso de batanees de energía co-mo condiciones de frontera en x = O y x = L (véase la ecuación 2.27), obtenga la distribución de tempera-turas dentro de la pared y el flujo de calor en términos de Too,¡ , T,,,, 2, h¡, h1, k Y L. 3.2 La ventana posterior de un automóvil se desempaña mediante el paso de aire caliente sobre su superficie in-.terna. (a) Si el aire caliente está a T,,, ¡ = 40ºC y el coefi-ciente de convección correspondiente es h¡ = 30 W/m2 • K, ¿cuáles son las temperaturas de las dings of the International Symposium on Cooling Tech-nology for Electronic Equipment, Honolulú, 1987, pp. 438-448. 6. Yovanovich, M. M. y M. Tuarze, "Experimental Eviden-ce of Thermal Resístance at Soldered :Toints", AIAA J. Spacecraft Rockets, 6, 1013, 1969. 7. Madhusudana, C. V. y L. S. Fletcher, "Contact Heat Transfer-The Last Decade'', AIAA J., 24, 510, 1986. 8. Yovanovich, M. M., "Recent Developments in Thermal Contact, Gap and Joint Conductance Theories and Expe-riment", en C. L. Tien, V. P. Carey y J. L. Ferre!, eds., Heat Transfer-1986, vol. l, Hemisphere, Nueva York, 1986, pp. 35-45. 9. Harper, D. R. y W. B. Brown, "Mathematical Equations for Heat Conduction in the Fins of Air Cooled Engines", Reporte NACA núm. 158, 1922. 10. Schneider, P. J., Conduction Heat Transfer, Adaisoi:J-Wesley, Reading, MA, 1955. superficies interna y externa de la ventana de vidrio de 4 mm de espesor, si la temperatura del aire am-biente del exterior es T "'· 0 = - lOºC y el coeficien-· te de convección asociado es ha= 65 W/m2 • K? ~ En la práctica, T x, a y ha varían de acuei;do con las condiCiones del clima y la velocidad del automóvil. Para valores de h0 = 2, 65, y 100 W/m2 • K, calcu-le y trace las temperaturas de las superficies inter-na y externa como función de T x, ª para - 30 ::s T "'·a :S OºC. 3.3 La ventana trasera de un automóvil se desempaña uniendo un elemento de calentamiento delgado de tipo película transparente a su superficie interior. Al calen-tar eléctricamente este elemento, se establece un flujo de calor uniforme en la superficie interna. .. i l 1 i 1 t i· l f 1 l l • Problemas (a) Para una ventana de vidrio de 4 mm, determine la potencia eléctrica que se requiere por unidad de área de la ventana para mantener una temperatura eri la superficie· interna de 15ºC cuando la tempera-tura del aire interior y el coeficiente de convección son T ,,,; = 25ºC y h; = 10 W/m2 • K, mientras la t¡;mperatura del aire exterior (ambiente) y el coefi-ciente de convección son T ""· 0 = - lOºC y h0 = 65 W/m2 • K. ~ En la práctica, T ""· 0 y h0 varían de acuerdo con las condiciones climáticas y la velocidad del automó-vil. Para valores de h0 = 2, 20, 65, y 100 W/m2 • K, determine y elabore una gráfica del requeri-miento de potencia eléctrica como función de T ""· 0 para -30 s Too, 0 s OºC. De sus resultados, ¿que concluye acerca de la necesidad de operar el calen-tador con valores bajos de h0? ¿Cómo resulta afec-tada esta conclusión por el valor de T ""· 0 ? Si h ix V", donde V es la velocidad del vehículo y n es un exponente positivo, ¿cómo afecta la velocidad del auto a la necesidad de la operación del calentador? 3.4 En un proceso de fabricación se unirá una película transparente a un sustrato como se muestra en el dia-grama. Para curar la unión a una temperatura T0, se uti-liza una fuente radiante que proporciona un flujo de calor q0 (W/m2) , la totalidad del cual es absorbido en la superficie unida. La parte posterior del sustrato se man-tiene a T1 mientras la superficie libre de la película se expone al aire a L , y a un coeficiente de transferencia de calor por convección h. --L¡ = 0.25 mm ~ ~,,...,..,,..,--====~~"-='""""""" k¡ = 0.025 W / m · K L, = 1.0 mm L¡ ks = 0.05 W / m · K T l---"'-"'~~ (a) Muestre el circuito térmico que represehte la situa-ción de transferencia de calor de estado estable. Asegúrese de etiquetar todos los elementos, nodos y flujos de calor. Déjelo en forma simbólica. (b) Suponga las siguientes condiciones: Too = 20ºC, h = 50 W/m2 • K, y T 1 = 30ºC. Calcule el flujo de calor q0 que se requiere para mantener la superfi-cie unida a T0 = 60ºC. ~ Calcule y trace el flujo de calor que se requiere .· como función del espesor de la película para O :S L¡ :S 1 mm. _ ~ Si la película no es transparente y la totalidad del flujo de calor radiante se absorbe en su superficie 135 superior, determine el fluj~ de calor que se requiere para lograr la unión. Elabore una gráfica de sus re-sultados como función de L¡ para O :S L¡ :S 1 mm. 3.5 Se consideran cobre y acero inoxidable (AISI 304) co-mo material para las paredes de la tobera de un cohete enfriada por líquido. El exterior enfriad.p de la pared se mantiene a 150ºC, mientras que los gases de combus-tión dentro de la tobera están a 2750ºC. El coeficiente de transferencia de calor del lado del gas es· h; = 2 X 104 W/m2 • K, y el radio de la tobera es. mucho mayor que el espesor de la pared. Limitaciones térmicas indi-can que la temperatura del cobre y la del acero no ex-ceden 540ºC y 980ºC, respectivamente. ¿Cuál es el espesor máximo de la pared que se podría emplear para cada uno de los dos materiales? Si la tobera se constru-ye con el espesor máximo de pared, ¿cuál material se preferiría? 3.6 Una técnica para medir coeficientes de transferencia de calor implica adherir una superficie de una hoja metáli-ca delgada a un material aislante y exponer la otra su-perficie a las condiciones de corriente del fluido de interés. r L l ---Hoja (P~1éc • T,) Aislante de espuma (k) Al hacer pasar una corriente eléctrica a través de la ho-ja, se disipa calor de manera uniforme dentro de la hoja y se infiere el flujo correspondiente, P;1 éc> a partir de las mediciones de voltaje y corriente relacionadas. Si se conocen el espesor L del aislante y la conductividad térmica k, y se miden las temperaturas del fluido, hoja y aislante (T "'' T,, Tb), es posible determinar el coefi-ciente de convección. Considere condiciones para las que Lo = Tb = 25ºC, P;léc = 2000 W/m2, L = 10 mm, y k = 0.040 W/m · K. (a) Con un flujo de agua sobre la superficie, la medi-ción de la temperatura de la hoja da T, = 27ºC. Determine el coeficiente de convección. ¿Qué error se cometería al suponer que toda la potencia disi-pada se transmite al agua por convección? (b) Si, en su lugar, fluye aire sobre la superficie y la medición de temperatura da Ts = 125ºC, ¿cuál es el coeficiente de convección? La hoja tiene una emisividad de 0.15 y se expone a alrededores a 25ºC. ¿Qué error se cometería al suponer que toda la potencia que se disipa se transfiere al aire por convección? --------------------;o----~ · 1 1 1 _ ( 136 Capítulo 3 • Conducción unidimensional de estado estable ~ Normalmente, los indicadores de flujo de calor se operan a temperatura fija (T5 ), en cuyo caso la disi-pación de potencia proporciona una medida directa . del coeficie.nte de convección. Para Ts = 27ºC, grafique P~Iéc como función de h0 para 10 ::5 h0 ::5 1000 W/m2 • K. ¿Qué efecto tiene h0 sobre el error asociado con que no se tome en cuenta la conduc-ción a través del aislante? 3.7 Lo helado de la brisa, que se experimenta en un día frío y con viento, se relacioria con el incremento de la trans-ferencia de calor de la piel humana expuesta a la atmós-fera circundante. Considere una capa de tejido adiposo de 3 mm de espesor y cuya superficie interior se man-tiene a una temperatura de 36ºC. En un día calmado el coeficíente de transferencia de calor por convección a la superficie externa es 25 W/m2 • K , pero con vientos de 30 km/h alcanza 65 W/m2 • K. En ambos casos, la tem-peratura del aire del ambiente es -15ºC. (a) ¿Cuál es la pérdida de c"alor por unidad de área de la piel que se produce de un día calmado a un día con viento? (b) ¿Cuál será la temperatura de la superficie externa de la piel en el día calmado? ¿Cuál en el día con viento? ( c) ¿Qué temperatura debería tener el aire en el día calmado para producir la misma pérdida de calor · que ocurre con una temperatura del aire de -15ºC en el día con viento? 3.8 Considere el transistor montado en superficie . que se ilustra en el problema 1.51. Constrnya el circuito térmi-co, escriba una expresión para una temperatura de caja Te y evalúe Te para dos situaciones, una en la que el hueco está lleno de aire estancado y otra en la que está lleno de una pasta conductora. 3.9 Una placa de acero de 1 m de largo (k = SO W lm · K) está bien aislada en sus lados, mientras que la superficie superior está a lOOºC y la superficie inferior se enfría por convección mediante un fluido a 20ºC. En condicio-nes de estado estable sin generación, un termopar en el punto medio de la placa r:evela una . temp~ratµ_ra de 85ºC. ¿Cuál es el valor _ del coeficiente de transferencia de ca-lor por convección en la superficie inferior? 3.10 Una ventana térmica de vidrio consiste en dos piezas de vidrio de 7 mm de espesor que encierran un espacio de aire de 7 mm de espesor. La ventana separa el aire del cuarto a 20ºC del aire ambiente del exterior a - lOºC. El coeficiente de convección asociado con la superficie interna (lado del cuarto) es 10 W/m2 • K. (a) Si el coeficiente de convección asociado con el aiie exterior (ambiente) es h0 = 80 W/m2 • K, ¿cuál es la pérdida de calor a través de una ventana que tie-ne 0.8 m de largo por 0.5 m de ancho? No tome en cuenta la radiación, y suponga que el aire encerra-do entre las hojas de vidrio está estancado. ~ Calcule y trace el efecto de h0 ·sobre la pérdida de calor para 10 ::5 h0 ::5 100 W/m2 • K. Repita este cálculo para una constrncción de tres vidrios en la que se agrega un tercer vidrio y un segundo ,espa-cio de aire de espesor equivalente. 3.11 La pared de un colector solar pasivo consiste en un ma-terial de cambio de fase (PCM) de espesor L encerrado entre dos superficies estructurales de soporte. Ts, 1 Ts, 2 t l l ólido, k5 Suponga una condición de estado estable para la que la absorción de radiación solar en una superficie mantiene su temperatura (Ts, 1) por arriba de la temperatura de fusión del PCN,f. Las porciones líquida y sólida del PCKf estáñ divi~idas por una interfaz vertical estrecha. El líquido tiene\una temperatura del núcleo de Tm y se caracteriza por ~n flujo recirculante movido por la flo-tación que maniiene el mismo coeficiente de convec-ción (hm) en sus\ interfaces con la superficie (s, 1) y el sólido. Considere condiciones para las que el flujo neto de radiación es q':act = 1000 W/m2, las temperaturas ambientes y los coeficientes de convección son T"', 1 = T,,, 2 = 20ºC y h1 = h2 = 20 W/m2 • K, la temperatura y coeficiente de convección del líquido PCM son T m = 50ºC y hm = 10 W/m2 · X y la conductividad térmica del sólido PCM es ks = 0.5 W/m · K. Evalúe la tempe-a de ca-tezas de espacio el aire erior a con la . K. el aire cuál es ¡ue tie-)me en lcerra-lida de ta este ; en la • .espa-m ma-~rrado "la :ne 'de le! a. '"'~ 1 • Problemas · ratura de la superficie, Ts, 1. Si el espesor total del PCM es L = 0.10 m, ¿cuál es el espesor de la capa líquida? Calcule la temperatura de la superficie Ts, 2. 3.12 La pared de un edificio es un compuesto que consiste en una capa de 100 mm de ladrillo común, una capa de 100 mm de fibra de vidrio (forrada, con papel, 28 kg/m3), una capa de 10 mm de revoque de yeso (vermiculita) y una capa de 6 mm de tabla de pino. Si el coeficiente de convección interior es 10 W/m2 · K y .el coeficiente de convección exterior es 70 W/m2 • K, ¿cuál es la resis-tencia total y el coeficiente global para la transferencia de calor? 3.13 La pared compuesta de un horno consiste en tres mate-riales, dos de los cuales son de conductividad térmica conocida, kA = 20 W/m · K y kc = 50 W/m · K, y de espesor conocido, LA = 0.30 m y Le = 0.15 m. El ter-cer material, B, que se intercala entre los materiales A y C, es de espesor conocido, L8 = 0.15 m, pero de con-ductividad térmica, k8 , desconocida. Ts, o En condiciones de operación de estado estable, las me-diciones revelan una temperatura de la superficie exter-na Ts, 0 = 20ºC, una temperatura de la superficie interna Ts, ; = 600ºC, y una temperatura del aire del horno T,, = 800ºC. Se sabe que el coeficiente de convección interior h es 25 W /m2 • K. ¿Cuál es el valor de k8 ? 3.14 Las paredes exteriores de un edificio son un compuesto que consiste en un tablero de yeso de 10 mm de espe-sor, espuma de uretano de 50 mm de espesor y 10 mm de madera blanda. En un típico día de invierno las tem-peraturas del aire exterior e interior son - 15ºC y 20ºC, respectivamente, con coeficientes de convección exter-no e interno de 15 W/m2 • K y 5 W/m2 • K, respectiva-mente. (a) ¿Cuál es la carga de calentamiento para una sec-ción de 1 m2 de pared? (b) ¿Cuál es la carga de calentamiento si la pared com-p~esta se reemplaza por una ventana de vidrio de 3 mm de espesor? / (c) ¿Cuál es la carga de calentamiento si la pared com-puesta se reemplaza con una ventana de doble vidrio que consiste en dos placas de vidrio de 3 mm . de espesor separadas por un hueco de aire estan-cado de 5 mm de espesor. ~~ Una casa tiene una pared compuesta de madera, aislan-1 /\ :¡ te de fibra de vidrio y tablero de yeso, c~mo se indica \y 137 en el esquemit, En un día frío de invierno los coeficien-tes de transferencia de calor por convección son ha = 60 W/m2 • K y h; = 30 W/m2 • K. El área total de la su-perficie de la pared es 350 m2. Capa de fibra de vidrio (28 kg/m3), kb Tablero de yeso, P • • h;, T "'·; =20ºC 3.16 3.17 t t t ttt 1 O mm --+! 1--100 mm--+! · 1-- 20 mm LP Lb L8 (a) DeteffilÍile una expresión simbólica para la resis-tencia térmica total de la pared, incluyendo los efectos de convección interior y exterior para las condiciones establecidas. (b) DeteffilÍile la pérdida total de calor a través de la pared. (c) Si el viento soplara de manera violenta, elevando h0 a 300 W/m2 • K, determine el porcentaje de aumento en la pérdida de calor. ( d) ¿Cuál es la resistencia controladora que determina la cantidad de flujo de calor a través de la pared? Considere la pared compuesta del problema 3.15 bajo condiciones para las que el aire interior aún se caracte-riza por Too,; = 20ºC y h; = 30 W/m2 • K. Sin embar-go, utilice las condiciones más realistas en las que el aire exterior se caracteriza por una temperatura que va-ría con el día (tiempo), de la forma Too aCK) = 273 + 5 sen ( 2:. '!!.. t) . 24 o :'.S t s; 12 h ' (2'Tf) T00 a(K) = 273 + 11 sen -- t ' 24 12 :'.S t s; 24 h con ha = 60 W/m2 • K. Suponiendo condiciones casi estables para las que es posible no tomar en cuenta los cambios en el almacenamiento de energía dentro de la pared, estime la pérdida diaria de calor a través de ésta si el área total d~ la superficie es 200 m2. Considere una pared compuesta que incluye un tablado de madera dura de 8 mm de espesor, travesaños de 40 mm por 130 mm de madera dura sobre centros de 0.65 m con aislante de fibra de viqno (recubierto con papel, 28 kg/m3) y una hoja de cartón de yeso (vermicu-lita) de 12 mm. ----- ·-- ··- --------,--~~------------------- -~!'\'" . ~ 133. Capítulo 3 •· C-0nducción unidi'm:ensional d'e estado estable 40:mm--l; {j;:~/;~:,.);:~;- ~~~~~~· de ~~~.;;...:.,~~-- Travesaño Cartón. de yeso ¿Cuál es la resistencia térmica asociada con una pared' que: tiene 2.5 m de altura por· 65 m de aneho· (y 1 O tra-vesaños, cada. uno de· 2.5 m de altura)?-3.1& Las características térmicas de un pequeño. refrigerador doméstico se. determinan. realizando dbs experimentos separados,. cada un.oo con la puerta· cerrada. y el refrigera-dór colocadb: en aii:e ambiente· a· Too = 25ºC. En un ca~ so,. un calentador eléctrico se suspende: en la cavidad del: refrigerador,.mientras el refrigerador está descorree~ tado. Con. el calentador disipando 20: W; se registra una temperatura: de estado estable de· 90ºC dentro de la ca-vidad.. Sin. el calentador y con el refrigerador ahora en operación; el· segundo experimento• implica mantener una temperatura de la. cavidad en e.stado estable de S?C para un intervalo de. tiempo. fijo• y registrar la• energía eléctrica que se: requiere: para. operar ef refrigerador: En este experimento, para el que. la operación de. estado-es-table se mantiene en un periodo de 12 horas,. la energía eléetrica de entrada es 125,000 J. Detennine el coefi-ciente de rendimiento del refrigerador (COP). 3.19· En et diseño-de: e.dificios,. el requerimiento de conserva-ción de la. energía dicta que el área de la superficie ex-terior, As, se minimice. Este requerimiento implica que, para. un espacio de piso deseado, hay valores óptimos asociados con el número. de pisos y con las. dimensio-nes horizontales del edificio. Considere un. diseño para el que se establecen- el espacio de piso;. A1, y la-distan-cia vertical entre pisos, H¡. (a) Si el edificio tiene una sección transversal cuadra-da de ancho· W en un fado, obtenga una expresión para el valor de W G¡ue minimice la pérdida de calor a los alrededores .. La pérdida de calor se supone que ocurre de las. cuatro· paredes verticales y de un tech o_ plano. Expres_ e sus resultados en términos de A¡Y H¡. (b) Si A¡= 32,768 m2 y H¡ = 4 m, ¿para qué valores de W y N1 (número de pisos) se mü:1imiza la pérdi-da de calor? Si el coeficiente global de transferen-. cia de calor promedio es U = 1 W/m2 • K y la diferencia entre las temperaturas del aire ambien-tal interior y exterior es 25-°C, ¿cuál es la pérdida de calor correspondiente?·¿ Cuál es el porcentaje de reducción en pérdida de calor comparado con un edificio de N¡ = 2? ResiStencia térmica de contacto 320' Una pared compuesta separa gases de c0rnbustión a 2600ºC de un Líquido refrigerante a 100ºC, con coefr-cientes de convección deI lado de gas y de.F líquido de 50 y 1000 W/rn2 • K. La pared se compone de una ca-pa de óxido de berilio de 1 O mm de espesor en el lado deI gas· y una placa• de acero inoxidable (AISI 304) de 20 mm de grosor en el lado del líquido. La resistencia: de contacto entre el óxido y ef acero es 0>.05 m2 • K/W. ¿Cuál es la perdida de ea:lor por área unitaria de super-ficie deI compuesto? Dibuje la distribución de tempe-raturas del gas al líquido. 3.21 Dos. placas de acero inoxidable. de 10· mm de espesor están sujetas. a una• presión de contacto· de 1 bar bajo condiciones de vacío para. las que hay una; caída gene-ral de temperatuva de lOOºC a lo. largo de. las. placas-. ¿Cuál es la caída de temperatura a. través del' plano de contacto? 3.22 Considere una pared plana compuesta integrada por dos materiales. de conductividades; térmicas kA. = o, 1 W/m · K y k8 = 0.04 W/m · K y espesores LA = 10 mm y L8 = 20 mm. Se sabe que la resistencia de contacto en la interfaz entre ros dos materiales es. 0.30 rn2 • KJW. El material A está al lado de un fluido a 200ºC pará el que h = 10 W/m2 • K, y el material B. a un fluido a 40ºC para el que h = 20 W/m2 • K. (a) ¿Cuál es la transferencia de cafor a través de una pared que tiene 2 rn de altura por 2.5 m: de ancho?· (b) Dibuje la distribución de temperaturas. l 3.23 1 El rendimiento de los motores de turbinas de gas. se mejora aumentando la tolerancia de las hojas de las tur-binas a los gases calientes.que. salen del combustor. Un método para. lograr altas temperaturas de operación im-plica la aplicación. de un revestimiento de barrera tér-mica (TBC) para la superficie externa de una. hoja, mientras pasa aire de enfriamiento a través.de la hoja Por lo común, la hoja está fabricada de una. superaleac ción de alta temperatura, como Inc0nel (k = 25 W/m · K), mientras una cerámica, como circonia (k = 1.3 W/rn · K), se usa como revestimiento de barra térmica TBC. Superaleación ___ __CMO.,¡¡¡¡,,--------------------~---- -- ···----·-1 .. tbusti0n a con coefi-'iquidode: le una ca-~n el lado I 304) de ~sistencia o.2 ·K/W .. :fe· super-e tempe-' espesor bar bajo ja, gene-; placas. Jlano de por dos lW/m· J mmy :acto ea C/W. Ei t el que a 40ºC de una ncho? gas. se astur-or. Un )n iin-a tér-un:a thoja raleac V!m· , 1.3 mica ·ación s -~ h¡ e 'ón ) • Problemas considere condiciones para las que gases cáliéntes . a r,,, 0 = 1700 K y aire de enfriairiiento a Too, ; = 4DO K pfdpcifcioriah coefici~htes de COhveccióri de la Si.Iperfi-Cie exterriá e interna de h0 = 1000 W/m2 · K y h¡ = 500 W/tri..2 . K, respectivamente. Si tirt TBC de circonio de 6.5 mm de espesor se uhe á lá pared de uha hoja de in-conel de 5 min de espesor por medio de uh agente de uh.i6h metálico, qtie proporciona iirta resistencia téi.-rnicá entre las interfaces dé R.~ e = ib-4 m 2 • K!w, ¿es posi-bie mantener el Iilconel a una temperatura que esté por debajo de su valor máximo permisible dé 12.Sb K? De-je de lado los efectos de radiación; y aproxime ia hoja de ia tui:bina como una pared pla.ria. Elabore una gráfi-ca de la distribución de temperaturas con y sin el tBC. ¿Existe algún líi:nite al espesor dei TBÓ 3.24 Uh chip de silicio se encapsulá de modo qúe; bajo con-didoiies de estado estable, la totálidad de la potencia que se disipa se transfiere por convección a una comen-te de fluido para ei que h = 1000 W/m 2 · k y T 00 = 2.SºC. El chip se separa del fi.tiido mediante una cu-bierta de placa de aluminio de 2 nim de espesor, y la resistencia de cohtácto de lá irttetfai chip/aiuffi.inio es -0.5 X l(j-4 m2 • KfW. Si ei área de la superficie del chip es 100 mm2 y la tem-peratura rrtáxima peiüi.isible es 85ºC, ¿cuál es la disi-pación de potencia máxima pertnisible en ei chip? 3.25 Aproximadamente 106 componentes eléctricos discre-tos se colocan en uri Sofo circuito integrado (chip), con disipación de calor eléctrico tan alta como 30,000 W/m2. El chip, que es muy delgado, se expone a un lí-quido dieléctrico en la superficie externa, con h0 = 1000 W/m2 • K y Too, 0 = 20ºC, y se tirie a uha tarjeta de circuitos en la superficie interior. La tésistencia térc mica de contacto entre el chip y la tarjeta es 10-4m2 • íO, a = O y a < O. Considere la pared de un tubo de radios interno y exter-no r¡ y r 0 , respectivamente. La conductividad térmica del cilindro depende de la temperatura y se representa me-diante una expresión de la forma.k = k0(l + aT), donde k0 y a son constantes. Obtenga una expresión para la transferencia de calor por unidad de longitud del tubo. ¿Cuál es la resistencia térmica de la pared del tubo? Ciertas mediciones muestran que la conducción de es-tado estable a través de una pared plana sin generación de calor produjeron una distribución de temperaturas convexa tal que la temperatura del punto medio fue 6.T0 más alta que la esperada para una-distribución-li-neal de temperaturas. T¡ T(x) i Suponiendo que la conductividad térmica tiene una de-pendencia lineal de la temperatura, k = k0(l + aT), donde a es una constante, desarrolle una relación para evaluar a en términos de 6.T0 , T¡ y Tz. 3.31 Use el método de análisis de conducción alternativo para derivar la expresión de la resistencia térmica de un cilindro hueco de conductividad térmica k, radios inter-no y externo r; y r 0 , respectivamente, y longitud L. Pared cilíndrica 3.32 Una tubería de vapor de 0.12 m de diámetro exterior se aísla con una capa de silicato de calcio. · (a) Si el aislante tiene 20 mm de espesor y las superfi-cies interna y externa se mantienen ·a Ts, 1 = 800 K y Ts, 2 = 490 K, respectivamente, ¿cuál es la pérdida de calor por unidad de longitud (q') de la tubería? [Q] Deseamos explorar el efecto del espesor de aislan-te sobre la pérdida de calor q' y la temperatura de la superficie externa Ts, 2, con la temperatura de la superficie interna fija a Ts, 1 = 800 K. La superficie externa se expone a un flujo de aire (T"' = 25 ºC) que mantiene un coeficiente de conveccióh' de h = 25 W/m2 • K y a grandes alrededores para los que T aJr = T"' = 25 ºC. La emisividad de la superfi-cie de silicato· de calcio es aproximadamente 0.8. Calcule y dibuje la distribución de temperaturas en el aislante co~o función de la coordenada radial adimensional (r - r 1)/(r2 -r 1), donde r 1 = 0.06 m y r2 es una variable (0.06 < r2 :S 0.20 m). Calcule Y dibuje la pérdida de calor como función del espe-sor del aislante para O :S (r2 - r1) :S 0.14 m. 3.33 Considere el calentador de agua que se describe en el problema 1.29. Deseamos ahora determinar la energía necesaria para compensar las pérdidas de calor que ocurren mientras el agua está almacenada a la tempera-tura establecida de 55ºC. El tanque cilíndrico de alma-cenamiento (con extremos planos) tiene una capacidad de 100 galones, y se usa uretano en espuma para aislar las paredes lateral y de los extremos del aire ambiental a una temperatura promedio anual de 20ºC. La resis-tencia a la tran~fere .ncia de calor está dominada por la . c~nd~cciÓn en el aislante y por la convección libre en el aire, para el que h = 2 W/m2 • K. Si se usa calenta-miento por resistencia eléctrica para compensar las pér-didas y el costo de la potencia eléctrica es $0.08/kWh, especifique las dimensiones del tanque y del aislante para las que Íos costos anuales asociados con las pérdi-das de calor son menores de $50. 3.34 Un calentador eléctrico delgado envuelve la superficie externa de un tubo cilíndrico largo cuya superficie inter-na se mantiene a una temperatura de 5ºC. La pared del tubo tiene radios interno y externo de 25 y 75 mm, res-" \ T 1 1 Cle-1 ¡ T), 1 f!Tª i ¡ YO lm :r-' :e 1 -( 1 a 1 -: ' f • Problemas pectivamente, y una conductividad térmica de 10 W/m · K. La resistencia térmica de contacto entre el calentador y la superficie externa del tubo (por unidad de longitud de tubo) es R ;'. e = Ó.01 m · KJW. La superficie externa del calentador se expone a un fluido con T,,, = - lOºC y un coeficiente de convección h = 100 W/m2 • K. Dec termine la potencia de calentamiento por unidad de tu-bo que se requiere para mantener el calentador a T0 = 25ºC. G.35] En el problema anterior, la ·potencia eléctrica que se re-quiere para mantener el calentador a T0 = 25ºC de-pende de la conductividad térmica del material de la pared k, la resistencia térmica de contacto R; e y el coe-ficiente de convección h. Calcule y dibuje p~r separado el efecto de cambios en k (1 :5 k :5 200 W/m · K), R;, e (O :5 R ;, e :5 0.1 m · KJW) y h (10 :5 h :5 1000 W/m2 • K) sobre el requerimiento de potencia total del calenta-dor, así como la transferencia de calor a la superficie interna del tubo y al fluido. 3.36 Uretano (k = 0.026 W/m · K) se usa para aislar lapa-red lateral y las partes superior e inferior de un tanque cilíndrico de agua caliente. El aislante tiene 40 mm de espesor y se intercala entre láminas de metal de pared delgada. La altura y el diámetro interior del tanque son 2 m y 0.80 m, respectivamente, y el tanque está ex-puesto al aire del ambiente para el que T,,, = lOºC y h = 10 W/m2 • K. Si el agua caliente mantiene la su-perficie interna a 55ºC y el costo de la energía ascien-de a $0.15/k:Wh, ¿cuál es el costo diario para mantener el agua almacenada? 3.37 Un calentador eléctrico delgado se inserta entre una varilla circular larga y un tubo concéntrico con radios interior y exterior de 20 y 40 mm. La varilla A tiene una conductividad térmica de. kA = 0.15 W/m · K, mientras el tubo B tiene una conductividad térmica de k8 = 1.5 W/m · K; la superficie externa está sujeta a convección con un fluido de temperatura T,,, = -15ºC y el coeficiente de transferencia de calor de 50 W/m2 • K. La resistencia térmica de contacto entre las superfi-cies del cilindro y el calentador es insignificante. (a) Determine la potencia eléctrica por. unidad de lon-gitud de los cilindros (W/m) que se requieren para mantener la superficie externa del cilindro B a 5 ºC. (b) ¿Cuál es la temperatura en el centro del cilindro A? 3.38 Una larga varilla cilíndrica de 100 mm de radio está he-cha de un material de reacción nuclear (k = .
0.5 W/m · K) que genera 24,000 W/m3 de manera uniforme a lo largo de su volumen. Esta varilla está encapsulada den-tro de un tubo que tiene un radio externo de 200 mm y una conductividad térmica de 4 W/m · K. La supei:ficie externa está rodeada por un fluido a lOOºC, y el coefi-141 ciente de convección entre la superficie y el fluido es 20 W/m2 • K. Encuentre las temperaturas en la interfaz entre los dos cilindros y la superficie externa. 3.39 Un recubrimiento especial, que se aplica a la superficie interior de un tubo de plástico, se cura colocando una fuente de calor por radiación cilíndrica dentro del tubo. El espacio entre el tubo y la fuente se vacía, y la fue~te entrega un flujo de calor uniforme q¡, que se absorbe en la superficie interna del tubo. La superficie externa del tubo se mantiene a una temperatura uniforme, Ts, 2. Tubo de plástico, k Ts, 1 Ts,2 Desarrolle una expresión para la distribución de tempe-raturas T(r) en la pared del tubo en términos de q'í, Ts, 2, r1, r2 y k. Si los radios interior y extenor del tubo son r 1 = 25 mm y r2 = 38 mm, ¿cuál es la potencia que se re-quiere por unidad de longitud de la fuente de radiación para mantener la superficie interna a Ts, 1 = 25ºC? La conductividad de la pared del tubo es k = 10 W/m · K. 3AO Considere un cilindro hueco largo de conductividad térmica k con radios interior y exterior r; y r 0 , respecti-vamente. La temperatura de la superficie interna se mantiene a T; mientras que la superficie externa experi-menta un flujo de calor uniforme q~. (a) Comenzando con la forma apropiada de la ecua-ción de difusión de calor, derive una expresióq pa-ra la distribución de temperatura, T(r), en términos de r¡, r 0 , k, T; y q~. ---- -----=--~~~-----~-----~----) 142 3.41 (P) ])ibuje la d:istrituc.ió{l cie temp7raturas en coo.
rqe-na,~as T-r. ( c) Escripa una expresión J?:a,J:ª la transferenci\l¡ <ie calor po~ ull.id<lci· de . íongi\Uc\ qel cilinclro ep. ~a Sl\perfi,cie ínterna~ q~(r;), en término.
s de q~ y 19s papímetros de ·\a geoi:netría del cili~qro. . . . La sección del evaporador de una qnidad ¡ie refrigera,-ción · consiste en tubos de pared delg~da d,e 10 inrn, de d,iá.m,etro a través de lqs que pasa el fl~id() refrige~ rante a una temperatura de -18ºC. Se enfría aire con-form,e fluye sobre los tubos, manteniendo un coeficiente de convección de superficie de ~00 Wlm2 • K, y en segui-da se dirige a la s .e~ción del i;efyig7ra,dor. · · (a) Para \as cop.dici9nes precedef\tes y una, temperatu-ra dei aire de. ~3ºC, ¿~uál es la 'rapicÚ;z a la que se ~xtrae c<µor del aire p~r unida~ de !ongituµ del tµ\Jo? [1!ill Si la unidad de descongel~ci<51'. f\1-qciona mal, len-tameµte se acumulará escarcha sobre la superficie externa del tubo. Evalúe el efec\o de la formación cie escarcha sobre la capacidad c\e enfriarnieNº de un tubo para espesores de la capa de escarcha en el r~go O :s 8 :s 4 mm. Se s~pone que la escar~ha tiene una conductivÚ:l~d tér:mi,ca de ÓA Wlm '. K. (c) Se desconecta el refrigerador después de que fajla l¡i. unidad de descongelamiento y de que se ha for-mado µna capa de escarcha de 2 mm de groso~. Si los tubos están en aire ambiente papel que T"' = 20ºC y una convección natural mantiene l\n coefi-ciente de convección de 2 W/m2 • K, ¿cuánto tiem-po tardará la escarcha en derretirse? Se supone que la escarcha tiene. una \iensid,ad de 700 kg/:in3 y una enta,lpía de fusión de 334 kJ/kg. 3.42 Una pared cilíndrica está compuesta por dos maxeriales de conciuctividad térmica kA y ks, separados por un ca-lentador de resistencia eléctrica muy delgado para el cual las resistencias térmicas de con\acto de las interfa-ses son insignificantes. Calentador de resistencia . (¡'!,; r;;-----· -· 1 "]µ líquido que se bombell: a través del tl.\P() está a una tei:npera~a f"', ¡ y prop()r~iona un coeficiente de con-vecéióf\ µe h¡ en la superficie intei;na, del cowpuesto. La superficie externa se expone al aire a,mbieµte, el cual está a T "'· 0 y proporcion~ \ln coeficiente <ie co~vección de ho. En C()ndicione~ de estacl,o esta.ple, el c<µentador disipa un flµjo qe calor uniforme q'f.. (a) Dibuje el circuito térmico equi\falente del sistema y exprese todas las resistencias en térniinos de va-riables relevantes. (b) Obtenga una expresión que sirva para determinar la temperatura del c~entador, Th-(c) ObJenga µna expresión para, la razón de los ftujos de cal()r a los fluicl,os externo e interno, q~ lqi. ¿Có-mo aju~t~ las variables del p.rol:>le!Il~ para, ~®i zar esta razpn? 3.43 Un alambre eléctrico que tiene un radio de r¡ =;oo 5 rru+i y una resistencia por unídad de longitud de 10-4 f!/m, se cubre con un aislante plástico de conductividad t~rmica k = 0.20 W/m · K. El aislante se expone al aire del 3!11-biente para el que T / = 300 K y h = 10 W/m2 • K Si el aislante tiene una temperatura máxima permisible de 450 :((,¿cuál, es la corrientt! máxima posible que se pue-de hacer pasar por el alambre? · 3.44 Una corriente eléctrica de 700 A fluye a través de un. ca-ble de acero inoxidable que tiene un diámetro de 5 mm y una resistencia eléctrica de 6 X 10- 4 DJm (por metro de longitud de cable). El cable está en un medio que tie-ne una temperatura de 30 ºC, y el qJeficiente to.tal as9-ciado con la convección y la radiación entre el cable y el medio es aproxim(\damente 25 W /m 2 '. K. . (a) Si el cable está expuesto, ¿cuál es la temperatura de la superficie? · (b) Si se aplica un recubrimiento muy delgaqo de ais-lante eléctrico al cable, con una resistencia de con-tacto de 0.02 m2 • K/W, ¿cuáles son las temperaturas superficiales del aislante y del cable? - (c) Hay cierta preocupación sobre-la capacic:laci del aislante para resistir temperaturas elevadas. ¿Cu ál espesor de este aislante (k = 0.5 W/rn · K) dará el valor más bajo de la temperatura m~xima del ais-lante? ¿Cuál es el valor de la temperatura máxima, cuando se usa dicho espesor? . . 3.45 Un tubo de acero de pared delgada de 0.20 m \fe ciiáme-tro se utiliza para transportar vapor saturado a una pre-sión d, e 20 bar en un cuarto p;ira, el que la temperatura del aire es 25ºC y el coefic\ente de transferencia de ca-lor en la superficie externa del tubo es 20. W/m2 • K. 1 1 1 1 1 1 \ 1 ! l \<1 ltll<\ ¡ con-to. La \ c~al ~ció~ ¡ ··. ' qqr ¡' 'bma t ~ . ; 'Ya-i !lar A .. . t 1 ps 5-t 1-, • Probwmµs (a) ¿Cuál es la pérclida de qlor por unidad .de lon~ítu<l del tubo expµesto (sin aislante)? .
Estime la pérclida de calor por unidad de longitud si ·, a~ga una capa de 50 mm de aislante (óxido de mag!lesio, 85%). Suponga que el ac.
ero y el óxido de magnesi<) tiene cada uno una emisividad de 0,8, y no tome en cuenta la resistencia de conv!'!cción del lado del v¡¡.por. (b) Se sabe que el costo ~sociado con la generación del vapor y la instalación del aislante son $4/109 J y $100/m de longitud de tubo, respectivamente. Si la línea de vapor operara 7500 h/año, ¿cuánfos años se necesitan para recuperar la inversión inicial en aislante? 3.46 A través de un tubo de acero (AISI 1010), de 60 mm de diámetro interior y 75 mm de diámetro exterior, fluye vapor a una temperatura de 250ºC. El coeficiente de convección entre el vapor y la superficie interna del tu-bo es 500 W/m2 • K, mientras que entre la superficie externa del tubo y los alrededores es 25 W/m2 • K. La emisividad del tubo es 0.8, y la temperatura. del aire y los alrededores es 20ºC. ¿Cuál es la pérdida de calor por unidad de longitud de tubo? Deseamos determinar el efecto de agregar una capa ais-lante de óxido de magnesio al tubo de vapor de¡(pro-blema anterior. Suponga que el coeficiente de convección en la superficie externa del aislante perma-nece a 25 W/m2 • K, y que la emisividad es e = 0.8. Determine y trace la pérdida de calor por unidad de longitud de tubo y la temperatura de la superficie exter-na como función del espesor del aislante. Si el costo de generación del vapor es $4/109 J y la línea de vapor opera 7000 h/año, recomiende un espesor de aislante y determine el ahorro anual correspondiente en costos de energía. Elabore una gráfica de la clistribución de tem-peraturas para el espesor recomendado. Un tubo de pared delgada de 100 mµi de diámetro sin aislar se usa para transportar agua a equipo que opera en el exterior y utiliza el agua como refrigerante. En condiciones de invierno Qar:tic~~~eBte adversas la pared del tubo alcanza una temperatura.de -15ºC y se forma una capa cilíndrica de hielo sobre la superficie interna de la pared. Si la temperatura media del agua es 3ºC y se mantiene un coeficiente de convección de 2000 W/m2 • K en la superficie interna del hielo, que está a O ºC, ¿cuál es el espesor de la capa de hielo? 3.49 El vapor que fluye a través de un tubo largo de pared delgada mantiene la pared del tubo a una temperatura uniforme de 500 K. El tubo está cubierto con una man-ta aislante compuesta con dos materiales diferentes, A y B. T,, 21AJ Ts, 218> T,, ¡ = 500 K 143 A=2W/m·K ~-ir" = 0.25 W/rn · K Se supone que la interfaz entre los dos materiales tiene una resistencia de contacto infinita, y que toda la super-ficie externa está expuesta al aire, para el cual T 00 = 300 K y h = 25 W/m2 • K. (a) Dibuje el circuito térmico del sistema. Usando los símbolos precedentes, marque todos los nodos y resistencias pertinentes. (b) Para las condiciones que se establecen, ¿cuál es la pérdida total de calor del tubo? ¿Cuáles son las temperaturas de la superficie externa T,, Z(A ) y T,, Z(B)? 3.50 Un recubrimiento de baquelita se usará con una varilla conductora de 10 mm de diámetro, cuya superficie se mantiene a 200ºC mediante el paso de una corriente eléctrica. La varilla está en un fluido a 25 ºC, y el coefi-ciente de convección es 140 W/rn2 • K. ¿Cuál es el ra-dio crítico asociado con el recubrimiento? ¿Cuál es la transferencia de calor por unidad de longitud para la va-rilla desnuda y para la varilla con un recubrimiento de baquelita que corresponde al radio crítico? ¿Cuánta ba-quelita debe agregarse para reducir en 25% la transfe-rencia de calor asociada con la varilla desnuda? Pared esférica 3.51 Un tanque de almacenamiento consiste en una sección cilíndrica que tiene una longitud y diámetro interior de L = 2 m y D; = 1 m, respectivamente, y dos secciones extremas. hemisféricas .
. El tanque se construye de vi-drio (Pyrex) de 20 mm de espesor y se expone al aire del ambiente para el que la temperatura es 300 K y el coeficiente de convección es 10 W/m2 • K. El tanque se utilíza para almacenar aceite. caliente,-que.-mantiene -la superficie interior a una temperatura de 400 K. Deter-mine la potencia eléctrica que debe suministrarse al calentador sumergido en el aceite para mantener las condiciones establecidas. Deje de lado los efectos de racliación y sup9nga que el Pyrex tiene una conductivi-dad térmica de t.4 W/m · K. 3.52 Considere el sistema de almacenamiento de oxígeno lí-quido y las condiciones ambientales del l aboratorio del problema 1.35. Para reducir la pérdida de oxígeno de-bida · ª la vaporización debe aplicarse una capa de ais-lante a la superficie externa el contenedor. Considere et i----- -----------------c--------------------- ----- -~- ~~·--- ~· ¡~ ' J ' ! 'j . ¡ '¡ ,. 1 ¡: '~ !! 1 j. 144 Capítulo 3 • Conducción unUliTnensional de estado estable 'uso de un aislante de hoja de aluminio laminado/vidrio mate, para el que la conductividad térmica y la ernisivi-dad superficial S<?n k = 0.00016 W/m · K y e = 0.20, respectivamente. (a) Si. el contenedor se cubre con una capa de aislante de 10 mm de espesor, ¿cuál es el porcentaje de re-ducción en la pérdida de oxígeno en relación con el contenedor sin recubrimiento? ~·Calcule y trace la masa de evaporación (kg/s) co-mo función del espesor del aislante t para O ::s t ::s 50mm. 3.53 En el ejemplo 3.4, se derivó una expresión para el ra-dio crítico de aislamiento de un tubo cilíndrico aislado. Derive la expresión apropiada para una esfera aislada. 3.54 Una esfera hueca de aluminio, con un calentador eléc-trico en el centro, se utiliza en pruebas para determinar la conductividad térmica de materiales aislantes. Los radios interior y exterior de la esfera son O .15 y O .18 m, respectivamente, y la prueba se hace en condiciones de estado estable, en las que la superficie interna del alu-minio se mantiene a 250ºC. ·En una prueba particular, una capa esférica de aislante se funde sobre la superfi-cie externa de la esfera y alcanza un espesor de 0.12 m. El sistema está en un cuarto para el que la temperatura del aire es 20ºC, y el coeficiente de convección en la superficie externa del aislante es 30 W/m2 • K. Si se disipan 80 W por el calentador bajo condiciones de es-tado estable, ¿cuál es la conductividad térmica del ais-lante? 3.55 Un tanque esférico para almacenar oxígeno líquido en un transbordador espacial se construye de acero inoxidable de 0.80 m de diámetro exterior y una pared de 5 mm de espesor. El punto de ebullición y la ental-pía de fusión del oxígeno líquido son 90 K y 213 k:J/kg, respectivamente. El tanque se instalará en un comparti-miento grande cuya temperatura se mantendrá a 240 K. Diseñe un sistema de aislamiento térmico que manten-ga las pérdidas de oxígeno debidas a la ebullición por debajo de 1 kg/día. 3.56 Una sonda esférica crioquirúrgica se incrusta en tejido enfermo con el propósito de congelarlo y, por tanto, destruirlo. Considere una sonda de 3 mm-de .diámetro cuya superficie se mantiene a - 30ºC cuando se incrus-ta en tejido que está a 37ºC. Una capa esférica de tejido congelado se forma alrededor de la sombra, con una tem-peratura de OºC en la fase frontal (interfaz) entre el tejido normal y el congelado. Si la conductividad térmica del tejido congela.
do es aproximadamente 1.5 W/m · K y la transferencia de calor en la fase frontal se caracteriza por un coeficiente de convección efectivo de 50 W/m2 • K, ¿cuál es el espesor de la capa del tejido congelado? 3.57 Una capa esférica compuesta de radio interior r1 0.25 m se construye de pfomo de radio exterior r2 = 0.3( y acero inoxidable AISI 302 de radio exterior r 3 0.31 m. La cavidad se llena de desechos radioacti\ que generan calor a una razón de q = 5 X 105 W/r Se propone sumergir el contenedor en aguas oceánic que están a una temperatura de T"' = lOºC y que pr porcionan un coeficiente de convección uniforme h 500 W/m2 • K en la superficie externa del contenedc ¿Hay algún problema asociado con esta propuesta? 3.59 Como una alternativa para almacenar materiales ra dioactivos en aguas oceánicas, se propone que el siste ma del problema 3.57 se coloque en un tanque gragdt en el cual se controle el flujo de agua· y, por consiguien-te, el coeficiente de convección h. Calcule y trace 12 temperatura máxima del plomo, T(r 1) , como función de h para 100 ::s h ::s 1000 W/m2 • K. Si la temperatura del plomo no deberá exceder 500 K, ¿cuál es el valor mínimo permisible de h? Para mejorar la seguridad del sistema, es deseable aumentar el espesor de la capa de acero inoxidable. Para h = 300, 500 y 1000 W/m2 • K, calcule y trace la temperatura máxima del plomo como función del espesor de la capa para r 3 2: 0.30 m. ¿Cuá-les son los valores correspondientes del espesor máxi-mo permisible? La energía que se transfiere de la cámara anterior del ojo a través de la córnea varía considerablemente de-pendiendo del uso de un lente de contacto. Trate al ojo como un sistema esférico y suponga que el sistema se encuentra en estado estable. El coeficiente de convec-ción h0 se mantiene inalterable con y sin el lente de contacto en su sitio. La córnea y el lente cubren un ter-cio del área de la superficie esférica. Lente de contacto Los valores de los parámetros que representan esta si-tuación son los siguientes: r1 = 10.2 mm r 3 = 16.5 mm T"'. ; = 37ºC r2 = 12.7 mm T"'·º = 2lºC ir r1 == 0.30m r r3 == activos W/m3. eánicas ue pro-ne h == enedor. ta? les ra-1 siste-grande guien--ace la ión de :ratura valor :id del pa de 2. K, como ,Cuá-náxi-r del : de-l ojo ta se vec-~ de ter-;i-3.60 k1 = 0.35 W/rn · K h¡ = 12 W/rn.2 • K • Problemas k2 = 0.80 W/m · K h0 = 6 W/m2 • K (a) Construya los circuitos térmicos, marcando todos los potenciales y flujos para los sistemas excluyen-do e incluyendo los lentes de contacto. Escriba los elementos de resistencia en términos de paráme1.-os apropiados. (b) Determine la pérdida de calor de la cámara ante-rior con los lentes de contacto y sin ellos. (c) Discuta la implicación de los resultados. La superficie externa de una esfera hueca de radio r2 se sujeta a un flujo de calor uniforme q¡. La superficie in-terna en r 1 se conserva a una temperatura constante Ts, 1. (a) ·Desarrolle una expresión para la distribución de temperaturas T(r) en la pared de la esfera en tér-minos de q2 , T5, 1, r 1, r2, y la conductividad térmica del material de la pared k. (b) Si los radios interno y externo son r1 = 50 mm y r2 = 100 mm, ¿qué flujo de calor q2 se requiere para mantener la superficie externa a Ts, 2 = 5Ó ºC, mientras que la superficie interna está a Ts, 1 = 20ºC? La conductividad térmica del material de la pared es k = 10 W/m · K. 3.61 Una capa esférica de radios interior y exterior r; y r 0 , respectivamente, se llena con un material generador de calor que proporciona una rapidez de generación volu-métrica uniforme (W/m3) de q. La superficie externa de la capa se expone a un fluido que tiene una temperatura T"' y un coeficiente de convef:ción h. Obtenga una ex-presión para la distribución de temperaturas de estado estable T(r) en la capa, y e~prese los resultados en tér-minos de r ¡, r 0 , q, h, Too, y la conductividad térmica k del material de la capa. 3.62 Un transistor, que se aproxima como una fuente de ca-lor hemisférica de radio r0 = 0.1 mm, se empotra en un sustrato de silicio grande (k = 125 W/m · K) y disipa calor a una velocidad q. Todas las fronteras del silicio se mantienen a una temperatura ambiente de T"' = 27ºC, excepto para una superficie plana que está bien aislada. Sustrato de silicio 145 Obtenga una expresión general para la distribución de temperaturas del sustrato y evalúe la temperatura su-perficial de la fuente de calor para q = 4 W. 3.63 Una modalidad para destruir tejido maligno implica in-crustar una pequeña fuente de calor esférica de radio r 0 dentro del tejido y mantener temperaturas locales por arriba de un valor crítico Te por un periodo extenso. Su-ponga que el tejido que se extirpa de la fuente permane-ce a la temperatura normal del cuerpo (Tb = 37ºC). Obtenga una expresión general para la distribución ra-dial de temperaturas en el tejido bajo condiciones de estado estable en las que se disipa calor a una velocidad q. Si r0 = 0.5 mm, ¿qué transferencia de calor debe su-ministrarse para mantener una temperatura del tejido de T 2: Te = 42ºC en el dominio 0.5 ~ r ~ 5 mm? La conductividad térmica del tejido es aproximadamente 0.5 W/m · K. Conducción con generació, n interna de calor 3.64 Considere corazas cilíndricas y esféricas con superfi-cies interior y exterior en r 1 y r2 que se mantienen a temperaturas uniformes Ts, 1 y Ts, 2, respectivamente. Si hay generación uniforme de calor dentro de las cora-zas, obtenga expresiones para las distribuciones radia-les unidimensionales de la temperatura, flujo de calor y transferencia de calor. Compare sus resultados con los que se resumen en el apéndice C. 3.65 La distribución de temperaturas de estado estable en una pared plana compuesta con tres diferentes materia-les, cada uno de conductividad térmica constante, se muestra a continuación. 2 3 4 X (a) Comente las magnitudes relativas de_ q2_ )'_ q) y de q] y q4 . (b) Haga comentarios sobre las magnitudes relativas de kA y k8 y de k8 y kc. (c) Dibuje el flujo de calor como función de x. 3.66 Una pared plana de espesor 0.1 m y conductividad tér-mica 25 W/m · K, con una generación de calor volu-métrica uniforme de 0.3 MW/m3, se aísla en uno de sus lados mientras que el otro lado se expone a un fluido a 92ºC. El coeficiente de transferencia de calor por con-vección entre la pared y el fluido es 500 W/m2 • K. De-termine la temperatura máxima en la pared. ·------------------~------------------------------~ -.- |
3755 | https://math.libretexts.org/Bookshelves/Calculus/The_Calculus_of_Functions_of_Several_Variables_(Sloughter)/01%3A_Geometry_of_R/1.05%3A_Linear_and_Affine_Functions | Skip to main content
1.5: Linear and Affine Functions
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: Sep 2, 2021
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1.4.E: Lines, Planes, and Hyperplanes (Exercises)
1.5.E: Linear and Affine Functions (Exercises)
Page ID
: 22922
Dan Sloughter
Furman University
( \newcommand{\kernel}{\mathrm{null}\,})
One of the central themes of calculus is the approximation of nonlinear functions by linear functions, with the fundamental concept being the derivative of a function. This section will introduce the linear and affine functions which will be key to understanding derivatives in the chapters ahead.
Linear functions
In the following, we will use the notation f:Rm→Rnf:Rm→Rn to indicate a function whose domain is a subset of RmRmand whose range is a subset of RnRn. In other words, fftakes a vector with mmcoordinates for input and returns a vector with nncoordinates. For example, the function
f(x,y,z)=(sin(x+y),2x2+z)
f(x,y,z)=(sin(x+y),2x2+z)
is a function from R3R3to R2R2.
Definition 1.5.11.5.1
We say a function L:Rm→RmL:Rm→Rmis linear if (1) for any vectors xxand yyin RmRm,
L(x+y)=L(x)+L(y),
L(x+y)=L(x)+L(y),(1.5.1)
and (2) for any vector xxin RmRmand scalar aa,
L(ax)=aL(x).
L(ax)=aL(x).(1.5.2)
Example 1.5.11.5.1
Suppose f:R→Rf:R→R is defined by f(x)=3xf(x)=3x. Then for any xxand yyin RR,
f(x+y)=3(x+y)=3x+3y=f(x)+f(y),
f(x+y)=3(x+y)=3x+3y=f(x)+f(y),
and for any scalar aa,
f(ax)=3ax=af(x).
f(ax)=3ax=af(x).
Thus ff is linear.
Example 1.5.21.5.2
Suppose L:R2→R3L:R2→R3 is defined by
L(x1,x2)=(2x1+3x2,x1−x2,4x2).
L(x1,x2)=(2x1+3x2,x1−x2,4x2).
Then if x=(x1,x2)x=(x1,x2) and y=(y1,y2)y=(y1,y2) are vectors in R2R2,
L(x+y)=L(x1+y1,x2+y2)=(2(x1+y1)+3(x2+y2),x1+y1−(x2+y2),4(x2+y2))=(2x1+3x2,x1−x2,4x2)+(2y1+3y2,y1−y2,4y2)=L(x1,x2)+L(y1,y2)=L(x)+L(y).
L(x+y)=L(x1+y1,x2+y2)=(2(x1+y1)+3(x2+y2),x1+y1−(x2+y2),4(x2+y2))=(2x1+3x2,x1−x2,4x2)+(2y1+3y2,y1−y2,4y2)=L(x1,x2)+L(y1,y2)=L(x)+L(y).
Also, for x=(x1,x2)x=(x1,x2) and any scalar aa, we have
L(ax)=L(ax1,ax2)=(2ax1+3ax2,ax1−ax2,4ax2)=a(2x2+3x2,x1−x2,4x2)=aL(x).
L(ax)=L(ax1,ax2)=(2ax1+3ax2,ax1−ax2,4ax2)=a(2x2+3x2,x1−x2,4x2)=aL(x).
Thus LLis linear.
Now suppose L:R→RL:R→R is a linear function and let a=L(1)a=L(1). Then for any real number xx,
L(x)=L(1x)=xL(1)=ax.
L(x)=L(1x)=xL(1)=ax.(1.5.3)
Since any function L:R→RL:R→R defined by L(x)=axL(x)=ax, where aais a scalar, is linear (see Exercise 1), it follows that the only functions L:R→RL:R→R which are linear are those of the form L(x)=axL(x)=ax for some real number aa. For example, f(x)=5xf(x)=5x is a linear function, but g(x)=sin(x)g(x)=sin(x) is not.
Next, suppose L:Rm→RL:Rm→R is linear and let a1=L(e1),a2=L(e2),…,am=L(em)a1=L(e1),a2=L(e2),…,am=L(em). If x=(x1,x2,…,xm)x=(x1,x2,…,xm) is a vector in RmRm, then we know that
x=x1e1+x2e2+⋯+xmem.
Thus
L(x)=L(x1e1+x2e2+⋯+xmem)=L(x1e1)+L(x2e2)+⋯+L(xmem)=x1L(e1)+x2L(e2)+⋯+xmL(em)=x1a1+x2a2+⋯+xmam=a⋅x,
where a=(a1,a2,…,am). Since for any vector ain Rm, the function L(x)=a⋅x is linear (see Exercise 1), it follows that the only functions L:Rm→R which are linear are those of the form L(x)=a⋅x for some fixed vector ain Rm. For example,
f(x,y)=(2,−3)⋅(x,y)=2x−3y
is a linear function from R2to R, but
f(x,y,z)=x2y+sin(z)
is not a linear function from R3to R.
Now consider the general case where L:Rm→Rnis a linear function. Given a vector xin Rm, let Lk(x) be the kth coordinate of L(x),k=1,2,…,n. That is,
L(x)=(L1(x),L2(x),…,Ln(x)).
Since Lis linear, for any xand yin Rmwe have
L(x+y)=L(x)+L(y),
or, in terms of the coordinate functions,
(L1(x+y),L2(x+y),…,Ln(x+y))=(L1(x),L2(x),…,Ln(x))+(L1(y),L2(y),…,Ln(y))=(L1(x)+L1(y),L2(x)+L2(y)…,Ln(x)+Ln(y)).
Hence Lk(x+y)=Lk(x)+Lk(y) for k=1,2,…,n. Similarly, if xis in Rmand ais a scalar, then L(ax)=aL(x), so
(L1(ax),L2(ax),…,Ln(ax)=a(L1(x),L2(x),…,Ln(x))=(aL1(x),aL2(x),…,aLn(x)).
Hence Lk(ax)=aLk(x) for k=1,2,…,n. Thus for each k=1,2,…,n,Lk:Rm→R is a linear function. It follows from our work above that, for each k=1,2,…,n, there is a fixed vector akin Rmsuch that Lk(x)=ak⋅xfor all xin Rm. Hence we have
L(x)=(a1⋅x,a2⋅x,…,an⋅x)
for all xin Rm. Since any function defined as in (1.5.5) is linear (see Exercise 1 again), it follows that the only linear functions from Rmto Rnmust be of this form.
Theorem 1.5.1
If L:Rm→Rnis linear, then there exist vectors a1,a2,…,anin Rmsuch that
L(x)=(a1⋅x,a2⋅x,…,an⋅x)
for all x in Rm.
Example 1.5.3
In a previous example, we showed that the function L:R2→R3defined by
L(x1,x2)=(2x1+3x2,x1−x2,4x2)
is linear. We can see this more easily now by noting that
L(x1,x2)=((2,3)⋅(x1,x2),(1,−1)⋅(x1,x2),(0,4)⋅(x1,x2)).
Example 1.5.4
The function
f(x,y,z)=(x+y,sin(x+y+z))
is not linear since it cannot be written in the form of (1.5.6). In particular, the function f2(x,y,z)=sin(x+y+z) is not linear; from our work above, it follows that fis not linear.
Matrix Notation
We will now develop some notation to simplify working with expressions such as (1.5.6). First, we define an n×mmatrix to be to be an array of real numbers with nrows and mcolumns. For example,
M=[231−104]
is a 3×2 matrix. Next, we will identify a vector x=(x1,x2,…,xm) in Rmwith the m×1 matrix
x=[x1x2⋮xm],
which is called a column vector. Now define the product Mxof an n×m matrix Mwith an m×1 column vector xto be the n×1 column vector whose kth entry, k=1,2,…,n, is the dot product of the kth row of Mwith x. For example,
[231−104]=[4+32−10+4]=.
In fact, for any vector x=(x1,x2) in R2,
[231−104][x1x2]=[2x1+3x2x1−x24x2].
In other words, if we let
L(x1,x2)=(2x1+3x2,x1−x2,4x2),
as in a previous example, then, using column vectors, we could write
L(x1,x2)=[231−104][x1x2].
In general, consider a linear function L:Rm→Rndefined by
L(x)=(a1⋅x,a2⋅x,…,an⋅x)
for some vectors a1,a2,…,anin Rm. If we let Mbe the n×mmatrix whose kth row is ak,k=1,2,…,n, then
L(x)=Mx
for any xin Rm. Now, from our work above,
ak=(Lk(e1),Lk(e2),…,Lk(em),
which means that the jth column of Mis
[L1(ej)L2(ej)⋮Ln(ej)],
j=1,2,…,m. But (1.5.10) is just L(ej) written as a column vector. Hence Mis the matrix whose columns are given by the column vectors L(e1),L(e2),…,L(em).
Theorem 1.5.2
Suppose L:Rm→Rnis a linear function and Mis the n×mmatrix whose jth column is L(ej),j=1,2,…,m. Then for any vector xin Rm,
L(x)=Mx.
Example 1.5.5
Suppose L:R3→R2is defined by
L(x,y,z)=(3x−2y+z,4x+y).
Then
L(e1)=L(1,0,0)=(3,4),L(e2)=L(0,1,0)=(−2,1),
and
L(e3)=L(0,0,1)=(1,0).
So if we let
M=[3−21410],
then
L(x,y,z)=[3−21410][xyz].
For example,
L(1,−1,3)=[3−21410][1−13]=[3+2+34−1+0]=.
Example 1.5.6
Let Rθ:R2→R2be the function that rotates a vector xin R2counterclockwise through an angle θ, as shown in Figure 1.5.1. Geometrically, it seems reasonable that Rθis a linear function; that is, rotating the vector x+ythrough an angle θ should give the same result as first rotating xand yseparately through an angle θ and then adding, and rotating a vector axthrough an angle θ should give the same result as first rotating xthrough an angle θ and then multiplying by a. Now, from the definition of cos(θ) and sin(θ),
Rθ(e1)=Rθ(1,0)=(cos(θ),sin(θ))
(see Figure 1.5.2), and, since e2is e1rotated, counterclockwise, through an angle π2,
Rθ(e2)=Rθ+π2(e1)=(cos(θ+π2),sin(θ+π2))=(−sin(θ),cos(θ)).
Hence
Rθ(x,y)=[cos(θ)−sin(θ)sin(θ)cos(θ)][xy].
You are asked in Exercise 9 to verify that the linear function defined in (1.5.12) does in fact rotate vectors through an angle θ in the counterclockwise direction. Note that, for example, when θ=π2, we have
Rπ2(x,y)=[0−110][xy].
In particular, note that Rπ2(1,0)=(0,1) and Rπ2(0,1)=(−1,0); that is, Rπ2 takes e1to e2and e2to −e1. For another example, if θ=π6, then
Rπ6(x,y)=[√32−1212√32][xy].
In particular,
Rπ6(1,2)=[√32−1212√32]=[√32−112+√3]=[√3−221+2√32].
Affine functions
Definition 1.5.2
We say a function A:Rm→Rn is affine if there is a linear function L:Rm→Rnand a vector bin Rnsuch that
A(x)=L(x)+b
for all x in Rm.
An affine function is just a linear function plus a translation. From our knowledge of linear functions, it follows that if A:Rm→Rnis affine, then there is an n×mmatrix Mand a vector bin Rnsuch that
A(x)=Mx+b
for all xin Rm. In particular, if f:R→R is affine, then there are real numbers mand bsuch that
f(x)=mx+b
for all real numbers x.
Example 1.5.7
The function
A(x,y)=(2x+3,y−4x+1)
is an affine function from R2to R2since we may write it in the form
A(x,y)=L(x,y)+(3,1),
where L is the linear function
L(x,y)=(2x,y−4x).
Note that L(1,0)=(2,−4) and L(0,1)=(0,1), so we may also write Ain the form
A(x,y)=[20−41][xy]+.
Example 1.5.8
The affine function
A(x,y)=[1√2−1√21√21√2][xy]+
first rotates a vector, counterclockwise, in R2through an angle of π4 and then translates it by the vector (1,2).
1.4.E: Lines, Planes, and Hyperplanes (Exercises)
1.5.E: Linear and Affine Functions (Exercises) |
3756 | https://math.stackexchange.com/questions/3216536/converting-an-equation-based-on-square-roots | algebra precalculus - Converting an equation based on square roots - Mathematics Stack Exchange
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So I was wondering how to convert an equation of the form √x 1+√x 2+√x 3+...√x n+k=0 x 1−−√+x 2−−√+x 3−−√+...x n−−√+k=0 into a polynomial equation based on each x i x i.
For example if the equation was √x 1+√x 2+k=0
x 1−−√+x 2−−√+k=0
, then subtracting √x 2 x 2−−√ from each side and squaring yields:x 1+k 2+2 k√x 1=x 2.
x 1+k 2+2 k x 1−−√=x 2.
This can then be rearranged to: 2 k√x 1=−x 1−k 2+x 2.
2 k x 1−−√=−x 1−k 2+x 2.
Squaring both sides yields: 4 k 2 x 1=x 2 1+k 4+x 2 2+2 k 2 x 1−2 x 1 x 2−2 k 2 x 2.
4 k 2 x 1=x 2 1+k 4+x 2 2+2 k 2 x 1−2 x 1 x 2−2 k 2 x 2.
Rearranging/simplifying yields: x 2 1+x 2 2+k 4−2 k 2 x 1−2 x 1 x 2−2 k 2 x 2=0.
x 2 1+x 2 2+k 4−2 k 2 x 1−2 x 1 x 2−2 k 2 x 2=0.
How can I find an equation of this form given that n n is greater than 4 4? I am most interested in when n=6 n=6.
algebra-precalculus
polynomials
roots
radicals
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edited May 6, 2019 at 23:52
NoChance
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asked May 6, 2019 at 23:24
Varun VejallaVarun Vejalla
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2 Expand a product with (k±√x 1±√x 2)(k±x 1−−√±x 2−−√) factors with all possible sign combinations.Somos –Somos 2019-05-06 23:52:05 +00:00 Commented May 6, 2019 at 23:52
Ok that's what I thought @RossMillikan. Thanks for confirming.Varun Vejalla –Varun Vejalla 2019-05-07 00:12:22 +00:00 Commented May 7, 2019 at 0:12
@Ross, but if you keep on squaring, you eventually keep getting the same set of square roots, and when you have enough equations, you can eliminate them all.Gerry Myerson –Gerry Myerson 2019-05-07 00:19:35 +00:00 Commented May 7, 2019 at 0:19
@GerryMyerson can u show an example of this? I didn't realize I could cancel parts out when I was working it out by hand.Varun Vejalla –Varun Vejalla 2019-05-07 00:28:05 +00:00 Commented May 7, 2019 at 0:28
3 Look at mathpages.com/home/kmath111/kmath111.htmAtaulfo –Ataulfo 2019-05-07 01:05:03 +00:00 Commented May 7, 2019 at 1:05
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Well, you may have to do a lot of squaring, and it may not be practical to do it by hand, but here's the theory: let's start with √u+√v+√w+√x+√y+√z=k u−−√+v√+w−−√+x−−√+y√+z√=k. Square both sides, transfer all the terms without square roots to the right, divide by two, and you get √u v+⋯+√y z=k 2+f(u,…,z)u v−−√+⋯+y z−−√=k 2+f(u,…,z) for some polynomial f f. Square again and move non-roots to the right. On the left, you get a sum with terms of the type √u v u v−−√ and √u v w x u v w x−−−−−√, on the right some new polynomial g(u,…,z)g(u,…,z). Do it again, on the left you'll have terms of the type √u v u v−−√, √u v w x u v w x−−−−−√, and √u v w x y z u v w x y z−−−−−−√, on the right some polynomial h(u,…,z)h(u,…,z).
Keep on doing this. You'll only ever get terms of those three types on the left, and polynomials on the right. Now there are only 15 15 different terms of type √u v u v−−√, another 15 15 of type √u v w x u v w x−−−−−√, and just one of type √u v w x y z u v w x y z−−−−−−√, making 31 31 different terms in all. So after you've done the procedure 32 32 times, you'll have 32 32 linear equations in these 31 31 terms, and you can use linear algebra to boil them down to a single equation with no square roots in it, and you win.
I hope you won't expect me to actually carry out this procedure here....
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answered May 7, 2019 at 6:23
Gerry MyersonGerry Myerson
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COMMENT (A replay to the last comment of the O.P.).-Your problem is actually finding the minimum polynomial of −k−k or k k (which, in principle, is an irrational of degree 2 n 2 n) and then the link I gave you offers you the solution you want.
But you post it in a way that could lead to think that it is not. I explain why the minimum polynomial of −k−k solves the problem. The easiest example is for two radicals and it is enough to understand what I want to say.
You can easily find out the minimal polynomial of x=√a+√b x=a−−√+b√ which is x 4−2(a+b)x 2+(b−a)2=0
x 4−2(a+b)x 2+(b−a)2=0
and certainly −k=√a+√b−k=a−−√+b√ is a root of it.
Well, in this polynomial you can note that the coefficients are rational functions of a a and b b then you can pose your problem the way you do in your post by replacing a a for x 1 x 1, b b for x 2 x 2 and x x for −k−k or k k. This gives the result k 4−2(x 1+x 2)k 2+(x 2−x 1)2=0
k 4−2(x 1+x 2)k 2+(x 2−x 1)2=0
what obviously answers your problem for the case n=2 n=2
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answered May 7, 2019 at 18:40
AtaulfoAtaulfo
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I see my silly misunderstanding - I didn't realize that I could plug in k k for x x. Thanks for the help Varun Vejalla –Varun Vejalla 2019-05-07 19:46:36 +00:00 Commented May 7, 2019 at 19:46
You are welcome.Ataulfo –Ataulfo 2019-05-07 19:56:26 +00:00 Commented May 7, 2019 at 19:56
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Just for fun, the polynomial drawn from √a+√b+√c+k=0 a−−√+b√+c√+k=0, using the method by Somos:
a 4+6 a 2 b 2+6 a 2 c 2+6 a 2 k 4−4 a b 3+4 a b 2 k 2+4 a b c 2−40 a b c k 2+4 a b k 4−4 a c 3+4 a c 2 k 2+4 a c b 2+4 a c k 4−4 a k 6+b 4+6 b 2 c 2+6 b 2 k 4−4 b a 3+4 b a 2 k 2−4 b c 3+4 b c 2 k 2+4 b c a 2+4 b c k 4−4 b k 6+c 4+6 c 2 k 4−4 c a 3+4 c a 2 k 2−4 c b 3+4 c b 2 k 2−4 c k 6+k 8−4 k 2 a 3−4 k 2 b 3−4 k 2 c 3=0.
a 4+6 a 2 b 2+6 a 2 c 2+6 a 2 k 4−4 a b 3+4 a b 2 k 2+4 a b c 2−40 a b c k 2+4 a b k 4−4 a c 3+4 a c 2 k 2+4 a c b 2+4 a c k 4−4 a k 6+b 4+6 b 2 c 2+6 b 2 k 4−4 b a 3+4 b a 2 k 2−4 b c 3+4 b c 2 k 2+4 b c a 2+4 b c k 4−4 b k 6+c 4+6 c 2 k 4−4 c a 3+4 c a 2 k 2−4 c b 3+4 c b 2 k 2−4 c k 6+k 8−4 k 2 a 3−4 k 2 b 3−4 k 2 c 3=0.
Then next challenge is to determine the number of terms as a function of the number of variables.
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edited May 7, 2019 at 7:05
answered May 7, 2019 at 7:00
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3757 | https://www.middleprofessor.com/files/applied-biostatistics_bookdown/_book/counts | Chapter 18 Linear models for counts, binary responses, skewed responses, and ratios – Generalized Linear Models | Statistics for the Bench Biologist
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Statistics for the Experimental Biologist
Preface
Linear Models
Aren’t t-tests and ANOVA good enough?
What is unusual about this book?
A Table of Models
including mapping between linear models and classical tests
1 Analyzing experimental data with a linear model
1.1 This text is about using linear models to estimate treatment effects and the uncertainty in our estimates. This, raises the question, what is “an effect”?
Part I: Getting Started
2 Getting Started – R Projects and R Markdown
2.1 R vs R Studio
2.2 Download and install R and R studio
2.3 Open R Studio and modify the workspace preference
2.4 If you didn’t modify the workspace preferences from the previous section, go back and do it
2.5 R Markdown in a nutshell
2.6 Install R Markdown
2.7 Importing Packages
2.8 Create an R Studio Project for this textbook
2.9 Working on a project, in a nutshell
2.10 Create and setup an R Markdown document (Rmd)
2.10.1 Modify the yaml header
2.10.2 Modify the “setup” chunk
2.11 Let’s play around with an R Markdown file
2.11.1 Create a “fake-data” chunk
2.11.2 Create a “plot” chunk
2.11.3 Knit the Rmd
Part II: R fundamentals
3 Data – Reading, Wrangling, and Writing
3.1 Long data format – the way data should be
3.2 Use the here function to construct the file path
3.3 Learning from this chapter
3.4 Working in R
3.4.1 Importing data
3.5 Data wrangling
3.5.1 Reshaping data – Wide to long
3.5.2 Reshaping data – Transpose (turning the columns into rows)
3.5.3 Combining data
3.5.4 Subsetting data
3.5.5 Wrangling columns
3.5.6 Missing data
3.6 Saving data
3.7 Exercises
4 Plotting Models
4.1 Pretty good plots show the model and the data
4.1.1 Pretty good plot component 1: Modeled effects plot
4.1.2 Pretty good plot component 2: Modeled mean and CI plot
4.1.3 Combining Effects and Modeled mean and CI plots – an Effects and response plot.
4.1.4 Some comments on plot components
4.2 Working in R
4.2.1 Source data
4.2.2 How to plot the model
4.2.3 Be sure ggplot_the_model is in your R folder
4.2.4 How to use the Plot the Model functions
4.2.5 How to generate a Response Plot using ggpubr
4.2.6 How to generate a Response Plot with a grid of treatments using ggplot2
4.2.7 How to generate an Effects Plot
4.2.8 How to combine the response and effects plots
4.2.9 How to add the interaction effect to response and effects plots
Part III: Some Fundamentals of Statistical Modeling
5 Variability and Uncertainty (Standard Deviations, Standard Errors, and Confidence Intervals)
5.1 Standard errors are used to compute p-values and confidence intervals
5.2 Background
5.2.1 Sample standard deviation
5.2.2 Standard error of the mean
5.3 Simulations – using fake data as an intuition pump
5.3.1 Using Google Sheets to generate fake data to explore the standard error
5.3.2 Using R to generate fake data to explore the standard error
5.4 Bootstrapped standard errors
5.4.1 An example of bootstrapped standard errors
5.5 Confidence Intervals
5.5.1 Just the math
5.5.2 Interpretation of a confidence interval
5.5.3 Bootstrap confidence interval
5.5.4 A plot of a “parametric” CI vs.bootstrap CI of the means
5.6 Standard error of a difference between means
5.6.1 The standard error of a difference between means is the standard deviation of the sampling distribution of the difference
5.7 Confidence limits of a difference between means
5.7.1 95% of 95% CIs of the difference include the true difference
5.8 Hidden code
5.8.1 Import exp3b
6 P-values
6.1 A p-value is the probability of sampling a value as or more extreme than the test statistic if sampling from a null distribution
6.2 Pump your intuition – Creating a null distribution
6.3 A null distribution of t-values – the t distribution
6.4 P-values from the perspective of permutation
6.5 Parametric vs.non-parametric statistics
6.6 frequentist probability and the interpretation of p-values
6.6.1 Background
6.6.2 This book covers frequentist approaches to statistical modeling and when a probability arises, such as the p-value of a test statistic, this will be a frequentist probability.
6.6.3 Two interpretations of the p-value
6.6.4 NHST
6.7 Some major misconceptions of the p-value
6.7.1 Misconception: p>0.05 p>0.05 means there is no effect of treatment
6.7.2 Misconception: a p-value is repeatable
6.7.3 Misconception: 0.05 is the lifetime rate of false discoveries
6.7.4 Misconception: a low p-value indicates an important effect
6.7.5 Misconception: a low p-value indicates high model fit or high predictive capacity
6.8 What the p-value does not mean
6.9 On using p-values in experimental biology
6.10 Better reproducibility
6.11 Multiple testing and controlling for false positives
6.11.1 Controlling the family-wise error rate when all tests are testing the same hypothesis
6.12 Recommendations
6.12.1 Primary sources for recommendations
6.13 Problems
7 Errors in inference
7.1 Classical NHST concepts of wrong
7.1.1 Type I error
7.1.2 Power
7.2 A non-Neyman-Pearson concept of power
7.2.1 Estimation error
7.2.2 Coverage
7.2.3 Type S error
7.2.4 Type M error
Part IV: Introduction to Linear Models
8 An introduction to linear models
8.1 Two specifications of a linear model
8.1.1 The “error draw” specification
8.1.2 The “conditional draw” specification
8.1.3 Comparing the error-draw and conditional-draw ways of specifying the linear model
8.1.4 ANOVA notation of a linear model
8.2 A linear model can be fit to data with continuous, discrete, or categorical X X variables
8.2.1 Fitting linear models to experimental data in which the X X variable is continuous or discrete
8.2.2 Fitting linear models to experimental data in which the X X variable is categorical
8.3 Statistical models are used for prediction, explanation, and description
8.4 What is the interpretation of a regression coefficient?
8.5 What do we call the X X and Y Y variables?
8.6 Modeling strategy
8.7 Predictions from the model
8.8 Inference from the model
8.8.1 Assumptions for inference with a statistical model
8.8.2 Specific assumptions for inference with a linear model
8.9 “linear model,”regression model”, or “statistical model”?
9 Linear models with a single, continuous X (“regression”)
9.1 A linear model with a single, continuous X is classical “regression”
9.1.1 Analysis of “green-down” data
9.1.2 Learning from the green-down example
9.1.3 Using a regression model for “explanation” – causal models
9.1.4 Using a regression model for prediction – prediction models
9.1.5 Using a regression model for creating a new response variable – comparing slopes of longitudinal data
9.1.6 Using a regression model for for calibration
9.2 Working in R
9.2.1 Fitting the linear model
9.2.2 Getting to know the linear model: the summary function
9.2.3 Inference – the coefficient table
9.2.4 How good is our model? – Model checking
9.2.5 Plotting models with continuous X
9.2.6 Creating a table of predicted values and 95% prediction intervals
9.3 Hidden code
9.3.1 Import and plot of fig2c (ecosystem warming experimental) data
9.3.2 Import and plot efig_3d (Ecosysem warming observational) data
9.3.3 Import and plot of fig1f (methionine restriction) data
9.4 Try it
9.4.1 A prediction model from the literature
9.5 Intuition pumps
9.5.1 Correlation and $R^2
10 Linear models with a single, categorical X (“t-tests” and “ANOVA”)
10.1 A linear model with a single factor (categorical X variable) estimates the effects of the levels of factor on the response
10.1.1 Example 1 (fig3d) – two treatment levels (“groups”)
10.1.2 Understanding the analysis with two treatment levels
10.1.3 Example 2 – three treatment levels (“groups”)
10.1.4 Understanding the analysis with three (or more) treatment levels
10.2 Working in R
10.2.1 Fit the model
10.2.2 Controlling the output in tables using the coefficient table as an example
10.2.3 Using the emmeans function
10.2.4 Using the contrast function
10.2.5 How to generate ANOVA tables
10.3 Hidden Code
10.3.1 Importing and wrangling the fig3d data for example 1
10.3.2 Importing and wrangling the fig2a data for example 2
11 Model Checking
11.1 All statistical analyses should be followed by model checking
11.2 Linear model assumptions
11.2.1 A bit about IID
11.3 Diagnostic plots use the residuals from the model fit
11.3.1 Residuals
11.3.2 A Normal Q-Q plot is used to check for characteristic departures from Normality
11.3.3 Mapping QQ-plot departures from Normality
11.3.4 Model checking homoskedasticity
11.4 Using R
11.5 Hidden Code
11.5.1 Normal Q-Q plots
12 Violations of independence, homogeneity, or Normality
12.1 Lack of independence
12.1.1 Example 1 (exp1b) – a paired t-test is a special case of a linear mixed model
12.1.2 Example 2 (diHOME exp2a) – A repeated measures ANOVA is a special case of a linear mixed model
12.2 Heterogeneity of variances
12.2.1 When groups of the focal test have >> variance
12.3 The conditional response isn’t Normal
12.3.1 Example 1 (fig6f) – Linear models for non-normal count data
12.3.2 My data aren’t normal, what is the best practice?
12.4 Hidden Code
12.4.1 Importing and wrangling the exp1b data
12.4.2 Importing and wrangling the exp2a data
12.4.3 Importing and wrangling the fig6f data
13 Issues in inference
13.1 Replicated experiments – include Experiment Experiment as a random factor (better than one-way ANOVA of means)
13.1.1 Multiple experiments Example 1 (wound healing Exp4d)
13.1.2 Models for combining replicated experiments
13.1.3 Understanding Model exp4d_m1
13.1.4 The univariate model is equivalent to a linear mixed model of the aggregated data (Model exp4d_m2)
13.1.5 A linear mixed model of the full data
13.1.6 Analysis of the experiment means has less precision and power
13.1.7 Don’t do this – a t-test/fixed-effect ANOVA of the full data
13.2 Comparing change from baseline (pre-post)
13.2.1 Pre-post example 1 (DPP4 fig4c)
13.2.2 What if the data in example 1 were from from an experiment where the treatment was applied prior to the baseline measure?
13.2.3 Pre-post example 2 (XX males fig1c)
13.2.4 Regression to the mean
13.3 Longitudinal designs with more than one-post baseline measure
13.3.1 Area under the curve (AUC)
13.4 Normalization – the analysis of ratios
13.4.1 Kinds of ratios in experimental biology
13.4.2 Example 1 – The ratio is a density (number of something per area)
13.4.3 Example 2 – The ratio is normalizing for size differences
13.5 Don’t do this stuff
13.5.1 Normalize the response so that all control values are equal to 1.
13.6 A difference in significance is not necessarily significant
13.7 Researcher degrees of freedom
13.8 Hidden code
13.8.1 Import exp4d vimentin cell count data (replicate experiments example)
13.8.2 Import Fig4c data
13.8.3 XX males fig1c
13.8.4 Generation of fake data to illustrate regression to the mean
13.8.5 Import fig3f
13.8.6 Import exp3b
13.8.7 Plot the model of exp3b (glm offset data)
Part V: More than one X X – Multivariable Models
14 Linear models with added covariates (“ANCOVA”)
14.1 Adding covariates can increases the precision of the effect of interest
14.2 Understanding a linear model with an added covariate – heart necrosis data
14.2.1 Fit the model
14.2.2 Plot the model
14.2.3 Interpretation of the model coefficients
14.2.4 Everything adds up
14.2.5 Interpretation of the estimated marginal means
14.2.6 Interpretation of the contrasts
14.2.7 Adding the covariate improves inference
14.3 Understanding interaction effects with covariates
14.3.1 Fit the model
14.3.2 Plot the model with interaction effect
14.3.3 Interpretation of the model coefficients
14.3.4 What is the effect of a treatment, if interactions are modeled? – it depends.
14.3.5 Which model do we use, M 1 M 1 or M 2 M 2?
14.4 Understanding ANCOVA tables
14.5 Working in R
14.5.1 Importing the heart necrosis data
14.5.2 Fitting the model
14.5.3 Using the emmeans function
14.5.4 ANCOVA tables
14.5.5 Plotting the model
14.6 Best practices
14.6.1 Do not use a ratio of part:whole as a response variable – instead add the denominator as a covariate
14.6.2 Do not use change from baseline as a response variable – instead add the baseline measure as a covariate
14.6.3 Do not “test for balance” of baseline measures
14.7 Best practices 2: Use a covariate instead of normalizing a response
15 Linear models with two categorical X X – Factorial linear models (“two-way ANOVA”)
15.1 A linear model with crossed factors estimates interaction effects
15.1.1 An interaction is a difference in simple effects
15.1.2 A linear model with crossed factors includes interaction effects
15.1.3 factorial experiments are frequently analyzed as flattened linear models in the experimental biology literature
15.2 Example 1 – Estimation of a treatment effect relative to a control effect (“Something different”) (Experiment 2j glucose uptake data)
15.2.1 Understand the experimental design
15.2.2 Fit the linear model
15.2.3 Inference
15.2.4 Plot the model
15.2.5 alternaPlot the model
15.3 Understanding the linear model with crossed factors 1
15.3.1 What the coefficients are
15.3.2 The interaction effect is something different
15.3.3 Why we want to compare the treatment effect to a control effect
15.3.4 The order of the factors in the model tells the same story differently
15.3.5 Power for the interaction effect is less than that for simple effects
15.3.6 Planned comparisons vs.post-hoc tests
15.4 Example 2: Estimation of the effect of background condition on an effect (“it depends”) (Experiment 3e lesian area data)
15.4.1 Understand the experimental design
15.4.2 Fit the linear model
15.4.3 Check the model
15.4.4 Inference from the model
15.4.5 Plot the model
15.5 Understanding the linear model with crossed factors 2
15.5.1 Conditional and marginal means
15.5.2 Simple (conditional) effects
15.5.3 Marginal effects
15.5.4 The additive model
15.5.5 Reduce models for the right reason
15.5.6 The marginal means of an additive linear model with two factors can be weird
15.6 Example 3: Estimation of synergy (“More than the sum of the parts”) (Experiment 1c JA data)
15.6.1 Examine the data
15.6.2 Fit the model
15.6.3 Model check
15.6.4 Inference from the model
15.6.5 Plot the model
15.6.6 Alternative plot
15.7 Understanding the linear model with crossed factors 3
15.7.1 Thinking about the coefficients of the linear model
15.8 Issues in inference
15.8.1 For pairwise contrasts, it doesn’t matter if you fit a factorial or a flattened linear model
15.8.2 For interaction contrasts, it doesn’t matter if you fit a factorial or a flattened linear model
15.8.3 Adjusting p-values for multiple tests
15.9 Two-way ANOVA
15.9.1 How to read a two-way ANOVA table
15.9.2 What do the main effects in an ANOVA table mean?
15.10 More issues in inference
15.10.1 Longitudinal experiments – include Time as a random factor (better than repeated measures ANOVA)
15.11 Working in R
15.11.1 Model formula
15.11.2 Using the emmeans function
15.11.3 Contrasts
15.11.4 Practice safe ANOVA
15.11.5 Better to avoid these
15.12 Hidden Code
15.12.1 Import exp2j (Example 1)
15.12.2 Import exp3e lesian area data (Example 2)
15.12.3 Import Exp1c JA data (Example 3)
Part VI – Expanding the Linear Model
16 Models for non-independence – linear mixed models
16.1 Liberal inference from pseudoreplication
16.2 Conservative inference from failure to identify blocks
16.3 Introduction to models for non-independent data (linear mixed models)
16.4 Experimental designs in experimental bench biology
16.4.1 Notation for models
16.4.2 Completely Randomized Design (CRD)
16.4.3 Completely Randomized Design with Subsampling (CRDS) or “Nested Design”
16.4.4 Randomized Complete Block Design (RCBD)
16.4.5 Randomized Split Plot Design (RSPD)
16.4.6 Generalized Randomized Complete Block Design (GRCBD)
16.4.7 Nested Randomized Complete Block Design (NRCBD)
16.4.8 Longitudinal Randomized Complete Block Design (LRCBD)
16.4.9 Variations due to multiple measures of the response variable
16.5 Building the linear (mixed) model for clustered data
16.6 Example 1 – A random intercepts and slopes explainer (demo1)
16.6.1 Batched measurements result in clustered residuals
16.6.2 Clustered residuals result in correlated error
16.6.3 In blocked designs, clustered residuals adds a variance component that masks treatment effects
16.6.4 Linear mixed models are linear models with added random factors
16.6.5 What the random effects are
16.6.6 In a blocked design, a linear model with added random effects increases precision of treatment effects
16.6.7 The correlation among random intercepts and slopes
16.6.8 Clustered residuals create heterogeneity among treatments
16.6.9 Linear mixed models are flexible
16.6.10 A random intercept only model
16.6.11 A model including an interaction intercept
16.6.12 AIC and model selection – which model to report?
16.6.13 The specification of random effects matters
16.6.14 Mixed Effect and Repeated Measures ANOVA
16.6.15 Pseudoreplication
16.7 Example 2 – experiments without subsampling replication (exp6g)
16.7.1 Understand the data
16.7.2 Model fit and inference
16.7.3 The model exp6g_m1 adds a random intercept but not a random slope
16.7.4 The fixed effect coefficients of model exp6g_m1
16.7.5 The random intercept coefficients of exp6g_m1
16.7.6 The random and residual variance and the intraclass correlation of model exp6g_m1
16.7.7 The linear mixed model exp6g_m1 increases precision of treatment effects, relative to a fixed effects model
16.7.8 Alternative models for exp6g
16.7.9 Paired t-tests and repeated measures ANOVA are special cases of linear mixed models
16.7.10 Classical (“univariate model”) repeated measures ANOVA of exp6g
16.7.11 “Multivariate model” repeated measures ANOVA
16.7.12 Linear mixed models vs repeated measures ANOVA
16.7.13 Modeling mouse_id mouse_id as a fixed effect
16.8 Example 3 – Factorial experiments and no subsampling replicates (exp5c)
16.8.1 Understand the data
16.8.2 Examine the data
16.8.3 Model fit and inference
16.8.4 Why we care about modeling batch in exp5c
16.8.5 The linear mixed model exp5c_m1 adds two random intercepts
16.8.6 The fixed effect coefficients of model exp5c_m1
16.8.7 The random effect coefficients of model exp5c_m1
16.8.8 Alternative models for exp5c
16.8.9 Classical (“univariate model”) repeated measures ANOVA
16.8.10 “Multivariate model” repeated measures ANOVA of exp5c
16.8.11 Modeling donor donor as a fixed effect
16.9 Example 4 – Experiments with subsampling replication (exp1g)
16.9.1 Understand the data
16.9.2 Examine the data
16.9.3 Fit the model
16.9.4 Inference from the model
16.9.5 Plot the model
16.9.6 Alternaplot the model
16.9.7 Understanding the alternative models
16.9.8 The VarCorr matrix of models exp1g_m1a and exp1g_m1b
16.9.9 The linear mixed model has more precision and power than the fixed effect model of batch means
16.9.10 Fixed effect models and pseudoreplication
16.9.11 Mixed-effect ANOVA
16.10 Statistical models for experimental designs
16.10.1 Models for Completely Randomized Designs (CRD)
16.10.2 Models for batched data (CRDS, RCBD, RSPD, GRCBD, NRCBD)
16.11 Which model and why?
16.11.1 CRD
16.11.2 CRDS
16.11.3 RCBD
16.11.4 RSPD
16.11.5 RCBDS
16.11.6 GRCBD
16.11.7 GRCBDS
16.12 Working in R
16.12.1 Plotting models fit to batched data
16.12.2 Repeated measures ANOVA (randomized complete block with no subsampling)
16.13 Hidden code
16.13.1 Import exp5c
16.13.2 Import exp1g
17 Linear models for longitudinal experiments – I. pre-post designs
17.1 Best practice models
17.2 Common alternatives that are not recommended
17.3 Advanced models
17.4 Understanding the alternative models
17.4.1 (M1) Linear model with the baseline measure as the covariate (ANCOVA model)
17.4.2 (M2) Linear model of the change score (change-score model)
17.4.3 (M3) Linear model of post-baseline values without the baseline as a covariate (post model)
17.4.4 (M4) Linear model with factorial fixed effects (fixed-effects model)
17.4.5 (M5) Repeated measures ANOVA
17.4.6 (M6) Linear mixed model
17.4.7 (M7) Linear model with correlated error
17.4.8 (M8) Constrained fixed effects model with correlated error (cLDA model)
17.4.9 Comparison table
17.5 Example 1 – a single post-baseline measure (pre-post design)
17.6 Working in R
17.7 Hidden code
17.7.1 Import and wrangle mouse sociability data
18 Linear models for counts, binary responses, skewed responses, and ratios – Generalized Linear Models
18.1 Introducing Generalized Linear Models using count data examples
18.1.1 The Generalized Linear Model (GLM)
18.1.2 Kinds of data that are modeled by a GLM
18.2 Example 1 – GLM models for count responses (“angiogenic sprouts” fig3a)
18.2.1 Understand the data
18.2.2 Model fit and inference
18.3 Understanding Example 1
18.3.1 Modeling strategy
18.3.2 Model checking fits to count data
18.3.3 Biological count data are rarely fit well by a Poisson GLM. Instead, fit a quasi-poisson or negative binomial GLM model.
18.3.4 A GLM is a linear model on the link scale
18.3.5 Coeffecients of a Generalized Linear Model with a log-link function are on the link scale.
18.3.6 Modeled means in the emmeans table of a Generalized Linear Model can be on the link scale or response scale – Report the response scale
18.3.7 Some consequences of fitting a linear model to count data
18.4 Example 2 – Use a GLM with an offset instead of a ratio of some measurement per total (“dna damage” data fig3b) #glm_offset
18.4.1 fig3b (“dna damage”) data
18.4.2 Understand the data
18.4.3 Model fit and inference
18.5 Understanding Example 2
18.5.1 An offset is an added covariate with a coefficient fixed at 1
18.5.2 A count GLM with an offset models the area-normalized means
18.5.3 Compare an offset to an added covariate with an estimated coefficient
18.5.4 Issues with plotting
18.6 Example 3 – GLM models for binary responses
18.7 Working in R
18.7.1 Fitting GLMs to count data
18.7.2 Fitting a GLM to a continuous conditional response with right skew.
18.7.3 Fitting a GLM to a binary (success or failure, presence or absence, survived or died) response
18.7.4 Fitting Generalized Linear Mixed Models
18.8 Model checking GLMs
18.9 Hidden code
18.9.1 Import Example 1 data (fig3a – “angiogenic sprouts”)
19 Linear models with heterogenous variance
Appendix: An example set of analyses of experimental data with linear models
20 Background physiology to the experiments in Figure 2 of “ASK1 inhibits browning of white adipose tissue in obesity”
21 Analyses for Figure 2 of “ASK1 inhibits browning of white adipose tissue in obesity”
21.1 Setup
21.2 Data source
21.3 control the color palette
21.4 useful functions
21.5 figure 2b – effect of ASK1 deletion on growth (body weight)
21.5.1 figure 2b – import
21.5.2 figure 2b – exploratory plots
21.6 Figure 2c – Effect of ASK1 deletion on final body weight
21.6.1 Figure 2c – import
21.6.2 Figure 2c – check own computation of weight change v imported value
21.6.3 Figure 2c – exploratory plots
21.6.4 Figure 2c – fit the model: m1 (lm)
21.6.5 Figure 2c – check the model: m1
21.6.6 Figure 2c – fit the model: m2 (gamma glm)
21.6.7 Figure 2c – check the model, m2
21.6.8 Figure 2c – inference from the model
21.6.9 Figure 2c – plot the model
21.6.10 Figure 2c – report
21.7 Figure 2d – Effect of ASK1 KO on glucose tolerance (whole curve)
21.7.1 Figure 2d – Import
21.7.2 Figure 2d – exploratory plots
21.7.3 Figure 2d – fit the model
21.7.4 Figure 2d – check the model
21.7.5 Figure 2d – inference
21.7.6 Figure 2d – plot the model
21.8 Figure 2e – Effect of ASK1 deletion on glucose tolerance (summary measure)
21.8.1 Figure 2e – message the data
21.8.2 Figure 2e – exploratory plots
21.8.3 Figure 2e – fit the model
21.8.4 Figure 2e – check the model
21.8.5 Figure 2e – inference from the model
21.8.6 Figure 2e – plot the model
21.9 Figure 2f – Effect of ASK1 deletion on glucose infusion rate
21.9.1 Figure 2f – import
21.9.2 Figure 2f – exploratory plots
21.9.3 Figure 2f – fit the model
21.9.4 Figure 2f – check the model
21.9.5 Figure 2f – inference
21.9.6 Figure 2f – plot the model
21.10 Figure 2g – Effect of ASK1 deletion on tissue-specific glucose uptake
21.10.1 Figure 2g – import
21.10.2 Figure 2g – exploratory plots
21.10.3 Figure 2g – fit the model
21.10.4 Figure 2g – check the model
21.10.5 Figure 2g – inference
21.10.6 Figure 2g – plot the model
21.11 Figure 2h
21.12 Figure 2i – Effect of ASK1 deletion on liver TG
21.12.1 Figure 2i – fit the model
21.12.2 Figure 2i – check the model
21.12.3 Figure 2i – inference
21.12.4 Figure 2i – plot the model
21.12.5 Figure 2i – report the model
21.13 Figure 2j
22 Simulations – Count data (alternatives to a t-test)
22.1 Use data similar to Figure 6f from Example 1
22.2 Functions
22.3 Simulations
22.3.1 Type I, Pseudo-Normal distribution
22.3.2 Type I, neg binom, equal n
22.3.3 Type I, neg binom, equal n, small theta
22.3.4 Type I, neg binom, unequal n
22.3.5 Power, Pseudo-Normal distribution, equal n
22.3.6 Power, neg binom, equal n
22.3.7 Power, neg binom, small theta
22.3.8 Power, neg binom, unequal n
22.3.9 Power, neg binom, unequal n, unequal theta
22.3.10 Type 1, neg binom, equal n, unequal theta
22.4 Save it, Read it
22.5 Analysis
Appendix 1: Getting Started with R
22.6 Get your computer ready
22.6.1 Start here
22.6.2 Install R
22.6.3 Install R Studio
22.6.4 Install R Markdown
22.6.5 (optional) Alternative LaTeX installations
22.7 Start learning R Studio
Appendix 2: Online Resources for Getting Started with Statistical Modeling in R
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Statistics for the Bench Biologist
Chapter 18 Linear models for counts, binary responses, skewed responses, and ratios – Generalized Linear Models
All updates to this book are now at
18.1 Introducing Generalized Linear Models using count data examples
Biologists frequently count stuff, and design experiments to estimate the effects of different factors on these counts. Count data can cause numerous problems with linear models that assume a normal, conditional distribution, including 1) counts are discrete, and can be zero or positive integers only, 2) counts tend to bunch up on the small side of the range, creating a distribution with a positive skew, 3) a sample of counts can have an abundance of zeros, and 4) the variance of counts increases with the mean (see Figure 18.1 for some of these properties). Some count data can be approximated by and reasonably modeled with a normal distribution. More often, count data should modeled with a Poisson distribution or negative binomial distribution using a generalized linear model. Poisson and negative binomial distributions are discrete probability distributions with two important properties: 1) the distribution contains only zero and positive integers and 2) the variance is a function of the mean. Back before modern computing and fast processors, count data were often analyzed by either transforming the response or by non-parametric hypothesis tests. One reason to prefer a statistical modeling approach with a GLM is that we can get interpretable parameter estimates. By contrast, both the analysis of transformed data and non-parametric hypothesis tests are really tools for computing “correct” p-values.
Figure 18.1: Histogram of the number of angiogenic sprouts in response to two of the four treatment combinations for the experiment in Example 1.
18.1.1 The Generalized Linear Model (GLM)
As outlined in Two specifications of a linear model, a common way that biological researchers are taught to think about a response variable is
r e s p o n s e=e x p e c t e d+e r r o r r e s p o n s e=e x p e c t e d+e r r o r
or, using the notation of this text,
y=β 0+β 1 treatment+ε ε∼Normal(0,σ 2)(18.1)y=β 0+β 1 treatment+ε(18.1)ε∼Normal(0,σ 2)
That is, we can think of a response as the sum of some systematic part (β 0+β 1 treatment β 0+β 1 treatment) and a stochastic (“random error”) part (ε ε), where the stochastic part is a random draw from a normal distribution with mean zero and variance σ 2 σ 2. This way of thinking about the generation of the response is useful for linear models, and model checking linear models, but is much less useful for thinking about generalized linear models or model checking generalized liner models. For example, if we want to model the number of angiogenic sprouts in response to some combination of GAS6 and treatment using a Poisson distribution, the following is the wrong way to think about the statistical model
sprouts=β 0+β 1 treatment GAS6+β 2 genotype FAK_ko+β 3 treatment GAS6:genotype FAK_ko+ε i ε∼Poisson(λ)(18.2)sprouts=β 0+β 1 treatment GAS6+β 2 genotype FAK_ko+β 3 treatment GAS6:genotype FAK_ko+ε i(18.2)ε∼Poisson(λ) That is, we should not think of a count as the sum of a systematic part and a random draw from a Poisson distribution. Why? Because it is the counts, conditional on the treatment treatment and genotype genotype, that are poisson distributed, not the residuals from the fit model.
Thinking about the distribution of count data using model (18.2) leads to absurd consequences. For example, if we set the mean of the Poisson “error” to zero (like with a normal distribution), then the error term for every observation would have to be zero (because the only way to get a mean of zero with non-negative integers is if every value is zero). Or, if the study is modeling the effect of a treatment on the counts (that is, the X X are dummy variables) then β 0 β 0 is the expected mean count of the control (or reference) group. But if we add non-zero Poisson error to this, then the mean of the control group would be larger than β 0 β 0. This doesn’t make sense. And finally, equation (18.2) generates a continuous response, instead of an integer, because β 0 β 0 and β 1 β 1 are continuous.
A better way to think about the data generation for a linear model that naturally leads to the correct way to think about data generation for a generalized linear model, is
y i∼N(μ i,σ 2)E(y i|treatment)=μ i μ i=β 0+β 1 treatment i(18.3)y i∼N(μ i,σ 2)E(y i|treatment)=μ i(18.3)μ i=β 0+β 1 treatment i
That is, a response is a random draw from a normal distribution with mean m u m u (not zero!) and variance σ 2 σ 2. Line 1 is the stochastic part of this specification. Line 3 is the systematic part.
The specification of a generalized linear model has both stochastic and systematic parts but adds a third part, which is a link function connecting the stochastic and systematic parts.
The stochastic part, which is a probability distribution from the exponential family (this is sometimes called the “random part”) y i∼Prob(μ i)y i∼Prob(μ i)
the systematic part, which is a linear predictor (I like to think about this as the deterministic part)
η=β 0+β 1 treatment i η=β 0+β 1 treatment i 3. a link function connecting the two parts
η i=g(μ i)η i=g(μ i)
μ μ (the Greek symbol mu) is the conditional mean (or expectation E(Y|X)E(Y|X)) of the response on the response scale and η η (the Greek symbol eta) is the conditional mean of the response on the link scale. The response scale is the scale of the raw measurements and has units of the raw measurements. The link scale is the scale of the transformed mean. A GLM models the response with a distribution specified in the stochastic part. The probability distributions introduced in this chapter are the Poisson and Negative Binomial. The natural link function, and default link function in R, for the Poisson and Negative Binomial is the “log link”, η=l o g(μ)η=l o g(μ). More generally, while each distribution has a natural (or, “canonical”) link function, one can use alternatives. Given this definition of a generalized linear model, a linear model is a GLM with a normal distribution and an Identity link (η=μ η=μ).
y i∼Normal(μ i,σ 2)E(y i|treatment)=μ i η=β 0+β 1 treatment μ i=η i y i∼Normal(μ i,σ 2)E(y i|treatment)=μ i η=β 0+β 1 treatment μ i=η i
Think about the link function and GLM more generally like this: the GLM constructs a linear model that predicts the conditional means on the link scale. If the model uses a “log link” (η i=l o g(μ i)η i=l o g(μ i)), then η i η i – the conditional mean on the link scale – is the log of the modeled mean on the response scale. The modeled mean on the response scale is the inverse link function.
μ i=g−1(η i)μ i=g−1(η i)
For a log link, a modeled mean is e x p(η i)e x p(η i))$.
Importantly, in a GLM, the individual data values are not transformed. A GLM with a log link is not the same as a linear model on log transformed data.
When modeling counts using the Poisson or negative binomial distributions with a log link, the link scale is linear, and so the effects are additive on the link scale, while the response scale is nonlinear (it is the exponent of the link scale), and so the effects are multiplicative on the response scale. If this doesn’t make sense now, an example is worked out below. The inverse of the link function backtransforms the parameters from the link scale back to the response scale. So, for example, a prediction on the response sale is e x p(^η)e x p(η^) and a coefficient on the response scale is e x p(b j)e x p(b j).
18.1.2 Kinds of data that are modeled by a GLM
If a response is a count then use a Poisson, quasi-Poisson or negative binomial family with a log link.
If a response is binary response, for example, presence/absence or success/failure or survive/die, then use the binomial family with a logistic link (other link functions are also useful). This is classically known as logistic regression.
If a response is a fraction that is a ratio of counts, for example the fraction of total cells that express some marker, then use the binomial family with a logistic link. Think of each count as a “success”.
If a response is a fraction of counts per “effort” (cells per area or volume or time), then use a Poisson, quasi-Poisson or negative binomial family with a log link on the raw count and include the measure of effort as an offset.
If a response is continuous but the variance increases with mean then use the gamma family
If a response is a fraction of a continuous measure per “effort” (tumor area per total area or volume or time), then use a gamma family on the raw measure and include the measure of effort as an offset.
18.2 Example 1 – GLM models for count responses (“angiogenic sprouts” fig3a)
18.2.1 Understand the data
Article source Lechertier, Tanguy, et al.“Pericyte FAK negatively regulates Gas6/Axl signalling to suppress tumour angiogenesis and tumour growth.” Nature communications 11.1 (2020): 1-14.
data source
The researchers designed a set of experiments to investigate the effects of pericyte derived FAK (focal adhesion kinase, a protein tyrosine kinase) on the Gas6/Axl pathway regulating angiogenesis promoting tumor growth. Pericytes are cells immediately deep to the endothelium (the epithelial lining) of the smallest blood vessels, including capillaries, arterioles, and venules. Angiogenesis is the growth of new blood vessels. GAS6 (growth arrest specific 6) is a protein commonly expressed in tumors.
The example data is from the experiment for Figure 3a. The design is 2×2 2×2 – two factors (treatment treatment and genotype genotype) each with two levels.
Factor 1: treatment treatment
reference level: “PBS”. Phosphate buffered saline added to tissue. This is the control treatment.
treatment level: “GAS6”. Added to tissue in solution. The experiment is designed to test its effect on promoting angiogenesis in the development of tumors.
Factor 2: genotype genotype
reference level: “FAK_wt”. The functional genotype.
treatment level: “FAK_ko”. Tissue-specific FAK deletion in pericytes. The experiment is designed to test the effect of pericyte-derived FAK on slowing angiogenesis in the development of tumors. If true, then its deletion should result in increased tumor development.
The four treatment by genotype combinations are
Control (“PBS FAK_wt”) – negative control
FAK_ko (“PBS FAK_ko”) – PBS control. Unknown response given deletion of putative angiogenesis inhibitor but no added putative angiogenesis promotor
GAS6 (“GAS6 FAK_wt”) – GAS6 control. GAS6 expected to promote angiogenesis but this the response is expected to be inhibited by some amount by FAK
GAS6+FAK_ko (“GAS6 FAK_ko”) – focal treatment. Expected positive angiogenesis
The planned contrasts are
(PBS FAK_ko) - (PBS FAK_wt) – the effect of FAK deletion given the control treatment
(GAS6 FAK_ko) - (GAS6 FAK_wt) – the effect of FAK deletion given the GAS6 treatment
((PBS FAK_ko) - (PBS FAK_wt)) - ((GAS6 FAK_ko) - (GAS6 FAK_wt)). The interaction effect giving the effect of the combined treatment relative to the individual effects.
18.2.2 Model fit and inference
18.2.2.1 Fit the models
fig3a_m1 <- lm(sprouts ~ treatment genotype, data = fig3a)
fig3a_m2 <- glm(sprouts ~ treatment genotype,
family = "poisson",
data = fig3a)
fig3a_m3 <- glm.nb(sprouts ~ treatment genotype,
data = fig3a)
18.2.2.2 Check the linear model
ggcheck_the_model(fig3a_m1)
Notes
left panel shows classic right skew conditional distribution with larger values much larger than expected with a normal distribution.
right panel shows heterogeneity and specifically the variance increasing with the mean.
18.2.2.3 Check the poisson model
Check shape and homogeneity
```
from the DHARMa package
fig3a_m2_simulation <- simulateResiduals(fittedModel = fig3a_m2, n = 250)
plot(fig3a_m2_simulation, asFactor = FALSE)
```
Notes
poisson glm fails to generated scaled residuals approximating uniform distribution.
Check dispersion
fig3a_m2_simulation_refit <- simulateResiduals(fittedModel = fig3a_m2,
n = 250,
refit = TRUE)
fig3a_m2_test_dispersion <- testDispersion(fig3a_m2_simulation_refit)
Notes
large overdispersion
Check zero inflation
fig3a_m2_test_zi <- testZeroInflation(fig3a_m2_simulation_refit)
Notes
The data has too many zeros relative to the expected number from a poisson GLM.
18.2.2.4 Check the negative binomial model
Check shape and homogeneity
```
from the DHARMa package
fig3a_m3_simulation <- simulateResiduals(fittedModel = fig3a_m3,
n = 250)
plot(fig3a_m3_simulation, asFactor = FALSE)
```
Notes
uniform q-q for negative binomial GLM looks good.
spread-location plot looks good.
Check dispersion
fig3a_m3_simulation_refit <- simulateResiduals(fittedModel = fig3a_m3,
n = 250,
refit = TRUE)
fig3a_m3_test_dispersion <- testDispersion(fig3a_m3_simulation_refit)
Notes
good
Check zero inflation
fig3a_m3_test_zi <- testZeroInflation(fig3a_m3_simulation_refit)
18.2.2.5 Inference from the model
fig3a_m3_coef <- cbind(coef(summary(fig3a_m3)),
confint(fig3a_m3))
| | Estimate | Std. Error | z value | Pr(>|z|) | 2.5 % | 97.5 % |
--- --- ---
| (Intercept) | 0.90 | 0.231 | 3.9 | 0.000 | 0.45 | 1.36 |
| treatmentGAS6 | 1.30 | 0.287 | 4.5 | 0.000 | 0.74 | 1.86 |
| genotypeFAK_ko | 0.04 | 0.357 | 0.1 | 0.900 | -0.65 | 0.75 |
| treatmentGAS6:genotypeFAK_ko | 0.46 | 0.425 | 1.1 | 0.275 | -0.37 | 1.30 |
fig3a_m3_emm <- emmeans(fig3a_m3,
specs = c("treatment", "genotype"),
type="response")
| treatment | genotype | response | SE | df | asymp.LCL | asymp.UCL |
--- --- ---
| PBS | FAK_wt | 2.47 | 0.6 | Inf | 1.57 | 3.88 |
| GAS6 | FAK_wt | 9.05 | 1.5 | Inf | 6.48 | 12.64 |
| PBS | FAK_ko | 2.58 | 0.7 | Inf | 1.52 | 4.40 |
| GAS6 | FAK_ko | 15.04 | 2.4 | Inf | 11.06 | 20.46 |
```
fig3a_m3_emm # print in console to get row numbers
set the mean as the row number from the emmeans table
pbs_fak_wt <- c(1,0,0,0)
gas6_fak_wt <- c(0,1,0,0)
pbs_fak_ko <- c(0,0,1,0)
gas6_fak_ko <- c(0,0,0,1)
1. (PBS FAK_ko) - (PBS FAK_wt)
2. (GAS6 FAK_ko) - (GAS6 FAK_wt)
3. ((PBS FAK_ko) - (PBS FAK_wt)) - (GAS6 FAK_ko) - (GAS6 FAK_wt).
fig3a_m3_planned <- contrast(
fig3a_m3_emm,
method = list(
"(PBS FAK_ko) - (PBS FAK_wt)" = c(pbs_fak_ko - pbs_fak_wt),
"(GAS6 FAK_ko) - (GAS6 FAK_wt)" = c(gas6_fak_ko - gas6_fak_wt),
"Interaction" = c(gas6_fak_ko - gas6_fak_wt) -
c(pbs_fak_ko - pbs_fak_wt)
),
adjust = "none"
) %>%
summary(infer = TRUE)
```
| contrast | ratio | SE | df | asymp.LCL | asymp.UCL | null | z.ratio | p.value |
--- --- --- ---
| (PBS FAK_ko) / (PBS FAK_wt) | 1.05 | 0.373 | Inf | 0.52 | 2.10 | 1 | 0.12516 | 0.9 |
| (GAS6 FAK_ko) / (GAS6 FAK_wt) | 1.66 | 0.385 | Inf | 1.06 | 2.62 | 1 | 2.19446 | 0.0 |
| Interaction | 1.59 | 0.676 | Inf | 0.69 | 3.66 | 1 | 1.09082 | 0.3 |
Notes
18.2.2.6 Plot the model
18.2.2.7 Alternaplot the model
18.3 Understanding Example 1
18.3.1 Modeling strategy
Instead of testing assumptions of a model using formal hypothesis tests before fitting the model, a better strategy is to 1) fit one or more models based on initial evaluation of the data, and then do 2) model checking using diagnostic plots, diagnostic statistics, and simulation (see Section All statistical analyses should be followed by model checking).
For the fig3a data, I fit a linear model, a Poisson GLM, and a negative binomial GLM. I use the diagnostic plots and statistics to help me decide which model to report.
18.3.2 Model checking fits to count data
We use the fit models to check
the compatibility between the quantiles of the observed residuals and the distribution of expected quantiles from the family in the model fit
if the observed distribution is over or under dispersed
if there are more zeros than expected by the theoretical distribution. If so, the observed distribution is zero-inflated
18.3.2.1 Checking the linear model fig3a_m1 – a Normal-QQ plot
Figure ??A shows a histogram of the residuals from the fit linear model. The plot shows that the residuals seem to be clumped at the negative end of the range, which suggests that a model with a normally distributed conditional outcome (or normal error) is not well approximated.
Figure 18.2: Diagnostic plots of angiogenic sprout data (fig3a). A) Distribution of the residuals of the fit linear model. B) Normal-QQ plot of the residuals of the fit linear model.
A better way to investigate this is with the Normal-QQ plot in Figure ??B, which plots the sample quantiles for a variable against their theoretical quantiles. If the conditional outcome approximates a normal distribution,
the points should roughly follow the robust regression line,
the points should be largely inside the 95% CI (gray) bounds for sampling a normal distribution with the variance estimated by the model, and
the robust regression line should be largely inside the CI bounds.
For the sprout data, the points are above the line at the positive end, hug the upper bound of the 95% CI at the negative end, are well above the 95% CI at the positive end, and the robust regression is distinctly shallower than a line bisecting the CI bounds. At the left (negative) end, the observed values are more positive than the theoretical values. Remembering that this plot is of residuals, if we think about this as counts, this means that our smallest counts are not as small as we would expect given the mean, the variance, and a normal distribution. This shouldn’t be surprising – the counts range down to zero and counts cannot be below zero. At the positive end, the sample values are again more positive than the theoretical values. Thinking about this as counts, this means that are largest counts are larger than expected given the mean, the variance, and a normal distribution. This pattern is what we’d expect of count data.
18.3.2.2 Checking the linear model fig3a_m1 – Spread-level plot for checking homoskedasticity
A linear model also assumes that the error is homoskedastic – the error variance is not a function of the value of the X X variables). Non-homoskedastic error is heteroskedastic. I will typically use “homogenous variance” and “heterogenous variance” since these terms are more familiar to biologists. The fit model can be checked for homogeneity using a spread-level (also known as a scale-location) plot, which comes in several forms. I like a spread-level plot that is a scatterplot of the positive square-root of the standardized residuals against the fitted values (remember that the fitted values are the values computed by the linear predictor of the model – they are the “predicted values” of the observed data). If the residuals approximate a normal distribution, then a regression line through the scatter should be close to horizontal. The regression line in the spread-level plot of the fit of the linear model to the sprout data shows a distinct increase in the “scale” (the square root of the standardized residuals) with increased fitted value, which is expected of data sampled from a distribution in which the variance increases the mean.
18.3.2.3 Two distributions for count data – Poisson and Negative Binomial
The pattern in the Normal-QQ plot in Figure ??B should discourage a researcher from modeling the data with a normal distribution and instead model the data with an alternative distribution using a Generalized Linear Model. There is no unique mapping between observed data and a data generating mechanism with a specific distribution, so this decision is not as easy as thinking about the data generation mechanism and then simply choosing the “correct” distribution. Section 4.5 in Bolker (xxx) is an excellent summary of how to think about the generating processes for different distributions in the context of ecological data. Since the response in the angiogenic sprouts data are counts, we need to choose a distribution that generates integer values, such as the Poisson or the negative binomial.
Poisson – A Poisson distribution is the probability distribution of the number of occurrences of some thing (a white blood cell, a tumor, or a specific mRNA transcript) generated by a process that generates the thing at a constant rate per unit effort (duration or space). This constant rate is the parameter λ λ, which is the expectation (the expected mean of the counts), so E(Y)=μ=λ E(Y)=μ=λ. Because the rate per effort is constant, the variance of a Poisson variable equals the mean, σ 2=μ=λ σ 2=μ=λ. Figure ?? shows three samples from a Poisson distribution with λ λ set to 1, 5, and 10. The plots show that, as the mean count (λ λ) moves away from zero, a Poisson distribution 1) becomes less skewed and more closely approximates a normal distribution and 2) has an increasingly low probability of including zero (less than 1% zeros when the mean is 5).
A Poisson distribution, then, is useful for count data in which the conditional variance is close to the conditional mean. Very often, biological count data are not well approximated by a Poisson distribution because the variance is either less than the mean, an example of underdispersion5, or greater than the mean, an example of overdispersion6. A useful distribution for count data with overdispersion is the negative binomial.
Negative Binomial – The negative binomial distribution is a discrete probability distribution of the number of successes that occur before a specified number of failures k k given a probability p of success. This isn’t a very useful way of thinking about modeling count data in biology. What is useful is that the Negative Binomial distribution can be used simply as way of modeling an “overdispersed” Poisson process. Using the parameterization in the MASS::glm.nb function, the mean of a negative binomial variable is μ μ and the variance is σ 2=μ+μ 2 θ σ 2=μ+μ 2 θ. As a method for modeling an overdispersed Poisson variable, θ θ functions as a dispersion parameter controlling the amount of overdispersion and can be any real, positive value (not simply a positive integer), including values less than 1. As θ θ approaches positive infinity, the “overdispersion” bit μ 2 θ μ 2 θ goes to zero and the variance goes to μ μ, which is the same as the Poisson.
18.3.2.4 Model checking a GLM I – the quantile-residual uniform-QQ plot
Normal-QQ plots were introduced in Section @ref{normal-qq} of the Model Checking chapter and applied to the linear model fit of the angiogenic sprout data in Section 18.3.2.1 above. We cannot use a Normal-QQ plot with a Poisson or negative binomial GLM fit because the residuals from this fit are not expected to be normally distributed. An alternative to a Normal-QQ plot for a GLM fit is a quantile-residual uniform-QQ plot of observed quantile residuals.
(#fig:glm-fig3a_m2-check-poisson-again)Quantile-residual uniform-QQ plot of the Poisson GLM fit to the angiogenic sprouts (fig3a) data.
Notes
The x-axis (“Expected”) contains the expected quantiles from a uniform distribution.
The y-axis (“Observed”) contains the observed quantile residuals from a GLM fit, which are the residuals from the fit model that are transformed in a way that the expected distribution is uniform under the fit model family. This means that we’d expect the quantile residuals to closely approximate the expected quantiles from a uniform distribution. If the approximation is close, the points will fall along the “y = x” line in the plot.
The gray shaded area is a 95% confidence interval computed using a parametric bootstrap. At any value of the expected quantile, the interval will include an observed quantile 95% of the time. This gray area gives us a sense of the variability we’d get when we fit models to random samples from the specified model.
In the quantile-residual QQ plot for Model fig3a_m2, the observed residuals are far outside the 95% boundary. The observed residuals are smaller than expected at the negative (left) end and larger than expected at the right (high) end. This means the residuals are more spread out than expected for a Poisson sample. The data are overdispersed for this model. Understand that overdispersion is not a property of data but of the residuals from a specific model fit to the data.
Misconceivable – A common misconception is that if the distribution of the response approximates a Poisson distribution, then the residuals of a GLM fit with a Poisson distribution should be normally distributed, which could then be checked with a Normal-QQ plot, and homoskedastic, which could be checked with a scale-location plot. Neither of these is true because a GLM does not transform the data and, in fact, the model definition does not specify anything about the distribution of an “error” term – there is no ε ε in the model definition above! This is why thinking about the definition of a linear model by specifying an error term with a normal distribution can be confusing and lead to misconceptions when learning GLMs.
18.3.2.5 Model checking a GLM II – Spread-level plot for checking homoskedasticity
(#fig:glm-fig3a_m2-spreadlevel-again)Spread-level plot of the Poisson GLM fit to the angiogenic sprouts (fig3a) data.
Notes
The three curves are quantile regressions fit to the 25% (the first quartile), 50% (the second quartile or median), and 75% (the third quartile) quantiles conditional on the fitted value (the x-axis). The p%p% quantile regression line goes through the p%p% quantile of the Y Y values at each value of X X. Think of it like this: at any location along the x-axis, there is a spread of points on the y-axis that contains a first quartile, a median, and a third quartile. The 25%, 50%, and 75% quantile regression lines go through these points.
Why fit the three quartiles? The classic spread-level plot typically fits a regression, which is the expected (mean) residual, conditional on the fitted value (for example, the plot in section @ref(glm-fig3a_m1-spreadlevel)). Researchers typically look at this and think, “the whole spread of points is becoming more variable as the fitted value increases” but the regression is only modeling the mean. By fitting the first, second, and third quartile regressions, this spread-level plot evaluates deviations from expected spread not just at the mean but in the lower and upper halves of the spread.
18.3.2.6 Model checking a GLM III – Checking dispersion
(#fig:glm-fig3a_m2-check-dispersion-again)Dispersion plot of the negative binomial GLM fit to the angiogenic sprouts (fig3a) data.
Notes
This plot is a histogram of the sum of squared Pearson residuals of fake data sampled from the fit model. Pearson residuals are the raw residuals divided by the square root of the fitted value. Remember that in the Poisson distribution, the variance is equal to the expectation (mean), so a Pearson residual is the raw residual divided by the standard deviation of the residual. A way to think about this is, Pearson residuals “correct” for the heterogeneity in variance that arises among groups with different mean counts.
The sum of squared Pearson residuals is a measure of the dispersion of the residuals.
The red line is the observed sum of the squared Pearson residuals of the fit model.
If the observed dispersion approximates that expected from sampling from the fit model, the red line will be within the histogram.
The red line here is far larger than expected given the histogram, which indicates that the residuals are overdispersed given the fit model.
Overdispersion will be common with Poisson GLM fits to biological data.
18.3.2.7 Model Checking a count GLM – Check zero inflation
Counts can have the value zero. Data that have more zeros than expected given a fit count GLM model (Poisson, quasi-Poisson, negative binomial) is zero-inflated.
zero_inflation_test <- testZeroInflation(simulation_output)
This plot is a histogram of the number of zeros in each of the fake data sets generated by the fit model. An observed number of zeros at the extremes of this distribution are unlikely given the fit model. The number of zeros in the observed data is greater than expected by the model.
18.3.3 Biological count data are rarely fit well by a Poisson GLM. Instead, fit a quasi-poisson or negative binomial GLM model.
Here are the diagnostic plots of the negative binomial GLM fit to the fig3a data
(#fig:glm-fig3a_m3-qq-again)Quantile-residual uniform-QQ plot of the negative binomial GLM fit to the angiogenic sprouts (fig3a) data.
(#fig:glm-fig3a_m3-spreadlevel-again)Spread-level plot of the negative binomial GLM fit to the angiogenic sprouts (fig3a) data.
(#fig:glm-fig3a_m3-check-dispersion-again)Dispersion plot of the negative binomial GLM fit to the angiogenic sprouts (fig3a) data.
18.3.4 A GLM is a linear model on the link scale
The negative-binomial GLM fit to the angiogenic sprout data (fig3a) is
sprouts i∼NB(μ i,θ)E(sprouts|treatment, genotype)=μ μ i=e x p(η i)η i=β 0+β 1 treatment GAS6+β 2 genotype FAK_ko+β 3 treatment GAS6:genotype FAK_ko(18.4)sprouts i∼NB(μ i,θ)E(sprouts|treatment, genotype)=μ μ i=e x p(η i)η i=β 0+β 1 treatment GAS6+β 2 genotype FAK_ko+(18.4)β 3 treatment GAS6:genotype FAK_ko
The first line of Model (18.4) is the stochastic part stating the response is modeled as a random Negative Binomial variable with conditional mean μ i μ i, and variance μ+μ 2 θ μ+μ 2 θ. Fitting the model to the data estimates μ i μ i for all i. μ i μ i will be the same for all mice within each treatment by genotype combination because they share the same conditions (the values of the indicator variables for treatment, genotype, and their interaction).
The second line states the μ μ is the mean conditional on the value of treatment treatment and genotype genotype
The third line connects the conditional mean on the link scale (η η) with the conditional mean on the response scale (μ μ). This is the backtransformation.
The fourth line is the linear predictor – it is a linear model on the link scale. The linear predictor includes three indicator variables.
Remember that a linear model is a model in which the coefficients are additive, meaning that the coefficients do not have exponents or are not multiplied by each other.
18.3.5 Coeffecients of a Generalized Linear Model with a log-link function are on the link scale.
The coefficients of the fit negative binomial model are
| | Estimate | Std. Error | z value | Pr(>|z|) | 2.5 % | 97.5 % |
--- --- ---
| (Intercept) | 0.90446 | 0.231 | 3.9 | 0.000 | 0.45 | 1.36 |
| treatmentGAS6 | 1.29805 | 0.287 | 4.5 | 0.000 | 0.74 | 1.86 |
| genotypeFAK_ko | 0.04462 | 0.357 | 0.1 | 0.900 | -0.65 | 0.75 |
| treatmentGAS6:genotypeFAK_ko | 0.46382 | 0.425 | 1.1 | 0.275 | -0.37 | 1.30 |
Notes
The coefficients are on the link scale, which is a linear (additive) scale.
18.3.5.1 The intercept of a Generalized Linear Model with a log-link function is the mean of the reference on the link scale
In the linear model fig3a_m1, the intercept is the modeled mean of the reference group. In a GLM with a log-link (including Models fig3a_m2 and fig3a_m3), the intercept is the mean of the reference group on the link scale. The modeled mean of the reference on the response scale is computed by the backtransformaiton exp(b 1)exp(b 1). The function “exp(x)” is the exponent, which is often written using the notation e x e x.
The transformation between link scale and response scale is part of the specification of a Generalized Linear Model. For the negative binomial GLM fit to the fig3a data, this specification is in line 3 of the specification (Model (18.4)).
Compare the computation here with the modeled mean in the emmeans table (Section 18.2.2.5 or with a computation of the sample mean.
b1 <- coef(fig3a_m3)
exp(b1)
```
(Intercept)
2.470588
```
mean(fig3a[t_by_g == "PBS FAK_wt", sprouts])
## 2.470588
18.3.5.2 The coefficients of the indicator variables of a Generalized Linear Model with a log-link function are effect-ratios on the response scale
In the linear model fig3a_m1, the coefficient of an indicator variable is a difference in means. In a GLM with a log-link, the coefficient of an indicator variable is the difference (group minus reference) of the modeled means on the link scale. Let’s check this with some computations.
The modeled means on the link scale are in the emmeans table
fig3a_m3_emm_link <- emmeans(fig3a_m3,
specs = c("treatment", "genotype")) %>%
summary() %>%
data.table()
Using this table, the difference between the link-scale mean of “GAS6 FAK_wt” and “PBS FAK_wt” (the reference) is
fig3a_m3_emm_link[treatment == "GAS6" &
genotype == "FAK_wt", emmean] -
fig3a_m3_emm_link[treatment == "PBS" & genotype == "FAK_wt", emmean]
## 1.298045
And the coefficient for the treatment_gas6 treatment_gas6 indicator variable is
```
coefficient for treatment_gas6
b2 <- coef(fig3a_m3)
b2
```
```
treatmentGAS6
1.298045
```
In a GLM with a log-link, the exponent of the coefficient of an indicator variable is the ratio of the modeled means on the response scale. Let’s check this with some computations.
The modeled means on the response scale are in the emmeans table
fig3a_m3_emm_response <- emmeans(fig3a_m3,
specs = c("treatment", "genotype"),
type = "response") %>%
summary() %>%
data.table()
Using this table, the ratio of the response-scale mean of “GAS6 FAK_wt” to the response-scale mean of “PBS FAK_wt” (the reference) is
fig3a_m3_emm_response[treatment == "GAS6" &
genotype == "FAK_wt", response]/
fig3a_m3_emm_response[treatment == "PBS" & genotype == "FAK_wt", response]
## 3.662132
And the coefficient for the treatment_gas6 treatment_gas6 indicator variable backtransformed to the response scale is
exp_b2 <- exp(b2)
exp_b2
```
treatmentGAS6
3.662132
```
Since this backtransformed coefficient is both an effect and a ratio, I call it an effect ratio. It’s value is how many times bigger (or smaller if less than one) the non-reference response is relative to the reference response. The response of the GAS6 treatment is 3.66×× that of the reference treatment.
18.3.6 Modeled means in the emmeans table of a Generalized Linear Model can be on the link scale or response scale – Report the response scale
18.3.6.1 The emmeans table on the link scale contains the log means
(#tab:fig3a_m3_emm_link)(#tab:fig3a_m3_emm_link)Emmeans table on the link scale | treatment | genotype | emmean | SE | df | asymp.LCL | asymp.UCL |
--- --- ---
| PBS | FAK_wt | 0.90446 | 0.23 | Inf | 0.45 | 1.36 |
| GAS6 | FAK_wt | 2.20250 | 0.17 | Inf | 1.87 | 2.54 |
| PBS | FAK_ko | 0.94908 | 0.27 | Inf | 0.42 | 1.48 |
| GAS6 | FAK_ko | 2.71094 | 0.16 | Inf | 2.40 | 3.02 |
The values are the statistics on the link scale. The mean and CI, but not the SE, can be meaningfully backtransformed to the response scale using the exponent function to get the mean and CI on the response scale.
The CIs are “asymptotic”, meaning they are computed using infinite degrees of freedom. The consequences is that the CIs will be too narrow, especially for small n.
Check the math! Given asymptotic CIs, the lower CI should be (mean - 1.96 ×× SE) and the upper should be (mean + 1.96 ×× SE).
0.90446 - 1.960.23
## 0.45366
1. A GLM is linear on the link scale. This means the model is additive on the link scale. Modeled means on the link scale are computed by adding the coefficients.
```
b <- coef(fig3a_m3) # the model coefficients
the way I typically compute these
pbs_fak_wt_link <- b
gas6_fak_wt_link <- b + b
pbs_fak_ko_link <- b + b
gas6_fak_ko_link <- b + b + b + b
but this is the algebra more consistent with the linear model math
pbs_fak_wt_link <- b + b0 + b0 + b0
gas6_fak_wt_link <- b + b1 + b0 + b0
pbs_fak_ko_link <- b + b0 + b1 + b0
gas6_fak_ko_link <- b + b1 + b1 + b1
combine into a table
fig3a_means <- data.table(
group = t_by_g_levels,
"mean (link scale)" = c(pbs_fak_wt_link,
gas6_fak_wt_link,
pbs_fak_ko_link,
gas6_fak_ko_link)
)
```
Table 18.1: Table 18.2: Table of group treatment by genotype means on the link scale | group | mean (link scale) |
--- |
| PBS FAK_wt | 0.90446 |
| GAS6 FAK_wt | 2.20250 |
| PBS FAK_ko | 0.94908 |
| GAS6 FAK_ko | 2.71094 |
Compare the values here to those in the emmeans table on the link scale (Table @ref(tab:fig3a_m3_emm_link)).
18.3.6.2 The emmeans table on the response scale contains more readily interpretable means
(#tab:fig3a_m3_emm_response)(#tab:fig3a_m3_emm_response)Emmeans table on the response scale | treatment | genotype | response | SE | df | asymp.LCL | asymp.UCL |
--- --- ---
| PBS | FAK_wt | 2.47059 | 0.57 | Inf | 1.57 | 3.88 |
| GAS6 | FAK_wt | 9.04762 | 1.54 | Inf | 6.48 | 12.64 |
| PBS | FAK_ko | 2.58333 | 0.70 | Inf | 1.52 | 4.40 |
| GAS6 | FAK_ko | 15.04348 | 2.36 | Inf | 11.06 | 20.46 |
The values in the column “response” are the modeled means on the response scale. These values are the exponent of the values in the “emmean” column of the emmeans table on the link scale.
The values in the columns “asymp.LCL” and “asymp.UCL” are the 95% confidence intervals on the response scale. These values are the exponent of the values in the same columns of the emmeans table on the link scale.
Don’t do additive math on the response scale! Remember, a GLM is linear on the link scale. The CIs on the response scale are not the mean plus or minus 1.96 ×× SE!
2.47059 - 1.960.57
## 1.35339
# oops
Modeled means on the response scale are computed by backtransforming the modeled means on the link scale. Since the fit model used a log-link, the backtransformation is the exponent.
```
pbs_fak_wt_response <- exp(pbs_fak_wt_link)
gas6_fak_wt_response <- exp(gas6_fak_wt_link)
pbs_fak_ko_response <- exp(pbs_fak_ko_link)
gas6_fak_ko_response <- exp(gas6_fak_ko_link)
add a column to the fig3a_means data.table
fig3a_means[, "mean (response scale)" := c(pbs_fak_wt_response,
gas6_fak_wt_response,
pbs_fak_ko_response,
gas6_fak_ko_response)]
```
Table 18.3: Table 18.4: Table of group treatment by genotype means on both the link and response scale. | group | mean (link scale) | mean (response scale) |
---
| PBS FAK_wt | 0.90446 | 2.47059 |
| GAS6 FAK_wt | 2.20250 | 9.04762 |
| PBS FAK_ko | 0.94908 | 2.58333 |
| GAS6 FAK_ko | 2.71094 | 15.04348 |
Compare the values here to those in the emmeans table on the response scale (Table @ref(tab:fig3a_m3_emm_response)).
18.3.7 Some consequences of fitting a linear model to count data
18.3.7.1 One – linear models can make absurd predictions
plot_grid(gg1, gg2, gg3, ncol=3, labels = "AUTO")
Notes
A prediction interval is a confidence interval of a prediction – using the fit model to predict future responses given the same conditions (here, assignment to one of the four different treatment combinations).
Left panel: The prediction intervals from the linear model imply that negative sprouts could be sampled. This is absurd.
Middle panel: The fit linear model is used to make 100 fake predictions in each group.
Right panel: The fit negative binomial GLM is used to make 100 fake predictions in each group. Nothing absurd here.
18.3.7.2 Two – linear models can perform surprisingly well if one is only interested in p-values
P-values are a function of the sampling distribution of group means and differences in means, and, due to the magic of the central limit theorem, linear models fit to count data perform surprisingly well in the sense of
Type I error that approximates the nominal value
Reasonable power compared to GLM models and many non-parametric tests.
18.4 Example 2 – Use a GLM with an offset instead of a ratio of some measurement per total (“dna damage” data fig3b) #glm_offset
A problem that often arises in count data (and other kinds of measures) are counts that are taken in samples with different areas or volumes of tissue. As a consequence, size and treatment response are confounded – samples with higher counts may have higher counts because of a different response to treatment, a larger amount of tissue, or some combination. The common practice in experimental biology is to adjust for tissue size variation by area-normalizing the count using the ratio c o u n t a r e a c o u n t a r e a and then testing for a difference in the normalized count using either a linear model NHST (t-test/ANOVA) or a non-parametric NHST (Mann-Whitney-Wilcoxan). These practices raise at least two statistical issues.
The ratio will have some kind of ratio distribution that violates the normal distribution assumption of the linear model.
A Mann-Whitney-Wilcoxan does not estimate meaningful effects, is less powerful than GLM models, and is not flexible for complex designs including factorial or covariate models.
A better practice is to model the count using a GLM that adds the denominator of the ratio (the area or volume) as an offset in the model. A count GLM with offset also models the ratio (see section 18.5.2 below), but in a way that is likely to be much more compatible with the data.
NHST of the ratios will perform okay in the sense of Type I error that is close to nominal but will have relatively low power compared to a generalized linear model with offset. If the researcher is interested in best practices including the reporting of uncertainty of estimated effects, a GLM with offset will have more useful confidence intervals – for example CIs from linear model assuming Normal error can often include absurd values such as ratios less than zero.
Source article (Fernández, Álvaro F., et al.“Disruption of the beclin 1–BCL2 autophagy regulatory complex promotes longevity in mice.” Nature 558.7708 (2018): 136-140.)
18.4.1 fig3b (“dna damage”) data
These data were first introduced in the Issues chapter, Section 13.4.2. There only the age 20 month data were analyzed. Here the full dataset is analyzed.
Public source
Source data for Fig. 3
The example here is from Fig 3b.
18.4.2 Understand the data
Response variable – number of TUNEL+ cells measured in kidney tissue, where a positive marker indicates nuclear DNA damage.
Background. The experiments in Figure 3 were designed to measure the effect of a knock-in mutation in the gene for the beclin 1 protein on autophagy and tissue health in the kidney and heart. The researchers were interested in autophagy because there is evidence in many non-mammalian model organisms that increased autophagy reduces age-related damage to tissues and increases health and lifespan. BCL2 is an autophagy inhibitor. Initial experiments showed that the knock-in mutation in beclin 1 inhibits BCL2. Inhibiting BCL2 with the knock-in mutation should increase autophagy and, as a consequence, reduce age-related tissue damage.
Design - 2×2 2×2 factorial with offset
Factor 1: age age with levels “Young” (two months) and “Old” (twenty months).
Factor 2: genotype genotype with levels “WT” (wildtype) and “KI” (knock-in).
Offset: area_mm2 area_mm2. The area of the kidney tissue containing the counted cells.
The planned contrasts are
(Young KI) - (Young WT) – the effect of the beclin 1 knock-in mutation in the 2 month old mice. We expect this effect to be negative (more damage in the WT) but small, assuming there is little BCL2-related DNA damage by 2 months.
(Old KI) - (Old WT) – the effect of the beclin 1 knock-in mutation in the 20 month old mice. We expect this effect to be negative (more damage in the WT) but large assuming there is substantial BCL2-related DNA damage by 20 months.
Interaction. The magnitude of the interaction will largely be a function of the difference in BCL2-related DNA damage between 2 and 20 months and not really a function with how the knockin functions at these two ages.
18.4.3 Model fit and inference
18.4.3.1 Fit the models
```
lm with ratio response
fig3b_m1 <- lm(count_per_area ~ age genotype,
data = fig3b)
poisson offset
fig3b_m2 <- glm(positive_nuclei ~ age genotype +
offset(log(area_mm2)),
family = "poisson",
data = fig3b)
nb offset
fig3b_m3 <- glm.nb(positive_nuclei ~ age genotype +
offset(log(area_mm2)),
data = fig3b)
non-parametric
fig3b_m4_2mo <- wilcox.test(count_per_area ~ genotype,
data = fig3b[age == "Young"])
fig3b_m4_20mo <- wilcox.test(count_per_area ~ genotype,
data = fig3b[age == "Old"])
```
18.4.3.2 Check the models
ggcheck_the_model(fig3b_m1)
fig3b_m2_simulation <- simulateResiduals(fittedModel = fig3b_m2, n = 250)
plot(fig3b_m2_simulation, asFactor = FALSE)
fig3b_m3_simulation <- simulateResiduals(fittedModel = fig3b_m3,
n = 250)
plot(fig3b_m3_simulation, asFactor = FALSE)
18.4.3.3 Inference from the model
fig3b_m3_coef <- cbind(coef(summary(fig3b_m3)),
confint(fig3b_m3))
| | Estimate | Std. Error | z value | Pr(>|z|) | 2.5 % | 97.5 % |
--- --- ---
| (Intercept) | -0.07 | 0.508 | -0.1 | 0.887 | -0.95 | 1.09 |
| ageOld | 0.86 | 0.575 | 1.5 | 0.133 | -0.39 | 1.91 |
| genotypeKI | -0.10 | 0.722 | -0.1 | 0.887 | -1.56 | 1.36 |
| ageOld:genotypeKI | -0.66 | 0.806 | -0.8 | 0.416 | -2.27 | 0.96 |
mean_area <- mean(fig3b[age == "Old" & genotype == "WT",
positive_nuclei])
fig3b_m3_emm <- emmeans(fig3b_m3,
specs = c("age", "genotype"),
type="response",
offset = log(mean_area))
| age | genotype | response | SE | df | asymp.LCL | asymp.UCL |
--- --- ---
| Young | WT | 58.20 | 29.6 | Inf | 21.51 | 157.48 |
| Old | WT | 137.86 | 37.0 | Inf | 81.42 | 233.40 |
| Young | KI | 52.53 | 27.0 | Inf | 19.20 | 143.72 |
| Old | KI | 64.57 | 15.3 | Inf | 40.55 | 102.80 |
```
fig3b_m3_pairs <- contrast(fig3b_m3_emm,
method = "revpairwise") %>%
summary(infer = TRUE)
fig3b_m3_emm # print in console to get row numbers
set the mean as the row number from the emmeans table
young_wt <- c(1,0,0,0)
old_wt <- c(0,1,0,0)
young_ki <- c(0,0,1,0)
old_ki <- c(0,0,0,1)
1. (Young KI) - (Young WT)
2. (Old KI) - (Old WT)
3. Interaction
fig3b_m3_planned <- contrast(
fig3b_m3_emm,
method = list(
"(Young KI) - (Young WT)" = c(young_ki - young_wt),
"(Old KI) - (Old WT)" = c(old_ki - old_wt),
"Interaction" = c((old_ki - old_wt) - (young_ki - young_wt))
# "Interaction" = c(old_ki - old_wt) -
# c(young_ki - young_wt)
),
adjust = "none"
) %>%
summary(infer = TRUE)
```
| contrast | ratio | SE | df | asymp.LCL | asymp.UCL | null | z.ratio | p.value |
--- --- --- ---
| (Young KI) / (Young WT) | 0.90 | 0.652 | Inf | 0.22 | 3.72 | 1 | -0.14200 | 0.9 |
| (Old KI) / (Old WT) | 0.47 | 0.168 | Inf | 0.23 | 0.95 | 1 | -2.11608 | 0.0 |
| Interaction | 0.52 | 0.418 | Inf | 0.11 | 2.52 | 1 | -0.81353 | 0.4 |
Notes
18.4.3.4 Plot the model
18.5 Understanding Example 2
18.5.1 An offset is an added covariate with a coefficient fixed at 1
The systematic part (the linear predictor) of the negative binomial model fit to the fig3b data is
positive_nuclei=β 0+β 1 age Old+β 2 genotype KI+β 3 age Old:genotype KI+1.0×log(area_mm2)positive_nuclei=β 0+β 1 age Old+β 2 genotype KI+β 3 age Old:genotype KI+1.0×log(area_mm2)
Notes
The offset variable is log(area_mm2)log(area_mm2). An offset is a covariate with a fixed coefficient of 1.0 – this coefficient is not estimated.
Figure 18.3 illustrates the offset on the response and the log scale.
Figure 18.3: What an offset looks like. A. The offset on the response scale. B. The offset on the log scale. In A and B, only the lines for the 20 month old mice are shown.
On the log scale (Figure 18.3B), the offset curves are parallel because each is forced to have a slope of one on the link scale. On the response scale, these curves have different slopes – this slope is e x p(b 0)e x p(b 0), that is, the exponent of the intercept on the log scale).
The plot on the log scale (Figure 18.3B) is easy to misinterpret as a regression with a slope of 1 fit to the log transformed values. It isn’t, because individual values are not transformed in a GLM. I’m specifically referring to this space as the log and not link scale to try to avoid this misinterpretation.
18.5.2 A count GLM with an offset models the area-normalized means
In the t-test or ANOVA of the difference in the area-normalized count of TUNEL+ nuclei, the systematic component of the equivalent linear model is (using only the 20 month old mice for simplicity)
μ area_mm2=β 0+β 1 genotype KI μ area_mm2=β 0+β 1 genotype KI
where μ μ is the mean response (positive_nuclei positive_nuclei) conditional on the level of genotype. Statisticians typically refer to the ratio μ area_mm2 μ area_mm2 as a rate – here, it is the expected rate of observing a TUNEL+ nucleus as more area is searched.
Let’s compare this to the GLM model recommended here. Recall that the conditional expecation on the link scale is η η and, using the inverse link function, l o g(μ)=η l o g(μ)=η, the systematic part of the count GLM model with offset is
η=β 0+β 1 genotype KI+l o g(area_mm2)l o g(μ)=β 0+β 1 genotype KI+l o g(area_mm2)l o g(μ)−l o g(area_mm2)=β 0+β 1 genotype KI l o g(μ area_mm2)=β 0+β 1 genotype KI μ area_mm2=e x p(β 0+β 1 genotype KI)η=β 0+β 1 genotype KI+l o g(area_mm2)l o g(μ)=β 0+β 1 genotype KI+l o g(area_mm2)l o g(μ)−l o g(area_mm2)=β 0+β 1 genotype KI l o g(μ area_mm2)=β 0+β 1 genotype KI μ area_mm2=e x p(β 0+β 1 genotype KI)
That is, the GLM with offset also models the conditional mean of the area-normalized count, but in a way where the data are much more compatible with the assumptions of the statistical model (in the sense that model checks show that the data look like a sample from the statistical model).
If you want the area-normalized means and their uncertainty under the assumptions of the count GLM offset model, here is some code.
```
mean_area <- mean(fig3b[, area_mm2])
link scale
compute rate or normalized count using emmeans
setting the offset will compute mean at the specified
value of the offset
be sure the stats are on the link scale
m3_emm <- emmeans(fig3b_m3,
specs = c("age", "genotype"),
offset = log(mean_area)) %>%
summary() %>%
data.table()
emmeans is returning the first line in equations above.
We want the last line. So subtract the log of the offset value.
m3_emm[, log_rate := emmean - log(mean_area)]
and compute the 95% asymptotic CIs
m3_emm[, log_rate_lcl := log_rate - 1.96SE]
m3_emm[, log_rate_ucl := log_rate + 1.96SE]
m3_emm
response scale
m3_emm_response <- emmeans(fig3b_m3,
specs = c("age", "genotype"),
offset = log(mean_area),
type = "response") %>%
summary() %>%
data.table()
replace the mean and CIs with the backtransformed values from the
link scale table
m3_emm_rates <- m3_emm_response
m3_emm_rates[, response := exp(m3_emm$log_rate)]
m3_emm_rates[, asymp.LCL := exp(m3_emm$log_rate_lcl)]
m3_emm_rates[, asymp.UCL := exp(m3_emm$log_rate_ucl)]
```
Table 18.5: Table 18.6: Modeled mean area-normalized counts of TUNEL+ cells, estimated using negative binomial GLM. | age | genotype | response | SE | df | asymp.LCL | asymp.UCL |
--- --- ---
| Young | WT | 0.93 | 10.79 | Inf | 0.34 | 2.5 |
| Old | WT | 2.20 | 13.52 | Inf | 1.30 | 3.7 |
| Young | KI | 0.84 | 9.85 | Inf | 0.31 | 2.3 |
| Old | KI | 1.03 | 5.59 | Inf | 0.65 | 1.6 |
Notes
Again, no ratios were computed. The mean area-normalized counts were computed using the coefficients of the GLM. Here, we used the emmeans package to do this. See section xxx below for code that does this withouth the emmeans package.
Figure 18.4: Area-normalized counts modeled using negative binomial GLM with offset, with the raw count as the response (A) and linear model with the area-normalized count as the response.
Notes
In Panel B, the CIs of the two Young treatment groups have lower bounds that extend below zero. This is absurd – negative area-normalized-count means cannot exist. But this inference is implied by an ANOVA of these data.
Panel A has the potential to lead to misconceptions because the GLM model with offset is not fit to the area-normalized data but to the raw counts. Regardless, the GLM model with offset is modeling the means of these ratios.
More than I need to know. The modeled means of a count GLM on the link scale are X b X b, where X X is the model matrix and b b is the vector of model coefficients. In a model with an offset, the values of the offset variable are not in the model matrix and the coefficient (1.0) is not in the vector of coefficients. Following the math above, this expectation is the modeled mean normalized-count (on the link scale) and not the modeled mean count conditional on some offset value (on the link scale). Again, following the math above, to compute the expected count conditional on some value of the offset, use X b+l o g(o f f s e t)X b+l o g(o f f s e t). But we want the rate not the mean, and we want the rate on the response scale, so simply use e x p(X b)e x p(X b).
```
compute rate or normalized count using matrix algebra using
final line of equations above
b <- coef(fig3b_m3)
X <- model.matrix(fig3b_m3)
fig3b[, rate := exp((X %% b)[,1])]
compare rate to area-normalized count
rates <- fig3b[, .(rate_manual = mean(rate),
norm_count = mean(count_per_area)),
by = .(age, genotype)]
m3_emm <- merge(m3_emm_rates, rates, by = c("age", "genotype"))
```
Table 18.7: Table 18.8: Mean area-normalized-counts computed using the emmeans package (response response), the matrix algebra code in this section (rate_manual rate_manual), and as the simple average of the normalized counts (norm_count norm_count) | age | genotype | response | SE | df | asymp.LCL | asymp.UCL | g_by_age | rate_manual | norm_count |
--- --- --- --- --- |
| Young | WT | 0.9304 | 10.79 | Inf | 0.34 | 2.52 | WT Young | 0.9304 | 0.9269 |
| Young | KI | 0.8397 | 9.85 | Inf | 0.31 | 2.30 | KI Young | 0.8397 | 0.8432 |
| Old | WT | 2.2039 | 13.52 | Inf | 1.30 | 3.73 | WT Old | 2.2039 | 2.2027 |
| Old | KI | 1.0322 | 5.59 | Inf | 0.65 | 1.64 | KI Old | 1.0322 | 1.0284 |
18.5.3 Compare an offset to an added covariate with an estimated coefficient
fig3b_m5 <- glm.nb(positive_nuclei ~ age genotype +
log(area_mm2),
data = fig3b)
Table 18.9: Table 18.10: Coefficients from model fig3b_m5 (covariate) and model fig3b_m3 (offset). | | Estimate | Std. Error | z value | Pr(>|z|) | 2.5 % | 97.5 % |
--- --- ---
| fig3b_m5 (covariate) |
| (Intercept) | -3.36 | 1.48 | -2.27 | 0.0232 | -6.27 | -0.33 |
| ageOld | -0.67 | 0.89 | -0.75 | 0.4528 | -2.33 | 1.00 |
| genotypeKI | 0.24 | 0.70 | 0.35 | 0.7299 | -1.20 | 1.69 |
| log(area_mm2) | 2.48 | 0.64 | 3.89 | 0.0001 | 1.19 | 3.74 |
| ageOld:genotypeKI | -1.16 | 0.78 | -1.48 | 0.1390 | -2.81 | 0.47 |
| fig3b_m3 (offset) |
| (Intercept) | -0.07 | 0.51 | -0.14 | 0.8871 | -0.95 | 1.09 |
| ageOld | 0.86 | 0.57 | 1.50 | 0.1334 | -0.39 | 1.91 |
| genotypeKI | -0.10 | 0.72 | -0.14 | 0.8871 | -1.56 | 1.36 |
| ageOld:genotypeKI | -0.66 | 0.81 | -0.81 | 0.4159 | -2.27 | 0.96 |
Notes
The coefficient table for Model fig3b_m5 has a row for the added covariate area_mm2 area_mm2. This row is absent for Model fig3b_m3 since the coefficient is fixed at 1.0.
The estimate of the coefficient of area_mm2 area_mm2 in Model fig3b_m5 is 2.48. A value of 1.0 – the value assumed by the offset model or an analysis of the ratio – is not very compatible with the data.
A coefficient greater than 1.0 means that the rate of DNA damage (the number of TUNEL positive cells per area) increases with the size of the area measured. Does this make any biological sense? The area measured should not have a biological component unless it is a function of the size of the organ. If the area measured is a function of the size of the organ, then we would want to use the ANCOVA model and not the offset model.
This increase in the rate of TUNEL positive cells is seen in the response-scale plot of the fit model in Figure ??A, where the slope of the regression line increases as area_mm2 area_mm2 increases.
Note that on the log scale (Figure ??B), the two regression lines are parallel, as they must be in an ANCOVA linear model.
18.5.4 Issues with plotting
18.6 Example 3 – GLM models for binary responses
18.7 Working in R
18.7.1 Fitting GLMs to count data
The Poisson family is specified with the base R glm() function. For negative binomial, use glm.nb from the MASS package
```
poisson - less likely to fit to real biological data well
because of overdispersion
fit <- glm(y ~ treatment, family = "poisson", data = dt)
two alternatives to overdispersed poisson fit
quasipoisson
fit <- glm(y ~ treatment, family = "quasipoisson", data=dt)
note that "family" is not an argument since this function is used only to fit a negative binomial distribution!
fit <- glm.nb(y ~ treatment, data = dt)
```
18.7.2 Fitting a GLM to a continuous conditional response with right skew.
The Gamma family is specified with the base R glm() function.
fit <- glm(y ~ treatment, family = Gamma(link = "log"), data = dt)
18.7.3 Fitting a GLM to a binary (success or failure, presence or absence, survived or died) response
The binomial family is specified with base R glm() function.
```
if the data includes a 0 or 1 for every observation of y
fit <- glm(y ~ treatment, family = "binomial", data = dt)
if the data includes the frequency of success AND there is a measure of the total n
dt[ , failure := n - success]
fit <- glm(cbind(success, failure) ~ treatment, family = "binomial", data = dt)
```
18.7.4 Fitting Generalized Linear Mixed Models
Generalized linear mixed models are fit with glmer from the lmer package.
```
random intercept of factor "id"
fit <- glmer(y ~ treatment + (1|id), family = "poisson", data = dt)
random intercept and slope of factor "id"
fit <- glmer(y ~ treatment + (treatment|id), family = Gamma(link = "log"), data = dt)
Again, negative binomial uses a special function
fit <- glmer.nb(y ~ treatment + (treatment|id), data = dt)
```
Another good package for GLMMs is glmmTMB from the glmmTMB package
```
negative binomial
fit <- glmmTMB(y ~ treatment + (1|id), family="nbinom2", data = dt)
nbinom1, the mean variance relationship is that of quasipoisson
fit <- glmmTMB(y ~ treatment + (1|id), family="nbinom1", data = dt)
```
18.8 Model checking GLMs
The DHARMa package has an excellent set of model checking tools. The DHARMa package uses simulation to generate fake data sampled from the fit model using the function simulateResiduals.
simulation_output <- simulateResiduals(fittedModel = fig3a_m3,
n = 250,
refit = FALSE)
simulation_output <- simulateResiduals(fittedModel = fig3a_m2,
n = 250,
refit = FALSE)
plot(simulation_output)
plotQQunif(simulation_output)
Three test statistics are superimposed. Use these p-values cautiously – they are guides and not thresholds of demarcation. The two we care about here are The KS statistic indicates that the quantile residuals are not very compatible with a Poisson model – think of this as having a very low probability of sampling these counts from a Poisson with the estimated λ λ. The dispersion statistic indicates that the value of the dispersion of the quantile residuals is not very compatible with a Poisson model – think of this as having a very low probability of sampling counts with this dispersion from a Poisson with the estimated λ λ.
18.9 Hidden code
18.9.1 Import Example 1 data (fig3a – “angiogenic sprouts”)
```
data_from <- "Pericyte FAK negatively regulates Gas6-Axl signalling to suppress tumour angiogenesis and tumour growth"
file_name <- "41467_2020_16618_MOESM3_ESM.xlsx"
file_path <- here(data_folder, data_from, file_name)
fig3a_wide <- read_excel(file_path,
sheet = "Figure 3",
range = "B4:E26",
col_names = FALSE) %>%
data.table()
input_labels <- c("PBS FAK_wt", "PBS FAK_ko", "GAS6 FAK_wt", "GAS6 FAK_ko")
colnames(fig3a_wide) <- input_labels
fig3a <- melt(fig3a_wide,
measure.vars = input_labels,
variable.name = "t_by_g",
value.name = "sprouts") %>%
na.omit()
change order of factor levels
t_by_g_levels <- c("PBS FAK_wt", "GAS6 FAK_wt", "PBS FAK_ko", "GAS6 FAK_ko")
fig3a[, c("treatment", "genotype"):= tstrsplit(t_by_g,
" ",
fixed = TRUE)]
fig3a[, t_by_g := factor(t_by_g, levels = t_by_g_levels)]
treatment_levels <- c("PBS", "GAS6")
fig3a[, treatment := factor(treatment, levels = treatment_levels)]
genotype_levels <- c("FAK_wt", "FAK_ko")
fig3a[, genotype := factor(genotype, levels = genotype_levels)]
```
the variance is less than that expected by the probability model↩︎
the variance is greater than that expected by the probability model↩︎ |
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3759 | https://www-mdp.eng.cam.ac.uk/web/library/enginfo/aerothermal_dvd_only/aero/fprops/poten/node35.html | | | | | | | | | | | | | | | | | | | | | | | | | | | | |
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| Next: Rankine Oval Up: Superposition of Elementary Flows Previous: Superposition of Elementary Flows Uniform Flow and a Source Let us now place a source in the path of a uniform flow. The stream function and the velocity potential for the resulting flow are given by adding the two stream functions and velocity potentials as follows, | | | | | | (4.89) | | | | | (4.90) | One of the interesting features to determine for the resulting force is the stagnation point of the flow, i.e., where the velocity goes to zero. One could calculate this from the equations. It is clear that for this flow the stagnation point will occur on the x-axis. The location can be arrived at purely intuitionally. The source produces a radial flow of magnitude | | | | while the uniform flow produces a velocity of U in the positive x-direction. When these two cancel out at a point we have the stagnation point. A negative radial flow that can cancel the uniform flow is possible only to the left of the x-axis, say at x = -b. Hence, | | | | | leading to | | | (4.91) | At x = -b, we have and r = b. Substituting these values in the expression for , i.e., Eqn. 4.89 we get the value of at the stagnation point to be | | | (4.92) | An equation to the streamline passing through the stagnation point, i.e., stagnation streamline is obtained as follows, | | | | | But hence | | | | | leading to | | | (4.93) | Figure 4.24: Flow about Rankine Half Body The streamlines for this flow are sketched in Fig.4.24. It is clear that we can make the stagnation streamline the solid body. In fact any streamline of a flow can be treated as a solid body since there is no flow across it. In the present example if we ignore the streamlines inside the "body" we have described the flow about a solid body given by Eqn. 4.25. This body is referred to as a Rankine Half Body as it is "open" at the right hand end. Limits of for this body are 0 and . At these values we have y approaching , which is called the Half Width of the body. Figure 4.25: Sketch of Flow about Rankine Half Body The velocity components for this flow are given by | | | (4.94) | | (4.95) | | The square of velocity reduces to | | | (4.96) | If the pressure in the free stream is it follows from Bernoulli Equation that, | | | (4.97) | which enables us to calculate the pressure. Usually in aerodynamic applications involving significant velocities and pressures any contribution due to elevation changes is negligible. The equation for pressure assumes a simple form, | | | (4.98) | It is left as an exercise for the student to show that the maximum velocity over the surface of the body occurs at the location and is approximately equal to . Next: Rankine Oval Up: Superposition of Elementary Flows Previous: Superposition of Elementary Flows (c) Aerospace, Mechanical & Mechatronic Engg. 2005 University of Sydney | |
3760 | https://www.rapidtables.com/math/calculus/laplace_transform.html | Laplace transform table ( F(s) = L{ f(t) } ) - RapidTables.com
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Home›Math›Calculus› Laplace transform
Laplace Transform
Laplace transform function
Laplace transform table
Laplace transform properties
Laplace transform examples
Laplace transform converts a time domain function to s-domain function by integration from zero to infinity
of the time domain function, multiplied by e-st.
The Laplace transform is used to quickly find solutions for differential equations and integrals.
Derivation in the time domain is transformed to multiplication by s in the s-domain.
Integration in the time domain is transformed to division by s in the s-domain.
Laplace transform function
The Laplace transform is defined with the L{} operator:
Inverse Laplace transform
The inverse Laplace transform can be calculated directly.
Usually the inverse transform is given from the transforms table.
Laplace transform table
| Function name | Time domain function | Laplace transform |
---
| f(t) | F(s) =L{f(t)} |
| Constant | 1 | |
| Linear | t | |
| Power | t n | |
| Power | t a | Γ(a+1) ⋅ s-(a+1) |
| Exponent | e at | |
| Sine | sin at | |
| Cosine | cos at | |
| Hyperbolic sine | sinh at | |
| Hyperbolic cosine | cosh at | |
| Growing sine | t sin at | |
| Growing cosine | t cos at | |
| Decaying sine | e -at sin ωt | |
| Decaying cosine | e -at cos ωt | |
| Delta function | δ(t) | 1 |
| Delayed delta | δ(t-a) | e-as |
Laplace transform properties
| Property name | Time domain function | Laplace transform | Comment |
--- --- |
| | f(t) | F(s) | |
| Linearity | a f(t)+bg(t) | aF(s) +bG(s) | a,b are constant |
| Scale change | f(at) | | a>0 |
| Shift | e-at f(t) | F(s+a) | |
| Delay | f(t-a) | e-as F(s) | |
| Derivation | | sF(s) - f(0) | |
| N-th derivation | | s n f(s) -s n-1 f(0) -s n-2 f'(0)-...-f (n-1)(0) | |
| Power | t n f(t) | | |
| Integration | | | |
| Reciprocal | | | |
| Convolution | f(t) g(t) | F(s) ⋅G(s) | is the convolution operator |
| Periodic function | f(t) = f(t+T) | | |
Laplace transform examples
Example #1
Find the transform of f(t):
f (t) = 3 t+ 2 t 2
Solution:
ℒ{t} = 1/s 2
ℒ{t 2} = 2/s 3
F(s) = ℒ{f (t)} = ℒ{3 t+ 2 t 2} = 3ℒ{t}+ 2ℒ{t 2} = 3/s 2 + 4/s 3
Example #2
Find the inverse transform of F(s):
F(s) = 3 / (s 2 + s - 6)
Solution:
In order to find the inverse transform, we need to change the s domain function to a simpler form:
F(s) = 3 / (s 2 + s - 6) = 3 / [(s-2)(s+3)] = a / (s-2) + b / (s+3)
[a(s+3) + b(s-2)] / [(s-2)(s+3)] = 3 / [(s-2)(s+3)]
a(s+3) + b(s-2) = 3
To find a and b, we get 2 equations - one of the s coefficients and second of the rest:
(a+b)s + 3 a-2 b = 3
a+b = 0 , 3 a-2 b = 3
a = 3/5 , b = -3/5
F(s) = 3 / 5(s-2) - 3 / 5(s+3)
Now F(s) can be transformed easily by using the transforms table for exponent function:
f (t) = (3/5)e 2 t - (3/5)e-3 t
See also
Derivative
Calculus symbols
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3761 | https://flexbooks.ck12.org/cbook/ck-12-interactive-geometry-for-ccss/section/2.4/primary/lesson/reflections-geo-ccss/ | Skip to content
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2.4
Reflections
Written by:CK-12 | Kaitlyn Spong
Fact-checked by:The CK-12 Editorial Team
Last Modified: Sep 01, 2025
Lesson
A reflection is one example of a rigid transformation. A reflection over line is a transformation in which each point of the original figure (the pre-image) has an image that is the same distance from the reflection line as the original point, but is on the opposite side of the line. In a reflection, the image is the same size and shape as the pre-image. A reflection across line moves each point to such that line is the perpendicular bisector of the segment connecting and
In other words, the line segment connecting a point in the pre-image to the matching point in the image is split in half by the mirror line.
Reflecting Shapes
Below, the quadrilateral (the pre-image) has been reflected across line to create a new quadrilateral, (the image).
Reflections move all points of a shape across a line called the line of reflection (sometimes informally called the mirror line). A point and its corresponding point in the image are each the same distance from the line.
Notice that the segments connecting each of the corresponding points are all perpendicular to line . Because each point and its corresponding point are the same distance from the line, line bisects each of these segments. This is why line is called the perpendicular bisector for each of these segments.
Keep in mind that you can perform reflections even when the line of reflection is “slanted” or the grid is not visible; however, it is much harder to do by hand.
Reflection
Select the direction in which you want to reflect the shape. Play with different shapes from triangle up by adding or removing points. Explore the reflection of various shapes by dragging any of the points.
Horizontal
CK-12 PLIX Interactive
Determining the Line of Reflection
Observe the line of reflection that created the reflected image below.
The fact that point is in the same location as point tells you that the line of reflection passes through point . Imagine folding the graph so that each point on the original parallelogram matched its point on the image. Where would the fold be?
The line of reflection is the line . When reflections are performed on graph paper with axes, you can define the lines of reflections with their equations.
INTERACTIVE
Reflections
Move the red points to manipulate the red preimage and the reflected, blue image.
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Determining Reflection
Is the following transformation a reflection?
Even though overall both the parallelogram and its image are 2 units from the -axis, each individual point is not the same distance from the -axis as its corresponding image.
Point is 3 units from the -axis and point is 2 units from the -axis.
The line of reflection for points and would be the line which is not the same line of reflection for points and
This is not a reflection (it is a translation).
Perform the reflection across line .
Draw a perpendicular line from each point that defines the parallelogram to line
Count the units between each point and line along the perpendicular lines.
Count the same number of units on the other side of line along the perpendicular lines to create the image.
INTERACTIVE
Reflections
Move the red points to manipulate the red preimage and the reflected, blue image. Move the purple point to change the line of reflection.
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Please check your internet connection and try again.
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Do you want to reset the PLIX?
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No
Examples
Example 1
Reflections are often informally called "flips". Why is this?
A reflection is informally called a “flip” because it's as if you are flipping the shape over the line of reflection.
Example 2
Consider the reflection of a triangle across the -axis.
What do you notice about the coordinates of each point and its image when a reflection is performed across the -axis?
The -coordinate of each point and its image are the same, but the -coordinate has changed sign. You could describe this as
Example 3
Describe the line of reflection that created the reflected image below.
Connect with This line segment has a slope of -1 and a midpoint at (2, 1).
The line of reflection is the perpendicular bisector of this segment. This means it passes through its midpoint and has an opposite reciprocal slope
The line of reflection is the line
| |
| Summary |
| A reflection is a rigid transformation where each point of the original figure has an image that is the same distance from the reflection line but on the opposite side of the line. The line of reflection is the line at which all points of a shape are moved across. Each point and its corresponding point are the same distance across the line of reflection. To determine the line of reflection imagine folding the graph so that each point on the original figure matches its point on the image. To perform a reflection, draw a perpendicular line from each point to the line of reflection, count the units between each point and the line, and count the same number of units on the other side of the line along the perpendicular lines to create the image. |
Review
Describe the line of reflection that created each of the reflected images below.
1.
2.
3.
4.
For questions 5 - 8, reflect each shape across line.
5.
6.
7.
8.
Questions 9 and 10: is the transformation a reflection? Explain.
9.
10.
Reflect a shape across the -axis. How are the points of the original shape related to the points of the image?
The point (7, 2) is reflected across the -axis. Can you find the coordinates of the image point using the relationship you found in the last question?
Reflect a shape across the line . How are the points of the original shape related to the points of the image?
The point (7, 2) is reflected across the line . Can you find the coordinates of the image point using the relationship you found in the last question?
Reflect a shape across the line . How are the points of the original shape related to the points of the image?
The point (7, 2) is reflected across the line . Can you find the coordinates of the image point using the relationship you found in the last question?
Given the diagram below, reflect across the y-axis and call the resulting image . What is the measure of ? What is the measure of ? Draw sketches to experiment with reflecting angles of different measures and describe your results.
If it hasn't been completed already, perform the reflection described above for . What is the relationship between and ? Explain.
The graph of the line is shown below. A vertical line is also shown. What is the measure of ? How do you know? Reflect the vertical line across the line . Describe the image. What is the relationship between the image and the original? Explain.
20. Look at the image below. The line is graphed. What are the coordinates of point A? How do those coordinates relate to the triangle shown? Reflect the triangle shown across the line . What are the coordinates of the image of A? How do those coordinates relate to the image triangle? How does this explain the rule for reflecting a point across the line ?
If a line is reflected across a line, is the image always parallel to the original? Why or why not? Is the image everparallel to the original? If so, under what conditions? Under what conditions is the image of a line reflected across a line coincident with the original?
Review (Answers)
Click HERE to see the answer key or go to the Table of Contents and click on the Answer Key under the 'Other Versions' option.
Asked by Students
Here are the top questions that students are asking Flexi for this concept:
Overview
A reflection is a rigid transformation where each point of the original figure has an image that is the same distance from the reflection line but on the opposite side of the line.
The line of reflection is the line at which all points of a shape are moved across.
Each point and its corresponding point are the same distance across the line of reflection.
To determine the line of reflection imagine folding the graph so that each point on the original figure matches its point on the image.
To perform a reflection, draw a perpendicular line from each point to the line of reflection, count the units between each point and the line, and count the same number of units on the other side of the line along the perpendicular lines to create the image.
Vocabulary
Reflection
rigid transformation
line
Point
distance
opposite
perpendicular bisector
line segment
Quadrilateral
Reflections
Corresponding
perpendicular
Triangle
parallelogram
perpendicular lines
midpoint
Test Your Knowledge
Question 1
A figure with vertices at and is reflected over the y-axis. What are the vertices of the reflected figure?
a
b
c
d
If is reflected over the y-axis, then the image is
The following graph shows the original figure and its reflection.
The vertices of the reflected figure are
Question 2
The point is reflected over the x-axis. What are the coordinates of the resulting point ?
a
b
c
d
None of the above
Given that,
The point is reflected over the -axis.
If is reflected over the -axis, then the image is .
The coordinates of the resulting point is .
Asked by Students
Here are the top questions that students are asking Flexi for this concept:
Related Content
Transformation: Reflection Principles - Basic
Transformation: Reflection Examples - Basic
Reflecting Reality
Squiggle Reflections
The Transformation
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| | | Credit: CK 12 Source: CK 12 License: CC BY-NC 3.0 |
| | | Credit: CK 12 Source: CK 12 License: CC BY-NC 3.0 |
| | | Credit: CK-12 Foundation Source: CK-12 Foundation License: CC BY-NC 3.0 |
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| | | Credit: Dennis Ashendorf Source: CK-12 Foundation License: CC BY-SA |
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| | | Credit: Dennis Ashendorf Source: CK-12 Foundation License: CC BY-SA |
| | | Credit: Dennis Ashendorf Source: CK-12 Foundation License: CC BY-SA |
| | | Credit: CK-12 Foundation Source: CK-12 Foundation License: CC BY-NC 3.0 |
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3762 | https://pmc.ncbi.nlm.nih.gov/articles/PMC10920024/ | Lead-Specific Performance for Atrial Fibrillation Detection in Convolutional Neural Network Models Using Sinus Rhythm Electrocardiography - PMC
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Circ Rep
. 2024 Feb 27;6(3):46–54. doi: 10.1253/circrep.CR-23-0068
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Lead-Specific Performance for Atrial Fibrillation Detection in Convolutional Neural Network Models Using Sinus Rhythm Electrocardiography
Shinya Suzuki
Shinya Suzuki, MD, PhD
1)Department of Cardiovascular Medicine, The Cardiovascular Institute, Tokyo, Japan
Find articles by Shinya Suzuki
1,✉, Jun Motogi
Jun Motogi
4)Nihon Kohden Corporation, Tokyo, Japan
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4, Takuya Umemoto
Takuya Umemoto
4)Nihon Kohden Corporation, Tokyo, Japan
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4, Naomi Hirota
Naomi Hirota, MD, PhD
1)Department of Cardiovascular Medicine, The Cardiovascular Institute, Tokyo, Japan
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1, Hiroshi Nakai
Hiroshi Nakai
2)Information System Division, The Cardiovascular Institute, Tokyo, Japan
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2, Wataru Matsuzawa
Wataru Matsuzawa
4)Nihon Kohden Corporation, Tokyo, Japan
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4, Tsuneo Takayanagi
Tsuneo Takayanagi
4)Nihon Kohden Corporation, Tokyo, Japan
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4, Akira Hyodo
Akira Hyodo
4)Nihon Kohden Corporation, Tokyo, Japan
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4, Keiichi Satoh
Keiichi Satoh
4)Nihon Kohden Corporation, Tokyo, Japan
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4, Takuto Arita
Takuto Arita, MD
1)Department of Cardiovascular Medicine, The Cardiovascular Institute, Tokyo, Japan
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1, Naoharu Yagi
Naoharu Yagi, MD
1)Department of Cardiovascular Medicine, The Cardiovascular Institute, Tokyo, Japan
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1, Mikio Kishi
Mikio Kishi, MD
1)Department of Cardiovascular Medicine, The Cardiovascular Institute, Tokyo, Japan
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1, Hiroaki Semba
Hiroaki Semba, MD, PhD
1)Department of Cardiovascular Medicine, The Cardiovascular Institute, Tokyo, Japan
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1, Hiroto Kano
Hiroto Kano, MD
1)Department of Cardiovascular Medicine, The Cardiovascular Institute, Tokyo, Japan
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1, Shunsuke Matsuno
Shunsuke Matsuno, MD
1)Department of Cardiovascular Medicine, The Cardiovascular Institute, Tokyo, Japan
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1, Yuko Kato
Yuko Kato, MD, PhD
1)Department of Cardiovascular Medicine, The Cardiovascular Institute, Tokyo, Japan
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1, Takayuki Otsuka
Takayuki Otsuka, MD, PhD
1)Department of Cardiovascular Medicine, The Cardiovascular Institute, Tokyo, Japan
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1, Takayuki Hori
Takayuki Hori, MD, PhD
3)Department of Cardiovascular Surgery, The Cardiovascular Institute, Tokyo, Japan
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3, Minoru Matsuhama
Minoru Matsuhama, MD, PhD
3)Department of Cardiovascular Surgery, The Cardiovascular Institute, Tokyo, Japan
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3, Mitsuru Iida
Mitsuru Iida, MD, PhD
3)Department of Cardiovascular Surgery, The Cardiovascular Institute, Tokyo, Japan
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3, Tokuhisa Uejima
Tokuhisa Uejima, MD, PhD
1)Department of Cardiovascular Medicine, The Cardiovascular Institute, Tokyo, Japan
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1, Yuji Oikawa
Yuji Oikawa, MD, PhD
1)Department of Cardiovascular Medicine, The Cardiovascular Institute, Tokyo, Japan
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1, Junji Yajima
Junji Yajima, MD, PhD
1)Department of Cardiovascular Medicine, The Cardiovascular Institute, Tokyo, Japan
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1, Takeshi Yamashita
Takeshi Yamashita, MD, PhD
1)Department of Cardiovascular Medicine, The Cardiovascular Institute, Tokyo, Japan
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1
Author information
Article notes
Copyright and License information
1)Department of Cardiovascular Medicine, The Cardiovascular Institute, Tokyo, Japan
2)Information System Division, The Cardiovascular Institute, Tokyo, Japan
3)Department of Cardiovascular Surgery, The Cardiovascular Institute, Tokyo, Japan
4)Nihon Kohden Corporation, Tokyo, Japan
Shinya Suzuki, MD, PhD
✉
Department of Cardiovascular Medicine, The Cardiovascular Institute, 3-2-19 Nishiazabu, Minato-ku, Tokyo 106-0031, Japan, Email: s-suzuki@cvi.or.jp
✉
Corresponding author.
Received 2023 Aug 20; Revised 2024 Jan 23; Accepted 2024 Jan 25; Collection date 2024 Mar 8.
Copyright © 2024, THE JAPANESE CIRCULATION SOCIETY
This article is licensed under a Creative Commons [Attribution-NonCommercial-NoDerivatives 4.0 International] license.
PMC Copyright notice
PMCID: PMC10920024 PMID: 38464990
Abstract
Background
We developed a convolutional neural network (CNN) model to detect atrial fibrillation (AF) using the sinus rhythm ECG (SR-ECG). However, the diagnostic performance of the CNN model based on different ECG leads remains unclear.
Methods and Results
In this retrospective analysis of a single-center, prospective cohort study, we identified 616 AF cases and 3,412 SR cases for the modeling dataset among new patients (n=19,170). The modeling dataset included SR-ECGs obtained within 31 days from AF-ECGs in AF cases and SR cases with follow-up ≥1,095 days. We evaluated the CNN model’s performance for AF detection using 8-lead (I, II, and V1–6), single-lead, and double-lead ECGs through 5-fold cross-validation. The CNN model achieved an area under the curve (AUC) of 0.872 (95% confidence interval (CI): 0.856–0.888) and an odds ratio of 15.24 (95% CI: 12.42–18.72) for AF detection using the eight-lead ECG. Among the single-lead and double-lead ECGs, the double-lead ECG using leads I and V1 yielded an AUC of 0.871 (95% CI: 0.856–0.886) with an odds ratio of 14.34 (95% CI: 11.64–17.67).
Conclusions
We assessed the performance of a CNN model for detecting AF using eight-lead, single-lead, and double-lead SR-ECGs. The model’s performance with a double-lead (I, V1) ECG was comparable to that of the 8-lead ECG, suggesting its potential as an alternative for AF screening using SR-ECG.
Key Words: Artificial intelligence, Atrial fibrillation, Electrocardiography
Atrial fibrillation (AF) is one of the most prevalent cardiac rhythm disorders and is associated with increased morbidity such as ischemic stroke, and mortality.1–3 A significant challenge in managing AF is its timely detection, given that it is asymptomatic nature in many patients. Various screening tools have been proposed as alternatives to the gold standard 12-lead ECG, including patient-initiated devices such as oscillometric blood pressure cuffs, intermittent ECG rhythm strips, or smartphone photoplethysmograms, as well as semi-continuous options such as smartwatch ECGs and continuous wearable devices such as long-term Holter monitors, wearable belts, or 1–2 week continuous ECG patches.4,5 Additionally, implanted devices have been used for continuous and repeated monitoring in patients with cryptogenic stroke, where the source of the blood clot is unknown, revealing a significant prevalence of undiagnosed AF.6,7
In practical terms, the development of simple methods to identify individuals at a high risk of AF would facilitate the selection of candidates for long-term monitoring devices. One such method involves the precise analysis of waveform patterns on the resting 12-lead ECG using artificial intelligence (AI), which has shown promise in identifying patients with AF on sinus rhythm ECG (SR-ECG).8 This approach is unique because, although the gold standard for diagnosing AF is the presence of AF on 12-lead ECG, AI-based methods can provide insight into the presence of AF even on ECGs where AF is not visually apparent. From the Mayo Clinic, Attia et al reported a landmark study using AI-enabled ECG analysis to predict AF on SR-ECG.8 Similarly, Raghunath et al reported on AI-enabled ECG analysis using a larger ECG database from the Geisinger Health System in the USA.9 Additionally, Gruwez et al in Belgium reported on AI-enabled ECG analysis.10 These studies have had a significant effect because of the high predictive ability, with an area under the curve (AUC) of approximately 0.9.
Recently, there have been advancements in both mobile and smartwatch ECG technologies incorporating AI.11,12 Mobile ECG systems offer a simpler and more accessible method of obtaining an ECG for individuals before visiting a clinical setting, and for physicians providing home-based medical care, both cases where obtaining a 12-lead ECG is not always feasible. These systems utilize fewer leads for recording the ECG, but in the context of AF screening in such circumstances, to date there have been no reports on AI-enabled ECG algorithms specifically designed to detect AF using single- or double-lead SR-ECG recordings.
We previously reported on a convolutional neural network (CNN) model for AF detection using SR-ECG,13 which achieved comparable AUC values to previous studies.8–10 In the present study, using the same dataset, we conducted an analysis to assess the lead-specific performance of the CNN model.
Methods
Ethics and Informed Consent
This study was performed in accordance with the Declaration of Helsinki (revised in 2013) and Ethical Guidelines for Medical and Health Research Involving Human Subjects (Public Notice of the Ministry of Education, Culture, Sports, Science and Technology, and the Ministry of Health, Labour and Welfare, Japan, issued in 2017). Written informed consent was given by all participants. The study protocol was reviewed and approved (IRB No. 424) by the Institutional Review Board of the Cardiovascular Institute.
Study Population
The Shinken database comprises all patients who newly visited the Cardiovascular Institute, Tokyo, Japan, excluding foreign travelers and patients with active cancer. This single-hospital database was established in June 2004, and further details have been described previously.13–15 For this study, we extracted data for 19,170 patients registered between February 2010 and March 2018, because a computerized ECG database has only been available since February 2010. We excluded 2,092 patients for at least 1 of the following reasons: presence of AF (n=1,601), atrial flutter (n=185; of which 8 were coincident with AF), atrial tachycardia (n=3), paroxysmal supraventricular tachycardia (n=190) on the initial-visit ECG, and insufficient follow-up data (n=121). The remaining dataset of 17,078 patients with a SR-ECG served as the main data source for CNN modeling in this study (Figure 1).
Figure 1.
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Flowchart of patient selection. AF, atrial fibrillation; CNN, convolutional neural network; ECG, electrocardiogram; SR, sinus rhythm.
Development of the CNN Model
The 12-lead ECGs were recorded for 10 s while the patient was supine, using an ECG machine (MAC 5500 HD with GE CardioSoft V6.71; GE Healthcare, Chicago, IL, USA) at a sampling rate of 500 Hz. The raw digital records were stored using the MUSE data management system. From the total study population of 17,078 patients, we identified those with a SR-ECG who met the criteria for the “AF label” and “SR label” to develop the CNN models for AF detection.13 Subsequently, the CNN models for AF detection were constructed using the dataset of SR-ECGs.8,13
Selecting “AF Label” SR-ECGs Patients with SR-ECGs were included in the “AF label” group if they met both of the following criteria: (1) at least 1 AF-ECG recorded in the ECG database during follow-up, and (2) at least 1 SR-ECG within 31 days before or after the first AF-ECG. A total of 616 patients were selected, and their corresponding SR-ECGs were used for the “AF label” dataset (Figure 1). In the case of multiple SR-ECGs with AF labels being available for the same patient, the SR-ECG taken on the nearest day to the first AF-ECG was chosen.
Selecting “SR Label” SR-ECGs Patients with SR-ECGs were included in the “SR label” group if they met all of the following criteria: (1) no recorded AF-ECG in the ECG database during follow-up, (2) did not have a previous diagnosis of AF before the initial visit to hospital, and (3) an observation period ≥1,095 days. A total of 3,412 patients were selected, and their corresponding SR-ECGs were used for the “SR label” dataset (Figure 1). The SR-ECG taken at the initial visit was chosen for analysis.
Dataset Management Given the small number of positive cases (AF) in the present study, we used the 5-fold cross-validation method to enable all data to be included in the testing dataset.16 Management of the dataset with this validation method is shown in Supplementary Figure 1 and described briefly. First, the dataset was randomly divided into 5 groups, and then 1 of the 5 groups was set as the testing dataset, and the others as the training dataset in which 12.5% of the data were used as the internal-validation dataset. Finally, the model was run 5 times using different combinations of training and testing datasets. Accordingly, model output was obtained from 5 testing datasets of 5 different models, in which all data were included in the testing dataset.
CNN Modeling We constructed the CNN using the Keras Framework with a Tensorflow backend (Google, Mountain View, CA, USA) and Python. From the 12-lead ECG recordings with a 10-s duration, we selected 8 independent leads (leads I, II, and V1–6) for analysis.
The CNN modeling was based on the model developed by Attia et al.8,13 The conceptual architecture is shown in Figure 2, and the detailed architecture is shown in Supplementary Figure 2. The model comprised layers for both temporal and lead axes. The temporal axis layers consisted of a convolution part and a residual part. The convolution part included a convolution layer, a batch-normalization layer, a non-linear Rectified Linear Unit (ReLU) activation layer, and a maximum pooling layer. The residual part comprised 2 residual blocks based on the Residual Network (ResNet) and average pooling, which were repeated X times (X was determined to achieve optimal performance, as outlined below). The lead axis layers consisted of paired batch-normalization layers, non-linear ReLU activation layers, and convolutional layers. Thereafter, a second paired batch-normalization layer and a layer for non-linear ReLU activation were included. The configuration of the lead axis layers were designed in the context of multiple-lead ECG models. Therefore, when we applied this structure to models using a single-lead ECG only, a 1×1 convolution filter was applied in the lead axis convolution. Finally, the data passed through a dropout layer with global average pooling and were fed into the final output layer, activated by the softmax function, which generated the probability of AF.
Figure 2.
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Convolutional neural network analysis.
The model was trained using the Keras software library on a computer with 128 GB RAM and a single Quadro P-2200 (NVIDIA) graphics processing unit. In the model training process, the Adam optimizer with categorical cross entropy as the loss function was used. The learning rate was set at 0.0001 and the maximum epochs was set at 500. Training was stopped if the loss did not decrease for 200 epochs in the internal-validation dataset, and the model with the lowest loss was selected. Considering the class imbalance between the positive and negative cases, we weighted the loss function n times higher for the positive class samples compared to the negative class samples, where n was determined based on the ratio of the number of negative to positive data in the training dataset (n=6).
Receiver operating characteristics (ROC) curves were generated, and the AUC was used to evaluate the performance of the CNN model in determining the presence or absence of AF using ECG data. By analyzing the ROC curve in the internal-validation dataset, we determined the number of repetitions (X) for the combination of the 2 residual blocks and average pooling described above. The probability threshold for classifying AF was determined as the point on the ROC curve closest to the (0,1) point17 in the internal-validation dataset for each of the 5 models in the 5-fold cross-validation method (thresholds are presented in Supplementary Table).
Outcome Measurement and Statistical Analysis
First, patient characteristics were summarized as mean±standard deviation [SD] for continuous variables and n (%) for categorical variables. Differences between the 2 groups were tested by unpaired t-tests for continuous variables and chi-squared tests for categorical variables. Second, the performance of the CNN models was assessed using 8-lead (I, II, V1–6), single-lead, and double-lead (I, II; I, V1 to I, V6; II, V1–II, V6) ECGs. The evaluation metrics included AUC, sensitivity, specificity, accuracy, and F1 score. The model performance data are presented as the mean (SD) of 5 model runs with 5-fold cross-validation. For AUC, 95% confidence intervals (CIs) were calculated considering 5-fold cross-validation.18 Third, the distribution of patients in the AF label and SR label groups, categorized by diagnostic probability levels determined by each CNN model, was described using the entire dataset. Fourth, odds ratios were calculated based on the CNN model’s diagnoses using 8-lead, single-lead, and double-lead ECGs. This calculation involved the ratio of true/false positives divided by the ratio of false/true negatives, utilizing the entire dataset. Fifth, the gradient-weighted class activation mapping (GradCAM) method was used for the multi-input models.19
The statistical analyses were performed using R version 4.0.3 (The R Foundation, Vienna, Austria) and SPSS version 28.0 (IBM Corp., Armonk, NY, USA).
Results
Patients’ Characteristics
The patients’ characteristics are presented in Table 1. The total dataset (n=4,028) included 616 AF-label and 3,412 SR-label patients. Among these, 404 (65.6%) and 2,125 (62.3%) were male in the AF-label and SR-label groups, respectively. The mean age was 67.0±12.4 years in the AF-label group and 61.4±13.2 years in the SR-label group (P<0.001). The mean left ventricular ejection fraction was 62.3±13.6% in the AF-label group and 65.5±10.7% in the SR-label group (P<0.001), and the respective mean left atrial diameters were 40.0±7.2 mm and 35.5±6.0 mm (P<0.001). The prevalence of congestive heart failure was 8.1% in the AF-label group and 2.0% in the SR-label group (P<0.001), and the respective prevalence of mitral regurgitation was 12.3% and 3.1% (P<0.001).
Table 1.
Patients’ Characteristics
| | Total (N=4,028) | AF label (N=616) | SR label (N=3,412) | P value |
:---: :---:
| Age, years | 62.2±13.3 | 67±12.4 | 61.4±13.2 | <0.001 |
| Male, n (%) | 2,529 (62.8) | 404 (65.6) | 2,125 (62.3) | 0.123 |
| Height, cm | 162.8±9.6 | 162.7±10.1 | 162.8±9.5 | 0.346 |
| Weight, kg | 62.6±13.6 | 61.6±13.1 | 62.8±13.6 | 0.016 |
| BMI, kg/m 2 | 23.5±3.9 | 23.1±3.6 | 23.5±3.9 | 0.005 |
| Systolic BP, mmHg | 131.2±20.1 | 130±20.6 | 131.4±20.0 | 0.077 |
| Diastolic BP, mmHg | 76.4±12.6 | 73.5±13.0 | 76.9±12.4 | <0.001 |
| IVST, mm | 9.8±2.3 | 10.5±2.4 | 9.7±2.2 | <0.001 |
| PWT, mm | 9.0±1.6 | 9.5±1.7 | 9.0±1.5 | <0.001 |
| LVDd, mm | 46.8±6.4 | 48.7±8.1 | 46.4±6.0 | <0.001 |
| LVDs, mm | 30.1±7.4 | 32.4±9.3 | 29.7±6.9 | <0.001 |
| LVEF, % | 65.0±11.3 | 62.3±13.6 | 65.5±10.7 | <0.001 |
| LAD, mm | 36.2±6.4 | 40.0±7.2 | 35.5±6.0 | <0.001 |
| Congestive HF, n (%) (HF admission within 90 days) | 119 (3.0) | 50 (8.1) | 69 (2.0) | <0.001 |
| HF with reduced EF, n (%) | 346 (8.6) | 89 (14.4) | 257 (7.5) | <0.001 |
| Ischemic heart disease, n (%) (PCI within 90 days) | 888 (22.0) | 85 (13.8) | 803 (23.5) | <0.001 |
| Asymptomatic ischemia, n (%) | 164 (4.1) | 24 (3.9) | 140 (4.1) | 0.911 |
| Old myocardial infarction, n (%) | 187 (4.6) | 26 (4.2) | 161 (4.7) | 0.677 |
| Acute coronary syndrome, n (%) | 308 (7.6) | 43 (7.0) | 265 (7.8) | 0.564 |
| Aortic stenosis, n (%) | 232 (5.8) | 87 (14.1) | 145 (4.2) | <0.001 |
| Aortic regurgitation, n (%) | 153 (3.8) | 45 (7.3) | 108 (3.2) | <0.001 |
| Mitral regurgitation, n (%) | 183 (4.5) | 76 (12.3) | 107 (3.1) | <0.001 |
| Mitral stenosis, n (%) | 21 (0.5) | 12 (1.9) | 9 (0.3) | <0.001 |
| Tricuspid regurgitation, n (%) | 58 (1.4) | 18 (2.9) | 40 (1.2) | 0.002 |
| Hypertrophic cardiomyopathy, n (%) | 54 (1.3) | 11 (1.8) | 43 (1.3) | 0.338 |
| Dilated cardiomyopathy, n (%) | 36 (0.9) | 6 (1.0) | 30 (0.9) | 0.815 |
| Dilated hypertrophic cardiomyopathy, n (%) | 6 (0.1) | 2 (0.3) | 4 (0.1) | 0.230 |
| Hypertensive cardiomyopathy, n (%) | 326 (8.1) | 84 (13.6) | 242 (7.1) | <0.001 |
| Ischemic cardiomyopathy, n (%) | 99 (2.5) | 30 (4.9) | 69 (2.0) | <0.001 |
| Aortic aneurysm, n (%) | 106 (2.6) | 26 (4.2) | 80 (2.3) | 0.012 |
| Aortic dissection, n (%) | 74 (1.8) | 33 (5.4) | 41 (1.2) | <0.001 |
| Hypertension, n (%) | 2,396 (59.5) | 420 (68.2) | 1,976 (57.9) | <0.001 |
| Diabetes, n (%) | 740 (18.4) | 150 (24.4) | 590 (17.3) | <0.001 |
| Smoking history, n (%) | 1,791 (44.5) | 280 (45.5) | 1,511 (44.3) | 0.597 |
| Chronic kidney disease, n (%) | 905 (22.5) | 241 (39.1) | 664 (19.5) | <0.001 |
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Data are presented as the mean±standard deviation unless otherwise stated. AF, atrial fibrillation; BMI, body mass index; BP, blood pressure; HF, heart failure; IVST, intraventricular septum thickness; LAD, left atrial diameter; LVDd, left ventricular end-diastolic diameter; LVDs, left ventricular end-systolic diameter; LVEF, left ventricular ejection fraction; PCI, percutaneous coronary intervention; PWT, posterior left ventricular wall thickness; SR, sinus rhythm.
Evaluation of the Utility of the CNN Models to Detect AF
Basic Performance of the CNN Models The basic performance of the CNN models for detecting AF using 8-lead, single-lead, and double-lead ECGs is summarized in Table 2 (detailed information of the 5-model runs is shown in Supplementary Table). The AUC (95% CI) for the 8-lead ECG was 0.872 (0.856–0.888). The AUCs (95% CI) for single-lead ECGs were generally lower than that for the 8-lead ECG, but relatively higher for single-lead V1 (0.843 [0.826–0.860]) and V6 (0.845 [0.827–0.862]). The AUCs (95% CI) for double-lead ECGs were generally higher than for the single-lead ECGs, especially in combinations using lead I, and were highest for double-lead I, V1 (0.871 [0.856–0.886]).
Table 2.
Performance of the Convolutional Neural Network Model for Detecting Atrial Fibrillation on 8-Lead, Single-Lead, and Double-Lead Electrocardiograms
| Model pattern / Leads | AUC (95% CI) | Sensitivity | Specificity | F1 score | Accuracy |
:---: :---: :---: |
| All leads (8 leads) | 0.872 (0.856–0.888) | 0.760 | 0.828 | 0.565 | 0.818 |
| Single lead |
| I | 0.801 (0.781–0.820) | 0.703 | 0.739 | 0.447 | 0.733 |
| II | 0.806 (0.787–0.825) | 0.692 | 0.754 | 0.454 | 0.744 |
| V1 | 0.843 (0.826–0.860) | 0.736 | 0.792 | 0.512 | 0.783 |
| V2 | 0.815 (0.795–0.835) | 0.721 | 0.772 | 0.488 | 0.764 |
| V3 | 0.805 (0.785–0.825) | 0.703 | 0.757 | 0.461 | 0.749 |
| V4 | 0.757 (0.734–0.779) | 0.674 | 0.724 | 0.420 | 0.716 |
| V5 | 0.808 (0.788–0.827) | 0.732 | 0.739 | 0.463 | 0.738 |
| V6 | 0.845 (0.827–0.862) | 0.792 | 0.743 | 0.493 | 0.750 |
| Double leads |
| I, II | 0.862 (0.845–0.879) | 0.778 | 0.806 | 0.545 | 0.801 |
| I, V1 | 0.871 (0.856–0.886) | 0.782 | 0.800 | 0.543 | 0.797 |
| I, V2 | 0.863 (0.846–0.879) | 0.782 | 0.790 | 0.533 | 0.789 |
| I, V3 | 0.863 (0.847–0.880) | 0.784 | 0.797 | 0.542 | 0.795 |
| I, V4 | 0.862 (0.846–0.879) | 0.787 | 0.793 | 0.539 | 0.792 |
| I, V5 | 0.864 (0.848–0.880) | 0.787 | 0.783 | 0.528 | 0.784 |
| I, V6 | 0.865 (0.849–0.881) | 0.797 | 0.792 | 0.544 | 0.793 |
| II, V1 | 0.851 (0.834–0.868) | 0.750 | 0.790 | 0.519 | 0.784 |
| II, V2 | 0.860 (0.843–0.877) | 0.769 | 0.789 | 0.523 | 0.786 |
| II, V3 | 0.830 (0.811–0.849) | 0.726 | 0.800 | 0.513 | 0.788 |
| II, V4 | 0.824 (0.805–0.843) | 0.703 | 0.802 | 0.501 | 0.786 |
| II, V5 | 0.848 (0.830–0.865) | 0.778 | 0.782 | 0.521 | 0.782 |
| II, V6 | 0.857 (0.841–0.873) | 0.768 | 0.794 | 0.532 | 0.790 |
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For each AUC, 95% CIs are calculated, taking into account the 5-fold cross-validation. AUC, area under the curve; CI, confidence interval; SD, standard deviation.
Distribution of Patients According to the CNN Model Outputs The distribution of patients in the CNN model outputs using the 8-lead ECG is shown in Figure 3. The proportion of patients in the AF-label group sharply increased with a high probability (model output >0.9), while the proportion of patients in the SR-label group sharply increased with a lower probability (model output <0.1).
Figure 3.
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Proportion of patients according to the model output in CNN-derived model using the 8-lead ECG. The vertical scale indicates the proportion of patients in the SR-label group (blue) and those in the AF-label group (orange). The horizontal scale indicates the diagnostic probability for AF yielded by the CNN model. AF, atrial fibrillation; CNN, convolutional neural network; ECG, electrocardiogram; SR, sinus rhythm.
The distribution of patients according to the CNN model outputs using single-lead and double-lead ECGs is shown in Supplementary Figure 3 and Supplementary Figure 4. In the models with single-lead ECGs, a sharp increase in the AF-label and SR-label patients in high and low probabilities, respectively, of the model output was observed in single-lead V1 and the double-leads I and II, which was similar to what was observed in the 8-lead ECG. Among the models with double-lead ECGs, a similar sharp increase was particularly observed when using lead I in the combination.
Odds Ratios Based on the Diagnosis of the CNN Models The odds ratio (95% CI) for the AF label, based on the diagnosis of the CNN model using the 8-lead ECG, was 15.24 (12.42–18.72) (Figure 4). Lower odds ratios were observed for the AF label in the CNN models using single-lead ECGs, whereas relatively higher odds ratios were observed in the model using double-lead ECGs, especially when using the combination with lead I. The odds ratio (95% CI) was 14.46 (11.75–17.81) for double-lead of I, II; 14.34 (11.64–17.67) for double-lead of I, V1; and 14.97 (12.10–18.53) for double-lead of I, V6, which were comparable to the 8-lead ECG (Figure 4).
Figure 4.
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Odds ratios for detecting atrial fibrillation with the CNN models using the 8-lead, single-lead, and double-lead ECG. CNN, convolutional neural network.
GradCAM for the Diagnosis of AF in the CNN Models The GradCAM images corresponding to the 8-lead, single-lead and double-lead ECGs are displayed in Supplementary Figure 5, depicting the specific areas of focus identified by the CNN models in a patient with true positive results for the AF label. As shown in Supplementary Figure 5A, the GradCAM on the 8-lead ECG revealed that the CNN model placed strong emphasis on various segments in the I and II leads, as well as the P wave in the V1 lead, and the QRS and ST-T segments in the V6 lead. On the other hand, for both the single-lead and double-lead ECGs, the CNN models primarily focused on the P wave, with some attention given to the QRS and ST-T segments (Supplementary Figure 5B,C).
Discussion
Major Findings
We developed a CNN-derived algorithm using digital ECG to predict AF and there were 2 major findings for the performance of the model. (1) The AUC with the 8-lead ECG was 0.872 (95% CI: 0.856–0.888) and the odds ratio was 15.24 (95% CI: 12.42–18.72). (2) Among the single- and double-lead ECGs, the model performance was highest when using the double-leads of I and V1, with an AUC of 0.871 (95% CI: 0.856–0.886) and an odds ratio of 14.34 (95% CI: 11.64–17.67).
Comparison With Previous Studies
Using AI-enabled ECG to predict AF using the 12-lead SR-ECG has already been reported by other study groups,8–10,20 which found a high predictive ability for AF using the AUC: 0.90 in the study by Attia et al,11 and 0.87 in the studies by Raghunath et al9 and Gruwez et al.10 It is quite surprising that the SR-ECG can predict AF with such high predictive capability. In our previous study, in which we excluded patients with structural heart diseases, we obtained an AUC of 0.86,13 and in the present study without any exclusion criteria, we obtained an AUC of 0.872.
Although our model predicted AF using the SR-ECG, it showed relatively high sensitivity (0.760) and specificity (0.828). Moreover, the positive predictive ratio was 0.436 (= 468 / [468 + 586] in Figure 4), resulting in an F1 score of 0.565. Of course, the SR-ECG cannot provide an absolute diagnosis of AF. However, the diagnostic values of the CNN model would be satisfactory for determining possible candidates for further screening with long-term ambulatory ECG recordings.
Model Performance According to Differences in Lead Application
The models for predicting AF using the SR-ECG have been based on the hypothesis that the AF signature, resulting from structural changes in the atria, can be identified by 12-lead ECG during SR,8,21 because structural changes in the atria predispose to atrial arrhythmia.22 Furthermore, in our previous study that utilized hundreds of ECG parameters analyzed with a random forest algorithm, the importance of the ECG parameters in predicting AF was similar in the P wave, QRS complex, and ST-T segment.23 This suggests that structural changes in the ventricle, likely due to aging or atherosclerosis, may be also of importance, which gives rise to another hypothesis that the predictive ability of the CNN model may differ according to each single-lead ECG; however, no reports have addressed this issue. To the best of our knowledge, our study is the first to report lead-specific predictive ability of a CNN model for predicting AF using the SR-ECG.
In our CNN models, among the single-lead ECGs, the AUC was relatively high in leads V1 and V6, with AUCs of 0.843 and 0.845, respectively. Among the double-lead ECGs, the model’s performance was highest when using leads I and V1, achieving an AUC of 0.871 and an odds ratio of 14.34, followed by double-leads I and V6, with an AUC of 0.865 and an odds ratio of 14.97. It is widely acknowledged that structural changes in the atria are prominently detected in leads II and V1. Consequently, it is assumed that the CNN models placed strong emphasis on the P wave in leads II and V1, which was supported by the finding of the GradCAM analysis. Moreover, the GradCAM analysis indicated that the CNN models placed strong emphasis on the QRS and ST-T segments in lead V6, consistent with our previous findings in s machine-learning model analysis.23
However, it was an unexpected result that the model’s performance, in terms of both AUCs and odds ratios, was generally higher when using lead I than lead II in the double-lead ECGs. There are 2 reports that the amplitude of the P wave in lead I, rather than in lead II, is associated with progression of electrical remodeling in the left atrium.24,25 Park et al reported a significant correlation between the mean left atrial voltage measured before pulmonary vein isolation and the P wave amplitude in lead I (β=2.510, P=0.010), but not with that in lead II (β=0.714, P=0.250).24 Moreover, Schreiber et al reported that left atrial low voltage areas (as a percentage of the total left atrial area) were more associated with P-wave amplitude in lead I (R=−0.578, P<0.001) than in lead II (R=−0.450, P<0.001).25
Clinical Implications of the CNN Model and Future Perspective
Wearable devices, such as an Apple Watch, could potentially aid in detecting AF, but not everyone is using such devices. Moreover, they are not yet capable of providing continuous monitoring specifically for AF. Therefore, diagnosing early-phase AF still relies on long-term ECG monitoring. Determining which individuals are at high risk for AF and should undergo such examination remains a significant challenge.
The core concept of “AI-ECG on SR-ECG” arises from the question: “Who is at high risk for AF and should undergo long-term ECG monitoring?” The CNN models using SR-ECG, taking into account previous findings8–10,13 and those from the present study, could identify individuals at high risk of AF. It is noteworthy that the best performance for such screening was achieved with the all-lead ECG, available only in the clinical setting. Alternatively, fewer-lead ECGs could provide screening at home using a portable device. Additionally, in the hospital setting, a fewer-lead ECG is easier to record and can be particularly useful for patients who have difficulty undressing or moving to a bed. Given its ease of recording, it also could be valuable for mass screening or repeated checks.
Of note, our model demonstrated that the AUC for the double-lead ECG (I, V1) was comparable to that for the 8-lead ECG. The ECG with lead I is easy to record, albeit somewhat inconvenient. Strictly speaking, the V1 lead is not considered a “single lead” because it requires Wilson’s central terminal (determined by 3 potentials) as a reference potential. For clinical use with this double-lead ECG, adopting a specific lead that mimics V1 lead may be necessary, necessitating further studies.
Study Limitations
There are several to highlight. First, our CNN-derived model was constructed using only data from a single cardiovascular center in Japan. Given that the CNN models may detect subtle ECG morphologic changes to identify AF, the model may not be generalizable to other populations. Second, although we used data from 19,170 patients, the number of cases with AF (n=616) was relatively small, which may limit generalizability. Third, although we restricted SR-ECG recordings with the SR label to patients followed up for ≥1,095 days, there remains a possibility of undetected AF in patients with the SR label. Fourth, our AI-enabled ECG model should be validated against external datasets to confirm the generalizability. Finally, we could not completely understand how the model makes predictions.
Conclusions
We evaluated the performance of a CNN models for detecting AF using 8-lead, single-lead, and double-lead SR-ECGs. The performance of the model with a double-lead (I, V1) ECG was comparable to that of the 8-lead ECG, suggesting that an ECG with fewer leads can serve as an alternative for AF screening using SR-ECG.
Funding
This study was partially supported by the Practical Research Project for Life-Style related Diseases including Cardiovascular Diseases and Diabetes Mellitus from the Japan Agency for Medical Research and Development, AMED (JP17ek0210082).
Disclosures
S.S. received lecture fees from Daiichi Sankyo and Bristol-Myers Squibb. T.Y. received research funds and/or lecture fees from Daiichi Sankyo, Bayer Yakuhin, Bristol-Myers Squibb, Pfizer, Nippon Boehringer Ingelheim, Eisai, Mitsubishi Tanabe Pharm, Ono Pharmaceutical, and Toa Eiyo. The remaining authors have nothing to disclose.
IRB Information
This study was approved by the Institutional Review Board of the Cardiovascular Institute (Reference number: 424).
Authors’ Contributions
S.S., J.M., and H.N. conceived the study concept and study design. W.M., T.T., and T. Umemoto analyzed the data. S.S., H.N., T.O., T.A., N.Y., and N.H. collected the data. S.S. and T. Umemoto drafted the manuscript. S.S., J.M., H.N., N.H., A.H., and K.S. checked the analyzed data. N.H., H.N., W.M., T.T., A.H., K.S., T.A., N.Y., M.K., H.S., H.K., S.M., Y.K., T.O., T.H., M.M., M.I., T. Uejima, Y.O., J.Y., and T.Y. checked the manuscript. All authors approved the final version of the manuscript.
Supplementary Files
Supplementary File 1
Supplementary Table. Performance of the Convolutional Neural Network Model for Detecting Atrial Fibrillation With Five-Fold Cross-Validation Supplementary Figure 1. Supplementary Figure 2. Supplementary Figure 3. Supplementary Figure 4. Supplementary Figure 5.
circrep-6-46-s001.pdf (1.9MB, pdf)
Acknowledgments
We express our gratitude to Shiro Ueda and Nobuko Ueda from Medical Edge Company, Ltd., for their assistance in assembling the database using the Clinical Study Supporting System. Our thanks also go to Yurika Hashiguchi, Hiroaki Arai, and Takashi Osada for their dedicated work in data management and system administration. Additionally, we are grateful to Professor Hideki Origasa from the Data Science and AI Innovation Research Promotion Center at Shiga University, and Professor Satoshi Teramukai from the Department of Biostatistics at Kyoto Prefectural University of Medicine, for their invaluable advice on statistical analysis.
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Associated Data
This section collects any data citations, data availability statements, or supplementary materials included in this article.
Supplementary Materials
Supplementary File 1
Supplementary Table. Performance of the Convolutional Neural Network Model for Detecting Atrial Fibrillation With Five-Fold Cross-Validation Supplementary Figure 1. Supplementary Figure 2. Supplementary Figure 3. Supplementary Figure 4. Supplementary Figure 5.
circrep-6-46-s001.pdf (1.9MB, pdf)
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3763 | https://ocw.mit.edu/courses/7-01sc-fundamentals-of-biology-fall-2011/resources/linkage-and-recombination-genetic-maps/ | Linkage and Recombination, Genetic maps | Fundamentals of Biology | Biology | MIT OpenCourseWare
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Meet the Instructors
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Biochemistry
Types of Organisms, Cell Composition
Covalent Bonds, Hydrogen Bonds
Macromolecules: Lipids, Carbohydrates, Nucleic Acid
Proteins, Levels of Structure, Non-Covalent Forces
Biochemical Reactions, Enzymes and ATP
Respiration and Fermentation
Chemiosmotic Principle, Photosynthesis
Exam 1
Molecular Biology
DNA Structure, Classic Experiments
DNA Replication
Transcription, Translation
Alternative Approaches to Molecular Biology
Gene Regulation and the Lac Operon
Exam 2
Genetics
Mendel's Laws
Linkage and Recombination, Genetic Maps
Pedigrees
Biochemical Genetics
Exam 3
Recombinant DNA
Development of Recombinant DNA
Basic Mechanics of Cloning
Constructing and Screening a Recombinant DNA Library
cDNA Libraries and Expression Libraries
Agarose Gel Electrophoresis, DNA Sequencing, PCR
Exam 4
Resource Index
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Prof. Eric Lander
Prof. Robert Weinberg
Prof. Tyler Jacks
Prof. Hazel Sive
Prof. Graham Walker
Prof. Sallie Chisholm
Dr. Michelle Mischke
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7.01SC | Fall 2011 | Undergraduate
Fundamentals of Biology
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More Info
Syllabus
Meet the Instructors
Meet the TAs
Biochemistry
Types of Organisms, Cell Composition
Covalent Bonds, Hydrogen Bonds
Macromolecules: Lipids, Carbohydrates, Nucleic Acid
Proteins, Levels of Structure, Non-Covalent Forces
Biochemical Reactions, Enzymes and ATP
Respiration and Fermentation
Chemiosmotic Principle, Photosynthesis
Exam 1
Molecular Biology
DNA Structure, Classic Experiments
DNA Replication
Transcription, Translation
Alternative Approaches to Molecular Biology
Gene Regulation and the Lac Operon
Exam 2
Genetics
Mendel's Laws
Linkage and Recombination, Genetic Maps
Pedigrees
Biochemical Genetics
Exam 3
Recombinant DNA
Development of Recombinant DNA
Basic Mechanics of Cloning
Constructing and Screening a Recombinant DNA Library
cDNA Libraries and Expression Libraries
Agarose Gel Electrophoresis, DNA Sequencing, PCR
Exam 4
Resource Index
Linkage and Recombination, Genetic Maps
Linkage and Recombination, Genetic maps
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0:06 ERIC LANDER: Good morning.
0:07 Good morning.
0:17 So last time, we ran into a problem.
0:27 We had Mendel, my hero Mendel, this MIT-like mathematical,
0:31 physical monk, had developed this gorgeous theory of
0:40 particles of inheritance.
0:41 He didn't use the word "gene" yet.
0:43 Gene doesn't get invented for much longer.
0:48 For every trait, you had two such particles.
0:53 You gave one to your offspring.
0:53 Each of the parents gave one to their offspring.
0:56 And that's how each offspring gets two of them.
1:00 That choice of which of the two alleles to transmit to
1:05 your offspring is a random draw.
1:07 And that explains beautifully, for example, the 3-to-1
1:11 segregation pattern that Mendel saw--
1:14 gorgeous.
1:16 We put that model, which was an ex post facto model, a
1:20 model made after the data were available, to a test.
1:24 You guys insisted we had to test it before you would
1:27 publish it.
1:28 And it holds up pretty well, making pretty surprising
1:31 predictions that you would otherwise
1:34 not have ever expected.
1:35 Like amongst that 3 to 1, the threes are not all the same.
1:39 Some of those round peas were homozygous
1:45 for the big-R allele.
1:48 And in selfing, they'll never produce wrinkled peas.
1:51 But 2/3 of those round peas were heterozygous.
1:56 And when you self them, you get 1/4 wrinkled peas.
2:00 So the wacky prediction that 1/3 of the rounds will give
2:04 rise to no wrinkleds, and 2/3 of the rounds will give rise
2:07 to 1/4 wrinkleds is a surprising prediction and,
2:11 therefore, something that bears stating as a scientific
2:14 prediction.
2:16 And so all those kind of predictions can emerge.
2:20 Then we turned to the question of two factors.
2:26 I'm practicing using our words here--
2:29 homozygous, heterozygous, alleles, et cetera.
2:32 We now turn to two traits, two phenotypes.
2:37 We have two phenotypes segregating.
2:41 We had round and wrinkled, and we had green and yellow.
2:45 And Mendel determined, rather brilliantly, that they
2:49 segregated, were transmitted independently of each other.
2:53 There was no correlation between which alleles you got
2:56 at round and which alleles you got at wrinkled.
2:59 That was pretty cool.
3:02 Let's just go over that, because there's a real tension
3:05 to be resolved because we have Mendel's second law versus the
3:23 chromosome theory.
3:32 So Mendel's second law--
3:34 let's just go back to it-- in the F0 generation, we had our
3:40 round green peas, genotype big R, big R, big G, big G. We had
3:50 our wrinkled yellow peas, genotype little r, little r,
3:59 little g, little g.
4:01 We cross them together, we get F1 double heterozygotes, big
4:08 R, little r, big G, little g.
4:11 We then perform a back cross or test cross to the doubly
4:18 homozygous parent with the two recessive phenotypes--
4:23 we're practicing our words here.
4:25 And what do we get?
4:26 Well, we get certain options.
4:29 As we said, the gametes that emerge from this parent on the
4:33 left could be of the following types.
4:41 The gametes from the parent on the right
4:44 are all those alleles--
4:49 Are those the recessive alleles?
4:54 No.
4:54 You'll forget this over time, but just to make me
4:57 comfortable--
4:58 they're actually the alleles associated with the recessive
5:00 phenotype that we're talking about.
5:02 Because you know-- but will forget, I assure you, because
5:04 all my colleagues in the biology department have
5:06 forgotten--
5:07 that they could also control multiple other phenotypes,
5:09 some of which could be dominant.
5:11 But that's OK.
5:12 I forgive you in advance that you'll call
5:14 them recessive alleles.
5:15 Anyway, you get this.
5:18 And then these should occur at equal frequency of one to one
5:23 to one to one.
5:28 All right.
5:29 Now, we had the chromosome theory.
5:36 The chromosome theory, the observation of the
5:39 choreography of chromosomes during meiosis--
5:43 chromosomes in meiosis--
5:46 looks like this.
5:50 We have chromosomes lining up in pairs.
5:56 Now, I'm not drawing it terribly well.
5:57 But the two members of each pair are of the same size.
6:04 But this pair could be bigger, and this
6:06 pair could be smaller.
6:07 It looks like they're really finding each other.
6:09 These chromosomes are visibly different in shape.
6:12 And so maybe I'll make this guy a little longer, just to
6:15 indicate that, that it actually knows.
6:18 Now an explanation, we said, for how Mendel's second law of
6:24 independent assortment of two different phenotypes could
6:27 occur is that, for example - round.
6:40 It could be that the gene for roundness is located on
6:44 chromosome number one.
6:46 These are two copies of chromosome one that have been
6:48 duplicated, each of which has the big-R allele.
6:52 Next to it, two copies of chromosome one that have been
6:55 duplicated, each of them having the little-r allele.
7:00 Over here on chromosome number two.
7:02 let's say, lies the gene for green or yellow.
7:07 And in this picture here, the big G's are on the two copies
7:17 of this chromosome number two that are here.
7:19 The little g's are on the two copies of chromosome number
7:21 two that are here.
7:24 When the cell undergoes meiosis one, we get to the
7:34 situation where we have big G, big G; little g, little g.
7:48 We've got big R, big R; little r, little r.
7:54 Could it have been the case that the big G was on the
7:56 right and the little g was on the left?
7:59 Yeah, of course.
8:00 It's totally independent which way.
8:01 I happened to draw it this way.
8:03 But with probability 50%, it's the other way.
8:06 That's why they're independent of each other.
8:07 And then when it undergoes meiosis two that looks just
8:11 like mitosis, we end up with our four gametes here.
8:21 Sorry, our four gametes with big G, big G; little g, little
8:41 g; big R, big R; little r, little r.
8:49 And that accounts for the big G, big G; big R, big G; little
8:54 r, little g gametes.
8:56 And then when they went the other way, the organism would
8:59 make a set of gametes that had the other combination of big
9:02 R's with little g's.
9:04 So that's perfectly fine.
9:06 And because the second chromosome is independently
9:09 ordered compared to the first chromosome-- when they line up
9:12 at the midline, they don't really care
9:13 which way they are--
9:14 that'll account for one to one to one to one.
9:16 It's so straightforward.
9:22 But what happens if, instead, both the roundness gene and
9:40 the greenness gene live on the same chromosome?
9:46 We'll have little r, little r there.
9:49 We'll have big G, big G here; little g, little g.
9:56 And then when they split, we'll end up with--
10:02 now, chromosome two has nothing that we
10:07 care about on it.
10:08 It has a lot of genes.
10:09 In fact, there could be chromosomes three, four, five.
10:12 I'm just not drawing them.
10:13 But we're really going to focus on
10:15 chromosome number one here.
10:17 And you'll notice that here on chromosome one, the big G's or
10:26 the little g's are coupled, physically coupled to each
10:34 other, the bigs with the bigs and the
10:36 smalls with the smalls.
10:37 So now the kind of gametes that can emerge from this will
10:46 only be big R, big G type or--
10:50 big G type--
10:51 or they'll be little r, little g type.
10:56 We can't get the reverse combination.
10:58 We can't get bigs and littles.
11:01 So this, because these are independent of each other,
11:10 will get us one to one to one to one.
11:19 These are the big, little, and then these other
11:27 combinations like that.
11:29 This will get us only, if we look at the big R, big G;
11:39 little r, little g; big R, little g; little r, big G;
11:43 will get us one to one to zero to zero.
11:49 Let's give a name to this type, the big R's and the
11:52 little g's.
11:53 Let's call that a recombinant type, OK?
11:58 I'm just going to use that word for the moment.
11:59 The recombinant types-- the bigs and the littles, and
12:01 littles and the bigs--
12:03 we're not going to see any of them.
12:05 This is a very strong difference between Mendel's
12:08 second law and the chromosome theory.
12:10 If the chromosome theory is right, and these chromosomes
12:13 are physical entities that have integrity, they can't
12:19 both be right.
12:22 So that's a great thing in science, when you have two
12:24 different models and they can't both be right, because
12:28 you learn things then.
12:29 You can test them.
12:30 Now, Mendel tested this without actually knowing the
12:35 chromosome theory.
12:37 And he always got one to one to one to one for the seven
12:39 traits he looked at.
12:43 Was he just lucky that they happened to lie on the seven
12:45 different chromosomes?
12:50 Or is there some problem with this
12:52 chromosome theory, or what?
12:55 It took a while.
12:56 And then, of course, everybody forgot about Mendel from 1865
12:59 until the year 1900.
13:00 In the year 1900, people begin to rediscover Mendel.
13:04 Cytology has come along.
13:06 In January of the year 1900, plant breeders start
13:10 rediscovering Mendel.
13:16 Sorry.
13:17 Thank you.
13:18 I see already.
13:20 Thank you.
13:22 I could tell by the look on your face that something must
13:24 be wrong there.
13:25 Good.
13:27 Plant breeders start rediscovering Mendel.
13:30 And in January of 1900, three different groups say, you
13:37 know, we found these laws.
13:38 And they're just like Mendel's laws--
13:40 which now everybody starts paying attention to.
13:42 But plant breeding-- and people tried to do mice, and
13:45 people tried to do rats.
13:48 What turned out to be the winner, the place to really
13:50 study genetics, was the fruit fly.
13:54 Thomas Hunt Morgan at Columbia University decided, after he
13:59 was really frustrated wasting years breeding mice and rats
14:02 that just took too long, around, oh, I don't know, 1906
14:07 or something like that, began to start breeding fruit flies.
14:12 Fruit flies are the teeny little flies that when you
14:15 open a banana or fruits or other kinds of
14:18 things, you'll see them.
14:20 And studying Drosophila gave us the answer to this question
14:26 of what the problem is, how can it be that either it's
14:29 independent or totally dependent?
14:31 So we're going to talk about Drosophila melanogaster, the
14:36 fruit fly, and the discovery of recombination.
14:47 So now, Morgan--
14:54 I'm going to now start using fruit fly notation to give you
14:56 some practice with fruit fly notation.
15:00 We're not going to use big G's and little g's.
15:02 They like to refer to the normal allele
15:04 and the mutant allele.
15:06 The normally allele is plus.
15:08 The mutant allele gets some kind of a name.
15:11 And so he had a female fly that was normal and normal at
15:19 two different loci, two different genes.
15:22 I haven't told you what the genes are.
15:24 And the male fly had a black-colored body-- black
15:30 across its whole body--
15:32 and its wings were shriveled and,
15:36 therefore, called vestigial.
15:38 So the phenotype here is black and vestigial, black body and
15:45 vestigial wings.
15:47 And this wasn't the normal appearance of a fly.
15:50 So he took these females by these males, and he crossed
15:54 them together.
15:55 And he got an F1.
15:58 And the F1 was black over vestigial, plus over plus.
16:05 And what was their phenotype?
16:09 Were they normal appearance, which is kind of a
16:12 sandy-colored body and normal wings?
16:15 Or were they this all black body and vestigial wings?
16:19 Turns out they were normal.
16:21 From that, what do we infer about these two traits, black
16:25 body and vestigial wings?
16:27 They're recessive traits.
16:29 So here, the phenotype was normal appearance.
16:37 So then he crosses them back, doing a test cross.
16:42 Let's say he'll take males here and females here, but it
16:46 actually works either way.
16:48 And what he does is just like we did there.
16:50 He could get gametes that were black, vestigial.
16:54 He could get gametes that were black, plus; plus, vestigial;
17:00 or plus, plus.
17:05 Those are the four possibilities that come out.
17:09 So when he does it-- let's keep score.
17:11 I'm going to write them now--
17:12 plus, plus; black, vestigial.
17:18 And from the other parent, you got this--
17:23 black, plus; black, vestigial; plus,
17:29 vestigial; black, vestigial.
17:32 Those are the four possibilities.
17:34 And if this was just like Mendel's traits, it would be
17:37 one to one to one to one.
17:39 These were the parental types that went in.
17:42 Plus, plus went in.
17:44 And black, vestigial went in.
17:46 Those were the combinations of traits here.
17:48 These were new combinations.
17:52 What he observed--
17:54 Let's see.
17:55 If Mendel's right, it'll be one to one to one to one .
17:57 If the chromosome theory's right, it'll be one to one to
17:59 zero to zero.
18:00 And who was right?
18:03 STUDENT: No one.
18:04 ERIC LANDER: No one.
18:05 The answer was 965 to 944 to 206 to 185.
18:18 Neither model was right.
18:21 Neither model's right.
18:24 The new combinations, the recombinant combinations, the
18:27 non-parental combinations--
18:29 we can call these recombinant combinations.
18:33 These were recombinant.
18:34 They recombined in some way.
18:36 They were a new combination, or they were the
18:38 non-parental types.
18:39 We use both of those words frequently--
18:43 were neither equal nor were they completely absent.
18:47 They occurred, but at a lower frequency.
18:50 What was the frequency?
18:51 Well, we could just add it up.
18:54 The frequency of recombinant types, of new types of were
19:04 different than the parents', is 17%.
19:22 What's going on?
19:26 Now, maybe this is some magic number like the 3-to-1 ratio.
19:31 And you should look at that 17% and say, ah, this is some
19:34 constant of the universe, that when you put in traits you get
19:37 17% percent of recombinant types.
19:41 But it takes a little judgment to say, I don't think so.
19:44 And he actually tried other traits.
19:46 And sometimes he got one to one to one to one.
19:49 But very often he got some funny number--
19:53 6%, 28%, 1%.
20:00 There was some funny business going on.
20:03 What's going on?
20:07 Recombination.
20:09 That's what's going on is there is
20:10 recombination occurring.
20:12 What do we mean by recombination?
20:16 Recombination is very important stuff, by the way.
20:20 At some point, I will tell you that understanding
20:23 recombination was actually the origin of the
20:26 Human Genome Project.
20:28 And it traces back to a good MIT story.
20:31 But that will be for a little later in the semester.
20:34 So what do we think's happening here?
20:42 I'm now going to draw a close-up of that chromosome.
20:46 And here's another chromosome, the other pair.
20:56 And what we think here might be happening is that you might
21:04 have plus and black; black and plus--
21:16 oh, sorry.
21:17 Black, right?
21:20 Black, black; plus, plus.
21:22 This would be plus.
21:24 This would be plus.
21:25 This is the normal chromosome.
21:26 This is plus.
21:28 So this chromosome here carries the pluses.
21:34 This guy here carries those alleles black and vestigial.
21:43 Well, what happens, the idea was that somehow these
21:48 chromosomes exchanged material here.
21:52 And the chromosome that was plus, plus; plus, plus now
21:56 somehow acquires that bit, and this chromosome
22:03 somehow gets that bit.
22:05 And we end up instead with a picture like this, where some
22:19 of this came from here.
22:27 And those two loci, black and vestigial, were separated from
22:34 each other such that the black allele moves over to that
22:38 chromosome.
22:39 Is that clear?
22:42 That's the notion.
22:43 Why did they think this was true?
22:45 Well, it turns out that in fruit flies, you can actually
22:48 look at eggs under the microscope.
22:50 And you can look at the chromosomes.
22:51 And if you take if you take the cells, and you take a
22:54 cover slip, and you squash it down with your finger, you can
22:57 actually see chromosomes lying right over each other, making
23:02 little crosses like I drew there, up there, called
23:05 chiasmata, which means crosses.
23:07 And so people said, see, in the microscope, you can see
23:11 they're lying on top of each other.
23:13 Are you impressed by that piece of evidence?
23:18 No.
23:19 You took the cover slip, you squished it
23:20 down with your fingers.
23:21 So they're lying on top of each other?
23:23 Big deal.
23:25 I'm not going to be impressed.
23:26 And calling it chiasmata doesn't make me any more
23:28 impressed, right?
23:30 Although it's always good to call things Greek names,
23:32 because people think they're more sophisticated if you call
23:34 them Greek names.
23:37 But this was the notion people had.
23:40 And the frequency 17% would indicate how often these
23:46 crossovers occur.
23:51 But if I were in the situation we talked about with Mendel,
23:53 and I wrote this up, and I said, see, it's 17% sometimes,
23:56 6% sometimes, 28% sometimes; and when I look in the
23:58 microscope, they lie on top of each other; therefore, it's
24:01 recombination--
24:05 There were actually other ideas floating around too.
24:07 Maybe it has something to do with developmental biology.
24:11 It was a puzzle.
24:13 When you have a really deep puzzle, the most important
24:18 thing in science is to find a young person.
24:22 Because young people come without prejudice.
24:25 They say, let me just look at all of this.
24:27 Stand back.
24:28 I don't come with any prejudice.
24:29 So at MIT, what is the solution when you have a
24:32 problem like this?
24:34 A UROP.
24:35 You want a UROP.
24:37 So even 1911, that was the solution at Columbia.
24:44 Thomas Hunt Morgan got a UROP.
24:47 I'm serious.
24:48 He was called Alfred Sturtevant.
24:51 Alfred Sturtevant was a sophomore at Columbia.
24:55 Everybody else was busy finding this recombination
24:58 data, how often this recombined with this, this
25:01 with this, this with this.
25:02 Sturtevant--
25:04 he's a sophomore--
25:05 he says, god, this stuff's interesting!
25:09 Professor Morgan, could I have all the data and
25:11 try to look at it?
25:13 Sturtevant took it home and actually pulled an
25:17 all-nighter, blew off his homework--
25:20 it actually says so in his autobiography.
25:22 He says, I blew off my homework and pulled an
25:25 all-nighter--
25:27 essentially in those words, he says. "To the detriment of my
25:30 undergraduate homework" is the way he puts it.
25:33 But in any case, genetic maps and Sturtevant's all-nighter.
25:45 What Sturtevant did was the following.
25:47 He said, how are we going to prove the chromosome theory?
25:54 I like this idea that recombination is about
25:57 distance on the chromosomes.
25:59 I like the concept that maybe 17% is how often
26:04 these things recombine.
26:07 Why would things only recombine 1% of the time?
26:10 What would that mean?
26:14 They've got to be pretty close together so that a crossover
26:16 between them happens not so often.
26:19 And what if things were far apart?
26:22 Well, it could be more likely.
26:23 So he likes the idea that recombination
26:26 frequency means distance.
26:29 But how are you going to prove that?
26:31 It could mean a zillion other things.
26:33 It could mean biochemical pathways,
26:35 developmental biology.
26:36 How are you going to prove that recombination frequency
26:39 means distance?
26:41 You've got to make predictions, right?
26:43 The only to do it would be to make predictions.
26:46 So Sturtevant takes the data, and he starts making
26:51 predictions.
26:58 I think, says Sturtevant, these things are alleles
27:04 living at genes with locations on the chromosome--
27:09 black, vestigial.
27:14 How often do black and vestigial
27:18 recombine with each other?
27:19 What is the frequency of recombinant
27:20 non-parental types?
27:23 17%.
27:26 So Sturtevant goes through the data and he says, what about
27:30 other crosses people did in the lab?
27:33 Well, it turns out people did crosses with another mutant
27:36 that produces a funny eye color called cinnabar, Cn.
27:41 So cinnabar.
27:47 It turns out that the recombination frequency
27:55 between cinnabar to vestigial was 8%.
28:07 Vestigial, cinnabar, 8% recombination.
28:12 If this chromosome business is right, where
28:14 should I put cinnabar?
28:18 Sorry?
28:19 Where do you want it?
28:21 STUDENT: [INAUDIBLE].
28:21 ERIC LANDER: Somewhere in between.
28:22 You'd like me to put that there.
28:26 STUDENT: On the other side.
28:27 ERIC LANDER: Oh, OK.
28:28 Wait a second.
28:28 On the other side.
28:31 OK, which is it?
28:32 How many vote for the left?
28:34 How many vote for the right?
28:37 How many conscientious abstainers are there?
28:40 Do we know?
28:41 STUDENT: No.
28:42 ERIC LANDER: No.
28:43 There are two possibilities.
28:44 It could be 8% this way, or it could be 8% that way.
28:52 How are we going to know?
28:53 Yes?
28:55 STUDENT: [INAUDIBLE].
28:57 ERIC LANDER: Black.
28:57 What if we knew the answer between cinnabar and black?
29:02 That would constrain the problem.
29:04 Can you give me two predictions for what the
29:06 answer might be?
29:07 STUDENT: Either 9% or--
29:08 ERIC LANDER: Either 9% or 25%.
29:12 So now we have a prediction.
29:13 We don't know where cinnabar is.
29:15 But the answer could be that black to vestigial should
29:20 either be about 9%--
29:23 that's what that would be here--
29:25 or about 25%.
29:32 The answer?
29:36 About 9%.
29:40 That's what Sturtevant found.
29:46 That's a prediction and kind of cool.
29:49 And you can imagine taking the data home and it's probably 9
29:53 o'clock, and you've now realized, wow,
29:55 freaky, it's 9%.
29:58 So then he looked at the mutation "lobe." Lobe was
30:04 another mutation.
30:06 The lobe mutation showed 5% recombination from vestigial.
30:15 Where should we put it?
30:20 Left or right, well, let's make some predictions.
30:27 Suppose it's over here.
30:31 Will it be very close to cinnabar?
30:34 Will it be closer to black?
30:37 But what if it was over here?
30:39 Well, it would be further.
30:41 So let's put lobe in.
30:45 And suppose we know that lob is 5%.
30:50 Then what's the prediction for black to lobe?
30:58 22%.
31:01 Answer, according to the notebooks, 21%, pretty close.
31:09 You'd like it to be exactly 22.
31:10 But life doesn't always turn out that way.
31:12 21's pretty close to 22.
31:15 What other predictions could we make?
31:17 If this is cinnabar here, could you give me a prediction
31:19 for cinnabar to lobe?
31:23 13%.
31:25 Yep, that works.
31:27 Curved wing.
31:33 Recombination distance from lobed, 3%.
31:38 So now you have some predictions.
31:39 You have this prediction here.
31:41 You predict 8%.
31:43 Answer, about 8%.
31:45 Over here you predict 16%.
31:50 Answer, about 16%.
31:53 Bingo.
31:56 Sturtivant says, if these genes--
32:00 we call them loci, often.
32:02 I'll use the word locus synonymous with gene.
32:05 Locus means a place.
32:07 And geneticists think about genes as a place on a
32:09 chromosome.
32:10 If these loci--
32:12 the plural of locus-- if these loci were really arrayed along
32:16 a linear structure, then it would have to be the case that
32:21 they would have certain additive
32:23 relationships between them.
32:25 And the chance that they would have these additive
32:27 relationships if they weren't part of a line is pretty
32:31 implausible.
32:33 That's a real prediction, a very remarkable prediction.
32:39 And it holds up with the data.
32:41 Sturtivant pulls the all-nighter.
32:43 By the time the sun comes up at Columbia University-- this
32:46 is Morningside Heights--
32:47 he's got the whole thing worked out.
32:50 Yes, this chromosome theory must be right.
32:53 It fits beautifully all of these data.
32:57 Pretty cool.
32:59 You are all authorized to blow off your homework anytime you
33:04 make a discovery like that.
33:07[LAUGHTER]
33:08 That's a course rule.
33:10 Any homework will be forgiven for
33:13 discoveries of that magnitude.
33:17 All right.
33:19 Tell your TAs.
33:23 So now, what does it really tell us?
33:25 It tells us that if genes are very close together, the
33:30 recombination frequency, R recombination frequency, or
33:35 RF, might be very little.
33:38 They could be as low as almost zero, which means you never
33:40 see a recombinant because they're right
33:42 next to each other.
33:44 Or it could be that they're further apart.
33:47 It might be 1%.
33:48 It might be 10%.
33:50 It could keep growing.
33:52 It might be 30%.
33:55 Suppose it's way, way, way, way, way far away.
34:00 What's the largest it ever gets to be?
34:06 Well, if they were on different chromosomes--
34:08 suppose there were totally independent segregation,
34:10 different chromosomes--
34:11 what would they be?
34:14 STUDENT: [INAUDIBLE]
34:15 ERIC LANDER: No, it's 50% because remember, one to one
34:17 to one to one says that the recombinant--
34:19 well, actually, this is an interesting question.
34:21 Let's come to that 100.
34:22 On different chromosomes, one to one to one to one means
34:25 that it's 50%, half of them are recombinant types.
34:28 It turns out on the same chromosome, as you get farther
34:31 and farther and farther, you might say there's going to be
34:33 100% chance of a crossover.
34:37 And you might say the recombination frequency could
34:39 keep growing past 50%.
34:42 It turns out it doesn't.
34:43 The reason is that multiple crossovers can happen.
34:49 So mathematical interjection here, if here's my gene and
34:53 here's my gene, there could be one crossover.
34:58 There could be two crossovers.
35:01 There could be three crossovers.
35:05 And so in fact, it turns out, as you get very far away, you
35:09 have to start paying attention to the probabilities of double
35:12 crossovers and triple crossovers.
35:15 And so it turns out that it's a Poisson process of the
35:18 number of crossovers that occurs, give or take.
35:21 And you see a recombination if there's an odd number.
35:25 And for that reason, it never gets above 50%.
35:28 So as the distance gets further and further and
35:30 further, it goes from zero to 50, which is the same number
35:33 you get for separate chromosomes.
35:35 Otherwise, you might think that if there was just one
35:37 crossover, it gets to 100% probability of recombination.
35:40 But it never does, because there are doubles.
35:42 And you can actually observe, if you make a cross that has
35:44 three different genes segregating in it, you can
35:47 actually see the double crossover type.
35:51 So you can see very nicely that if you make a cross
35:55 involving black, cinnabar, and vestigial over plus, plus,
36:03 plus, you can see that cinnabar is right in the
36:06 middle because this recombination happens at a
36:10 pretty good frequency.
36:12 This recombination happens at a good frequency, giving you
36:17 plus, cinnabar, vestigial or black, plus, plus.
36:21 Sometimes you will get that recombination.
36:26 You'll get out gametes that are black, plus, vestigial,
36:30 but at a much lower frequency because they take two
36:33 crossovers.
36:34 What will be the probability of seeing a
36:36 black, cinnabar, vestigial?
36:38 Well, we said that this one was about 9%, this
36:42 one was about 8%.
36:45 What's the product of a 9% chance and an 8% chance?
36:48 It's about a 1% chance, a little less than a 1% chance.
36:52 That how frequently you see black, plus, vestigial.
36:55 You can even predict the probability of a double
36:58 crossover event by multiplying the two events
37:01 that have to happen.
37:02 So your bottom-line rules here is that because of these
37:05 double crossovers our recombination frequency can go
37:09 from zero to about 50%.
37:15 This is independent assortment.
37:19 And it either occurs if you're on different chromosomes or,
37:25 if you're very far away on the same chromosome, they behave
37:29 as if they're independent of each other.
37:31 Any questions about any of that?
37:33 Yes?
37:34 STUDENT: Why didn't Mendel ever see recombination?
37:36 ERIC LANDER: Why didn't Mendel ever see recombination?
37:38 It turns out that with seven chromosomes, and they're
37:40 biggish in length, he never actually ran into two loci
37:44 that were close enough, to notice it.
37:47 That's why.
37:48 Flies actually only have three major chromosomes.
37:52 There's a fourth, but it's a puny little thing.
37:55 And because they were much more intensively collecting
37:58 mutations in Morgan's fly room, they began having a lot
38:01 of them, and they had to bump into
38:03 recombination pretty early.
38:05 Mendel simply didn't have enough that were close enough.
38:08 Think about what would have happened if just by chance,
38:10 somebody were selling a strain of peas in the market which
38:14 had a mutation in a locus that had 10% recombination
38:19 distance, and it screwed up Mendel's law of independent
38:22 assortment of two loci?
38:24 Mendel might not have published the paper.
38:26 Sometimes in science it's actually valuable to first get
38:30 the oversimplification out , like his second law, and then
38:34 deal with the complexity that sits on top of the
38:36 oversimplification.
38:38 It's kind of lucky that Mendel didn't have enough of them to
38:40 be bothered, in the first paper.
38:43 So in fact, all of Mendel's seven loci
38:45 have now been mapped.
38:46 Most have been cloned molecularly.
38:48 And so we actually know where they are, et cetera.
38:50 It's a really good question.
38:51 That's sort of why he didn't.
Course Info
Instructors
Prof. Eric Lander
Prof. Robert Weinberg
Prof. Tyler Jacks
Prof. Hazel Sive
Prof. Graham Walker
Prof. Sallie Chisholm
Dr. Michelle Mischke
Departments
Biology
As Taught In
Fall 2011
Level
Undergraduate
Topics
Science
Biology
Learning Resource Types
assignment_turned_in Problem Sets with Solutions
grading Exams with Solutions
theaters Lecture Videos
assignment Problem Sets
grading Exams
theaters Tutorial Videos
notes Lecture Notes
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3764 | https://www.spotsylvania.va.us/DocumentCenter/View/34522/For-Web---FY-2025-Adopted-to-FY-2026-Adopted-Budget-Comparison-PDF?bidId= | FY 2025 Adopted to FY 2026 Adopted Budget Comparison
Account Account Description FY 2025 Adopted Budget
FY 2026 Adopted Budget $ Change % Change
Board of Supervisors:
110-1101-401.11-01 REGULAR 172,182 170,979 (1,203) -1%
110-1101-401.21-01 FICA 10,022 9,449 (573) -6%
110-1101-401.21-02 MEDICARE 2,344 2,210 (134) -6%
110-1101-401.23-10 HEALTH INSURANCE 55,046 64,172 9,126 17%
110-1101-401.31-90 OTHER PROFESSIONAL SER 1,760 3,000 1,240 70%
110-1101-401.35-01 PRINTING & BINDING 200 200 0 0%
110-1101-401.36-01 ADVERTISING 20,000 20,000 0 0%
110-1101-401.52-30 TELEPHONE SERVICES 9,199 9,199 0 0%
110-1101-401.55-10 MILEAGE 1,820 1,754 (66) -4%
110-1101-401.55-30 SUBSISTENCE & LODGING 5,920 5,808 (112) -2%
110-1101-401.55-40 EDUCATION & TRAINING 5,375 6,425 1,050 20%
110-1101-401.58-10 DUES & ASSOC MEMBERSHIPS 33,000 33,000 0 0%
110-1101-401.58-40 MEETING EXPENSES 2,500 10,000 7,500 300%
110-1101-401.58-99 RECOGNITNS/AWARDS/SYMPTHY 1,500 1,500 0 0%
110-1101-401.60-01 OFFICE SUPPLIES 3,000 3,000 0 0%
Board of Supervisors 323,868 340,696 16,828 5.2%
County Administration:
110-1210-402.11-01 REGULAR 717,582 773,746 56,164 8%
110-1210-402.21-01 FICA 34,502 36,971 2,469 7%
110-1210-402.21-02 MEDICARE 10,189 11,046 857 8%
110-1210-402.22-10 RETIREMENT 94,621 101,419 6,798 7%
110-1210-402.23-10 HEALTH INSURANCE 70,135 64,172 (5,963) -9%
110-1210-402.24-01 INSURANCE 9,827 9,512 (315) -3%
110-1210-402.27-10 SELF INSURED 430 155 (275) -64%
110-1210-402.33-11 AUTO REPAIRS & MAINT 1,133 1,840 707 62%
110-1210-402.33-20 MAINTENANCE SVC CONTRACTS 5,884 5,884 0 0%
110-1210-402.35-01 PRINTING & BINDING 250 250 0 0%
110-1210-402.52-10 POSTAL SERVICES 622 900 278 45%
110-1210-402.52-30 TELEPHONE SERVICES 3,000 3,000 0 0%
110-1210-402.53-05 MOTOR VEHICLE INSURANCE 867 457 (410) -47%
110-1210-402.55-10 MILEAGE 500 1,000 500 100%
110-1210-402.55-30 SUBSISTENCE & LODGING 1,190 1,190 0 0%
110-1210-402.55-40 EDUCATION & TRAINING 2,560 2,830 270 11%
110-1210-402.58-10 DUES & ASSOC MEMBERSHIPS 4,417 4,417 0 0%
110-1210-402.58-45 EMPLOYEE RELATIONS 8,500 8,500 0 0%
110-1210-402.58-99 RECOGNITNS/AWARDS/SYMPTHY 800 800 0 0%
110-1210-402.60-01 OFFICE SUPPLIES 3,765 3,765 0 0%
110-1210-402.60-08 VEHICLE & EQUIPMENT FUELS 2,027 2,027 0 0%
110-1210-402.60-12 BOOKS & SUBSCRIPTIONS 6,500 13,000 6,500 100%
110-1210-402.80-01 MACHINERY & EQUIPMENT 0 20,000 20,000 n/a
County Administration 979,301 1,066,881 87,580 8.9%
County Attorney:
110-1221-402.11-01 REGULAR 1,207,719 1,317,067 109,348 9%
110-1221-402.21-01 FICA 69,636 72,582 2,946 4%
110-1221-402.21-02 MEDICARE 17,433 19,059 1,626 9%
110-1221-402.22-10 RETIREMENT 179,749 195,700 15,951 9%
110-1221-402.23-10 HEALTH INSURANCE 94,519 89,642 (4,877) -5%
110-1221-402.23-15 HEALTH INSURANCE (HSA) 3,612 3,686 74 2%
110-1221-402.24-01 INSURANCE 19,785 20,916 1,131 6%
110-1221-402.27-10 SELF INSURED 849 265 (584) -69%
110-1221-402.28-35 VEHICLE ALLOWANCE 5,973 5,973 0 0%
110-1221-402.28-60 401A BENEFIT PLAN 2,987 4,487 1,500 50%
110-1221-402.31-50 LEGAL SERVICES 20,000 20,000 0 0%
110-1221-402.31-51 LITIGATION COSTS 5,000 5,000 0 0%
110-1221-402.33-20 MAINTENANCE SVC CONTRACTS 2,400 2,600 200 8%
1FY 2025 Adopted to FY 2026 Adopted Budget Comparison
Account Account Description FY 2025 Adopted Budget
FY 2026 Adopted Budget $ Change % Change
110-1221-402.35-01 PRINTING & BINDING 150 150 0 0%
110-1221-402.39-10 SOFTWARE APPLICATIONS 8,740 8,740 0 0%
110-1221-402.52-10 POSTAL SERVICES 900 1,200 300 33%
110-1221-402.52-30 TELEPHONE SERVICES 6,145 6,205 60 1%
110-1221-402.54-10 LEASE/RENTAL EQUIPMENT 2,990 1,931 (1,059) -35%
110-1221-402.55-10 MILEAGE 7,062 6,332 (730) -10%
110-1221-402.55-30 SUBSISTENCE & LODGING 6,052 8,904 2,852 47%
110-1221-402.55-40 EDUCATION & TRAINING 7,650 8,560 910 12%
110-1221-402.58-10 DUES & ASSOC MEMBERSHIPS 3,610 3,650 40 1%
110-1221-402.58-99 RECOGNITNS/AWARDS/SYMPTHY 0 400 400 n/a
110-1221-402.60-01 OFFICE SUPPLIES 6,625 6,625 0 0%
110-1221-402.60-02 FOOD SUPPLIES & SERVICE 550 550 0 0%
110-1221-402.60-12 BOOKS & SUBSCRIPTIONS 10,000 11,000 1,000 10%
110-1221-402.80-01 MACHINERY & EQUIPMENT 0 15,000 15,000 n/a
110-1221-402.80-02 FURNITURE & FIXTURES 500 0 (500) -100%
County Attorney 1,690,636 1,836,224 145,588 8.6%
Human Resources:
110-1222-402.11-01 REGULAR 694,837 797,931 103,094 15%
110-1222-402.21-01 FICA 40,962 46,375 5,413 13%
110-1222-402.21-02 MEDICARE 9,816 11,253 1,437 15%
110-1222-402.22-10 RETIREMENT 108,021 122,626 14,605 14%
110-1222-402.23-10 HEALTH INSURANCE 74,927 71,753 (3,174) -4%
110-1222-402.23-15 HEALTH INSURANCE (HSA) 1,500 1,500 0 0%
110-1222-402.24-01 INSURANCE 12,969 15,263 2,294 18%
110-1222-402.27-10 SELF INSURED 3,017 1,034 (1,983) -66%
110-1222-402.28-20 EDUC TUITION ASSISTANCE 50,000 50,000 0 0%
110-1222-402.31-50 LEGAL SERVICES 7,000 7,000 0 0%
110-1222-402.31-90 OTHER PROFESSIONAL SER 0 0 0 n/a
110-1222-402.33-20 MAINTENANCE SVC CONTRACTS 67,827 131,176 63,349 93%
110-1222-402.35-01 PRINTING & BINDING 4,200 4,200 0 0%
110-1222-402.36-01 ADVERTISING 4,000 14,000 10,000 250%
110-1222-402.39-10 SOFTWARE APPLICATIONS 600 800 200 33%
110-1222-402.52-10 POSTAL SERVICES 256 256 0 0%
110-1222-402.52-30 TELEPHONE SERVICES 2,020 2,620 600 30%
110-1222-402.54-10 LEASE/RENTAL EQUIPMENT 4,800 4,800 0 0%
110-1222-402.55-10 MILEAGE 573 1,418 845 147%
110-1222-402.55-30 SUBSISTENCE & LODGING 5,931 1,782 (4,149) -70%
110-1222-402.55-40 EDUCATION & TRAINING 3,495 4,410 915 26%
110-1222-402.58-10 DUES & ASSOC MEMBERSHIPS 1,500 2,300 800 53%
110-1222-402.58-45 EMPLOYEE RELATIONS 30,000 30,000 0 0%
110-1222-402.58-46 PRE-EMPLOYMENT EXPENSES 12,000 12,000 0 0%
110-1222-402.58-47 IN-HOUSE TRAINING 29,067 29,067 0 0%
110-1222-402.60-01 OFFICE SUPPLIES 5,000 5,000 0 0%
110-1222-402.80-02 FURNITURE & FIXTURES 0 0 0 n/a
110-1222-402.80-07 COMPUTER EQUIPMENT 0 0 0 n/a
Human Resources 1,174,318 1,368,564 194,246 16.5%
Independent Auditor:
110-1224-402.31-20 AUDIT SERVICES 188,430 188,430 0 0%
Independent Auditor 188,430 188,430 0 0.0%
Internal Auditor:
110-1225-402.31-20 AUDIT SERVICES 150,000 90,000 (60,000) -40%
110-1225-402.31-25 HOTLINE MONITORING 31,500 16,500 (15,000) -48%
110-1225-402.33-20 MAINTENANCE SVC CONTRACTS 1,500 1,500 0 0%
Internal Auditor 183,000 108,000 (75,000) -41.0%
2FY 2025 Adopted to FY 2026 Adopted Budget Comparison
Account Account Description FY 2025 Adopted Budget
FY 2026 Adopted Budget $ Change % Change
Commissioner of the Revenue:
110-1231-403.11-01 REGULAR 1,279,850 1,141,641 (138,209) -11%
110-1231-403.12-01 OVERTIME 1,096 0 (1,096) -100%
110-1231-403.21-01 FICA 76,647 68,411 (8,236) -11%
110-1231-403.21-02 MEDICARE 18,112 16,206 (1,906) -11%
110-1231-403.22-10 RETIREMENT 175,450 157,039 (18,411) -10%
110-1231-403.23-10 HEALTH INSURANCE 171,004 163,025 (7,979) -5%
110-1231-403.24-01 INSURANCE 20,153 17,314 (2,839) -14%
110-1231-403.27-10 SELF INSURED 2,127 711 (1,416) -67%
110-1231-403.31-90 OTHER PROFESSIONAL SER 1,000 1,000 0 0%
110-1231-403.33-11 AUTO REPAIRS & MAINT 977 977 0 0%
110-1231-403.33-20 MAINTENANCE SVC CONTRACTS 1,800 2,200 400 22%
110-1231-403.36-01 ADVERTISING 1,375 1,375 0 0%
110-1231-403.52-10 POSTAL SERVICES 12,500 13,600 1,100 9%
110-1231-403.52-30 TELEPHONE SERVICES 1,644 2,100 456 28%
110-1231-403.53-05 MOTOR VEHICLE INSURANCE 891 1,122 231 26%
110-1231-403.54-10 LEASE/RENTAL EQUIPMENT 900 980 80 9%
110-1231-403.55-10 MILEAGE 1,046 1,370 324 31%
110-1231-403.55-30 SUBSISTENCE & LODGING 3,906 3,840 (66) -2%
110-1231-403.55-40 EDUCATION & TRAINING 2,440 2,590 150 6%
110-1231-403.58-10 DUES & ASSOC MEMBERSHIPS 940 940 0 0%
110-1231-403.60-01 OFFICE SUPPLIES 29,697 31,197 1,500 5%
110-1231-403.60-12 BOOKS & SUBSCRIPTIONS 87,911 93,371 5,460 6%
110-1231-403.60-24 NON STATE FUNDED BUS EXPS 0 0 0 n/a
110-1231-403.80-02 FURNITURE & FIXTURES 600 1,500 900 150%
Commissioner of the Revenue 1,892,066 1,722,509 (169,557) -9.0%
Assessment:
110-1232-403.11-01 REGULAR 977,286 1,073,948 96,662 10%
110-1232-403.13-01 PART-TIME 88,764 77,908 (10,856) -12%
110-1232-403.14-01 BOARDS & COMMISSIONS 5,906 5,904 (2) 0%
110-1232-403.21-01 FICA 64,660 69,375 4,715 7%
110-1232-403.21-02 MEDICARE 15,123 16,225 1,102 7%
110-1232-403.22-10 RETIREMENT 137,325 151,814 14,489 11%
110-1232-403.23-10 HEALTH INSURANCE 152,583 137,614 (14,969) -10%
110-1232-403.23-15 HEALTH INSURANCE (HSA) 0 2,100 2,100 n/a
110-1232-403.24-01 INSURANCE 16,031 16,799 768 5%
110-1232-403.27-10 SELF INSURED 19,428 6,558 (12,870) -66%
110-1232-403.31-90 OTHER PROFESSIONAL SER 0 52,950 52,950 n/a
110-1232-403.33-11 AUTO REPAIRS & MAINT 5,992 6,001 9 0%
110-1232-403.33-20 MAINTENANCE SVC CONTRACTS 880 880 0 0%
110-1232-403.36-01 ADVERTISING 1,600 2,600 1,000 63%
110-1232-403.52-10 POSTAL SERVICES 2,000 5,000 3,000 150%
110-1232-403.52-30 TELEPHONE SERVICES 2,500 2,500 0 0%
110-1232-403.53-05 MOTOR VEHICLE INSURANCE 3,246 3,537 291 9%
110-1232-403.55-30 SUBSISTENCE & LODGING 5,495 6,640 1,145 21%
110-1232-403.55-40 EDUCATION & TRAINING 7,050 8,550 1,500 21%
110-1232-403.58-10 DUES & ASSOC MEMBERSHIPS 1,680 1,680 0 0%
110-1232-403.60-01 OFFICE SUPPLIES 8,400 9,400 1,000 12%
110-1232-403.60-08 VEHICLE & EQUIPMENT FUELS 7,761 8,000 239 3%
110-1232-403.60-11 UNIFORMS 900 1,200 300 33%
110-1232-403.60-12 BOOKS & SUBSCRIPTIONS 1,500 1,500 0 0%
110-1232-403.80-02 FURNITURE & FIXTURES 0 0 0 n/a
Assessment 1,526,110 1,668,683 142,573 9.3%
Treasurer:
110-1241-404.11-01 REGULAR 1,497,390 1,582,856 85,466 6%
110-1241-404.12-01 OVERTIME 27,123 26,308 (815) -3%
3FY 2025 Adopted to FY 2026 Adopted Budget Comparison
Account Account Description FY 2025 Adopted Budget
FY 2026 Adopted Budget $ Change % Change
110-1241-404.21-01 FICA 90,099 94,555 4,456 5%
110-1241-404.21-02 MEDICARE 21,465 22,541 1,076 5%
110-1241-404.22-10 RETIREMENT 205,160 223,387 18,227 9%
110-1241-404.23-10 HEALTH INSURANCE 199,758 192,684 (7,074) -4%
110-1241-404.24-01 INSURANCE 23,086 24,150 1,064 5%
110-1241-404.27-10 SELF INSURED 869 322 (547) -63%
110-1241-404.33-10 REPAIRS & MAINTENANCE 3,000 5,000 2,000 67%
110-1241-404.33-20 MAINTENANCE SVC CONTRACTS 11,050 11,050 0 0%
110-1241-404.35-01 PRINTING & BINDING 102,100 110,000 7,900 8%
110-1241-404.36-01 ADVERTISING 2,200 2,200 0 0%
110-1241-404.39-10 SOFTWARE APPLICATIONS 500 500 0 0%
110-1241-404.52-10 POSTAL SERVICES 216,700 239,700 23,000 11%
110-1241-404.52-30 TELEPHONE SERVICES 2,300 2,300 0 0%
110-1241-404.54-10 LEASE/RENTAL EQUIPMENT 37,163 37,163 0 0%
110-1241-404.55-10 MILEAGE 2,793 2,050 (743) -27%
110-1241-404.55-30 SUBSISTENCE & LODGING 3,740 3,404 (336) -9%
110-1241-404.55-40 EDUCATION & TRAINING 4,790 5,400 610 13%
110-1241-404.58-10 DUES & ASSOC MEMBERSHIPS 2,115 2,115 0 0%
110-1241-404.60-01 OFFICE SUPPLIES 18,000 18,000 0 0%
110-1241-404.60-11 UNIFORMS 800 800 0 0%
110-1241-404.80-01 MACHINERY & EQUIPMENT 1,500 0 (1,500) -100%
110-1241-404.80-02 FURNITURE & FIXTURES 2,395 0 (2,395) -100%
110-1241-404.80-07 COMPUTER EQUIPMENT 0 7,000 7,000 n/a
Treasurer 2,476,096 2,613,485 137,389 5.5%
Accounting:
110-1242-404.11-01 REGULAR 1,087,860 1,252,739 164,879 15%
110-1242-404.12-01 OVERTIME 5,993 1,052 (4,941) -82%
110-1242-404.21-01 FICA 64,707 75,242 10,535 16%
110-1242-404.21-02 MEDICARE 15,133 17,597 2,464 16%
110-1242-404.22-10 RETIREMENT 159,439 178,957 19,518 12%
110-1242-404.23-10 HEALTH INSURANCE 154,635 150,048 (4,587) -3%
110-1242-404.23-15 HEALTH INSURANCE (HSA) 1,200 3,600 2,400 200%
110-1242-404.24-01 INSURANCE 19,041 22,155 3,114 16%
110-1242-404.27-10 SELF INSURED 656 250 (406) -62%
110-1242-404.31-30 MGT CONSULTING SERVICES 775 775 0 0%
110-1242-404.31-60 OTH FINANCIAL CONSULTANTS 22,000 16,416 (5,584) -25%
110-1242-404.33-20 MAINTENANCE SVC CONTRACTS 139,272 144,867 5,595 4%
110-1242-404.35-01 PRINTING & BINDING 69,481 74,976 5,495 8%
110-1242-404.39-10 SOFTWARE APPLICATIONS 41,329 41,829 500 1%
110-1242-404.52-10 POSTAL SERVICES 287,345 303,560 16,215 6%
110-1242-404.52-30 TELEPHONE SERVICES 800 1,968 1,168 146%
110-1242-404.54-10 LEASE/RENTAL EQUIPMENT 1,241 1,241 0 0%
110-1242-404.55-10 MILEAGE 3,402 1,815 (1,587) -47%
110-1242-404.55-30 SUBSISTENCE & LODGING 4,161 9,346 5,185 125%
110-1242-404.55-40 EDUCATION & TRAINING 6,894 8,556 1,662 24%
110-1242-404.58-10 DUES & ASSOC MEMBERSHIPS 2,712 2,087 (625) -23%
110-1242-404.60-01 OFFICE SUPPLIES 11,840 11,775 (65) -1%
Accounting 2,099,916 2,320,851 220,935 10.5%
Budget :
110-1244-404.11-01 REGULAR 857,005 987,699 130,694 15%
110-1244-404.13-01 PART-TIME 15,457 0 (15,457) -100%
110-1244-404.21-01 FICA 50,983 57,548 6,565 13%
110-1244-404.21-02 MEDICARE 12,363 14,066 1,703 14%
110-1244-404.22-10 RETIREMENT 113,430 133,908 20,478 18%
110-1244-404.23-10 HEALTH INSURANCE 105,384 101,557 (3,827) -4%
110-1244-404.24-01 INSURANCE 12,353 13,459 1,106 9%
4FY 2025 Adopted to FY 2026 Adopted Budget Comparison
Account Account Description FY 2025 Adopted Budget
FY 2026 Adopted Budget $ Change % Change
110-1244-404.27-10 SELF INSURED 524 197 (327) -62%
110-1244-404.31-30 MGT CONSULTING SERVICES 690 690 0 0%
110-1244-404.31-60 OTH FINANCIAL CONSULTANTS 47,000 47,000 0 0%
110-1244-404.31-90 OTHER PROFESSIONAL SER 25,000 0 (25,000) -100%
110-1244-404.33-20 MAINTENANCE SVC CONTRACTS 1,485 1,485 0 0%
110-1244-404.35-01 PRINTING & BINDING 2,759 2,800 41 1%
110-1244-404.52-30 TELEPHONE SERVICES 516 516 0 0%
110-1244-404.55-10 MILEAGE 498 302 (196) -39%
110-1244-404.55-30 SUBSISTENCE & LODGING 2,090 5,505 3,415 163%
110-1244-404.55-40 EDUCATION & TRAINING 3,230 3,510 280 9%
110-1244-404.58-10 DUES & ASSOC MEMBERSHIPS 750 870 120 16%
110-1244-404.60-01 OFFICE SUPPLIES 4,768 4,768 0 0%
110-1244-404.60-02 FOOD SUPPLIES & SERVICE 1,012 1,162 150 15%
Budget 1,257,297 1,377,042 119,745 9.5%
Grants:
110-1246-404.11-01 REGULAR 255,959 271,141 15,182 6%
110-1246-404.21-01 FICA 15,793 16,584 791 5%
110-1246-404.21-02 MEDICARE 3,693 3,878 185 5%
110-1246-404.22-10 RETIREMENT 34,853 38,130 3,277 9%
110-1246-404.23-10 HEALTH INSURANCE 33,221 23,588 (9,633) -29%
110-1246-404.24-01 INSURANCE 4,205 4,331 126 3%
110-1246-404.27-10 SELF INSURED 154 54 (100) -65%
110-1246-404.52-30 TELEPHONE SERVICES 0 540 540 n/a
110-1246-404.55-10 MILEAGE 355 385 30 8%
110-1246-404.55-30 SUBSISTENCE & LODGING 2,688 4,042 1,354 50%
110-1246-404.55-40 EDUCATION & TRAINING 1,950 1,950 0 0%
110-1246-404.58-10 DUES & ASSOC MEMBERSHIPS 210 230 20 10%
110-1246-404.60-01 OFFICE SUPPLIES 4,000 3,420 (580) -15%
110-1246-404.60-12 BOOKS & SUBSCRIPTIONS 2,560 765 (1,795) -70%
Grants 359,641 369,038 9,397 2.6%
Procurement:
110-1247-404.11-01 REGULAR 437,317 469,231 31,914 7%
110-1247-404.12-01 OVERTIME 1,381 526 (855) -62%
110-1247-404.21-01 FICA 26,439 28,442 2,003 8%
110-1247-404.21-02 MEDICARE 6,184 6,652 468 8%
110-1247-404.22-10 RETIREMENT 62,508 69,492 6,984 11%
110-1247-404.23-10 HEALTH INSURANCE 46,692 42,745 (3,947) -8%
110-1247-404.24-01 INSURANCE 7,474 7,987 513 7%
110-1247-404.27-10 SELF INSURED 263 94 (169) -64%
110-1247-404.39-10 SOFTWARE APPLICATIONS 500 500 0 0%
110-1247-404.52-10 POSTAL SERVICES 50 50 0 0%
110-1247-404.52-30 TELEPHONE SERVICES 0 480 480 n/a
110-1247-404.54-10 LEASE/RENTAL EQUIPMENT 2,042 1,500 (542) -27%
110-1247-404.55-10 MILEAGE 1,542 1,177 (365) -24%
110-1247-404.55-30 SUBSISTENCE & LODGING 2,903 4,179 1,276 44%
110-1247-404.55-40 EDUCATION & TRAINING 5,001 7,055 2,054 41%
110-1247-404.58-10 DUES & ASSOC MEMBERSHIPS 285 980 695 244%
110-1247-404.60-01 OFFICE SUPPLIES 3,200 1,500 (1,700) -53%
110-1247-404.80-07 COMPUTER EQUIPMENT 3,900 0 (3,900) -100%
Procurement 607,681 642,590 34,909 5.7%
Information Services:
110-1251-405.11-01 REGULAR 4,004,808 4,323,973 319,165 8%
110-1251-405.12-01 OVERTIME 45,686 38,936 (6,750) -15%
110-1251-405.12-02 ON-CALL 22,824 18,942 (3,882) -17%
110-1251-405.13-01 PART-TIME 0 26,266 26,266 n/a
5FY 2025 Adopted to FY 2026 Adopted Budget Comparison
Account Account Description FY 2025 Adopted Budget
FY 2026 Adopted Budget $ Change % Change
110-1251-405.21-01 FICA 245,715 265,323 19,608 8%
110-1251-405.21-02 MEDICARE 57,584 62,248 4,664 8%
110-1251-405.22-10 RETIREMENT 596,034 649,502 53,468 9%
110-1251-405.23-10 HEALTH INSURANCE 460,163 406,083 (54,080) -12%
110-1251-405.23-15 HEALTH INSURANCE (HSA) 6,000 8,100 2,100 35%
110-1251-405.24-01 INSURANCE 71,001 76,818 5,817 8%
110-1251-405.27-10 SELF INSURED 2,442 881 (1,561) -64%
110-1251-405.28-98 PERSONNEL BUDGET REDUCTIO 0 0 0 n/a
110-1251-405.31-30 MGT CONSULTING SERVICES 155,000 155,000 0 0%
110-1251-405.31-70 DISASTER RECOVERY SERVICE 9,511 9,511 0 0%
110-1251-405.31-80 GIS DEVELOPMENT SERVICES 55,000 60,000 5,000 9%
110-1251-405.31-90 OTHER PROFESSIONAL SER 353,500 353,500 0 0%
110-1251-405.33-10 REPAIRS & MAINTENANCE 156,500 156,500 0 0%
110-1251-405.33-11 AUTO REPAIRS & MAINT 2,411 2,531 120 5%
110-1251-405.33-12 RADIO REPAIRS & MAINT 70,000 70,000 0 0%
110-1251-405.33-20 MAINTENANCE SVC CONTRACTS 4,366,399 5,604,722 1,238,323 28%
110-1251-405.35-01 PRINTING & BINDING 500 500 0 0%
110-1251-405.36-01 ADVERTISING 1,450 1,000 (450) -31%
110-1251-405.39-11 IS STRATEGIC PLAN 25,000 0 (25,000) -100%
110-1251-405.52-10 POSTAL SERVICES 560 560 0 0%
110-1251-405.52-30 TELEPHONE SERVICES 17,880 17,880 0 0%
110-1251-405.53-05 MOTOR VEHICLE INSURANCE 2,427 3,790 1,363 56%
110-1251-405.54-20 LEASE/RENTAL BUILDINGS 25,000 25,000 0 0%
110-1251-405.55-10 MILEAGE 20,120 19,332 (788) -4%
110-1251-405.55-30 SUBSISTENCE & LODGING 31,133 42,155 11,022 35%
110-1251-405.55-40 EDUCATION & TRAINING 65,434 170,612 105,178 161%
110-1251-405.58-10 DUES & ASSOC MEMBERSHIPS 9,118 9,313 195 2%
110-1251-405.58-99 RECOGNITNS/AWARDS/SYMPTHY 2,500 0 (2,500) -100%
110-1251-405.60-01 OFFICE SUPPLIES 5,000 10,000 5,000 100%
110-1251-405.60-02 FOOD SUPPLIES & SERVICE 2,012 2,512 500 25%
110-1251-405.60-08 VEHICLE & EQUIPMENT FUELS 2,483 2,483 0 0%
110-1251-405.60-12 BOOKS & SUBSCRIPTIONS 6,586 4,586 (2,000) -30%
110-1251-405.60-22 EDP SUPPLIES 10,900 7,900 (3,000) -28%
110-1251-405.80-02 FURNITURE & FIXTURES 3,500 0 (3,500) -100%
110-1251-405.80-07 COMPUTER EQUIPMENT 10,000 10,000 0 0%
Information Services 10,922,181 12,616,459 1,694,278 15.5%
Central Supply:
110-1253-405.33-20 MAINTENANCE SVC CONTRACTS 930 930 0 0%
110-1253-405.54-10 LEASE/RENTAL EQUIPMENT 16,000 16,000 0 0%
110-1253-405.60-01 OFFICE SUPPLIES 800 800 0 0%
Central Supply 17,730 17,730 0 0.0%
Risk Management:
110-1255-405.11-01 REGULAR 104,439 109,604 5,165 5%
110-1255-405.21-01 FICA 6,559 6,885 326 5%
110-1255-405.21-02 MEDICARE 1,534 1,610 76 5%
110-1255-405.22-10 RETIREMENT 17,080 17,950 870 5%
110-1255-405.24-01 INSURANCE 1,929 2,080 151 8%
110-1255-405.26-01 UNEMPLOYMENT INSURANCE 25,000 25,000 0 0%
110-1255-405.27-10 SELF INSURED 63 22 (41) -65%
110-1255-405.52-30 TELEPHONE SERVICES 500 540 40 8%
110-1255-405.53-06 SURETY BONDS 3,366 3,366 0 0%
110-1255-405.53-07 PUBLIC OFFICIAL LIAB INS 9,896 9,898 2 0%
110-1255-405.53-08 GENERAL LIABILITY INS 35,180 35,190 10 0%
110-1255-405.55-10 MILEAGE 831 835 4 0%
110-1255-405.55-30 SUBSISTENCE & LODGING 1,328 1,328 0 0%
110-1255-405.55-40 EDUCATION & TRAINING 1,535 1,535 0 0%
6FY 2025 Adopted to FY 2026 Adopted Budget Comparison
Account Account Description FY 2025 Adopted Budget
FY 2026 Adopted Budget $ Change % Change
110-1255-405.60-01 OFFICE SUPPLIES 1,245 1,245 0 0%
Risk Management 210,485 217,088 6,603 3.1%
Safety Division:
110-1256-405.11-01 REGULAR 329,472 345,699 16,227 5%
110-1256-405.21-01 FICA 19,899 21,002 1,103 6%
110-1256-405.21-02 MEDICARE 4,654 4,912 258 6%
110-1256-405.22-10 RETIREMENT 44,986 50,979 5,993 13%
110-1256-405.23-10 HEALTH INSURANCE 30,179 20,594 (9,585) -32%
110-1256-405.24-01 INSURANCE 5,545 5,787 242 4%
110-1256-405.27-10 SELF INSURED 198 69 (129) -65%
110-1256-405.31-30 CONSULTING SERVICES 75,000 75,000 0 0%
110-1256-405.31-90 OTHER PROFESSIONAL SERVICES 25,000 35,540 10,540 42%
110-1256-405.33-11 AUTO REPAIRS & MAINT 0 2,000 2,000 n/a
110-1256-405.39-10 SOFTWARE APPLICATIONS 0 2,200 2,200 n/a
110-1256-405.52-30 TELEPHONE SERVICES 1,800 3,600 1,800 100%
110-1256-405.53-05 MOTOR VEHICLE INSURANCE 0 673 673 n/a
110-1256-405.55-30 SUBSISTENCE & LODGING 1,926 3,816 1,890 98%
110-1256-405.55-40 EDUCATION & TRAINING 5,700 10,089 4,389 77%
110-1256-405.58-10 DUES & ASSOC MEMBERSHIPS 0 4,200 4,200 n/a
110-1256-405.60-01 OFFICE SUPPLIES 2,000 4,000 2,000 100%
110-1256-405.60-08 VEHICLE & EQUIPMENT FUELS 4,000 4,000 0 0%
110-1256-405.60-11 UNIFORMS 1,500 1,500 0 0%
Safety Division 551,859 595,660 43,801 7.9%
Registrar/Electoral Board:
110-1320-406.11-01 REGULAR 401,449 420,537 19,088 5%
110-1320-406.12-01 OVERTIME 14,848 15,625 777 5%
110-1320-406.13-01 PART-TIME 42,156 47,725 5,569 13%
110-1320-406.14-01 BOARDS & COMMISSIONS 15,712 16,184 472 3%
110-1320-406.21-01 FICA 28,942 30,402 1,460 5%
110-1320-406.21-02 MEDICARE 6,769 7,110 341 5%
110-1320-406.22-10 RETIREMENT 57,279 60,238 2,959 5%
110-1320-406.23-10 HEALTH INSURANCE 43,216 46,342 3,126 7%
110-1320-406.24-01 INSURANCE 6,224 6,253 29 0%
110-1320-406.27-10 SELF INSURED 275 97 (178) -65%
110-1320-406.31-90 OTHER PROFESSIONAL SER 70,320 10,620 (59,700) -85%
110-1320-406.33-10 REPAIRS & MAINTENANCE 500 500 0 0%
110-1320-406.33-20 MAINTENANCE SVC CONTRACTS 40,428 45,428 5,000 12%
110-1320-406.35-01 PRINTING & BINDING 80,157 141,157 61,000 76%
110-1320-406.36-01 ADVERTISING 1,750 3,000 1,250 71%
110-1320-406.38-55 ELECT OFFL/BD STIPENDS 95,130 55,680 (39,450) -41%
110-1320-406.51-10 ELECTRICAL SERVICES 8,360 8,592 232 3%
110-1320-406.51-20 HEATING SERVICES 8,759 8,759 0 0%
110-1320-406.51-30 WATER & SEWER SERVICES 0 0 0 n/a
110-1320-406.52-10 POSTAL SERVICES 53,830 53,830 0 0%
110-1320-406.52-30 TELEPHONE SERVICES 420 750 330 79%
110-1320-406.54-10 LEASE/RENTAL EQUIPMENT 4,758 5,100 342 7%
110-1320-406.54-20 LEASE/RENTAL BUILDINGS 14,000 14,000 0 0%
110-1320-406.55-10 MILEAGE 1,193 2,764 1,571 132%
110-1320-406.55-30 SUBSISTENCE & LODGING 2,172 2,271 99 5%
110-1320-406.55-40 EDUCATION & TRAINING 900 1,000 100 11%
110-1320-406.58-10 DUES & ASSOC MEMBERSHIPS 630 630 0 0%
110-1320-406.60-01 OFFICE SUPPLIES 5,772 6,600 828 14%
110-1320-406.60-02 FOOD SUPPLIES & SERVICE 420 504 84 20%
110-1320-406.60-14 OPERATING SUPPLIES 11,019 18,675 7,656 69%
110-1320-406.80-01 MACHINERY & EQUIPMENT 15,620 1,060 (14,560) -93%
110-1320-406.80-02 FURNITURE & FIXTURES 0 1,000 1,000 n/a
7FY 2025 Adopted to FY 2026 Adopted Budget Comparison
Account Account Description FY 2025 Adopted Budget
FY 2026 Adopted Budget $ Change % Change
110-1320-406.80-07 COMPUTER EQUIPMENT 0 0 0 n/a
Registrar/Electoral Board 1,033,008 1,032,433 (575) -0.1%
Circuit Court -1:
110-2110-411.11-01 REGULAR 99,060 104,016 4,956 5%
110-2110-411.21-01 FICA 5,977 6,325 348 6%
110-2110-411.21-02 MEDICARE 1,398 1,479 81 6%
110-2110-411.22-10 RETIREMENT 12,921 13,583 662 5%
110-2110-411.23-10 HEALTH INSURANCE 18,349 13,798 (4,551) -25%
110-2110-411.24-01 INSURANCE 1,328 1,228 (100) -8%
110-2110-411.27-10 SELF INSURED 59 21 (38) -64%
110-2110-411.52-30 TELEPHONE SERVICES 1,830 1,830 0 0%
110-2110-411.55-60 JUROR & WITNESS EXPENSES 1,500 1,500 0 0%
110-2110-411.60-01 OFFICE SUPPLIES 2,500 2,500 0 0%
110-2110-411.60-12 BOOKS & SUBSCRIPTIONS 2,700 2,700 0 0%
Circuit Court -1 147,622 148,980 1,358 0.9%
Circuit Court - 2:
110-2111-411.11-01 REGULAR 154,825 165,640 10,815 7%
110-2111-411.21-01 FICA 9,522 10,185 663 7%
110-2111-411.21-02 MEDICARE 2,227 2,382 155 7%
110-2111-411.22-10 RETIREMENT 20,584 16,886 (3,698) -18%
110-2111-411.23-10 HEALTH INSURANCE 7,436 6,796 (640) -9%
110-2111-411.23-15 HEALTH INSURANCE (HSA) 0 0 0 n/a
110-2111-411.24-01 INSURANCE 2,347 2,708 361 15%
110-2111-411.27-10 SELF INSURED 93 33 (60) -65%
110-2111-411.39-10 SOFTWARE APPLICATIONS 0 0 0 n/a
110-2111-411.52-30 TELEPHONE SERVICES 1,510 1,510 0 0%
110-2111-411.54-10 LEASE/RENTAL EQUIPMENT 500 500 0 0%
110-2111-411.55-30 SUBSISTENCE & LODGING 0 0 0 n/a
110-2111-411.55-60 JUROR & WITNESS EXPENSES 2,043 2,500 457 22%
110-2111-411.58-10 DUES & ASSOC MEMBERSHIPS 25 25 0 0%
110-2111-411.60-01 OFFICE SUPPLIES 2,500 2,500 0 0%
110-2111-411.60-12 BOOKS & SUBSCRIPTIONS 3,100 3,175 75 2%
Circuit Court - 2 206,712 214,840 8,128 3.9%
General District Court:
110-2120-411.31-50 LEGAL SERVICES 8,000 8,000 0 0%
110-2120-411.31-90 OTHER PROFESSIONAL SER 2,000 2,500 500 25%
110-2120-411.33-20 MAINTENANCE SVC CONTRACTS 1,500 1,900 400 27%
110-2120-411.52-10 POSTAL SERVICES 425 425 0 0%
110-2120-411.52-30 TELEPHONE SERVICES 3,000 3,000 0 0%
110-2120-411.54-10 LEASE/RENTAL EQUIPMENT 8,126 8,426 300 4%
110-2120-411.55-40 EDUCATION & TRAINING 2,500 0 (2,500) -100%
110-2120-411.56-01 NEW REGIONAL AGENCY REQ 0 0 0 n/a
110-2120-411.58-10 DUES & ASSOC MEMBERSHIPS 1,540 1,540 0 0%
110-2120-411.60-01 OFFICE SUPPLIES 2,500 2,700 200 8%
110-2120-411.60-12 BOOKS & SUBSCRIPTIONS 5,000 6,500 1,500 30%
General District Court 34,591 34,991 400 1.2%
Magistrates:
110-2130-411.52-30 TELEPHONE SERVICES 1,200 1,200 0 0%
110-2130-411.54-10 LEASE/RENTAL EQUIPMENT 2,760 2,760 0 0%
110-2130-411.60-01 OFFICE SUPPLIES 1,500 1,500 0 0%
110-2130-411.80-02 FURNITURE & FIXTURES 700 0 (700) -100%
Magistrates 6,160 5,460 (700) -11.4%
Juvenile & Domestic Relations Court:
8FY 2025 Adopted to FY 2026 Adopted Budget Comparison
Account Account Description FY 2025 Adopted Budget
FY 2026 Adopted Budget $ Change % Change
110-2150-411.31-50 LEGAL SERVICES 1,500 1,500 0 0%
110-2150-411.31-90 OTHER PROFESSIONAL SER 4,000 4,000 0 0%
110-2150-411.33-20 MAINTENANCE SVC CONTRACTS 4,008 6,000 1,992 50%
110-2150-411.52-10 POSTAL SERVICES 400 400 0 0%
110-2150-411.52-30 TELEPHONE SERVICES 2,400 4,400 2,000 83%
110-2150-411.54-10 LEASE/RENTAL EQUIPMENT 14,500 14,500 0 0%
110-2150-411.55-10 MILEAGE 5,391 4,500 (891) -17%
110-2150-411.55-30 SUBSISTENCE & LODGING 5,828 3,028 (2,800) -48%
110-2150-411.55-40 EDUCATION & TRAINING 3,630 3,630 0 0%
110-2150-411.58-10 DUES & ASSOC MEMBERSHIPS 1,850 2,000 150 8%
110-2150-411.60-01 OFFICE SUPPLIES 11,000 11,000 0 0%
110-2150-411.60-02 FOOD SUPPLIES & SERVICE 0 2,500 2,500 n/a
110-2150-411.60-12 BOOKS & SUBSCRIPTIONS 6,000 6,000 0 0%
Juvenile & Domestic Relations Court 60,507 63,458 2,951 4.9%
Clerk of the Circuit Court:
110-2160-412.11-01 REGULAR 1,484,764 1,595,926 111,162 7%
110-2160-412.12-01 OVERTIME 12,564 13,221 657 5%
110-2160-412.13-01 PART-TIME 0 0 0 n/a
110-2160-412.21-01 FICA 87,061 93,604 6,543 8%
110-2160-412.21-02 MEDICARE 21,021 22,610 1,589 8%
110-2160-412.22-10 RETIREMENT 206,335 231,507 25,172 12%
110-2160-412.23-10 HEALTH INSURANCE 213,012 211,660 (1,352) -1%
110-2160-412.24-01 INSURANCE 24,912 26,464 1,552 6%
110-2160-412.27-10 SELF INSURED 898 322 (576) -64%
110-2160-412.31-20 AUDIT SERVICES 3,000 3,000 0 0%
110-2160-412.31-50 LEGAL SERVICES 2,000 2,000 0 0%
110-2160-412.31-75 MICROFILMING 16,000 16,000 0 0%
110-2160-412.31-90 OTHER PROFESSIONAL SER 2,400 2,400 0 0%
110-2160-412.33-20 MAINTENANCE SVC CONTRACTS 122,655 124,480 1,825 1%
110-2160-412.35-01 PRINTING & BINDING 11,000 11,000 0 0%
110-2160-412.52-10 POSTAL SERVICES 30,500 20,500 (10,000) -33%
110-2160-412.52-30 TELEPHONE SERVICES 2,766 2,766 0 0%
110-2160-412.54-10 LEASE/RENTAL EQUIPMENT 8,986 8,980 (6) 0%
110-2160-412.55-10 MILEAGE 1,503 2,948 1,445 96%
110-2160-412.55-30 SUBSISTENCE & LODGING 7,140 5,937 (1,203) -17%
110-2160-412.55-40 EDUCATION & TRAINING 4,000 2,800 (1,200) -30%
110-2160-412.55-60 JUROR & WITNESS EXPENSES 42,600 42,600 0 0%
110-2160-412.58-10 DUES & ASSOC MEMBERSHIPS 980 980 0 0%
110-2160-412.60-01 OFFICE SUPPLIES 19,000 19,000 0 0%
110-2160-412.60-02 FOOD SUPPLIES & SERVICE 1,382 1,640 258 19%
110-2160-412.60-12 BOOKS & SUBSCRIPTIONS 750 750 0 0%
Clerk of the Circuit Court 2,327,229 2,463,095 135,866 5.8%
Sheriff - Courts/Civil Process:
110-2170-412.11-01 REGULAR 3,859,511 3,230,073 (629,438) -16%
110-2170-412.12-01 OVERTIME 238,233 245,779 7,546 3%
110-2170-412.12-02 ON-CALL 44,816 76,550 31,734 71%
110-2170-412.13-01 PART-TIME 520,243 425,671 (94,572) -18%
110-2170-412.21-01 FICA 278,838 241,977 (36,861) -13%
110-2170-412.21-02 MEDICARE 66,222 56,591 (9,631) -15%
110-2170-412.22-10 RETIREMENT 498,380 415,268 (83,112) -17%
110-2170-412.23-10 HEALTH INSURANCE 447,368 299,203 (148,165) -33%
110-2170-412.23-15 HEALTH INSURANCE (HSA) 0 1,200 1,200 n/a
110-2170-412.24-01 INSURANCE 51,426 37,790 (13,636) -27%
110-2170-412.27-10 SELF INSURED 72,286 58,566 (13,720) -19%
110-2170-412.28-10 LINE OF DUTY BENEFITS 37,851 38,233 382 1%
110-2170-412.31-10 HEALTH SERVICES 810 810 0 0%
9FY 2025 Adopted to FY 2026 Adopted Budget Comparison
Account Account Description FY 2025 Adopted Budget
FY 2026 Adopted Budget $ Change % Change
110-2170-412.33-11 AUTO REPAIRS & MAINT 0 0 0 n/a
110-2170-412.33-20 MAINTENANCE SVC CONTRACTS 8,200 3,450 (4,750) -58%
110-2170-412.35-01 PRINTING & BINDING 500 0 (500) -100%
110-2170-412.52-30 TELEPHONE SERVICES 3,000 24,600 21,600 720%
110-2170-412.53-05 MOTOR VEHICLE INSURANCE 2,430 1,126 (1,304) -54%
110-2170-412.55-50 EXTRADITION OF PRISONERS 20,000 22,000 2,000 10%
110-2170-412.58-46 PRE EMPLOYMENT EXPENSES 1,700 1,700 0 0%
110-2170-412.60-01 OFFICE SUPPLIES 3,800 4,500 700 18%
110-2170-412.60-10 POLICE OPERATING SUPPLIES 3,850 7,500 3,650 95%
110-2170-412.60-11 UNIFORMS 22,750 14,750 (8,000) -35%
110-2170-412.80-01 MACHINERY & EQUIPMENT 22,700 8,350 (14,350) -63%
110-2170-412.80-02 FURNITURE & FIXTURES 0 750 750 n/a
Sheriff - Courts/Civil Process 6,204,914 5,216,437 (988,477) -15.9%
Victim/Witness Program:
110-2190-412.11-01 REGULAR 279,788 81,315 (198,473) -71%
110-2190-412.21-01 FICA 16,193 20,470 4,277 26%
110-2190-412.21-02 MEDICARE 3,787 4,787 1,000 26%
110-2190-412.22-10 RETIREMENT 36,959 45,535 8,576 23%
110-2190-412.23-10 HEALTH INSURANCE 61,782 40,186 (21,596) -35%
110-2190-412.24-01 INSURANCE 3,995 4,826 831 21%
110-2190-412.27-10 SELF INSURED 179 69 (110) -61%
110-2190-412.31-90 OTHER PROFESSIONAL SER 1,000 1,000 0 0%
110-2190-412.52-10 POSTAL SERVICES 0 500 500 n/a
110-2190-412.55-10 MILEAGE 672 1,635 963 143%
110-2190-412.55-30 SUBSISTENCE & LODGING 1,282 4,148 2,866 224%
110-2190-412.55-40 EDUCATION & TRAINING 700 2,250 1,550 221%
110-2190-412.56-63 RAPP COUNCIL AG SEX ASSLT 12,000 12,000 0 0%
110-2190-412.58-10 DUES & ASSOC MEMBERSHIPS 250 250 0 0%
110-2190-412.60-01 OFFICE SUPPLIES 5,784 7,800 2,016 35%
110-2190-412.80-07 COMPUTER EQUIPMENT 0 2,600 2,600 n/a
Victim/Witness Program 424,371 229,371 (195,000) -46.0%
Commonwealth's Attorney:
110-2210-413.11-01 REGULAR 2,644,454 2,699,185 54,731 2%
110-2210-413.12-01 OVERTIME 5,993 6,307 314 5%
110-2210-413.13-01 PART-TIME 3,159 3,157 (2) 0%
110-2210-413.21-01 FICA 156,344 162,417 6,073 4%
110-2210-413.21-02 MEDICARE 37,273 38,729 1,456 4%
110-2210-413.22-10 RETIREMENT 363,201 381,699 18,498 5%
110-2210-413.23-10 HEALTH INSURANCE 278,862 259,233 (19,629) -7%
110-2210-413.23-15 HEALTH INSURANCE (HSA) 9,300 7,200 (2,100) -23%
110-2210-413.24-01 INSURANCE 39,631 40,075 444 1%
110-2210-413.27-10 SELF INSURED 1,846 550 (1,296) -70%
110-2210-413.33-20 MAINTENANCE SVC CONTRACTS 4,948 44,948 40,000 808%
110-2210-413.35-01 PRINTING & BINDING 1,000 1,500 500 50%
110-2210-413.52-10 POSTAL SERVICES 2,280 1,780 (500) -22%
110-2210-413.52-30 TELEPHONE SERVICES 16,600 18,301 1,701 10%
110-2210-413.53-05 MOTOR VEHICLE INSURANCE 223 0 (223) -100%
110-2210-413.55-10 MILEAGE 8,184 10,232 2,048 25%
110-2210-413.55-30 SUBSISTENCE & LODGING 32,178 41,642 9,464 29%
110-2210-413.55-40 EDUCATION & TRAINING 8,450 12,400 3,950 47%
110-2210-413.55-60 JUROR & WITNESS EXPENSES 7,500 3,000 (4,500) -60%
110-2210-413.56-56 RAPP LEGAL SERVICES INC 28,684 28,684 0 0%
110-2210-413.58-10 DUES & ASSOC MEMBERSHIPS 12,500 13,900 1,400 11%
110-2210-413.60-01 OFFICE SUPPLIES 31,000 33,000 2,000 6%
110-2210-413.60-08 VEHICLE & EQUIPMENT FUELS 196 0 (196) -100%
110-2210-413.60-12 BOOKS & SUBSCRIPTIONS 30,837 32,157 1,320 4%
10 FY 2025 Adopted to FY 2026 Adopted Budget Comparison
Account Account Description FY 2025 Adopted Budget
FY 2026 Adopted Budget $ Change % Change
110-2210-413.80-02 FURNITURE & FIXTURES 0 0 0 n/a
110-2210-413.80-07 COMPUTER EQUIPMENT 0 4,929 4,929 n/a
Commonwealth's Attorney 3,724,643 3,845,025 120,382 3.2%
Communications:
110-3140-421.11-01 REGULAR 2,328,109 3,409,964 1,081,855 46%
110-3140-421.12-01 OVERTIME 266,051 429,561 163,510 61%
110-3140-421.12-02 ON-CALL 38,930 40,163 1,233 3%
110-3140-421.13-01 PART-TIME 92,955 177,858 84,903 91%
110-3140-421.21-01 FICA 162,989 244,591 81,602 50%
110-3140-421.21-02 MEDICARE 38,118 57,202 19,084 50%
110-3140-421.22-10 RETIREMENT 316,697 456,201 139,504 44%
110-3140-421.23-10 HEALTH INSURANCE 387,493 468,902 81,409 21%
110-3140-421.23-15 HEALTH INSURANCE (HSA) 0 2,400 2,400 n/a
110-3140-421.24-01 INSURANCE 35,735 40,533 4,798 13%
110-3140-421.27-10 SELF INSURED 736 20,461 19,725 2680%
110-3140-421.31-10 HEALTH SERVICES 700 500 (200) -29%
110-3140-421.31-90 OTHER PROFESSIONAL SER 4,500 4,500 0 0%
110-3140-421.33-10 REPAIRS & MAINTENANCE 2,700 1,300 (1,400) -52%
110-3140-421.33-20 MAINTENANCE SVC CONTRACTS 108,600 67,600 (41,000) -38%
110-3140-421.35-01 PRINTING & BINDING 400 1,400 1,000 250%
110-3140-421.52-10 POSTAL SERVICES 100 300 200 200%
110-3140-421.52-30 TELEPHONE SERVICES 321,000 252,000 (69,000) -21%
110-3140-421.54-10 LEASE/RENTAL EQUIPMENT 360 360 0 0%
110-3140-421.55-10 MILEAGE 6,200 5,370 (830) -13%
110-3140-421.55-30 SUBSISTENCE & LODGING 26,088 30,940 4,852 19%
110-3140-421.55-40 EDUCATION & TRAINING 23,340 40,134 16,794 72%
110-3140-421.58-10 DUES & ASSOC MEMBERSHIPS 1,100 1,500 400 36%
110-3140-421.58-46 PRE-EMPLOYMENT EXPENSES 1,200 1,200 0 0%
110-3140-421.58-99 RECOGNITNS/AWARDS/SYMPTHY 1,500 1,500 0 0%
110-3140-421.60-01 OFFICE SUPPLIES 5,200 5,200 0 0%
110-3140-421.60-02 FOOD SUPPLIES & SERVICE 1,300 1,700 400 31%
110-3140-421.60-11 UNIFORMS 1,500 1,500 0 0%
110-3140-421.60-12 BOOKS & SUBSCRIPTIONS 400 400 0 0%
110-3140-421.60-14 OPERATING SUPPLIES 0 0 0 n/a
110-3140-421.80-07 COMPUTER EQUIPMENT 7,200 0 (7,200) -100%
Communications 4,181,201 5,765,240 1,584,039 37.9%
Sheriff:
110-3160-421.11-01 REGULAR 16,934,871 18,451,411 1,516,540 9%
110-3160-421.12-01 OVERTIME 1,598,398 1,950,985 352,587 22%
110-3160-421.12-02 ON-CALL 82,383 63,964 (18,419) -22%
110-3160-421.13-01 PART-TIME 523,146 536,698 13,552 3%
110-3160-421.21-01 FICA 1,149,820 1,205,061 55,241 5%
110-3160-421.21-02 MEDICARE 269,751 282,802 13,051 5%
110-3160-421.22-10 RETIREMENT 2,163,565 2,229,253 65,688 3%
110-3160-421.23-10 HEALTH INSURANCE 2,344,319 2,151,043 (193,276) -8%
110-3160-421.23-15 HEALTH INSURANCE (HSA) 4,800 6,000 1,200 25%
110-3160-421.24-01 INSURANCE 227,239 203,178 (24,061) -11%
110-3160-421.27-10 SELF INSURED 286,490 290,693 4,203 1%
110-3160-421.28-10 LINE OF DUTY BENEFITS 133,929 142,332 8,403 6%
110-3160-421.28-25 CLOTHING ALLOWANCE 30,000 30,000 0 0%
110-3160-421.31-10 HEALTH SERVICES 2,465 4,465 2,000 81%
110-3160-421.31-50 LEGAL SERVICES 10,000 10,000 0 0%
110-3160-421.31-90 OTHER PROFESSIONAL SER 22,500 22,500 0 0%
110-3160-421.33-11 AUTO REPAIRS & MAINT 608,214 616,260 8,046 1%
110-3160-421.33-12 RADIO REPAIRS & MAINT 10,000 10,000 0 0%
110-3160-421.33-20 MAINTENANCE SVC CONTRACTS 432,603 590,171 157,568 36%
11 FY 2025 Adopted to FY 2026 Adopted Budget Comparison
Account Account Description FY 2025 Adopted Budget
FY 2026 Adopted Budget $ Change % Change
110-3160-421.35-01 PRINTING & BINDING 16,000 16,000 0 0%
110-3160-421.36-01 ADVERTISING 3,550 3,550 0 0%
110-3160-421.39-10 SOFTWARE APPLICATIONS 175,080 233,612 58,532 33%
110-3160-421.51-10 ELECTRICAL SERVICES 4,004 4,004 0 0%
110-3160-421.52-10 POSTAL SERVICES 9,700 9,700 0 0%
110-3160-421.52-30 TELEPHONE SERVICES 208,712 220,816 12,104 6%
110-3160-421.53-05 MOTOR VEHICLE INSURANCE 105,992 115,915 9,923 9%
110-3160-421.54-10 LEASE/RENTAL EQUIPMENT 868 1,220 352 41%
110-3160-421.54-30 POLICE MALL SUBSTATION 5,000 2,520 (2,480) -50%
110-3160-421.55-10 MILEAGE 30,316 60,291 29,975 99%
110-3160-421.55-30 SUBSISTENCE & LODGING 177,748 377,940 200,192 113%
110-3160-421.55-40 EDUCATION & TRAINING 115,448 245,077 129,629 112%
110-3160-421.58-10 DUES & ASSOC MEMBERSHIPS 16,050 21,344 5,294 33%
110-3160-421.58-39 EMERGENCY RESPONSE TEAM 65,381 84,365 18,984 29%
110-3160-421.58-46 PRE-EMPLOYMENT EXPENSES 7,485 7,380 (105) -1%
110-3160-421.58-51 DRUG ENFORCEMENT 40,000 40,000 0 0%
110-3160-421.58-53 DARE PROGRAM 27,750 30,000 2,250 8%
110-3160-421.58-54 NEIGHBORHOOD WATCH PROG 3,900 3,900 0 0%
110-3160-421.58-57 SPECIALTY TEAMS 86,650 80,098 (6,552) -8%
110-3160-421.58-99 RECOGNITNS/AWARDS/SYMPTHY 4,000 4,000 0 0%
110-3160-421.60-01 OFFICE SUPPLIES 31,200 31,200 0 0%
110-3160-421.60-02 FOOD SUPPLIES & SERVICE 4,000 4,600 600 15%
110-3160-421.60-03 AGRICULTURAL/ANIMAL SUPPL 27,243 25,875 (1,368) -5%
110-3160-421.60-08 VEHICLE & EQUIPMENT FUELS 729,660 729,660 0 0%
110-3160-421.60-10 POLICE OPERATING SUPPLIES 163,525 278,168 114,643 70%
110-3160-421.60-11 UNIFORMS 137,692 144,805 7,113 5%
110-3160-421.60-12 BOOKS & SUBSCRIPTIONS 1,845 1,845 0 0%
110-3160-421.60-17 BODY ARMOR & PROTECT GEAR 98,800 117,500 18,700 19%
110-3160-421.70-01 RAPP CRIMINAL JUSTICE ACD 148,809 163,689 14,880 10%
110-3160-421.80-01 MACHINERY & EQUIPMENT 152,734 104,325 (48,409) -32%
110-3160-421.80-03 COMMUNICATION EQUIPMENT 11,800 750 (11,050) -94%
110-3160-421.80-05 MOTOR VEHICLES & EQUIP 0 0 0 n/a
110-3160-421.80-07 COMPUTER EQUIPMENT 26,740 3,300 (23,440) -88%
110-3160-421.80-09 EMERG RESP TEAM EQUIPMENT 86,880 54,582 (32,298) -37%
Sheriff 29,559,055 32,018,847 2,459,792 8.3%
Fire, Rescue & Emergency Services:
110-3210-422.11-01 REGULAR 23,669,871 25,525,312 1,855,441 8%
110-3210-422.12-01 OVERTIME 3,156,621 3,377,532 220,911 7%
110-3210-422.12-02 ON-CALL 46,486 33,908 (12,578) -27%
110-3210-422.13-01 PART-TIME 133,598 288,209 154,611 116%
110-3210-422.21-01 FICA 1,618,264 1,753,284 135,020 8%
110-3210-422.21-02 MEDICARE 379,077 410,723 31,646 8%
110-3210-422.22-10 RETIREMENT 2,912,556 3,150,240 237,684 8%
110-3210-422.23-10 HEALTH INSURANCE 3,067,204 2,962,554 (104,650) -3%
110-3210-422.23-15 HEALTH INSURANCE (HSA) 33,600 34,800 1,200 4%
110-3210-422.24-01 INSURANCE 301,453 288,879 (12,574) -4%
110-3210-422.27-10 SELF INSURED 934,883 575,277 (359,606) -38%
110-3210-422.28-10 LINE OF DUTY BENEFITS 98,710 103,646 4,936 5%
110-3210-422.31-10 HEALTH SERVICES 218,810 279,545 60,735 28%
110-3210-422.31-90 OTHER PROFESSIONAL SER 2,361 949 (1,412) -60%
110-3210-422.33-10 REPAIRS & MAINTENANCE 67,070 104,796 37,726 56%
110-3210-422.33-20 MAINTENANCE SVC CONTRACTS 472,136 531,986 59,850 13%
110-3210-422.35-01 PRINTING & BINDING 4,025 4,025 0 0%
110-3210-422.39-10 SOFTWARE APPLICATIONS 280,810 291,144 10,334 4%
110-3210-422.39-20 VOPEX EXERCISE CAREER DEV 8,000 0 (8,000) -100%
110-3210-422.39-27 SANITATION SERVICES 1,560 1,560 0 0%
110-3210-422.51-10 ELECTRICAL SERVICES 2,230 2,607 377 17%
12 FY 2025 Adopted to FY 2026 Adopted Budget Comparison
Account Account Description FY 2025 Adopted Budget
FY 2026 Adopted Budget $ Change % Change
110-3210-422.52-10 POSTAL SERVICES 3,500 5,000 1,500 43%
110-3210-422.53-02 PROPERTY INSURANCE 3,562 4,710 1,148 32%
110-3210-422.54-10 LEASE/RENTAL EQUIPMENT 20,840 20,840 0 0%
110-3210-422.55-10 MILEAGE 16,400 25,350 8,950 55%
110-3210-422.55-30 SUBSISTENCE & LODGING 84,856 99,534 14,678 17%
110-3210-422.55-40 EDUCATION & TRAINING 217,328 251,642 34,314 16%
110-3210-422.58-10 DUES & ASSOC MEMBERSHIPS 11,883 12,713 830 7%
110-3210-422.58-46 PRE-EMPLOYMENT EXPENSES 68,282 88,052 19,770 29%
110-3210-422.58-47 IN-HOUSE TRAINING 20,000 20,000 0 0%
110-3210-422.58-59 PUBLIC EDUCATION 20,900 16,600 (4,300) -21%
110-3210-422.60-01 OFFICE SUPPLIES 26,000 26,000 0 0%
110-3210-422.60-02 FOOD SUPPLIES & SERVICE 1,400 4,648 3,248 232%
110-3210-422.60-05 JANITORIAL SUPPLIES 53,235 55,735 2,500 5%
110-3210-422.60-07 REPAIRS & MAINT SUPPLIES 30,000 40,000 10,000 33%
110-3210-422.60-12 BOOKS & SUBSCRIPTIONS 1,640 1,240 (400) -24%
110-3210-422.60-14 OPERATING SUPPLIES 4,853 8,399 3,546 73%
110-3210-422.60-20 INSTRUCTIONAL MATERIALS 71,500 71,500 0 0%
110-3210-422.60-27 SMALL TOOL/EQUIP REPLCMNT 35,103 40,840 5,737 16%
110-3210-422.60-29 HAZMAT REPLACEMENT 6,500 8,500 2,000 31%
110-3210-422.80-01 MACHINERY & EQUIPMENT 351,197 594,774 243,577 69%
110-3210-422.80-02 FURNITURE & FIXTURES 81,260 75,090 (6,170) -8%
110-3210-422.80-07 COMPUTER EQUIPMENT 34,330 12,600 (21,730) -63%
Fire, Rescue & Emergency Services 38,573,894 41,204,743 2,630,849 6.8%
Volunteer Fire & Rescue Services:
110-3220-422.22-20 LENGTH SVC PGM VOLUNTEERS 50,000 50,000 0 0%
110-3220-422.56-40 RAPP EMERG MED SERVICES 12,000 12,000 0 0%
110-3220-422.56-41 FOREST FIRE EXT SERVICE 11,996 11,996 0 0%
Volunteer Fire & Rescue Services 73,996 73,996 0 0.0%
Consolidated Fire & Rescue:
110-3240-422.28-10 LINE OF DUTY BENEFITS 21,780 22,095 315 1%
110-3240-422.31-90 OTHER PROFESSIONAL SER 246,000 247,000 1,000 0%
110-3240-422.33-10 REPAIRS & MAINTENANCE 7,400 12,000 4,600 62%
110-3240-422.33-11 AUTO REPAIRS & MAINT 664,294 691,697 27,403 4%
110-3240-422.36-01 ADVERTISING 5,000 5,000 0 0%
110-3240-422.38-53 VOL FIRE/RESCUE PER DIEMS 125,000 125,000 0 0%
110-3240-422.51-10 ELECTRICAL SERVICES 20,000 10,500 (9,500) -48%
110-3240-422.51-20 HEATING SERVICES 9,160 2,000 (7,160) -78%
110-3240-422.52-30 TELEPHONE SERVICES 115,379 115,379 0 0%
110-3240-422.52-31 SATELLITE SERVICES 20,180 20,180 0 0%
110-3240-422.53-05 MOTOR VEHICLE INSURANCE 152,966 178,495 25,529 17%
110-3240-422.53-08 GENERAL LIABILITY INS 105,855 107,972 2,117 2%
110-3240-422.55-41 TRAINING FOR VOLUNTEERS 110,270 130,270 20,000 18%
110-3240-422.58-61 FOUR FOR LIFE FUNDS 130,000 0 (130,000) -100%
110-3240-422.58-88 STATE FIRE PROGRAMS 450,000 0 (450,000) -100%
110-3240-422.60-04 MEDICAL AND LAB SUPPLIES 402,141 419,884 17,743 4%
110-3240-422.60-08 VEHICLE & EQUIPMENT FUELS 566,971 566,971 0 0%
110-3240-422.60-11 UNIFORMS 296,012 412,470 116,458 39%
110-3240-422.60-14 OPERATING SUPPLIES 87,666 60,162 (27,504) -31%
Consolidated Fire & Rescue 3,536,074 3,127,075 (408,999) -11.6%
Regional Detention Facilities:
110-3320-423.70-02 RAPPAHANNOCK SECURITY CTR 9,032,995 9,393,279 360,284 4%
110-3320-423.70-03 RAPP JUVENILE DETENTION 2,455,733 2,870,547 414,814 17%
Regional Detention Facilities 11,488,728 12,263,826 775,098 6.7%
Court Services Unit:
13 FY 2025 Adopted to FY 2026 Adopted Budget Comparison
Account Account Description FY 2025 Adopted Budget
FY 2026 Adopted Budget $ Change % Change
110-3330-423.52-30 TELEPHONE SERVICES 200 200 0 0%
110-3330-423.54-20 LEASE/RENTAL BUILDINGS 18,424 18,424 0 0%
Court Services Unit 18,624 18,624 0 0.0%
CSU - Outreach Detention:
110-3334-423.11-01 REGULAR 59,112 60,817 1,705 3%
110-3334-423.21-01 FICA 3,332 3,774 442 13%
110-3334-423.21-02 MEDICARE 779 882 103 13%
110-3334-423.22-10 RETIREMENT 8,133 8,381 248 3%
110-3334-423.23-10 HEALTH INSURANCE 0 0 0 n/a
110-3334-423.24-01 INSURANCE 1,082 1,144 62 6%
110-3334-423.27-10 SELF INSURED 35 12 (23) -66%
110-3334-423.31-90 OTHER PROFESSIONAL SER 960 960 0 0%
110-3334-423.52-10 POSTAL SERVICES 340 340 0 0%
110-3334-423.52-30 TELEPHONE SERVICES 850 1,200 350 41%
110-3334-423.55-40 EDUCATION & TRAINING 150 150 0 0%
110-3334-423.58-10 DUES & ASSOC MEMBERSHIPS 75 75 0 0%
110-3334-423.60-01 OFFICE SUPPLIES 500 500 0 0%
110-3334-423.60-08 VEHICLE & EQUIPMENT FUELS 705 705 0 0%
CSU - Outreach Detention 76,053 78,940 2,887 3.8%
CSU - VJCCCA Crime Control Programs:
110-3335-423.31-90 OTHER PROFESSIONAL SER 84,296 94,296 10,000 12%
110-3335-423.70-05 GROUP HOME COMMISSION 40,000 30,000 (10,000) -25%
CSU - VJCCCA Crime Control Programs 124,296 124,296 0 0.0%
Correction & Detention:
110-3336-423.13-01 PART-TIME 36,747 36,726 (21) 0%
110-3336-423.21-01 FICA 2,278 2,277 (1) 0%
110-3336-423.21-02 MEDICARE 533 533 0 0%
110-3336-423.27-10 SELF INSURED 22 7 (15) -68%
110-3336-423.38-59 DETENTION COST/NON RAPP 500 500 0 0%
110-3336-423.52-30 TELEPHONE SERVICES 1,080 1,080 0 0%
110-3336-423.56-60 RAPP AREA OFFICE ON YOUTH 110,806 160,664 49,858 45%
110-3336-423.58-10 DUES & ASSOC MEMBERSHIPS 150 150 0 0%
110-3336-423.58-99 RECOGNITNS/AWARDS/SYMPTHY 1,150 1,150 0 0%
110-3336-423.60-01 OFFICE SUPPLIES 250 250 0 0%
110-3336-423.70-06 CHAPLIN YOUTH CTR-LOCAL 91,622 105,283 13,661 15%
110-3336-423.80-02 FURNITURE & FIXTURES 0 0 0 n/a
Correction & Detention 245,138 308,620 63,482 25.9%
Animal Control:
110-3510-425.11-01 REGULAR 1,326,205 1,374,127 47,922 4%
110-3510-425.12-01 OVERTIME 77,462 79,915 2,453 3%
110-3510-425.12-02 ON-CALL 18,591 19,180 589 3%
110-3510-425.13-01 PART-TIME 72,298 90,834 18,536 26%
110-3510-425.21-01 FICA 89,295 94,249 4,954 6%
110-3510-425.21-02 MEDICARE 20,883 22,042 1,159 6%
110-3510-425.22-10 RETIREMENT 173,103 179,304 6,201 4%
110-3510-425.23-10 HEALTH INSURANCE 264,437 187,518 (76,919) -29%
110-3510-425.23-15 HEALTH INSURANCE (HSA) 2,400 2,400 0 0%
110-3510-425.24-01 INSURANCE 18,589 17,804 (785) -4%
110-3510-425.27-10 SELF INSURED 19,091 17,260 (1,831) -10%
110-3510-425.28-10 LINE OF DUTY BENEFITS 715 751 36 5%
110-3510-425.31-10 HEALTH SERVICES 1,000 1,000 0 0%
110-3510-425.31-77 VETERINARIAN CARE 100,000 121,000 21,000 21%
110-3510-425.33-11 AUTO REPAIRS & MAINT 42,853 43,711 858 2%
110-3510-425.33-12 RADIO REPAIRS & MAINT 300 0 (300) -100%
14 FY 2025 Adopted to FY 2026 Adopted Budget Comparison
Account Account Description FY 2025 Adopted Budget
FY 2026 Adopted Budget $ Change % Change
110-3510-425.33-20 MAINTENANCE SVC CONTRACTS 8,672 1,172 (7,500) -86%
110-3510-425.35-01 PRINTING & BINDING 2,000 2,000 0 0%
110-3510-425.36-01 ADVERTISING 1,500 3,600 2,100 140%
110-3510-425.52-10 POSTAL SERVICES 100 100 0 0%
110-3510-425.52-30 TELEPHONE SERVICES 7,908 8,092 184 2%
110-3510-425.53-05 MOTOR VEHICLE INSURANCE 5,570 5,292 (278) -5%
110-3510-425.54-10 LEASE/RENTAL EQUIPMENT 250 250 0 0%
110-3510-425.55-30 SUBSISTENCE & LODGING 3,312 3,660 348 11%
110-3510-425.55-40 EDUCATION & TRAINING 2,120 2,850 730 34%
110-3510-425.58-10 DUES & ASSOC MEMBERSHIPS 1,400 1,400 0 0%
110-3510-425.58-46 PRE-EMPLOYMENT EXPENSES 500 500 0 0%
110-3510-425.58-62 RABIES CLINIC 4,400 4,400 0 0%
110-3510-425.58-63 SPAY/NEUTER PROGRAM 49,000 49,000 0 0%
110-3510-425.60-01 OFFICE SUPPLIES 3,600 3,600 0 0%
110-3510-425.60-02 FOOD SUPPLIES & SERVICE 1,000 1,000 0 0%
110-3510-425.60-03 AGRICULTURAL/ANIMAL SUPPL 26,000 26,000 0 0%
110-3510-425.60-04 MEDICAL AND LAB SUPPLIES 3,500 3,500 0 0%
110-3510-425.60-08 VEHICLE & EQUIPMENT FUELS 66,392 66,392 0 0%
110-3510-425.60-11 UNIFORMS 6,500 3,500 (3,000) -46%
110-3510-425.60-14 OPERATING SUPPLIES 26,000 26,000 0 0%
110-3510-425.60-17 BODY ARMOR & PROTECT GEAR 0 0 0 n/a
110-3510-425.80-01 MACHINERY & EQUIPMENT 0 1,100 1,100 n/a
110-3510-425.80-07 COMPUTER EQUIPMENT 0 500 500 n/a
Animal Control 2,446,946 2,465,003 18,057 0.7%
Medical Examiner:
110-3530-425.38-45 CORONER 1,280 1,280 0 0%
Medical Examiner 1,280 1,280 0 0.0%
Facilities Management:
110-4210-431.11-01 REGULAR 491,269 536,174 44,905 9%
110-4210-431.21-01 FICA 29,465 31,924 2,459 8%
110-4210-431.21-02 MEDICARE 6,891 7,466 575 8%
110-4210-431.22-10 RETIREMENT 70,452 77,917 7,465 11%
110-4210-431.23-10 HEALTH INSURANCE 70,135 64,172 (5,963) -9%
110-4210-431.24-01 INSURANCE 8,061 8,635 574 7%
110-4210-431.27-10 SELF INSURED 7,747 3,068 (4,679) -60%
110-4210-431.31-10 HEALTH SERVICES 250 250 0 0%
110-4210-431.31-90 OTHER PROFESSIONAL SER 0 900 900 n/a
110-4210-431.33-11 AUTO REPAIRS & MAINT 1,484 1,558 74 5%
110-4210-431.35-01 PRINTING & BINDING 50 250 200 400%
110-4210-431.52-10 POSTAL SERVICES 118 258 140 119%
110-4210-431.52-30 TELEPHONE SERVICES 2,300 3,240 940 41%
110-4210-431.53-05 MOTOR VEHICLE INSURANCE 1,090 1,098 8 1%
110-4210-431.54-10 LEASE/RENTAL EQUIPMENT 0 3,000 3,000 n/a
110-4210-431.55-10 MILEAGE 1,469 229 (1,240) -84%
110-4210-431.55-30 SUBSISTENCE & LODGING 3,198 1,316 (1,882) -59%
110-4210-431.55-40 EDUCATION & TRAINING 1,215 6,225 5,010 412%
110-4210-431.58-10 DUES & ASSOC MEMBERSHIPS 1,595 1,595 0 0%
110-4210-431.58-46 PRE-EMPLOYMENT EXPENSES 50 50 0 0%
110-4210-431.58-55 INDUSTRIAL SAFETY PROGRAM 2,500 2,500 0 0%
110-4210-431.60-01 OFFICE SUPPLIES 3,500 5,500 2,000 57%
110-4210-431.60-08 VEHICLE & EQUIPMENT FUELS 3,117 3,117 0 0%
110-4210-431.60-11 UNIFORMS 2,000 1,000 (1,000) -50%
110-4210-431.60-12 BOOKS & SUBSCRIPTIONS 150 150 0 0%
110-4210-431.80-02 FURNITURE & FIXTURES 0 4,000 4,000 n/a
Facilities Management 708,106 765,592 57,486 8.1%
15 FY 2025 Adopted to FY 2026 Adopted Budget Comparison
Account Account Description FY 2025 Adopted Budget
FY 2026 Adopted Budget $ Change % Change
Refuse Collection:
110-4230-431.11-01 REGULAR 1,420,340 1,582,446 162,106 11%
110-4230-431.12-01 OVERTIME 57,108 60,095 2,987 5%
110-4230-431.12-02 ON-CALL 6,123 6,443 320 5%
110-4230-431.13-01 PART-TIME 1,047,332 1,139,519 92,187 9%
110-4230-431.21-01 FICA 154,283 169,851 15,568 10%
110-4230-431.21-02 MEDICARE 36,083 39,724 3,641 10%
110-4230-431.22-10 RETIREMENT 195,688 217,307 21,619 11%
110-4230-431.23-10 HEALTH INSURANCE 275,832 244,277 (31,555) -11%
110-4230-431.24-01 INSURANCE 22,571 25,051 2,480 11%
110-4230-431.27-10 SELF INSURED 95,328 43,935 (51,393) -54%
110-4230-431.31-10 HEALTH SERVICES 7,000 5,000 (2,000) -29%
110-4230-431.31-90 OTHER PROFESSIONAL SER 402,416 402,416 0 0%
110-4230-431.33-11 AUTO REPAIRS & MAINT 4,025 4,025 0 0%
110-4230-431.33-13 TRUCK REPAIRS & MAINT 130,000 160,000 30,000 23%
110-4230-431.33-14 HEAVY EQUIP REP & MAINT 151,000 191,000 40,000 26%
110-4230-431.33-20 MAINTENANCE SVC CONTRACTS 750 3,250 2,500 333%
110-4230-431.35-01 PRINTING & BINDING 0 1,200 1,200 n/a
110-4230-431.36-01 ADVERTISING 350 350 0 0%
110-4230-431.39-27 SANITATION SERVICES 9,261 10,200 939 10%
110-4230-431.52-10 POSTAL SERVICES 200 200 0 0%
110-4230-431.52-30 TELEPHONE SERVICES 13,524 18,804 5,280 39%
110-4230-431.53-05 MOTOR VEHICLE INSURANCE 12,575 12,240 (335) -3%
110-4230-431.54-10 LEASE/RENTAL EQUIPMENT 3,300 3,300 0 0%
110-4230-431.55-30 SUBSISTENCE & LODGING 1,260 2,496 1,236 98%
110-4230-431.55-40 EDUCATION & TRAINING 6,710 44,745 38,035 567%
110-4230-431.58-10 DUES & ASSOC MEMBERSHIPS 792 792 0 0%
110-4230-431.58-46 PRE-EMPLOYMENT EXPENSES 2,500 2,500 0 0%
110-4230-431.58-55 INDUSTRIAL SAFETY PROGRAM 21,350 28,900 7,550 35%
110-4230-431.58-64 STONE & HAULING 1,000 1,000 0 0%
110-4230-431.58-65 HOUSEHOLD HAZARD WASTE PR 40,068 42,000 1,932 5%
110-4230-431.58-72 RECYCLING OPERATIONS 131,250 131,250 0 0%
110-4230-431.60-01 OFFICE SUPPLIES 3,000 3,000 0 0%
110-4230-431.60-07 REPAIRS & MAINT SUPPLIES 18,000 24,000 6,000 33%
110-4230-431.60-08 VEHICLE & EQUIPMENT FUELS 381,672 381,672 0 0%
110-4230-431.60-09 VEHICLE & EQUIPMENT SUPP 72,089 72,089 0 0%
110-4230-431.60-11 UNIFORMS 42,980 42,980 0 0%
110-4230-431.60-14 OPERATING SUPPLIES 900 900 0 0%
110-4230-431.60-27 SMALL TOOL/EQUIP REPLCMNT 3,200 3,200 0 0%
Refuse Collection 4,771,860 5,122,157 350,297 7.3%
Refuse Disposal:
110-4240-431.11-01 REGULAR 880,711 923,187 42,476 5%
110-4240-431.12-01 OVERTIME 49,113 67,373 18,260 37%
110-4240-431.12-02 ON-CALL 12,495 8,548 (3,947) -32%
110-4240-431.13-01 PART-TIME 169,952 197,983 28,031 16%
110-4240-431.21-01 FICA 66,906 72,926 6,020 9%
110-4240-431.21-02 MEDICARE 15,647 17,056 1,409 9%
110-4240-431.22-10 RETIREMENT 112,607 131,394 18,787 17%
110-4240-431.23-10 HEALTH INSURANCE 140,415 135,538 (4,877) -3%
110-4240-431.24-01 INSURANCE 15,217 16,802 1,585 10%
110-4240-431.27-10 SELF INSURED 41,829 19,750 (22,079) -53%
110-4240-431.31-10 HEALTH SERVICES 2,396 2,396 0 0%
110-4240-431.31-30 MGT CONSULTING SERVICES 199,300 199,300 0 0%
110-4240-431.31-90 OTHER PROFESSIONAL SER 141,650 141,650 0 0%
110-4240-431.33-11 AUTO REPAIRS & MAINT 8,289 8,289 0 0%
110-4240-431.33-14 HEAVY EQUIP REP & MAINT 224,000 244,000 20,000 9%
110-4240-431.33-15 SCALE MAINTENANCE 3,500 3,500 0 0%
16 FY 2025 Adopted to FY 2026 Adopted Budget Comparison
Account Account Description FY 2025 Adopted Budget
FY 2026 Adopted Budget $ Change % Change
110-4240-431.33-20 MAINTENANCE SVC CONTRACTS 4,780 9,280 4,500 94%
110-4240-431.52-10 POSTAL SERVICES 300 300 0 0%
110-4240-431.52-30 TELEPHONE SERVICES 2,820 2,820 0 0%
110-4240-431.53-05 MOTOR VEHICLE INSURANCE 3,169 3,701 532 17%
110-4240-431.55-30 SUBSISTENCE & LODGING 4,380 1,664 (2,716) -62%
110-4240-431.55-40 EDUCATION & TRAINING 9,600 25,895 16,295 170%
110-4240-431.58-10 DUES & ASSOC MEMBERSHIPS 2,112 2,112 0 0%
110-4240-431.58-46 PRE-EMPLOYMENT EXPENSES 280 1,000 720 257%
110-4240-431.58-55 INDUSTRIAL SAFETY PROGRAM 19,000 24,730 5,730 30%
110-4240-431.58-64 STONE & HAULING 43,000 50,000 7,000 16%
110-4240-431.58-71 GROUNDWATER MONITORING 322,000 322,000 0 0%
110-4240-431.58-87 POST CLOSURE MAINTENANCE 48,000 48,000 0 0%
110-4240-431.60-01 OFFICE SUPPLIES 5,264 6,000 736 14%
110-4240-431.60-02 FOOD SUPPLIES & SERVICE 570 570 0 0%
110-4240-431.60-07 REPAIRS & MAINT SUPPLIES 31,800 31,800 0 0%
110-4240-431.60-08 VEHICLE & EQUIPMENT FUELS 278,503 278,503 0 0%
110-4240-431.60-09 VEHICLE & EQUIPMENT SUPP 30,000 30,000 0 0%
110-4240-431.60-11 UNIFORMS 21,845 25,750 3,905 18%
110-4240-431.60-27 SMALL TOOL/EQUIP REPLCMNT 2,060 2,060 0 0%
110-4240-431.60-28 ALT DAILY COVER CONSUMABL 65,000 65,000 0 0%
110-4240-431.80-07 COMPUTER EQUIPMENT 1,000 1,300 300 30%
Refuse Disposal 2,979,510 3,122,177 142,667 4.8%
Maintenance:
110-4320-432.11-01 REGULAR 800,766 854,578 53,812 7%
110-4320-432.12-01 OVERTIME 41,043 23,151 (17,892) -44%
110-4320-432.12-02 ON-CALL 17,707 18,633 926 5%
110-4320-432.13-01 PART-TIME 165,221 176,476 11,255 7%
110-4320-432.21-01 FICA 62,074 64,489 2,415 4%
110-4320-432.21-02 MEDICARE 14,517 15,082 565 4%
110-4320-432.22-10 RETIREMENT 111,916 119,697 7,781 7%
110-4320-432.23-10 HEALTH INSURANCE 142,177 133,063 (9,114) -6%
110-4320-432.24-01 INSURANCE 13,154 13,745 591 4%
110-4320-432.27-10 SELF INSURED 13,155 6,088 (7,067) -54%
110-4320-432.31-10 HEALTH SERVICES 2,500 2,500 0 0%
110-4320-432.31-90 OTHER PROFESSIONAL SER 92,840 100,900 8,060 9%
110-4320-432.33-11 AUTO REPAIRS & MAINT 15,871 15,871 0 0%
110-4320-432.33-17 INDUSTRIAL PARK EXPENSES 175,000 175,000 0 0%
110-4320-432.33-18 HVAC SYSTEM REPAIR &MAINT 273,337 250,000 (23,337) -9%
110-4320-432.33-20 MAINTENANCE SVC CONTRACTS 656,651 785,838 129,187 20%
110-4320-432.39-28 JANITORIAL SERVICES 279,250 263,819 (15,431) -6%
110-4320-432.52-30 TELEPHONE SERVICES 7,320 7,320 0 0%
110-4320-432.53-05 MOTOR VEHICLE INSURANCE 5,283 5,847 564 11%
110-4320-432.55-10 MILEAGE 284 0 (284) -100%
110-4320-432.55-30 SUBSISTENCE & LODGING 1,101 0 (1,101) -100%
110-4320-432.55-40 EDUCATION & TRAINING 6,954 6,500 (454) -7%
110-4320-432.58-10 DUE & MEMBERSHIPS 834 834 0 0%
110-4320-432.58-46 PRE-EMPLOYMENT EXPENSES 500 500 0 0%
110-4320-432.58-55 INDUSTRIAL SAFETY PROGRAM 5,000 5,000 0 0%
110-4320-432.60-03 AGRICULTURAL/ANIMAL SUPPL 8,500 8,500 0 0%
110-4320-432.60-05 JANITORIAL SUPPLIES 50,000 50,000 0 0%
110-4320-432.60-07 REPAIRS & MAINT SUPPLIES 300,800 275,000 (25,800) -9%
110-4320-432.60-08 VEHICLE & EQUIPMENT FUELS 19,864 19,864 0 0%
110-4320-432.60-09 VEHICLE & EQUIPMENT SUPP 4,578 4,378 (200) -4%
110-4320-432.60-11 UNIFORMS 16,750 17,750 1,000 6%
110-4320-432.60-14 OPERATING SUPPLIES 17,500 17,500 0 0%
110-4320-432.80-01 MACHINERY & EQUIPMENT 0 19,000 19,000 n/a
110-4320-432.80-07 COMPUTER EQUIPMENT 0 0 0 n/a
17 FY 2025 Adopted to FY 2026 Adopted Budget Comparison
Account Account Description FY 2025 Adopted Budget
FY 2026 Adopted Budget $ Change % Change
Maintenance 3,322,447 3,456,923 134,476 4.0%
General Buildings & Grounds:
110-4340-432.51-10 ELECTRICAL SERVICES 1,405,350 1,548,069 142,719 10%
110-4340-432.51-20 HEATING SERVICES 407,455 440,963 33,508 8%
110-4340-432.51-30 WATER & SEWER SERVICES 152,727 195,731 43,004 28%
110-4340-432.52-30 TELEPHONE SERVICES 268,498 250,000 (18,498) -7%
110-4340-432.53-02 PROPERTY INSURANCE 118,824 117,975 (849) -1%
110-4340-432.54-20 LEASE/RENTAL BUILDINGS 2,850 2,850 0 0%
General Buildings & Grounds 2,355,704 2,555,588 199,884 8.5%
Health Department:
110-5110-441.56-10 LOCAL HEALTH DEPARTMENT 721,324 742,964 21,640 3%
Health Department 721,324 742,964 21,640 3.0%
Rappahannock Area Comm Svcs Board (RACSB):
110-5250-442.56-20 RAPP AREA COMM SVCS BOARD 682,605 682,605 0 0%
Rappahannock Area Comm Svcs Board (RACSB) 682,605 682,605 0 0.0%
Social Services:
110-5310-443.11-01 REGULAR 9,694,561 10,745,802 1,051,241 11%
110-5310-443.12-01 OVERTIME 235,389 247,702 12,313 5%
110-5310-443.12-02 ON-CALL 69,278 65,243 (4,035) -6%
110-5310-443.13-01 PART-TIME 157,265 271,658 114,393 73%
110-5310-443.14-01 BOARDS & COMMISSIONS 38,786 30,533 (8,253) -21%
110-5310-443.21-01 FICA 614,062 678,802 64,740 11%
110-5310-443.21-02 MEDICARE 143,669 160,187 16,518 11%
110-5310-443.22-10 RETIREMENT 1,381,401 1,522,518 141,117 10%
110-5310-443.23-10 HEALTH INSURANCE 1,419,564 1,376,541 (43,023) -3%
110-5310-443.23-15 HEALTH INSURANCE (HSA) 7,500 10,800 3,300 44%
110-5310-443.24-01 INSURANCE 167,045 181,888 14,843 9%
110-5310-443.27-10 SELF INSURED 34,875 15,462 (19,413) -56%
110-5310-443.28-98 PERSONNEL BUDGET REDUCTIO (525,000) (934,764) (409,764) 78%
110-5310-443.31-50 LEGAL SERVICES 200,000 135,000 (65,000) -33%
110-5310-443.31-90 OTHER PROFESSIONAL SER 109,850 109,850 0 0%
110-5310-443.33-10 REPAIRS & MAINTENANCE 500 500 0 0%
110-5310-443.33-11 AUTO REPAIRS & MAINT 17,859 17,871 12 0%
110-5310-443.33-20 MAINTENANCE SVC CONTRACTS 12,164 18,764 6,600 54%
110-5310-443.35-01 PRINTING & BINDING 6,800 6,800 0 0%
110-5310-443.36-01 ADVERTISING 425 5,000 4,575 1076%
110-5310-443.38-10 VA DEPT OF HEALTH 4,110 4,110 0 0%
110-5310-443.39-28 JANITORIAL SERVICES 78,150 79,380 1,230 2%
110-5310-443.51-10 ELECTRICAL SERVICES 55,580 60,817 5,237 9%
110-5310-443.52-10 POSTAL SERVICES 20,000 20,500 500 3%
110-5310-443.52-30 TELEPHONE SERVICES 77,000 77,440 440 1%
110-5310-443.53-05 MOTOR VEHICLE INSURANCE 6,643 6,389 (254) -4%
110-5310-443.54-10 LEASE/RENTAL EQUIPMENT 3,316 3,316 0 0%
110-5310-443.54-20 LEASE/RENTAL BUILDINGS 0 0 0 n/a
110-5310-443.55-10 MILEAGE 20,322 12,217 (8,105) -40%
110-5310-443.55-30 SUBSISTENCE & LODGING 35,838 94,630 58,792 164%
110-5310-443.55-31 DAY MEALS DSS STAFF ONLY 0 3,000 3,000 n/a
110-5310-443.55-40 EDUCATION & TRAINING 50,055 72,505 22,450 45%
110-5310-443.56-01 NEW REGIONAL AGENCY REQ 502,943 20,000 (482,943) -96%
110-5310-443.56-15 MICAH MINISTRIES 42,500 52,500 10,000 24%
110-5310-443.56-52 COUNCIL ON DOM VIOLENCE 81,049 81,049 0 0%
110-5310-443.56-53 RAPP REFUGE HOPE HOUSE 20,000 45,000 25,000 125%
110-5310-443.56-55 AREA AGENCY ON AGING 27,642 32,642 5,000 18%
110-5310-443.56-61 BRISBEN HOMELESS SHELTER 50,000 40,000 (10,000) -20%
18 FY 2025 Adopted to FY 2026 Adopted Budget Comparison
Account Account Description FY 2025 Adopted Budget
FY 2026 Adopted Budget $ Change % Change
110-5310-443.56-62 RAPPAHANNOCK CASA 22,000 22,000 0 0%
110-5310-443.56-64 RAPP BIG BROTHERS/SISTERS 3,000 3,300 300 10%
110-5310-443.56-66 RAPP UN WAY VOL/INFO PRG 0 0 0 n/a
110-5310-443.56-69 FBG AREA FOOD RELIEF 12,000 25,000 13,000 108%
110-5310-443.56-71 SECA 13,250 13,250 0 0%
110-5310-443.56-77 MOSS FREE CLINIC 33,150 33,150 0 0%
110-5310-443.56-82 HEALTHY FAMILIES RAPP 10,000 10,000 0 0%
110-5310-443.56-88 MENTAL HEALTH ASSOC FRED 31,083 31,083 0 0%
110-5310-443.56-93 DISABILITY RESOURCE CTR 30,100 34,766 4,666 16%
110-5310-443.57-02 AUXILIARY GRANTS AGED 40,713 95,500 54,787 135%
110-5310-443.57-04 AUXILIARY GRANTS DISABLED 82,660 68,000 (14,660) -18%
110-5310-443.57-05 AID DEP CHILD/TANF MAN CK 2,000 2,000 0 0%
110-5310-443.57-06 AID DEP CHLDN-FOSTER CARE 477,505 522,072 44,567 9%
110-5310-443.57-07 EMERGENCY ASSISTANCE 11,500 11,500 0 0%
110-5310-443.57-09 REFUGEE ASSISTANCE 56,282 67,538 11,256 20%
110-5310-443.57-10 DAY CARE/VIEW PURCH SERVS 283,386 173,180 (110,206) -39%
110-5310-443.57-11 OTHER PURCHASED SERVICES 10,000 5,000 (5,000) -50%
110-5310-443.57-14 SPECIAL NEEDS ADOPTIONS 308,205 320,000 11,795 4%
110-5310-443.57-15 ADOPTION SUBSIDY PYMTS 2,999,310 2,876,048 (123,262) -4%
110-5310-443.57-16 ADULT SERV/HOME BASE COMP 32,000 29,500 (2,500) -8%
110-5310-443.57-17 ADULT PROTECTIVE SERVICES 15,530 15,530 0 0%
110-5310-443.57-23 RESPITE PROGRAM 6,100 9,000 2,900 48%
110-5310-443.57-25 CHILD WELFARE 15,012 15,012 0 0%
110-5310-443.57-34 FAMILY PRESERVATION 12,600 12,600 0 0%
110-5310-443.57-35 FAMILY SUPPORT SERVICES 92,310 87,698 (4,612) -5%
110-5310-443.57-39 PROVIDER TRAINING 53,765 79,765 26,000 48%
110-5310-443.57-42 SPOTSY INDEPENDENT LIVING 20,152 18,800 (1,352) -7%
110-5310-443.57-44 KINSHIP GUARDIAN PROGRAM 20,980 0 (20,980) -100%
110-5310-443.58-10 DUES & ASSOC MEMBERSHIPS 8,355 8,905 550 7%
110-5310-443.58-38 SAFE HARBOR 7,586 10,200 2,614 34%
110-5310-443.58-99 RECOGNITNS/AWARDS/SYMPTHY 3,000 3,000 0 0%
110-5310-443.60-01 OFFICE SUPPLIES 61,000 51,000 (10,000) -16%
110-5310-443.60-02 FOOD SUPPLIES & SERVICE 7,154 21,920 14,766 206%
110-5310-443.60-08 VEHICLE & EQUIPMENT FUELS 12,968 12,968 0 0%
110-5310-443.60-12 BOOKS & SUBSCRIPTIONS 1,850 2,050 200 11%
110-5310-443.60-14 OPERATING SUPPLIES 18,000 13,000 (5,000) -28%
110-5310-443.80-01 MACHINERY & EQUIPMENT 8,000 0 (8,000) -100%
110-5310-443.80-02 FURNITURE & FIXTURES 21,500 5,000 (16,500) -77%
110-5310-443.80-03 COMMUNICATION EQUIPMENT 0 0 0 n/a
110-5310-443.80-05 MOTOR VEHICLES & EQUIP 90,000 150,000 60,000 67%
110-5310-443.80-07 COMPUTER EQUIPMENT 1,000 855 (145) -15%
110-5310-443.80-08 COMPUTER EQUIP - STATE 0 12,860 12,860 n/a
Social Services 19,796,147 20,244,702 448,555 2.3%
Children's Services Act (CSA)
110-5360-443.11-01 REGULAR 142,666 221,470 78,804 55%
110-5360-443.21-01 FICA 8,503 13,522 5,019 59%
110-5360-443.21-02 MEDICARE 1,988 3,162 1,174 59%
110-5360-443.22-10 RETIREMENT 19,731 30,421 10,690 54%
110-5360-443.23-10 HEALTH INSURANCE 25,785 22,755 (3,030) -12%
110-5360-443.24-01 INSURANCE 2,307 3,722 1,415 61%
110-5360-443.27-10 SELF INSURED 86 44 (42) -49%
110-5360-443.31-90 OTHER PROFESSIONAL SER 30,000 30,000 0 0%
110-5360-443.33-20 MAINTENANCE CONTRACTS 750 1,550 800 107%
110-5360-443.35-01 PRINTING & BINDING 1,200 700 (500) -42%
110-5360-443.39-10 SOFTWARE APPLICATIONS 3,600 3,600 0 0%
110-5360-443.52-30 TELEPHONE SERVICES 0 2,080 2,080 n/a
110-5360-443.55-10 MILEAGE 198 201 3 2%
19 FY 2025 Adopted to FY 2026 Adopted Budget Comparison
Account Account Description FY 2025 Adopted Budget
FY 2026 Adopted Budget $ Change % Change
110-5360-443.55-30 SUBSISTENCE & LODGING 2,715 6,969 4,254 157%
110-5360-443.55-40 EDUCATION & TRAINING 1,150 3,175 2,025 176%
110-5360-443.57-30 CSA MANDATED 12,200,000 11,200,000 (1,000,000) -8%
110-5360-443.57-31 CSA NON-MANDATED 59,000 29,000 (30,000) -51%
110-5360-443.60-01 OFFICE SUPPLIES 4,180 5,180 1,000 24%
110-5360-443.60-02 FOOD SUPPLIES & SERVICE 2,678 4,973 2,295 86%
110-5360-443.80-02 FURNITURE & FIXTURES 1,200 1,200 0 0%
110-5360-443.80-07 COMPUTER EQUIPMENT 0 0 0 n/a
110-5360-443.80-08 COMPUTER EQUIP - STATE 0 1,900 1,900 n/a
Children's Services Act (CSA) 12,507,737 11,585,624 (922,113) -7.4%
Tax Relief:
110-5380-445.58-86 TAX RELIEF FOR ELDERLY 1,400,000 1,400,000 0 0%
Tax Relief 1,400,000 1,400,000 0 0.0%
Regional Agencies:
110-5390-444.33-20 MAINTENANCE SVC CONTRACTS 1,674 1,674 0 0%
110-5390-444.56-01 NEW REGIONAL AGENCY REQ 12,000 13,000 1,000 8%
110-5390-444.56-58 GWRC 107,324 108,579 1,255 1%
110-5390-444.56-91 LAKE ANNA CIVIC ASSN 5,000 10,000 5,000 100%
Regional Agencies 125,998 133,253 7,255 5.8%
Germanna Community College:
110-6810-451.56-49 GERMANNA COMM COLLEGE 89,171 89,171 0 0%
Germanna Community College 89,171 89,171 0 0.0%
Parks & Recreation:
110-7110-461.11-01 REGULAR 1,803,943 1,960,765 156,822 9%
110-7110-461.12-01 OVERTIME 43,743 56,554 12,811 29%
110-7110-461.12-02 ON-CALL 8,424 427 (7,997) -95%
110-7110-461.13-01 PART-TIME 587,983 718,559 130,576 22%
110-7110-461.21-01 FICA 147,300 165,219 17,919 12%
110-7110-461.21-02 MEDICARE 34,449 38,640 4,191 12%
110-7110-461.22-10 RETIREMENT 247,413 267,937 20,524 8%
110-7110-461.23-10 HEALTH INSURANCE 223,418 207,182 (16,236) -7%
110-7110-461.23-15 HEALTH INSURANCE (HSA) 1,200 0 (1,200) -100%
110-7110-461.24-01 INSURANCE 27,445 28,037 592 2%
110-7110-461.27-10 SELF INSURED 38,101 16,968 (21,133) -55%
110-7110-461.31-10 HEALTH SERVICES 2,500 2,500 0 0%
110-7110-461.31-90 OTHER PROFESSIONAL SER 360,661 48,600 (312,061) -87%
110-7110-461.33-10 REPAIRS & MAINTENANCE 13,430 14,820 1,390 10%
110-7110-461.33-11 AUTO REPAIRS & MAINT 33,835 32,220 (1,615) -5%
110-7110-461.33-20 MAINTENANCE SVC CONTRACTS 7,055 8,870 1,815 26%
110-7110-461.33-27 WEED/DBRIS REMVL ORDINANC 15,000 15,000 0 0%
110-7110-461.35-01 PRINTING & BINDING 8,860 8,860 0 0%
110-7110-461.36-01 ADVERTISING 11,800 10,400 (1,400) -12%
110-7110-461.39-27 SANITATION SERVICES 17,820 19,095 1,275 7%
110-7110-461.52-10 POSTAL SERVICES 2,054 2,054 0 0%
110-7110-461.52-30 TELEPHONE SERVICES 57,136 65,207 8,071 14%
110-7110-461.53-05 MOTOR VEHICLE INSURANCE 6,869 6,838 (31) 0%
110-7110-461.54-10 LEASE/RENTAL EQUIPMENT 6,765 6,765 0 0%
110-7110-461.55-10 MILEAGE 106 107 1 1%
110-7110-461.55-30 SUBSISTENCE & LODGING 4,416 3,942 (474) -11%
110-7110-461.55-40 EDUCATION & TRAINING 26,674 33,260 6,586 25%
110-7110-461.58-10 DUES & ASSOC MEMBERSHIPS 2,520 2,410 (110) -4%
110-7110-461.58-46 PRE-EMPLOYMENT EXPENSES 6,870 4,900 (1,970) -29%
110-7110-461.58-55 INDUSTRIAL SAFETY PROGRAM 30,840 51,380 20,540 67%
110-7110-461.58-59 PUBLIC EDUCATION 7,000 0 (7,000) -100%
20 FY 2025 Adopted to FY 2026 Adopted Budget Comparison
Account Account Description FY 2025 Adopted Budget
FY 2026 Adopted Budget $ Change % Change
110-7110-461.58-80 SELF SUPPORTING PROGRAMS 113,854 119,158 5,304 5%
110-7110-461.58-83 SENIOR CITIZEN/TEEN CENTR 3,500 3,500 0 0%
110-7110-461.58-90 SPORTS PROGRAMS 294,841 331,231 36,390 12%
110-7110-461.60-01 OFFICE SUPPLIES 18,700 24,500 5,800 31%
110-7110-461.60-02 FOOD SUPPLIES & SERVICE 3,000 1,450 (1,550) -52%
110-7110-461.60-05 JANITORIAL SUPPLIES 24,041 15,916 (8,125) -34%
110-7110-461.60-07 REPAIRS & MAINT SUPPLIES 178,810 196,980 18,170 10%
110-7110-461.60-08 VEHICLE & EQUIPMENT FUELS 53,245 53,245 0 0%
110-7110-461.60-09 VEHICLE & EQUIPMENT SUPP 8,980 11,855 2,875 32%
110-7110-461.60-11 UNIFORMS 34,507 16,470 (18,037) -52%
110-7110-461.60-12 BOOKS & SUBSCRIPTIONS 410 410 0 0%
110-7110-461.60-13 RECREATION SUPPLIES 42,290 29,845 (12,445) -29%
110-7110-461.60-14 OPERATING SUPPLIES 367,620 382,805 15,185 4%
110-7110-461.80-01 MACHINERY & EQUIPMENT 96,570 191,480 94,910 98%
110-7110-461.80-02 FURNITURE & FIXTURES 54,060 40,975 (13,085) -24%
110-7110-461.80-07 COMPUTER EQUIPMENT 1,600 1,600 0 0%
110-7110-461.80-10 CAPITAL IMPROVEMENTS 117,660 0 (117,660) -100%
Parks & Recreation 5,199,318 5,218,936 19,618 0.4%
Museum:
110-7220-462.13-01 PART-TIME 37,944 42,805 4,861 13%
110-7220-462.21-01 FICA 2,353 2,654 301 13%
110-7220-462.21-02 MEDICARE 550 621 71 13%
110-7220-462.27-10 SELF INSURED 23 9 (14) -61%
110-7220-462.31-90 OTHER PROFESSIONAL SER 4,580 300 (4,280) -93%
110-7220-462.33-11 AUTO REPAIRS & MAINT 2,788 1,922 (866) -31%
110-7220-462.35-01 PRINTING & BINDING 6,000 6,000 0 0%
110-7220-462.36-01 ADVERTISING 2,000 2,000 0 0%
110-7220-462.39-10 SOFTWARE APPLICATIONS 1,183 1,348 165 14%
110-7220-462.53-05 MOTOR VEHICLE INSURANCE 446 228 (218) -49%
110-7220-462.58-10 DUES & ASSOC MEMBERSHIPS 750 1,168 418 56%
110-7220-462.60-01 OFFICE SUPPLIES 1,700 1,700 0 0%
110-7220-462.60-08 VEHICLE & EQUIPMENT FUELS 45 45 0 0%
110-7220-462.60-14 OPERATING SUPPLIES 2,000 2,000 0 0%
110-7220-462.80-01 MACHINERY & EQUIPMENT FUELS 15,000 31,744 16,744 112%
Museum 77,362 94,544 17,182 22.2%
Regional Library:
110-7320-463.70-04 RAPPAHANNOCK REGIONAL LIB 4,307,827 4,774,352 466,525 11%
Regional Library 4,307,827 4,774,352 466,525 10.8%
Planning:
110-8110-471.11-01 REGULAR 1,149,497 1,446,998 297,501 26%
110-8110-471.13-01 PART-TIME 35,971 0 (35,971) -100%
110-8110-471.21-01 FICA 68,668 84,326 15,658 23%
110-8110-471.21-02 MEDICARE 16,631 20,464 3,833 23%
110-8110-471.22-10 RETIREMENT 157,122 198,746 41,624 26%
110-8110-471.23-10 HEALTH INSURANCE 147,199 154,140 6,941 5%
110-8110-471.23-15 HEALTH INSURANCE (HSA) 1,200 2,400 1,200 100%
110-8110-471.24-01 INSURANCE 17,222 21,318 4,096 24%
110-8110-471.27-10 SELF INSURED 712 289 (423) -59%
110-8110-471.31-90 OTHER PROFESSIONAL SER 500 207,600 207,100 41420%
110-8110-471.33-11 AUTO REPAIRS & MAINT 2,688 2,688 0 0%
110-8110-471.33-20 MAINTENANCE SVC CONTRACTS 0 0 0 n/a
110-8110-471.35-01 PRINTING & BINDING 7,000 10,000 3,000 43%
110-8110-471.36-01 ADVERTISING 43,000 43,000 0 0%
110-8110-471.39-28 JANITORIAL SERVICES 18,100 19,089 989 5%
110-8110-471.51-10 ELECTRICAL SERVICES 13,460 14,680 1,220 9%
21 FY 2025 Adopted to FY 2026 Adopted Budget Comparison
Account Account Description FY 2025 Adopted Budget
FY 2026 Adopted Budget $ Change % Change
110-8110-471.52-10 POSTAL SERVICES 37,000 32,000 (5,000) -14%
110-8110-471.52-30 TELEPHONE SERVICES 2,663 2,663 0 0%
110-8110-471.53-05 MOTOR VEHICLE INSURANCE 0 453 453 n/a
110-8110-471.54-10 LEASE/RENTAL EQUIPMENT 15,500 15,500 0 0%
110-8110-471.55-10 MILEAGE 2,394 2,937 543 23%
110-8110-471.55-30 SUBSISTENCE & LODGING 4,879 3,508 (1,371) -28%
110-8110-471.55-40 EDUCATION & TRAINING 6,855 5,905 (950) -14%
110-8110-471.58-10 DUES & ASSOC MEMBERSHIPS 4,465 4,465 0 0%
110-8110-471.58-99 RECOGNITNS/AWARDS/SYMPTHY 0 500 500 n/a
110-8110-471.60-01 OFFICE SUPPLIES 7,100 7,100 0 0%
110-8110-471.60-08 VEHICLE & EQUIPMENT FUELS 864 864 0 0%
110-8110-471.60-12 BOOKS & SUBSCRIPTIONS 483 483 0 0%
Planning 1,761,173 2,302,116 540,943 30.7%
Planning Commission:
110-8120-471.14-01 BOARDS & COMMISSIONS 52,500 52,500 0 0%
110-8120-471.21-01 FICA 3,263 3,263 0 0%
110-8120-471.21-02 MEDICARE 763 763 0 0%
110-8120-471.35-01 PRINTING & BINDING 300 400 100 33%
110-8120-471.55-10 MILEAGE 1,096 1,113 17 2%
110-8120-471.55-30 SUBSISTENCE & LODGING 0 1,896 1,896 n/a
110-8120-471.55-40 EDUCATION & TRAINING 0 2,000 2,000 n/a
110-8120-471.58-40 MEETING EXPENSES 500 1,000 500 100%
110-8120-471.58-99 RECOGNITNS/AWARDS/SYMPTHY 100 100 0 0%
110-8120-471.60-01 OFFICE SUPPLIES 100 500 400 400%
Planning Commission 58,622 63,535 4,913 8.4%
Planning Comm/Committees:
110-8130-471.55-10 MILEAGE 647 1,005 358 55%
110-8130-471.55-30 SUBSISTENCE & LODGING 0 1,424 1,424 n/a
110-8130-471.55-40 EDUCATION & TRAINING 0 800 800 n/a
110-8130-471.58-40 MEETING EXPENSES 0 200 200 n/a
110-8130-471.60-01 OFFICE SUPPLIES 0 100 100 n/a
Planning Comm/Committees 647 3,529 2,882 445.4%
Zoning Division:
110-8140-471.11-01 REGULAR 621,169 634,465 13,296 2%
110-8140-471.14-01 BOARDS & COMMISSIONS 8,400 8,400 0 0%
110-8140-471.21-01 FICA 37,681 38,254 573 2%
110-8140-471.21-02 MEDICARE 8,813 8,946 133 2%
110-8140-471.22-10 RETIREMENT 88,871 89,776 905 1%
110-8140-471.23-10 HEALTH INSURANCE 93,505 91,743 (1,762) -2%
110-8140-471.23-15 HEALTH INSURANCE (HSA) 2,400 0 (2,400) -100%
110-8140-471.23-20 RETIREES 30,000 0 (30,000) -100%
110-8140-471.24-01 INSURANCE 9,804 9,592 (212) -2%
110-8140-471.27-10 SELF INSURED 6,400 2,015 (4,385) -69%
110-8140-471.31-90 OTHER PROFESSIONAL SER 6,000 6,000 0 0%
110-8140-471.33-11 AUTO REPAIRS & MAINT 3,028 3,185 157 5%
110-8140-471.35-01 PRINTING & BINDING 1,450 1,450 0 0%
110-8140-471.36-01 ADVERTISING 4,700 4,700 0 0%
110-8140-471.39-28 JANITORIAL SERVICES 11,200 10,080 (1,120) -10%
110-8140-471.51-10 ELECTRICAL SERVICES 6,750 7,340 590 9%
110-8140-471.52-10 POSTAL SERVICES 2,000 2,000 0 0%
110-8140-471.52-30 TELEPHONE SERVICES 6,852 6,852 0 0%
110-8140-471.53-05 MOTOR VEHICLE INSURANCE 1,289 1,098 (191) -15%
110-8140-471.54-10 LEASE/RENTAL EQUIPMENT 5,591 5,591 0 0%
110-8140-471.55-10 MILEAGE 2,375 2,351 (24) -1%
110-8140-471.55-30 SUBSISTENCE & LODGING 4,344 1,278 (3,066) -71%
22 FY 2025 Adopted to FY 2026 Adopted Budget Comparison
Account Account Description FY 2025 Adopted Budget
FY 2026 Adopted Budget $ Change % Change
110-8140-471.55-40 EDUCATION & TRAINING 2,400 2,600 200 8%
110-8140-471.58-10 DUES & ASSOC MEMBERSHIPS 1,412 1,412 0 0%
110-8140-471.60-01 OFFICE SUPPLIES 3,000 3,000 0 0%
110-8140-471.60-08 VEHICLE & EQUIPMENT FUELS 2,979 2,979 0 0%
110-8140-471.60-11 UNIFORMS 2,000 2,000 0 0%
110-8140-471.60-12 BOOKS & SUBSCRIPTIONS 460 460 0 0%
110-8140-471.60-14 OPERATING SUPPLIES 181 181 0 0%
110-8140-471.80-01 MACHINERY & EQUIPMENT 500 0 (500) -100%
110-8140-471.80-07 COMPUTER EQUIPMENT 1,500 0 (1,500) -100%
Zoning Division 977,054 947,748 (29,306) -3.0%
Economic Development:
110-8150-471.11-01 REGULAR 496,141 473,655 (22,486) -5%
110-8150-471.13-01 PART-TIME 0 17,363 17,363 n/a
110-8150-471.21-01 FICA 29,457 29,195 (262) -1%
110-8150-471.21-02 MEDICARE 6,889 6,828 (61) -1%
110-8150-471.22-10 RETIREMENT 65,797 62,180 (3,617) -5%
110-8150-471.23-10 HEALTH INSURANCE 65,958 53,549 (12,409) -19%
110-8150-471.24-01 INSURANCE 7,220 5,933 (1,287) -18%
110-8150-471.27-10 SELF INSURED 298 98 (200) -67%
110-8150-471.33-11 AUTO REPAIRS & MAINT 1,550 1,545 (5) 0%
110-8150-471.33-20 MAINTENANCE SVC CONTRACTS 0 0 0 n/a
110-8150-471.35-01 PRINTING & BINDING 4,000 0 (4,000) -100%
110-8150-471.39-10 SOFTWARE APPLICATIONS 28,380 19,940 (8,440) -30%
110-8150-471.39-28 JANITORIAL SERVICES 12,600 12,600 0 0%
110-8150-471.52-10 POSTAL SERVICES 200 200 0 0%
110-8150-471.52-30 TELEPHONE SERVICES 4,600 4,000 (600) -13%
110-8150-471.53-05 MOTOR VEHICLE INSURANCE 644 657 13 2%
110-8150-471.54-10 LEASE/ RENTAL EQUIPMENT 3,410 3,410 0 0%
110-8150-471.55-10 MILEAGE 0 101 101 n/a
110-8150-471.56-17 BAY CNSRTIUM WKFC DEV BD 35,919 37,397 1,478 4%
110-8150-471.56-76 FRED REGIONAL ALLIANCE 135,000 135,000 0 0%
110-8150-471.58-10 DUES & ASSOC MEMBERSHIPS 2,500 2,600 100 4%
110-8150-471.58-76 PROSPECT DEVELOPMENT 73,000 50,000 (23,000) -32%
110-8150-471.58-79 REGIONAL MARKETING 7,700 5,000 (2,700) -35%
110-8150-471.60-01 OFFICE SUPPLIES 5,000 4,000 (1,000) -20%
110-8150-471.60-08 VEHICLE & EQUIPMENT FUELS 269 269 0 0%
110-8150-471.60-12 BOOKS & SUBSCRIPTIONS 300 300 0 0%
Economic Development 986,832 925,820 (61,012) -6.2%
Tourism:
110-8160-471.31-90 OTHER PROFESSIONAL SER 3,696 5,000 1,304 35%
110-8160-471.35-01 PRINTING & BINDING 2,408 2,408 0 0%
110-8160-471.36-01 ADVERTISING 126,000 201,002 75,002 60%
110-8160-471.52-10 POSTAL SERVICES 100 100 0 0%
110-8160-471.53-05 MOTOR VEHICLE INSURANCE 0 661 661 n/a
110-8160-471.55-10 MILEAGE 0 279 279 n/a
110-8160-471.55-30 SUBSISTENCE & LODGING 4,034 6,648 2,614 65%
110-8160-471.55-40 EDUCATION & TRAINING 3,000 2,677 (323) -11%
110-8160-471.56-02 CONFEDERATE CEMETARY 4,800 4,800 0 0%
110-8160-471.56-85 REGIONAL TOURISM 74,000 74,000 0 0%
110-8160-471.58-10 DUES & ASSOC MEMBERSHIPS 2,505 2,505 0 0%
110-8160-471.58-73 DISPLAYS 3,000 1,500 (1,500) -50%
110-8160-471.58-74 CIVIL WAR TRAILS 1,800 1,800 0 0%
110-8160-471.60-08 VEHICLE & EQUIPMENT FUELS 535 535 0 0%
110-8160-471.60-12 BOOKS & SUBSCRIPTIONS 7,220 16,068 8,848 123%
110-8160-471.60-15 MERCHANDISE FOR RESALE 8,000 5,000 (3,000) -38%
Tourism 241,098 324,983 83,885 34.8%
23 FY 2025 Adopted to FY 2026 Adopted Budget Comparison
Account Account Description FY 2025 Adopted Budget
FY 2026 Adopted Budget $ Change % Change
Visitor Center:
110-8162-471.11-01 REGULAR 219,501 230,704 11,203 5%
110-8162-471.13-01 PART-TIME 166,182 197,259 31,077 19%
110-8162-471.21-01 FICA 23,242 25,833 2,591 11%
110-8162-471.21-02 MEDICARE 5,436 6,041 605 11%
110-8162-471.22-10 RETIREMENT 29,913 33,341 3,428 11%
110-8162-471.23-10 HEALTH INSURANCE 46,161 37,385 (8,776) -19%
110-8162-471.24-01 INSURANCE 3,619 3,718 99 3%
110-8162-471.27-10 SELF INSURED 231 85 (146) -63%
110-8162-471.33-20 MAINTENANCE SVC CONTRACTS 2,200 2,200 0 0%
110-8162-471.35-01 PRINTING & BINDING 3,000 3,000 0 0%
110-8162-471.51-10 ELECTRICAL SERVICES 4,290 4,495 205 5%
110-8162-471.51-20 HEATING SERVICES 3,727 3,727 0 0%
110-8162-471.51-30 WATER & SEWER SERVICES 0 0 0 n/a
110-8162-471.52-10 POSTAL SERVICES 2,000 2,000 0 0%
110-8162-471.52-30 TELEPHONE SERVICES 4,569 3,451 (1,118) -24%
110-8162-471.53-05 MOTOR VEHICLE INSURANCE 0 0 0 n/a
110-8162-471.54-10 LEASE/RENTAL EQUIPMENT 5,924 5,924 0 0%
110-8162-471.58-73 DISPLAYS 500 500 0 0%
110-8162-471.60-01 OFFICE SUPPLIES 3,500 3,500 0 0%
110-8162-471.60-11 UNIFORMS 1,300 1,300 0 0%
110-8162-471.60-12 BOOKS & SUBSCRIPTIONS 500 500 0 0%
110-8162-471.80-07 COMPUTER EQUIPMENT 900 0 (900) -100%
Visitor Center 526,695 564,963 38,268 7.3%
Tourism Projects - ED:
110-8163-471.58-73 DISPLAYS 5,250 6,230 980 19%
110-8163-471.58-75 TOURISM/HERITAGE PROJS 36,000 36,000 0 0%
110-8163-471.58-99 RECOGNITNS/AWARDS/SYMPTHY 500 0 (500) -100%
Tourism Projects - ED 41,750 42,230 480 1.1%
Tourism Projects - P&R:
110-8164-471.56-96 FILM FESTIVAL 3,000 3,000 0 0%
110-8164-471.58-75 TOURISM/HERITAGE PROJS 750 750 0 0%
110-8164-471.58-89 4TH JULY FIREWORKS FESTIV 81,500 81,500 0 0%
Tourism Projects - P&R 85,250 85,250 0 0.0%
Community Engagement:
110-8165-471.11-01 REGULAR 324,043 344,288 20,245 6%
110-8165-471.21-01 FICA 19,697 20,951 1,254 6%
110-8165-471.21-02 MEDICARE 4,607 4,900 293 6%
110-8165-471.22-10 RETIREMENT 47,110 50,050 2,940 6%
110-8165-471.23-10 HEALTH INSURANCE 28,344 25,954 (2,390) -8%
110-8165-471.24-01 INSURANCE 5,082 5,157 75 1%
110-8165-471.27-10 SELF INSURED 194 69 (125) -64%
110-8165-471.31-90 OTHER PROFESSIONAL SER 8,000 0 (8,000) -100%
110-8165-471.33-20 MAINTENANCE SVC CONTRACTS 66,620 96,195 29,575 44%
110-8165-471.39-10 SOFTWARE APPLICATIONS 7,750 777 (6,973) -90%
110-8165-471.52-30 TELEPHONE SERVICES 3,024 3,024 0 0%
110-8165-471.55-40 EDUCATION & TRAINING 0 0 0 n/a
110-8165-471.60-01 OFFICE SUPPLIES 1,500 1,500 0 0%
110-8165-471.60-11 UNIFORMS 0 400 400 n/a
Community Engagement 515,971 553,265 37,294 7.2%
Extension Office:
110-8350-473.11-01 REGULAR 38,113 39,959 1,846 5%
110-8350-473.13-01 PART-TIME 34,836 0 (34,836) -100%
24 FY 2025 Adopted to FY 2026 Adopted Budget Comparison
Account Account Description FY 2025 Adopted Budget
FY 2026 Adopted Budget $ Change % Change
110-8350-473.21-01 FICA 4,524 2,479 (2,045) -45%
110-8350-473.21-02 MEDICARE 1,058 580 (478) -45%
110-8350-473.22-10 RETIREMENT 5,184 5,449 265 5%
110-8350-473.23-10 HEALTH INSURANCE 0 0 0 n/a
110-8350-473.24-01 INSURANCE 690 744 54 8%
110-8350-473.27-10 SELF INSURED 44 8 (36) -82%
110-8350-473.31-90 OTHER PROFESSIONAL SER 108,892 116,095 7,203 7%
110-8350-473.52-10 POSTAL SERVICES 1,250 1,250 0 0%
110-8350-473.52-30 TELEPHONE SERVICES 625 625 0 0%
110-8350-473.56-59 4-H EDUCATIONAL CENTER 2,900 2,900 0 0%
110-8350-473.60-01 OFFICE SUPPLIES 1,500 4,200 2,700 180%
Extension Office 199,616 174,289 (25,327) -12.7%
General Fund Non Departmental:
110-9110-491.23-10 HEALTH INSURANCE (720,000) 0 720,000 -100%
110-9110-491.23-20 RETIREES 2,132,000 2,718,155 586,155 27%
110-9110-491.23-25 COBRA IN EXCESS OF CONTRI 10,000 45,000 35,000 350%
110-9110-491.23-35 HCA - PCORI FEE 7,560 8,392 832 11%
110-9110-491.23-40 RETIREE HRA ADMIN FEES 720 660 (60) -8%
110-9110-491.23-45 RETIREE HRA STIPENDS 167,198 175,661 8,463 5%
110-9110-491.24-03 OPEB COSTS 3,409,526 3,438,246 28,720 1%
110-9110-491.28-98 PERSONNEL BUDGET REDUCTIO (948,000) (1,538,236) (590,236) 62%
110-9110-491.31-82 BANK SERVICE CHARGES 65,000 65,000 0 0%
110-9110-491.59-01 CONTINGENCY 506,121 557,811 51,690 10%
110-9110-491.59-10 OTHER / EMPL INCENTIVE/SALRY INCR 0 1,577,831 1,577,831 n/a
110-9110-491.59-15 OPERATING RESERVE 912,494 480,179 (432,315) -47%
General Fund Non Departmental 5,542,619 7,528,699 1,986,080 35.8%
General Fund Transfers:
110-9210-492.99-21 TO SCHOOL OPERATING FUND 155,518,759 170,849,521 15,330,762 10%
110-9210-492.99-22 TO GRANTS & OTH ASST FUND 0 284,747 284,747 n/a
110-9210-492.99-23 TO EDO FUND 760,000 632,000 (128,000) -17%
110-9210-492.99-24 TO LEE HILL TENANT BLDING 0 331,110 331,110 n/a
110-9210-492.99-26 TO CODE COMPLIANCE FUND 759,263 851,805 92,542 12%
110-9210-492.99-28 TO TRANSPORTATION FUND 3,312,421 4,050,464 738,043 22%
110-9210-492.99-31 TO CAPITAL PROJECTS 18,147,381 15,071,685 (3,075,696) -17%
110-9210-492.99-32 TO SCHOOL CAP PROJECTS 0 0 0 n/a
General Fund Transfers 178,497,824 192,071,332 13,573,508 7.6%
General County Debt:
110-9510-494.90-02 BOND ADMINISTRATIVE FEES 23,450 23,450 0 0%
110-9510-494.91-23 2019 GO BONDS 880,680 892,070 11,390 1%
110-9510-494.91-24 2014 GO BONDS 2,530,270 2,603,829 73,559 3%
110-9510-494.91-25 2014 EDA BONDS 355,000 355,000 0 0%
110-9510-494.91-26 2015 GO BONDS 426,155 426,155 0 0%
110-9510-494.91-27 2016 GO BONDS 360,000 360,000 0 0%
110-9510-494.91-28 2017 GO BONDS 120,000 120,000 0 0%
110-9510-494.91-29 2018 GO BONDS 95,000 95,000 0 0%
110-9510-494.91-69 2020 GO BONDS 700,000 710,000 10,000 1%
110-9510-494.91-70 2021 GO BONDS 705,816 708,371 2,555 0%
110-9510-494.91-71 2021 EDA BONDS 486,472 491,894 5,422 1%
110-9510-494.91-72 2022 GO BONDS 225,000 225,000 0 0%
110-9510-494.91-73 2023 GO BONDS 260,000 260,000 0 0%
110-9510-494.91-75 2024 GO BONDS 540,000 1,620,000 1,080,000 200%
110-9510-494.92-23 2019 GO BONDS 281,297 237,264 (44,033) -16%
110-9510-494.92-24 2014 GO BONDS 254,800 178,892 (75,908) -30%
110-9510-494.92-25 2014 EDA BONDS 87,795 77,146 (10,649) -12%
110-9510-494.92-26 2015 GO BONDS 157,205 135,898 (21,307) -14%
25 FY 2025 Adopted to FY 2026 Adopted Budget Comparison
Account Account Description FY 2025 Adopted Budget
FY 2026 Adopted Budget $ Change % Change
110-9510-494.92-27 2016 GO BONDS 84,388 66,388 (18,000) -21%
110-9510-494.92-28 2017 GO BONDS 35,850 29,850 (6,000) -17%
110-9510-494.92-29 2018 GO BONDS 13,877 10,314 (3,563) -26%
110-9510-494.92-69 2020 GO BONDS 213,500 178,500 (35,000) -16%
110-9510-494.92-70 2021 GO BONDS 312,306 275,970 (36,336) -12%
110-9510-494.92-71 2021 EDA BONDS 123,632 99,308 (24,324) -20%
110-9510-494.92-72 2022 GO BONDS 118,350 107,100 (11,250) -10%
110-9510-494.92-73 2023 GO BONDS 243,000 230,000 (13,000) -5%
110-9510-494.92-75 2024 GO BONDS 144,885 722,529 577,644 399%
General County Debt 9,778,728 11,239,928 1,461,200 14.9%
Total General Fund 389,214,652 $ 416,590,785 $ 27,376,133 $ 7.0%
School Grant Fund:
208-6100-452.61-01 INSTRUCTION 0 18,640,190 18,640,190 n/a
208-6100-452.61-02 ADMINISTRATION 0 150,000 150,000 n/a
208-6100-452.61-03 TRANSPORTATION 0 200,000 200,000 n/a
208-6100-452.61-04 MAINTENANCE 0 925,000 925,000 n/a
208-6100-452.61-09 EDUCATIONAL TECHNOLOGY 0 908,000 908,000 n/a
Total School Textbook Fund -$ 20,823,190 $ 20,823,190 $ n/a
School Textbook Fund:
209-6100-452.61-01 INSTRUCTION 0 3,747,276 3,747,276 n/a
Total School Textbook Fund -$ 3,747,276 $ 3,747,276 $ n/a
School Operating Fund:
210-6100-452.61-01 INSTRUCTION 273,927,006 266,936,665 (6,990,341) -3%
210-6100-452.61-02 ADMINISTRATION 14,841,319 16,855,603 2,014,284 14%
210-6100-452.61-03 TRANSPORTATION 22,923,287 27,969,218 5,045,931 22%
210-6100-452.61-04 MAINTENANCE 25,486,981 29,245,888 3,758,907 15%
210-6100-452.61-07 DEBT SERVICE 31,032,536 34,649,922 3,617,386 12%
210-6100-452.61-09 EDUCATIONAL TECHNOLOGY 15,823,940 16,480,367 656,427 4%
210-6100-452.99-13 TO SCHOOL TEXTBOOK FUND 0 1,387,242 1,387,242 n/a
210-6100-452.99-62 TO SCHOOL GRANT FUND 0 167,073 167,073 n/a
Total School Operating Fund 384,035,069 $ 393,691,978 $ 9,656,909 $ 2.5%
School Food Service Fund:
212-6100-452.61-05 FOOD SERVICE 15,743,958 17,382,186 1,638,228 10%
Total School Food Service Fund 15,743,958 $ 17,382,186 $ 1,638,228 $ 10.4%
Grant & Other Assistance - Victim/Witness Program:
215-2190-831.11-01 REGULAR 0 159,370 159,370 n/a
215-2190-833.11-01 REGULAR 0 100,883 100,883 n/a
Grant & Other Assistance - Victim/Witness Program 0 260,253 260,253 n/a
Grant & Other Assistance - Commonwealth's Attorney:
215-2210-832.11-01 REGULAR 0 121,016 121,016 n/a
215-2210-832.21-01 FICA 0 7,400 7,400 n/a
215-2210-832.21-02 MEDICARE 0 1,731 1,731 n/a
215-2210-832.22-10 RETIREMENT 0 18,224 18,224 n/a
215-2210-832.23-10 HEALTH INSURANCE 0 14,594 14,594 n/a
215-2210-832.23-15 HEALTH INSURANCE (HSA) 0 2,400 2,400 n/a
215-2210-832.24-01 INSURANCE 0 2,322 2,322 n/a
215-2210-832.27-10 SELF INSURED 0 24 24 n/a
215-2210-833.11-01 REGULAR 0 45,000 45,000 n/a
Grant & Other Assistance - Commonwealth's Attorney 0 212,711 212,711 n/a
Grant & Other Assistance - Sheriff:
26 FY 2025 Adopted to FY 2026 Adopted Budget Comparison
Account Account Description FY 2025 Adopted Budget
FY 2026 Adopted Budget $ Change % Change
215-3160-831.11-01 REGULAR 0 186,694 186,694 n/a
215-3160-831.21-01 FICA 0 11,575 11,575 n/a
215-3160-831.21-02 MEDICARE 0 2,707 2,707 n/a
215-3160-831.22-10 RETIREMENT 0 24,380 24,380 n/a
215-3160-831.23-10 HEALTH INSURANCE 0 30,384 30,384 n/a
215-3160-831.24-01 INSURANCE 0 2,203 2,203 n/a
215-3160-831.27-10 SELF INSURED 0 2,819 2,819 n/a
215-3160-833.11-01 REGULAR 0 247,743 247,743 n/a
215-3160-833.21-01 FICA 0 14,673 14,673 n/a
215-3160-833.21-02 MEDICARE 0 3,432 3,432 n/a
215-3160-833.22-10 RETIREMENT 0 32,156 32,156 n/a
215-3160-833.23-10 HEALTH INSURANCE 0 50,374 50,374 n/a
215-3160-833.24-01 INSURANCE 0 2,906 2,906 n/a
215-3160-833.27-10 SELF INSURED 0 3,741 3,741 n/a
Grant & Other Assistance - Sheriff 0 615,787 615,787 n/a
Grant & Other Assistance - Fire, Rescue & Emer Svcs:
215-3210-831.11-01 REGULAR 0 18,079 18,079 n/a
215-3210-834.39-10 SOFTWARE APPLICATIONS 0 22,900 22,900 n/a
215-3210-834.39-20 VOPEX EVERCISE CAREER DEV 0 7,100 7,100 n/a
215-3210-831.80-01 MACHINERY & EQUIPMENT 0 22,500 22,500 n/a
Grant & Other Assistance - Fire, Rescue & Emer Svcs 0 70,579 70,579 n/a
Grant & Other Assistance - Consolidated Fire & Rescue:
215-3240-833.58-61 FOUR FOR LIFE FUNDS 0 130,000 130,000 n/a
215-3240-833.58-88 STATE FIRE PROGRAMS 0 700,000 700,000 n/a
Grant & Other Assistance - Consolidated Fire & Rescue 0 830,000 830,000 n/a
Grant & Other Assistance - Parks & Recreation:
215-7110-833.31-90 OTHER PROFESSIONAL SER 0 4,961 4,961 n/a
215-7110-833.58-59 PUBLIC EDUCATION 0 7,000 7,000 n/a
215-7110-833.60-13 RECREATION SUPPLIES 0 4,000 4,000 n/a
215-7110-833.60-14 OPERATING SUPPLIES 0 1,000 1,000 n/a
Grant & Other Assistance - Parks & Recreation 0 16,961 16,961 n/a
Total Grant and Other Assistance Fund -$ 2,006,291 $ 2,006,291 $ n/a
Economic Development Authority Fund:
221-8151-471.31-50 LEGAL SERVICES 41,000 41,000 0 0%
221-8151-471.38-52 EDA MEMBER STIPENDS 6,300 6,300 0 0%
221-8151-471.58-77 EC DEV INITIATIVE PROGRAM 760,000 632,000 (128,000) -17%
221-8151-471.58-99 RECOGNITNS/AWARDS/SYMPTHY 9,000 9,000 0 0%
221-8151-471.97-05 SMALL BUSINESS GRANTS 5,000 5,000 0 0%
Total Economic Development Authority Fund 821,300 $ 693,300 $ (128,000) $ -15.6%
Lee Hill Tenant Building Fund:
225-4330-432.11-01 REGULAR 0 138,904 138,904 n/a
225-4330-432.21-01 FICA 0 8,612 8,612 n/a
225-4330-432.21-02 MEDICARE 0 2,014 2,014 n/a
225-4330-432.22-10 RETIREMENT 0 19,528 19,528 n/a
225-4330-432.23-10 HEALTH INSURANCE 0 30,589 30,589 n/a
225-4330-432.24-01 INSURANCE 0 2,665 2,665 n/a
225-4330-432.27-10 SELF INSURED 0 849 849 n/a
225-4330-432.31-55 SECURITY SERVICES 0 8,670 8,670 n/a
225-4330-432.31-90 OTHER PROFESSIONAL SER 0 27,583 27,583 n/a
225-4330-432.33-10 REPAIRS & MAINTENANCE 0 17,853 17,853 n/a
225-4330-432.33-18 HVAC SYSTEM REPAIR &MAINT 0 37,132 37,132 n/a
225-4330-432.33-30 ELECTRICAL REPAIR & MAINT 0 19,096 19,096 n/a
27 FY 2025 Adopted to FY 2026 Adopted Budget Comparison
Account Account Description FY 2025 Adopted Budget
FY 2026 Adopted Budget $ Change % Change
225-4330-432.33-31 ELEVATOR REPAIR & MAINT 0 13,155 13,155 n/a
225-4330-432.33-32 PLUMBING REPAIR & MAINT 0 15,914 15,914 n/a
225-4330-432.33-33 EXTERIOR REPAIR & MAINT 0 5,305 5,305 n/a
225-4330-432.33-34 FIRE SYST REPAIR & MAINT 0 21,218 21,218 n/a
225-4330-432.33-35 GROUNDS MAINTENANCE 0 37,132 37,132 n/a
225-4330-432.33-37 PEST CONTROL SERVICES 0 15,914 15,914 n/a
225-4330-432.33-38 SNOW REMOVAL 0 8,323 8,323 n/a
225-4330-432.39-27 SANITATION SERVICES 0 5,808 5,808 n/a
225-4330-432.39-28 JANITORIAL SERVICES 0 171,866 171,866 n/a
225-4330-432.51-10 ELECTRICAL SERVICES 0 309,717 309,717 n/a
225-4330-432.51-20 HEATING SERVICES 0 67,759 67,759 n/a
225-4330-432.51-30 WATER & SEWER SERVICES 0 37,556 37,556 n/a
225-4330-432.52-30 TELEPHONE SERVICES 0 64,112 64,112 n/a
225-4330-432.53-02 PROPERTY INSURANCE 0 38,192 38,192 n/a
225-4330-432.58-15 COMMON AREA ASSESSMENT 0 23,820 23,820 n/a
225-9510-494.91-76 2024 EDA BONDS 0 630,000 630,000 n/a
225-9510-494.92-76 2024 EDA BONDS 0 1,202,400 1,202,400 n/a
Total Lee Hill Tenant Building Fund -$ 2,981,686 $ 2,981,686 $ n/a
Fire/EMS Service Fee Fund:
240-9210-492.99-10 TO GENERAL FUND 3,600,000 3,600,000 0 0%
Total Fire/EMS Service Fee Fund 3,600,000 $ 3,600,000 $ -$ 0.0%
Code Compliance - Building:
260-3410-424.11-01 REGULAR 2,462,142 2,643,016 180,874 7%
260-3410-424.12-01 OVERTIME 64,173 36,831 (27,342) -43%
260-3410-424.13-01 PART-TIME 0 0 0 n/a
260-3410-424.21-01 FICA 150,745 159,671 8,926 6%
260-3410-424.21-02 MEDICARE 35,255 37,343 2,088 6%
260-3410-424.22-10 RETIREMENT 343,930 368,307 24,377 7%
260-3410-424.23-10 HEALTH INSURANCE 409,246 388,000 (21,246) -5%
260-3410-424.23-20 RETIREES 60,000 77,949 17,949 30%
260-3410-424.24-01 INSURANCE 40,862 41,866 1,004 2%
260-3410-424.26-01 UNEMPLOYMENT INSURANCE 0 0 0 n/a
260-3410-424.27-10 SELF INSURED 34,679 11,717 (22,962) -66%
260-3410-424.31-90 OTHER PROFESSIONAL SER 1,000 2,000 1,000 100%
260-3410-424.33-11 AUTO REPAIRS & MAINT 14,737 15,457 720 5%
260-3410-424.33-20 MAINTENANCE SVC CONTRACTS 1,610 1,610 0 0%
260-3410-424.35-01 PRINTING & BINDING 300 100 (200) -67%
260-3410-424.36-01 ADVERTISING 16,700 16,700 0 0%
260-3410-424.39-10 SOFTWARE APPLICATIONS 1,555 4,750 3,195 205%
260-3410-424.39-28 JANITORIAL SERVICES 9,100 8,820 (280) -3%
260-3410-424.51-10 ELECTRICAL SERVICES 7,412 7,412 0 0%
260-3410-424.52-10 POSTAL SERVICES 300 500 200 67%
260-3410-424.52-30 TELEPHONE SERVICES 24,488 24,488 0 0%
260-3410-424.53-05 MOTOR VEHICLE INSURANCE 5,381 6,165 784 15%
260-3410-424.54-10 LEASE/RENTAL EQUIPMENT 8,269 8,269 0 0%
260-3410-424.55-10 MILEAGE 4,417 2,824 (1,593) -36%
260-3410-424.55-30 SUBSISTENCE & LODGING 5,339 3,810 (1,529) -29%
260-3410-424.55-40 EDUCATION & TRAINING 22,590 14,750 (7,840) -35%
260-3410-424.58-10 DUES & ASSOC MEMBERSHIPS 2,565 2,245 (320) -12%
260-3410-424.58-46 PRE-EMPLOYMENT EXPENSES 500 350 (150) -30%
260-3410-424.60-01 OFFICE SUPPLIES 3,200 3,300 100 3%
260-3410-424.60-08 VEHICLE & EQUIPMENT FUELS 24,887 24,887 0 0%
260-3410-424.60-11 UNIFORMS 6,000 6,080 80 1%
260-3410-424.60-12 BOOKS & SUBSCRIPTIONS 1,700 1,700 0 0%
260-3410-424.60-14 OPERATING SUPPLIES 1,003 1,200 197 20%
260-3410-424.80-02 FURNITURE & FIXTURES 2,850 10,250 7,400 260%
28 FY 2025 Adopted to FY 2026 Adopted Budget Comparison
Account Account Description FY 2025 Adopted Budget
FY 2026 Adopted Budget $ Change % Change
260-3410-424.80-05 MOTOR VEHICLES & EQUIP 0 80,000 80,000 n/a
260-3410-424.80-07 COMPUTER EQUIPMENT 9,440 11,140 1,700 18%
260-3410-424.99-10 TO GENERAL FUND 789,457 668,634 (120,823) -15%
260-3410-424.99-31 TO CAPITAL PROJECTS 119,773 45,122 (74,651) -62%
Code Compliance - Building 4,685,605 4,737,263 51,658 1.1%
Code Compliance - Environmental Codes:
260-3440-424.11-01 REGULAR 993,840 1,233,649 239,809 24%
260-3440-424.12-01 OVERTIME 17,979 8,418 (9,561) -53%
260-3440-424.21-01 FICA 61,911 74,573 12,662 20%
260-3440-424.21-02 MEDICARE 14,480 17,441 2,961 20%
260-3440-424.22-10 RETIREMENT 147,851 182,353 34,502 23%
260-3440-424.23-10 HEALTH INSURANCE 125,736 175,651 49,915 40%
260-3440-424.23-20 RETIREES 30,000 35,203 5,203 17%
260-3440-424.24-01 INSURANCE 17,930 22,949 5,019 28%
260-3440-424.26-01 UNEMPLOYMENT INSURANCE 10,000 10,000 0 0%
260-3440-424.27-10 SELF INSURED 16,840 5,487 (11,353) -67%
260-3440-424.31-30 MGT CONSULTING SERVICES 1,100,000 512,144 (587,856) -53%
260-3440-424.31-50 LEGAL SERVICES 12,000 12,000 0 0%
260-3440-424.31-90 OTHER PROFESSIONAL SER 1,440 0 (1,440) -100%
260-3440-424.33-11 AUTO REPAIRS & MAINT 6,953 6,457 (496) -7%
260-3440-424.35-01 PRINTING & BINDING 3,405 3,405 0 0%
260-3440-424.36-01 ADVERTISING 8,500 8,500 0 0%
260-3440-424.39-10 SOFTWARE APPLICATIONS 1,946 3,950 2,004 103%
260-3440-424.39-28 JANITORIAL SERVICES 9,100 8,820 (280) -3%
260-3440-424.51-10 ELECTRICAL SERVICES 5,790 6,291 501 9%
260-3440-424.52-10 POSTAL SERVICES 1,500 1,800 300 20%
260-3440-424.52-30 TELEPHONE SERVICES 9,483 12,000 2,517 27%
260-3440-424.53-05 MOTOR VEHICLE INSURANCE 3,668 2,672 (996) -27%
260-3440-424.54-10 LEASE/RENTAL EQUIPMENT 4,791 4,791 0 0%
260-3440-424.55-30 SUBSISTENCE & LODGING 1,475 1,662 187 13%
260-3440-424.55-40 EDUCATION & TRAINING 7,839 16,912 9,073 116%
260-3440-424.56-51 TRI-COUNTY SWCD 55,000 55,000 0 0%
260-3440-424.58-10 DUES & ASSOC MEMBERSHIPS 8,747 13,747 5,000 57%
260-3440-424.58-46 PRE-EMPLOYMENT EXPENSES 180 230 50 28%
260-3440-424.60-01 OFFICE SUPPLIES 1,500 2,000 500 33%
260-3440-424.60-08 VEHICLE & EQUIPMENT FUELS 9,344 9,344 0 0%
260-3440-424.60-11 UNIFORMS 1,625 2,875 1,250 77%
260-3440-424.60-14 OPERATING SUPPLIES 2,000 900 (1,100) -55%
260-3440-424.80-02 FURNITURE & FIXTURES 0 7,250 7,250 n/a
260-3440-424.80-05 MOTOR VEHICLES & EQUIP 33,000 134,000 101,000 306%
260-3440-424.80-07 COMPUTER EQUIPMENT 5,396 6,660 1,264 23%
260-3440-424.99-10 TO GENERAL FUND 325,348 280,929 (44,419) -14%
260-3440-424.99-31 TO CAPITAL PROJECTS 50,227 20,378 (29,849) -59%
Code Compliance - Environmental Codes 3,106,824 2,900,441 (206,383) -6.6%
Code Compliance Non Departmental:
260-9110-491.24-03 OPEB COSTS 157,961 157,961 0 0%
260-9110-491.59-15 OPERATING RESERVE 0 0 0 n/a
260-9110-491.59-10 EMPL INCENTIVE/SALRY INCR 0 50,504 50,504 n/a
Code Compliance - Non Departmental 157,961 208,465 50,504 32.0%
Total Code Compliance Fund 7,950,390 $ 7,846,169 $ (104,221) $ -1.3%
Transportation:
280-4110-433.11-01 REGULAR 586,576 689,178 102,602 17%
280-4110-433.21-01 FICA 35,080 41,046 5,966 17%
280-4110-433.21-02 MEDICARE 8,205 9,600 1,395 17%
29 FY 2025 Adopted to FY 2026 Adopted Budget Comparison
Account Account Description FY 2025 Adopted Budget
FY 2026 Adopted Budget $ Change % Change
280-4110-433.22-10 RETIREMENT 79,584 91,842 12,258 15%
280-4110-433.23-10 HEALTH INSURANCE 101,207 75,260 (25,947) -26%
280-4110-433.24-01 INSURANCE 9,053 9,509 456 5%
280-4110-433.27-10 SELF INSURED 2,042 704 (1,338) -66%
280-4110-433.31-90 OTHER PROFESSIONAL SER 58,783 345,150 286,367 487%
280-4110-433.33-11 AUTO REPAIRS & MAINT 3,000 3,000 0 0%
280-4110-433.35-01 PRINTING & BINDING 0 150 150 n/a
280-4110-433.36-01 ADVERTISING 3,600 3,600 0 0%
280-4110-433.39-10 SOFTWARE APPLICATIONS 21,750 22,950 1,200 6%
280-4110-433.52-30 TELEPHONE SERVICES 4,900 7,300 2,400 49%
280-4110-433.53-05 MOTOR VEHICLE INSURANCE 219 1,118 899 411%
280-4110-433.54-10 LEASE/ RENTAL EQUIPMENT 1,000 3,600 2,600 260%
280-4110-433.55-10 MILEAGE 2,939 4,690 1,751 60%
280-4110-433.55-30 SUBSISTENCE & LODGING 3,386 6,562 3,176 94%
280-4110-433.55-40 EDUCATION & TRAINING 5,975 10,850 4,875 82%
280-4110-433.56-72 FRED BUS SYSTEM 185,486 546,035 360,549 194%
280-4110-433.58-10 DUES & ASSOC MEMBERSHIPS 4,000 5,500 1,500 38%
280-4110-433.60-01 OFFICE SUPPLIES 3,205 5,005 1,800 56%
280-4110-433.60-08 VEHICLE & EQUIPMENT FUELS 2,173 2,173 0 0%
280-4110-433.60-11 UNIFORMS 4,035 5,035 1,000 25%
280-4110-433.60-12 BOOKS & SUBSCRIPTIONS 1,000 1,000 0 0%
280-4110-433.60-14 OPERATING SUPPLIES 1,800 1,800 0 0%
280-4110-433.80-01 MACHINERY & EQUIPMENT 2,800 0 (2,800) -100%
280-4110-433.80-02 FURNITURE & FIXTURES 800 5,000 4,200 525%
280-4110-433.80-05 MOTOR VEHICLES & EQUIP 0 9,640 9,640 n/a
280-4110-433.80-07 COMPUTER EQUIPMENT 1,600 6,600 5,000 313%
Transportation 1,134,198 1,913,897 779,699 68.7%
Transportation - Litter Control:
280-4261-431.11-01 REGULAR 78,840 0 (78,840) -100%
280-4261-431.13-01 PART-TIME 0 0 0 n/a
280-4261-431.21-01 FICA 4,888 0 (4,888) -100%
280-4261-431.21-02 MEDICARE 1,143 0 (1,143) -100%
280-4261-431.22-10 RETIREMENT 11,073 0 (11,073) -100%
280-4261-431.23-10 HEALTH INSURANCE 14,872 0 (14,872) -100%
280-4261-431.24-01 INSURANCE 1,473 0 (1,473) -100%
280-4261-431.27-10 SELF INSURED 1,797 0 (1,797) -100%
280-4261-431.31-10 HEALTH SERVICES 506 0 (506) -100%
280-4261-431.31-90 OTHER PROFESSIONAL SER 300 0 (300) -100%
280-4261-431.33-11 AUTO REPAIRS & MAINTENANCE 3,000 0 (3,000) -100%
280-4261-431.55-40 EDUCATION & TRAINING 1,000 0 (1,000) -100%
280-4261-431.58-46 PRE-EMPLOYMENT EXPENSES 200 0 (200) -100%
280-4261-431.60-08 VEHICLE & EQUIPMENT FUELS 8,000 0 (8,000) -100%
280-4261-431.60-11 UNIFORMS 2,400 0 (2,400) -100%
Transportation - Litter Control: 129,492 0 (129,492) -100.0%
Transportation - Non Departmental:
280-9110-491.24-03 OPEB COSTS 32,055 32,055 0 0%
280-9110-491.59-10 EMPL INCENTIVE/SALRY INCR 0 8,215 8,215 n/a
Transportation - Non Departmental: 32,055 40,270 8,215 25.6%
Harrison Crossing Special Service District:
280-9172-494.91-66 2012A GO BONDS - HC SSD 110,000 120,000 10,000 9%
280-9172-494.91-74 2023 GO BONDS - HC SSD 177,806 177,806 0 0%
280-9172-494.92-66 2015 GO BONDS - HC SSD 50,750 45,250 (5,500) -11%
280-9172-494.92-74 2023 GO BONDS - HC SSD 168,916 160,026 (8,890) -5%
Harrison Crossing Special Service District 507,472 503,082 (4,390) -0.9%
30 FY 2025 Adopted to FY 2026 Adopted Budget Comparison
Account Account Description FY 2025 Adopted Budget
FY 2026 Adopted Budget $ Change % Change
Lee Hill East Special Service District:
280-9173-494.91-67 19 LEE HILL EAST REF 105,032 106,842 1,810 2%
280-9173-494.92-67 19 LEE HILL EAST REF 55,658 50,406 (5,252) -9%
Lee Hill East Special Service District 160,690 157,248 (3,442) -2.1%
Lee Hill West Special Service District:
280-9174-494.91-68 19 LEE HILL WEST REF 178,299 181,373 3,074 2%
280-9174-494.92-68 19 LEE HILL WEST REF 94,056 85,140 (8,916) -9%
Lee Hill West Special Service District 272,355 266,513 (5,842) -2.1%
Transportation Fund Transfers:
280-9210-492.99-31 TO CAPITAL PROJECTS 0 509,900 509,900 n/a
280-9210-492.99-51 TO UTIL OPERATING FUND 20,000 0 (20,000) -100%
Transportation Fund Transfers 20,000 509,900 489,900 2449.5%
VRE:
280-9400-493.34-10 VRE SUBSIDY 2,218,752 2,658,250 439,498 20%
280-9400-493.34-11 PRTC SUBSIDY 180,700 156,100 (24,600) -14%
280-9400-493.59-15 OPERATING RESERVE 13,871 0 (13,871) -100%
VRE 2,413,323 2,814,350 401,027 16.6%
Transportation Debt Service:
280-9510-494.90-02 BOND ADMINISTRATIVE FEES 0 0 0 n/a
280-9510-494.91-23 2019 GO BONDS 576,983 563,106 (13,877) -2%
280-9510-494.91-24 2014 GO BONDS 1,181,771 1,209,732 27,961 2%
280-9510-494.91-26 2015 GO BONDS 344,583 344,583 0 0%
280-9510-494.91-27 2016 GO BONDS 50,000 45,000 (5,000) -10%
280-9510-494.91-29 2018 GO BONDS 265,000 265,000 0 0%
280-9510-494.91-39 2012B GO BONDS 65,000 65,000 0 0%
280-9510-494.91-69 2020 GO BONDS 65,000 65,000 0 0%
280-9510-494.91-70 2021 GO BONDS 152,016 153,450 1,434 1%
280-9510-494.91-72 2022 GO BONDS 280,000 280,000 0 0%
280-9510-494.91-73 2023 GO BONDS 427,194 427,194 0 0%
280-9510-494.91-75 2024 GO BONDS 520,000 685,000 165,000 32%
280-9510-494.92-23 2019 GO BONDS 225,080 193,754 (31,326) -14%
280-9510-494.92-24 2014 GO BONDS 169,454 134,000 (35,454) -21%
280-9510-494.92-26 2015 GO BONDS 131,803 114,574 (17,229) -13%
280-9510-494.92-27 2016 GO BONDS 15,044 12,544 (2,500) -17%
280-9510-494.92-29 2018 GO BONDS 154,668 144,731 (9,937) -6%
280-9510-494.92-39 2012B GO BONDS 19,365 17,090 (2,275) -12%
280-9510-494.92-69 2020 GO BONDS 37,850 34,600 (3,250) -9%
280-9510-494.92-70 2021 GO BONDS 265,429 259,286 (6,143) -2%
280-9510-494.92-72 2022 GO BONDS 232,400 218,400 (14,000) -6%
280-9510-494.92-73 2023 GO BONDS 405,835 384,474 (21,361) -5%
280-9510-494.92-75 2024 GO BONDS 201,541 303,406 101,865 51%
Transportation Debt Service 5,786,016 5,919,924 133,908 (0)
Reservation of Service District Funds:
280-9600-496.98-20 MASSAPONAX SPEC TAX DISTR 3,672 20,706 17,034 464%
280-9600-496.98-21 HARRISON CROSSNG SPEC TAX 438,957 502,183 63,226 14%
280-9600-496.98-22 LEE HILL EAST SPEC TAX DI 32,835 98,837 66,002 201%
280-9600-496.98-23 LEE HILL WEST SPEC TAX DI 77,827 186,653 108,826 140%
Reservation of Service District Funds 553,291 808,379 255,088 46.1%
Total Transportation Fund 11,008,892 $ 12,933,563 $ 1,924,671 $ 2469.7%
Capital Projects - Non-Departmental:
310-9110-491.24-03 OPEB COSTS 6,789 6,789 0 0%
31 FY 2025 Adopted to FY 2026 Adopted Budget Comparison
Account Account Description FY 2025 Adopted Budget
FY 2026 Adopted Budget $ Change % Change
310-9110-491.59-10 EMPL INCENTIVE/SALRY INCR 0 1,082 1,082 n/a
310-9110-491.59-15 OPERATING RESERVE 113,145 110,705 (2,440) -2%
310-9110-493.99-28 TO TRANSPORTATION FUND 14,728 12,905 (1,823) -12%
310-9110-493.99-51 TO UTIL OPERATING FUND 65,000 0 (65,000) -100%
Capital Projects - Non-Departmental 199,662 131,481 (68,181) -34.1%
Capital Projects - Construction Management:
310-9115-493.11-01 REGULAR 74,776 70,452 (4,324) -6%
310-9115-493.21-01 FICA 4,367 4,343 (24) -1%
310-9115-493.21-02 MEDICARE 1,021 1,016 (5) 0%
310-9115-493.22-10 RETIREMENT 10,502 9,905 (597) -6%
310-9115-493.23-10 HEALTH INSURANCE 15,089 5,698 (9,391) -62%
310-9115-493.23-15 HEALTH INSURANCE (HSA) 0 1,200 1,200 n/a
310-9115-493.24-01 INSURANCE 1,398 1,352 (46) -3%
310-9115-493.27-10 SELF INSURED 1,503 388 (1,115) -74%
310-9115-493.31-10 HEALTH SERVICES 200 200 0 0%
310-9115-493.33-11 AUTO REPAIRS & MAINT 435 435 0 0%
310-9115-493.52-30 TELEPHONE SERVICES 2,500 2,500 0 0%
310-9115-493.53-05 MOTOR VEHICLE INSURANCE 422 437 15 4%
310-9115-493.55-30 SUBSISTENCE & LODGING 0 1,545 1,545 n/a
310-9115-493.55-40 EDUCATION & TRAINING 0 1,225 1,225 n/a
310-9115-493.58-46 PRE-EMPLOYMENT EXPENSES 500 500 0 0%
310-9115-493.60-08 VEHICLE & EQUIPMENT FUELS 221 221 0 0%
310-9115-493.60-11 UNIFORMS 150 200 50 33%
310-9115-493.80-02 FURNITURE & FIXTURES 0 0 0 n/a
310-9116-493.11-01 REGULAR 91,933 91,551 (382) 0%
310-9116-493.21-01 FICA 5,833 5,676 (157) -3%
310-9116-493.21-02 MEDICARE 1,364 1,327 (37) -3%
310-9116-493.22-10 RETIREMENT 14,980 12,871 (2,109) -14%
310-9116-493.23-10 HEALTH INSURANCE 0 16,791 16,791 n/a
310-9116-493.24-01 INSURANCE 1,691 1,757 66 4%
310-9116-493.27-10 SELF INSURED 1,830 568 (1,262) -69%
310-9116-493.55-40 EDUCATION & TRAINING 0 0 0 n/a
310-9116-493.60-11 UNIFORMS 200 200 0 0%
Capital Projects - Construction Management 230,915 232,358 1,443 0.6%
Capital Projects - General:
310-9120-802.31-90 OTHER PROFESSIONAL SER 250,000 250,000 0 0%
310-9120-802.39-10 SOFTWARE APPLICATIONS 1,775,000 163,000 (1,612,000) -91%
310-9120-802.80-01 MACHINERY & EQUIPMENT 2,643,895 1,203,000 (1,440,895) -54%
310-9120-802.80-05 MOTOR VEHICLES & EQUIP 1,946,876 439,050 (1,507,826) -77%
310-9120-802.80-07 COMPUTER EQUIPMENT 2,259,000 2,006,000 (253,000) -11%
310-9120-802.89-02 DESIGN 0 0 0 n/a
310-9120-802.89-03 CONSTRUCTION 1,545,000 3,064,900 1,519,900 98%
310-9120-806.89-02 DESIGN 0 0 0 n/a
Capital Projects - General 10,419,771 7,125,950 (3,293,821) -31.6%
Capital Projects -Fire & Rescue:
310-9130-801.80-01 MACHINERY & EQUIPMENT 4,629,999 6,500,000 1,870,001 40%
310-9130-801.89-03 CONSTRUCTION 5,927,214 9,277,746 3,350,532 57%
310-9130-802.39-10 SOFTWARE APPLICATIONS 0 227,000 227,000 n/a
310-9130-802.80-01 MACHINERY & EQUIPMENT 0 6,491,376 6,491,376 n/a
310-9130-802.80-05 MOTOR VEHICLES & EQUIP 134,683 396,900 262,217 195%
310-9130-806.89-03 CONSTRUCTION 72,786 78,194 5,408 7%
310-9130-808.80-01 MACHINERY & EQUIPMENT 0 0 0 n/a
310-9130-808.80-05 MOTOR VEHICLES & EQUIP 327,211 0 (327,211) -100%
Capital Projects - Fire & Rescue 11,091,893 22,971,216 11,879,323 107.1%
32 FY 2025 Adopted to FY 2026 Adopted Budget Comparison
Account Account Description FY 2025 Adopted Budget
FY 2026 Adopted Budget $ Change % Change
Capital Projects -Solid Waste:
310-9140-802.80-01 MACHINERY & EQUIPMENT 2,629,111 44,268 (2,584,843) -98%
310-9140-802.80-05 MOTOR VEHICLES & EQUIP 200,000 0 (200,000) -100%
310-9140-802.89-02 DESIGN 157,841 1,700,000 1,542,159 977%
310-9140-802.89-03 CONSTRUCTION 0 535,000 535,000 n/a
310-9140-806.89-02 DESIGN 20,159 0 (20,159) -100%
310-9140-806.89-03 CONSTRUCTION 0 0 0 n/a
Capital Projects - Solid Waste 3,007,111 2,279,268 (727,843) -24.2%
Capital Projects -Parks & Recreation:
310-9150-802.89-02 DESIGN 150,000 0 (150,000) -100%
310-9150-802.89-03 CONSTRUCTION 5,447,767 2,788,000 (2,659,767) -49%
310-9150-806.89-03 CONSTRUCTION 68,483 0 (68,483) -100%
Capital Projects - Parks & Recreation 5,666,250 2,788,000 (2,878,250) -50.8%
Capital Projects -Transportation:
310-9160-801.31-90 OTHER PROFESSIONAL SERVICES 500,000 0 (500,000) -100%
310-9160-802.31-90 OTHER PROFESSIONAL SERVICES 0 500,000 500,000 n/a
310-9160-801.89-01 LAND ACQUISITION 0 4,756,748 4,756,748 n/a
310-9160-801.89-02 DESIGN 200,000 9,728,077 9,528,077 4764%
310-9160-801.89-03 CONSTRUCTION 14,154,634 2,857,073 (11,297,561) -80%
310-9160-802.89-03 CONSTRUCTION 0 0 0 n/a
310-9160-803.89-03 CONSTRUCTION 0 500,000 500,000 n/a
310-9160-806.89-02 DESIGN 50,000 1,957,927 1,907,927 3816%
310-9160-806.89-01 LAND ACQUISITION 0 35,511 35,511 n/a
310-9160-806.89-03 CONSTRUCTION 692,235 0 (692,235) -100%
Capital Projects - Transportation 15,596,869 20,335,336 4,738,467 30.4%
Total Capital Projects Fund 46,212,471 $ 55,863,609 $ 9,651,138 $ 20.9%
Schools Capital Projects Fund:
320-9199-493.61-08 FACILITIES 36,807,856 47,691,578 10,883,722 30%
Total Schools Capital Projects Fund 36,807,856 $ 47,691,578 $ 10,883,722 $ 29.6%
Utilities - Administration:
510-4510-501.11-01 REGULAR 2,349,523 2,600,025 250,502 11%
510-4510-501.21-01 FICA 140,682 155,037 14,355 10%
510-4510-501.21-02 MEDICARE 33,464 36,995 3,531 11%
510-4510-501.22-10 RETIREMENT 335,770 371,164 35,394 11%
510-4510-501.23-10 HEALTH INSURANCE 198,080 217,998 19,918 10%
510-4510-501.23-15 HEALTH INSURANCE (HSA) 3,300 3,300 0 0%
510-4510-501.23-20 RETIREES 350,000 440,000 90,000 26%
510-4510-501.24-01 INSURANCE 38,355 41,218 2,863 7%
510-4510-501.24-03 OPEB COSTS 631,811 631,811 0 0%
510-4510-501.26-01 UNEMPLOYMENT INSURANCE 2,500 2,500 0 0%
510-4510-501.27-10 SELF INSURED 22,622 8,253 (14,369) -64%
510-4510-501.28-20 EDUC TUITION ASSISTANCE 5,000 5,000 0 0%
510-4510-501.28-98 PERSONNEL BUDGET REDUCTIO (400,000) (514,641) (114,641) 29%
510-4510-501.31-10 HEALTH SERVICES 1,889 889 (1,000) -53%
510-4510-501.31-80 GIS DEVELOPMENT SERVICES 150,000 150,000 0 0%
510-4510-501.31-90 OTHER PROFESSIONAL SER 96,500 96,500 0 0%
510-4510-501.33-10 REPAIRS & MAINTENANCE 2,500 3,500 1,000 40%
510-4510-501.33-11 AUTO REPAIRS & MAINT 29,741 29,742 1 0%
510-4510-501.33-20 MAINTENANCE SVC CONTRACTS 313,665 240,500 (73,165) -23%
510-4510-501.35-01 PRINTING & BINDING 200 200 0 0%
510-4510-501.36-01 ADVERTISING 9,100 9,100 0 0%
510-4510-501.39-10 SOFTWARE APPLICATIONS 0 500 500 n/a
510-4510-501.39-28 JANITORIAL SERVICES 25,700 4,095 (21,605) -84%
33 FY 2025 Adopted to FY 2026 Adopted Budget Comparison
Account Account Description FY 2025 Adopted Budget
FY 2026 Adopted Budget $ Change % Change
510-4510-501.51-10 ELECTRICAL SERVICES 20,470 25,760 5,290 26%
510-4510-501.51-20 HEATING SERVICES 14,155 14,155 0 0%
510-4510-501.52-10 POSTAL SERVICES 4,300 4,300 0 0%
510-4510-501.52-30 TELEPHONE SERVICES 19,730 19,730 0 0%
510-4510-501.53-02 PROPERTY INSURANCE 73,332 80,417 7,085 10%
510-4510-501.53-05 MOTOR VEHICLE INSURANCE 34,661 35,072 411 1%
510-4510-501.53-07 PUBLIC OFFICIAL LIAB INS 5,786 5,785 (1) 0%
510-4510-501.53-08 GENERAL LIABILITY INS 20,572 20,580 8 0%
510-4510-501.55-10 MILEAGE 5,000 5,100 100 2%
510-4510-501.55-30 SUBSISTENCE & LODGING 21,444 18,346 (3,098) -14%
510-4510-501.55-40 EDUCATION & TRAINING 28,426 30,720 2,294 8%
510-4510-501.56-79 RAPP RIVER BASIN COMMSN 1,000 1,000 0 0%
510-4510-501.58-10 DUES & ASSOC MEMBERSHIPS 11,548 11,548 0 0%
510-4510-501.58-46 PRE-EMPLOYMENT EXPENSES 200 500 300 150%
510-4510-501.58-55 INDUSTRIAL SAFETY PROGRAM 126,600 185,350 58,750 46%
510-4510-501.58-99 RECOGNITNS/AWARDS/SYMPTHY 5,300 5,300 0 0%
510-4510-501.59-01 CONTINGENCY 150,000 150,000 0 0%
510-4510-501.59-10 EMPL INCENTIVE/SALRY INCR 0 159,715 159,715 n/a
510-4510-501.59-15 OPERATING RESERVE 3,913 0 (3,913) -100%
510-4510-501.59-30 CNTRL SERV COST ALLOCATN 1,878,684 2,682,515 803,831 43%
510-4510-501.59-60 BUDGET REDUCTIONS 0 (425,000) (425,000) n/a
510-4510-501.60-01 OFFICE SUPPLIES 10,000 10,000 0 0%
510-4510-501.60-07 REPAIRS & MAINT SUPPLIES 3,625 3,625 0 0%
510-4510-501.60-08 VEHICLE & EQUIPMENT FUELS 10,826 10,826 0 0%
510-4510-501.60-11 UNIFORMS 3,524 4,124 600 17%
510-4510-501.60-14 OPERATING SUPPLIES 3,000 3,000 0 0%
510-4510-501.80-02 FURNITURE & FIXTURES 0 17,000 17,000 n/a
510-4510-501.80-07 COMPUTER EQUIPMENT 14,200 0 (14,200) -100%
Utilities - Administration 6,810,698 7,613,154 802,456 11.8%
Utilities - Customer Service:
510-4530-501.11-01 REGULAR 307,385 334,625 27,240 9%
510-4530-501.12-01 OVERTIME 1,175 1,236 61 5%
510-4530-501.21-01 FICA 17,752 19,203 1,451 8%
510-4530-501.21-02 MEDICARE 4,152 4,491 339 8%
510-4530-501.22-10 RETIREMENT 41,731 45,422 3,691 9%
510-4530-501.23-10 HEALTH INSURANCE 62,482 57,170 (5,312) -9%
510-4530-501.24-01 INSURANCE 4,913 5,123 210 4%
510-4530-501.27-10 SELF INSURED 185 67 (118) -64%
510-4530-501.31-90 OTHER PROFESSIONAL SER 51,240 51,240 0 0%
510-4530-501.33-20 MAINTENANCE SVC CONTRACTS 143,000 162,180 19,180 13%
510-4530-501.35-01 PRINTING & BINDING 14,834 16,170 1,336 9%
510-4530-501.51-10 ELECTRICAL SERVICES 210 125 (85) -40%
510-4530-501.52-10 POSTAL SERVICES 1,000 1,000 0 0%
510-4530-501.52-30 TELEPHONE SERVICES 1,040 1,040 0 0%
510-4530-501.55-30 SUBSISTENCE & LODGING 1,500 1,560 60 4%
510-4530-501.55-40 EDUCATION & TRAINING 500 1,250 750 150%
510-4530-501.58-46 PRE-EMPLOYMENT EXPENSES 200 200 0 0%
510-4530-501.59-30 CNTRL SERV COST ALLOCATN 721,784 874,595 152,811 21%
510-4530-501.60-01 OFFICE SUPPLIES 1,300 1,300 0 0%
510-4530-501.80-07 COMPUTER EQUIPMENT 0 0 0 n/a
Utilities - Customer Service 1,376,383 1,577,997 201,614 14.6%
Utilities - Ni River Water Plant:
510-4540-502.11-01 REGULAR 826,520 806,731 (19,789) -2%
510-4540-502.12-01 OVERTIME 121,176 127,515 6,339 5%
510-4540-502.12-02 ON-CALL 4,264 0 (4,264) -100%
510-4540-502.21-01 FICA 57,573 55,863 (1,710) -3%
34 FY 2025 Adopted to FY 2026 Adopted Budget Comparison
Account Account Description FY 2025 Adopted Budget
FY 2026 Adopted Budget $ Change % Change
510-4540-502.21-02 MEDICARE 13,465 13,065 (400) -3%
510-4540-502.22-10 RETIREMENT 112,813 112,279 (534) 0%
510-4540-502.23-10 HEALTH INSURANCE 140,053 121,342 (18,711) -13%
510-4540-502.24-01 INSURANCE 12,552 12,240 (312) -2%
510-4540-502.27-10 SELF INSURED 16,099 6,696 (9,403) -58%
510-4540-502.31-10 HEALTH SERVICES 5,275 5,275 0 0%
510-4540-502.31-78 LABORATORY SERVICES 8,000 8,000 0 0%
510-4540-502.31-79 REGULATORY AGENCY FEE 105,566 107,700 2,134 2%
510-4540-502.31-90 OTHER PROFESSIONAL SER 10,000 24,900 14,900 149%
510-4540-502.33-10 REPAIRS & MAINTENANCE 118,157 112,000 (6,157) -5%
510-4540-502.33-11 AUTO REPAIRS & MAINT 1,511 1,547 36 2%
510-4540-502.33-20 MAINTENANCE SVC CONTRACTS 29,000 33,900 4,900 17%
510-4540-502.51-10 ELECTRICAL SERVICES 250,000 376,370 126,370 51%
510-4540-502.51-20 HEATING SERVICES 5,609 6,709 1,100 20%
510-4540-502.52-10 POSTAL SERVICES 200 200 0 0%
510-4540-502.52-30 TELEPHONE SERVICES 8,000 8,000 0 0%
510-4540-502.54-10 LEASE/RENTAL EQUIPMENT 2,000 1,000 (1,000) -50%
510-4540-502.55-40 EDUCATION & TRAINING 7,101 7,101 0 0%
510-4540-502.58-10 DUES & ASSOC MEMBERSHIPS 1,700 1,700 0 0%
510-4540-502.58-55 INDUSTRIAL SAFETY PROGRAM 30,000 26,000 (4,000) -13%
510-4540-502.60-01 OFFICE SUPPLIES 1,100 2,000 900 82%
510-4540-502.60-04 MEDICAL AND LAB SUPPLIES 56,000 56,000 0 0%
510-4540-502.60-07 REPAIRS & MAINT SUPPLIES 25,000 25,000 0 0%
510-4540-502.60-08 VEHICLE & EQUIPMENT FUELS 3,014 3,014 0 0%
510-4540-502.60-11 UNIFORMS 8,294 8,294 0 0%
510-4540-502.60-14 OPERATING SUPPLIES 19,000 19,000 0 0%
510-4540-502.60-26 CHEMICAL TREATMENT SUPPLY 852,809 852,809 0 0%
510-4540-502.80-01 MACHINERY & EQUIPMENT 110,000 118,000 8,000 7%
510-4540-502.80-02 FURNITURE & FIXTURES 1,500 1,500 0 0%
Utilities - Ni River Water Plant 2,963,351 3,061,750 98,399 3.3%
Utilities - Motts Run WTP:
510-4541-502.11-01 REGULAR 929,026 966,717 37,691 4%
510-4541-502.12-01 OVERTIME 150,578 158,455 7,877 5%
510-4541-502.12-02 ON-CALL 3,186 0 (3,186) -100%
510-4541-502.21-01 FICA 65,372 68,320 2,948 5%
510-4541-502.21-02 MEDICARE 15,289 15,978 689 5%
510-4541-502.22-10 RETIREMENT 123,444 129,432 5,988 5%
510-4541-502.23-10 HEALTH INSURANCE 158,401 126,183 (32,218) -20%
510-4541-502.23-15 HEALTH INSURANCE (HSA) 0 2,100 2,100 n/a
510-4541-502.24-01 INSURANCE 13,635 13,555 (80) -1%
510-4541-502.27-10 SELF INSURED 16,483 7,801 (8,682) -53%
510-4541-502.31-10 HEALTH SERVICES 9,092 4,392 (4,700) -52%
510-4541-502.31-78 LABORATORY SERVICES 10,000 10,000 0 0%
510-4541-502.31-79 REGULATORY AGENCY FEE 4,000 4,000 0 0%
510-4541-502.31-90 OTHER PROFESSIONAL SER 15,600 28,100 12,500 80%
510-4541-502.33-10 REPAIRS & MAINTENANCE 135,000 150,000 15,000 11%
510-4541-502.33-11 AUTO REPAIRS & MAINT 3,024 3,025 1 0%
510-4541-502.33-20 MAINTENANCE SVC CONTRACTS 37,300 43,592 6,292 17%
510-4541-502.51-10 ELECTRICAL SERVICES 964,310 920,031 (44,279) -5%
510-4541-502.51-20 HEATING SERVICES 20,824 20,824 0 0%
510-4541-502.52-10 POSTAL SERVICES 3,507 3,507 0 0%
510-4541-502.52-30 TELEPHONE SERVICES 10,800 10,800 0 0%
510-4541-502.54-10 LEASE/RENTAL EQUIPMENT 1,000 1,000 0 0%
510-4541-502.55-40 EDUCATION & TRAINING 7,348 8,137 789 11%
510-4541-502.58-10 DUES & ASSOC MEMBERSHIPS 2,166 2,166 0 0%
510-4541-502.58-46 PRE-EMPLOYMENT EXPENSES 500 500 0 0%
510-4541-502.58-55 INDUSTRIAL SAFETY PROGRAM 42,000 34,000 (8,000) -19%
35 FY 2025 Adopted to FY 2026 Adopted Budget Comparison
Account Account Description FY 2025 Adopted Budget
FY 2026 Adopted Budget $ Change % Change
510-4541-502.60-01 OFFICE SUPPLIES 2,000 2,200 200 10%
510-4541-502.60-04 MEDICAL AND LAB SUPPLIES 62,307 62,307 0 0%
510-4541-502.60-07 REPAIRS & MAINT SUPPLIES 43,100 38,100 (5,000) -12%
510-4541-502.60-08 VEHICLE & EQUIPMENT FUELS 15,599 15,599 0 0%
510-4541-502.60-11 UNIFORMS 8,797 9,797 1,000 11%
510-4541-502.60-14 OPERATING SUPPLIES 29,590 29,590 0 0%
510-4541-502.60-26 CHEMICAL TREATMENT SUPPLY 1,367,350 1,367,350 0 0%
510-4541-502.80-01 MACHINERY & EQUIPMENT 51,500 89,400 37,900 74%
510-4541-502.80-02 FURNITURE & FIXTURES 2,000 2,000 0 0%
510-4541-502.80-05 MOTOR VEHICLES & EQUIP 0 0 0 n/a
Utilities - Motts Run WTP 4,324,128 4,348,958 24,830 0.6%
Utilities - Water Conservation:
510-4549-502.31-90 OTHER PROFESSIONAL SER 15,000 15,000 0 0%
Utilities - Water Conservation 15,000 15,000 0 0.0%
Utilities - Massaponax WWTP:
510-4550-503.11-01 REGULAR 1,556,557 1,627,640 71,083 5%
510-4550-503.12-01 OVERTIME 62,727 66,008 3,281 5%
510-4550-503.12-02 ON-CALL 33,917 28,412 (5,505) -16%
510-4550-503.21-01 FICA 99,474 103,763 4,289 4%
510-4550-503.21-02 MEDICARE 23,264 24,267 1,003 4%
510-4550-503.22-10 RETIREMENT 219,674 228,735 9,061 4%
510-4550-503.23-10 HEALTH INSURANCE 238,097 208,667 (29,430) -12%
510-4550-503.23-15 HEALTH INSURANCE (HSA) 0 0 0 n/a
510-4550-503.24-01 INSURANCE 25,048 25,462 414 2%
510-4550-503.27-10 SELF INSURED 24,827 11,367 (13,460) -54%
510-4550-503.31-10 HEALTH SERVICES 5,844 4,244 (1,600) -27%
510-4550-503.31-78 LABORATORY SERVICES 51,500 46,200 (5,300) -10%
510-4550-503.31-79 REGULATORY AGENCY FEE 12,600 12,600 0 0%
510-4550-503.31-90 OTHER PROFESSIONAL SER 13,400 29,400 16,000 119%
510-4550-503.33-10 REPAIRS & MAINTENANCE 472,000 407,000 (65,000) -14%
510-4550-503.33-20 MAINTENANCE SVC CONTRACTS 56,200 61,000 4,800 9%
510-4550-503.33-11 AUTO REPAIRS & MAINT 23,946 23,446 (500) -2%
510-4550-503.51-10 ELECTRICAL SERVICES 724,390 776,390 52,000 7%
510-4550-503.52-10 POSTAL SERVICES 2,000 2,000 0 0%
510-4550-503.52-30 TELEPHONE SERVICES 8,000 9,000 1,000 13%
510-4550-503.55-10 MILEAGE 0 0 0 n/a
510-4550-503.55-30 SUBSISTENCE & LODGING 4,934 17,046 12,112 245%
510-4550-503.55-40 EDUCATION & TRAINING 19,572 19,962 390 2%
510-4550-503.58-10 DUES & ASSOC MEMBERSHIPS 11,315 12,515 1,200 11%
510-4550-503.58-46 PRE-EMPLOYMENT EXPENSES 2,400 2,400 0 0%
510-4550-503.58-55 INDUSTRIAL SAFETY PROGRAM 54,000 48,000 (6,000) -11%
510-4550-503.60-01 OFFICE SUPPLIES 3,000 3,000 0 0%
510-4550-503.60-07 REPAIRS & MAINT SUPPLIES 19,000 21,000 2,000 11%
510-4550-503.60-08 VEHICLE & EQUIPMENT FUELS 34,814 34,814 0 0%
510-4550-503.60-11 UNIFORMS 32,494 32,494 0 0%
510-4550-503.60-14 OPERATING SUPPLIES 19,263 19,263 0 0%
510-4550-503.60-26 CHEMICAL TREATMENT SUPPLY 991,690 991,690 0 0%
510-4550-503.80-02 FURNITURE & FIXTURES 0 18,200 18,200 n/a
510-4550-503.80-07 COMPUTER EQUIPMENT 1,800 1,200 (600) -33%
Utilities - Massaponax WWTP 4,847,747 4,917,185 69,438 1.4%
Utilities - FMC WWTP:
510-4552-503.11-01 REGULAR 630,666 639,669 9,003 1%
510-4552-503.12-01 OVERTIME 17,928 18,866 938 5%
510-4552-503.12-02 ON-CALL 6,769 7,123 354 5%
510-4552-503.21-01 FICA 40,202 40,606 404 1%
36 FY 2025 Adopted to FY 2026 Adopted Budget Comparison
Account Account Description FY 2025 Adopted Budget
FY 2026 Adopted Budget $ Change % Change
510-4552-503.21-02 MEDICARE 9,402 9,497 95 1%
510-4552-503.22-10 RETIREMENT 86,930 88,731 1,801 2%
510-4552-503.23-10 HEALTH INSURANCE 74,094 84,560 10,466 14%
510-4552-503.24-01 INSURANCE 10,292 10,229 (63) -1%
510-4552-503.27-10 SELF INSURED 9,560 4,327 (5,233) -55%
510-4552-503.31-10 HEALTH SERVICES 2,733 2,373 (360) -13%
510-4552-503.31-78 LABORATORY SERVICES 46,500 46,500 0 0%
510-4552-503.31-79 REGULATORY AGENCY FEE 15,000 15,000 0 0%
510-4552-503.31-90 OTHER PROFESSIONAL SER 1,150 900 (250) -22%
510-4552-503.33-10 REPAIRS & MAINTENANCE 205,000 186,000 (19,000) -9%
510-4552-503.33-11 AUTO REPAIRS & MAINT 1,221 1,521 300 25%
510-4552-503.33-20 MAINTENANCE SVC CONTRACTS 37,000 37,000 0 0%
510-4552-503.38-57 SLUDGE DISPOSAL 33,000 33,000 0 0%
510-4552-503.51-10 ELECTRICAL SERVICES 356,480 361,478 4,998 1%
510-4552-503.52-30 TELEPHONE SERVICES 3,200 3,800 600 19%
510-4552-503.54-10 LEASE/RENTAL EQUIPMENT 5,500 5,500 0 0%
510-4552-503.55-10 MILEAGE 0 0 0 n/a
510-4552-503.55-30 SUBSISTENCE & LODGING 2,430 2,430 0 0%
510-4552-503.55-40 EDUCATION & TRAINING 14,564 15,744 1,180 8%
510-4552-503.58-10 DUES & ASSOC MEMBERSHIPS 3,200 1,000 (2,200) -69%
510-4552-503.58-46 PRE-EMPLOYMENT EXPENSES 1,800 1,800 0 0%
510-4552-503.58-55 INDUSTRIAL SAFETY PROGRAM 32,000 23,000 (9,000) -28%
510-4552-503.60-01 OFFICE SUPPLIES 820 1,320 500 61%
510-4552-503.60-07 REPAIRS & MAINT SUPPLIES 1,300 1,300 0 0%
510-4552-503.60-08 VEHICLE & EQUIPMENT FUELS 3,684 3,684 0 0%
510-4552-503.60-11 UNIFORMS 13,000 13,000 0 0%
510-4552-503.60-14 OPERATING SUPPLIES 12,350 13,850 1,500 12%
510-4552-503.60-26 CHEMICAL TREATMENT SUPPLY 340,000 340,000 0 0%
510-4552-503.80-07 COMPUTER EQUIPMENT 2,220 0 (2,220) -100%
Utilities - FMC WWTP 2,019,995 2,013,808 (6,187) -0.3%
Utilities - Thornburg WWTP:
510-4553-503.11-01 REGULAR 295,000 305,583 10,583 4%
510-4553-503.12-01 OVERTIME 7,704 8,107 403 5%
510-4553-503.21-01 FICA 18,065 18,715 650 4%
510-4553-503.21-02 MEDICARE 4,225 4,377 152 4%
510-4553-503.22-10 RETIREMENT 39,429 40,704 1,275 3%
510-4553-503.23-10 HEALTH INSURANCE 51,569 47,175 (4,394) -9%
510-4553-503.24-01 INSURANCE 4,447 4,310 (137) -3%
510-4553-503.27-10 SELF INSURED 4,419 2,039 (2,380) -54%
510-4553-503.31-10 HEALTH SERVICES 1,128 1,128 0 0%
510-4553-503.31-90 OTHER PROFESSIONAL SER 8,360 25,600 17,240 206%
510-4553-503.31-78 LABORATORY SERVICES 21,000 36,000 15,000 71%
510-4553-503.31-79 REGULATORY AGENCY FEE 7,548 7,548 0 0%
510-4553-503.33-10 REPAIRS & MAINTENANCE 42,200 43,200 1,000 2%
510-4553-503.33-20 MAINTENANCE SVC CONTRACTS 4,600 8,600 4,000 87%
510-4553-503.33-11 AUTO REPAIRS & MAINT 3,160 3,170 10 0%
510-4553-503.51-10 ELECTRICAL SERVICES 42,760 55,588 12,828 30%
510-4553-503.52-30 TELEPHONE SERVICES 4,885 5,185 300 6%
510-4553-503.54-10 LEASE/RENTAL EQUIPMENT 0 6,000 6,000 n/a
510-4553-503.55-10 MILEAGE 0 0 0 n/a
510-4553-503.55-30 SUBSISTENCE & LODGING 1,215 1,215 0 0%
510-4553-503.55-40 EDUCATION & TRAINING 3,676 3,836 160 4%
510-4553-503.58-10 DUES & ASSOC MEMBERSHIPS 320 720 400 125%
510-4553-503.58-46 PRE-EMPLOYMENT EXPENSES 2,400 1,600 (800) -33%
510-4553-503.58-55 INDUSTRIAL SAFETY PROGRAM 11,000 12,000 1,000 9%
510-4553-503.60-01 OFFICE SUPPLIES 1,302 1,302 0 0%
510-4553-503.60-07 REPAIRS & MAINT SUPPLIES 700 500 (200) -29%
37 FY 2025 Adopted to FY 2026 Adopted Budget Comparison
Account Account Description FY 2025 Adopted Budget
FY 2026 Adopted Budget $ Change % Change
510-4553-503.60-08 VEHICLE & EQUIPMENT FUELS 3,633 3,633 0 0%
510-4553-503.60-11 UNIFORMS 8,101 8,101 0 0%
510-4553-503.60-14 OPERATING SUPPLIES 2,273 2,773 500 22%
510-4553-503.60-26 CHEMICAL TREATMENT SUPPLY 50,687 51,444 757 1%
510-4553-503.80-05 MOTOR VEHICLES & EQUIP 0 50,000 50,000 n/a
510-4553-503.80-07 COMPUTER EQUIPMENT 1,400 0 (1,400) -100%
Utilities - Thornburg WWTP 647,206 760,153 112,947 17.5%
Utilities - Composting Operations:
510-4555-503.11-01 REGULAR 394,466 412,965 18,499 5%
510-4555-503.12-01 OVERTIME 11,642 12,251 609 5%
510-4555-503.21-01 FICA 24,505 25,131 626 3%
510-4555-503.21-02 MEDICARE 5,731 5,877 146 3%
510-4555-503.22-10 RETIREMENT 52,105 54,714 2,609 5%
510-4555-503.23-10 HEALTH INSURANCE 48,093 70,763 22,670 47%
510-4555-503.24-01 INSURANCE 5,548 5,280 (268) -5%
510-4555-503.27-10 SELF INSURED 11,547 4,497 (7,050) -61%
510-4555-503.31-10 HEALTH SERVICES 848 848 0 0%
510-4555-503.31-30 MGT CONSULTING SERVICES 0 0 0 n/a
510-4555-503.31-78 LABORATORY SERVICES 14,500 14,500 0 0%
510-4555-503.31-79 REGULATORY AGENCY FEE 110 110 0 0%
510-4555-503.31-90 OTHER PROFESSIONAL SER 6,460 6,500 40 1%
510-4555-503.33-10 REPAIRS & MAINTENANCE 173,000 83,000 (90,000) -52%
510-4555-503.33-11 AUTO REPAIRS & MAINT 2,489 2,489 0 0%
510-4555-503.33-13 TRUCK REPAIRS & MAINT 27,500 33,500 6,000 22%
510-4555-503.33-14 HEAVY EQUIP REP & MAINT 211,034 211,034 0 0%
510-4555-503.35-01 PRINTING & BINDING 0 2,500 2,500 n/a
510-4555-503.36-01 ADVERTISING 7,200 5,200 (2,000) -28%
510-4555-503.51-10 ELECTRICAL SERVICES 136,580 129,618 (6,962) -5%
510-4555-503.52-10 POSTAL SERVICES 3,000 3,000 0 0%
510-4555-503.52-30 TELEPHONE SERVICES 936 936 0 0%
510-4555-503.53-05 MOTOR VEHICLE INSURANCE 1,909 1,534 (375) -20%
510-4555-503.54-10 LEASE/RENTAL EQUIPMENT 2,000 2,000 0 0%
510-4555-503.55-40 EDUCATION & TRAINING 500 1,000 500 100%
510-4555-503.58-10 DUES & ASSOC MEMBERSHIPS 1,323 1,323 0 0%
510-4555-503.58-55 INDUSTRIAL SAFETY PROGRAM 15,000 14,000 (1,000) -7%
510-4555-503.58-64 STONE & HAULING 15,000 15,000 0 0%
510-4555-503.60-01 OFFICE SUPPLIES 1,563 1,563 0 0%
510-4555-503.60-07 REPAIRS & MAINT SUPPLIES 18,000 18,000 0 0%
510-4555-503.60-08 VEHICLE & EQUIPMENT FUELS 87,990 87,990 0 0%
510-4555-503.60-09 VEHICLE & EQUIPMENT SUPP 6,100 11,100 5,000 82%
510-4555-503.60-11 UNIFORMS 15,169 13,194 (1,975) -13%
510-4555-503.60-14 OPERATING SUPPLIES 6,000 6,000 0 0%
510-4555-503.60-27 SMALL TOOL/EQUIP REPLCMNT 2,000 2,000 0 0%
510-4555-503.80-07 COMPUTER EQUIPMENT 0 0 0 n/a
Utilities - Composting Operations 1,309,848 1,259,417 (50,431) -3.9%
Utilities - W/S Transmissions:
510-4560-504.11-01 REGULAR 1,150,615 1,235,534 84,919 7%
510-4560-504.12-01 OVERTIME 147,404 168,369 20,965 14%
510-4560-504.12-02 ON-CALL 63,173 76,819 13,646 22%
510-4560-504.21-01 FICA 83,085 90,162 7,077 9%
510-4560-504.21-02 MEDICARE 19,431 21,087 1,656 9%
510-4560-504.22-10 RETIREMENT 157,256 169,429 12,173 8%
510-4560-504.23-10 HEALTH INSURANCE 152,341 127,704 (24,637) -16%
510-4560-504.24-01 INSURANCE 18,557 19,211 654 4%
510-4560-504.27-10 SELF INSURED 28,704 11,567 (17,137) -60%
510-4560-504.31-10 HEALTH SERVICES 6,255 4,755 (1,500) -24%
38 FY 2025 Adopted to FY 2026 Adopted Budget Comparison
Account Account Description FY 2025 Adopted Budget
FY 2026 Adopted Budget $ Change % Change
510-4560-504.31-90 OTHER PROFESSIONAL SER 27,830 27,830 0 0%
510-4560-504.33-10 REPAIRS & MAINTENANCE 330,000 430,000 100,000 30%
510-4560-504.33-11 AUTO REPAIRS & MAINT 169,933 170,375 442 0%
510-4560-504.33-20 MAINTENANCE SVC CONTRACTS 3,000 3,000 0 0%
510-4560-504.33-22 TANK MAINTENANCE PROGRAM 20,400 20,400 0 0%
510-4560-504.35-01 PRINTING & BINDING 0 500 500 n/a
510-4560-504.52-10 POSTAL SERVICES 100 100 0 0%
510-4560-504.52-30 TELEPHONE SERVICES 15,700 15,700 0 0%
510-4560-504.54-10 LEASE/RENTAL EQUIPMENT 2,500 2,500 0 0%
510-4560-504.55-40 EDUCATION & TRAINING 1,400 4,700 3,300 236%
510-4560-504.58-46 PRE-EMPLOYMENT EXPENSES 500 500 0 0%
510-4560-504.58-55 INDUSTRIAL SAFETY PROGRAM 32,000 63,000 31,000 97%
510-4560-504.60-01 OFFICE SUPPLIES 1,500 1,500 0 0%
510-4560-504.60-07 REPAIRS & MAINT SUPPLIES 1,000 1,000 0 0%
510-4560-504.60-08 VEHICLE & EQUIPMENT FUELS 68,210 68,210 0 0%
510-4560-504.60-11 UNIFORMS 26,835 26,835 0 0%
510-4560-504.60-14 OPERATING SUPPLIES 856,300 1,081,300 225,000 26%
510-4560-504.60-27 SMALL TOOL/EQUIP REPLCMNT 3,700 5,000 1,300 35%
510-4560-504.80-05 MOTOR VEHICLES & EQUIP 340,000 85,000 (255,000) -75%
510-4560-504.80-07 COMPUTER EQUIPMENT 4,100 0 (4,100) -100%
Utilities - W/S Transmissions 3,731,829 3,932,087 200,258 5.4%
Utilities - Infiltration & Inflow:
510-4561-504.11-01 REGULAR 1,013,052 1,079,240 66,188 7%
510-4561-504.12-01 OVERTIME 73,155 76,982 3,827 5%
510-4561-504.12-02 ON-CALL 37,780 50,954 13,174 35%
510-4561-504.21-01 FICA 68,349 73,370 5,021 7%
510-4561-504.21-02 MEDICARE 15,985 17,159 1,174 7%
510-4561-504.22-10 RETIREMENT 143,031 151,228 8,197 6%
510-4561-504.23-10 HEALTH INSURANCE 149,698 132,085 (17,613) -12%
510-4561-504.23-15 HEALTH INSURANCE (HSA) 1,500 1,500 0 0%
510-4561-504.24-01 INSURANCE 17,498 18,344 846 5%
510-4561-504.27-10 SELF INSURED 22,850 9,287 (13,563) -59%
510-4561-504.31-10 HEALTH SERVICES 4,960 4,960 0 0%
510-4561-504.31-90 OTHER PROFESSIONAL SER 15,200 19,700 4,500 30%
510-4561-504.33-10 REPAIRS & MAINTENANCE 70,000 75,000 5,000 7%
510-4561-504.33-11 AUTO REPAIRS & MAINT 14,446 44,045 29,599 205%
510-4561-504.33-20 MAINTENANCE SVC CONTRACTS 25,880 29,880 4,000 15%
510-4561-504.33-24 ROAD/EASEMENT MAINTENANCE 6,000 6,000 0 0%
510-4561-504.33-26 MANHOLE REHABILITATION 20,000 20,000 0 0%
510-4561-504.35-01 PRINTING & BINDING 0 500 500 n/a
510-4561-504.52-10 POSTAL SERVICES 500 500 0 0%
510-4561-504.52-30 TELEPHONE SERVICES 16,860 16,860 0 0%
510-4561-504.54-10 LEASE/RENTAL EQUIPMENT 1,500 1,500 0 0%
510-4561-504.55-40 EDUCATION & TRAINING 7,180 9,150 1,970 27%
510-4561-504.58-10 DUES & ASSOC MEMBERSHIPS 295 295 0 0%
510-4561-504.58-21 SEWAGE BACKUP CLEANING 20,000 20,000 0 0%
510-4561-504.58-46 PRE-EMPLOYMENT EXPENSES 200 800 600 300%
510-4561-504.58-55 INDUSTRIAL SAFETY PROGRAM 38,500 34,000 (4,500) -12%
510-4561-504.60-08 VEHICLE & EQUIPMENT FUELS 45,324 45,324 0 0%
510-4561-504.60-11 UNIFORMS 22,042 22,042 0 0%
510-4561-504.60-14 OPERATING SUPPLIES 45,000 55,000 10,000 22%
510-4561-504.60-27 SMALL TOOL/EQUIP REPLCMNT 6,000 11,000 5,000 83%
510-4561-504.80-01 MACHINERY & EQUIPMENT 0 40,000 40,000 n/a
510-4561-504.80-05 MOTOR VEHICLES & EQUIP 80,000 0 (80,000) -100%
Utilities - Infiltration & Inflow 1,982,785 2,066,705 83,920 4.2%
Utilities - Line Location:
39 FY 2025 Adopted to FY 2026 Adopted Budget Comparison
Account Account Description FY 2025 Adopted Budget
FY 2026 Adopted Budget $ Change % Change
510-4562-504.11-01 REGULAR 516,497 614,394 97,897 19%
510-4562-504.12-01 OVERTIME 12,560 16,805 4,245 34%
510-4562-504.12-02 ON-CALL 25,586 26,395 809 3%
510-4562-504.21-01 FICA 32,975 38,982 6,007 18%
510-4562-504.21-02 MEDICARE 7,712 9,117 1,405 18%
510-4562-504.22-10 RETIREMENT 71,839 86,213 14,374 20%
510-4562-504.23-10 HEALTH INSURANCE 92,660 121,982 29,322 32%
510-4562-504.24-01 INSURANCE 8,017 9,782 1,765 22%
510-4562-504.27-10 SELF INSURED 11,392 4,660 (6,732) -59%
510-4562-504.31-10 HEALTH SERVICES 1,655 1,655 0 0%
510-4562-504.31-90 OTHER PROFESSIONAL SER 2,880 1,500 (1,380) -48%
510-4562-504.33-10 REPAIRS & MAINTENANCE 4,000 500 (3,500) -88%
510-4562-504.33-11 AUTO REPAIRS & MAINT 10,055 11,816 1,761 18%
510-4562-504.33-20 MAINTENANCE SVC CONTRACTS 23,166 23,166 0 0%
510-4562-504.39-31 MISS UTILITY FEES 30,200 30,200 0 0%
510-4562-504.52-30 TELEPHONE SERVICES 9,010 9,010 0 0%
510-4562-504.55-30 SUBSISTENCE & LODGING 0 432 432 n/a
510-4562-504.55-40 EDUCATION & TRAINING 2,960 2,215 (745) -25%
510-4562-504.58-46 PRE-EMPLOYMENT EXPENSES 200 500 300 150%
510-4562-504.58-55 INDUSTRIAL SAFETY PROGRAM 10,500 16,500 6,000 57%
510-4562-504.60-08 VEHICLE & EQUIPMENT FUELS 28,864 28,864 0 0%
510-4562-504.60-11 UNIFORMS 12,068 10,618 (1,450) -12%
510-4562-504.60-14 OPERATING SUPPLIES 3,896 9,046 5,150 132%
510-4562-504.60-27 SMALL TOOL/EQUIP REPLCMNT 500 7,500 7,000 1400%
510-4562-504.80-05 MOTOR VEHICLES & EQUIP 0 134,339 134,339 n/a
510-4562-504.80-07 COMPUTER EQUIPMENT 0 0 0 n/a
Utilities - Line Location 919,192 1,216,191 296,999 32.3%
Utilities - Pump Station Maintenance:
510-4563-504.11-01 REGULAR 229,299 259,705 30,406 13%
510-4563-504.12-01 OVERTIME 21,957 17,363 (4,594) -21%
510-4563-504.12-02 ON-CALL 15,793 18,198 2,405 15%
510-4563-504.21-01 FICA 15,990 17,741 1,751 11%
510-4563-504.21-02 MEDICARE 3,739 4,149 410 11%
510-4563-504.22-10 RETIREMENT 30,672 34,685 4,013 13%
510-4563-504.23-10 HEALTH INSURANCE 35,780 32,750 (3,030) -8%
510-4563-504.24-01 INSURANCE 3,476 3,679 203 6%
510-4563-504.27-10 SELF INSURED 4,590 2,054 (2,536) -55%
510-4563-504.31-10 HEALTH SERVICES 3,509 2,509 (1,000) -28%
510-4563-504.31-90 OTHER PROFESSIONAL SER 14,900 14,900 0 0%
510-4563-504.33-10 REPAIRS & MAINTENANCE 344,600 310,600 (34,000) -10%
510-4563-504.33-11 AUTO REPAIRS & MAINT 10,639 10,639 0 0%
510-4563-504.33-20 MAINTENANCE SVC CONTRACTS 26,000 33,000 7,000 27%
510-4563-504.33-24 ROAD/EASEMENT MAINTENANCE 2,000 2,000 0 0%
510-4563-504.51-10 ELECTRICAL SERVICES 252,390 263,495 11,105 4%
510-4563-504.52-30 TELEPHONE SERVICES 47,140 47,140 0 0%
510-4563-504.54-10 LEASE/RENTAL EQUIPMENT 3,150 3,150 0 0%
510-4563-504.55-40 EDUCATION & TRAINING 7,200 7,200 0 0%
510-4563-504.58-22 ODOR CONTROL PROGRAM 150,848 171,048 20,200 13%
510-4563-504.58-46 PRE-EMPLOYMENT EXPENSES 200 200 0 0%
510-4563-504.58-55 INDUSTRIAL SAFETY PROGRAM 17,500 7,500 (10,000) -57%
510-4563-504.60-08 VEHICLE & EQUIPMENT FUELS 22,405 22,405 0 0%
510-4563-504.60-11 UNIFORMS 9,687 9,687 0 0%
510-4563-504.60-14 OPERATING SUPPLIES 6,413 13,913 7,500 117%
510-4563-504.60-27 SMALL TOOL/EQUIP REPLCMNT 3,000 8,000 5,000 167%
510-4563-504.80-05 MOTOR VEHICLES & EQUIP 0 148,000 148,000 n/a
Utilities - Pump Station Maintenance 1,282,877 1,465,710 182,833 14.3%
40 FY 2025 Adopted to FY 2026 Adopted Budget Comparison
Account Account Description FY 2025 Adopted Budget
FY 2026 Adopted Budget $ Change % Change
Utilities - Laboratory Services:
510-4564-501.11-01 REGULAR 720,869 826,493 105,624 15%
510-4564-501.12-01 OVERTIME 5,993 5,262 (731) -12%
510-4564-501.12-02 ON-CALL 14,546 15,048 502 3%
510-4564-501.21-01 FICA 43,501 50,181 6,680 15%
510-4564-501.21-02 MEDICARE 10,173 11,736 1,563 15%
510-4564-501.22-10 RETIREMENT 103,551 119,940 16,389 16%
510-4564-501.23-10 HEALTH INSURANCE 124,046 116,576 (7,470) -6%
510-4564-501.24-01 INSURANCE 12,373 13,980 1,607 13%
510-4564-501.27-10 SELF INSURED 11,373 5,590 (5,783) -51%
510-4564-501.31-10 HEALTH SERVICES 200 200 0 0%
510-4564-501.31-30 MGT CONSULTING SERVICES 75,000 75,000 0 0%
510-4564-501.31-78 LABORATORY SERVICES 49,500 64,500 15,000 30%
510-4564-501.31-79 REGULATORY AGENCY FEE 4,700 5,700 1,000 21%
510-4564-501.31-90 OTHER PROFESSIONAL SERVICES 1,200 2,080 880 73%
510-4564-501.33-10 REPAIRS & MAINTENANCE 6,000 7,000 1,000 17%
510-4564-501.33-11 AUTO REPAIRS & MAINT 5,204 4,773 (431) -8%
510-4564-501.33-20 MAINTENANCE SVC CONTRACTS 45,000 56,900 11,900 26%
510-4564-501.35-01 PRINTING & BINDING 50 50 0 0%
510-4564-501.39-10 SOFTWARE APPLICATIONS 0 700 700 n/a
510-4564-501.52-10 POSTAL SERVICES 3,000 3,000 0 0%
510-4564-501.52-30 TELEPHONE SERVICES 6,720 7,420 700 10%
510-4564-501.55-30 SUBSISTENCE & LODGING 9,672 9,064 (608) -6%
510-4564-501.55-40 EDUCATION & TRAINING 6,915 8,120 1,205 17%
510-4564-501.58-10 DUES & ASSOC MEMBERSHIPS 2,025 2,225 200 10%
510-4564-501.58-46 PRE-EMPLOYMENT EXPENSES 200 400 200 100%
510-4564-501.58-55 INDUSTRIAL SAFETY PROGRAM 14,500 15,500 1,000 7%
510-4564-501.60-01 OFFICE SUPPLIES 7,500 8,250 750 10%
510-4564-501.60-07 REPAIRS & MAINT SUPPLIES 6,000 6,000 0 0%
510-4564-501.60-08 VEHICLE & EQUIPMENT FUELS 7,184 9,264 2,080 29%
510-4564-501.60-11 UNIFORMS 20,800 21,454 654 3%
510-4564-501.60-12 BOOKS & SUBSCRIPTIONS 1,300 1,500 200 15%
510-4564-501.60-14 OPERATING SUPPLIES 160,000 176,000 16,000 10%
510-4564-501.80-01 MACHINERY & EQUIPMENT 123,900 77,100 (46,800) -38%
510-4564-501.80-02 FURNITURE & FIXTURES 0 9,500 9,500 n/a
510-4564-501.80-05 MOTOR VEHICLES & EQUIP 58,500 55,000 (3,500) -6%
510-4564-501.80-07 COMPUTER EQUIPMENT 900 2,500 1,600 178%
Utilities - Laboratory Services 1,662,395 1,794,006 131,611 7.9%
Utilities Fund Transfers:
510-9190-501.99-31 TO CAPITAL PROJECTS 350,000 165,500 (184,500) -53%
Utilities Fund Transfers 350,000 165,500 (184,500) -52.7%
Utilities Debt Service:
510-9510-506.90-02 BOND ADMINISTRATIVE FEES 20,500 24,340 3,840 19%
510-9510-506.91-41 2022 W/S REVENUE BONDS 160,000 165,000 5,000 3%
510-9510-506.91-42 2023 W/S REVENUE BONDS 210,000 220,000 10,000 5%
510-9510-506.91-43 2024 W/S REVENUE BONDS 0 685,000 685,000 n/a
510-9510-506.91-55 2015 W/S REVENUE BONDS 3,855,000 4,040,000 185,000 5%
510-9510-506.91-57 2019 W/S REVENUE BONDS 1,350,000 1,420,000 70,000 5%
510-9510-506.91-58 2020 W/S REVENUE BONDS 2,260,000 2,365,000 105,000 5%
510-9510-506.92-41 2022 W/S REVENUE BONDS 2,970,025 2,961,901 (8,124) 0%
510-9510-506.92-42 2023 W/S REVENUE BONDS 1,745,750 1,735,000 (10,750) -1%
510-9510-506.92-43 2024 W/S REVENUE BONDS 978,908 3,523,050 2,544,142 260%
510-9510-506.92-55 15 W/S REF BONDS 1,082,919 890,170 (192,749) -18%
510-9510-506.92-57 2019 W/S REVENUE BONDS 1,000,100 930,850 (69,250) -7%
510-9510-506.92-58 2020 W/S REVENUE BONDS 1,523,900 1,408,275 (115,625) -8%
Utilities Debt Service 17,157,102 20,368,586 3,211,484 18.7%
41 FY 2025 Adopted to FY 2026 Adopted Budget Comparison
Account Account Description FY 2025 Adopted Budget
FY 2026 Adopted Budget $ Change % Change
Total Utilities Operating Fund 51,400,536 $ 56,576,207 $ 5,175,671 $ 10.1%
Utilities Capital Projects - Water Projects:
520-9170-802.89-03 CONSTRUCTION 7,612,000 28,198,384 20,586,384 270%
520-9170-803.89-03 CONSTRUCTION 3,000,000 0 (3,000,000) -100%
520-9170-805.89-02 DESIGN 600,000 0 (600,000) -100%
520-9170-805.89-03 CONSTRUCTION 17,158,000 51,571,616 34,413,616 201%
Utilities Capital Projects - Water Projects 28,370,000 79,770,000 51,400,000 181.2%
Utilities Capital Projects - Sewer Projects:
520-9180-802.80-01 MACHINERY & EQUIPMENT 2,315,000 1,090,000 (1,225,000) -53%
520-9180-802.89-01 LAND ACQUISITION 0 750,000 750,000 n/a
520-9180-802.89-03 CONSTRUCTION 2,000,000 1,000,000 (1,000,000) -50%
520-9180-805.89-03 CONSTRUCTION 25,030,000 27,000,000 1,970,000 8%
Utilities Capital Projects - Sewer Projects 29,345,000 29,840,000 495,000 1.7%
Utilities Capital Projects - General:
520-9190-802.31-90 OTHER PROFESSIONAL SER 1,005,000 575,000 (430,000) -43%
520-9190-802.80-01 MACHINERY & EQUIPMENT 0 700,000 700,000 n/a
520-9190-802.89-03 CONSTRUCTION 0 250,000 250,000 n/a
Utilities Capital Projects - General 1,005,000 1,525,000 520,000 51.7%
Total Utilities Capital Projects Fund 58,720,000 $ 111,135,000 $ 52,415,000 $ 89.3%
Joint Fleet Maintenance Fund:
610-4610-601.33-11 AUTO REPAIRS & MAINT 3,766,329 4,037,370 271,041 7%
Total Joint Fleet Maintenance Fund 3,766,329 $ 4,037,370 $ 271,041 $ 7.2%
Grand Total less Joint Fleet Maintenance Fund 1,005,515,124 $ 1,153,562,818 $ 148,047,694 $ 14.7%
Excludes the Joint Fleet Maintenance Fund so as not to double-count the revenues and expenditures associated with fleet maintenance. The Joint Fleet Maintenance Fund charges other funds for costs, and the expenditures show in the funds being charged.
42 |
3765 | https://www.cs.cornell.edu/courses/cs381/2000FA/Assignments/ho30.pdf |
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3766 | https://www.cantab.net/users/henry.liu/Yaping%20Mao%2028-6-19%20GZDM%20seminar%20slides.pdf | Gallai-Ramsey Number of Graphs Yaping Mao School of Mathematics and Statistics Qinghai Normal University, Xining, Qinghai, China May 2019 Introduction Main results Outline 1 Introduction Classical Ramsey number Gallai Ramsey number under K3-coloring Gallai Ramsey number under S+ 3 -coloring 2 Main results Books Wheels Fans Stars with extra independent edges Odd cycles Two Classes of Unicyclic Graphs Results for S+ 3 -coloring Yaping Mao June, 2019 Gallai-Ramsey Number of Graphs Introduction Main results Co-authors Joint work with Colton Magnant: Clayton State University, USA; Ingo Schiermeyer: Technische Universit¨ at Bergakademie Freiberg, Germany; Zhao Wang: China Jiliang University, China; Jinyu Zou: Qinghai Normal University, China.
Yaping Mao June, 2019 Gallai-Ramsey Number of Graphs Introduction Main results Classical Ramsey number Gallai Ramsey number under K3-coloring Gallai Ramsey number under S+ 3 -coloring Outline 1 Introduction Classical Ramsey number Gallai Ramsey number under K3-coloring Gallai Ramsey number under S+ 3 -coloring 2 Main results Books Wheels Fans Stars with extra independent edges Odd cycles Two Classes of Unicyclic Graphs Results for S+ 3 -coloring Yaping Mao June, 2019 Gallai-Ramsey Number of Graphs Introduction Main results Classical Ramsey number Gallai Ramsey number under K3-coloring Gallai Ramsey number under S+ 3 -coloring Rainbow coloring A coloring of a graph is called rainbow if no two edges have the same color.
Colorings of complete graphs that contain no rainbow triangle have very interesting and somewhat surprising structure.
T. Gallai, Transitiv orientierbare Graphen, Acta Math. Acad.
Sci. Hungar 18 (1967), 25–66 first examined this structure under the guise of transitive orientations.
Yaping Mao June, 2019 Gallai-Ramsey Number of Graphs Introduction Main results Classical Ramsey number Gallai Ramsey number under K3-coloring Gallai Ramsey number under S+ 3 -coloring Rainbow coloring The result was reproven in A. Gy´ arf´ as and G. Simonyi, Edge colorings of complete graphs without tricolored triangles, J.
Graph Theory 46(3) (2004), 211–216 in the terminology of graphs and can also be traced to K. Cameron and J. Edmonds, Lambda composition, J. Graph Theory 26(1) (1997), 9–16.
Yaping Mao June, 2019 Gallai-Ramsey Number of Graphs Introduction Main results Classical Ramsey number Gallai Ramsey number under K3-coloring Gallai Ramsey number under S+ 3 -coloring Outline 1 Introduction Classical Ramsey number Gallai Ramsey number under K3-coloring Gallai Ramsey number under S+ 3 -coloring 2 Main results Books Wheels Fans Stars with extra independent edges Odd cycles Two Classes of Unicyclic Graphs Results for S+ 3 -coloring Yaping Mao June, 2019 Gallai-Ramsey Number of Graphs Introduction Main results Classical Ramsey number Gallai Ramsey number under K3-coloring Gallai Ramsey number under S+ 3 -coloring Classical Ramsey number Given two graphs G and H, the graph Ramsey number R(G, H) is the minimum integer n such that every (red/blue)-coloring of the edges of Kn contains either a red copy of G or a blue copy of H.
K5 Colored K5 Yaping Mao June, 2019 Gallai-Ramsey Number of Graphs Introduction Main results Classical Ramsey number Gallai Ramsey number under K3-coloring Gallai Ramsey number under S+ 3 -coloring Classical Ramsey number R(K3, K3) = 6; R(K4, K4) = 18; 43 ≤R(K5, K5) ≤49; · · · See the dynamic survey S. P. Radziszowski, Small Ramsey numbers, Electron. J. Combin., Dynamic Survey 1, 30, 1994.
Yaping Mao June, 2019 Gallai-Ramsey Number of Graphs Introduction Main results Classical Ramsey number Gallai Ramsey number under K3-coloring Gallai Ramsey number under S+ 3 -coloring Outline 1 Introduction Classical Ramsey number Gallai Ramsey number under K3-coloring Gallai Ramsey number under S+ 3 -coloring 2 Main results Books Wheels Fans Stars with extra independent edges Odd cycles Two Classes of Unicyclic Graphs Results for S+ 3 -coloring Yaping Mao June, 2019 Gallai-Ramsey Number of Graphs Introduction Main results Classical Ramsey number Gallai Ramsey number under K3-coloring Gallai Ramsey number under S+ 3 -coloring Gallai-Ramsey number Definition 1 Given two graphs G and H and an integer k, the Gallai-Ramsey number grk(G : H) is defined to be the minimum integer n such that any k coloring of the complete graph Kn contains either a rainbow colored G or a monochromatic H.
We generally assume G = K3, and therefore k ≥3.
Note that these numbers are bounded by multicolor Ramsey numbers so existence is obvious.
Yaping Mao June, 2019 Gallai-Ramsey Number of Graphs Introduction Main results Classical Ramsey number Gallai Ramsey number under K3-coloring Gallai Ramsey number under S+ 3 -coloring Rainbow triangle free coloring For the following statement, a trivial partition is a partition into only one part.
Theorem 1.1 The vertices of every rainbow triangle free coloring of a complete graph can be partitioned such that all edges between a pair of parts have a single color and all edges between the parts come from only two colors.
Here by “rainbow” we mean all different colors.
Yaping Mao June, 2019 Gallai-Ramsey Number of Graphs Introduction Main results Classical Ramsey number Gallai Ramsey number under K3-coloring Gallai Ramsey number under S+ 3 -coloring Gallai Partition under K3-coloring What a partition might look like.
Yaping Mao June, 2019 Gallai-Ramsey Number of Graphs Introduction Main results Classical Ramsey number Gallai Ramsey number under K3-coloring Gallai Ramsey number under S+ 3 -coloring Gallai Partition under K3-coloring But we don’t know how many parts there are...
Yaping Mao June, 2019 Gallai-Ramsey Number of Graphs Introduction Main results Classical Ramsey number Gallai Ramsey number under K3-coloring Gallai Ramsey number under S+ 3 -coloring Gallai Partition under K3-coloring But we don’t know how many parts there are...or big they are...
Yaping Mao June, 2019 Gallai-Ramsey Number of Graphs Introduction Main results Classical Ramsey number Gallai Ramsey number under K3-coloring Gallai Ramsey number under S+ 3 -coloring Gallai Partition under K3-coloring But we don’t know how many parts there are... or big they are...
or what happens inside the parts...
Yaping Mao June, 2019 Gallai-Ramsey Number of Graphs Introduction Main results Classical Ramsey number Gallai Ramsey number under K3-coloring Gallai Ramsey number under S+ 3 -coloring Gallai Partition under K3-coloring But we don’t know how many parts there are... or big they are...
or what happens inside the parts...
?
?
?
?
Yaping Mao June, 2019 Gallai-Ramsey Number of Graphs Introduction Main results Classical Ramsey number Gallai Ramsey number under K3-coloring Gallai Ramsey number under S+ 3 -coloring Gallai Partition under K3-coloring But we don’t know how many parts there are... or big they are...
or what happens inside the parts... but we do know the inside is rainbow triangle free.
Yaping Mao June, 2019 Gallai-Ramsey Number of Graphs Introduction Main results Classical Ramsey number Gallai Ramsey number under K3-coloring Gallai Ramsey number under S+ 3 -coloring Gallai-coloring under K3-coloring For ease of notation, we refer to a colored complete graph with no rainbow triangle as a Gallai-coloring and the partition provided by Theorem 1.1 as a Gallai-partition.
The induced subgraph of a Gallai colored complete graph constructed by selecting a single vertex from each part of a Gallai partition is called the reduced graph of that partition.
By Theorem 1.1, the reduced graph is a 2-colored complete graph.
Yaping Mao June, 2019 Gallai-Ramsey Number of Graphs Introduction Main results Classical Ramsey number Gallai Ramsey number under K3-coloring Gallai Ramsey number under S+ 3 -coloring Survey papers The theory of Gallai-Ramsey numbers has grown by leaps and bounds in recent years, especially for the case where G = K3.
We refer the interested reader to the survey S. Fujita, C.
Magnant, K. Ozeki, Rainbow generalizations of Ramsey theory: a survey, Graphs Combin. 26(1) (2010), 1–30.
An updated version at S. Fujita, C. Magnant, and K. Ozeki.
Rainbow generalizations of Ramsey theory–a dynamic survey, Theo. Appl. Graphs 0(1), 2014 for more general information.
Yaping Mao June, 2019 Gallai-Ramsey Number of Graphs Introduction Main results Classical Ramsey number Gallai Ramsey number under K3-coloring Gallai Ramsey number under S+ 3 -coloring Outline of a Simple Proof Theorem 1.2 grk(K3 : C4) = k + 4.
Yaping Mao June, 2019 Gallai-Ramsey Number of Graphs Introduction Main results Classical Ramsey number Gallai Ramsey number under K3-coloring Gallai Ramsey number under S+ 3 -coloring Outline of a Simple Proof For the lower bound, consider the lex. coloring, starting at a 2-colored K5.
No rainbow triangle and no monochromatic C4.
Yaping Mao June, 2019 Gallai-Ramsey Number of Graphs Introduction Main results Classical Ramsey number Gallai Ramsey number under K3-coloring Gallai Ramsey number under S+ 3 -coloring Outline of a Simple Proof For the lower bound, consider the lex. coloring, starting at a 2-colored K5.
No rainbow triangle and no monochromatic C4.
Yaping Mao June, 2019 Gallai-Ramsey Number of Graphs Introduction Main results Classical Ramsey number Gallai Ramsey number under K3-coloring Gallai Ramsey number under S+ 3 -coloring Outline of a Simple Proof For the lower bound, consider the lex. coloring, starting at a 2-colored K5.
No rainbow triangle and no monochromatic C4.
Yaping Mao June, 2019 Gallai-Ramsey Number of Graphs Introduction Main results Classical Ramsey number Gallai Ramsey number under K3-coloring Gallai Ramsey number under S+ 3 -coloring Outline of a Simple Proof n = k + 3, no rainbow triangle, no monochromatic C4.
Yaping Mao June, 2019 Gallai-Ramsey Number of Graphs Introduction Main results Classical Ramsey number Gallai Ramsey number under K3-coloring Gallai Ramsey number under S+ 3 -coloring Outline of a Simple Proof Now consider an arbitrary coloring of Kk+4 using k colors with no rainbow triangle.
Yaping Mao June, 2019 Gallai-Ramsey Number of Graphs Introduction Main results Classical Ramsey number Gallai Ramsey number under K3-coloring Gallai Ramsey number under S+ 3 -coloring Outline of a Simple Proof There is a Gallai partition, but we don’t know how big the parts are.
Recall: all one color between each pair.
Yaping Mao June, 2019 Gallai-Ramsey Number of Graphs Introduction Main results Classical Ramsey number Gallai Ramsey number under K3-coloring Gallai Ramsey number under S+ 3 -coloring Outline of a Simple Proof If all parts are single vertices, then we simply have a 2-coloring with n ≥6.
Applying R(C4, C4) = 6, we have ?
Yaping Mao June, 2019 Gallai-Ramsey Number of Graphs Introduction Main results Classical Ramsey number Gallai Ramsey number under K3-coloring Gallai Ramsey number under S+ 3 -coloring Outline of a Simple Proof There must be parts with at least 2 vertices.
But if two parts have at least 2 vertices each, then we’ve easily gotten a monochromatic C4.
Yaping Mao June, 2019 Gallai-Ramsey Number of Graphs Introduction Main results Classical Ramsey number Gallai Ramsey number under K3-coloring Gallai Ramsey number under S+ 3 -coloring Outline of a Simple Proof Then there must be at most one big part (≥2), and all others are one vertex.
Yaping Mao June, 2019 Gallai-Ramsey Number of Graphs Introduction Main results Classical Ramsey number Gallai Ramsey number under K3-coloring Gallai Ramsey number under S+ 3 -coloring Outline of a Simple Proof If there are three vertices outside, then pigeonhole gives us a C4.
Yaping Mao June, 2019 Gallai-Ramsey Number of Graphs Introduction Main results Classical Ramsey number Gallai Ramsey number under K3-coloring Gallai Ramsey number under S+ 3 -coloring Outline of a Simple Proof Finally there are at most two vertices outside, induct on the number of colors (with maximum degree at least 2) in the big part.
Yaping Mao June, 2019 Gallai-Ramsey Number of Graphs Introduction Main results Classical Ramsey number Gallai Ramsey number under K3-coloring Gallai Ramsey number under S+ 3 -coloring Outline of a Simple Proof Finally there are at most two vertices outside, induct on the number of colors (with maximum degree at least 2) in the big part.
Yaping Mao June, 2019 Gallai-Ramsey Number of Graphs Introduction Main results Classical Ramsey number Gallai Ramsey number under K3-coloring Gallai Ramsey number under S+ 3 -coloring Outline of a Simple Proof Finally there are at most two vertices outside, induct on the number of colors (with maximum degree at least 2) in the big part.
Yaping Mao June, 2019 Gallai-Ramsey Number of Graphs Introduction Main results Classical Ramsey number Gallai Ramsey number under K3-coloring Gallai Ramsey number under S+ 3 -coloring Outline 1 Introduction Classical Ramsey number Gallai Ramsey number under K3-coloring Gallai Ramsey number under S+ 3 -coloring 2 Main results Books Wheels Fans Stars with extra independent edges Odd cycles Two Classes of Unicyclic Graphs Results for S+ 3 -coloring Yaping Mao June, 2019 Gallai-Ramsey Number of Graphs Introduction Main results Classical Ramsey number Gallai Ramsey number under K3-coloring Gallai Ramsey number under S+ 3 -coloring Gallai Partition under S+ 3 -coloring Let S+ 3 be the graph on 4 vertices consisting of a triangle and a pendant edge.
S. Fujita, C. Magnant, Extensions of Gallai-Ramsey results, J.
Graph Theory 70(4) (2012), 404–426 proved a decomposition theorem for rainbow S+ 3 -free colorings of a complete graph.
S+ 3 Yaping Mao June, 2019 Gallai-Ramsey Number of Graphs Introduction Main results Classical Ramsey number Gallai Ramsey number under K3-coloring Gallai Ramsey number under S+ 3 -coloring Gallai Partition under S+ 3 -coloring Theorem 1.3 (Fujita, Magnant) In any rainbow S+ 3 -free coloring G of a complete graph, one of the following holds: (1) V (G) can be partitioned such that there are 2 colors on the edges among the parts, and at most 2 colors on the edges between each pair of parts; or (2) There are three (different colored) monochromatic spanning trees, and moreover, there exists a partition of V (G) with exactly 3 colors on edges between parts and between each pair of parts, the edges have only one color.
Yaping Mao June, 2019 Gallai-Ramsey Number of Graphs Introduction Main results Classical Ramsey number Gallai Ramsey number under K3-coloring Gallai Ramsey number under S+ 3 -coloring Gallai Partition under S+ 3 -coloring H4 H2 H1 H3 Hr H4 H2 H1 H3 Hr Type 1 Type 2 Yaping Mao June, 2019 Gallai-Ramsey Number of Graphs Introduction Main results Books Wheels Fans Stars with extra independent edges Odd cycles Two Classes of Unicyclic Graphs Results for S+ 3 -coloring Outline 1 Introduction Classical Ramsey number Gallai Ramsey number under K3-coloring Gallai Ramsey number under S+ 3 -coloring 2 Main results Books Wheels Fans Stars with extra independent edges Odd cycles Two Classes of Unicyclic Graphs Results for S+ 3 -coloring Yaping Mao June, 2019 Gallai-Ramsey Number of Graphs Introduction Main results Books Wheels Fans Stars with extra independent edges Odd cycles Two Classes of Unicyclic Graphs Results for S+ 3 -coloring Outline 1 Introduction Classical Ramsey number Gallai Ramsey number under K3-coloring Gallai Ramsey number under S+ 3 -coloring 2 Main results Books Wheels Fans Stars with extra independent edges Odd cycles Two Classes of Unicyclic Graphs Results for S+ 3 -coloring Yaping Mao June, 2019 Gallai-Ramsey Number of Graphs Introduction Main results Books Wheels Fans Stars with extra independent edges Odd cycles Two Classes of Unicyclic Graphs Results for S+ 3 -coloring The book graph The book graph with m pages is denoted by Bm, where Bm = K2 + Km.
Note that B1 = K3 and B2 = K4{e} where e is an edge of the K4.
In this work, we prove bounds on the Gallai-Ramsey number of all books, with sharp results for several small books.
B 4 Yaping Mao June, 2019 Gallai-Ramsey Number of Graphs Introduction Main results Books Wheels Fans Stars with extra independent edges Odd cycles Two Classes of Unicyclic Graphs Results for S+ 3 -coloring Small graphs Theorem 2.1 (Gyarfas, Simonyi (2004)) In any G-coloring of a complete graph, there is a vertex with at least 2n 5 incident edges in a single color.
Theorem 2.2 (Chv´ atal and Harary (1976), Rousseau and Sheehan (1978)) R(B2, B2) = 10, R(B3, B3) = 14, R(B4, B4) = 18, R(B5, B5) = 21.
Yaping Mao June, 2019 Gallai-Ramsey Number of Graphs Introduction Main results Books Wheels Fans Stars with extra independent edges Odd cycles Two Classes of Unicyclic Graphs Results for S+ 3 -coloring General Lower Bound In this section, we prove a lower bound on the Gallai-Ramsey number for books by a straightforward inductive construction.
Yaping Mao June, 2019 Gallai-Ramsey Number of Graphs Introduction Main results Books Wheels Fans Stars with extra independent edges Odd cycles Two Classes of Unicyclic Graphs Results for S+ 3 -coloring General Lower Bound In this section, we prove a lower bound on the Gallai-Ramsey number for books by a straightforward inductive construction.
Theorem 2.3 (Zou, Mao, Wang, Magnant, Ye) If Bm is the book with m pages, Bm = K2 + Km, then for k ≥2, grk(K3 : Bm) ≥ ( (R(Bm, Bm) −1) · 5(k−2)/2 + 1 if k is even, 2 · (R(Bm, Bm) −1) · 5(k−3)/2 + 1 if k is odd.
Yaping Mao June, 2019 Gallai-Ramsey Number of Graphs Introduction Main results Books Wheels Fans Stars with extra independent edges Odd cycles Two Classes of Unicyclic Graphs Results for S+ 3 -coloring General Upper Bound Let Rm = R(Bm : Bm) and define R′ m = m−1 X i=1 [R⌈m/i⌉−1].
This quantity provides a bound on a type of restricted Ramsey number as seen in the following lemma.
Lemma 2 For m ≥2, the largest number of vertices in a G-coloring of a complete graph with no monochromatic Bm in which all parts of the G-partition have order at most m −1 is at most R′ m.
Yaping Mao June, 2019 Gallai-Ramsey Number of Graphs Introduction Main results Books Wheels Fans Stars with extra independent edges Odd cycles Two Classes of Unicyclic Graphs Results for S+ 3 -coloring General Upper Bound Note that for m large, since Rm ∼(4 + o(1))m (see [Rousseau and Sheehan (1978)]), we get R′ m ∼(4 + o(1))m ln[(4 + o(1))m].
For small values of m, we compute R′ 2 = 9, R′ 3 = 22, and R′ 4 = 35.
Call a color m-admissible if it induces a subgraph with maximum degree at least m, and m-inadmissible otherwise.
Yaping Mao June, 2019 Gallai-Ramsey Number of Graphs Introduction Main results Books Wheels Fans Stars with extra independent edges Odd cycles Two Classes of Unicyclic Graphs Results for S+ 3 -coloring General Upper Bound Lemma 3 (Zou, Mao, Wang, Magnant, Ye) Given integers m ≥2 and k ≥2, let n be the largest number of vertices in a k-coloring of a complete graph in which there is no rainbow triangle, no monochromatic Bm, a G-partition with all parts having order at most m −1, and only one m-admissible color.
Then n ≤ ( 3m −1 if k = 2, 5m −5 otherwise.
Yaping Mao June, 2019 Gallai-Ramsey Number of Graphs Introduction Main results Books Wheels Fans Stars with extra independent edges Odd cycles Two Classes of Unicyclic Graphs Results for S+ 3 -coloring General Upper Bound Let ℓ= ℓ(m) be the number of colors that are m-inadmissible and define the quantity grk,ℓ(K3 : H) to be the minimum integer n such that every k coloring of Kn with at least ℓdifferent m-inadmissible colors contains either a rainbow triangle or a monochromatic copy of H.
We may now state our main result, which provides a general upper bound on the Gallai-Ramsey numbers for any book with any number of colors.
Yaping Mao June, 2019 Gallai-Ramsey Number of Graphs Introduction Main results Books Wheels Fans Stars with extra independent edges Odd cycles Two Classes of Unicyclic Graphs Results for S+ 3 -coloring General Upper Bound Theorem 2.4 (Zou, Mao, Wang, Magnant, Ye) Given positive integers k ≥1, m ≥3, and 0 ≤ℓ≤k, let grk,ℓ,m = m + 2 −ℓ if k = 1, R′ m · 5 k−2 2 + 1 −(m −1)ℓ if k is even, 2 · R′ m · 5 k−3 2 + 1 −(m −1)ℓ if k ≥3 is odd.
Then grk,ℓ(K3 : Bm) ≤grk,ℓ,m.
Yaping Mao June, 2019 Gallai-Ramsey Number of Graphs Introduction Main results Books Wheels Fans Stars with extra independent edges Odd cycles Two Classes of Unicyclic Graphs Results for S+ 3 -coloring Some Small Cases In this section, we provide the sharp Gallai-Ramsey number for several small books. The proof of this result follows the proof of Theorem 2.23 except each step is improved in order to produce the sharp result.
Yaping Mao June, 2019 Gallai-Ramsey Number of Graphs Introduction Main results Books Wheels Fans Stars with extra independent edges Odd cycles Two Classes of Unicyclic Graphs Results for S+ 3 -coloring Some Small Cases In this section, we provide the sharp Gallai-Ramsey number for several small books. The proof of this result follows the proof of Theorem 2.23 except each step is improved in order to produce the sharp result.
Lemma 4 Let k, ℓ, m be integers with k ≥3, 0 ≤ℓ≤k −2, and 2 ≤m ≤5. If G is a G-coloring of Kp with p ≥grk,ℓ,m using k colors in which all parts of a G-partition have order at most m −1 and ℓcolors are m-inadmissible, then G contains a monochromatic copy of Bm.
Yaping Mao June, 2019 Gallai-Ramsey Number of Graphs Introduction Main results Books Wheels Fans Stars with extra independent edges Odd cycles Two Classes of Unicyclic Graphs Results for S+ 3 -coloring Some Small Cases For the values of m in question, we have p ≥ 18 if m = 2, 25 if m = 3, 32 if m = 4, and 37 if m = 5.
Let t be the number of parts in the partition. When m = 2, all parts of the assumed G-partition have order 1, meaning that G is simply a 2-coloring. Since |G| = p > 10 = R2, the claim is immediate. We consider cases for the remaining values of m.
Yaping Mao June, 2019 Gallai-Ramsey Number of Graphs Introduction Main results Books Wheels Fans Stars with extra independent edges Odd cycles Two Classes of Unicyclic Graphs Results for S+ 3 -coloring Outline 1 Introduction Classical Ramsey number Gallai Ramsey number under K3-coloring Gallai Ramsey number under S+ 3 -coloring 2 Main results Books Wheels Fans Stars with extra independent edges Odd cycles Two Classes of Unicyclic Graphs Results for S+ 3 -coloring Yaping Mao June, 2019 Gallai-Ramsey Number of Graphs Introduction Main results Books Wheels Fans Stars with extra independent edges Odd cycles Two Classes of Unicyclic Graphs Results for S+ 3 -coloring Gallai-Ramsey number for wheels Let Wn be a wheel of order n, that is, Wn = K1 ∨Cn−1 where Cn−1 is the cycle on n −1 vertices.
Theorem 2.5 (Mao, Wang, Magnant, Schiermeyer) (1) R(W5, W5) = 15; (2) R(W6, W6) = 17.
Yaping Mao June, 2019 Gallai-Ramsey Number of Graphs Introduction Main results Books Wheels Fans Stars with extra independent edges Odd cycles Two Classes of Unicyclic Graphs Results for S+ 3 -coloring Gallai-Ramsey number for wheels As far as we are aware, for n ≥7, the classical diagonal Ramsey number for the wheel is yet unknown.
We give upper and lower bounds for classical Ramsey number of the general wheel Wn.
Theorem 2.6 (Mao, Wang, Magnant, Schiermeyer) For k ≥1 and n ≥7, ( 3n −3 ≤R(Wn, Wn) ≤8n −10, if n is even; 2n −2 ≤R(Wn, Wn) ≤6n −8 if n is odd.
Yaping Mao June, 2019 Gallai-Ramsey Number of Graphs Introduction Main results Books Wheels Fans Stars with extra independent edges Odd cycles Two Classes of Unicyclic Graphs Results for S+ 3 -coloring Gallai-Ramsey number for wheels We obtain the exact value of the Gallai Ramsey number for W5.
Theorem 2.7 (Mao, Wang, Magnant, Schiermeyer) For k ≥1, grk(K3 : W5) = 5 if k = 1, 14 · 5 k−2 2 + 1 if k is even, 28 · 5 k−3 2 + 1 if k ≥3 is odd.
Yaping Mao June, 2019 Gallai-Ramsey Number of Graphs Introduction Main results Books Wheels Fans Stars with extra independent edges Odd cycles Two Classes of Unicyclic Graphs Results for S+ 3 -coloring Gallai-Ramsey number for wheels We provide general lower bounds on the Gallai-Ramsey numbers for all wheels.
Theorem 5 (Mao, Wang, Magnant, Schiermeyer) For k ≥2 and n ≥6, we have grk(K3 : Wn) ≥ (3n −4)5 k−2 2 + 1 if n is even and k is even; (6n −8)5 k−3 2 + 1 if n is even and k is odd; (2n −3)5 k−2 2 + 1 if n is odd and k is even; (4n −6)5 k−3 2 + 1 if n is odd and k is odd.
Yaping Mao June, 2019 Gallai-Ramsey Number of Graphs Introduction Main results Books Wheels Fans Stars with extra independent edges Odd cycles Two Classes of Unicyclic Graphs Results for S+ 3 -coloring Gallai-Ramsey number for wheels We provide general upper bounds on the Gallai-Ramsey numbers for all wheels.
Theorem 6 (Mao, Wang, Magnant, Schiermeyer) For k ≥3 and n ≥6, we have grk(K3 : Wn) ≤(n −4)2 · 30k + k(n −1).
Yaping Mao June, 2019 Gallai-Ramsey Number of Graphs Introduction Main results Books Wheels Fans Stars with extra independent edges Odd cycles Two Classes of Unicyclic Graphs Results for S+ 3 -coloring Outline 1 Introduction Classical Ramsey number Gallai Ramsey number under K3-coloring Gallai Ramsey number under S+ 3 -coloring 2 Main results Books Wheels Fans Stars with extra independent edges Odd cycles Two Classes of Unicyclic Graphs Results for S+ 3 -coloring Yaping Mao June, 2019 Gallai-Ramsey Number of Graphs Introduction Main results Books Wheels Fans Stars with extra independent edges Odd cycles Two Classes of Unicyclic Graphs Results for S+ 3 -coloring Gallai-Ramsey number for fans The fan graph of order n is denoted by Fn, where Bn = K1 + nK2.
Note that F1 = K3 and F2 is a graph obtained from two triangles by sharing one vertex.
Theorem 2.8 (1) R(F2, F2) = 9; (2) R(F3, F3) = 13; (3) 4n + 1 ≤R(Fn, Fn) ≤6n for large sufficiently n, 6n < n2 + 1.
Yaping Mao June, 2019 Gallai-Ramsey Number of Graphs Introduction Main results Books Wheels Fans Stars with extra independent edges Odd cycles Two Classes of Unicyclic Graphs Results for S+ 3 -coloring Gallai-Ramsey number for fans First our sharp result for F2.
Theorem 2.9 (Mao, Wang, Magnant, Schiermeyer) grk(K3; F2) = 9, if k = 2; 83 2 · 5 k−4 2 + 1 2, if k is even, k ≥4; 4 · 5 k−1 2 + 1, if k is odd.
Yaping Mao June, 2019 Gallai-Ramsey Number of Graphs Introduction Main results Books Wheels Fans Stars with extra independent edges Odd cycles Two Classes of Unicyclic Graphs Results for S+ 3 -coloring Gallai-Ramsey number for fans Next our general bounds (and sharp result for any even number of colors) for F3.
Theorem 2.10 (Mao, Wang, Magnant, Schiermeyer) For k ≥2, grk(K3; F3) = 14 · 5 k−2 2 −1, if k is even; grk(K3; F3) = 33 · 5 k−3 2 , if k = 3, 5; 33 · 5 k−3 2 ≤grk(K3; F3) ≤33 · 5 k−3 2 + 3 4 · 5 k−5 2 −3 4, if k is odd, k ≥7.
Yaping Mao June, 2019 Gallai-Ramsey Number of Graphs Introduction Main results Books Wheels Fans Stars with extra independent edges Odd cycles Two Classes of Unicyclic Graphs Results for S+ 3 -coloring Gallai-Ramsey number for fans In particular, we conjecture the following, which claims that the lower bound in above theorem is the sharp result.
Conjecture 2.1 (Mao, Wang, Magnant, Schiermeyer) For k ≥2, grk(K3; F3) = ( 14 · 5 k−2 2 −1, if k is even; 33 · 5 k−3 2 , if k is odd.
Yaping Mao June, 2019 Gallai-Ramsey Number of Graphs Introduction Main results Books Wheels Fans Stars with extra independent edges Odd cycles Two Classes of Unicyclic Graphs Results for S+ 3 -coloring Gallai-Ramsey number for fans Finally our general bound for all fans.
Theorem 2.11 (Mao, Wang, Magnant, Schiermeyer) For k ≥2, 4n · 5 k−2 2 + 1 ≤grk(K3; Fn) ≤10n · 5 k−2 2 −5 2n + 1, if k is even; 2n · 5 k−1 2 + 1 ≤grk(K3; Fn) ≤9 2n · 5 k−1 2 −5 2n + 1, if k is odd.
Yaping Mao June, 2019 Gallai-Ramsey Number of Graphs Introduction Main results Books Wheels Fans Stars with extra independent edges Odd cycles Two Classes of Unicyclic Graphs Results for S+ 3 -coloring Outline 1 Introduction Classical Ramsey number Gallai Ramsey number under K3-coloring Gallai Ramsey number under S+ 3 -coloring 2 Main results Books Wheels Fans Stars with extra independent edges Odd cycles Two Classes of Unicyclic Graphs Results for S+ 3 -coloring Yaping Mao June, 2019 Gallai-Ramsey Number of Graphs Introduction Main results Books Wheels Fans Stars with extra independent edges Odd cycles Two Classes of Unicyclic Graphs Results for S+ 3 -coloring Stars with extra independent edges Let Sr t be a star of order t by adding extra r independent edges for 0 ≤r ≤t−1 2 .
For r = 0 we obtain Sr t = K1,t−1, which are called stars.
For r = t−1 2 if t is odd we obtain Sr t = F t−1 2 , which are called fans.
Yaping Mao June, 2019 Gallai-Ramsey Number of Graphs Introduction Main results Books Wheels Fans Stars with extra independent edges Odd cycles Two Classes of Unicyclic Graphs Results for S+ 3 -coloring Results for Ramsey number We deal with those graphs Sr t , where 0 < r < t−1 2 , i.e. where Sr t is neither a star nor a fan.
Theorem 2.12 (Mao, Wang, Magnant, Schiermeyer) (1) For t ≥7, R(S2 t , S2 t ) = 2t −1.
(2) For t ≥15, R(S3 t , S3 t ) = 2t −1.
(3) For t ≥6r −5, R(Sr t , Sr t ) = 2t −1.
Yaping Mao June, 2019 Gallai-Ramsey Number of Graphs Introduction Main results Books Wheels Fans Stars with extra independent edges Odd cycles Two Classes of Unicyclic Graphs Results for S+ 3 -coloring Results for Ramsey number For graph S2 t , we have the following.
Theorem 2.13 (Mao, Wang, Magnant, Schiermeyer) (1) For k ≥1, grk(K3; S2 6) = 2 × 5 k 2 + 1 4 × 5 k−2 2 + 3 4, if k is even; ⌈51 10 × 5 k−1 2 + 1 2⌉, if k is odd.
(2) For k ≥3, grk(K3; S2 8) = 14 × 5 k−2 2 + 1 2 × 5 k−4 2 + 1 2, if k is even; 7 × 5 k−1 2 + 1 4 × 5 k−3 2 + 3 4, if k is odd.
Yaping Mao June, 2019 Gallai-Ramsey Number of Graphs Introduction Main results Books Wheels Fans Stars with extra independent edges Odd cycles Two Classes of Unicyclic Graphs Results for S+ 3 -coloring Results for Gallai-Ramsey number For graph S2 t , we have the following.
Theorem 2.14 (Mao, Wang, Magnant, Schiermeyer) (3) For k ≥1 and t ≥6, 2(t −1) × 5 k−2 2 + 1 ≤grk(K3; S2 t ) ≤2t × 5 k−2 2 , if k is even; (t −1) × 5 k−1 2 + 1 ≤grk(K3; S2 t ) ≤t × 5 k−1 2 , if k is odd.
Yaping Mao June, 2019 Gallai-Ramsey Number of Graphs Introduction Main results Books Wheels Fans Stars with extra independent edges Odd cycles Two Classes of Unicyclic Graphs Results for S+ 3 -coloring Results for Gallai-Ramsey number For graph Sr t , we have the following.
Theorem 2.15 (Mao, Wang, Magnant, Schiermeyer) For t ≥6r −5, if k is even, then 2(t −1) × 5 k−2 2 + 1 ≤grk(K3; Sr t ) ≤[2t + 8(r −1)] × 5 k−2 2 −4(r −1); If k is odd, then (t −1) × 5 k−1 2 + 1 ≤grk(K3; Sr t ) ≤[t + 4(r −1)] × 5 k−1 2 −4(r −1).
Yaping Mao June, 2019 Gallai-Ramsey Number of Graphs Introduction Main results Books Wheels Fans Stars with extra independent edges Odd cycles Two Classes of Unicyclic Graphs Results for S+ 3 -coloring Outline 1 Introduction Classical Ramsey number Gallai Ramsey number under K3-coloring Gallai Ramsey number under S+ 3 -coloring 2 Main results Books Wheels Fans Stars with extra independent edges Odd cycles Two Classes of Unicyclic Graphs Results for S+ 3 -coloring Yaping Mao June, 2019 Gallai-Ramsey Number of Graphs Introduction Main results Books Wheels Fans Stars with extra independent edges Odd cycles Two Classes of Unicyclic Graphs Results for S+ 3 -coloring Upper and lower bound S. Fujita, C. Magnant, Gallai-Ramsey numbers for cycles, Discrete Math. 311(13)(2011), 1247–1254 and M. Hall, C.
Magnant, K. Ozeki, and M. Tsugaki. Improved upper bounds for Gallai-Ramsey numbers of paths and cycles, J. Graph Theory 75(1)(2014), 59–74 derived the following result.
Theorem 2.16 ℓ2k + 1 ≤grk(K3 : C2ℓ+1) ≤ℓ(2k+3 −3) log ℓ.
Sadly, we were not clever enough to get closer. We believe the lower bounds to be the truth...
Yaping Mao June, 2019 Gallai-Ramsey Number of Graphs Introduction Main results Books Wheels Fans Stars with extra independent edges Odd cycles Two Classes of Unicyclic Graphs Results for S+ 3 -coloring For C3 For the triangle C3 = K3, M. Axenovich, P. Iverson, Edge-colorings avoiding rainbow and monochromatic subgraphs, Discrete Math. 308(20)(2008), 4710–4723 and F. R. K. Chung, R.
L. Graham. Edge-colored complete graphs with precisely colored subgraphs, Combinatorica 3(3-4)(1983), 315–324 and A. Gy´ arf´ as, G. S´ ark¨ ozy, A. Seb˝ o, S. Selkow, Ramsey-type results for gallai colorings, J. Graph Theory 64(3)(2010), 233–243 obtained the following result.
Theorem 2.17 For k ≥2, grk(K3 : K3) = ( 5k/2 + 1 if k is even, 2 · 5(k−1)/2 + 1 if k is odd.
Yaping Mao June, 2019 Gallai-Ramsey Number of Graphs Introduction Main results Books Wheels Fans Stars with extra independent edges Odd cycles Two Classes of Unicyclic Graphs Results for S+ 3 -coloring For C5 For C5, S. Fujita, C. Magnant, Gallai-Ramsey numbers for cycles, Discrete Math. 311(13)(2011), 1247–1254 obtained the following result.
Theorem 2.18 For any positive integer k ≥2, we have grk(K3 : C5) = 2k+1 + 1.
Yaping Mao June, 2019 Gallai-Ramsey Number of Graphs Introduction Main results Books Wheels Fans Stars with extra independent edges Odd cycles Two Classes of Unicyclic Graphs Results for S+ 3 -coloring Our Main Result Theorem 2.19 (Wang, Mao, Magnant, Schiermeyer, Zou) For integers ℓ≥3 and k ≥1, we have grk(K3 : C2ℓ+1) = ℓ· 2k + 1.
The lower bound is sharp for odd cycles!
Yaping Mao June, 2019 Gallai-Ramsey Number of Graphs Introduction Main results Books Wheels Fans Stars with extra independent edges Odd cycles Two Classes of Unicyclic Graphs Results for S+ 3 -coloring General Lower Bound for Odd Cycles no H Yaping Mao June, 2019 Gallai-Ramsey Number of Graphs Introduction Main results Books Wheels Fans Stars with extra independent edges Odd cycles Two Classes of Unicyclic Graphs Results for S+ 3 -coloring Proof Outline - Setup Upper bound proof setup: Induction on the number of colors k.
First set aside vertices with (almost) all one color on their incident edges, call the set T .
Claim: T is not too big.
If T is big, then there is a large set of vertices Bi ⊆T (say |Bi| ≥ℓ) with all color i to what remains (A).
That color must be missing from A, so apply induction.
Yaping Mao June, 2019 Gallai-Ramsey Number of Graphs Introduction Main results Books Wheels Fans Stars with extra independent edges Odd cycles Two Classes of Unicyclic Graphs Results for S+ 3 -coloring Main Lemma Preview T A k′ colors Bk′−1 B2 B1 Yaping Mao June, 2019 Gallai-Ramsey Number of Graphs Introduction Main results Books Wheels Fans Stars with extra independent edges Odd cycles Two Classes of Unicyclic Graphs Results for S+ 3 -coloring Main Lemma Preview T A k′ colors Bk′−1 B2 B1 color k′ color k′ + 1 Yaping Mao June, 2019 Gallai-Ramsey Number of Graphs Introduction Main results Books Wheels Fans Stars with extra independent edges Odd cycles Two Classes of Unicyclic Graphs Results for S+ 3 -coloring Main Lemma Preview T A k′ colors Bk′−1 B2 B1 color k′ change to color k′ Yaping Mao June, 2019 Gallai-Ramsey Number of Graphs Introduction Main results Books Wheels Fans Stars with extra independent edges Odd cycles Two Classes of Unicyclic Graphs Results for S+ 3 -coloring Proof Outline - Main Lemma Lemma 7 Let k ≥3, 2 ≤k′ ≤k and let G be a Gallai coloring of the complete graph Kn containing no monochromatic copy of C2ℓ+1.
If G = A ∪B1 ∪B2 ∪· · · ∪Bk′−1 where A uses at most k′ colors, |Bi| ≤2ℓfor all i, and all edges between A and Bi have color i, then n ≤grk′(K3 : H) −1.
Note that this lemma uses the assumed structure to provide a bound on |G| even if G itself uses more than k′ colors.
Yaping Mao June, 2019 Gallai-Ramsey Number of Graphs Introduction Main results Books Wheels Fans Stars with extra independent edges Odd cycles Two Classes of Unicyclic Graphs Results for S+ 3 -coloring Proof Outline - Reductions Proof outline: Claims: The largest part of partition is small (|H1| ≤ℓ/2).
This means k = 3 since no C2ℓ+1 could fit inside a part, so n = 8ℓ+ 1.
Finally consider structure of H1 relative to the rest of G.
Yaping Mao June, 2019 Gallai-Ramsey Number of Graphs Introduction Main results Books Wheels Fans Stars with extra independent edges Odd cycles Two Classes of Unicyclic Graphs Results for S+ 3 -coloring Sketch of the Proof Broad structure of G. Suppose GR is the larger side.
H1 GR GB Yaping Mao June, 2019 Gallai-Ramsey Number of Graphs Introduction Main results Books Wheels Fans Stars with extra independent edges Odd cycles Two Classes of Unicyclic Graphs Results for S+ 3 -coloring Sketch of the Proof Let P be a longest red path within GR.
H1 GR GB P Yaping Mao June, 2019 Gallai-Ramsey Number of Graphs Introduction Main results Books Wheels Fans Stars with extra independent edges Odd cycles Two Classes of Unicyclic Graphs Results for S+ 3 -coloring Sketch of the Proof The ends cannot have more red edges to the rest of GR.
H1 GR GB P Yaping Mao June, 2019 Gallai-Ramsey Number of Graphs Introduction Main results Books Wheels Fans Stars with extra independent edges Odd cycles Two Classes of Unicyclic Graphs Results for S+ 3 -coloring Sketch of the Proof Within this remaining set F, there can be no long red path and no long blue path.
H1 GR GB P F Yaping Mao June, 2019 Gallai-Ramsey Number of Graphs Introduction Main results Books Wheels Fans Stars with extra independent edges Odd cycles Two Classes of Unicyclic Graphs Results for S+ 3 -coloring Sketch of the Proof The two sides are roughly balanced: |GR| ∼|GB|.
H1 GR GB Yaping Mao June, 2019 Gallai-Ramsey Number of Graphs Introduction Main results Books Wheels Fans Stars with extra independent edges Odd cycles Two Classes of Unicyclic Graphs Results for S+ 3 -coloring Sketch of the Proof Symmetry and similar ideas show we have this structure.
H1 GR GB Yaping Mao June, 2019 Gallai-Ramsey Number of Graphs Introduction Main results Books Wheels Fans Stars with extra independent edges Odd cycles Two Classes of Unicyclic Graphs Results for S+ 3 -coloring Sketch of the Proof Reserve several key vertices for later use. (Absorbing argument) H1 GR GB Yaping Mao June, 2019 Gallai-Ramsey Number of Graphs Introduction Main results Books Wheels Fans Stars with extra independent edges Odd cycles Two Classes of Unicyclic Graphs Results for S+ 3 -coloring Sketch of the Proof Apply the known even cycles result to obtain a mono-chromatic copy of C2ℓ−2 avoiding the reserved set.
H1 GR GB Yaping Mao June, 2019 Gallai-Ramsey Number of Graphs Introduction Main results Books Wheels Fans Stars with extra independent edges Odd cycles Two Classes of Unicyclic Graphs Results for S+ 3 -coloring Sketch of the Proof Construct a monochromatic copy of C2ℓ+1 using the reserved set.
H1 GR GB Yaping Mao June, 2019 Gallai-Ramsey Number of Graphs Introduction Main results Books Wheels Fans Stars with extra independent edges Odd cycles Two Classes of Unicyclic Graphs Results for S+ 3 -coloring Outline 1 Introduction Classical Ramsey number Gallai Ramsey number under K3-coloring Gallai Ramsey number under S+ 3 -coloring 2 Main results Books Wheels Fans Stars with extra independent edges Odd cycles Two Classes of Unicyclic Graphs Results for S+ 3 -coloring Yaping Mao June, 2019 Gallai-Ramsey Number of Graphs Introduction Main results Books Wheels Fans Stars with extra independent edges Odd cycles Two Classes of Unicyclic Graphs Results for S+ 3 -coloring Two Classes of Unicyclic Graphs In this work, we consider the Gallai-Ramsey numbers for finding either a rainbow triangle or monochromatic graph coming from two classes of unicyclic graphs: a star with an extra edge that forms a triangle, and a path with an extra edge from an end vertex to an internal vertex formaing a triangle.
Yaping Mao June, 2019 Gallai-Ramsey Number of Graphs Introduction Main results Books Wheels Fans Stars with extra independent edges Odd cycles Two Classes of Unicyclic Graphs Results for S+ 3 -coloring Graphs S+ t and P + t Let S+ t denote graph consisting of the star St with the addition of an edge between two of the pendant vertices, forming a triangle.
Let P + t denote the graph consisting of the path Pt with the addition of an edge between one end and the vertex at distance 2 along the path from that end, forming a triangle.
S+ t P + t Yaping Mao June, 2019 Gallai-Ramsey Number of Graphs Introduction Main results Books Wheels Fans Stars with extra independent edges Odd cycles Two Classes of Unicyclic Graphs Results for S+ 3 -coloring Two Classes of Unicyclic Graphs These graphs are particularly interesting because although they are not bipartite, they are very close to being a tree (and therefore bipartite).
The dichotomy between bipartite and non-bipartite graphs is critical in the study of Gallai-Ramsey numbers in light of the following result.
Yaping Mao June, 2019 Gallai-Ramsey Number of Graphs Introduction Main results Books Wheels Fans Stars with extra independent edges Odd cycles Two Classes of Unicyclic Graphs Results for S+ 3 -coloring General Upper Bound A. Gy´ arf´ as, G. S´ ark¨ ozy, A. Seb˝ o, and S. Selkow. Ramsey-type results for gallai colorings, J. Graph Theory 64(3)(2010), 233–243 obtained the following result.
Theorem 2.20 Let H be a fixed graph with no isolated vertices. If H is not bipartite, then grk(K3 : H) is exponential in k. If H is bipartite, then grk(K3 : H) is linear in k.
Yaping Mao June, 2019 Gallai-Ramsey Number of Graphs Introduction Main results Books Wheels Fans Stars with extra independent edges Odd cycles Two Classes of Unicyclic Graphs Results for S+ 3 -coloring Two Classes of Unicyclic Graphs In order to produce sharp results for the Gallai-Ramsey numbers of these graphs, we first proved the 2-color Ramsey numbers for these graphs.
Theorem 2.21 (Wang, Mao, Magnant, Zou) For t ≥3, R(S+ t , S+ t ) = 2t −1.
Yaping Mao June, 2019 Gallai-Ramsey Number of Graphs Introduction Main results Books Wheels Fans Stars with extra independent edges Odd cycles Two Classes of Unicyclic Graphs Results for S+ 3 -coloring Two Classes of Unicyclic Graphs In order to produce sharp results for the Gallai-Ramsey numbers of these graphs, we first proved the 2-color Ramsey numbers for these graphs.
Theorem 2.22 (Wang, Mao, Magnant, Zou) For t ≥4, R(P + t , P + t ) = 2t −1.
Yaping Mao June, 2019 Gallai-Ramsey Number of Graphs Introduction Main results Books Wheels Fans Stars with extra independent edges Odd cycles Two Classes of Unicyclic Graphs Results for S+ 3 -coloring Gallai-Ramsey number for S+ t The precise Gallai-Ramsey number for S+ t are obtained.
Theorem 2.23 (Wang, Mao, Magnant, Zou) For k ≥1, grk(K3 : S+ t ) = ( 2(t −1) · 5 k−2 2 + 1 if k is even, (t −1) · 5 k−1 2 + 1 if k is odd.
Yaping Mao June, 2019 Gallai-Ramsey Number of Graphs Introduction Main results Books Wheels Fans Stars with extra independent edges Odd cycles Two Classes of Unicyclic Graphs Results for S+ 3 -coloring Gallai-Ramsey number for path with an extra edge The precise Gallai-Ramsey number for P + t are also obtained.
Theorem 2.24 (Wang, Mao, Magnant, Zou) For t ≥4 and k ≥1, grk(K3 : P + t ) = ( 2(t −1) · 5 k−2 2 + 1 if k is even, (t −1) · 5 k−1 2 + 1 if k is odd.
Yaping Mao June, 2019 Gallai-Ramsey Number of Graphs Introduction Main results Books Wheels Fans Stars with extra independent edges Odd cycles Two Classes of Unicyclic Graphs Results for S+ 3 -coloring Outline 1 Introduction Classical Ramsey number Gallai Ramsey number under K3-coloring Gallai Ramsey number under S+ 3 -coloring 2 Main results Books Wheels Fans Stars with extra independent edges Odd cycles Two Classes of Unicyclic Graphs Results for S+ 3 -coloring Yaping Mao June, 2019 Gallai-Ramsey Number of Graphs Introduction Main results Books Wheels Fans Stars with extra independent edges Odd cycles Two Classes of Unicyclic Graphs Results for S+ 3 -coloring Stars S. Fujita, C. Magnant, Extensions of Gallai-Ramsey results, J.
Graph Theory 70(4) (2012), 404–426 proposed the following conjecture for star graphs.
Conjecture 2.2 For k ≥4, grk(S+ 3 ; K1,t) = 3t −2k + 4.
Yaping Mao June, 2019 Gallai-Ramsey Number of Graphs Introduction Main results Books Wheels Fans Stars with extra independent edges Odd cycles Two Classes of Unicyclic Graphs Results for S+ 3 -coloring A tree of order five Theorem 2.25 For k ≥1, grk(K3; T1) = k + 4, where T1 is a tree obtained from a path P4 by adding a pendant edge on one of the internal vertices in P4.
Theorem 2.26 (Mao, Su, Wang, Magnant) For k ≥1, grk(S+ 3 ; T1) = k + 4, if k ̸= 3; 9, if k = 3, Yaping Mao June, 2019 Gallai-Ramsey Number of Graphs Introduction Main results Books Wheels Fans Stars with extra independent edges Odd cycles Two Classes of Unicyclic Graphs Results for S+ 3 -coloring Sketch of the Proof To show gr3(S+ 3 ; T1) ≥9, we have the following examples.
v6 (a) v2 v1 v5 v4 v3 v8 v7 v6 (b) v2 v1 v5 v4 v3 v8 v7 v6 (c) v2 v1 v5 v4 v3 v8 v7 Yaping Mao June, 2019 Gallai-Ramsey Number of Graphs Introduction Main results Books Wheels Fans Stars with extra independent edges Odd cycles Two Classes of Unicyclic Graphs Results for S+ 3 -coloring The End Thank you for your attention !
Yaping Mao June, 2019 Gallai-Ramsey Number of Graphs |
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How does the speed of sound compare between dry air and water?
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The speed of sound is faster in water than in dry air. In dry air at 20 degrees Celsius, the speed of sound is approximately 343 meters per second. In water at the same temperature, the speed of sound is approximately 1482 meters per second. This is because water is denser than air, and sound waves travel more quickly through denser mediums.
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Study Guides>Intermediate Algebra
Multi-Step Linear Equations
Learning Objectives
Use properties of equality to isolate variables and solve algebraic equations
Solve equations containing absolute values
Use the properties of equality and the distributive property to solve equations containing parentheses
Clear fractions and decimals from equations to make them easier to solve
Solve equations that have one solution, no solution, or an infinite number of solutions
Steps With an End In Sight
Use properties of equality to isolate variables and solve algebraic equations
There are some equations that you can solve in your head quickly, but other equations are more complicated. Multi-step equations, ones that takes several steps to solve, can still be simplified and solved by applying basic algebraic rules such as the multiplication and addition properties of equality. In this section we will explore methods for solving multi-step equations that contain grouping symbols and several mathematical operations. We will also learn techniques for solving multi-step equations that contain absolute values. Finally, we will learn that some equations have no solutions, while others have an infinite number of solutions. First, let's define some important terminology:
variables: variables are symbols that stand for an unknown quantity, they are often represented with letters, like x, y, or z.
coefficient:Sometimes a variable is multiplied by a number. This number is called the coefficient of the variable. For example, the coefficient of 3 x is 3.
term:a single number, or variables and numbers connected by multiplication. -4, 6x and x 2 x^2 x 2 are all terms
expression:groups of terms connected by addition and subtraction.2 x 2−5 2x^2-5 2 x 2−5 is an expression
equation:an equation is a mathematical statement that two expressions are equal. An equation will always contain an equal sign with an expression on each side.Think of an equal sign as meaning "the same as." Some examples of equations are y=m x+b y = mx +b\y=m x+b, 3 4 r=v 3−r\frac{3}{4}r = v^{3} - r\4 3r=v 3−r, and 2(6−d)+f(3+k)=1 4 d 2(6-d) + f(3 +k) = \frac{1}{4}d\2(6−d)+f(3+k)=4 1d
The following figure shows how coefficients, variables, terms, and expressions all come together to make equations. In the equation 2 x−3 2=10 x 2x-3^2=10x 2 x−3 2=10 x, the variable is x x x, a coefficient is 10 10 10, a term is 10 x 10x 10 x, an expression is 2 x−3 2 2x-3^2 2 x−3 2. There are some equations that you can solve in your head quickly. For example—what is the value of y in the equation 2 y=6 2y=6 2 y=6? Chances are you didn’t need to get out a pencil and paper to calculate that y=3 y=3 y=3. You only needed to do one thing to get the answer: divide 6 by 2. Other equations are more complicated. Solving 4(1 3 t+1 2)=6\displaystyle 4\left( \frac{1}{3}t+\frac{1}{2}\right)=6 4(3 1t+2 1)=6 without writing anything down is difficult! That’s because this equation contains not just a variable but also fractions and terms inside parentheses. This is a multi-step equation, one that takes several steps to solve. Although multi-step equations take more time and more operations, they can still be simplified and solved by applying basic algebraic rules. Remember that you can think of an equation as a balance scale, with the goal being to rewrite the equation so that it is easier to solve but still balanced. The addition property of equality and the multiplication property of equality explain how you can keep the scale, or the equation, balanced. Whenever you perform an operation to one side of the equation, if you perform the same exact operation to the other side, you’ll keep both sides of the equation equal. If the equation is in the form a x+b=c ax+b=c a x+b=c, where x is the variable, you can solve the equation as before. First “undo” the addition and subtraction, and then “undo” the multiplication and division.
Example
Solve 3 y+2=11 3y+2=11 3 y+2=11.
Answer: Subtract 2 from both sides of the equation to get the term with the variable by itself.
3 y+2=11−2−2‾3 y=9 \displaystyle \begin{array}{r}3y+2\,\,\,=\,\,11\\underline{\,\,\,\,\,\,\,-2\,\,\,\,\,\,\,\,-2}\3y\,\,\,\,=\,\,\,\,\,9\end{array}3 y+2=11−2−23 y=9
Divide both sides of the equation by 3 to get a coefficient of 1 for the variable.
3 y‾=9‾3 9 y=3\begin{array}{r}\,\,\,\,\,\,\underline{3y}\,\,\,\,=\,\,\,\,\,\underline{9}\3\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,9\\,\,\,\,\,\,\,\,\,\,y\,\,\,\,=\,\,\,\,3\end{array}3 y=93 9 y=3
Answer
y=3 y=3 y=3
Example
Solve 3 x+5 x+4−x+7=88 3x+5x+4-x+7=88 3 x+5 x+4−x+7=88.
Answer: There are three like terms 3 x 3x 3 x, 5 x 5x 5 x,and–x–x–x involving a variable.Combine these like terms.4 and 7 are also like terms and can be added.
3 x+5 x+4−x+7=88 7 x+4+7=88\begin{array}{r}\,\,3x+5x+4-x+7=\,\,\,88\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,7x+4+7=\,\,\,88\end{array}3 x+5 x+4−x+7=88 7 x+4+7=88
The equation is now in the form a x+b=c ax+b=c a x+b=c, so we can solve as before.
7 x+11=88 7x+11\,\,\,=\,\,\,88 7 x+11=88
Subtract 11 from both sides.
7 x+11=88−11−11‾7 x=77\begin{array}{r}7x+11\,\,\,=\,\,\,88\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline{-11\,\,\,\,\,\,\,-11}\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,7x\,\,\,=\,\,\,77\end{array}7 x+11=88−11−117 x=77
Divide both sides by 7.
7 x‾=77‾7 7 x=11\begin{array}{r}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline{7x}\,\,\,=\,\,\,\underline{77}\7\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,7\,\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x\,\,\,=\,\,\,11\end{array}7 x=777 7 x=11
Answer
x=11 x=11 x=11
Some equations may have the variable on both sides of the equal sign, as in this equation: 4 x−6=2 x+10 4x-6=2x+10 4 x−6=2 x+10. To solve this equation, we need to “move” one of the variable terms.This can make it difficult to decide which side to work with. It doesn’t matter which term gets moved, 4 x 4x 4 x or 2 x 2x 2 x, however, to avoid negative coefficients, you can move the smaller term.
Examples
Solve:4 x−6=2 x+10 4x-6=2x+10 4 x−6=2 x+10
Answer: Choose the variable term to move—to avoid negative terms choose 2 x 2x 2 x
4 x−6=2 x+10−2 x−2 x‾2 x−6=10\,\,\,4x-6=2x+10\\underline{-2x\,\,\,\,\,\,\,\,\,\,-2x}\\,\,\,2x-6=10 4 x−6=2 x+10−2 x−2 x2 x−6=10
Now add 6 to both sides to isolate the term with the variable.
2 x−6=10+6+6‾2 x=16\begin{array}{r}2x-6=10\\underline{\,\,\,\,+6\,\,\,+6}\2x=16\end{array}2 x−6=10+6+62 x=16
Now divide each side by 2 to isolate the variable x.
2 x 2=16 2 x=8\begin{array}{c}\frac{2x}{2}=\frac{16}{2}\\x=8\end{array}2 2 x=2 16x=8
In this video, we show an example of solving equations that have variables on both sides of hte equal sign.
The Distributive Property
As we solve linear equations, we often need to do some work to write the linear equations in a form we are familiar with solving.This section will focus on manipulating an equation we are asked to solve in such a way that we can use the skills we learned for solving multi-step equations to ultimately arrive at the solution. Parentheses can make solving a problem difficult, if not impossible. To get rid of these unwanted parentheses we have the distributive property. Using this property we multiply the number in front of the parentheses by each term inside of the parentheses.
The Distributive Property of Multiplication
For all real numbers a, b, and c,a(b+c)=a b+a c a(b+c)=ab+ac a(b+c)=ab+a c. What this means is that when a number multiplies an expression inside parentheses, you can distribute the multiplication to each term of the expression individually. Then, you can follow the steps we have already practiced to isolate the variableand solve the equation.
Example
Solve for a a a. 4(2 a+3)=28 4\left(2a+3\right)=28 4(2 a+3)=28
Answer: Apply the distributive property to expand 4(2 a+3)4\left(2a+3\right)4(2 a+3) to 8 a+12 8a+12 8 a+12
4(2 a+3)=28 8 a+12=28\begin{array}{r}4\left(2a+3\right)=28\ 8a+12=28\end{array}4(2 a+3)=28 8 a+12=28
Subtract 12 from both sides to isolate the variable term.
8 a+12=28−12−12‾8 a=16\begin{array}{r}8a+12\,\,\,=\,\,\,28\ \underline{-12\,\,\,\,\,\,-12}\ 8a\,\,\,=\,\,\,16\end{array}8 a+12=28−12−128 a=16
Divide both terms by 8 to get a coefficient of 1.
8 a‾=16‾8 8 a=2\begin{array}{r}\underline{8a}=\underline{16}\8\,\,\,\,\,\,\,\,\,\,\,\,8\a\,=\,\,2\end{array}8 a=168 8 a=2
Answer
a=2 a=2 a=2
In the video that follows, we show another example of how to use the distributive property to solve a multi-step linear equation. In the next example, you will see that there are parentheses on both sides of the equal sign, so you will need to use the distributive property twice. Notice that you are going to need to distribute a negative number, so be careful with negative signs!
Example
Solve for t t t.2(4 t−5)=−3(2 t+1)2\left(4t-5\right)=-3\left(2t+1\right)2(4 t−5)=−3(2 t+1)
Answer: Apply the distributive property to expand 2(4 t−5)2\left(4t-5\right)2(4 t−5) to 8 t−10 8t-10 8 t−10 and −3(2 t+1)-3\left(2t+1\right)−3(2 t+1) to−6 t−3-6t-3−6 t−3. Be careful in this step—you are distributing a negative number, so keep track of the sign of each number after you multiply.
2(4 t−5)=−3(2 t+1)8 t−10=−6 t−3\begin{array}{r}2\left(4t-5\right)=-3\left(2t+1\right)\,\,\,\,\,\, \ 8t-10=-6t-3\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\end{array}2(4 t−5)=−3(2 t+1)8 t−10=−6 t−3
Add −6 t-6t−6 t to both sides to begin combining like terms.
8 t−10=−6 t−3+6 t+6 t‾14 t−10=−3\begin{array}{r}8t-10=-6t-3\ \underline{+6t\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+6t}\,\,\,\,\,\,\,\ 14t-10=\,\,\,\,-3\,\,\,\,\,\,\,\end{array}8 t−10=−6 t−3+6 t+6 t14 t−10=−3
Add 10 to both sides of the equation to isolate t.
14 t−10=−3+10+10‾14 t=7\begin{array}{r}14t-10=-3\ \underline{+10\,\,\,+10}\ 14t=\,\,\,7\,\end{array}14 t−10=−3+10+1014 t=7
The last step is to divide both sides by 14 to completely isolate t.
14 t=7 14 t 14=7 14\begin{array}{r}14t=7\,\,\,\,\\frac{14t}{14}=\frac{7}{14}\end{array}\14 t=7 14 14 t=14 7
Answer
t=1 2 t=\frac{1}{2}\t=2 1 We simplified the fraction 7 14\frac{7}{14}\14 7 into 1 2\frac{1}{2}\2 1
In the following video, we solve another multi-step equation with two sets of parentheses. Sometimes, you will encounter a multi-step equation with fractions. If you prefer not working with fractions, you can use the multiplication property of equality to multiply both sides of the equation by a common denominator of all of the fractions in the equation. This will clear all the fractions out of the equation. See the example below.
Example
Solve 1 2 x−3=2−3 4 x\frac{1}{2}x-3=2-\frac{3}{4}x 2 1x−3=2−4 3x by clearing the fractions in the equation first.
Answer: Multiply both sides of the equation by 4, the common denominator of the fractional coefficients.
1 2 x−3=2−3 4 x 4(1 2 x−3)=4(2−3 4 x)\begin{array}{r}\frac{1}{2}x-3=2-\frac{3}{4}x\,\,\,\,\,\,\,\,\,\,\,\,\,\\ 4\left(\frac{1}{2}x-3\right)=4\left(2-\frac{3}{4}x\right)\end{array}2 1x−3=2−4 3x 4(2 1x−3)=4(2−4 3x)
Use the distributive property to expand the expressions on both sides.Multiply.
4(1 2 x)−4(3)=4(2)−4(−3 4 x)4 2 x−12=8−12 4 x 2 x−12=8−3 x\begin{array}{r}4\left(\frac{1}{2}x\right)-4\left(3\right)=4\left(2\right)-4\left(-\frac{3}{4}x\right)\\ \frac{4}{2}x-12=8-\frac{12}{4}x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \\ 2x-12=8-3x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \end{array}4(2 1x)−4(3)=4(2)−4(−4 3x)2 4x−12=8−4 12x 2 x−12=8−3 x
Add 3 x to both sides to move the variable terms to only one side. Add 12 to both sides to move the variable terms to only one side.
2 x−12=8−3 x+3 x+3 x‾5 x−12=8\begin{array}{r}2x-12=8-3x\, \\underline{+3x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+3x}\ 5x-12=8\,\,\,\,\,\,\,\,\,\,\,\,\,\end{array}2 x−12=8−3 x+3 x+3 x5 x−12=8
Add 12 to both sides to move the constant terms to the other side.
5 x−12=8+12+12‾5 x=20\begin{array}{r}5x-12=8\,\,\ \underline{\,\,\,\,\,\,+12\,+12} \5x=20\end{array}5 x−12=8+12+125 x=20
Divide to isolate the variable.
5 x‾=5‾5 5 x=4\begin{array}{r}\underline{5x}=\underline{5}\ 5\,\,\,\,\,\,\,\,\,5\ x=4\end{array}5 x=55 5 x=4
Answer
x=4 x=4 x=4
Of course, if you like to work with fractions, you can just apply your knowledge of operations with fractions and solve. In the following video, we show how to solve a multi-step equation with fractions. Regardless of which method you use to solve equations containing variables, you will get the same answer. You can choose the method you find the easiest! Remember to check your answer by substituting your solution into the original equation. Sometimes, you will encounter a multi-step equation with decimals. If you prefer not working with decimals, you can use the multiplication property of equality to multiply both sides of the equation by a a factor of 10 that will help clear the decimals. See the example below.
Example
Solve 3 y+10.5=6.5+2.5 y 3y+10.5=6.5+2.5y 3 y+10.5=6.5+2.5 y by clearing the decimals in the equation first.
Answer: Since the smallest decimal place represented in the equation is 0.10, we want to multiply by 10 to make 1.0 and clear the decimals from the equation.
3 y+10.5=6.5+2.5 y 10(3 y+10.5)=10(6.5+2.5 y)\begin{array}{r}3y+10.5=6.5+2.5y\,\,\,\,\,\,\,\,\,\,\,\,\\ 10\left(3y+10.5\right)=10\left(6.5+2.5y\right)\end{array}3 y+10.5=6.5+2.5 y 10(3 y+10.5)=10(6.5+2.5 y)
Use the distributive property to expand the expressions on both sides.
10(3 y)+10(10.5)=10(6.5)+10(2.5 y)\begin{array}{r}10\left(3y\right)+10\left(10.5\right)=10\left(6.5\right)+10\left(2.5y\right)\end{array}\10(3 y)+10(10.5)=10(6.5)+10(2.5 y)
Multiply.
30 y+105=65+25 y 30y+105=65+25y 30 y+105=65+25 y
Move the smaller variable term, 25 y 25y 25 y, by subtracting it from both sides.
30 y+105=65+25 y−25 y−25 y‾5 y+105=65\begin{array}{r}30y+105=65+25y\,\,\ \underline{-25y\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-25y} \5y+105=65\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\end{array}30 y+105=65+25 y−25 y−25 y5 y+105=65
Subtract 105 from both sides to isolate the term with the variable.
5 y+105=65−105−105‾5 y=−40\begin{array}{r}5y+105=65\,\,\,\ \underline{\,\,\,\,\,\,-105\,-105} \5y=-40\end{array}5 y+105=65−105−1055 y=−40
Divide both sides by 5 to isolate the y.
5 y‾=−40‾5 5 x=−8\begin{array}{l}\underline{5y}=\underline{-40}\ 5\,\,\,\,\,\,\,\,\,\,\,\,\,5\ \,\,\,x=-8\end{array}5 y=−405 5 x=−8
Answer
x=−8 x=-8 x=−8
In the following video, we show another example of clearing decimals first to solve a multi-step linear equation. Here are some steps to follow when you solve multi-step equations.
Solving Multi-Step Equations
(Optional) Multiply to clear any fractions or decimals. 2. Simplify each side by clearing parentheses and combining like terms. 3. Add or subtract to isolate the variable term—you may have to move a term with the variable. 4. Multiply or divide to isolate the variable. 5. Check the solution.
Classify Solutions to Linear Equations
There are three cases that can come up as we are solving linear equations. We have already seen one, where an equation has one solution. Sometimes we come across equations that don't have any solutions, and even some that have an infinite number of solutions. The case where an equation has no solution is illustrated in the next examples.
Equations with no solutions
Example
Solve for x.12+2 x–8=7 x+5–5 x 12+2x–8=7x+5–5x 12+2 x–8=7 x+5–5 x
Answer: Combine like terms on both sides of the equation.
12+2 x−8=7 x+5−5 x 2 x+4=2 x+5 \displaystyle \begin{array}{l}12+2x-8=7x+5-5x\\,\,\,\,\,\,\,\,\,\,\,\,\,\,2x+4=2x+5\end{array}12+2 x−8=7 x+5−5 x 2 x+4=2 x+5
Isolate the x term by subtracting 2 x from both sides.
2 x+4=2 x+5−2 x−2 x‾4=5\begin{array}{l}\,\,\,\,\,\,\,\,\,\,\,\,2x+4=2x+5\\,\,\,\,\,\,\,\,\underline{-2x\,\,\,\,\,\,\,\,\,\,-2x\,\,\,\,\,\,\,\,}\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,4= \,5\end{array}2 x+4=2 x+5−2 x−2 x4=5
This false statement implies there are no solutions to this equation. Sometimes, we say the solution does not exist, or DNE for short.
This is not a solution! You did not find a value for x. Solving for x the way you know how, you arrive at the false statement 4=5 4=5 4=5. Surely 4 cannot be equal to 5! This may make sense when you consider the second line in the solution where like terms were combined. If you multiply a number by 2 and add 4 you would never get the same answer as when you multiply that same number by 2 and add 5. Since there is no value of x that will ever make this a true statement, the solution to the equation above is “no solution.” Be careful that you do not confuse the solution x=0 x=0 x=0 with “no solution.” The solution x=0 x=0 x=0 means that the value 0 satisfies the equation, so there is a solution. “No solution” means that there is no value, not even 0, which would satisfy the equation. Also, be careful not to make the mistake of thinking that the equation 4=5 4=5 4=5 means that 4 and 5 are values for x that are solutions. If you substitute these values into the original equation, you’ll see that they do not satisfy the equation. This is because there is truly no solution—there are no values for x that will make the equation 12+2 x–8=7 x+5–5 x 12+2x–8=7x+5–5x 12+2 x–8=7 x+5–5 x true.
Think About It
Try solving these equations. How many steps do you need to take before you can tell whether the equation has no solution or one solution? a) Solve 8 y=3(y+4)+y 8y=3(y+4)+y 8 y=3(y+4)+y Use the textbox below to record how many steps you think it will take before you can tell whether there is no solution or one solution. [practice-area rows="1"][/practice-area]
Answer:
Solve 8 y=3(y+4)+y 8y=3(y+4)+y 8 y=3(y+4)+y
First, distribute the 3 into the parentheses on the right-hand side.
8 y=3(y+4)+y=8 y=3 y+12+y 8y=3(y+4)+y=8y=3y+12+y 8 y=3(y+4)+y=8 y=3 y+12+y
Next, begin combining like terms.
8 y=3 y+12+y=8 y=4 y+12 8y=3y+12+y = 8y=4y+12 8 y=3 y+12+y=8 y=4 y+12
Now move the variable terms to one side. Moving the 4 y 4y 4 y will help avoid a negative sign.
8 y=4 y+12−4 y−4 y‾4 y=12\begin{array}{l}\,\,\,\,8y=4y+12\\underline{-4y\,\,-4y}\\,\,\,\,4y=12\end{array}8 y=4 y+12−4 y−4 y4 y=12
Now, divide each side by 4 y 4y 4 y.
4 y 4=12 4 y=3\begin{array}{c}\frac{4y}{4}=\frac{12}{4}\y=3\end{array}4 4 y=4 12y=3
Because we were able to isolate y on one side and a number on the other side, we have one solution to this equation. b) Solve 2(3 x−5)−4 x=2 x+7 2\left(3x-5\right)-4x=2x+7 2(3 x−5)−4 x=2 x+7 Use the textbox below to record how many steps you think it will take before you can tell whether there is no solution or one solution. [practice-area rows="1"][/practice-area]
Answer: Solve 2(3 x−5)−4 x=2 x+7 2\left(3x-5\right)-4x=2x+7 2(3 x−5)−4 x=2 x+7. First, distribute the 2 into the parentheses on the left-hand side.
2(3 x−5)−4 x=2 x+7 6 x−10−4 x=2 x+7\begin{array}{r}2\left(3x-5\right)-4x=2x+7\6x-10-4x=2x+7\end{array}2(3 x−5)−4 x=2 x+7 6 x−10−4 x=2 x+7
Now begin simplifying. You can combine the x terms on the left-hand side.
6 x−10−4 x=2 x+7 2 x−10=2 x+7\begin{array}{r}6x-10-4x=2x+7\2x-10=2x+7\end{array}6 x−10−4 x=2 x+7 2 x−10=2 x+7
Now, take a moment to ponder this equation. It says that 2 x−10 2x-10 2 x−10 is equal to 2 x+7 2x+7 2 x+7. Can some number times two minus 10 be equal to that same number times two plus seven? Let's pretend x=3 x=3 x=3. Is it true that 2(3)−10=−4 2\left(3\right)-10=-4 2(3)−10=−4 is equal to 2(3)+7=13 2\left(3\right)+7=13 2(3)+7=13. NO! We don't even really need to continue solving the equation, but we can just to be thorough. Add 10 10 10 to both sides.
2 x−10=2 x+7+10+10‾2 x=2 x+17\begin{array}{r}2x-10=2x+7\,\,\\,\,\underline{+10\,\,\,\,\,\,\,\,\,\,\,+10}\2x=2x+17\end{array}2 x−10=2 x+7+10+102 x=2 x+17
Now move 2 x 2x 2 x from the right hand side to combine like terms.
2 x=2 x+17−2 x−2 x‾0=17\begin{array}{l}\,\,\,\,\,2x=2x+17\\,\,\underline{-2x\,\,-2x}\\,\,\,\,\,\,\,0=17\end{array}2 x=2 x+17−2 x−2 x0=17
We know that 0 and 17 0\text{ and }17 0 and 17 are not equal, so there is no number that x could be to make this equation true. This false statement implies there are no solutions to this equation, or DNE (does not exist) for short.
Algebraic Equations with an Infinite Number of Solutions
You have seen that if an equation has no solution, you end up with a false statement instead of a value for x. It is possible to have an equation where any value for x will provide a solution to the equation. In the example below, notice how combining the terms 5 x 5x 5 x and −4 x-4x−4 x on the left leaves us with an equation with exactly the same terms on both sides of the equal sign.
Example
Solve for x.5 x+3–4 x=3+x 5x+3–4x=3+x 5 x+3–4 x=3+x
Answer: Combine like terms on both sides of the equation.
5 x+3−4 x=3+x x+3=3+x \displaystyle \begin{array}{r}5x+3-4x=3+x\x+3=3+x\end{array}5 x+3−4 x=3+x x+3=3+x
Isolate the x term by subtracting x from both sides.
x+3=3+x−x−x‾3=3\begin{array}{l}\,\,\,\,\,\,\,\,\,\,\,\,\,x+3=3+x\\,\,\,\,\,\,\,\,\underline{\,-x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-x\,}\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,3\,\,=\,\,3\end{array}x+3=3+x−x−x3=3
This true statement implies there are an infinite number of solutions to this equation, or we can also write the solution as "All Real Numbers"
You arrive at the true statement “3=3 3=3 3=3.” When you end up with a true statement like this, it means that the solution to the equation is “all real numbers.” Try substituting x=0 x=0 x=0 into the original equation—you will get a true statement! Try x=−3 4 x=-\frac{3}{4}x=−4 3, and it also will check! This equation happens to have an infinite number of solutions. Any value for x that you can think of will make this equation true. When you think about the context of the problem, this makes sense—the equation x+3=3+x x+3=3+x x+3=3+x means “some number plus 3 is equal to 3 plus that same number.” We know that this is always true—it’s the commutative property of addition!
Example
Solve for x.3(2 x−5)=6 x−15 3\left(2x-5\right)=6x-15 3(2 x−5)=6 x−15
Answer: Distribute the 3 through the parentheses on the left-hand side.
3(2 x−5)=6 x−15 6 x−15=6 x−15 \begin{array}{r}3\left(2x-5\right)=6x-15\6x-15=6x-15\end{array}3(2 x−5)=6 x−15 6 x−15=6 x−15
Wait! This looks just like the previous example. You have the same expression on both sides of an equal sign. No matter what number you choose for x, you will have a true statement. We can finish the algebra:
6 x−15=6 x−15−6 x−6 x‾−15=−15\begin{array}{l}\,\,\,\,\,\,\,\,\,\,\,\,\,6x-15=6x-15\\,\,\,\,\,\,\,\,\underline{\,-6x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-6x\,}\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-15\,\,=\,\,-15\end{array}6 x−15=6 x−15−6 x−6 x−15=−15
This true statement implies there are an infinite number of solutions to this equation.
In the following video, we show more examples of attempting to solve a linear equation with either no solution or many solutions. In the following video, we show more examples of solving linear equations with parentheses that have either no solution or many solutions.
Summary
Complex, multi-step equations often require multi-step solutions. Before you can begin to isolate a variable, you may need to simplify the equation first. This may mean using the distributive property to remove parentheses or multiplying both sides of an equation by a common denominator to get rid of fractions. Sometimes it requires both techniques. If your multi-step equation has an absolute value, you will need to solve two equations, sometimes isolating the absolute value expression first. We have seen that solutions to equations can fall into three categories:
One solution
No solution, DNE (does not exist)
Many solutions, also called infinitely many solutions or All Real Numbers
And sometimes, we don't need to do much algebra to see what the outcome will be.
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Solving Two Step Equations (Basic).Authored by:James Sousa (Mathispower4u.com) for Lumen Learning.License:CC BY: Attribution.
Solving an Equation that Requires Combining Like Terms.Authored by:James Sousa (Mathispower4u.com) for Lumen Learning.License:CC BY: Attribution.
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3769 | https://mathworld.wolfram.com/MultipleRoot.html | Multiple Root -- from Wolfram MathWorld
TOPICS
AlgebraApplied MathematicsCalculus and AnalysisDiscrete MathematicsFoundations of MathematicsGeometryHistory and TerminologyNumber TheoryProbability and StatisticsRecreational MathematicsTopologyAlphabetical IndexNew in MathWorld
Calculus and Analysis
Roots
Multiple Root
A multiple root is a root with multiplicity, also called a multiple point or repeated root. For example, in the equation , 1 is multiple (double) root.
If a polynomial has a multiple root, its derivative also shares that root.
See also
Multiplicity, Root, Simple Root
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References
Krantz, S.G. "Zero of Order n." §5.1.3 in Handbook of Complex Variables. Boston, MA: Birkhäuser, p.70, 1999.
Referenced on Wolfram|Alpha
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Inradius of the triangle (Formula and Tricks)
Rajnikant Mishra
2 min readNov 27, 2022
The point of intersection of the angle bisectors of a triangle is called its Incenter. The distance of the incenter from all the sides of the triangle is equal. The altitude subtended from the incenter of a triangle to any of its sides is called the Inradius of the triangle. The Incenter is the center of the inscribed circle (Incircle) of the triangle.
Let a △ABC in which BC=a, AC=b, and AB=c. The length of its semi-perimeter is ‘s’ and its area is represented by △.
Then, the length of its Inradius (r)=△/s
s=(a+b+c)/2
Here is the formula of the length of the Inradius of a Right angle triangle, with proof.
∆ABC is a Right angle triangle, ∠B=90°. AB=p, BC=b & hypotenuse AC=h. The length of the Inradius of the triangle is r then find the length of AI, where I is the Incenter of the triangle.
In quadrilateral ALIN, ∠ALI+∠ANI=180° (Sum of opposite angles =180°)
So, ALIN is the Cyclic Quadrilateral.
Similarly, CMIN is also a Cyclic Quadrilateral.
Here, AI and CI are the diameters of the respective circles.
∆ALI ⩭ ∆ANI (Congruent Triangles)
AL=AN=p-r
Similarly, CM=CN=b-r
AC=AN+CN
h=p-r+b-r
h=p+b-2r
r=(p+b-h)/2
For more concepts and tricks, Click the link given below:
## Inradius And Circumradius Of A Triangle (Effective Geometry Hack 2022) - Logicxonomy
Concepts of Inradius and Circumradius of a triangle (Right angle). Here is the formula of the length of the Inradius…
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Radian: The value of the angle subtended by an arc of the length equal to the radius of the circle, at the center of the circle, is 1…
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3771 | https://cdn.mdedge.com/files/s3fs-public/issues/articles/70216_main_28.pdf | 22 PRACTICE TRENDS J U LY 2 0 0 9 • C L I N I C A L E N D O C R I N O L O G Y N E W S FDA Launches New Transparency Task Force B Y J O Y C E F R I E D E N I n one of her first public acts at the Food and Drug Administration, new commissioner Margaret Hamburg announced that the agency aims to be more transparent about its daily work and decision making process.
“Over the years, complaints have been made about FDA’s lack of transparency,” Dr. Hamburg said June 2 in announcing the launch of a transparency task force.
“The agency has been referred to as a ‘black box’ that makes important deci-sions without disclosing them. The agency can and should communicate in a way that provides more transparency, not less.” “On President Obama’s first day in of-fice, he pledged to strengthen democra-cy ... by creating an unprecedented lev-el of openness” in government, noted Dr. Hamburg, who took over at FDA on May 22. “This will be an agency-wide ef-fort charged with figuring out how to make the FDA and its processes more transparent to the public.” The transparency task force will in-clude the directors of all FDA centers as well as the agency’s associate commis-sioner for regulatory affairs, its chief counsel, and its chief scientist. Its first public meeting was held June 24; another will take place in the fall. The task force “expects to submit a written report to the commissioner about 6 months from now,” according to Dr.
Joshua Sharfstein, FDA principal deputy com-missioner and task force chair.
Clarity is one area of interest for Dr.
Sharfstein. “People don’t understand why the FDA may have done something or not done something,” he said. “In many cases, the agency has an explana-tion, but you don’t necessarily hear that explanation very clearly.” Dr. Hamburg said she expects that a wide range of recommendations could emerge from the task force’s work. Some recommendations “will be in areas that we can implement swiftly, but there may be other types of information that will take more time, and there may be some areas where we have limitations within the current law and need to examine whether appropriate changes can and should be made,” she said.
Dr. Hamburg and Dr. Sharfstein empha-sized that a balance will need to be struck between providing more information and the appropriate use of confidentiality. “There are other policy goals besides transparency, and one of the other ques-tions is what information should remain confidential,” Dr. Sharfstein said. “The secret formula for how to make X pill may be legitimately confidential infor-mation.” Another balancing act will come in terms of clinical trials, Dr. Sharfstein said. “What is the argument for different amounts of data [being disclosed] at dif-ferent points in the drug development process, and on the other side, what are the confidentiality concerns and the rea-sons for them?” The call for transparency comes at a time when the FDA has a backlog of re-quests under the Freedom of Informa-tion Act. Asked how she planned to han-dle personnel needs while the agency is behind in its work, Dr. Hamburg said, “When the recommendations come in, I will work with the task force and oth-ers on implementation. Some activity may result in more work, and some may result in decreased work. If we make more information available, there may be fewer Freedom of Information Act re-quests and citizen petitions.” ■ The Federal Register notice announcing the task force’s formation is available online at www.federalregister.gov/OFRUpload/OFR Data/2009-12902_PI.pdf. Comments on the task force’s mission are being accepted through Aug. 7.
Propylthiouracil Use Associated With Hepatotoxicity in Graves’ Disease B Y E L I Z A B E T H M E C H C AT I E T he antithyroid drug propylthiouracil (PTU) has been associated with severe liver injuries and deaths in patients and should be reserved for those in the first trimester of pregnancy or those who are intolerant of or allergic to methimazole, according to a warning by the Food and Drug Administration. “If PTU therapy is chosen, the patient should be close-ly monitored for symptoms and signs of liver injury, es-pecially during the first 6 months after initiating thera-py,” Dr. Amy Egan, deputy director for safety at the FDA’s division of metabolism and endocrinology prod-ucts at the Center for Drug Evaluation and Research, said in a statement that was posted on the agency’s Med-Watch Web site. From 1969, when the agency’s adverse event reporting program was established, through Oc-tober 2008, 32 cases of serious liver injury associated with PTU were reported to the FDA. Of these cases, 22 were in adults, and included 12 fatalities and 5 liver transplants.
Among the 10 pediatric cases, there was 1 fatality and 6 reports of liver transplants. Based on an analysis of these reports, the FDA de-termined that the risk of hepatotoxicity is greater with PTU than with methimazole (MMI). Like PTU, MMI is indicated for the treatment of hyperthyroidism caused by Graves’ disease. The FDA received only five reports of serious liver injury associated with MMI, which was approved in 1950. PTU, approved in 1947, is considered a second-line therapy for patients with Graves’ disease, with the ex-ception of those who are intolerant of or allergic to MMI. But because of rare cases of embryopathy re-ported in association with MMI treatment during preg-nancy, PTU “may be more appropriate” for treating women with Graves’ disease in the first trimester of pregnancy, according to the FDA.
The FDA plans to change the prescribing informa-tion for PTU to reflect the hepatotoxicity warning, and said that the American Thyroid Association (ATA) plans to update its Graves’ disease treatment guidelines.
In April, PTU-induced liver failure was among the topics on the agenda at a public meeting on the role of PTU in managing Graves’ disease in adults. The ATA/American Association of Clinical Endocrinolo-gists Hyperthyroidism Guidelines Task Force is final-izing recommendations on the use of antithyroid drugs, including during pregnancy and childhood, and will cover the role of monitoring hepatic function in patients on PTU, according to the ATA. Hepatoxicity has been recognized as one of the more serious but rare side effects associated with PTU, Dr. David Cooper, professor of medicine at Johns Hopkins University, Baltimore, said in an interview.
MMI also can be hepatotoxic, but is usually associated with cholestatic dysfunction, while PTU causes hepa-tocellular necrosis.
With input from the ATA, the FDA decided to issue the warning, said Dr. Cooper, who directs the Johns Hopkins thyroid clinic. He and Dr. Scott Rivkees of the Yale Pediatric Thyroid Center, New Haven, Conn., coauthored an editorial in which they wrote, “one could reasonably conclude that PTU should never be used as a first line agent in either children or adults, with the possible exception of pregnant women and pa-tients with life-threatening thyrotoxicosis” (J. Clin. En-docrinol. Metab. 2009; 94:1881-2). Dr. Cooper said that most patients who can’t take MMI should be treated with radioactive iodine or surgery. Patients who develop jaundice, fatigue, or other symptoms should stop PTU immediately, and contact their physicians. White blood cell count and bilirubin, alkaline phosphatase, and transaminase lev-els should be checked in such patients, he said.
■ A link to the FDA’s alert is available at www.fda.gov/Drugs/DrugSafety/PostmarketDrugSafety InformationforPatientsandProviders/DrugSafetyInforma-tionforHeathcareProfessionals/ucm162701.htm. Serious adverse events associated with PTU products can be reported to MedWatch at 800-332-1088 or www.fda.gov/MedWatch/report.htm.
Stolen Insulin Vials Pose Health Threat B Y M I R I A M E . T U C K E R T he Food and Drug Administration has issued an advisory about stolen vials of Levemir that have now resurfaced.
Three separate lots totaling approximately 129,000 10-mL vials of the long-acting basal insulin analogue, made by Novo Nordisk, were stolen in North Car-olina and are now being sold in the U.S. market.
Some vials from one of the lots were discovered at a medical center in Houston. These stolen insulin vials may not have been stored or handled properly and may be dangerous to patients. The FDA has received one report of a pa-tient who suffered an adverse event due to poor glu-cose control after using a vial from one of the lots. The agency advises the following for patients who use Levemir: Determine if you have Levemir from one of the following lots: XZF0036, XZF0037, or XZF0038. Lot numbers are located on the side of the box of insulin and also on the side of the vial.
If the Levemir is from one of these lots, replace it with a vial of Levemir from another lot. Do not switch to a different brand of insulin without first contacting your health care provider, because an-other insulin product may require dosing adjust-ments.
Always visually inspect your insulin before using it. Levemir is a clear and colorless solution. Contact the Novo Nordisk Customer Care Cen-ter at 800-727-6500 for instructions on what to do with vials from these lots, or if you have any other questions. “The safety of our patients is of paramount con-cern and we are working with our partners, the pharmacy, the FDA, and law enforcement authori-ties to investigate the situation and take immediate steps to maintain the highest standard of safety and quality for our products,” Novo Nordisk said in a statement. ■ ‘This will be an agency-wide effort charged with figuring out how to make the FDA and its processes more transparent to the public.’ |
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2011 AIME II Problems/Problem 4
Contents
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1 Problem 4
2 Solution 1
3 Solution 2 (mass points)
4 Solution 3
5 Solution 4
6 Solution 5
7 Solution 6 (quick Menelaus)
8 Solution 7 (Visual)
9 Solution 8 (Cheese)
10 Solution 9 (Menelaus + Ceva's + Angle Bisector Theorem)
11 Video Solution by OmegaLearn
12 See also
Problem 4
In triangle , and . The angle bisector of intersects at point , and point is the midpoint of . Let be the point of the intersection of and . The ratio of to can be expressed in the form , where and are relatively prime positive integers. Find .
Solution 1
Let be on such that . It follows that , so by the Angle Bisector Theorem. Similarly, we see by the Midline Theorem that . Thus, and .
Solution 2 (mass points)
Assign mass points as follows: by Angle-Bisector Theorem, , so we assign . Since , then , and , so .
Solution 3
By Menelaus' Theorem on with transversal, So .
Solution 4
We will use barycentric coordinates. Let , , . By the Angle Bisector Theorem, . Since is the midpoint of , . Therefore, the equation for line BM is . Let . Using the equation for , we get Therefore, so the answer is .
Solution 5
Let . Then by the Angle Bisector Theorem, . By the Ratio Lemma, we have that Notice that since their bases have the same length and they share a height. By the sin area formula, we have that Simplifying, we get that Plugging this into what we got from the Ratio Lemma, we have that
Solution 6 (quick Menelaus)
First, we will find . By Menelaus on and the line , we have This implies that . Then, by Menelaus on and line , we have Therefore, The answer is . -brainiacmaniac31
Solution 7 (Visual)
vladimir.shelomovskii@gmail.com, vvsss
Solution 8 (Cheese)
Assume is a right triangle at . Line and . These two lines intersect at which have coordinates and thus has coordinates . Thus, the line . When , has coordinate equal to = which equals giving an answer of
Solution 9 (Menelaus + Ceva's + Angle Bisector Theorem)
We start by using Menelaus' theorem on and . So, we see that . By Angle Bisector theorem, , and therefore after plugging in our values we get . Then, by Ceva's on the whole figure, we have . Plugging in our values, we get , giving an answer of . ~ESAOPS
Video Solution by OmegaLearn
~ pi_is_3.14
See also
2011 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 3Followed by
Problem 5
1•2•3•4•5•6•7•8•9•10•11•12•13•14•15
All AIME Problems and Solutions
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DISTANCE d(z,w) in the COMPLEX Plane Defined with Modulus (Absolute Value/Norm)
What’s the DISTANCE Between Two COMPLEX Numbers? | Modulus, Norms, Metrics, Circles, & Lines · @billkinneymath What’s the DISTANCE Between Two COMPLEX Numbers? | Modulus, Norms, Metrics, Circles, & Lines · @billkinneymath
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DISTANCE d(z,w) in the COMPLEX Plane Defined with Modulus (Absolute Value/Norm)
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Anytime you have a norm, you can define a distance (or metric) — and the complex plane ℂ is no exception. In this video, we define the distance between two complex numbers z and w using the modulus ("absolute value" or "norm"). The function notation d(z,w) represents the distance (and is also called a metric, making ℂ a metric space). Distance is just the modulus of z - w, written d(z,w) = |z - w|, giving a geometric measure of separation in ℂ. This is a fundamental building block for deeper ideas in complex analysis, geometry, and vector spaces. 📖 Visual Complex Analysis, by Tristan Needham: #ComplexNumbers#ComplexAnalysis#MathShorts#LearnMath#MathEducation#Mathematics#Algebra#VectorSpaces#Norm#DistanceFormula#ComplexPlane#ImaginaryNumbers#MathVideo#EducationalShorts#stem Links and resources =============================== 🔴 Subscribe to Bill Kinney Math: 🔴 Subscribe to my Math Blog, Infinity is Really Big: 🔴 Follow me on Twitter: 🔴 Follow me on Instagram: 🔴 You can support me by buying "Infinite Powers, How Calculus Reveals the Secrets of the Universe", by Steven Strogatz, or anything else you want to buy, starting from this link: 🔴 Check out my artist son Tyler Kinney's YouTube Channel: 🔴 Desiring God website: AMAZON ASSOCIATE As an Amazon Associate I earn from qualifying purchases. …...more
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What’s the DISTANCE Between Two COMPLEX Numbers? | Modulus, Norms, Metrics, Circles, & Lines
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Mesenteric ischemia: Recognizing an uncommon disorder and distinguishing among its causes
ABSTRACT
Mesenteric ischemia occurs because of inadequate intestinal blood flow. Its severity depends on the vessels involved and whether collateral blood vessels are available to prevent malperfusion. Mesenteric ischemia is an uncommon cause of abdominal pain, but it is associated with high mortality and often poses a diagnostic challenge to clinicians because its symptoms are nonspecific. Early recognition and treatment are imperative to improve patient outcomes.
Mesenteric ischemia is classified into acute or chronic subtypes according to the timing of vessel occlusion and onset of symptoms.
Diagnosis requires a high index of suspicion with focused evaluation.
Early recognition and intervention are key to preventing morbidity and mortality.
Mesenteric ischemia is an uncommon cause of abdominal pain that occurs from inadequate intestinal blood flow. It is associated with high mortality owing to the challenge of diagnosis. Mesenteric ischemia is classified as acute or chronic based on the timing of vessel occlusion and onset of symptoms. Early recognition and treatment are imperative to improve patient outcomes. This article reviews key features of common and uncommon causes of mesenteric ischemia.
MESENTERIC CIRCULATION
The mesenteric circulation is a complex vascular network supplied by 3 primary vessels: the celiac artery, superior mesenteric artery (SMA), and inferior mesenteric artery. The celiac artery perfuses the gastric, splenic, and hepatic organs, as well as the intestines through the gastroduodenal artery; the SMA supplies the midgut organs from the duodenal papillae to the proximal two-thirds of the colon; and the inferior mesenteric artery perfuses the distal one-third of the transverse colon, descending colon, and proximal two-thirds of the rectum. The celiac artery and SMA are primarily connected through the pancreaticoduodenal arcades, and the SMA and inferior mesenteric artery are connected by the marginal artery of Drummond and the arc of Riolan. Such collateral pathways ensure that the intestines are protected from transient periods of malperfusion.
TYPES OF MESENTERIC ISCHEMIA
Mesenteric ischemia can be classified as acute or chronic according to the timing of blood flow compromise and symptom onset. Acute mesenteric ischemia is a potentially fatal vascular emergency characterized by sudden intestinal hypoperfusion after abrupt obstruction of arterial or venous blood flow.1 Symptoms of acute mesenteric ischemia are typically profound owing to a lack of available collateral blood vessels. Chronic mesenteric ischemia refers to episodic intestinal hypoperfusion caused by multivessel stenosis or occlusion, usually due to atherosclerosis. Symptomatic patients typically present with abdominal angina, characterized by postprandial pain, weight loss, and food aversion.2–4
ACUTE MESENTERIC ISCHEMIA: CLINICAL RECOGNITION
Acute mesenteric ischemia is uncommon, accounting for less than 1.5% of all emergency department visits for abdominal pain, but its overall mortality exceeds 60%, owing to complications of intestinal infarction and sepsis.5–7 Clinical presentation varies depending on the underlying pathologic process (Table 1).6,8 The classic clinical presentation involves severe abdominal pain that is “out of proportion” to the physical examination.7 However, patients may present with atypical symptoms such as nausea, vomiting, and diarrhea, or complications such as peritonitis or sepsis, which often contribute to diagnostic delay.6,7
Causes and clinical features of acute mesenteric ischemia
The nonspecific nature of symptoms makes it difficult to differentiate acute mesenteric ischemia from other intra-abdominal pathologies such as acute cholecystitis, pancreatitis, and small-bowel obstruction. A high index of suspicion is critical to making the diagnosis and restoring blood flow, thereby improving patient outcomes. Morbidity depends on how long the vessel has been occluded and whether collateral circulation is present. Patients who present with sepsis are more likely to have poorer outcomes.7
CAUSES OF ACUTE MESENTERIC ISCHEMIA
Mesenteric arterial occlusion from embolism or thrombosis is the most common cause of acute mesenteric ischemia (49% and 29%, respectively), followed by nonocclusive mesenteric ischemia (20%–22%) from splanchnic hypoperfusion and vasoconstriction and venous thrombosis (10%).6,8
Arterial embolism
Acute arterial embolism causing partial or complete occlusion of the vessel lumen accounts for most (49%) cases of acute mesenteric ischemia.8 The SMA is often affected owing to its large diameter and oblique take-off angle, with most emboli lodging distal to the origin of the middle colic artery.9,10 Most emboli originate from an intracardiac source of thrombus, such as the left atrium in atrial tachyarrhythmia or left ventricle in cardiomyopathy or myocardial ischemia.11,12 Less commonly, atheroembolism or thromboembolism originates from a proximal aortic segment.13 Rarely, mesenteric artery embolism has been described as a sequela of infective or nonbacterial thrombotic endocarditis.14,15
The symptoms of embolic mesenteric artery occlusion are usually acute and dramatic because of a lack of collateral blood vessels. The typical patient is older, and women are more likely to be affected.9 Abdominal pain is the predominant symptom in 50% to 80% of patients and may be poorly localized.16 Nausea, vomiting, or diarrhea are observed in approximately 50% of patients.16 As bowel infarction develops, signs of peritonitis such as abdominal rebound and guarding may be seen on examination. Bloody bowel movements are uncommon until advanced stages of ischemia, and only one-third of patients present with the classic triad of abdominal pain, fever, and bloody stools.2 Clinicians should be aware of more atypical presentations, such as mental status change in patients age 65 or older.17
Risk factors that should raise suspicion for acute arterial occlusion include history of cardiac arrhythmia, valvular disease, recent myocardial infarction, or aortic atherosclerosis. Roughly one-third of patients report a prior embolic event, and approximately 50% have a history of atrial fibrillation.7
Computed tomography angiography is 82% to 96% sensitive and 94% specific in the diagnosis of acute mesenteric ischemia and can additionally indicate a proximal source of embolus.12,18 Evaluation of these patients should also include prompt echocardiography. Select patients may benefit from ambulatory cardiac monitoring at a later time.
Arterial thrombosis
Mesenteric artery thrombosis accounts for 25% to 30% of acute mesenteric ischemia events.6 A majority occur at the origin of the SMA or celiac arteries in patients with preexisting atherosclerotic disease.
Acute thrombosis of a stenotic vessel results in symptoms like those of acute mesenteric embolism, also referred to as acute-on-chronic mesenteric ischemia.19 In some cases, however, progression from stenosis to occlusion and bowel infarction may be insidious owing to the ability of extensive collaterals to maintain gut viability.20 This often leads to delays in seeking medical care. Many patients report prodromal symptoms of postprandial pain, weight loss, or food aversion suggestive of chronic mesenteric ischemia.19 Acute abdominal pain in a patient who has a history of such symptoms should raise suspicion for acute mesenteric artery thrombosis.
Less commonly, mesenteric artery thrombosis may occur from vessel injury following abdominal trauma or dissection or an underlying hypercoagulable state from infection or malignancy.21 Hereditary or acquired thrombophilia are rare causes of mesenteric artery thrombosis. Vasculitis, typically of small to medium vessels, can infrequently result in acute mesenteric ischemia by way of arterial occlusion or vasospasm.22,23
Nonocclusive ischemia
Nonocclusive mesenteric ischemia accounts for approximately 20% of acute mesenteric ischemia cases and has an in-hospital mortality of up to 50%.6,9,24 This form of mesenteric ischemia occurs in the setting of low blood flow states, such as low cardiac output, hypovolemia, or septic shock, leading to splanchnic arterial vasoconstriction, intestinal hypoxia, and necrosis.25 The extent of ischemic injury is dependent on the number of vessels affected, collateral circulation available, and duration of hypoperfusion.
Nonocclusive mesenteric ischemia is typically seen in patients with severe preexisting disease such as heart failure, aortic insufficiency, and renal impairment, or in patients who are critically ill and receiving vasoconstrictive medications.25 Symptoms may be absent, as these patients are usually intubated and sedated, and diagnosis is often further delayed by other overshadowing conditions such as hypovolemia and hypotension. As a result, the diagnosis may not be established until complications such as peritonitis or sepsis have developed.26 Unexplained clinical deterioration with biomarkers of tissue ischemia should raise suspicion for nonocclusive mesenteric ischemia.27 Computed tomography angiography is used for initial screening and to exclude other causes of acute mesenteric ischemia, and the diagnosis is confirmed on catheter-directed angiography or surgical exploration.28
Treatment is focused on hemodynamic support and correcting the underlying cause. Transcatheter infusion of vasodilators such as papaverine and nitroglycerin may be used to relieve mesenteric vasoconstriction in cases where bowel necrosis has not occurred, and laparotomy is indicated when acute peritoneal signs are present.24
Venous thrombosis
Mesenteric venous thrombosis is the least common cause of acute mesenteric ischemia, accounting for 10% of cases, with the superior mesenteric vein affected in approximately 95% of cases.6,29 Other factors that can impact blood flow include inflammation caused by pancreatitis, inflammatory bowel disease, infection, or trauma including surgery.2,30 Malignancy is present in up to 16% of patients diagnosed with acute mesenteric venous thrombosis.29 Cirrhosis and hereditary or acquired thrombophilia can also increase the risk for mesenteric venous thrombosis.30 Approximately 20% of cases are idiopathic.2,30
Mean age at presentation is 40 to 60 years, and mesenteric venous thrombosis is slightly more common in men.31 The severity of ischemic symptoms depends on the timing of thrombotic occlusion, with acute thrombotic venous occlusion resulting in more profound symptoms because collateral circulation has not developed. Patients may describe middle abdominal pain that is vague and colicky. The onset of pain is usually less abrupt than with acute arterial mesenteric ischemia, and patients typically present with nausea, vomiting, diarrhea, and abdominal cramping. Approximately 75% of patients have symptoms for at least 48 hours before seeking medical attention, and up to 29% of patients are hemodynamically unstable on presentation.31
Computed tomography with and without oral contrast is an appropriate initial screening test, and computed tomography or magnetic resonance angiography may be pursued if the initial computed tomography is nondiagnostic and clinical suspicion remains high.2,30 Doppler ultrasonography is highly specific but less sensitive for mesenteric venous thrombosis, and assessment of the smaller vessels is limited.29 All patients should be assessed for history of malignancy, liver disease, and recent surgery. Anticoagulation is recommended in cases of acute mesenteric venous thrombosis; the duration of anticoagulation is 6 months or longer depending on the underlying cause.29
CAUSES OF CHRONIC MESENTERIC ISCHEMIA
Chronic mesenteric ischemia describes intermittent or continuous intestinal hypoperfusion caused by occlusive disease of the mesenteric vessels. Most cases of chronic mesenteric ischemia are due to atherosclerosis. Less common causes include fibromuscular dysplasia, vasculitis, and retroperitoneal fibrosis.
Atherosclerosis
More than 90% of cases of chronic mesenteric ischemia result from atherosclerotic disease affecting the proximal segments of the visceral vessels.4,9 Risk factors include smoking, hypertension, diabetes, and the presence of atherosclerosis in other arterial beds. Most patients with chronic mesenteric ischemia are female.3 The reason is not entirely clear but appears to be related to more acutely angulated mesenteric vessels in women compared with men.32
Mesenteric artery stenosis is common, with post-mortem and duplex ultrasonography studies reporting an overall prevalence of 6% to 29% and as high as 67% in patients older than 80.4 Despite this, clinical manifestations of chronic mesenteric ischemia are rare because extensive collateral vessels develop over time, protecting against visceral malperfusion. Because of these collateral networks, symptoms and the need for revascularization are often delayed until at least 2 of the mesenteric vessels are stenosed.33 The likelihood of mesenteric artery stenosis progressing to symptomatic chronic mesenteric ischemia is higher in multivessel disease.4 In a retrospective analysis of 77 patients with asymptomatic SMA stenosis, patients with stenosis of 2 or more mesenteric vessels had a higher incidence of chronic mesenteric ischemia compared with patients with single-vessel disease (15.1% vs 0%).34 Approximately 20% to 50% of cases of symptomatic chronic mesenteric ischemia will progress to acute mesenteric ischemia, or acute-on-chronic mesenteric ischemia.35
More than 70% of patients with symptomatic chronic mesenteric ischemia report abdominal angina, a postprandial abdominal pain often described as dull and crampy that usually begins within 30 minutes of eating and lasts 1 to 2 hours.4 As abdominal pain progresses over time, many patients turn to adaptive eating patterns, eating smaller portions or, in advanced cases, avoiding food (ie, food fear).35 Weight loss is a key feature and is present in more than 60% of patients.4,34 Less typical symptoms include nausea, vomiting, diarrhea, or constipation.36
Physical examination is often nonspecific but may reveal signs of malnutrition or cachexia. An abdominal bruit may be present; however, the classic triad of abdominal bruit, postprandial pain, and weight loss is present in only approximately 22% of cases.4
The nonspecific nature of symptoms makes it challenging to differentiate chronic mesenteric ischemia from common abdominal pathologies such as gallstone disease and peptic ulcer disease. Again, a high index of suspicion is crucial to promptly establish the diagnosis. A careful patient history should aim to identify patients with atherosclerotic risk factors and those reporting weight loss. The diagnosis is further supported by radiographic findings of high-grade stenosis or occlusion of at least 2 mesenteric vessels. Computed tomography angiography is recommended as the initial study of choice for mesenteric ischemia by the Society for Vascular Surgery, American College of Radiology, and European Society of Vascular Surgery, with close to 100% sensitivity. However, duplex ultrasonography is an effective, low-cost alternative that is more than 90% sensitive and specific in detecting high-grade stenosis.37
Fibromuscular dysplasia
Rarely, chronic mesenteric ischemia has been reported as a complication of fibromuscular dysplasia, a nonatherosclerotic, noninflammatory disorder leading to stenosis, aneurysm, dissection, or occlusion of arteries that predominantly occurs in young and middle-aged women.38 The US Registry for Fibromuscular Dysplasia reported mesenteric involvement in 15.1% of cases; however, symptoms of mesenteric ischemia were rare (1.3%).39,40
Symptomatic patients present with severe abdominal pain or signs of acute arterial dissection.41,42 Because fibromuscular dysplasia can affect nearly any vascular bed, many patients will have multivessel involvement, which can result in other signs and symptoms, including pulsatile tinnitus and hypertension.
The classic angiographic appearance of beading (medial fibroplasia) or focal stenosis (intimal fibroplasia) supports the diagnosis.39 Histopathology has shown proliferation of the arterial smooth muscle cells and destruction of elastic fibers.41
Vasculitis
Mesenteric ischemia is a rare but severe, life-threatening manifestation of systemic vasculitis.43 Chronic inflammation can cause arterial wall thickening leading to stenosis or occlusion, or can weaken the arterial media and lead to aneurysm formation.44 Gastrointestinal involvement is mostly seen in vasculitis affecting the medium and large arteries, such as polyarteritis nodosa, giant cell arteritis, and Takayasu arteritis.43 Cases of mesenteric vasculitis have also been reported in patients with systemic lupus erythematosus and Behçet disease.45,46 Inflammatory markers may be elevated but can be nonspecific. Further serologic testing is often necessary, including a viral hepatitis panel, antineutrophil cytoplasmic antibodies, and antinuclear antibodies. Diagnosing the underlying condition is important, as these patients may require immunosuppression in addition to other therapies for ischemia.
Retroperitoneal fibrosis
Retroperitoneal fibrosis is a rare inflammatory disease of the retroperitoneum that occurs predominantly in middle-aged men. The fibrosis characteristically encases the infrarenal abdominal aorta and iliac arteries, and may compress visceral vessels, resulting in ischemia.47,48 More than 50% of cases are idiopathic; other causes include malignancy and infection.49
Patients typically present with dull abdominal or low back pain. Other symptoms include diarrhea, weight loss, jaundice, and leg swelling. Renal impairment resulting from ureteral obstruction is seen in up to 25% of cases.50 Diagnosis is made based on a high index of suspicion and computed tomography and magnetic resonance imaging showing retroperitoneal perivascular soft-tissue masses.
CONCLUSION
Mesenteric ischemia remains a diagnostic challenge to many clinicians because it is uncommon and its symptoms are nonspecific. Early recognition and focused evaluation are crucial for timely diagnosis and prevention of catastrophic complications.
DISCLOSURES
Dr. Wu reports no relevant financial relationships which, in the context of their contributions, could be perceived as a potential conflict of interest. Dr. Nanjundappa has disclosed teaching and speaking for Abbott, Medtronic, and Phillips Healthcare, consulting for Argon and Phillips Healthcare, and ownership interest (stock, stock options in a publicly owned company) for Zoll.
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3775 | https://www.stannescet.ac.in/cms/staff/qbank/MECH/Notes/ME3451-THERMAL%20ENGINEERING-1892254488-ME3451%20Thermal%20Engineering%20Anna%20university%20NOTES.pdf | ME3451-Thermal Engineering Department of Mechanical Engineering 2023-2024 SANCET 1 ST. ANNE’S COLLEGE OF ENGINEERING AND TECHNOLOGY (Approved by AICTE, New Delhi. Affiliated to Anna University, Chennai) ANGUCHETTYPALAYAM, PANRUTI – 607 106. DEPARTMENT OF MECHANICAL ENGINEERING ME3451 THERMAL ENGINEERING L T P C 4 0 0 4 COURSE OBJECTIVES: 1. To learn the concepts and laws of thermodynamics to predict the operation of thermodynamic cycles and performance of Internal Combustion(IC) engines and Gas Turbines. 2. To analysing the performance of steam nozzle, calculate critical pressure ratio 3. To Evaluating the performance of steam turbines through velocity triangles, understand the need for governing and compounding of turbines 4. To analysing the working of IC engines and various auxiliary systems present in IC engines 5. To evaluating the various performance parameters of IC engines UNIT I THERMODYNAMIC CYCLES Air Standard Cycles – Carnot, Otto, Diesel, Dual, and Brayton – Cycle Analysis, Performance and Comparison, Basic Rankine Cycle, modified, reheat and regenerative cycles UNIT II STEAM NOZZLE AND INJECTOR Types and Shapes of nozzles, Flow of steam through nozzles, Critical pressure ratio, Variation of mass flow rate with pressure ratio. Effect of friction. Metastable flow. UNIT III STEAM AND GAS TURBINES 12 Types, Impulse and reaction principles, Velocity diagrams, Work done and efficiency – optimal operating conditions. Multi-staging, compounding and governing. Gas turbine cycle analysis – open and closed cycle. Performance and its improvement - Regenerative, Intercooled, Reheated cycles and their combination. UNIT IV INTERNAL COMBUSTION ENGINES – FEATURES AND COMBUSTION IC engine – Classification, working, components and their functions. Ideal and actual : Valve and port timing diagrams, p-v diagrams- two stroke & four stroke, and SI & CI engines – comparison. Geometric, operating, and performance comparison of SI and CI engines. Desirable properties and qualities of fuels. Air-fuel ratio calculation – lean and rich mixtures. Combustion in SI & CI Engines – Knocking – phenomena and control. UNIT V INTERNAL COMBUSTION ENGINE PERFORMANCE AND AUXILIARY SYSTEMS Performance and Emission Testing, Performance parameters and calculations. Morse and Heat Balance tests. Multipoint Fuel Injection system and Common rail direct injection systems. Ignition systems – Magneto, Battery and Electronic. Lubrication and Cooling systems. Concepts of Supercharging and Turbocharging – Emission Norms TOTAL: 60 PERIODS TEXT BOOKS: 1. Mahesh. M. Rathore, “Thermal Engineering”, 1st Edition, Tata McGraw Hill, 2010. 2. Ganesan. V, "Internal Combustion Engines" 4th Edition, Tata McGraw Hill, 2012. REFERENCES: 1. Ballaney. P, “Thermal Engineering”, 25th Edition, Khanna Publishers, 2017. 2. Domkundwar & Kothandaraman, “A Course in Thermal Engineering”, 6th Edition, Dhanpat Rai& Sons, 2011. ME3451-Thermal Engineering Department of Mechanical Engineering 2023-2024 SANCET 2 COURSE OUTCOMES (COs) CO 1 Apply thermodynamic concepts to different air standard cycles and solve problems. CO 2 To solve problems in steam nozzle and calculate critical pressure ratio. CO 3 Explain the flow in steam turbines, draw velocity diagrams, flow in Gas turbines and solve problems CO 4 Explain the functioning and features of IC engine, components and auxiliaries. CO 5 Calculate the various performance parameters of IC engines MAPPING BETWEEN COs, POs AND PSOs COs PROGRAMME OUTCOMES (POs) PSOs PO1 PO2 PO3 PO4 PO5 PO6 PO7 PO8 PO9 PO10 PO11 PO12 PSO1 PSO2 PSO3 CO 1 3 2 1 1 1 2 1 CO 2 2 2 2 1 1 2 1 CO 3 2 2 2 1 1 2 1 CO 4 2 2 1 1 1 2 1 CO 5 2 2 1 1 1 2 1 Low (1); Medium (2); High (3) ME3451-Thermal Engineering Department of Mechanical Engineering 2023-2024 SANCET 3 UNIT I THERMODYNAMIC CYCLES PART-A 1. What are the assumptions made in air standard cycle? (May 2019) The working fluid is a perfect gas. It follows the law of Pv=mRT the compression and expansion process are reversible adiabatic the working fluid should not undergo any chemical change throughout the cycle Kinetic and potential energies of the working fluid are neglected No heat loss between the system and surroundings 2. Define mean effective pressure and comment its application in internal combustion Engines? (Apr/May 2019) Mean effective pressure is defined as the constant pressure acting in the piston during working stroke. It is also defined as the ratio of work done to the stroke volume or piston Displacement volume. Mean effective pressure (MEP) pm= work done/ stroke volume or piston displacement volume 3. What are the factors influencing the ideal Brayton cycle efficiency? (April/May 2019) The following factors influencing the ideal Brayton cycle efficiency The network output of the cycle Heat supplied 4. Define air standard cycle efficiency? (Nov/Dec 2018) Air standard efficiency is the ratio of work done during the process to the heat supplied During the process. Air standard efficiency = work done / heat supplied Air standard efficiency I taken as the ideal efficiency of an internal combustion engine. In This case we imagine air is used instead of petrol or fuel oil mixed with air to form a gas. 5. Define cut-off ratio. (Nov/Dec 2018) It is defined as the ratio of volume after the expansion to the volume before the expansion. 6. Write any four differences between Otto and Diesel cycle? Otto cycle Diesel cycle It consists of two isentropic and two constant volume processes It consists of two adiabatic, one constant volume and one constant pressure processes Heat addition takes place in constant volume process Heat addition takes place in constant pressure process Efficiency is high Efficiency is less Compression ratio is equal to expansion ratio Compression ratio is greater than expansion ratio 7. Define mean effective pressure as applied to gas power cycles. Mean effective pressure is defined as the constant pressure acting on the piston during the Working stroke. It is also defined as the ratio of work done to the stroke volume of the Piston displacement volume. ME3451-Thermal Engineering Department of Mechanical Engineering 2023-2024 SANCET 4 8. What is thermodynamic cycle? Thermodynamic cycle is defined as the series of processes performed on the system, such that the system attains its original state. 9. What is an air standard cycle/ why such cycles are used? Air standard cycle is a thermodynamic cycle, which used air, as the working fluid is known as air standard cycle. To carry out the analysis of heat engines, the concept of air Standard cycles are used. 10. Define air standard efficiency. Air standard efficiency is an ideal efficiency. It is defined as the ratio of work done by the heat supplied. Actual efficiency of an engine will always less than the ideal or air standard Efficiency. 11. What are the various types of gas power cycles? Otto cycle Diesel cycle Brayton cycle Carnot cycle Dual combustion cycle 12. Define compression ratio. It is defined as the ratio of total cylinder volume to the clearance volume. 13. Define mean effective pressure. (April/May 2019) It the constant (or) average pressure acting on the piston during the working stroke. It is defined as the ratio of work done to the swept (or) stroke volume. 14. What is the effect of compression ratio on air standard efficiency of an ideal Otto Cycle? The efficiency of Otto cycle increases with the increase in compression ratio. 15. What is meant by steam (or) vapour power cycles? The cycle used to convert heat into work are called the power cycle. If the working fluid in the cycle is vapour, it is called as vapour power cycle. 16. Define heat rate. It is defined as the rate of heat input required to produce unit work output (1 kW) Heat rate = Heat input in kJ / s x 3600 s / h --------------------------------------- Net power output in kW 17. Define the term relative efficiency (or) efficiency ratio. It is defined as the ratio of the actual thermal efficiency of steam power plant to the Rankine efficiency. Relative efficiency = actual thermal efficiency / Rankine efficiency 18. Mention the processes involved in Diesel cycle. 1→2 : isentropic compression of the fluid (blue) 2→3 : constant pressure heating (red) 3→4 : isentropic expansion (yellow) ME3451-Thermal Engineering Department of Mechanical Engineering 2023-2024 SANCET 5 19. What is mean effective pressure of an engine? () and comment the application in internal combustion engine. The mean effective pressure (MEP) is defined as the average pressure required to act on the piston as it moves one displacement to give the work W. Mean Effective Pressure is a valuable parameter in internal combustion engines, providing insights into the overall performance and efficiency of the engine. It plays a crucial role in design, optimization, and evaluation processes in the automotive and engineering industries. 20. What constitute an engine? An engine is a machine that burns fuel and converts it into mechanical power. Most modern vehicles use internal combustion engines (ICE), which ignite fuel and use the reaction to move mechanical parts. 21. What is a two stroke engine? A two-stroke engine is a type of internal combustion engine that completes a power cycle in only two strokes of the piston, as opposed to the more common four-stroke engines that require four strokes (intake, compression, power, and exhaust) for each cycle. The two-stroke engine is known for its simplicity and high power-to-weight ratio but is also associated with certain drawbacks, such as higher emissions and less fuel efficiency. 22. Define Coefficient of nozzle. The ratio of the actual discharge to the ideal discharge 23. Draw the PV diagram of Diesel, Otto cycle and mark the various processes. () ME3451-Thermal Engineering Department of Mechanical Engineering 2023-2024 SANCET 6 24. Write down the air standard efficiency equation of Diesel cycle. () 25. State the function of flywheel, connecting rod, piston and crankshaft. Not only rotates the engine, the function of the flywheel is to store mechanical energy to balance the engine so that it continues to have good performance. Mechanical power is the energy created when the engine is running. The flywheel works to balance the mechanical power by storing it. The up-down motion of each piston is transferred to the crankshaft via connecting rods. A flywheel is often attached to one end of the crankshaft, in order to smoothen the power delivery and reduce vibration. ME3451-Thermal Engineering Department of Mechanical Engineering 2023-2024 SANCET 7 26. Draw an actual valve timing diagram of a four stroke diesel engine. 27. Write any four major difference between Otto cycle and diesel cycle. The major difference between these two is how energy is created. In diesel engines, the air is compressed before the fuel is injected. In petrol engines, gas and air are mixed, and then compressed and ignited. Another difference is, of course, in the type of fuel used. PART-B 1. In a compression ignition engine, working on a dual combustion cycle, pressure and temperature at the start of compression are 1 bar and 300 K respectively. At the end of compression, pressure reaches a value of 25 bar. The heat is supplied at 420 kJ per kg of air during constant volume heating and pressure becomes 2.8 bar at the end of isentropic expansion. Estimate the ideal cycle efficiency. Take Cp = 1005 J/(kg K) and Cv = 712 J/(kg K). ME3451-Thermal Engineering Department of Mechanical Engineering 2023-2024 SANCET 8 ME3451-Thermal Engineering Department of Mechanical Engineering 2023-2024 SANCET 9 ME3451-Thermal Engineering Department of Mechanical Engineering 2023-2024 SANCET 10 ME3451-Thermal Engineering Department of Mechanical Engineering 2023-2024 SANCET 11 2. A certain quantity of air at a pressure of 1 bar and temperature of 70°C is compressed adiabatically until the pressure is 7 bar in Otto cycle engine. 465 kJ of heat per kg of air is now added at constant volume. Determine: (i) Compression ratio of the engine. (ii) Temperature at the end of compression. (iii) Temperature at the end of heat addition. Take for air cp = 1.0 kJ/kg K, cv = 0.706 kJ/kg K. Initial pressure, p1 = 1 bar Initial temperature, T1 = 70 + 273 = 343 K Pressure after adiabatic compression, p2 = 7 bar Heat addition at constant volume, Qs = 465 kJ/kg of air Specific heat at constant pressure, Cp = 1.0 kJ/kg K Specific heat at constant volume, Cv = 0.706 kJ/kg K ME3451-Thermal Engineering Department of Mechanical Engineering 2023-2024 SANCET 12 Hence compression ratio of the engine = 3.97. (ii) Temperature at the end of compression, T2 : In case of adiabatic compression 1-2, ME3451-Thermal Engineering Department of Mechanical Engineering 2023-2024 SANCET 13 Hence temperature at the end of compression = 330.7°C. (iii) Temperature at the end of heat addition, T3: According to constant volume heating operation 2-3 Hence temperature at the end of heat addition = 989.3°C. 3. The mean effective pressure of an ideal diesel cycle is 8 bar. If the initial pressure is 1.03 bar and the compression ratio is 12, determine the cutoff ratio and the air standard efficiency. Assume ratio of specific heat for air to be 1.4 ME3451-Thermal Engineering Department of Mechanical Engineering 2023-2024 SANCET 14 ME3451-Thermal Engineering Department of Mechanical Engineering 2023-2024 SANCET 15 4. A gas engine operating on the ideal cycle Otto cycle has a compression ratio of 6. The pressure and pressure and temperature at commencement of compression are 1 bar 27°C. Heat added during the constant volume combustion process is 1170kj/kg. Determine: the peak pressure and Temperature, work output per kg of air and air standard efficiency assume Take for air Cv = 0.717 kJ/kg K and γ =1.4 for air. ME3451-Thermal Engineering Department of Mechanical Engineering 2023-2024 SANCET 16 ME3451-Thermal Engineering Department of Mechanical Engineering 2023-2024 SANCET 17 5. Air is used as the working fluid in a simple ideal Brayton cycle that has a pressure ratio of 12, a compressor inlet temperature of 300K, and a turbine temperature of 1000K. Determine the required mass flow rate of air for a net power output of 70MW, assuming both the compressor and the turbine have an isentropic efficiency of 85%. ME3451-Thermal Engineering Department of Mechanical Engineering 2023-2024 SANCET 18 6. The swept volume of a diesel engine working on dual cycle is 0.0053m3 and clearance volume is 0.00035m3. The maximum pressure is 65bar. Fuel injection ends at 5 percent of the stroke. The temperature and pressure at the start of the compression are 800C and 0.9 bar determine the air standard efficiency of the cycle. Take for air = 1.4 ME3451-Thermal Engineering Department of Mechanical Engineering 2023-2024 SANCET 19 ME3451-Thermal Engineering Department of Mechanical Engineering 2023-2024 SANCET 20 7. In a gas turbine plant working on the Brayton cycle the air at the inlet is at 270C, 0.1 MPa. The pressure ratio is 6.25 and the maximum temperature is 8000C, the turbine and compressor efficiencies are each 80% Find i) The compressor work per kg of air ii) The turbine work per kg of air iii) The heat supplied per kg of air iv) The cycle efficiency and v) The turbine exhaust temperature. ME3451-Thermal Engineering Department of Mechanical Engineering 2023-2024 SANCET 21 ME3451-Thermal Engineering Department of Mechanical Engineering 2023-2024 SANCET 22 8. An oil engine works on the dual cycle, the heat liberated at constant pressure being twice that liberated at constant volume. The compression ratio of the engine is 8 and the expansion ratio is 5.3. But the compression and expansion processes follow the law pV1.3 =C. the pressure and temperature at beginning of compression are 1 bar and 27°C respectively. Assuming Cp = 1.004 kJ/kg K and Cv = 0.717 kJ/kg K for air, find the air standard efficiency and the mean effective pressure. ME3451-Thermal Engineering Department of Mechanical Engineering 2023-2024 SANCET 23 9. In a gas turbine plant working on the Brayton cycle. The inlet is at 250C and 1 bar. The maximum pressure and temperature 3bar and 6500C, determine the following i) The cycle efficiency and ii) The heat supplied and heat rejected per kg of air iii) Work output iv) Exhaust temperature. ME3451-Thermal Engineering Department of Mechanical Engineering 2023-2024 SANCET 24 ME3451-Thermal Engineering Department of Mechanical Engineering 2023-2024 SANCET 25 10. Brief the working of Otto cycle with the help of p-V diagram, T-s diagram and derive the air standard efficiency of the cycle. ME3451-Thermal Engineering Department of Mechanical Engineering 2023-2024 SANCET 26 UNIT-II STEAM NOZZLE AND INJECTOR PART-A 1. What is steam nozzle? A steam nozzle is defined as a passage of varying cross section, through which heat energy of steam is converted to kinetic energy. Its major function is to produce steam jet with high velocity to drive steam turbines. 2. Write about the function of nozzle The major function of nozzle is to produce jet of steam or gas of high velocity to produce thrust for the propulsion of rocket motors and jet engines and drive steam or gas turbines. 3. List the types of nozzle. 1. Convergent Nozzle, 2. Divergent Nozzle, 3. Convergent-Divergent Nozzle 4. Define Convergent nozzle. In a convergent nozzle, the cross sectional area decreases continuously from its entrance to exit. It is used in a case where the back pressure is equal to or greater than the critical pressure ratio 5. Define divergent nozzle. The cross sectional area of divergent nozzle increases continuously from its entrance to exit. It is used in a case, where the back pressure is less than the critical pressure ratio 6. Define Convergent-Divergent nozzle. In this case, the cross sectional area first decreases from its entrance to throat, and then increases from throat to exit.it is widely used in many type of steam turbines 7. Draw the shape of supersonic nozzle. [May 2016] 8. Define critical pressure ratio. Give its expression. [Nov 2017,Nov 2018,Jul 2021] The critical pressure ratio is the pressure ratio which will accelerate the flow to a velocity equal to the local velocity of sound in the fluid. The maximum gas flow through a nozzle is determined by the critical pressure. The pressure at which the area is minimum and discharge per unit area is maximum is called critical pressure ratio. 9. Define nozzle efficiency or coefficient of nozzle [Dec 2013, Nov 2018] The nozzle efficiency is therefore defined as the ratio of the actual enthalpy drop to the isentropic enthalpy drop between the same pressures. Nozzle efficiency = (actual enthalpy drop) / (isentropic enthalpy drop) ME3451-Thermal Engineering Department of Mechanical Engineering 2023-2024 SANCET 27 10. List the effects of friction in nozzle. [May 2014, Dec 2015, May 2018] In practice, there is friction produced between the steam and the sides of the nozzle; this friction causes a resistance to the flow which is converted into heat. The heat formed tends drying the steam. i) The expansion is no more isentropic and enthalpy drop is reduced ii) The final dryness fraction of steam is increased as the kinetic energy gets converted in to heat due to friction and is absorbed by steam. iii) The specific volume of steam is increased as the steam becomes more dry due to this frictional reheating. 11. List the factors which influence nozzle efficiency. When the steam flows through a nozzle the final velocity of steam for given pressure drop is reduced due to the following reasons i) The friction between the nozzle surface and steam ii) The internal friction of steam itself iii) The shock losses. 12. Define degree of undercooling and degree of super saturation. [Jul 2021] The difference of supersaturated temperature and saturation temperature at that pressure is known as degree of under cooling. The ratio of super saturation pressures corresponding to the temperature between super saturated region is known as the degree of super saturation. 13. Define coefficient of velocity in nozzle. [Dec 2014] The ratio of the actual velocity of gas emerging from a nozzle to the velocity calculated under ideal conditions; it is less than 1 because of friction losses. 14. Define coefficient of Discharge. The ratio of the actual discharge to maximum discharge is known as coefficient of discharge. 15. What is meant by carry over loss? The velocity of steam at exit is sufficiently high thereby resulting in a kinetic energy loss called Carry over loss or Leading velocity loss. 16. If the enthalpy drop in a stem nozzle of efficiency 92% is 100 kJ/kg determine the exit velocity of steam. [May2017] C2 = √η(∆h) C2 = √0.92 × 10 0 𝑪𝟐 = 𝟗. 𝟓𝟗 𝒎/𝒔 17. Write the equation of maximum discharge through a nozzle. 18. Mention the values of maximum discharge for various steam. Types of Steam Index number Maximum Discharge Critical Pressure ratio Dry saturated n=1.135 p1 mmax = 0.637 A√v 1 p2 = 0.577 p1 Superheated n=1.3 p1 mmax = 0.666 A√v 1 p2 = 0.546 p1 Gas n=1.4 p1 mmax = 0.685 A√v 1 p2 = 0.582 p1 Wet steam n= ME3451-Thermal Engineering Department of Mechanical Engineering 2023-2024 SANCET 28 19. What is meant by metastable flow? Equilibrium between the liquid and vapour phase is delayed and the steam Continues to expand in a dry state. The steam in such a set of conditions is said to be supersaturated or in metastable state as its temperature at any pressure is less than the saturation temperature corresponding to the pressure. The flow of supersaturated steam, through the nozzle is called supersaturated flow or metastable flow. 20. What are the effects of super saturation or supersaturated flow? [Nov 2016] There is an increase in the entropy and specific volume of steam The heat drop is reduced below that for thermal equilibrium as an consequence the exit velocity is reduced. The density of supersaturated steam will be more than for the equilibrium conditions which gives the increase in the mass of steam discharged. The dryness fraction of steam is improved. 21. Differentiate supersaturated flow and isentropic flow. S.No Supersaturated flow Isentropic flow 1 Entropy is not constant Entropy remains constant 2 Super saturation reduces the heat drop therefore exit velocity is reduced No reduction in enthalpy drop. 3 Mollier diagram cannot be used Mollier diagram can be used. 22. What is meant by steam injector? A steam injector is a device employed to force water in to the boiler under pressure. 23. List the applications of steam nozzle i) To rotate steam turbine ii) Thermal power plant iii) To produce a very fine jet spray iv) It is also used for cleaning purpose. 24. What is the effect friction in steam nozzle? () The kinetic energy of the steam increases at the expense of its pressure energy in a steam nozzle. Some kinetic energy gets lost to overcome the friction in the nozzle. Therefore, the exit velocity of steam decreases due to nozzle friction . 25. What is the effect of supersaturation in the nozzles? Thus the effect of supersaturation is to reduce the enthalpy drop slightly during the expansion and consequently a corresponding reduction in final velocity. The final dryness fraction and entropy are also increased and the measured discharge is greater than that theoretically calculated. PART – B 1. Derive the condition for maximum flow rate in steam nozzle. [May 2018] 2. Define critical pressure ratio of a nozzle and discuss why attainment of sonic velocity. Also determines the maximum discharge through steam nozzle. 3. Derive the equation for critical pressure ratio in steam nozzle. [Nov 2017] 4. Derive the following expression for nozzle flow: 𝑑𝐴 = 1 𝑑𝑝 [ 1−𝑀2 ] where the 𝐴 𝛾 𝑝 𝑀2 Symbols are having usual meanings. [Jul 2021] 5. In a steam nozzle, the steam expands from 4 bar to 1 bar. The initial velocity is 60m/s ME3451-Thermal Engineering Department of Mechanical Engineering 2023-2024 SANCET 29 and initial temperature is 200ºC. Determine the exit velocity if nozzle efficiency is 92%. [Nov/Dec 2018] 6. Steam expands isentropic ally in a nozzle from 1 MPa, 250º C to 10 Kpa. The flow rate of the steam is 1 kg/s. Find the following when the inlet velocity is neglected, (i) Quality of steam, (ii) Velocity of steam at the exit of the nozzle, (iii) Exit area of the nozzle. [Dec 2013] 7. The flow rate through steam nozzle with isentropic flow from pressure of 13 bar Was found 60 kg/min. steam is initially saturated. Determine the throat area. If the flow is super saturated, determine the increase in flow rate. [May 2014] 8. Dry saturated steam at a pressure of 11 bar enters a convergent-divergent nozzle and leaving at a pressure of 2 bar. If the flow is adiabatic and frictionless, determine i) the exit velocity of a steam, ii) Ratio of cross section of exit and that at throat. Assume the index of adiabatic expansion to be 1.135 [May2015,Jul2021] 9. Steam at a pressure of 10.5 bar and 0.95 dry is expanded through a CD nozzle. The pressure of steam leaving the nozzle is 0.85 bar. Find the velocity of steam at throat for maximum discharge. Take n=1.35. Also find the area at the exit and steam discharge if the throat area is 1.2square cm. Assume the flow is isentropic and there are no friction losses. [Dec 2014] 10. (a) What are the effects of friction in a nozzle? Explain 11. (b) A convergent-divergent nozzle is required to discharge 2 kg/s of steam. The nozzle is supplied with steam at 7 bar and 180⁰C and discharge takes place against a back pressure of 1 bar. The expansion up to the throat is isentropic and the frictional resistance between the throat and the exit is equivalent to 63 kJ/kg of steam. Taking approach velocity of 75m/s and throat pressure of 4 bar estimate suitable areas for throat and exit and overall of the nozzle based on the enthalpy drop between the actual inlet pressure and temperature and the exit pressure.
[May 2013,Nov 16] 12. In a test on a steam nozzle, the issuing steam jet impinges on a stationary flat Plate which is perpendicular to the direction of flow and the force on the plate is measured. With convergent-divergent nozzle supplied with steam at 10 bar dry saturated and discharging at 1 bar; the force is experimentally measured to be 600N. The area of the nozzle at throat measures 5 cm2 and that exit area is such that complete expansion is achieved under these conditions. Determine: (i) flow ME3451-Thermal Engineering Department of Mechanical Engineering 2023-2024 SANCET 30 rate of the steam, and (ii) the efficiency of the nozzle assuming that all losses occur after the throat. Assume n = 1.135 for isentropic expansion. [May 2017] 13. The dry and saturated steam at a pressure of 10.5 bar is expanded isentropically in a nozzle to a pressure of 0.7 bar. Determine the final velocity of the steam issuing from the nozzle, when (a) friction is neglected, and (h) 10% of the heat drop is lost in friction. The initial velocity of steam may be neglected. 14. Gases expand in a convergent divergent nozzle from 3.6 bar and 425°C to aback pressure of I bar, at the rate of 18kg/s. If the nozzle efficiency is 0.92, calculate the required throat and exit areas of the nozzle. Neglect inlet velocity and friction in the convergent part. For the gases, take Cp = 1.113 kJ/kg K and γ= 1.33 15. Dry saturated steam at 2.8bar is expanded through a convergent nozzle to 1.7 bar. The exit area is 3cm3. Calculate the exit velocity and mass flow rate for, (i) Isentropic expansion, (ii) Super saturated flow. [Nov/Dec 2018] 16. Explain the supersaturated or metastable flow of steam through nozzle and the Significance of Wilson’s line. [May 2016] 17. What are the effects of super saturation on discharge and heat drop? 18. The dry saturated steam is expanded in a nozzle from pressure of 10 bar to a pressure of 5 bar if she expansion is supersaturated, find: 1. the degree of undercooling and 2. The degree of supersaturation. PART – C 1. Dry saturated steam at a pressure of 8bar enters a convergent divergent nozzle and leaves it at a pressure of 1.5bar. If the flow is isentropic and if the corresponding expansion index is 1.133, find the ratio of cross-sectional area at exit and throat for maximum discharge. [AU Nov/Dec 2015] 2. Steam turbine develops 185 kW with a consumption of 16.5 kg/kW/h. The Pressure and temperature of the steam entering the nozzle are 12 bar and 220º C. The steam leaves the nozzle at 1.2 bar. The diameter of the nozzle at throat is 7mm, Find the number of nozzles. If 8% of the total enthalpy drop is lost in friction in the diverging part of the nozzle, determine the diameter at the exit of the nozzle and the exit velocity of the leaving steam. Sketch the skeleton Mollier diagram and show on it the values of pressure, temperature or dryness fraction, enthalpy and specific volume at inlet, throat and exit. [Nov/Dec 2018] 3. Calculate the throat and exit diameters of a convergent divergent nozzle which will discharge 820kg of steam per hour from a pressure of 8bar superheated to 220⁰ C into a chamber having a pressure of 1.5bar. The friction loss in the divergent part of the nozzle maybe taken as 0.15 of the total enthalpy drop. 4. State the relation between the velocity of steam and heat during any part of a steam nozzle. 5. Find the percentage increase in discharge from a convergent-divergent nozzle expanding steam from 8.75 bar dry to 2 bar. When; 1. The expansion is taking place under thermal equilibrium, and 2. The steam is in metastable state during pan of its expansion. Take area of nozzle as 2500 mm2 ME3451-Thermal Engineering Department of Mechanical Engineering 2023-2024 SANCET 31 1. Dry saturated steam at a pressure of 11 bar enters a steam nozzle and leaves at a pressure of 2 bar. If the flow is adiabatic and frictionless, determine: (i) the exit velocity of steam (ii) ratio of cross-section at exit and that at throat. Assume the index of adiabatic expansion to be 1.135 2. Derive the following expression for nozzle flow: 𝐝𝐀 𝐀= 𝟏𝐝𝐩 𝛄[1-M2/M2] where, the symbols having usual meanings. ME3451-Thermal Engineering Department of Mechanical Engineering 2023-2024 SANCET 32 ME3451-Thermal Engineering Department of Mechanical Engineering 2023-2024 SANCET 33 ME3451-Thermal Engineering Department of Mechanical Engineering 2023-2024 SANCET 34 ME3451-Thermal Engineering Department of Mechanical Engineering 2023-2024 SANCET 35 3. Describe the flow of steam through nozzles and hence deduce the expression for a critical pressure ratio. (8) Critical pressure ratio is the ratio of pressure at which the system gives maximum mass flow rate and it cannot be increased further by adjusting the system pressure. The critical pressure ratio of any fluid depends on the polytropic index (n) of that fluid. This maximum mass flow rate condition is reached when the match number at minimum cross-section becomes equal to 1. Critical pressure ratio = Pc/Po Where Pc = Pressure at the minimum cross-section Po = Inlet pressure Critical pressure ratio = Pc/Po= (2/n+1)n−1 Where n = polytropic index of the fluid The significance of the critical pressure ratio is given below 1. The critical pressure ratio helps to attain maximum mass flow rate through the nozzle. 2. It also helps us to avoid choking of nozzle. Critical pressure ratio for air: For air, the value of the polytropic index is given by, n = 1.41 By using the formula of critical pressure ratio, Pc/Po=(2/n+1)n−1==(2/(1.41+1))1..41−1= 0.526 Hence for air, the critical pressure ratio is 0.526 Critical pressure ratio for steam: For saturated steam, the value of the polytropic index is given by, n = 1.135 By using the formula of critical pressure ratio, Pc/Po=(2n+1)n−1=(2/1.135+1).135−1= 0.577 For steam, the critical pressure ratio is 0.577 ME3451-Thermal Engineering Department of Mechanical Engineering 2023-2024 SANCET 36 4. Explain the supersaturated flow in nozzles and their effects. (5) Supersaturated flow in a nozzle occurs when the flow of a liquid is at a temperature and pressure such that it is on the verge of vaporization, but has not yet vaporized. This can occur when the liquid is under high pressure and then passes through a constriction, such as a nozzle, causing a rapid drop in pressure. 5. A convergent-divergent nozzle is required to discharge 350 kg of steam per hour. The nozzle is supplied with steam at 8.5 bar and 90% dry and discharges against a back pressure of 0.4 bar. Neglecting the effect of friction, find the throat and exit diameters. (8) ME3451-Thermal Engineering Department of Mechanical Engineering 2023-2024 SANCET 37 ME3451-Thermal Engineering Department of Mechanical Engineering 2023-2024 SANCET 38 ME3451-Thermal Engineering Department of Mechanical Engineering 2023-2024 SANCET 39 ME3451-Thermal Engineering Department of Mechanical Engineering 2023-2024 SANCET 40 6. Derive the condition for maximum discharge and expression for maximum discharge in steam nozzle. Determine the throat area, exit area and exit velocity for a steam nozzle to pass a mass flow of 0.2 kg/s when inlet conditions are 10 bar and 250°C and the final pressure is 2 bar. Assume expansion is isentropic and that the inlet velocity is negligible. Use pV1.3 = Constant. Do not calculate from h-s chart. ME3451-Thermal Engineering Department of Mechanical Engineering 2023-2024 SANCET 41 ME3451-Thermal Engineering Department of Mechanical Engineering 2023-2024 SANCET 42 7. Steam at a pressure of 10.5 bar and 0.95 dry is expanded convergent divergent nozzle. The pressure of steam leaving nozzle is 0.85 bar. Find the velocity of steam at the throat for maximum discharge taken = 1.135. Also find the area at the exit and steam discharge if the throat area is 1.2 cm2. Assume flow is isentropic and there are no friction losses. ME3451-Thermal Engineering Department of Mechanical Engineering 2023-2024 SANCET 43 ME3451-Thermal Engineering Department of Mechanical Engineering 2023-2024 SANCET 44 8. The inlet conditions to a steam nozzle are 10bar and 2500C. the exit pressure is 2 bar. Assuming isentropic expansion and negligible velocity determine i) the throat area ii) The exit velocity iii) the exit area of the nozzle. ME3451-Thermal Engineering Department of Mechanical Engineering 2023-2024 SANCET 45 ME3451-Thermal Engineering Department of Mechanical Engineering 2023-2024 SANCET 46 ME3451-Thermal Engineering Department of Mechanical Engineering 2023-2024 SANCET 47 UNIT-III STEAM AND GAS TURBINES PART-A 1. Define Steam turbine. A steam turbine is a prime mover in1iich rotary motion is obtained by the gradual change of momentum of the steam. The force exerted on the blades is due to the velocity of steam. This is due to the fact that the curved blades by changing the direction of steam receive a force or impulse. 2. Advantage of steam turbine over reciprocating steam engines. Steam turbine may develop higher speeds and a greater steam range is possible. The efficiency of a steam turbine is higher. The steam consumption is less. Since all the moving pails are enclosed in a casing, the steam turbine is comparatively safe. A steam turbine requires less space and lighter foundations, as there are little vibrations. There is less frictional loss due to fewer sliding parts. The applied torque is more uniform to the driven shaft. A steam turbine requires less attention during running. Moreover, the repair Costs are generally less. 3. Classify steam turbine according to the classification of flow. i) Impulse turbine ii) Reaction turbine iii) combination of impulse and reaction 4. Classification of steam Turbine The steam turbines may be classified into the following types: According to the mode of steam action: (i) Impulse turbine, and (ii) Reaction turbine. ME3451-Thermal Engineering Department of Mechanical Engineering 2023-2024 SANCET 48 According to the direction of steam flow: (i) Axial flow turbine, and (ii) Radial flow turbine. According to the exhaust condition of steam: (i) Condensing turbine, and (ii) Non-condensing turbine. According to the pressure of steam: (i) High pressure turbine, (ii) Medium pressure turbine, and (iii) Low pressure turbine. According to the number of stages: (i) Single stage turbine, and (ii) Multi-stage turbine. 5. Define Impulse turbine. An impulse turbine, as the name indicates, is a turbine which runs by the impulse of steam jet In this turbine, the steam is first made to flow through a nozzle. Then the steam jet impinges on the turbine blades (which are curved like buckets) and are mounted on the circumference of the wheel. The steam jet after impinging glides over the concave surface of the blades and finally leaves the turbine. This is also known as De-Level Impulse. 6. How impulse turbine is classified? Impulse turbines are a type of hydraulic turbine that extracts energy from the kinetic energy of a fluid (usually water) in motion. These turbines are classified based on various factors, including the direction of the water flow, the number of jets or nozzles, and the specific design features. 7. Define two stages Impulse turbine. The steam after leaving the moving blade is made to flow through a fixed blade ring (in order to make the steam to flow at a designed angle and again impinges on second moving blade. This type of turbine is called two-stage impulse turbine. 8. Define Reaction turbine. [Jul 2021] In a reaction turbine, the steam enters the wheel under pressure and flows over the blades. The steam while gliding propels the blades and makes them to move. As a matter of fact, the turbine runner is rotated by the reactive forces of steam jets. The backward motion of the blades is similar to the recoil of a gun. This is also known as Parson's Reaction Turbine 9. Differentiate impulse turbine and reaction turbine. [MAY 2018] S.No Particulars Impulse Turbine Reaction Turbine 1 Pressure drop Only in nozzles and not in moving blades. In fixed blades (nozzles) as well as in moving blades. 2 Area of blade channels Constant Varying 3 Blades Profile type Aerofoil type. 4 Admission of steam Not all round All round or complete 5 Nozzles Diaphragm contains the nozzle Fixed blades similar to moving blades attached to the casing serve as nozzles and guide the steam. 6 Power Not much power can be developed. Much power can be developed. 7 Efficiency Low High ME3451-Thermal Engineering Department of Mechanical Engineering 2023-2024 SANCET 49 10. Define blade efficiency or diagram efficiency. It is the ratio of work done on the blade per second to the energy entering the blade per second. 11. Define stage efficiency. The stage efficiency covers all the losses in the nozzles, blades, diaphragms and discs that associated with that stage. Network done on shaft per kg of steam flowing stage Adiabatic heat dropper stage 12. Define blade velocity coefficient or coefficient of velocity or Friction factor. [Jul 2021] The blade velocity coefficient is defined as the ratio of relative velocity of steam as is passes over the blades without frictional resistance to relative velocity of steam with friction resistance. K C r 0 where K is blade velocity coefficient C r1 13. Define blade speed ratio Blade speed ratio is defined as the ratio of blade speed to steam speed Cbl C1 14. Define degree of reaction. [May/June 2014] [Nov/Dec 2014] Degree of reaction or reaction ratio (R) is defined as the ratio of static pressure drop in the rotor to the static pressure drop in the stage or as the ratio of static enthalpy drop in the rotor to the static enthalpy drop in the stage. 15. Define coefficient of velocity in nozzle? [Nov/Dec 2014] The ratio of the actual velocity of gas emerging from a nozzle to the velocity calculated under ideal conditions; it is less than 1 because of friction losses. 16. What is meant by carry over loss? The velocity of steam at exit is sufficiently high thereby resulting in a kinetic energy loss called Carry over loss or Leading velocity loss. 17. What are the methods adopted to prevent erosion in steam turbines? i) By raising the temperature of steam at inlet, so that at exit of turbine the wetness does not exceed 10% ii) By adopting reheat cycle; so that wetness at exit remains in limit. iii) Drainage belts are provided on the turbine, so that the water droplets are on outer periphery, due to centrifugal force are drained. The drained amount is about 25 percent of total water particles present. 18. What do you mean by bleeding in steam turbine? Bleeding is the process of draining steam from the turbine at certain points during its expansion and using this steam for heating the feed water supplied to the boiler. 19. What is meant by stage in turbine? In an impulse turbine, stage means set of nozzles outside the turbine + moving blades on the rotor. In a reaction turbine, stage means one set of fixed blades + one set of moving blades. ME3451-Thermal Engineering Department of Mechanical Engineering 2023-2024 SANCET 20. What are the losses in steam turbine? Residual velocity losses, Loss due to friction, Radiation losses, Loss due to moisture. 21. What are the possible causes of excessive vibration or noise in steam turbine? Misalignment, worn bearings, unbalanced wheel, unbalanced coupling, bent shaft, piping strain. 22. Define compounding of turbine and classify it. [NOV 2017] The steam is expanded from the boiler pressure to condenser pressure in one stage the speed of the rotor becomes tremendously high which crops up practical complicacies. There are several methods of reducing this speed to lower value all these methods utilize a multiple system of rotor in series keyed on a common shaft and the steam pressure or jet velocity is absorbed in stages as the steam flows over the blades. This is known as compounding. The different methods of compounding are i) Velocity compounding ii) Pressure compounding iii) Pressure velocity compounding. 23. What is the purpose of compounding? Compounding is the method in which multiple system or rotors are keyed to common shaft in series and the steam pressure or jet velocity is absorbed in stages as it flows over the rotor blades. Purpose of compounding: Reduction of pressure (from boiler pressure to condenser pressure) in single results in the very high velocity entering the turbine blades. Therefore, the turbine rotor will run at a high speed about 30,000 rpm which is not useful for practical purpose. In order to reduce the rotor speed up to about 400 m/sec, compounding of steam turbine is necessary. 24. What is pressure compounding? [April/May 2015] The Steam Pressure compounding is the method in which pressure in a steam turbine is made to drop in a number of stages rather than in a single nozzle. This method of compounding is used in Rateau and Zoelly turbines. 25. What are the advantages of velocity compounded impulse turbine? i) Owing to relatively large heat drop, a velocity compounded impulse turbine requires a comparatively small number of stages. ii) Due to number of stages being small, its cost is less iii) The steam temperature is sufficiently low in a two or three row wheel; therefore cast iron cylinder may be used. This will cause saving in material cost. 26. What do you mean by governing of steam turbine? Classify it Governing of steam is to control the rotational speed of turbine by controlling the flow of steam into turbine irrespective of varying load on turbine. Classification i) Throttle governing ii) Nozzle governing iii) By-pass governing iv) combination of Throttle Nozzle By-pass governing. 27. What is the remedy for a bent steam turbine shaft causing excessive vibration? i) The run-out of the shaft near the centre as well as the shaft extension should be checked. ii) If the run-out is excessive, the shaft is to be replaced 28. What is gas turbine? Gas turbine is an axial flow rotary turbine in which gas is used as working medium. ME3451-Thermal Engineering Department of Mechanical Engineering 2023-2024 SANCET 29. Differentiate open and closed cycle gas turbine? Open cycle gas turbine Closed cycle gas turbine Gas is exhausted to the atmosphere after each cycle Gas is recirculated again and again Size is small Size is large Weight is less Weight is high High quality fuels are used Low quality fuels are used Intercooler is not required Intercooler is required to cool the exhaust gas to the original temperature 30. Distinguish between impulse and reaction turbine. Impulse turbines are typically used for high-head, low-flow applications such as hydroelectric power plants, where the water is delivered under high pressure. Reaction turbines are typically used for low-head, high-flow applications such as hydropower plants, where the water is delivered at a lower pressure. PART – B 1. Explain the pressure and velocity compounding diagram of multistage turbine with neat sketch. [Nov/Dec 2014] [Jul 2021] 2. Elucidate the working of velocity, pressure and velocity pressure compounding methods with neat sketch. [May 2018] 3. Explain the pressure band velocity compounding of a multi stage turbine. 4. In a single stage impulse turbine, nozzle angle is 20º and blade angles are equal. The velocity coefficient for blade is 0.85. Find maximum blade efficiency possible. If the actual blade efficiency is 92% of the maximum blade efficiency, find the possible ratio of blade speed to steam speed. [Jul 2021] 5. In a De-lavel turbine, the steam enters the wheel through a nozzle with a velocity of 500 m/s and at an angle of 20º to the direction of motion of the blade. The blade speed is 200 m/s and the exit angle of the moving blade is 25°. Find the inlet angle of the moving blade, exit velocity of steam and its direction and work done per kg of steam. 6. In a De Laval Turbine steam issues from the nozzle with a velocity of 1200 m/s. The nozzle angle is 20˚C, the mean blade velocity is 400 m/s and the inlet and outlet angles are equal. The mass of steam flowing through the turbine per hour is 1000 kg. Calculate i) Blade angles ii) Relative velocity of steam entering the blades iii) Tangential force on the blades iv) Power developed v) Blade efficiency. Take blade velocity co-efficient as 0.8 [April/May 2015] 7. A steam jet enters the row of blades with a velocity of 375 m/s at an angle of 20º with the direction of motion of the moving blades. If the blade speed is 165 m/s, find the suitable inlet and outlet blade angles assuming that there is no thrust on the blades. The velocity of steam passing over the blades is reduced by 15%. Also determine power developed by the turbine per kg of steam flowing aver the blades per second. 8. In a single stage impulse turbine the isentropic enthalpy drop of 200 kJ/kg occurs in the nozzle having efficiency of 96% and nozzle angle of 15°. The blade velocity coefficient is 0.96 and ratio of blade speed to steam velocity is 0.5. The steam mass flow rate is 20 kg/s and velocity of steam entering is 50 m/s. Determine (a) the blade angles at inlet and outlet if the steam enters blades smoothly and leaves axially, (b) the blade efficiency, (c) ME3451-Thermal Engineering Department of Mechanical Engineering 2023-2024 SANCET the power developed in kW and (d) the axial thrust. 9. Steam enters the blade row of an impulse turbine with the velocity of 600 m/s at an angle of 25°C to the plane of rotation of the blades the blade mean speed is 250 m/s. The blade angle at the exit side is 30°C. The blade friction loss is 10 %. Determine blade angle inlet, blade efficiency and work done per kg of steam [May/ June 2014] 10. (a) The velocity of steam leaving the nozzle of an impulse turbine is 10000m/s and the nozzle angle is 20oC. The blade velocity is 350m/s and the blade velocity of co-efficient is 0.85.Assuming no losses due to shock at inlet ,calculate for a mass flow of 1.5kg/s ,and symmetrical blading, (i) Blade inlet angle, (ii) Driving force on the wheel, (iii) Axial thrust on the wheel and (iv) Power developed by the turbine. (b) Differentiate between impulse and reaction turbine. [April/May 2013] 11. In a single stage impulse turbine the blade angles are equal and nozzle angle is 20o.the velocity coefficient for the blade is 0.83 find the maximum blade efficiency possible. If the actual blade efficiency is 90% of maximum blade efficiency, find the possible ratio of blade speed to steam speed. [Nov/Dec 2017] 12. In one stage of a reaction steam turbine, both the fixed and moving blades have Inlet and outlet blade tip angles of 35º and 20º respectively. The mean blade speed is 80 m/s and the steam consumption is 22 500 kg per hour. Determine the power developed In the pair, if the isentropic heat drop for the pair is 23.5 kJ per kg. 13. A Parson's reaction turbine, while running a 1400 r.p.m. consumes 30 tonnes of steam per hour. The steam at a certain stage is at 6 bar with dryness fraction of 0.9 and the stage develops 10 kW. The axial velocity of flow is constant and equal to 0.75 of the blade velocity. Find mean diameter of the drum and the volume of steam flowing per second. Take blade tip angles at inlet and exit as 35º and 20º respectively. 14. A Parson’s reaction turbine has mean diameter of blades as 1.6 m and rotor moving at 1500 rpm. The inlet and outlet angles are 80°C and 20°C respectively. Turbine receives steam at 12 bar, 200°C and has isentropic heat drop of 26 kJ/kg. 5% of steam supplied is lost through leakage. Determine the following considering horse power developed in stage to be 600 hp.(a) the stage efficiency and (b) the blade height. PART – C 1. A convergent-divergent nozzle for a steam turbine has to deliver steam under a supply condition of 11 bar with 100ºC superheat and a back pressure of 0.15bar. if the outlet area of the nozzle is 9.7cm2, determine using steam tables, the mass of steam discharged per hour. If the turbine converts 60% of the total enthalpy drop into useful work, determine the power delivered by the turbine. Neglect the effect of friction in the nozzle. Take CP of superheated steam as 2.3 kJ/kg.k [Nov/Dec 2018] 2. A 50% reaction turbine (with symmetrical velocity triangles) running at 400 r.p.m. has the exit angle of the blades as 20º and the velocity of steam relative to the blades at the exit is 1.35 times the mean blade speed. The steam flow rate is 8.33 kg/s and at a particular stage the specific volume is 1.381m3 /kg. Calculate for this stage. A suitable blade height, assuming the rotor mean diameter to be 12 times the ME3451-Thermal Engineering Department of Mechanical Engineering 2023-2024 SANCET blade height. 3. In a reaction turbine, the blade tips are inclined at 35º and 20º in direction of motion. The guide blades are of the same shape as the moving blades, but reversed in direction. At a certain place in the turbine, the drum diameter is 1 meter and the blades are 100 mm high. At this place, steam has a pressure of 1.7 bar and dryness 0.935. If the speed of the turbine is 250 r.p.m. and the steam passes through the blades without shock, find the mass of steam flow and the power developed in the ring of the moving blades 4. A reaction turbine runs at 300 r.p.m. and its steam consumption is 15400 kg/hr. The pressure of steam at certain pair is 1.9 bar; its dryness 0.93 and power developed by the pair is 3.5 kW. The discharging blade tip angle is 20° for both fixed and moving blades and the axial velocity of flow is 0.72 of the blade velocity. Find the drum diameter and blade height. Take the tip leakage steam as 8%, but neglect blade thickness. 5. (a) List the advantages of steam turbines over gas turbines. (b) Determine the isentropic enthalpy drop in the stage of Parson's reaction turbine which has the following particulars: speed=1500 rpm, mean diameter of the rotor = 1m, stage efficiency =80%, speed ratio = 0.7, blade outlet angle =20o. ME3451-Thermal Engineering Department of Mechanical Engineering 2023-2024 SANCET 1. Draw a schematic of closed cycle gas turbine plant and discuss its function. Also suggest fuels that are especially required for closed cycle gas turbine plant. A closed-cycle gas turbine method is adopted to overcome the disadvantages of an open cycle gas turbine method. The corrosion and erosion of turbine blades is the main drawback in an open cycle. This drawback can be overcome by using superior quality of working medium (air or helium, argon, hydrogen or neon) where it doesn’t mix with the fuel in the combustion chamber. The other advantage of using a closed cycle method is, the rejection of heat of exhaust gases takes place in a re-cooler or re-heaters or heat exchangers. This article discusses an overview of this turbine, working, advantages, and disadvantages. What is a Closed-cycle Gas Turbine? A closed-cycle gas turbine can be defined as a gas turbine, which overcomes the drawbacks of the open cycle gas turbine. In this type of turbine, the air is circulated continuously within the gas turbine with the help of a compressor, heat chamber, gas turbine, and cooling chamber. The ratios of pressure, temperature, and air velocities will be constant in this type. It performs a thermodynamic cycle, which means working fluid is circulated and used continuously again and again without leaving the system. The gas is compressed in the compressor. The compressed gas is heated in the heating chamber. The gas turbine helps to generate electricity. Electricity is generated by the generator with the use of gas turbine The cooling of gases passed from the turbine gets cooled in the cooling chamber. Efficiency The efficiency of a closed cycle gas turbine can be explained with the help of the T- S diagram as shown below. ME3451-Thermal Engineering Department of Mechanical Engineering 2023-2024 SANCET T-S Diagram The efficiency of this can be given as, n = (available network) / input heat n = Cp(Wt – Wc) / input heat n= 1 – [(T4-T1) /(T3-T2)] Where ‘Wt’ = work is done by the gas turbine per kg of air = Cp(T2-T3) ‘Wc’ = work is done by the compressor per kg of air = Cp(T1-T4) ‘Cp’ constant pressure is taken in kJ/ Kg.K ‘T’ = temperature Input heat = Cp (T3-T2) The efficiency of this turbine is higher than the open cycle gas turbine Closed Cycle Gas Turbine Working Principle The closed-cycle gas turbine working principle is based on the Brayton cycle or Joule’s cycle. In this type of gas turbine, the compressor is used to compress the gas isotropically and the resultant compressed gas flows into the heating chamber. The rotor type compressor is preferred in this turbine. An external source is utilized to heat the compressed air and then passed over the turbine blades. When the gas is flowing over the turbine blades, it gets expanded and it is allowed to pass into the cooling chamber and gets cooled down. The gas gets cooled by using the circulation of water at constant pressure to its initial temperature. Again the gas is passed into the compressor and the process is repeated. In this turbine, the same gas is circulated repeatedly. The complexity of the system and the cost would increase if the working fluid/medium used in the turbine is other than air. This may lead to problems and it is difficult to resolve. ME3451-Thermal Engineering Department of Mechanical Engineering 2023-2024 SANCET Difference between Open Cycle and Closed Cycle Gas Turbine Heat source, type of fluid used for working, circulated air, turbine blades capacity, cost of maintenance and installation, it gives the difference between the open cycle and closed gas turbine. The circulation of working fluid is the main difference. Open Cycle Gas Turbine Closed Cycle Gas Turbine In this type, the combustion chamber is used for heating compressed air. Due to the mixing of products in the combustion chamber and heated air, the gas doesn’t remain constant. In this type, the heating chamber heats the compressed air, which is compressed firstly before heating. When an external source heats the air, then the gas remains constant. The amount of gas that came out from the turbine is exhausted in the atmosphere The amount of gas came out from the gas turbine is allowed to pass into the cooling chamber. Replacement of working fluid is continues Circulation of working fluid continues. The working fluid is air For better thermodynamic properties, helium is used as a working fluid As the air in the combustion chamber gets contaminated, results in the earlier wearing of turbine blades As there is no contamination of enclosed gas while passing through the heating chamber, results in no earlier wearing of turbine blades Mainly used for moving vehicles Mainly used for stationary installation and marine applications. The cost of maintenance is low The cost of maintenance is high Installation mass per KW is less Installation mass per KW is more. Advantages The closed-cycle gas turbine advantages are High thermal efficiency at any temperature limit and pressure ratio Any type of working fluid can be used with low caloric value. For example helium. No corrosion. Internal cleaning is not required. Re-heaters can be used to heat the water for the supply of hot water for domestic and industrial purposes. The size of the gas turbine is small An increase in pressure gives a better heat transmission coefficient in the exchanger Fluid friction loss is less. Disadvantages The closed-cycle gas turbine disadvantages are As the whole system work under high pressure with a working fluid (medium), it increases the cost. It requires a large air heater and it is not enough when the combustion chamber is used in the open cycle. Not used in aeronautical engines because this type of gas turbine uses cooling water. Complex system and should resist at high pressure. Applications The closed-cycle gas turbine applications include the following. Used in the generation of electric power Used in many industrial applications Used in marine propulsion, locomotive propulsion, automotive propulsion Used in aviation to provide power to the jet Propulsion ME3451-Thermal Engineering Department of Mechanical Engineering 2023-2024 SANCET 2. A constant pressure open cycle gas turbine plant works between temperature range of 15 deg.C and 700 deg.C and pressure ratio of 6. Find the mass of air circulating in the installation, if it develops 1100 kW. Also find the heat supplied by the heating chamber. ME3451-Thermal Engineering Department of Mechanical Engineering 2023-2024 SANCET ME3451-Thermal Engineering Department of Mechanical Engineering 2023-2024 SANCET 3. A gas turbine plant of 800 kW capacities takes the air at 1.01 bar and 15°C. The pressure ratio of the cycle is 6 and maximum temperature is limited to 700°C. A regenerator of 75% effectiveness is added in the plant to increase the overall efficiency of the plant. The pressure drop in the combustion chamber is 0.15 bars as well as in the regenerator is also 0.15 bars. Assuming the isentropic efficiency of the compressor 80% and of the turbine 85%, determine the plant thermal efficiency. Neglect the mass of the fuel. ME3451-Thermal Engineering Department of Mechanical Engineering 2023-2024 SANCET 4. In a single stage impulse turbine, nozzle angle is 20º and blade angles are equal. The velocity coefficient for blade is 0.85. Find maximum blade efficiency possible. If the actual blade efficiency is 92% of the maximum blade efficiency, find the possible ratio of blade speed to steam speed. ME3451-Thermal Engineering Department of Mechanical Engineering 2023-2024 SANCET ME3451-Thermal Engineering Department of Mechanical Engineering 2023-2024 SANCET ME3451-Thermal Engineering Department of Mechanical Engineering 2023-2024 SANCET 5. Explain various type of compounding in steam turbine. ME3451-Thermal Engineering Department of Mechanical Engineering 2023-2024 SANCET 6. A textile factory requires 10t/h of steam for process heating at 3 bar saturated and 1000 kW of power, for which a back pressure turbine of 70% internal efficiency is used. Find the steam condition required at inlet of the turbine. ME3451-Thermal Engineering Department of Mechanical Engineering 2023-2024 SANCET ME3451-Thermal Engineering Department of Mechanical Engineering 2023-2024 SANCET 7. The blade speed of a single ring of an impulse turbine is 300 m/s and the nozzle angle is 20°. The isentropic heat drop is 473 kJ/kg and the nozzle efficiency is 0.85. Given that the blade velocity co-efficient is 0.7 and the blades are symmetrical, draw the vector diagrams and calculate for a mass flow of 1 kg/s. (i) Axial thrust on the blading (ii) Steam consumption per B.P hour if the mechanical efficiency is 90% (iii) Blade efficiency, stage efficiency and maximum blade efficiency (iv) Heat equivalent of the friction of blading. ME3451-Thermal Engineering Department of Mechanical Engineering 2023-2024 SANCET ME3451-Thermal Engineering Department of Mechanical Engineering 2023-2024 SANCET ME3451-Thermal Engineering Department of Mechanical Engineering 2023-2024 SANCET 8. Distinguish between impulse and reaction turbines. (4) Impulse turbines are typically used for high-head, low-flow applications such as hydroelectric power plants, where the water is delivered under high pressure. Reaction turbines are typically used for low-head, high-flow applications such as hydropower plants, where the water is delivered at a lower pressure. 9. Consider a Parson’s stage with a rotor (at mid-height of blades) diameter of 1.2 m, operating at a speed of 3000 rpm, with the steam entry angle of steam be 20°.Steam enters the stator at 12 bar, 3000C and an isentropic enthalpy drop of 50 kJ/kg is chosen per row of blades. The isentropic efficiency of each row is assumed as 0.84. ME3451-Thermal Engineering Department of Mechanical Engineering 2023-2024 SANCET 10. Derive the value of blade speed ratio for maximum efficiency of impulse turbine. (8) ME3451-Thermal Engineering Department of Mechanical Engineering 2023-2024 SANCET 11. Describe the various methods of compounding with suitable diagrams.(13) Compounding of steam turbines is a method of extracting steam energy in multiple stages rather than in a single stage in a steam turbine. A compounded steam turbine has multiple stages with more than one set of nozzles and rotors. These are arranged in series, either keyed to the common shaft or fixed to the casing. The result of this arrangement allows either the steam pressure or the jet velocity to be absorbed by the turbine in a number of stages. Compounded steam turbines are used to reduce rotor speeds to achieve optimal operating revolutions per minute. The steam produced in the boiler has sufficiently high enthalpy when superheated. In all turbines the blade velocity is directly proportional to the velocity of the steam passing over the blade. Now, if the entire energy of the steam is extracted in one stage, i.e. if the steam is expanded from the boiler pressure to the condenser pressure in a single stage, then its velocity will be very high. Hence the velocity of the rotor (to which the blades are keyed) can reach to about 30,000 rpm, which is too high for practical uses due to very high vibration. Moreover, at such high speeds the centrifugal forces are immense, and can damage the structure. Hence, compounding is needed. The high velocity steam just strikes on a single ring of rotor that causes wastage of steam ranging 10% to 12%. To overcome the wastage of steam, compounding of steam turbines are used. Types of steam turbines 1. Impulse: There is no change in the pressure of the steam as it passes through the moving blades. There is change only in the velocity of the steam flow. 2. Reaction: There is change in both pressure and velocity as the steam flows through the moving blades. Types of compounding In an Impulse steam turbine compounding can be achieved in the following three ways: 1. Velocity compounding 2. Pressure compounding 3. Pressure-Velocity Compounding In a reaction turbine compounding can be achieved only by pressure compounding. Velocity compounding of Impulse Turbine Fig-1:Schematic Diagram of Curtis Stage Impulse Turbine ME3451-Thermal Engineering Department of Mechanical Engineering 2023-2024 SANCET The velocity compounded Impulse turbine was first proposed by C.G. Curtis to solve the problem of single stage Impulse turbine for use of high pressure and temperature steam. The rings of moving blades are separated by rings of fixed blades. The moving blades are keyed to the turbine shaft and the fixed blades are fixed to the casing. The high pressure steam coming from the boiler is expanded in the nozzle first. The Nozzle converts the pressure energy of the steam into kinetic energy. The total enthalpy drop and hence the pressure drop occurs in the nozzle. Hence, the pressure thereafter remains constant. This high velocity steam is directed on to the first set (ring) of moving blades. As the steam flows over the blades, due to the shape of the blades, it imparts some of its momentum to the blades and loses some velocity. Only a part of the high kinetic energy is absorbed by these blades. The remainder is exhausted on to the next ring of fixed blade. The function of the fixed blades is to redirect the steam leaving from the first ring of moving blades to the second ring of moving blades. There is no change in the velocity of the steam as it passes through the fixed blades. The steam then enters the next ring of moving blades; this process is repeated until practically all the energy of the steam has been absorbed. A schematic diagram of the Curtis stage impulse turbine, with two rings of moving blades one ring of fixed blades is shown in figure 1. The figure also shows the changes in the pressure and the absolute steam velocity as it passes through the stages. where, = pressure of steam at inlet = velocity of steam at inlet = pressure of steam at outlet = velocity of steam at outlet In the above figure there are two rings of moving blades separated by a single of ring of fixed blades. As discussed earlier the entire pressure drop occurs in the nozzle, and there are no subsequent pressure losses in any of the following stages. Velocity drop occurs in the moving blades and not in fixed blades. Velocity Diagram As shown in the above diagram there are two rings of moving blades separated by a ring of fixed blades. The velocity diagram in figure 2, shows the various components of steam velocity and the blade velocity of the moving blades. where, = absolute velocity of steam = relative velocity of steam = Blade velocity = Nozzle angle ME3451-Thermal Engineering Department of Mechanical Engineering 2023-2024 SANCET = Blade entrance angle = Blade exit angle = fluid exit angle From the above figure it can be seen that the steam, after exiting from the moving blades, enters into the fixed blades. The fixed blades redirect the steam into the next set of moving blades. Hence, steam loses its velocity in multiple stages rather than in a single stage. Optimum Velocit It is the velocity of the blades at which maximum power output can be achieved. Hence, the optimum blade velocity for this case is, where n is the number of stages. This value of optimum velocity is 1/n times that of the single stage turbine. This means that maximum power can be produced at much lower blade velocities. However, the work produced in each stage is not the same. The ratio of work produced in a 2-stage turbine is 3:1 as one move from higher to lower pressure. This ratio is 5:3:1 in three stage turbine and changes to 7:5:3:1 in a four-stage turbine. Disadvantages of Velocity Compounding Due to the high steam velocity there are high friction losses. Work produced in the low-pressure stages is much less. The designing and fabrication of blades that can withstand such high velocities is difficult. Pressure compounding of Impulse Turbine Fig-3:Schematic Diagram of Pressure compounded Impulse Turbine The pressure compounded Impulse turbine is also called a Rateau turbine, after its inventor. This is used to solve the problem of high blade velocity in the single-stage impulse turbine. ME3451-Thermal Engineering Department of Mechanical Engineering 2023-2024 SANCET It consists of alternate rings of nozzles and turbine blades. The nozzles are fitted to the casing and the blades are keyed to the turbine shaft. In this type of compounding, the steam is expanded in a number of stages, instead of just one (nozzle) in the velocity compounding. It is done by the fixed blades which act as nozzles. The steam expands equally in all rows of fixed blade. The steam coming from the boiler is fed to the first set of fixed blades i.e. the nozzle ring. The steam is partially expanded in the nozzle ring. Hence, there is a partial decrease in pressure of the incoming steam. This leads to an increase in the velocity of the steam. Therefore, the pressure decreases and velocity increases partially in the nozzle. This is then passed over the set of moving blades. As the steam flows over the moving blades, nearly all its velocity is absorbed. However, the pressure remains constant during this process. After this it is passed into the nozzle ring and is again partially expanded. Then it is fed into the next set of moving blades, and this process is repeated until the condenser pressure is reached. This process has been illustrated in figure 3 where the symbols have the same meaning as given above. It is a three-stage pressure compounded impulse turbine. Each stage consists of one ring of fixed blades, which act as nozzles, and one ring of moving blades. As shown in the figure, pressure drop takes place in the nozzles and is distributed in many stages. An important point to note here is that the inlet steam velocities to each stage of moving blades are essentially equal. It is because the velocity corresponds to the lowering of the pressure. Since, in a pressure compounded steam turbine, only a part of the steam is expanded in each nozzle. The steam velocity is lower than in the previous case. It can be explained mathematically from the following formula i.e. where, = absolute exit velocity of fluid = enthalpy of fluid at exit = absolute entry velocity of fluid = enthalpy of fluid at entry One can see from the formula that only a fraction of the enthalpy is converted into velocity in the fixed blades. Hence, velocity is less as compared to the previous case. Velocity Diagram[edit] Fig-4:Velocity Diagram of Pressure compounded Impulse Turbine ME3451-Thermal Engineering Department of Mechanical Engineering 2023-2024 SANCET The velocity diagram shown in figure 4 gives detail about the various components of steam velocity and Blade velocity. where, symbols have the same meaning as given above. An important point to note from the above velocity diagram is that the fluid exit angle (δ) is 90⁰. This indicates that the whirl velocity of fluid at exit of all stages is zero, which is in compliance with the optimum velocity concept (as discussed earlier). The ratio of work produced in different stages is similar to the above type. Disadvantages of Pressure Compounding Since there is pressure drop in the nozzles, it has to be made air-tight. They are much larger at 34 inches Pressure-Velocity compounded Impulse Turbine Fig-5:Schematic Diagram of Pressure-Velocity compounded Impulse Turbine It is a combination of the above two types of compounding. The total pressure drop of the steam is divided into a number of stages. Each stage consists of rings of fixed and moving blades. Each set of rings of moving blades is separated by a single ring of fixed blades. In each stage there is one ring of fixed blades and 3-4 rings of moving blades. Each stage acts as a velocity compounded impulse turbine. The fixed blades act as nozzles. The steam coming from the boiler is passed to the first ring of fixed blades, where it gets partially expanded. The pressure partially decreases and the velocity rises correspondingly. The velocity is absorbed by the following rings of moving blades until it reaches the next ring of fixed blades and the whole process is repeated once again. This process is shown diagrammatically in figure 5. where, symbols have their usual meaning. ME3451-Thermal Engineering Department of Mechanical Engineering 2023-2024 SANCET Pressure compounding of Reaction Turbine Fig-6:Schematic Diagram of Pressure compounded Reaction Turbine As explained earlier a reaction turbine is one in which there is pressure and velocity loss in the moving blades. The moving blades have a converging steam nozzle. Hence when the steam passes over the fixed blades, it expands with decrease in steam pressure and increase in kinetic energy. This type of turbine has a number of rings of moving blades attached to the rotor and an equal number of fixed blades attached to the casing. In this type of turbine the pressure drops take place in a number of stages. The steam passes over a series of alternate fixed and moving blades. The fixed blades act as nozzles i.e. they change the direction of the steam and also expand it. Then steam is passed on the moving blades, which further expand the steam and also absorb its velocity. This is explained in figure 6. where symbols have the same meaning as above. Velocity Diagram Fig-7: Velocity Diagram of Pressure Compounded Reaction turbine The velocity diagram given in figure 7 gives a detail about the various components of steam velocity and blade velocity (symbols have the same meaning as above). ME3451-Thermal Engineering Department of Mechanical Engineering 2023-2024 SANCET UNIT IV INTERNAL COMBUSTION ENGINES – FEATURES AND COMBUSTION IC engine – Classification, working, components and their functions. Ideal and actual : Valve and port timing diagrams, p-v diagrams- two stroke & four stroke, and SI & CI engines – comparison. Geometric, operating, and performance comparison of SI and CI engines. Desirable properties and qualities of fuels. Air-fuel ratio calculation – lean and rich mixtures. Combustion in SI & CI Engines – Knocking – phenomena and control. 1. Define mean effective pressure and comment its application in internal combustion Engines? (Apr/May 2019) Mean effective pressure is defined as the constant pressure acting in the piston during working stroke. It is also defined as the ratio of work done to the stroke volume or piston Displacement volume. Mean effective pressure (MEP) pm= work done/ stroke volume or piston displacement volume 2. Define cut-off ratio. (Nov/Dec 2018) It is defined as the ratio of volume after the expansion to the volume before the expansion. 3. Define compression ratio. It is defined as the ratio of total cylinder volume to the clearance volume. 4. Define mean effective pressure. (April/May 2019) It the constant (or) average pressure acting on the piston during the working stroke. It is defined as the ratio of work done to the swept (or) stroke volume. 5. What is mean effective pressure of an engine? () and comment the application in internal combustion engine. The mean effective pressure (MEP) is defined as the average pressure required to act on the piston as it moves one displacement to give the work W. Mean Effective Pressure is a valuable parameter in internal combustion engines, providing insights into the overall performance and efficiency of the engine. It plays a crucial role in design, optimization, and evaluation processes in the automotive and engineering industries. 6. What constitute an engine? An engine is a machine that burns fuel and converts it into mechanical power. Most modern vehicles use internal combustion engines (ICE), which ignite fuel and use the reaction to move mechanical parts. 7. What is a two stroke engine? A two-stroke engine is a type of internal combustion engine that completes a power cycle in only two strokes of the piston, as opposed to the more common four-stroke engines that require four strokes (intake, compression, power, and exhaust) for each cycle. The two-stroke engine is known for its simplicity and high power-to-weight ratio but is also associated with certain drawbacks, such as higher emissions and less fuel efficiency. 8. State the function of flywheel, connecting rod, piston and crankshaft. Not only rotates the engine, the function of the flywheel is to store mechanical energy to balance the engine so that it continues to have good performance. Mechanical power is the energy created when the engine is running. The flywheel works to balance the mechanical power by storing it. The up-down motion of each piston is transferred to the crankshaft via connecting rods. A flywheel ME3451-Thermal Engineering Department of Mechanical Engineering 2023-2024 SANCET is often attached to one end of the crankshaft, in order to smoothen the power delivery and reduce vibration. 9. List the main parts of a lubrication system () Oil sump, Engine oil filter, Piston cooling nozzles, Oil pump, The oil galleries, Oil cooler, The oil pressure indicator/light. 10. What is known as pre ignition? State its effect. () Pre-ignition is the ignition of the air- fuel charge while the piston is still compressing the charge. The ignition source can be caused by a cracked spark plug tip, carbon or lead deposits in the combustion chamber, or a burned exhaust valve, anything that can act as a glow plug to ignite the charge prematurely. 11. What is cutoff ratio? () The ratio of the volume at the end of constant-pressure energy addition process to the volume at the beginning of the energy addition process. 12. Mention the use of a camshaft. The camshaft is a mechanical component of an internal combustion engine. It opens and closes the inlet and exhaust valves of the engine at the right time, with the exact stroke and in a precisely defined sequence. The camshaft is driven by the crankshaft by way of gearwheels, a toothed belt or a timing chain. 13. Mention the use of a carburetor. Carburetors are the devices which help in mixing air-fuel in an engine to help it work efficiently. In other words, a carburetor is a clever gadget which mixes the fuel and air in correct proportions as per the demand of the engine. 14. What are homogeneous and heterogeneous mixtures? In which engines these mixtures are used? Homogeneous compression ignition is a form of internal combustion in which air and fuel are well mixed at the point of combustion. On the other hand, in heterogeneous combustion engine, the air and fuel are not mixed till the point of combustion. 15. What are the advantages of air cooling system? Air-cooled ICEs are simpler, lighter, and cheaper than liquid-cooled ICEs, and they do not require a radiator, water pump, hoses, or antifreeze. However, they also have some drawbacks, such as lower thermal efficiency, higher noise levels, and more sensitivity to ambient temperature and altitude 16. What is the antifreeze solution used in water cooling systems? Most antifreeze is made by mixing distilled water with additives and a base product, usually MEG (mono ethylene glycol) or MPG (mono propylene glycol). Antifreeze is a tinted liquid that you put (along with water) in your radiator to help regulate engine temperature. 17. What is meant by motoring test? The test in which the engine runs at a constant speed using the motor and the engine is connected to the electric motor is called the Motoring test. PART B 1. How reciprocating internal combustion engines are classified? Discuss. (5) Reciprocating internal combustion engines are classified based on various criteria, including their design, application, fuel type, and operational characteristics. Here are some common classifications: 1. Number of Cylinders: Single-Cylinder Engines: These engines have only one cylinder and are commonly found in small, portable equipment like lawnmowers and motorcycles. ME3451-Thermal Engineering Department of Mechanical Engineering 2023-2024 SANCET Multi-Cylinder Engines: These engines have more than one cylinder. Common configurations include inline, V-shaped, and horizontally opposed layouts. 2. Arrangement of Cylinders: Inline Engines: Cylinders are arranged in a straight line. V-Shaped Engines: Cylinders are arranged in a V shape. Horizontally Opposed Engines: Cylinders are arranged opposite each other in a flat configuration. 3. Cycle Type: Two-Stroke Engines: Complete combustion and power cycles occur in two strokes of the piston (one upstroke and one down stroke). Four-Stroke Engines: Complete combustion and power cycles occur in four strokes of the piston (intake, compression, power, and exhaust). 4. Aspiration: Naturally Aspirated Engines: Rely on atmospheric pressure for air intake. Turbocharged Engines: Use a turbocharger to compress incoming air, increasing the air-fuel mixture density. Supercharged Engines: Use a supercharger, a mechanically driven compressor, to increase air intake. 5. Ignition Type: Spark-Ignition Engines (SI): Ignition is initiated by a spark plug. Common in gasoline engines. Compression-Ignition Engines (CI): Ignition is achieved through the high temperature resulting from compressing the air-fuel mixture. Common in diesel engines. 6. Application: Automotive Engines: Designed for use in cars, trucks, and other road vehicles. Industrial Engines: Used in various non-automotive applications, such as generators, pumps, and construction equipment. 7. Fuel Type: Gasoline Engines: Use gasoline as the primary fuel. Diesel Engines: Use diesel fuel for combustion. Dual-Fuel Engines: Can run on two different types of fuel, typically diesel and natural gas. 8. Cooling Method: Air-Cooled Engines: Use air to dissipate heat from the engine. Liquid-Cooled Engines: Use a liquid coolant (usually water and antifreeze) circulated through a cooling system. 9. Power Output: High-Speed Engines: Designed for high rotational speeds, common in automotive applications. Low-Speed Engines: Designed for lower rotational speeds, common in marine and stationary applications. These classifications help engineers and manufacturers tailor engines to specific applications, optimize performance, and meet regulatory and efficiency standards. Additionally, advancements in technology and increasing emphasis on environmental sustainability continue to influence the classification and design of reciprocating internal combustion engines. 2. With a neat sketch discuss the essential parts of an IC engine. (8) 1. Cylinder: Function: Provides the chamber in which the combustion of fuel and air takes place. It houses the piston and forms the basic working unit of the engine. 2. Piston: Function: Moves up and down inside the cylinder in response to the combustion process. The reciprocating motion of the piston is converted into rotational motion to drive the crankshaft. 3. Crankshaft: ME3451-Thermal Engineering Department of Mechanical Engineering 2023-2024 SANCET Function: Converts the reciprocating motion of the piston into rotary motion. It is connected to the piston through a connecting rod and transfers power to the transmission. 4. Connecting Rod: Function: Connects the piston to the crankshaft and transmits the reciprocating motion of the piston to the rotary motion of the crankshaft. 5. Crankcase: Function: Encloses and protects the crankshaft and connecting rods. It also contains the engine oil to lubricate moving parts. 6. Cylinder Head: Function: Covers the top of the cylinder, forming the combustion chamber. It often houses the valves, spark plugs (in spark-ignition engines), and other components necessary for the combustion process. 7. Valves (Intake and Exhaust): Function: Control the flow of air (intake valve) and exhaust gases (exhaust valve) into and out of the combustion chamber. The opening and closing of valves are synchronized with the piston's movement. 8. Camshaft: Function: Responsible for operating the valves by translating the rotational motion into the linear motion required to open and close them. It is driven by the crankshaft. 9. Timing Gear/Belt/Chain: Function: Synchronizes the rotation of the crankshaft and camshaft to ensure precise timing of valve opening and closing. 10. Combustion Chamber: Function: The space enclosed by the cylinder head and piston where the combustion of fuel and air mixture occurs, leading to the generation of power. 11. Fuel Injector (in Diesel Engines) or Carburettor (in Gasoline Engines): Function: Delivers the fuel into the combustion chamber in the correct proportion with air for combustion. 12. Spark Plug (in Spark-Ignition Engines): Function: Generates sparks to ignite the air-fuel mixture in the combustion chamber. 13. Piston Rings: Function: Seal the gap between the piston and the cylinder wall to prevent leakage of gases and to help transfer heat away from the piston. 14. Cooling System (Radiator and Water Pump or Air-Cooling Fins): Function: Regulates the temperature of the engine by dissipating heat generated during the combustion process. 15. Exhaust System (Exhaust Manifold, Catalytic Converter, Muffler): Function: Collects and directs the exhaust gases away from the engine, reduces emissions, and minimizes noise. These components work together to convert fuel into mechanical energy, powering the vehicle or equipment in which the internal combustion engine is installed. 3. Draw a theoretical and actual valve timing diagram of a 4-stroke petrol engine. (5) ME3451-Thermal Engineering Department of Mechanical Engineering 2023-2024 SANCET Theoretical Valve Timing Diagram: Intake Stroke (Suction Stroke): Inlet Valve Opens: Slightly before the piston reaches the top dead center (TDC) on the exhaust stroke. Inlet Valve Closes: After the piston reaches the bottom dead center (BDC) and starts moving upward. Compression Stroke: Both Valves Closed: Piston moves upward, compressing the air-fuel mixture. Power Stroke: Both Valves Closed: Spark plug ignites the compressed air-fuel mixture, pushing the piston down. Exhaust Stroke: ME3451-Thermal Engineering Department of Mechanical Engineering 2023-2024 SANCET Exhaust Valve Opens: Slightly before the piston reaches the bottom dead canter (BDC) on the power stroke. Exhaust Valve Closes: After the piston reaches the top dead canter (TDC) and starts moving downward. Actual Valve Timing Diagram: In an actual engine, there may be deviations from the theoretical timing due to factors like valve train dynamics, camshaft design, and manufacturing tolerances. Intake Stroke (Suction Stroke): Inlet Valve Opens: Deviates from theoretical timing due to factors like valve lift, duration, and clearance. Inlet Valve Closes: May close slightly earlier or later than the theoretical timing. Compression Stroke: Both Valves Closed: Deviations may occur in valve closure timing. Power Stroke: Both Valves Closed: Spark plug ignition and combustion occur, but variations in valve timing may exist. Exhaust Stroke: Exhaust Valve Opens: Deviates from theoretical timing. Exhaust Valve Closes: May close earlier or later than the theoretical timing. Actual valve timing diagrams are determined through experimentation and tuning during the engine design and development process. Engineers optimize the valve timing to achieve better performance, fuel efficiency, and emissions control under real-world conditions. The specifics of the actual valve timing depend on the engine design, camshaft profile, and other factors unique to the particular engine model. 4. With a neat sketch explain the operation of 2-stroke diesel engine. (8) 1. Intake Stroke: The piston starts at the top of the cylinder, and as it moves downward, it uncovers the intake ports. Simultaneously, fresh air is drawn into the cylinder through the intake ports due to the vacuum created by the descending piston. ME3451-Thermal Engineering Department of Mechanical Engineering 2023-2024 SANCET 2. Compression Stroke: As the piston reaches the bottom of its stroke, it starts to move back up, compressing the air that was drawn in during the intake stroke. Fuel is injected into the highly compressed air at the end of this stroke. 3. Combustion/Power Stroke: The compressed air-fuel mixture is ignited by the heat generated during compression. Combustion forces the piston down the cylinder, producing power. This stroke is the primary source of energy for the engine. 4. Exhaust Stroke: As the piston reaches the bottom of its power stroke, exhaust ports are uncovered. The burned gases are expelled from the cylinder as the piston moves back up. This stroke completes the 2-stroke cycle, and the engine is ready to begin the intake stroke again. Key Points: Unlike 4-stroke engines, which have a separate intake and exhaust stroke, 2-stroke engines combine these functions into a single revolution of the crankshaft. Lubrication is crucial in 2-stroke engines because the same chamber is used for both intake and exhaust. Oil is often mixed with the fuel or injected separately to lubricate moving parts. The simplicity of design makes 2-stroke engines lighter and more compact, but they tend to be less fuel-efficient and produce more emissions than 4-stroke engines. Common applications of 2-stroke diesel engines include small boats, chainsaws, motorcycles, and certain industrial equipment. 5. Discuss the construction and working principle of a four stroke engine with sketch. (16) ME3451-Thermal Engineering Department of Mechanical Engineering 2023-2024 SANCET ME3451-Thermal Engineering Department of Mechanical Engineering 2023-2024 SANCET ME3451-Thermal Engineering Department of Mechanical Engineering 2023-2024 SANCET 6. Explain the construction and working principle of Battery coil ignition system with neat sketch. ME3451-Thermal Engineering Department of Mechanical Engineering 2023-2024 SANCET ME3451-Thermal Engineering Department of Mechanical Engineering 2023-2024 SANCET UNIT V INTERNAL COMBUSTION ENGINE PERFORMANCE AND AUXILIARY SYSTEMS Performance and Emission Testing, Performance parameters and calculations. Morse and Heat Balance tests. Multipoint Fuel Injection system and Common rail direct injection systems. Ignition systems – Magneto, Battery and Electronic. Lubrication and Cooling systems. Concepts of Supercharging and Turbocharging – Emission Norms 1. Which air standard cycle (Otto/Diesel/Dual) is more efficient for the same heat input? Justify. In the dual cycle, combustion takes place in two stages: constant volume and constant pressure. It's a term that can be applied to internal combustion engines. Ans: The Otto cycle is more efficient than the dual and diesel cycles. 2. State the merits of a diesel engine over a petrol engine. Diesels are more fuel efficient than petrol engines and emit less CO2, which makes them better for the environment. Diesels produce considerably more torque (pulling power) than their petrol counterparts, which makes them good engines for towing or carrying heavy loads. 3. What is meant by valve overlap? Valve overlap is the period during engine operation when both intake and exhaust valves are open at the same time. Valve overlap occurs when the piston nears TDC between the exhaust event and the intake event. Duration of valve overlap is between 10° - 20° of crankshaft rotation, depending on the engine design. 4. How are SI and CI engine fuels rated? Knock rating of a CI engine fuel is found by comparing it with a reference fuel under prescribed working conditions. Reference fuel: normal Cetane (C16H34) which is assigned a Cetane number ME3451-Thermal Engineering Department of Mechanical Engineering 2023-2024 SANCET of 100 and α-methyl naphthalene (C11H10) which is assigned a Cetane number of zero. According to standard practice, the antiknock value of an SI engine fuel is determined by comparing its antiknock property with a mixture of two reference fuels, normal heptane (C7H18) and iso-octane (C8H18). Iso-octane chemically being a very good antiknock fuel, is arbitrarily assigned a rating of 100 octane numbers. 5. What is meant by ignition delay? The ignition delay (ID) is the time between start of injection and start of combustion. It defines the quality of ignition in terms of the cetane number. The ID is divided in physical delay and chemical delay. Physical delay is the time required for atomization of fuel, air-fuel mixing, and vaporization. 6. What is the necessity of cooling of an IC engine? First, it removes excess heat from the engine; second, it maintains the engine operating temperature where it works most efficiently; and finally, it brings the engine up to the right operating temperature as quickly as possible. 7. What is turbo charging? Turbocharging is a method of increasing an internal combustion engine's power output and efficiency by forcing compressed air into the engine. A turbocharged engine uses a turbocharger, a turbine-driven forced induction device that runs on hot exhaust gas from the engine. 8. How the efficiency of diesel engine varies for different cutoff ratios and for which cutoff ratio the efficiencies of Otto and Diesel cycles become identical. The efficiencies of the Otto and Diesel cycles become identical when the cutoff ratio (r) is equal to the compression ratio (V1/V2). The choice of the cutoff ratio affects the engine's performance, and it is often optimized based on factors such as fuel efficiency, power output, and other design considerations. 9. What is swept volume? Swept volume - It is the area covered by the piston while it is moves from TDC (Top Dead Centre) to BDC (Bottom Dead Centre) inside the engine cylinder. Clearance volume- It is the clearance at the top of the engine cylinder, above the TDC. Clearance volume= Total volume-Swept Volume 10. Why petrol engines and diesel engines are called as SI and CI engines respectively? Spark plugs are used to ignite the air fuel mixture. That's why petrol engines are called spark ignition or is engines. Where as in diesel engines compressing the air and spraying fuel to this mixture ignites the engine. So diesel engines are also called compression ignition or ci engines. 11. Write any two disadvantages of 2-stroke cycle engines over 4-stroke cycle engines. Disadvantages: Thermal efficiency of a two stroke cycle engine is less than that a four stroke cycle engine. Overall efficiency of a two stroke cycle engine is also less than that of four stroke cycle engine. The consumption of lubricating oil is large in a two stroke cycle engine because of high operating temperature. 12. What is indicated thermal efficiency of IC engine? Indicated thermal efficiency (ηit): Indicated thermal efficiency is the ratio of energy in the indicated power to the fuel energy. 13. Why specific fuel consumption in petrol engine is higher than diesel engine? Diesel is more combustible than petrol, therefore less fuel is required to generate the same amount of power. Secondly, petrol doesn't undergo complete combustion and hence there is a small wastage of fuel. This is why diesel engines are more efficient than petrol engines. ME3451-Thermal Engineering Department of Mechanical Engineering 2023-2024 SANCET 14. Plot the general diesel cycle efficiency as a function of compression ratio for various cut off ratios. The x-axis represents the compression ratio (V1/V2), and the y-axis represents the Diesel cycle efficiency. The plot allows you to observe how changing the cutoff ratio influences the efficiency at different compression ratios. 15. What is unit injection system? The electronically controlled unit injector generates a high fuel pressure using the integrated piston pump and injects exact quantities of fuel into the combustion chamber. The advantage of the unit injector system is that there is no high-pressure line between the high-pressure pump and injection nozzle. 16. What do you mean by short circuiting in two-stroke engines? An inherent loss in two-stroke engines results when some fresh charge escapes through the exhaust ports without participating in the process of scavenging residual gas. This phenomenon is often referred to as short-circuiting. PART B 1. An I.C. Engine uses 6 kg of fuel having calorific value 44000 kJ/kg in one hour. The IP developed is 18 kW. The temperature of 11.5 kg of cooling water was found to rise through 25°C per minute. The temperature of 4.2 kg of exhaust gas with specific heat 1 kJ/kg.k was found to rise through 220°C. Draw the heat balance sheet for the engine. ME3451-Thermal Engineering Department of Mechanical Engineering 2023-2024 SANCET ME3451-Thermal Engineering Department of Mechanical Engineering 2023-2024 SANCET 2. The following data are obtained while testing a 4 stroke, 4 cylinder, petrol engine : Air fuel ratio (by weight) = 15:1 Calorific value of fuel = 45000 kJ/kg Mechanical efficiency = 85% Air standard efficiency = 53% Relative efficiency = 65% Volumetric efficiency = 80% Stroke bore ratio = 1.3 Suction conditions = 1 bar, 30°C Engine speed = 3000 rpm Power at brakes = 75 kW Calculate i) compression ratio (ii) indicated thermal efficiency (iii) brake Specific fuel consumption (iv) bore and stroke of the engine. ME3451-Thermal Engineering Department of Mechanical Engineering 2023-2024 SANCET 3. What is the need of Multi Point Fuel Injection system (MPFI)? Discuss the function of basic arrangements of MPFI system. ME3451-Thermal Engineering Department of Mechanical Engineering 2023-2024 SANCET Multi-Point Fuel Injection (MPFI) is an advanced fuel delivery system used in internal combustion engines. The primary need for MPFI arises from its ability to enhance engine performance, fuel efficiency, and emissions control compared to traditional carburetion or single-point fuel injection systems. Here are some reasons for the adoption of Multi-Point Fuel Injection: 1. Better Fuel Distribution: MPFI systems use multiple fuel injectors, each spraying fuel directly into the intake port of its respective cylinder. This ensures more even fuel distribution among all cylinders, leading to improved combustion efficiency. 2. Precision Fuel Metering: MPFI systems allow precise control over the amount of fuel delivered to each cylinder. This precision helps optimize the air-fuel mixture for various driving conditions, resulting in better fuel efficiency and performance. 3. Enhanced Combustion Efficiency: By injecting fuel directly into each cylinder during the intake stroke, MPFI systems promote better mixing of air and fuel. This results in more complete combustion, improving power output and reducing emissions. 4. Cold Start Performance: MPFI systems contribute to better cold start performance by delivering the right amount of fuel to each cylinder based on temperature and engine conditions. This helps in minimizing engine hesitation and rough idling during cold starts. 5. Adaptability to Variable Driving Conditions: MPFI systems are equipped with sensors that monitor various parameters such as engine speed, load, temperature, and throttle position. The electronic control unit (ECU) can adjust the fuel injection timing and duration in real-time, optimizing performance for different driving conditions. Now, let's discuss the basic arrangements and functions of an MPFI system: 1. Fuel Injectors: Multiple fuel injectors are positioned near the intake valves, with each injector supplying fuel to a specific cylinder. The injectors are controlled by the ECU, which determines the timing and duration of fuel injection. 2. Sensors: ME3451-Thermal Engineering Department of Mechanical Engineering 2023-2024 SANCET Various sensors, such as the throttle position sensor, oxygen sensor, engine speed sensor, and temperature sensor, provide input to the ECU. These sensors help the ECU adjust the air-fuel mixture based on driving conditions. 3. Electronic Control Unit (ECU): The ECU is the brain of the MPFI system, processing input from sensors and determining the optimal fuel injection timing and duration. It adjusts the fuel delivery in real-time to optimize performance, fuel efficiency, and emissions. 4. Fuel Rail: The fuel rail is a pipe that distributes pressurized fuel to each injector. It ensures a consistent supply of fuel to all injectors. 5. Throttle Body: The throttle body contains the throttle valve that controls the amount of air entering the engine. The ECU adjusts the fuel injection based on the position of the throttle valve. Overall, the combination of precise fuel metering, adaptability to different conditions, and improved combustion efficiency makes Multi-Point Fuel Injection a preferred choice for modern internal combustion 4. A six cylinder 4 stroke SI engine having a piston displacement of 700cc per cylinder develops 78kw at 3200 rpm and consumed 27kg/h of petrol. The calorific value of petrol is 44 MJ/kg determine i) volumetric efficiency it A/F ratio is 12 and intake air is at 0.9 bar and 320C R air 287J/kg.K ii) Brake thermal efficiency iii) brake torque. ME3451-Thermal Engineering Department of Mechanical Engineering 2023-2024 SANCET ME3451-Thermal Engineering Department of Mechanical Engineering 2023-2024 SANCET ME3451-Thermal Engineering Department of Mechanical Engineering 2023-2024 SANCET ME3451-Thermal Engineering Department of Mechanical Engineering 2023-2024 SANCET 5. An air-standard Diesel cycle has a compression ratio of 18, and the heat transferred to the working fluid per cycle is 1800 kJ/kg. At the beginning of the compression stroke, the pressure is 1 bar and the temperature is 300 K. Calculate: (i) Thermal efficiency, (ii) The mean effective pressure. |Ans. (i) 61%; (ii) 13.58 bar] ME3451-Thermal Engineering Department of Mechanical Engineering 2023-2024 SANCET 6. Explain with neat sketches the various stages of combustion in CI engines. Stages of Combustion in CI engine: There are four different stages of combustion in CI engine where proper combustion of air and fuel takes place as follows: 1. Ignition Delay Period 2. Period of Uncontrolled Combustion 3. Period of Controlled Combustion 4. After Burning ME3451-Thermal Engineering Department of Mechanical Engineering 2023-2024 SANCET 1. Ignition Delay Period At this first stage of combustion in the CI engine, the fuel from the injection system sprayed in the combustion chamber in the form of a jet. Due to atomization and vaporization, this fuel disintegrates at the core which is surrounded by a spray of air and fuel particles. In this vaporization process, the fuel gets heat from the compressed and hot surrounding air. It causes some pressure drop in the cylinder. You can see this pressure drop (curve AB) in the above figure. After completion of the vaporization process, the preflame reaction of the mixture in the combustion chamber starts. During the preflame reaction, pressure into the cylinder starts increasing with the release of energy at a slow rate. This preflame reaction starts slowly and then speeds up until the ignition of the fuel takes place. You can see this process at point C on the diagram. This time interval between the starting of the fuel injection and the beginning of the combustion is called the delay period. This delay period can further be divided into two parts – Physical delay and chemical delay. The period between the time of injection of the fuel and its achievement of self-ignition temperature during vaporization is called physical delay. When physical delay completes, the time interval up to the fuel ignites and the flame of the combustion appears is called chemical delay. Preflame reaction we discussed above is taking place during the chemical delay. Due to the complex process of combustion is a CI engine, it’s difficult to separate these two delay periods. If this delay period performs longer than usual, then we can here knocking in CI engine. 2. Period of Uncontrolled Combustion This is the second stage of combustion in the CI engine. After the above-mentioned delay period is over, the air and fuel mixture will auto-ignite as they have achieved their self-ignition temperature. The mixture of air and fuel in CI engines is heterogeneous unlike homogeneous in the SI engines. Due to this heterogeneous mixture, flames appear at more than one location where the concentration of the mixture is high. When the flame formed the mixture in the other low concentration starts burning by the propagation of flames or due to auto-ignition, because of the process of heat transfer. The accumulated fuel during the delay is now started burning at an extremely rapid rate. It causes a rise in in-cylinder pressure and temperature. So, the higher the delay period, the higher would be the rate of pressure rise. During this stage, you can’t control the amount of fuel burning, that’s why this period is called a period of uncontrolled combustion. This period is represented by the curve CD in the above figure. 3. Period of Controlled Combustion When the accumulated fuel during the delay period completely burned in the period uncontrolled combustion, the temperature and pressure of the mixture in the cylinder are so high that new injected fuel from the nozzle will burn rapidly due to the presence of sufficient oxygen in the combustion chamber. That’s the reason we can control the rise of pressure into the cylinder by controlling the fuel injection rate. Therefore, this period of combustion is called a period of controlled combustion. 4. After Burning This is the last stage out of the four stages of combustion in CI engine. Naturally, the combustion process is completed at the point when the maximum pressure is obtained in the combustion chamber at point E as shown in the figure. Practically, the burning of the fuel in the combustion chamber remains to continue during the expansion stroke. The main reason behind it is the reassociation of dissociated gases and unburnt fuel. Therefore, this last phase of combustion is called After Burning. These are the four different stages of combustion in CI engine. ME3451-Thermal Engineering Department of Mechanical Engineering 2023-2024 SANCET 7. Explain with neat sketches the various stages of combustion in SI engines. The combustion process will be completed in the three stages in an actual engine. 1. Ignition Lag 2. Flame Propagation 3. After burning 1. Ignition Lag The time interval between the passage of the spark and the inflammation of the air-fuel mixture is known as ignition lag or Ignition delay. It is also referred to as the preparation phase. There are two chances that can cause the ignition delay. Physical delay and chemical delay. Physical delay due to the atomisation, vaporization and mixing of air fuel. The chemical delay due to pre-combustion reactions. The ignition lag depends on the heat, pressure, the nature of the fuel and the proportion of the exhaust gas residuals. 2. Flame Propagation The flame propagation means that the propagation of combustion wave through a combustible mixture. Or simply the spread of the flame throughout the combustion chamber. When the ignition initiated, the adjacent layer of the reaction zone also ignites and propagated to the next layer. This continued throughout the mixture in the combustion chamber. This process takes some time to spread the flame throughout the combustion chamber. During this stage the pressure rises with very little change in the volume. But it can not be instantaneous as we claimed to be in the actual cycle. 3. After burning This After Burning stage begins where the cylinder pressure reaches a maximum point(c) in the cylinder. Also, flame propagation gradually decreases due to the flame velocity will reduce. The expansion stroke will starts at or before this stage. so there will be no pressure rise in this stage. ME3451-Thermal Engineering Department of Mechanical Engineering 2023-2024 SANCET 8. Explain the pressure lubrication system with a neat sketch. Pressure lubrication (also known as injection lubrication) is a form of lubrication that uses one or more pumps to deliver oil to the lubrication points. The lubricant is distributed throughout the oil circuit. It is the most commonly used lubrication in engines. However, it is also used in other components such as gearboxes or compressors. In the case of pressure lubrication, wet sump lubrication, in which the oil supply is collected and stored in the oil sump, is the most common design. Pressure lubrication can be divided into two systems, wet sump lubrication and dry sump lubrication. Both are possible, but differ in the application, storage of the oil supply and delivery of the oil. How does wet sump lubrication work? In a four-stroke engine, the oil is drawn from an oil sump by a pump, passed through a filter and fed into the engine compartment. If the main flow filter should become clogged, the overflow valve ensures that the oil flow is maintained. The lubricant reaches all relevant lubrication points such as connecting rods, crankshaft and camshaft through sufficient pressure. The rotation of the crankshaft ensures that oil is distributed to the cylinder liners and connecting rod bearings. The rotation of the crankshaft creates an additional oil mist in the crankcase. This helps to cool the piston crown. When the oil has passed through the interior of the engine, it flows back into the oil pan, cools down and is pumped into the oil circuit again. ME3451-Thermal Engineering Department of Mechanical Engineering 2023-2024 SANCET Advantages of pressure lubrication The pressure from the oil pump ensures that even the most distant lubrication point can be supplied with oil. The design of the system is efficient and comparatively cheap. As a result, a smaller amount of oil is needed in the engine. At the same time, a compact engine design reduces the overall weight of the vehicle. Regardless of the operating status of the unit, almost constant, reliable lubrication is ensured (in industrial systems, lubrication is switched on before the unit is started). Disadvantages of pressure lubrication Pressure lubrication involves general problems which can be solved by certain design adjustments or the optimum lubricant. When the oil enters the engine compartment, it reaches the crankshaft or the cylinder head on its way back to the oil pan. The lubrication points between these areas are not lubricated at first. This applies to connecting rod bearings, connecting rods and cylinder liners. In order to cover these lubrication points with a lubricating film, drill holes must be made in the crankshaft or special spray nozzles must be installed. When the engine is cold, the oil pressure can become very high because the oil is still very viscous and difficult to pump. Pressure peaks can occur that cause damage to the oil circuit. These pressure peaks are to be mitigated by the built-in pressure relief valves. In modern industrial systems, the performance and compactness of the circulation system are measured in terms of how many minutes it takes to circulate the total volume of oil. In the past, it often took hours in turbine systems to pump the total amount of oil (large oil volume). Nowadays, it takes only two to three minutes (small volume, high pump output) for the entire oil volume to be pumped into the lubrication system. The disadvantage of this is the short settling time, which can lead to foam, cavitation and/or vibrations. 9. Explain the Bosch fuel injector with a neat sketch. The Bosch fuel injection system delivers the metered quantity at high pressure to mix with compressed air inside the combustion chamber for efficient combustion. ME3451-Thermal Engineering Department of Mechanical Engineering 2023-2024 SANCET 10. Air consumption for a four stroke petrol engine is measured by means of a circular orifice of diameter 3.5cm. The coefficient of discharge for the orifice is 0.6 and the pressure across the orifice is 14cm of water. The barometer reads 760mm of Hg. The temperature of air in the room is 240C. the piston displacement volume is 1800cm3. The compression ratio is 6.5. the fuel consumption is 0.13 kg/min and calorific value is 44000kj/kg. The brake power developed at 2500 rpm is 28kw. Determine i) Air fuel ratio ii) Volumetric efficiency on the basis of air alone. iii) Brake mean effective pressure iv) Relative efficiency on brake thermal efficiency basis. ME3451-Thermal Engineering Department of Mechanical Engineering 2023-2024 SANCET ME3451-Thermal Engineering Department of Mechanical Engineering 2023-2024 SANCET ME3451-Thermal Engineering Department of Mechanical Engineering 2023-2024 SANCET 11. Discuss the convergence of state of art of fuel supply system of spark ignited a engine from carburetor to MPFI fuel supply system. Fuel Supply System of S.I Engine In a S.I engine, a measured quantity of petrol or gas is already mixed with air in a calculated proportion before being sucked by engine. Then this combustible charge having right quantity of fuel-air is ignited at the appropriate time at the end of compression stroke in the engine with the help of a spark plug. The operation of spark plug is timed along with the moment of crank shaft with the help of ignition system. Low voltage of battery is given to primary winding of ignition coil through an ignition switch and contact breaker. The secondary winding is connected to spark plugs through distributor (in case of multi-cylinder engines). A cam on cam shaft breaks the contact and causes the electric primary circuit to open and close. When the circuit is broken or the current flow is stopped in the primary winding, the magnetic field collapses inducing a high voltage in the secondary winding due to which a spark occurs instantaneously in the spark plug. This spark initiates the combustion of compressed air-fuel mixture in the engine cylinder. Depending on the method of mixing petrol or gas in the air in right proportion as per the requirement of engine, the fuel supply system of S.I engine is conventionally is of two types as discussed below Carburetion ME3451-Thermal Engineering Department of Mechanical Engineering 2023-2024 SANCET In the carburetion method, fuel stored in the fuel tank is supplied to carburetor by means of a simple diaphragm pump through a fuel filter. The job of pump is only to supply fuel from fuel tank to float chamber of carburetor. Sometimes when the fuel tank is above the carburetor like in 2-wheeler engine, fuel will come by gravity and fuel pump is not needed. The design & working of a simple carburetor fitted in the suction line or Inlet manifold of engine is very simple. As shown in fig 15.1 there is a venturi in the flow passage of air being sucked by engine? A jet is situated in the venturi and connected to the float chamber of carburetor, where fuel is stored at atmospheric pressure. The float keeps the fuel at a constant level in the float chamber. While passing through the venturi, pressure of air reduces and a pressure difference is created across the fuel jet connecting float chamber with venturi. Due to this pressure difference fuel is continuously supplied to air flowing through venturi. As jet is of very small inner diameter, fuel i.e. petrol is sprayed in the flowing air and due to its high volatility, it vapourizes and forms a combustible mixture of fuel vapour plus air. The fuel air ratio is automatically controlled by the speed of air through the venturi which eventually depends on speed/rpm of engine. The size of venturi and jet are designed on the basis of desired fuel air ratio. In an actual carburetor, some additional systems are there to satisfy the demands of engine under varying conditions like cold starting, engine idling, requirement of additional power at high speed & load etc. (Fig. 15.1) Electronic fuel injection (MPFI system) In this system an electrically driven fuel pump draws fuel from fuel tank and supplies it to a common header or tube. A pressure regulator fitted at the end maintains a constant pressure of fuel approx. 3 bars in the header. The header is connected to different branches of inlet manifold through fuel injectors. For each cylinder of engine there is separate fuel injector which injects fuel in the corresponding air passage of that cylinder. Due to this the system is called multi-point fuel injection (MPFI) system. The fuel injectors are precision built solenoid valves having single or multiple orifices. Due to constant pressure of fuel maintained in the common header, the quantity of fuel injected depends only on the time period for which the solenoid valve type fuel injectors are kept in open position. An on-board ECU (Electronic control unit) i.e. microprocessor controls the quantity of fuel injected to each cylinder individually and also the ignition timing of each cylinder. The data input to the ECU comes from a number of sensors located all over the engine. These sensors collect the following data continuously. 1. Ambient temperature 2. Inlet manifold vacuum or Air Velocity 3. Exhaust gases temperature 4. Exhaust O2 content 5. Throttle position 6. Engine r.p.m. 7. Crankshaft & position 8. Engine coolant temperature ME3451-Thermal Engineering Department of Mechanical Engineering 2023-2024 SANCET Electronic fuel injection system Based on programmed interpretation or processing of this data, ECU calculates the amount of fuel needed to maintain stoichiometry i.e. air/fuel ratio of 14.7:1 and converts it into required pulse width i.e. time period for which it keeps the solenoid injector energized. ECU also gives command to spark ignition system. In this way ECU ensures overall satisfactory performance of the engine from start to shut down including emission control by sending right quantity and quality of fuel air mixture to each cylinder of engine at right time based on requirement of engine and also ignites it at right time. Fuel Supply System of C.I Engine The primary requirement of a C.I engine is to inject the right quantity of fuel at a very high pressure either directly over the piston in the cylinder of engine or indirectly in a combustion chamber in the cylinder head which is connected to cylinder of engine. In any method the fuel injection system has to control injection timing, injection period and injection pressure. Now in most of the diesel engines direct injection is used with improved injection technology. The diesel injection systems used nowadays are of two types: (1) Mechanical Injection System (2) Electronic Injection System Mechanical injection system This system was universally being used in Diesel Engines until the introduction of new fuel injection technology like CRDI etc. But still due to the reason of more initial cost involved in adopting newer technology, it is being used in Diesel Engines of different sizes. In this type, there are two basic components as one pressurizing unit (High Pressure Pump) and other atomizing unit (High pressure nozzle or injector). Depending on the design, manner of operation and control of these two basic components, mechanical fuel injection system are of three types. (a) Individual pump and injector system In this system there is a separate pump and separate injector for each cylinder of engine. The pump creates high pressure of fuel and also meters and times the injection of fuel through injector. It is a plunger type pump driven by ME3451-Thermal Engineering Department of Mechanical Engineering 2023-2024 SANCET engine power itself. (b) Common rail system Here a high pressure pump keeps constant high pressure of fuel in an accumulator with the help of a pressure regulating valve. This accumulator is connected to common rail which is extended to different distributing elements of each cylinder. A separate metering & timing element controls the supply of metered & timed pressurized fuel to each injector of cylinder. This system is self-governing and more smooth operation is there. (c) Distributor system Like the first system, here also the pump pressurizes, times and meters the fuel. But which quantum of fuel is to be supplied to which injector, is decided by a rotating distributor. This system is cheaper than first system in case of a multi-cylinder engine. The diagram of a plunger type pump and fuel injector is shown in Fig. 15.3. Electronic injection system (with CRDI technology) In this system of diesel injection, a common rail diesel injection (CRDI) technology is used. It is more or less same as Common Rail System of Mechanical Injection but the difference comes in the control over metering & timing of injectors which is done by an on-board computer system or electronic control unit. Thus it resembles with the operation of MPFI system of S.I engine. The basic system is same; there is a high pressure fuel pump which maintains high pressure in a common rail (steel tube) through high pressure regulator. But here the pressure maintained is very high of the order of 2000 bar as compared to 3-5 bars in MPFI system. The fuel injectors are very special either solenoid type or piezzo electric type which control the fuel injection from common rail to each cylinder very precisely. The opening time, pulse width etc of fuel injectors can be electrically controlled by the E.C.U. Here is the main advantage of system that the fuel can be injected in more than one pulse in a very controlled manner unlike only one pulse or one injection per cycle in the mechanical system. Here a pilot injection is done before the main injection for fast burning and less ignition delay of the fuel. It reduces the noise level very much and also ensures complete burning of fuel, high efficiency, low emission and good cold start. This new technology has considerably removed the demerits of diesel engines like high noise level, high pollution, difficulty in starting etc and improved fuel efficiency a lot. A schematic diagram of CRDI system is given here in fig.15.4 ME3451-Thermal Engineering Department of Mechanical Engineering 2023-2024 SANCET Common rail diesel injection system 12. Explain the working of magneto ignition system with neat sketch. The magneto ignition system is a robust and self-sustaining ignition technology used in internal combustion engines to generate the high-voltage spark needed for engine ignition. Unlike battery-powered systems, the magneto ignition system generates its own electrical energy, eliminating the need for an external power source. This ingenious design employs the principles of electromagnetic induction to produce a continuous flow of electricity through the rotation of a permanent magnet within a coil assembly. This discussion shall unravel all the information related to Magneto Ignition System, including its Diagram, Working, Parts, Advantages and Disadvantages. ME3451-Thermal Engineering Department of Mechanical Engineering 2023-2024 SANCET Magneto Ignition System The magneto ignition system revolutionized internal combustion engines by generating high-voltage sparks without relying on external electrical sources. Employed in early automobiles and aircraft, it functions through a rotating magnet that induces electrical pulses, igniting the fuel-air mixture. Its reliability and self-sufficiency made it an engineering breakthrough. An internal combustion engine characterized by high speed and high internal compressions necessitates a powerful ignition system capable of producing extremely high ignitions from spark plugs. The ignition system, which employs spark plugs as its source, receives electrical energy input to generate the required sparks for ignition. Parts of Magneto Ignition System A magneto ignition system comprises several key components that work together to generate spark for an internal combustion engine. These parts include a magneto, which produces electrical current through rotational motion, a breaker points mechanism to interrupt the current flow, a condenser to prevent arcing at the points, and high-tension wires leading to the spark plugs. Together, these components ensure reliable ignition in various engines, especially in older vehicles and small engines. Here is a list of the key parts utilised in the Magneto Ignition System: Magneto The Magneto serves as the source of energy generation in the ignition system. It acts as a small generator that operates on electricity. When the engine rotates the magneto, it produces voltage. The more rapid the rotation, the higher the voltage generated. Unlike other systems, the magneto does not require an external power source like a battery for kick starting, as it itself generates energy. There are two types of winding in the magneto: primary winding and secondary winding. Additionally, the magneto can be classified based on its engine rotation: o Armature rotating type o Magnet rotating type o Polar inductor type Distributor The distributor used in the Magneto Ignition System is also employed in multi-cylinder engines. In multi-cylinder engines, the distributor regulates spark distribution in the correct sequence among the spark plugs. It uniformly distributes the ignition surge to all the spark plugs. There are two types of distributors: o Carbon brush-type distributor o Gap-type distributor Spark Plug The spark plug in the Magneto Ignition System has two electrodes separated from each other. A high voltage flows through it, leading to the generation of a spark used to ignite the cylinder's combustion mixture, such as oil. The spark plug consists of a steel shell and an insulator. The central electrode is connected to the ignition coil's supply, while the outer steel shell is grounded, effectively insulating both. A small air gap between the central electrode and the steel shell generates the spark. The central electrode is constructed with a high nickel alloy to withstand high temperatures and resistances. ME3451-Thermal Engineering Department of Mechanical Engineering 2023-2024 SANCET Capacitor The capacitor utilised in the Magneto Ignition System is a simple electrical capacitor consisting of two metal plates separated by an insulating material at a distance. Generally, air is used as the insulating material, but in some cases, a high-quality insulating material is employed for specific technical requirements. 13. Explain the construction and working principle of diesel reciprocating pump and fuel injector with neat sketch. Fuel injection is the introduction of fuel in an internal combustion engine, most commonly automotive engines, by the means of an injector. The fuel injection system lies at the very heart of the diesel engine. By pressurizing and injecting the fuel, the system forces it into air that has been compressed to high pressure in the combustion chamber. Fuel injector is an electronically controlled mechanical device that is responsible for spraying (injecting) the right amount of fuel into the engine so that a suitable air/fuel mixture is created for optimal combustion. The electronic control unit (ECU at engine management system) determines the precise amount and specific timing of required gasoline (petrol) dose for every cycle, by collecting information from various engine sensors. So, the ECU sends a command electrical signal of the correct duration and timing to the fuel injector coil. In that way opens the injector and allows petrol to pass through it into the engine. The one terminal of the injector coil is directly supplied by 12 volts which are controlled by the ECU, and the other terminal of the injector coil is open. When ECU determined the exact amount of fuel and when to inject it, activates the appropriate injector by switching the other terminal to the ground (mass, i.e. negative pole). ME3451-Thermal Engineering Department of Mechanical Engineering 2023-2024 SANCET FUNCTIONS The diesel fuel injection system has four main functions: 1. Feeding fuel Pump elements such as the cylinder and plunger are built into the injection pump body. The fuel is compressed to high pressure when the cam lifts the plunger, and is then sent to the injector. 2. Adjusting fuel quantity In diesel engines the intake of air is almost constant, irrespective of the rotating speed and load. If the injection quantity is changed with the engine speed and the injection timing is constant, the output and fuel consumption change. Since the engine output is almost proportional to the injection quantity, this is adjusted by the accelerator pedal. 3. Adjusting injection timing Ignition delay is the period of time between the point when the fuel is injected, ignited and combusted and when maximum combustion pressure is reached. As this period of time is almost constant, irrespective of engine speed, a timer is used to adjust and change injection timing – enabling optimum combustion to be achieved. 4. Atomizing fuel When fuel is pressurized by the injection pump and then atomized from the injection nozzle, it mixes thoroughly with air, thus improving ignition. The result is complete combustion. COMPONENTS The objectives of the fuel injection system are to meter, atomize and distribute the fuel throughout the air mass in the cylinder. At the same time, it must maintain the required air-fuel ratio as per the load and speed demand on the engine. The fuel injection system consists of: fuel injection pump - pressurizes fuel to high pressure high-pressure pipe - sends fuel to the injection nozzle injection nozzle - injects the fuel into the cylinder feed pump – sucks fuel from the fuel tank fuel filter - filtrates the fuel TYPES OF FUEL INJECTORS 1. Top-Feed – Fuel enters from the in the top and exits the bottom. 2. Side-Feed – Fuel enters on the side on the injector fitting inside the fuel rail. 3. Throttle Body Injectors – (TBI) Located directly in the throttle body. ME3451-Thermal Engineering Department of Mechanical Engineering 2023-2024 SANCET TYPES OF FUEL INJECTION SYSTEMS 1. Single-Point OR Throttle Body Fuel Injection Also referred to as a single port, this was the earliest type of fuel injection to hit the market. All vehicles have an air intake manifold where clean air first enters the engine. TBFI works by adding the correct amount of fuel to the air before it is distributed to the individual cylinders. The advantage of TBFI is that it’s inexpensive and easy to maintain. If you ever have an issue with your injector, you’ve only got one to replace. Additionally, since this injector has a fairly high flow rate, it’s not as easy to clog up. Technically, throttle body systems are very robust and require less maintenance. That being said, throttle body injection is rarely used today. The vehicles that still use it are old enough that maintenance will be more of an issue than it would with a newer, lower mileage car. Another disadvantage to TBFI is the fact that it’s inaccurate. If you let off the accelerator, there will still be a lot of fuel in the air mixture that is being sent to your cylinders. This can result in a slight lag before you decelerate, or in some vehicles, it can result in unburned fuel being sent out through the exhaust. This means that TBFI systems are not nearly as fuel-efficient as modern systems. 2. Multi-port Injection Multi-port injection simply moved the injectors further down towards the cylinders. Clean air enters the primary manifold and is directed out towards each cylinder. The injector is located at the end of this port, right before it’s sucked through the valve and into your cylinder. The advantage of this system is that fuel is distributed more accurately, with each cylinder receiving its own spray of fuel. Each injector is smaller and more accurate, offering an improvement in fuel economy. The downside is that all injectors spray at the same time, while the cylinders fire one after the other. This means that you may have leftover fuel in between intake periods, or you may have a cylinder fire before the injector has had a chance to deliver additional fuel. Multi-port systems work great when you are traveling at a consistent speed. But when you are quickly accelerating or removing your foot from the throttle, this design reduces either fuel economy or performance. 3. Sequential Injection Sequential fuel delivery systems are very similar to multi-port systems. That being said, there is one key difference. Sequential fuel delivery is times. Instead of all injectors firing at the same time, they deliver fuel one after the other. The timing is matched to your cylinders, allowing the engine to mix the fuel right before the valve opens to suck it in. This design allows for improved fuel economy and performance. Because fuel only remains in the port for a short amount of time, sequential injectors tend to last longer and remain cleaner than other systems. Because of these advantages, sequential systems are the most common type of fuel injection in vehicles today. the one small downside to this platform is that it leaves less room for error. The fuel/air mixture is sucked into the cylinder only moments after the injector opens. If it is dirty, clogged, or unresponsive, your engine will be starved of fuel. Injectors need to be kept at their peak performance, or your vehicle will start to run rough. ME3451-Thermal Engineering Department of Mechanical Engineering 2023-2024 SANCET 4. Direct Injection If you’ve started to notice the pattern, you can probably guess what direct injection is. In this system, fuel is squirted right into the cylinder, bypassing the air intake altogether. Premium automobile manufacturers like Audi and BMW would have you believe that direct injection is the latest and greatest. With regards to the performance of gasoline vehicles, they’re absolutely right! But this technology is far from new. It’s been used in aircraft engines since the second world war, and diesel vehicles are almost all direct injection because the fuel is so much thicker and heavier. In diesel engines, direct injection is very robust. Fuel delivery can take a lot of abuse, and maintenance issues are kept to a minimum. With gasoline engines, direct injection is found almost exclusively in performance vehicles. Because these vehicles operate with very precise parameters, it’s especially important to maintain your fuel delivery system. Although the car will continue to run for a long time when neglected, the performance will quickly decline. METHODS OF FUEL INJECTION There are two methods of fuel injection in the compression ignition system 1. Air-blast injection 2. Air-less or solid injection 1. Air-blast injection This method was originally used in large stationary and marine engines. But now it is obsolete. In this method, the air is first compressed to very high pressure. A blast of this air is then injected carrying the fuel along with it into the cylinders. The rate of fuel injection is controlled by varying the pressure of the air. The high-pressure air requires a multi-stage compressor so as to keep the air bottles charged. The fuel ignites by the high temperature of the air caused by the high compression. The compressor consumes about 10% of the power developed by the engine, decreasing the net output of the engine. 2. Airless or solid injector In this method, the fuel under high pressure is directly injected into the combustion chamber. It burns due to the heat of compression of the air. This method requires a fuel pump to deliver the fuel at high pressure around 300kg/cm^2. This method is used for all types of small and big diesel engines. It can be divided into two systems 1. Individual pump system: in this system each cylinder has its own individual high-pressure pump and a measuring unit. 2. Common rail system: in this system the fuel is pumped by a multi-cylinder pump into a common rail, the pressure in the rail is controlled by a relief valve. A measured quantity of fuel is supplied to each cylinder from the common rail. This is all about the fuel injection system. If you have any query regarding this article, ask by commenting. If you like this article, don’t forget to share it on social networks. Subscribe our website for more informative articles. Thanks for reading it. ME3451-Thermal Engineering Department of Mechanical Engineering 2023-2024 SANCET WORKING PRINCIPLES The injectors are controlled by the Engine Control Unit (ECU). First, the ECU obtains information about the engine conditions and requirements using different internal sensors. Once the state and requirements of the engine have been determined, the fuel is drawn from the fuel tank, transported through the fuel lines and then pressurized with fuel pumps. Proper pressure is checked by a fuel pressure regulator. In many cases, the fuel is also divided using a fuel rail in order to supply the different cylinders of the engine. Finally, the injectors are ordered to inject the necessary fuel for the combustion. The exact fuel/air mixture required depends on the engine, the fuel used and the current requirements of the engine (power, fuel economy, exhaust emission levels, etc.) |
3776 | https://texasgop.org/wp-content/uploads/2024/06/2024-RPT-Platform.pdf | Table of Contents 1 2024 Platform and Resolutions of the Republican Party of Texas Platform Committee Members Matt Patrick, Chairman Vergel Cruz, Convention Secretary Carson Cheshire, Assistant to the Chair SD 1: Darren York SD 11: Gina Smith SD 22: Jon Ker SD 2: Dennis London SD 12: Cindi Castilla SD 23: Elaine Cook SD 3: Tony Robertson SD 13: Milinda Morris SD 24: Andy Eller SD 4: Dale Inman SD 14: David Armstrong SD 25: Patrick Von Dohlen SD 5: Michelle Evans SD 15: Brian Vachris SD 26: Melinda Roberts SD 6: Marga Matthews SD 16: Brian Bodine SD 27: Isela Lindquist SD 7: JR Haas SD 17: Peter Batura SD 28: Dan Pickens SD 8: Stephen Kallas SD 18: Carter Thomas SD 29: Mark Dunham SD 9: Tamma Gunn SD 19: Warrington Lee Austerman SD 30: Christine Whitmore SD 10: Cary Cheshire SD 20: Andrew Duarte SD 31: Thomas Warren SD 21: Terry Harper Editorial Committee Rick Townsend, Team Lead Heath Bell Karen Marshall Jett Burns Sharon Nations Orlando Dona Lauren Pena Olga Farnam Ruth Potts Lauren Langas Justin Rorick Joe Leatherwood Debbie Wolgemuth Table of Contents 2 Table of Contents Platform Committee Members .................................................................................................................. 1 Editorial Committee .................................................................................................................................. 1 Preamble.................................................................................................................................................... 3 Principles ................................................................................................................................................... 3 Constitutional Issues ................................................................................................................................. 3 Preservation of Constitution ................................................................................................................. 3 Citizen Rights........................................................................................................................................ 5 State Sovereignty .................................................................................................................................. 6 Business, Commerce, and Transportation ................................................................................................. 7 Markets and Regulation ........................................................................................................................ 7 Retirement, Savings, Unions................................................................................................................. 9 Energy and Environment ...................................................................................................................... 9 Transportation ...................................................................................................................................... 11 COVID Response................................................................................................................................. 11 Privacy, Information Freedom, Internet ............................................................................................... 11 Finance .................................................................................................................................................... 12 Spending Restraint .............................................................................................................................. 12 School Finance and Property Taxation ............................................................................................... 13 Opposition to Market-Distorting Tax and Fiscal Subsidies ................................................................ 14 Transparency and Oversight ............................................................................................................... 15 Education ................................................................................................................................................ 15 Parents’ Rights .................................................................................................................................... 15 Curriculum .......................................................................................................................................... 16 Governance ......................................................................................................................................... 18 Higher Education ................................................................................................................................ 19 Health and Human Services .................................................................................................................... 20 Parents’ Rights .................................................................................................................................... 20 Healthcare Independence .................................................................................................................... 21 Government-funded Health Programs ................................................................................................ 22 Mental Health...................................................................................................................................... 23 Homosexuality and Gender Issues ...................................................................................................... 24 Substance Abuse and Addiction .......................................................................................................... 24 Life-Affirming Health Care Concepts ................................................................................................ 24 Environmental Health ......................................................................................................................... 26 Criminal and Civil Justice ....................................................................................................................... 26 Rights and Protections ........................................................................................................................ 26 Courts, Prosecutions, Restitution ........................................................................................................ 27 Law Enforcement ................................................................................................................................ 28 Family Law ......................................................................................................................................... 28 State Affairs ............................................................................................................................................. 28 Heritage Preservation .......................................................................................................................... 28 Individual Rights and Freedoms ......................................................................................................... 29 Family and Gender Issues ................................................................................................................... 30 Pro-Life Issues .................................................................................................................................... 32 Land Use ............................................................................................................................................. 32 State Governance ................................................................................................................................ 34 Government and Election Integrity ......................................................................................................... 34 Government Operations ...................................................................................................................... 34 Elections .............................................................................................................................................. 35 National Defense and Foreign Affairs .................................................................................................... 38 Veterans Affairs ................................................................................................................................... 38 Border Security and Immigration ....................................................................................................... 39 Foreign Affairs .................................................................................................................................... 40 Resolutions .............................................................................................................................................. 42 Index........................................................................................................................................................ 43 Table of Contents 3 Preamble Affirming our belief in God, we still hold these truths to be self-evident: that all men are created equal, that they are endowed by their Creator with certain unalienable Rights, that among these are Life, Liberty, and the Pursuit of Happiness. Throughout the world, people dare to dream of freedom and opportunity. The Republican Party of Texas unequivocally defends that dream. We strive to preserve the freedom given to us by God, implemented by our Founding Fathers, and embodied in the Constitution. We recognize that human nature is immutable. We further recognize that the traditional family is the strength of our nation. It is our solemn duty to protect innocent life and develop responsible citizens. We understand that our economic success depends upon free market principles. If we fail to maintain our sovereignty, we risk losing the freedom to live these ideals. Principles We, the 2024 Republican Party of Texas, believe in this platform and expect our elected leaders to uphold these truths through acknowledgment and action. We believe in: 1. “The laws of nature and nature’s God,” and we support the strict adherence to the original language and intent of the Declaration of Independence and the Constitutions of the United States and of Texas. 2. The sanctity of innocent human life, created in the image of God, which should be equally protected from fertilization to natural death. 3. Preserving individual, Texan, and American sovereignty and freedom. 4. Limiting government power to those items enumerated in the United States and Texas Constitutions. 5. Personal accountability and responsibility. 6. Self-sufficient families, founded on the traditional marriage of a natural man and a natural woman. 7. Having an educated population, with parents having the freedom of choice for the education of their children. 8. The inalienable right of all people to defend themselves and their property. 9. A free enterprise society unencumbered by government interference or subsidies. 10. Honoring all of those that serve and protect our freedom. Constitutional Issues Preservation of Constitution 1. Keep Oath to the Constitutions: We call for all who swear the oath to support and maintain the limitations and clear meaning of the United States and Texas Constitutions. 2. Amendments to the Texas Constitution: For the preservation of our Constitutional Republic as set forth in the Texas Bill of Rights we: a) Support an amendment to the Texas Constitution that will require a majority of the voters in at least 170 counties (two-thirds), instead of a simple majority of the votes, to pass amendments. b) Support limiting the terms of the Texas Speaker of the House to two consecutive terms, after which the Speaker must vacate the office for the same number of terms served, after which that Speaker is eligible to run again. Table of Contents 4 c) Support an amendment to Article 4, Section 8, of the Texas Constitution to allow the Texas Legislature to convene itself upon petition from 55 percent of each house’s elected and serving members between Regular Sessions. d) Support an amendment for term limits of twelve (12) years for state and county offices. e) Support an amendment to allow the Attorney General of Texas to prosecute cases anywhere in Texas where county district attorneys refuse to enforce state law, and to prosecute those who are accused with prima facie evidence of violating state laws. f) Support amending the language of the Texas Constitution to mirror the language of the 2nd Amendment of the United States Constitution, ensuring clarity and consistency in the protection of the people’s right to keep and bear arms. In addition, licensed concealed carry holders may legally carry firearms in public sporting venues, including, but not limited to, ball parks, arenas, sporting events, airshows, car shows, rodeos, or if financed in part with taxpayer funds or hold tax exemptions. g) Pass unrestricted Constitutional Carry by amending Article 1, Section 23, of the Texas Constitution by removing, “but the Legislature shall have power, by law, to regulate the wearing of arms, with a view to prevent crime.” h) Support an amendment that would allow for the recall of any elected officials in the state of Texas for misconduct while in office, or failure to properly represent their constituents. 3. Enforce the Constitution Article 4, Section 4: The sovereignty of this State requires the protections afforded under Article 4, Section 4, of the Constitution, and any failing thereof authorizes the Governor of this State or the Legislature to declare an invasion, which shall be met with the full force of this State. 4. No Foreign Law: No foreign law, contract, or judgment arising from any foreign state whose laws violate fundamental constitutional rights shall be honored or enforced by any Texas court. 5. Judicial Overreach: All attempts by the judiciary to rule in areas not constitutionally granted to the judiciary, including abuses of the “Commerce Clause,” the “General Welfare Clause,” and the “Supremacy Clause,” should be nullified. Any federal enforcement activities that occur in Texas shall be conducted under the authority of the county sheriff (SCOTUS ruling in 1997 Mack and Printz v. US). 6. Limiting the Power of the Supreme Court: Judges do not define the roles of judges; Article 3, Section 2, defines the roles of judges. We support Congress’s limiting the power of the Supreme Court by invoking Article 3, Section 2, Clause 2, by “legislating exceptions and creating regulations“ to limit the cases for which the Supreme Court has jurisdiction. 7. Amendments to the United States Constitution: We: a) Support term limits of twelve (12) years for federal offices of US Senate and US House of Representatives. b) Oppose “packing” (or enlarging) the US Supreme Court and support the pending “Keep Nine Amendment” as filed in the US Senate and the House of Representatives with bipartisan support. c) Support the repeal or the nullification of the 16th Amendment (Federal Income Tax). d) Support restoring state sovereignty with the repeal of the 17th Amendment of the US Constitution and the appointment of US Senators by the state legislatures. e) Support a change to the 14th Amendment to eliminate “birth tourism” or anchor babies by granting citizenship only to those with at least one biological parent who is a US citizen. f) Support a constitutional amendment making English the official language of the United States, and one of no more than two official languages of all US territories and other possessions. 8. Executive Orders: We oppose all executive orders, whether by a president, a governor, or a local official, that go beyond administration of executive authority and have the effect of legislation. We call upon the Texas Legislature or local lawmakers to nullify such executive orders. Table of Contents 5 9. Limiting Overreaching State Government: We recognize that the sovereignty of this State and its citizenry has been imperiled and threatened by the ongoing overreach of state elected officials and agencies. We therefore call for the enforcement of Article 2 of the Texas Constitution and restoration of our liberty by the following: a) We must repeal and replace Texas Code 418. b) No form of government shall ever again implement mass lockdowns on the people, our businesses, or our churches. c) We oppose funding or implementation of any form of contact tracing. d) The State Constitution must be changed to require the Legislature come to session after a declared emergency lasting 30 days or more in five or more counties. e) The Texas Governor shall no longer use executive orders to create public policy or law and shall no longer have the power to close businesses or declare some as “essential” or “non-essential.” f) The Governor‘s authority during an emergency shall not be delegated. 10. Dereliction of Duty: The failure by a public official to discharge any duty shall be a violation of the terms of his or her oath of office, which shall constitute a crime, and upon conviction this crime shall be punishable by a fine or imprisonment, depending on the nature of the offense. Any entity or person who acts under the color of federal or state law to deprive a Texan of the rights or privileges insured by the federal and state constitutions shall be liable to the injured parties for redress, including monetary damages and injunctive relief, notwithstanding any preexisting immunities. Citizen Rights 11. The Rights of a Sovereign People: We support the historic concept, established by our nation’s Founders, of limited civil government jurisdiction under the natural laws of God, and we oppose the concept that the state is sovereign over the affairs of men, the family, or the church. We believe that government properly exists by the consent of the governed and must be restrained from intruding into the freedoms of its citizens to include due process. The function of government is not to grant rights, but to protect the unalienable, God-given rights of life, liberty, property, and the pursuit of happiness of all, including the unborn. Texas shall keep the Texas Citizens Participation Act (TCPA) intact and preserve its broad scope and essential protections for 1st Amendment Rights. 12. Protecting Constitutional Rights Regarding Age: There should be a single age of majority upon which, when reached, all citizens will be guaranteed their rights, duties, and privileges. 13. National Popular Vote: The National Popular Vote Interstate Compact is a direct violation of Article 1, Section 10, and Article 2, Section 1, of the Constitution and shall be rejected by Texas and all its officials. We support the Electoral College. 14. Habeas Corpus: Any federal suspension of the writ of habeas corpus against a Texas citizen shall be violative of the 10th Amendment, Texas sovereignty, sovereignty of the individual, and actionable by the state or the citizen. 15. Census: In accordance with the United States Constitution, we support an actual count of United States citizens, and we oppose Census Bureau estimates and the collection of all other data. Illegal aliens must not be included in any census. 16. Equal Rights Amendment: We call upon the Texas Legislature to adopt a resolution clarifying that the 1972 ratification by the 62nd Texas Legislature of the proposed Equal Rights Amendment to the United States Constitution was valid only through March 22, 1979. 17. Parental Rights: The rights of parents are foundational to Western society and shall be respected, affirmed, and protected by the Texas Constitution and Texas Law. Furthermore, it is imperative that the Texas Legislature pass a Parental Rights Amendment to be added to the Texas Constitution in order to secure these rights for future generations. We call upon the Legislature to properly recognize and affirm the fundamental right of parents to make all decisions regarding the upbringing and control of their Table of Contents 6 children in all aspects, especially in light of the grievous violations of the Texas education system. Any failure to recognize, protect, or honor these fundamental rights shall be actionable. No parent exercising any of these fundamental rights shall be prosecuted as domestic terrorists. 18. Prayer, Bible, and Ten Commandments in Schools: We support affirmation of God, including prayer, the Bible, and the Ten Commandments being returned to our schools, courthouses, and other government buildings. 19. The Right to Keep and Bear Arms: State and Federal Legislatures shall: a) Repeal and/or nullify the National Firearms Act of 1934 and the Gun Control Act of 1968. b) Nullify any gun laws that violate the 2nd Amendment or rights of due process. c) Support national reciprocity for gun ownership rights and the right to carry. d) Recognize the right of License to Carry holders to carry anywhere, off-duty or retired law enforcement personnel to carry anywhere, including any federal property accessible to the general public, including, but not limited to waterways, lakes, dams, and postal offices. e) Ensure that any of the above legislation shall not be construed as impinging on private property rights. f) Ensure individuals who have acted in justified defense should presumptively be immune from civil liability. g) Ensure tenants of leased premises should not be evicted for acting in justified self-defense. h) Require that financial service firms doing business in Texas do not record financial transactions using Merchant Category Codes (MCC) for purchases related to guns, ammunition, gun accessories, gun safes, gun range payments, gun range memberships, and other 2nd Amendment protected activities. The use of any such services for the tracking of these purchases shall be banned. i) Require that businesses or commercial property owners that prohibit licensed permit holders from carrying a firearm into their establishments assume liability for their safety since the owners are denying those persons the right to protect themselves. j) Republicans believe armor piercing ammunition and chemical dispensing devices (OC spray) are useful for defense and must be legalized in Texas. State Sovereignty 20. State Sovereignty: Pursuant to Article 1, Section 1, of the Texas Constitution, the federal government has impaired our right of local self-government. Therefore, federally mandated legislation that infringes upon the 10th Amendment rights of Texas shall be ignored, opposed, refused, and nullified. Texas retains the right to secede from the United States, and the Texas Legislature should be called upon to pass a referendum consistent thereto and pass the Texas Sovereignty Act as filed in the 88th Legislative Regular Session as HB 384. 21. Concurrent Majority: The Legislature shall cause to be enacted a State Constitutional Amendment to add the additional criteria for election to a statewide office to include the majority vote of the counties with each individual county being assigned one vote allocated to the popular majority vote winner of each individual county. 22. Unfunded and Under-Funded Mandates: Unfunded mandates and under-funded mandates are unacceptable. The State of Texas must fully fund, at a minimum, the following additional costs to local governments provided to all legal residents: a) Indigent criminal defense b) Inmate healthcare in jails c) Indigent burials and autopsies d) Veteran services offices 23. Equal Protection for the Preborn: We urge lawmakers to enact legislation to abolish abortion by immediately securing the right to life and equal protection of the laws to all preborn children from the moment of fertilization, because abortion violates the United States Constitution by denying such persons the equal protection of the law. Table of Contents 7 24. Article 5 Convention of States: The Texas Legislature shall extend the call for a Convention of States to impose fiscal restraints on the federal government, limit the power and jurisdiction of the federal government, and limit the terms of office of federal officials and members of Congress. Business, Commerce, and Transportation Markets and Regulation 25. Municipal Preemption: We encourage the Legislature to preempt local government efforts to interfere with the state’s sovereignty over business, employees, and property rights. This includes but is not limited to burdensome regulations on short-term rentals, plastic bags, sick leave, trees, and employee criminal screening. We support preemption of city ordinances which prohibit or penalize private charity to United States citizens. This excludes the handling of emergency orders. 26. Licensing: We call upon the Texas Legislature to review all Texas Department of Licenses and Regulations (TDLR), business/professional licensing programs, and associated licensing for the purpose of abolishing or removing as many as reasonably possible and repealing those laws, rules, and regulations. a) Practice of Medicine: We support allowing any board-licensed medical graduate to practice medicine under the supervision of a full physician, just as Nurse Practitioners (NPs) and Physicians Assistants (PAs) are permitted to practice medicine under the supervision of a full physician. b) Practice of Law: We oppose mandatory State Bar membership. The Texas Bar may not collect any more dues unless it conducts a membership referendum and with more than two-thirds of all dues-paying members approve of the mandatory membership requirement. We support sanctions for those who weaponize and politicize the legal disciplinary process of Texas in attorney disciplinary matters. Texas should adopt a “clear and convincing” evidentiary standard instead of its current “preponderance of the evidence.” The State Bar of Texas may not try to influence judges, justices, judicial candidates, or associations. 27. Reduce Business Regulations: We believe that the following businesses shall be minimally regulated at all levels. a) Federal Laws: i. Repeal Minimum Wage Law and prevailing wage laws. ii. Repeal Dodd-Frank. iii. Repeal Sarbanes-Oxley. iv. Repeal The Lacey Act. v. Texans should be able to buy beef and farm products directly from farmers without any FDA and USDA or other governmental agency regulations. vi. Mandatory sick or family leave. b) State Laws: i. Eliminate Blue laws. ii. Eliminate the three-tier alcohol system (production, distribution, and retail). iii. Mandatory sick or family leave. iv. Business licensing. v. Professional licensing. vi. Purchase of edible products from small farms. vii. Use of hemp as an agricultural commodity. 28. Trade Agreements: We support free trade as a necessary component of American capitalism and of the United States’ influence in the world. However, all trade agreements between the Federal Government and other nations shall strictly adhere to the United States Constitution and require approval by two-thirds (2/3) of the Senate. Table of Contents 8 29. Origin Labeling: We urge that all food products entering the United States show not only the country of origin, but also the country that processed it, and the country that packaged it. a) All products created to be consumed by people or animals shall be labeled as “lab-grown” in “state” and “country.” b) Any products created for consumption by people or animals with insect parts/liquid shall be labeled as “insects included.” c) Texas Legislature should amend shipping laws for distilleries and breweries to match that of wine shipments that are allowed to and from Texas. All shipped wine must be in packages that are clearly labeled and show that the packages contain wine. It must be delivered by a business holding a Carrier’s Permit. Wine can be delivered to: i. The person who purchased it. ii. A recipient chosen in advance by the purchaser. iii. A person at the delivery address who is 21 or older, after the recipient presents a valid ID. 30. Municipal Permitting: We call on the Texas Legislature to continue to streamline the building permitting process to ease burdens and costs on developers and consumers. We encourage the Legislature to monitor implementation of existing legislation that protects property development rights and to close loopholes that cities are using to circumvent these laws. We oppose retroactive rulemaking and changing the rules on developments that have already been issued permits. Unused landfill permits issued by the Texas Commission of Environmental Quality shall automatically expire after 10 years. 31. Texas Resistance to The Great Reset: Texas should continue to pass legislation needed to protect the citizens of Texas from corporate violation of their rights in the Great Reset/Environmental, Social, and Governance (ESG) scheme as follows: a) Ban companies and corporations that attempt to suppress funding for Texas industries from doing business with the State of Texas and its subdivisions. b) Make it illegal for banks and financial institutions operating in Texas to make lending decisions based on anything other than financial concerns. The model for such legislation shall be the Fair Access to Financial Services rule promulgated by President Trump’s Office of the Comptroller of the Currency, a rule that was immediately suspended by the current occupant of the White House. c) Add penalties in Texas law for corporations operating in Texas that lead or participate in boycotts against Texas due to legislative action to protect the rights of Texans to decline vaccination, protect the unborn, stop the teaching of Critical Race Theory in schools, compete in sports with only those of their own biological gender, or to protect children and juveniles against sexual organ mutilation, hormones, and puberty blockers designed to fake transition from one gender to another. d) Consider the expansion or use of existing Texas anti-trust law to prohibit collusion between woke corporations to cancel/drive others out of business. e) Enact legislation protecting the property, rights, and freedoms of citizens and businesses in Texas by eliminating the priority given to the largest financial institutions to investor assets under the Uniform Commercial Code. 32. Texas Currencies: Texas should recognize the natural right to use currency of choice in the Texas Bill of Rights and provide choices for ordinary Texans to use gold and silver in everyday transactions. The State of Texas shall prohibit the use of Central Bank Digital Currencies (CBDC) for use as legal tender to conduct private and commercial transactions or to settle debts, public or private. The State of Texas does not recognize the authority of the Federal Reserve Bank or the United States Treasury to create or implement Central Bank Digital Currencies as legal tender consistent with the provisions delegated to the Federal Government under Article 1, Section 8, of the Constitution of the United States of America. 33. Patent Protection: We support reversing State and Federal legislation and court decisions that have damaged patent rights for Texas independent inventors and startups. These actions include abolishing administrative tribunals and restoring injunctive relief, thus restoring the United States patent system and driving America back to leading the world technologically, economically, and militarily. Table of Contents 9 Retirement, Savings, Unions 34. Government Accountability: We call upon the Texas Legislature to eliminate all special collective bargaining statutes for public employees and to hold all public servants accountable to taxpayers through existing civil statutes. We oppose any distribution of taxpayer dollars to unions. 35. Social Security Privatization: We support privatization of the Social Security system to lift the rate of return workers obtain on retirement contributions and boost national saving and economic growth. We also support removal of the “Windfall Elimination Provision” which punishes teachers and government workers (legislation pending), and encourage process simplification for Teacher Retirement System members to file for, and receive, benefits. 36. Rural and Volunteer Fire Departments: We urge Congress to overturn the rules of the United States Department of Labor restricting volunteerism by paid firefighters and emergency medical technician personnel and to support protections similar to those provided to National Guardsmen for service during declared emergencies. 37. Employee Stock Ownership Plans (ESOPs): We support maximum economic opportunity for all citizens and believe that legal limits on employee ownership of Texas firms by Employee Stock Ownership Plan (ESOP) trusts must be eliminated. We also believe that changes in ownership from private individuals to a majority ownership by an ESOP must not create disadvantages when doing business with the State of Texas or political subdivisions of the State. We believe the State of Texas should encourage the creation of more ESOPs by making information easily available to businesses located in Texas. 38. Unions: We support legislation requiring labor unions to obtain consent of the union member before that member’s dues can be used for political purposes. We oppose card check. Texas should prohibit governmental entities from collecting dues for labor unions through deductions from public employee paychecks. We also encourage the adoption of a National Right to Work Act. Energy and Environment 39. Utilities: We support free-market solutions for providing resilient and reliable utilities. The State of Texas shall require electric companies that buy electrical production from private citizens on their homes to provide equitable credits for overproduction, to be applied during periods of underproduction; such credits should be allowed to accumulate indefinitely. 40. Proposed Water District (MUD): In addition to those who have lived inside a proposed water district for the statutorily required minimum six months prior to an election to approve water districts, the Texas Legislature should amend the standards for approval of a water district, which is the precursor to a Municipal Utility District (MUD), to include any registered voter who has resided on a property within one thousand (1,000) feet from the proposed property line for at least the same amount of time. 41. Power Grid: We urge the Texas Legislature to pass legislation to harden the Texas Electric Grid increasing capacity and preparedness for all hazards, including: a) Cyberattacks on the grid‘s computerized command and control system. b) Physical attacks on substations and major high-voltage transformers. c) Geomagnetic storms created by solar flares. d) Electromagnetic Pulse (EMP). e) Extreme weather events, both cold and hot. f) Texas Electric Grid shall remain independent from the United States power grid system. 42. Carbon Tax: We oppose any and all efforts to implement a carbon tax. 43. Environment: We oppose environmentalism, or “climate change“ initiatives, that obstruct legitimate business interests and private property use, including the regulatory use limitation and confiscation by governmental agencies. We support the reclassification of carbon dioxide as a non-Table of Contents 10 pollutant, abolition of the Environmental Protection Agency, and repeal of the Endangered Species Act. We oppose President Biden’s Executive Order 14008, “Tackling the Climate Crisis at Home and Abroad,” the “America the Beautiful” Initiative, also known as the 30 x 30 program, and all “climate justice“ initiatives. We support Texas Land Application Permits (TLAP) and oppose TLAP conversion to Waste Water Treatment Plants (WWTP) that allow for partially treated sewage to be dumped into our creeks and lakes. We oppose cloud seeding and geoengineering of any kind. 44. Flooding Mitigation, Hurricane, and Early Warning of Impending Disaster: We support the immediate study, implementation, and construction of projects that will: a) Address river, bayou, reservoir, and other flood threats to public and private property. b) Seek the input of those most likely to be affected by a casualty to public and private property, to include infrastructure and facilities that affect national security. c) Address the risk of storm or tidal surge that affects critical industries along the Texas Gulf Coast. d) Create an early warning system that will immediately alert residents to an impending flood, wind, or casualty weather event. e) Establish regional flood control districts where necessary for counties to resolve joint flooding issues. f) Provide funds to complete these projects from federal, state, and local funds. g) Government boards responsible for flood mitigation shall be elected and not appointed. Governments at all levels must work together to avoid the historical bent to push projects, safety, and implementation into the future. Projects must be of most urgent priority now to avoid further trauma; loss of life; loss of personal, government, and business wealth; and diminution of the tax base. 45. Tidelands and Resources: We assert that the State of Texas and all coastal states shall enjoy and maintain jurisdiction and control of their offshore waters up to the international water boundaries, as well as state inland waterways, regarding all natural resources therein, and that the federal government shall not set limits on harvesting or taking natural resources therein, nor allow foreign entities to harvest or take such natural resources therein, including minerals, game, fisheries, and hydrocarbons. Also, we demand that no entity shall usurp Texas’ original tideland boundaries. 46. Water Resources: We support tying surface rights of ownership to groundwater rights of ownership. We support regulations that may put limits on a person’s capture and use of groundwater, if such use will negatively impact adjoining owners’ use of their groundwater for private wells, their water supply, or agricultural use. We support innovative solutions to meet growing water demand in Texas, including development and desalination of seawater and both private and Texas-owned brackish waters. 47. Energy Production: We support free-market solutions and the immediate removal of government barriers and direct subsidies to the production, transportation, reformulation, refining, and distribution of energy. We oppose federally directed plans and proposals that favor renewable energy sources that may constitute a nuisance, or otherwise have a substantially negative impact on neighboring landowners, including harming property values of our neighborhoods, farms, and ranch areas. We fully support and encourage greater energy production in the Permian Basin and throughout Texas, as well as the necessary policies and infrastructure investment in roads, pipelines, and ports. These steps will support Texas workers and help America’s energy security, instead of increasing our reliance on foreign governments that do not benefit America or our allies. We urge the state to require minimum firebreak setbacks on solar and battery facilities for wildfire prevention mitigation. 48. Energy Innovation: Texas should take advantage of its independent grid and mines to build and operate traditional and next-generation nuclear energy plants. To keep the federal government from interfering, the Texas Legislature shall establish a nuclear regulatory authority and ensure that the facilities and their supply chain of components and fuel remains in state and off federally claimed land. The Texas Legislature shall begin the process of creating a Strategic Petroleum Reserve. The Texas Legislature must ensure the safety of these reserves from enemies, foreign and domestic. Table of Contents 11 Transportation 49. Freedom to Travel: Lawmakers must take action immediately to protect drivers from government-mandated remote kill switches that are currently scheduled to be in every vehicle starting in 2026, to prevent this unconstitutional surveillance of drivers and the weaponization of this technology against Texas drivers. We oppose climate mandates like Net Zero, Vision Zero, or declarations of a climate emergency that threaten our freedom to travel, impose any sort of state or federal mileage tax, or institute diversity, equity, and inclusion policies on taxpayers and drivers. We oppose anti-car measures that punish those who choose to travel alone in their vehicles. We also oppose mandates such as autonomous vehicle technology, digital IDs, digital license plates, or other mandates that restrict, control, or prohibit driving (particularly gas-powered vehicles), shrink auto capacity, or intentionally clog vehicle lanes to force deference to pedestrian, bike, and mass transit options. 50. Vehicle Taxes and Inspections: At the point of sale, taxes shall be based on the actual price of the vehicle or trailer. The Legislature shall implement a method and structure so that owners of electric vehicles pay their ongoing fair share of highway and road construction, maintenance, and usage. Only commercial vehicles will be required to complete a vehicle inspection. 51. Toll Roads: We call on the Texas Legislature to abolish existing toll roads and prohibit future construction, returning responsibility for road construction and maintenance to the appropriate jurisdiction. 52. High-Speed Rail: Taxpayer money shall not fund or subsidize high-speed rail, nor shall eminent domain be used in the construction of high-speed rail. 53. Prohibit Abortion Transportation Across State Lines: We support legislation to prohibit the use of any government funds, as well as the transportation of pregnant women across Texas’ state lines, for the purpose of procuring an elective abortion and for the provision of a private right of action against all persons and organizations who aid and abet in the harming of the woman, and the killing of her pre-born child. COVID Response 54. Jobs Are Essential: We urge the Texas Legislature to adopt legislation that recognizes and establishes all businesses and jobs as essential and a fundamental right. Governments do not have the authority to determine what entities are essential during an emergency. Businesses shall not be held liable for any customer who frequents their business and later is confirmed to test positive for a contagious illness. Privacy, Information Freedom, Internet 55. Doxing and Swatting: We recommend the Legislature consider modifying existing state law to take into account the implications of doxing, which results in harm to one’s person or business, and swatting, a criminal harassment tactic of deceiving an emergency service into sending a police or emergency response team to another person’s address. 56. Regulating the Internet: We oppose efforts to implement net neutrality on internet service providers. We oppose all efforts to further regulate the internet in the United States or internationally, or to impose taxation upon internet digital goods and services. 57. Artificial Intelligence Protections: We believe that each person is the rightful owner of his/her name, likeness, voice, knowledge, opinions, etc., and we support the individual’s right to protect his/her identity, name, image, data, information, and likeness from being collected or otherwise used by data brokers, or artificial intelligence applications without written consent. Each person has the right to know who is collecting information, when and how the information is being collected, and how the information is being used or sold. Each person has the right to opt out of any data collection or data use. Table of Contents 12 58. Personal Data Privacy: We demand that all rights to privacy that individuals have in their homes shall be extended to all digital data via the use of strong public key encryption technologies. We call upon Texas to prohibit vendors of the State of Texas and its subdivisions from selling or sharing data captured in providing services to Texans. We support laws limiting the ways in which internet providers, data brokers, electronic applications, websites, schools, government entities, and others may access the electronic communications or documents of all Texans. We support a no-cost smart meter opt out for utility customers, or the provision of an auditable option to limit storing data to only once per monthly bill cycle. 59. Social Media Freedom and Responsibility: We call on our Congressional Delegation to push for reform of Section 230 of the Communications Decency Act to limit the ability of online social media platforms to censor the speech of citizens in the new digital town square, which these media platforms currently control. We support Texas legislative efforts, such as HB 20 of the 87th Second Called Session, that afford Texas residents the power to sue big tech companies for targeting and censorship. We also support stripping Section 230 immunity from sites that knowingly publish obscene and indecent material, particularly advertisement, promotion, presentation, distribution, or solicitation of child sexual abuse material, thus opening these sites to criminal and civil liability for the content posted on their platforms. 60. Cyber Security Self-Defense: We support “hack-backs,” defined as counterattacks aimed at disabling or collecting evidence against a perpetrator, as a legitimate form of self-defense of persons and organizations to ensure their cyber security. The right to defend oneself in our current era must be expanded to preserve the safety, property, and livelihood of Texans. 61. American Workers First: Prioritize American workers in the workforce. Finance Spending Restraint Note: Letters “F,” “S” and “L” at the end of each plank in this section denote financial impact related to Federal, State, and Local factors. 62. Government Spending and Taxation: We believe in the principles of constitutionally limited government based on federalist principles. Government spending is out of control at the federal, state, and local levels, and action is needed to reduce spending. We urge the Legislature to amend the Texas Constitution and State statute with a stricter spending limitation based on US Census population growth plus inflation and apply the new limit to Texas’ total government budget. We call on the Texas Legislature to freeze state spending. Any budget surplus shall be applied to a Texas Health Savings Account (HSA) until such time as we have exited Medicaid. Additionally, we support imposing a spending cap on all local government jurisdictions to be bound by the metric of population and inflation. (F, S, L) 63. Economic Stabilization Fund: Use of the Economic Stabilization Fund shall be limited to its intended purposes of preventing tax increases during economic downturns and responding to unforeseen disasters. Six months after such events occur, excess funds shall first be deposited into the Texas Health Savings Account (HSA), and then go towards property tax relief. (S) 64. Government Pensions: The Texas Legislature shall enact new rules to begin to transition government pensions for public sector employees from a defined benefit pension to a defined contribution retirement plan similar to a 403(b). (S, L) 65. Gambling: We oppose any expansion of gambling, including legalized casino gambling. We oppose and call for a veto of any budget that relies on expansion of legalized casino gambling of any type or size, whether as a standalone business or partnered with any other business or resort, as a method of finance. We call on all Republican legislators to decline campaign contributions from Table of Contents 13 gambling PACs and lobbyists and oppose any effort from the House leadership or members of the House Calendars Committee to pressure members to vote for expanded gambling. We also call for the repeal of the vague “fuzzy animal” exception to Texas’ anti-gambling laws that allow non-cash prizes to be awarded based on use of eight-liner slot machines. (S) 66. Truth in Taxation: We urge that taxes established for a particular purpose shall not be used for any other purpose. Tax revenue derived from gasoline taxes and all other taxes or fees on our vehicles (including vehicle sales tax) shall only be used for road construction and maintenance, and shall not be diverted to any other use, including mass transit, rail, restrictive lanes, and bicycle paths. (S, L) 67. Public Posting: We support all government entities compiling and publicly posting their current debt, future obligations, financial statements, check registers, and all government contracts on their official websites. (F, S, L) a) Compiled information shall be presented in a table with the principal, estimated interest, and estimated combined principal, interest required for full payment of the proposed bonds, the principal, estimated interest, estimated combined principal, and interest required for full payment of all outstanding bonds, as of the date the political subdivision adopted the debt obligation order. b) Voter approval is for a maximum spend, not a blank check for all projects on a list. An approved bond amount will include principal, interest, and a contingency amount for overruns. 68. Repeal Taxes: We support abolishing the following: a) Estate tax (commonly known as the Death Tax. (F) b) Inventory taxes. (S, L) c) Business franchise taxes. (F, S, L) d) Taxes on phone and internet services. (F, S, L) e) Affordable Care Act Home Sales Tax. (F) f) Personal property tax. (S, L) 69. Federal Taxes: We support elimination of all federal income taxes. Until such time, we support the repeal of the Required Minimum Distribution from qualified retirements accounts and institutions. (F) 70. Let Texans Run Texas: The Texas Legislature shall pass legislation that prohibits political subdivisions of Texas and State agencies from accepting federal funds that violate Texas law. Texas government should resist unconstitutional federal acts, violations of federal border laws, and the use of financial pressure by globalist institutions that seek to destroy everything Texans hold dear and usurp the political power of Texas. Such resistance should include stopping unconstitutional federal acts in Texas, defense of the border by Texas without interference by the federal government, and the prohibition of the use of globalist financial pressure to destroy Texas businesses. (S) 71. Prevent Hotel Occupancy Tax (HOT) Abuse: The Texas Legislature is requested to restrict the Hotel Occupancy Tax for Texas cities to prevent the display, promulgation, and promotion of obscene or pornographic materials and acts. (S, L) School Finance and Property Taxation 72. Elected Officials’ Salaries: State, county, or local elected officials shall not be eligible to receive salary or benefit increases that they vote on. Voters must otherwise approve such benefits by a two-thirds (2/3) majority of those voting, and only if 20% of all registered voters in the district cast ballots. (S, L) 73. Impact Fees and Management Districts: We oppose the creation of management or special purpose districts with the authority to impose taxes and bonded debt, and we oppose the use of eminent domain by these districts. (S, L) 74. Permanent School Fund: We oppose any effort to remove State Board of Education authority over the Permanent School Fund, whose constitutionally intended purpose is to fund SBOE-approved instructional materials. (S) Table of Contents 14 75. Ax the Property Tax: We support replacing the property tax system for businesses and individuals with an alternative other than the income tax, and requiring voter approval to increase the overall tax burden. We demand the Legislature to immediately develop and implement a transition plan that is a net tax cut. (S, L) 76. Property Tax Relief: We support these incremental steps toward the ultimate abolition of property tax: a) Amend the school finance formulas to reduce the Local Fund Assignment so that it is phased out completely in twelve (12) years, funding Tier I entirely from state revenue, and ending the Robin Hood system of school finance. (S, L) b) Replace the appraisal system with a system that values property at the purchase price, thus negating the need for an appraisal district. (S, L) c) Require appraisal districts to publish the amount of property taxes and appraisals attributable to each rental unit. (S, L) d) Limit annual increase on Texas ISD property taxes that Texas ISD property tax revenues shall increase no more than the no-new-revenue rate of 2% per year. This increase shall include the addition of any and all school bond issuances, existing and future, within the school district. (S, L) e) Close the loophole called the “Unused Increment Rate,” which allows taxing entities to bypass recently added limits to increases in property taxes. (S, L) f) The Texas Legislature should require the Texas Comptroller and the Texas Education Agency (TEA) to provide a report, as ordered by the courts, of the spending of funds earmarked for accountability, and to disclose any surplus. (S, L) g) Account for New Minerals so that the New Minerals are treated in the same fashion as New Improvements (that are added to the tax base) when calculating the no-new-revenue tax rate for counties. (S, L) h) No property taxes shall be collected on private properties, including homesteads, fees simple, and land patented lands, and excluding leases, licenses, and permits on publicly owned lands. (S, L) 77. Administrative Bloat Is Not Transformative: We call on Texas school administrators to deliver more education for our dollars, instead of nonstop lobbying for more dollars for education. At a minimum, 80% “Pareto Principle” of revenue should be spent in the classroom. We oppose the underhanded strategy of making cuts to visible frontline teaching positions instead of to administrators and overhead. Total compensation for all administrative staff should be posted publicly. (S, L) Opposition to Market-Distorting Tax and Fiscal Subsidies 78. No Corporate Welfare: We encourage government to divest its ownership of all businesses that should be run in the private sector. We oppose all bailouts of and subsidies to domestic and foreign government entities, states, and for all businesses, public and private. We agree with the Texas Constitution’s requirement for fair and uniform taxation and oppose special treatment or tax breaks for favored industries or companies. We call for repeal or sunset of existing subsidy or special-interest tax exemptions, including the Special Events Trust Fund program, the Texas Enterprise Fund, Moving Image Industry Incentive Program, and lab-grown meat incentives, and now request repeal of Chapter 403.601 of the Texas Tax Code. Tax dollars shall not be used to fund the building of stadiums for professional or semi-professional sports teams, unless otherwise approved by a two-thirds (2/3) majority of those voting and only if 20% of all registered voters in the district cast ballots. (S, L) 79. Ban Public Facility Corporations: The Texas Legislature should eliminate the ability of local governments to set up public facility corporations. These corporations allow local governments to spend taxpayer money without voter approval, thereby increasing debt and taxpayer burden. Table of Contents 15 80. Eliminate Federal Activity: We call upon the Federal Government to stop the following: a) Community Reinvestment Act. (F) b) Funding for the Corporation for Public Broadcasting. (F) c) Ownership of or insurance related to Federal National Mortgage Association (Fannie Mae), Federal Home Loan Mortgage Corporation (Freddie Mac), and SLM Corporation (Sallie Mae). (F) d) College and university loan forgiveness. Failure of loan repayment shall fall to the colleges and universities or their respective endowment fund(s). (F, S) Transparency and Oversight 81. End the Fed: We support abolishing the Federal Reserve. Until that is accomplished, we support additional accountability and transparency for the Federal Reserve System, including regular independent performance audits. (F) 82. Robin Hood Accounting: We direct the Texas Legislature to have the Texas Comptroller of Public Accounts and Texas Education Agency provide a full accounting for the funds collected by recapture “Robin Hood” are spent to comply with the Texas Supreme Court-ordered mandate for Wealth Equalization. 83. Right to Use Gold, Silver, Cash, and Cash Substitutes: We support Texas in exercising its constitutional right (Article 1, Section 10) to declare gold and silver as legal tender, and to authorize the ability to transact, transmit, or exchange such gold and silver bullion by physical and/or electronic means, or written instruction. To further that right, we support legislation for Texas to utilize the Texas Bullion Depository for persons to buy and hold gold and silver bullion and electronically transact with it as money, essentially allowing debit card transactions to spend gold and silver as functional legal tender. The State of Texas should protect that money (and its use) from capital gains taxation at both the state and federal level. No government shall prohibit or encumber the ownership or holding of any form or amount of money or other currency. We support a ban on any form of social credit system based on programmable/ trackable digital currencies including Central Bank Digital Currencies (CBDCs). (S) Education Parents’ Rights 84. School Choice: We support further empowering all Texas families to choose from public, private, charter, or homeschool options for their children’s education and funding which shall follow the student with no strings attached. We oppose regulations on homeschooling or the curriculum of private or religious schools and believe a constitutional amendment should be adopted accordingly. In lieu of funding, citizens may use property tax exemptions. 85. Parents’ Rights in Education: Parents are the primary educators and disciplinarians of their children, to which all other entities are inferior. The fundamental parental rights of parents to make decisions regarding the upbringing and control of their children in all aspects, but especially in all aspects of the Texas education system, shall be recognized, affirmed, and protected by changes to the Texas Constitution and Texas law, including codifying the protections currently existing in the Texas Family and Education Codes. No public service entity nor its agents, district personnel, community partners, guest speakers, or District Board of Trustees shall infringe upon these rights. We call for the development and dissemination of the Parent’s Right to Know and Consent booklet that contains pertinent state and federal law. Table of Contents 16 86. Enforcement of Parental Rights and Students’ Rights: We implore the Legislature to create enforcement measures for violations of Texas laws, while prioritizing parental rights, especially the right to protect one’s own children from harm. Enforcement mechanisms must include investigation and referral authority for criminal, civil, and other disciplinary actions. We urge the Legislature to create an independent Office of Inspector General of Education, appointed by a majority vote of the elected State Board of Education, to investigate waste, fraud, and abuse, particularly sexual abuse of students, violations of parental rights, and student due process rights when they occur in a school setting. Such position shall have authority to refer any matter to the Texas Attorney General for further investigation and prosecution, if warranted. 87. Sexual Education, Health-Related Education, and the Classroom to Clinic Pipeline: We demand that the Legislature pass a law prohibiting the teaching of sex education, sexual health, or sexual choice or identity in any government school in any grade whatsoever, or disseminating or permitting the dissemination by any party of any material regarding the same. All government schools are prohibited from contracting with or making any payment to any third party for material concerning any of the above topics. Until this prohibition goes into effect, sexual education shall only utilize sexual risk avoidance programs and promote abstinence outside of marriage. Before a student may be provided with any health-related instruction, human sexuality, or family planning instruction, the district must obtain the written consent of the student’s parent or guardian. Written consent of the student’s parent or guardian must include the district’s full disclosure of all guest speakers and referral resources that students will be exposed to (opt-in status). 88. Prohibiting Grooming of Minors: We request that the Texas Legislature pass legislation that requires Texas schools, libraries, and contracted vendors to filter and vet inappropriate and harmful content, such as pornography. We support the passage of a law even more comprehensive than the Florida law that prohibits instruction in sexual orientation and gender identity ideology in government schools and libraries. We believe any school employees or contractors responsible for exposing a student to inappropriate material or engaging in inappropriate conduct with students should have their teaching license revoked, forfeit their pension, and be criminally prosecuted with enhanced penalties. We call on the Legislature to prohibit school districts from abusing their District of Innovation status to avoid parental rights’ laws or restrictions on human sexuality education. 89. Healthcare in Public Schools: Legislators shall prohibit reproductive healthcare services, including mental health counseling or referrals to outside health services or clinics, and distribution or demonstration of condoms and contraception through public schools. We support parents’ right to know, and consent to choose, without penalty, which healthcare medications and mental health services are administered to their minor children. We support requiring informed consent of parents before any school-based mental health assessments or interventions are performed; the district must obtain written parental consent. Written parental consent must include the district’s full disclosure of the business or clinic students will be referred to for healthcare or mental health services (opt-in status). 90. Religious Freedom and Government Schools: We demand school administrators and officials protect the rights of students and staff to pray and engage in religious speech, individually or in groups, on school property, without government interference. We urge the Legislature and the State Board of Education to require instruction on the Bible, servant leadership, and Christian self-governance. We support the use of chaplains in schools to counsel and give guidance from a traditional biblical perspective based on Judeo-Christian principles with the informed consent of a parent. Curriculum 91. Instructional Excellence: The educational system should focus on imparting essential academic knowledge, qualities of good citizenship, practical skills, and self-sufficiency. We support, without exception, requiring Texas public schools to be fully transparent with parents regarding everything to Table of Contents 17 which their child is or will be exposed. Curriculum of Instructional Excellence shall include the following: a) Language arts with phonics-based reading instruction, spelling, grammar, classical literature, and cursive writing. b) Civics, which includes passing the US Citizenship and Immigration Services test prior to graduation. c) In mathematics, we support stronger emphasis on the TEKS 2nd and 3rd grade math standards to recall addition/multiplication facts “with automaticity,” which significantly enables students to better comprehend subsequent material. d) In science, we support objective teaching of scientific method, practices, and theories including the complexity of life origins and the changing climate throughout geologic history. These concepts shall be taught as challengeable scientific theories subject to change as new data is produced. Teachers and students shall discuss the strengths and weaknesses of these theories openly, without fear of retribution or discrimination. e) In computer science, we support instruction in programming and introducing discrete math (logic, counting, probability, number theory, recursion, graph theory) to supplement it. f) Social studies, including geography, economics, and US and world history. Such instruction shall focus on American exceptionalism, the benefits of the free-enterprise system, and the consistent failures of socialism and communism. g) We support education in the arts and music and building critical thinking skills, including logic, rhetoric, and analytical sciences. We support quality vocational educational training that imparts skills needed for meaningful post-graduation employment. h) We encourage more participation in academic competitions to complement the curriculum, to think critically and creatively, and to motivate students as they acquire higher levels of knowledge. 92. Founding Documents in High School: We support a high school-level curriculum for the study of American history that is heavily weighted toward the study of original founding documents, including the Declaration of Independence, the United States Constitution, the Constitutional Convention, the Federalist Papers, the Anti-Federalist Papers, Jefferson’s Letter to the Danbury Baptists, and other Founders’ writings. 93. American Identity and Heritage: We favor strengthening our common American identity and preserving our heritage and culture. We reject Critical Race Theory as a Marxist ideology that seeks to undermine the system of law and order itself and reduce individuals to a group identity. To facilitate the appreciation of our American identity, the contrast between freedom and the tyrannical history of socialism/communism throughout history must be taught. Government schools must display the United States and Texas flags, and require the Pledges of Allegiance daily to instill patriotism. Other flags promoting progressive agendas should be prohibited from being displayed at government schools. Students shall have the right to display patriotic items on their person as well as on school property. 94. National Core Curriculum: We oppose the use of national or international standards in the State of Texas (i.e., International Baccalaureate, Common Core, any remnants of CSCOPE, United Nations Inclusion, National Sexuality Education Standards, and SIECUS, etc.). We strongly urge the Texas Legislature to replace the STAAR accountability test with one of the dozens of group-administered achievement tests that meet the Texas statutory and State Board of Education (SBOE) requirements related to the relevance of data used to compute state and national norms for the assessments. We also oppose the modification of college entrance exams to reflect any national core philosophies as well as use of Social Emotional Learning (SEL) programs, and other quasi-therapeutic programs in schools. Any school district that violates state law banning the use of a national core curriculum or standards shall lose all state funding until said curricula or standards are removed and no longer used in classrooms. 95. Illegal Aliens and Foreign Students: We believe the role of the public education system is to educate American citizens. We support the instruction of lawfully present residents and foreign Table of Contents 18 exchange students at cost, but we oppose allowing illegal aliens into government schools and oppose the instruction of core classes in languages other than English. 96. Oversight of Instructional Materials: All instructional materials and services paid for with state funds shall be vetted by the elected State Board of Education. We oppose appropriation of state funds for instructional content that has not been approved by the SBOE. This approval process must include public review, hearings, and the right to have factual errors corrected. We call on local districts to hold public hearings before deciding which instructional materials they will use including supplemental materials and programs. We support policies limiting students’ screen time and focusing on direct instruction. 97. Education on Humanity and Dignity of the Preborn Child: We support requiring Texas students to learn about the humanity of the preborn child, including life-affirming definitions of life and the study of life, the concept that life begins at fertilization, milestones of fetal development at two-week gestational intervals, use of fetal baby models, witnessing of a live ultrasound, viewing the following videos: Meet Baby Olivia, A Glimpse Inside, and Miracle of Life, and (for high school students) the contents of the Woman’s Right to Know booklet. In addition, students should receive instruction on the dignity of human life and the principles of equal protection that were instituted in the Declaration of Independence and the United States Constitution that “We hold these truths to be self-evident, that all men are created equal, that they are endowed by their Creator with certain unalienable Rights, that among these are Life, Liberty and the pursuit of Happiness,” as well as that “no State shall deprive any person of life, liberty, or property, without due process of law; nor deny to any person within its jurisdiction the equal protection of the laws.” Governance 98. Abolish Department of Education: The Department of Education should be abolished because education is not an enumerated power of the federal government. We believe, therefore, the transfer of any of its functions to any other federal agency should be prohibited. 99. Elected State Board of Education: We believe that the SBOE should continue to be an elected body consisting of fifteen members. The SBOE shall be staffed out of general revenue. Their responsibilities must include: a) Overseeing the Commissioner of Education. b) Maintaining constitutional authority over the Permanent School Fund. c) Maintaining authority over curriculum. d) The state adoption of all educational materials. This process must include public hearings. e) Granting, revoking, or amending open enrollment school charters. The SBOE should require charter school operators and board members to be American citizens. f) Teacher and administrator certification. We call for the abolition of the State Board for Educator Certification. 100. School Security and Safety: We support passage of legislation encouraging local law enforcement to provide handgun safety and proficiency training for all educators and allowing LTC (License to Carry) holders to carry concealed firearms on all school campuses for security and protection purposes. 101. Withdraw from Taxpayer-Funded Lobby Groups Like TASB: We believe locally elected school boards have a duty to ensure that the education provided reflects traditional Texas values, and have purview over policy, curriculum, and budget. Local independent school districts should sever all ties with taxpayer-funded lobby groups, including the Texas Association of School Boards (TASB), the Texas Association of School Administrators (TASA), and the Texas Association of Community Schools (TACS). We support prohibiting taxpayer dollars from being used to pay dues or any fees for these groups. Required training now provided by groups like the above-named organizations would instead be under the auspices of the State Board of Education, with funds appropriated for that purpose. 102. Enforcement of Open Meetings: We demand requiring audio or video recordings and publication of minutes for closed sessions with employee and student identifiers redacted and allowing taxpayers to Table of Contents 19 seek civil penalties for government school officials who violate the Texas Open Meetings Act. We believe an open meetings violation should be an affirmative defense to a charge of disrupting a public meeting. 103. Gender Identity Ideology in Schools: The official position of the Texas schools shall be that there are only two genders: biological male and biological female, which are immutable and cannot be changed. We support the total prohibition of so-called social transitioning. We oppose transgender normalizing curriculum, library materials, and pronoun use. We support the passage of legislation that prohibits any course of instruction, unit of study, library materials, instructional materials, or any other curricular or extracurricular offering that adopts, supports, or promotes gender fluidity or transgender ideology in Texas government schools. We support the passage of legislation prohibiting school staff from engaging in sexualized drag activities, crossdressing, or transgenderism. We hold that biological men should compete only against other biological men and that biological women should compete only against other biological women in all school athletics. 104. Parents’ Right to Local Control of Health Education: The Texas Department of State Health Services School Health program, the Centers for Disease Control “Whole School, Whole Community, Whole Child” Coordinated School Health program, and the Texas Education Agency undermine parental influence in education as authorized in Texas Education Code 28.004. The Legislature should defund and abolish Texas Department of State Health Services’ School Health Program, the Texas School Health Advisory Council (SHAC), and repeal Texas Education Code 38.013, Texas Education Code 38.014, and Texas Administrative Code 102.1031. Texas law should remove and prohibit websites, marketing, training, and policy recommendations from Texas Department of State Health Services and Texas Education Agency Coordinated School Health programs to local schools, teachers, and local School Health Advisory Councils. Until the legislature removes sex education from the curriculum of public schools, the State of Texas should adopt changes to Texas Education Code 28.004 requiring: a) At least 50% of the SHAC appointees to be parents of students within the district and who are not themselves or related to district employees. b) At least 50% of the parent officials to be present for any and all business to be conducted and be open to the public. Higher Education 105. College Tuition and Student Loan Reform: College costs are out of control, and reform is urgently needed. We support freezing public spending on higher education until waste and administrative costs are reduced, and we support withholding public funding to state institutions that violate or promote the violation of state law. We support implementation of the following reforms: a) We call for the Texas Legislature to freeze tuition and fees at state colleges and universities. b) We oppose mass cancellation of student loan obligations. We call upon Congress to end the federal student loan program. c) We support requiring universities to eliminate tuition set-asides and any other programs that redistribute the costs of education among students or charge students different tuition rates based on household income. d) We oppose in-state tuition and financial aid for illegal aliens and support the elimination of the TEXAS grant. e) We support limiting international undergraduate and graduate student enrollment at Texas colleges and universities to no more than 5% of the student body. We support a total prohibition on students who are members of the Chinese Communist Party. 106. Defund Political Correctness, Fund and Support Western Civilization Instruction: We believe colleges and universities should reject diversity as understood by the Left and instead promote diversity of thought in order to reverse the Left-wing politicization of higher education. We support the following perspectives and actions at Texas colleges, universities, and secondary schools: Table of Contents 20 a) Like Hillsdale College, we agree that state universities “should value the merit of each unique individual, rather than succumbing to the discriminatory trend of so-called social justice and multicultural diversity, which judges individuals not as individuals, but as members of a group which pits one group against other competing groups in divisive power struggles.” b) We oppose any state funding or graduation requirements for divisive curricula inconsistent with the above, including Marxist, anti-American, Critical Race Theory (CRT), multiculturalism, Social and Emotional Learning (SEL), or diversity, equity, and inclusion courses. c) We oppose using public funds for homosexuality, transgender, or diversity, equity, and inclusion centers, employees, or programs. d) Public universities should be required to create a comprehensive program of instruction in Western Civilization, American Institutions, national and state heritage, Keynesian versus Austrian economics, and free-market liberty principles. Successful participation in this program should be required for graduation. 107. Campus Speech: We urge the Texas Legislature to protect the 1st Amendment rights of those on college campuses in the practice their faith, the formation and governance of their organizations, and in speech. We recognize that vandalism and physical threats to individuals are not protest, and we adamantly oppose riots, encampments, and other activities intended to destroy property or endanger the safety of students and staff. 108. Eliminate Tenure: We support abolishing the system of tenure in academia and advocate replacing it with a merit-based system for teacher retention. 109. Equal Access: All Texas students shall have equal access to all state-supported university admissions, grants, scholarships, and loans, based upon measurable academic criteria. We support the suspension of federal funding from universities that prohibit military recruitment on campus. We support allowing homeschool and private school students to compete as individuals in UIL academic competitions and to be eligible for associated scholarships. The Classic Learning Test (CLT) should be accepted for purposes of admission and weighed in the same manner as institutions weigh scores achieved on the ACT or SAT. 110. Medical Students’ Religious Liberty: All persons have the liberty of conscience and shall be protected under Texas law if they conscientiously object to participating in practices that conflict with their moral or religious beliefs. This includes, but is not limited to, abortion, including any requirement for a medical resident or physician to perform an elective abortion on an opt-out basis instead of an opt-in basis, the prescription for and dispensing of drugs with abortifacient potential, human cloning, IVF, medical contraception, embryonic stem cell research, eugenic screenings, genetic engineering, euthanasia, assisted suicide, harmful futile procedures, vaccines, and the withdrawal of nutrition and hydration. Any state agency, educational institution, or local entity in Texas which desires to send its employees, students, or their representatives out of state to conduct an activity that is illegal in Texas must first submit a request and obtain approval from the office of the Attorney General of the State of Texas using a form developed and approved by the Attorney General’s office. The form shall be used to assist in the determination of whether an activity is legal, and must provide the full identification of the sources, addresses, and amounts of the funding by the individual or entity. Health and Human Services Parents’ Rights 111. Parental Rights and Responsibilities: We support the fundamental God-given right and responsibility of parents to direct and guide their children’s care and moral upbringing. Therefore, we Table of Contents 21 support the constitutional rights of parents to raise and educate their children, including their rights to direct the care, custody, control, upbringing, moral and religious training, and medical care of their children. Local, state, or federal laws, regulations, or policies that limit parental rights in the rearing of both biological and adopted children shall not be enacted. 112. Parental Rights of Dependent Adult Children: As long as parents are responsible for an adult child, through college or the age of 26 when children are on the parents’ insurance, the parents must have access to medical information, grades, and other information normally afforded to parents of minor children. 113. Parental Consent: We insist on informed parental consent for all medical care, counseling, etc., for all minors. Healthcare Independence 114. Medical Freedom: We call for an addition to the Texas Bill of Rights that explicitly states that Texans have the natural, inalienable right to refuse vaccination or other medical treatment. Our personal healthcare decisions are private, and informed consent is a basic human right. Therefore, the following are expressly forbidden even in an emergency or in a pandemic: a) Any attempt to mandate, force, or coerce any medical test, procedure, blood, or product, including vaccines or masks. b) Any attempt to use a citizen’s vaccination status as a condition to maintain or obtain housing or employment or employee benefits, attend school or childcare, or access state services. c) The denial of any public service or benefit based on vaccination status. d) Any mandates by public, private, government, or medical entities for treatment, vaccination, vaccine passports, mask requirements, or health insurance surcharges. e) Any involuntary isolation or quarantine of anyone not experiencing an active contagious infection. f) Any withholding of the risks and benefits of a proposed intervention, including quantifiable adverse effects, that must be equally communicated and accessible to the patient or to a minor patient’s parents or guardian. g) Any prevention of visitation to the ill when risks are acknowledged and mitigated according to patient and visitor choice. h) Any Nuremberg Code violations, including but not limited to the requirement that experimental-use medications, vaccines, or other treatments must provide full knowledgeable consent and be free from any form of coercion or inducement. i) Any tracing of individuals by cell phones or another means for any reason without an individual court-issued warrant. j) Any requirement that a nurse practitioner can only provide healthcare to Texans under a delegation agreement with a physician in the State of Texas. k) Any holding of an individual against their will (or that of their parent or guardian) in a hospital or residential care facility, or preventing an individual from changing their healthcare provider. l) Any jeopardization of the statutorily protected right of Texans to utilize the Texas Exemption from Immunizations for Reasons of Conscience Affidavit without scrutiny or adverse action. m) Any denial of directed or autologous blood donation. 115. Texas Medical Practice Act: To protect the rights of both patients and physicians, the Texas Medical Board (TMB) should adopt the following provisions in the Texas Medical Practice Act: a) Protect the right of patients to choose natural solutions, including chiropractic care, to their health problems, as well as the physician’s right to provide natural solutions for health problems. b) Protect physicians from interference by the TMB or the Texas State Board of Pharmacy in the physician’s treatment plans or prescriptions. Table of Contents 22 c) Eliminate confidential complaints against physicians. d) Eliminate anonymous medical witnesses against physicians. e) Mandate legal due process in all TMB proceedings. f) Allow physicians the right to have a complaint against them tried in a state district court, rather than in an administrative law court. g) Prohibit TMB members from working for insurance, pharmaceutical companies, or hospitals while serving on the board, to prevent conflict of interest. h) Prohibit intimidation tactics by TMB lawyers against physicians. 116. Medication Manufacturing: Medications and prescription drugs consumed in the United States should be manufactured in the United States for security, consistency, and reliability of the drug. We strongly encourage our state to promote private entities to initiate and sustain the buildup of the supply chain and manufacturing of the medical and health products in this state to help reduce the costs and increase the availability of medical products to its constituents. 117. Informative Labels on Medications: We require informative labelling on: a) All prescription and over-the-counter drugs, supplements, and medical supplies. In addition, these medical products must be required to show the country of manufacturing or production. b) All medications and vaccines which contain mRNA or aborted fetal material. c) All medications, therapeutics, and vaccines which have not completed human trial studies. These should be considered to be experimental and potentially harmful, and may not be therapeutic. 118. Secure Medical Records and Informed Consent: We oppose any state or federal medical record computer databases that store personal identifiable records on citizens without their written consent. All medical service providers, including insurance providers, must implement and maintain a high-level technology security safeguard for consumers and patients to prevent financial losses and life-threatening delays in treatments due to breaches in one or more databases, third-party communications, or simultaneous hacks causing egregious delays and financial hardship to doctors and clinics. 119. Right to Try: We urge the Texas Legislature and Governor to enact laws that protect patients’ (or parents’ if the patient is a minor) and their doctors’ rights to have access to experimental or off-label medications and procedures that can potentially be lifesaving or improve quality of life, consistent with the RPT Platform, without the interference of the Medical Board, Pharmacy Board, or hospital boards. 120. Healthcare Savings Accounts: All individuals shall be allowed to establish health savings accounts. Individuals shall be allowed higher annual contributions to health savings accounts. 121. Texas HSA: We recommend the creation of the State of Texas Health Savings Account, with funds in excess of those needed in the Economic Stabilization Fund (Rainy Day Fund) set aside for the purpose of enabling the state to develop reserves sufficient to exit the federal Medicaid program. Funds in this Texas HSA shall not expire nor be utilized for any other purpose. Government-funded Health Programs 122. Parental Safeguard: We support abolishing the Texas Child Mental Health Care Consortium, the trauma-informed care policy, school-based mental health providers, school-based or school-connected mental health interventions, and any other public school programs that serve to expand access to minors. Legislators shall prohibit all reproductive healthcare services in public schools. 123. Welfare Reform: We support the abolition of all federal welfare programs, as they are not an appropriate role of the federal government. Until such time, welfare reform should encourage partnerships with (without providing any taxpayer funding to) faith-based institutions, community, and business organizations to temporarily assist individuals in need. We encourage welfare reform in the following areas: a) Denying benefits to individuals who cannot prove citizenship. b) Reforming welfare programs to require recipients to work, learn, and train to move toward self-sufficiency. Table of Contents 23 c) Reforming welfare programs to require recipients to remain substance abuse free and willing to submit to and pass random drug testing. d) Requiring that money provided through the Supplemental Nutrition Assistance Program (SNAP) be used only for nutritious foods consistent with those included under the WIC program, and be released only with a photo ID of the approved user. e) Implementing a non-monetary-based assistance program for providing supplemental food benefits. f) Removing prisoners from welfare rolls. 124. Child Support Related to Welfare: Mothers applying for government financial support, exempting rape victims, shall provide the verifiable name and any known contact information of the birth father, which information shall be turned over to the State of Texas Attorney General’s Office within 30 days for collection of child support. 125. Oversight of Disability Claims: We call for stronger and more stringent reviews of disability claims to ensure that assistance is provided only to those truly in need. 126. Medicaid Reform: Until we establish the Texas HSA in order to eliminate reliance on federal Medicaid, we support Medicaid block grants to the states and returning Medicaid to its original purpose to be a temporary assistance program for US citizens. We oppose any further expansion of Medicaid and stand in opposition to Medicaid for non-citizens. Medicaid funds shall only be used for genuine biological medical ailments. Medicaid 1115 Waiver allocations must be not be used to pay for the chemical sterilization of teenage girls or to cause abortions, and the funds must be redirected to genuine biological medical needs that will truly improve the lives of young women and authentically contribute to good preconception care, good birth outcomes, and good maternal health in Texas. 127. Medicare Reform: Medicare should be provided only for United States citizens and legal residents and there should be a non-penalized opt-out for those who have health insurance through their employers and continue to work. 128. Patient Protection and Affordable Care Act (“Obamacare”): We demand the immediate repeal of the Patient Protection and Affordable Care Act, which we believe to be unconstitutional. 129. Home and Community-Based Services: In order to avoid costly institutional care and preserve families, we call on the Texas Legislature to enact policies that would support invest in and fully fund home and community-based services (HCBS) as a pro-life policy, and to enact associated policy solutions to protect, preserve, and defend the sanctity and dignity of human life. We urge the Legislature to address any loopholes that fail to protect or provide appropriate home and community-based supports and access to care for children and people with disabilities. In addition, we ask that families be provided with information about life-affirming social and medical services available to them in Texas as alternatives to abortion and costly institutional care. Mental Health 130. Caring for Citizens who Are Mentally Disabled: We urge the Legislature to continue funding and operating all state-supported living centers for mentally disabled legal Texas residents, and to continually seek common sense improvements to increase efficiency. 131. Mental Health: We oppose all mental, emotional, or well-being surveys, screenings, check-ins, assessments, and similar such instruments in public schools and demand that the Texas Legislature ban all psychological or mental health questions, instruction, activities, surveys, and check-ins in any capacity in public schools. If an employee or contractor of the district has the opinion that a child needs to be referred to a mental health professional, he or she shall make such a recommendation to the parent or guardian of the child. We implore the Legislature to require informed parental consent prior to these and any other psychological questions being presented to a student. The Legislature shall enact strict penalties for violation of parental rights regarding school health and mental health services and order the strict enforcement of such penalties. Table of Contents 24 Homosexuality and Gender Issues 132. Homosexuality: Homosexuality is an abnormal lifestyle choice. We believe there should be no granting of special legal entitlements or creation of special status for homosexual behavior, regardless of state of origin, and we oppose any criminal or civil penalties against those who oppose homosexuality out of faith, conviction, or belief in traditional values. No one should be granted special legal status based on their LGBTQ+ identification. 133. Gender Identity: We oppose all efforts to validate transgender identity. We believe gender modification and any form of gender affirming care for minors does not constitute medical care and is, in fact, child abuse. Further, there shall be no attempt to engage in so-called “gender affirming” medical or mental health intervention for persons between the ages of 18 and 26, including: a) Intervening in any way to prevent natural progression of puberty. b) Administering or providing opposite sex hormones. c) Performing any surgery on healthy body parts of that person. d) Assigning name and/or pronoun changes. Any agency, individual, or other entity promoting, performing, or facilitating gender-transitioning or gender-modification of a minor child shall be criminally prosecuted for child abuse and exposed to civil actions, enjoying no immunity regardless of profession, relation, or standing. 134. No Taxpayer Funding for Sex Change: We oppose the use of taxpayer funds for any type of medical treatments or interventions intended to alter sex characteristics. This includes but is not limited to military personnel and inmates in federal, state, or local prisons or jails. No federal, state, insurance, or probate monies may be allocated for the use of such treatment. 135. Counseling Methods: Therapists, psychologists, and counselors practicing in the State of Texas shall not be forbidden or penalized by any licensing board for practicing authentic reparative therapy or other counseling methods when counseling clients of any age with identity disorder or unwanted same-sex attraction. Substance Abuse and Addiction 136. Addiction: We oppose legalization and decriminalization of illicit natural or illegal synthetic drugs, and we support the exercise of a zero-tolerance policy with maximum penalty for manufacturers and distributors of illegal drugs or their precursors. We also oppose any needle exchange programs or supervised drug consumption sites. Faith-based rehabilitation programs shall be considered a part of an overall rehabilitation program. 137. Cannabis Classification: The State of Texas shall retain cannabis as a Schedule I drug. Life-Affirming Health Care Concepts 138. Pornography Crisis: The State of Texas shall recognize that pornography and pedophilia are public health hazards. We call on legislators to: a) Expand protective measures to block incidental or unwanted exposure to inappropriate pornographic material by viewers or users. Violators shall be subject to civil and/or criminal penalties. b) Pass a complete ban on production and public display of pornography within the state. We support restricting the access by minors to internet websites and social media platforms that contain or promote any sexually explicit material. We oppose website regulations that specify fractional judgments regarding content. c) Prohibit the possession, sale, and distribution of “child-like sex dolls,” which are obscene anatomically correct dolls, mannequins, or robots that are used for sexual stimulation or gratification and that have the features of or have features that resemble those of minors. d) Prohibit the advertising, promotion, distribution, and sale of obscene devices to minors or displayed within minor’s reach. Table of Contents 25 139. Conscience Clause: All persons and legal entities have the right of conscience and shall be protected under Texas law if they conscientiously object to participate in practices that conflict with their moral or religious beliefs. This includes, but is not limited to, abortion, the prescription for and dispensing of drugs with abortifacient potential, human cloning, embryonic stem cell research, eugenic screenings, genetic engineering, euthanasia, assisted suicide, harmful futile procedures, vaccines, and the withdrawal of nutrition and hydration. We call on the Texas Legislature to enact additional conscience protections for all healthcare professionals, including medical students, that are all-encompassing, enforceable at the state level, and protect against adverse action and retaliation taken against an individual. 140. Fetal Tissue Harvesting and Stem Cell Research: We support legislation prohibiting and criminalizing the harvesting, sale, and experimentation, or commercial use of human fetal tissue, including those intended for use in vaccines, which requires or is dependent upon the destruction of human life. We encourage adult stem cell research using cells from umbilical cords, adults, and any other means that does not kill human embryos. We also support elimination of public funding for embryonic stem cell research, research on fetal tissue, or human cloning. All products that use embryonic and fetal tissue in their production shall be labeled in the State of Texas to inform consumers, promote alternatives, and affirm the value of human life. 141. Support for the Sanctity of Life: The Republican Party of Texas supports programs that provide assistance to pregnant women and promote the sanctity of life. 142. Preventing Any Death by Abortion: We support legislation such as the Preborn Non-Discrimination Act (pre-NDA) to close existing discriminatory loopholes that fail to protect preborn children suspected of having “fetal anomalies” or disabilities, and we support legislation to enact anti-discriminatory language to apply additional protections to preborn children at risk of being aborted because of their sex, race, disability, or age of gestation. Such legislation should provide families with information about life-affirming social and medical services available to them in Texas, such as perinatal palliative care. We support protecting preborn children and their mothers by stopping abortion pill distributors from sending and trafficking these lethal and illegal drugs into Texas and holding those accountable who break state Pro-Life laws by selling and trafficking illegal abortion pills. 143. Medical Emergencies within Pro-Life Laws: We support the current medical emergency exception laws which include the management of confirmed ectopic pregnancies, which are not to be considered abortions. We do not support inaccurate arguments against abortion which occur due to false and misleading rhetoric. The abortion law does not need to be altered, but implementation does need to be addressed. We urge the Legislature and health agencies to educate and inform medical professionals and the public about the law of medical emergency exceptions. The mother’s life remains the primary consideration in providing emergency care exceptions in the management of ectopic pregnancies and complicated preterm premature rupture of the membrane (PPROM). 144. Planned Parenthood: We support completely eliminating public funding for, or contracts with, Planned Parenthood, other abortion providers, or any of their affiliates. We oppose their digital or physical presence in our schools and other public institutions, and the expansion of their facilities in our neighborhoods. We call for a state law prohibiting governmental contracts with abortion providers and their affiliates. 145. Preserving the Dignity of Human Embryos: We support continued efforts to preserve the inherent dignity of human embryos, including adoption of human embryos and the banning of human embryo trafficking. 146. Life-Affirming Patient Protection: We call for the Texas Legislature to secure due process and the right for vulnerable Texas patients by continuing to reform Chapter 166 of the Health and Safety Code (Texas Advance Directives Act) by: Table of Contents 26 a) Repealing the unethical, unconstitutional, unprecedented, and anti-life 25-Day Rule in Section 166.046, Health and Safety Code, and replacing it with a truly life-affirming law that requires physicians to adhere to a patient’s or surrogate’s medical decision about life-sustaining treatment, and that provides for physicians who disagree with the patient’s decision to transfer the patient to another physician or facility that will honor the decision to continue life-sustaining treatment. b) Improving language that protects Texas patients with disabilities to clarify and strengthen that disability should not be a considered factor. c) Guaranteeing judicial review, ensuring the ability to appeal a hospital committee’s decision and provide impartial legal recourse. Environmental Health 147. Toxic Exposure: We support the immediate implementation of the Toxic Exposure Research Act of 2016, which will ensure that the federal government will establish a database on all exposed veterans and their families and descendants. The State of Texas shall prohibit the addition of sodium fluoride or any other chemicals deemed dangerous, carcinogenic, harmful, or poisonous to community water systems. 148. Less Tech for Little Texans: Any government agency with responsibility for children under ten years of age must limit screen time to thirty minutes for every six-hour span on one-to-one devices and VR headsets. Children ages ten to fourteen must be limited to two hours per day. Many medical research studies have shown significant health adverse reactions to EMF emissions and abnormal behavior disorders. 149. Prohibition of Biocultured Food: The Republican Party of Texas supports the prohibition or manufacture, sale, or distribution of food products made from cultured animal cells, bioengineered, or mesenchymal cell lines. Criminal and Civil Justice Rights and Protections 150. Obscenity Exemption: We urge the Texas Legislature to repeal all laws based on the fraudulent research by Dr. Alfred Kinsey, repeal the Texas Penal Code affirmative defenses in the Harmful Material to Minor’s Statue and sexual performance of a child. Texas should modify the Miller Test to close the loopholes exploited by publishers and digital resources and prohibit taxpayer funding to any entity that permits the presentation or encouragement of sexuality, pornography, or transgender ideology to minor children. 151. Civil Asset Forfeiture: We call upon the Texas Legislature to abolish civil asset forfeiture, independently or in partnership with federal authorities, and to ensure that private property only be forfeited upon a criminal conviction. 152. Government Surveillance: We oppose all forms of warrantless government surveillance of United States citizens and businesses. 153. Location and Data Privacy: We call upon the Texas Legislature to protect citizens’ current and historic technologically available location data by requiring a warrant based on probable cause or a legally obtained subpoena. 154. Hate Crimes: We urge the complete repeal of the hate crime laws, since ample laws are currently in effect to punish criminal behavior towards other persons. 155. Marriage Officiation: We believe religious institutions have the freedom to recognize and perform only those marriages that are consistent with their doctrine. Table of Contents 27 156. Warrant Validity: The filers of search warrants shall be held responsible for the validity of the information used to obtain the warrants. 157. State of Emergency and Pandemic Business Fines: We support prohibiting fines or imprisonment of business owners for operating their business during a state of emergency or during pandemics, and we call for the Legislature to pass these protections into law. 158. Fraudulent Lien Filings: We support legislation designed to reduce fraud by requiring that the identity of the debtor on all liens be confirmed prior to filing by at least three methods, including but not necessarily limited to state ID, county records, and personal contact. 159. Gain-of-Function Accountability: We call for the banning of gain-of-function research in Texas. We support investigations and indictments of those who participate in funding, developing, introducing, or releasing gain-of-function pathogens. We call on a ban of mRNA technology for vaccines in humans, animals, and food. We urge the Legislature to authorize the Attorney General and appropriate law enforcement agencies to investigate and prosecute for treason and murder any and all participants who knowingly release biological weapons or vaccines that cause harm. Courts, Prosecutions, Restitution 160. Penalty For False or Illegal Accusation of Impeachment: If someone brings articles of impeachment and the charges are found to be false, then the accusers shall be held personally liable for legal fees and past wages which shall be reimbursed from the accusers’ personal funds to the defendant. 161. Court Accountability: We support the right to inform the jurors of their common law power to judge law (jury nullification) as well as the evidence, and to vote on the verdict according to their conscience. We believe district attorneys have a duty to seek justice for victims of all forms of crime and oppose policies that systematically decline to prosecute crimes. All judges shall be required to state their final decisions on record, immediately validating and producing a mandatory Finding of Facts and Conclusions of Law for every substantive decision or final ruling in a case, and preserving the complete record in all transactions in court and in chambers, including audio and video. We support providing the Texas Attorney General a process to investigate and remove district attorneys that fail to enforce Texas law. Texas judges, attorneys, and court actors in Texas will no longer have any immunity when violating the law. Strong penalties should be implemented for elected officials found reducing criminal cases to skew records. We urge the Texas Legislature to enact legislation to hold judges and prosecutors accountable for violating their oaths of office or the laws of the State of Texas and to end any explicit or implicit judicial or prosecutorial immunity for their actions. 162. Frivolous Lawsuits: We support further reform to discourage frivolous lawsuits. We oppose the abusive use of class action lawsuits and any law that allows government agencies to collect lawyer fees from the plaintiff when they win, but not have to pay the plaintiff fees when they lose. We call for the legislature to keep the Texas Citizens Participation Act and essential protections for Texans’ 1st Amendment rights. 163. Bail Reform: We call upon the Texas Legislature to ensure bail in Texas is based only on a person’s danger to society, risk of flight, and criminal history. District attorneys and judges will stop setting bails that are egregiously low for the nature of the crime or release violent offenders into the community. If said offender commits a violent crime while out on bail, and bail or release was not granted based on the merits of the alleged violent crime, the district attorney and judge involved in the case are not exempt from penalties or civil suits for enabling offenders. Personal Recognizance (PR) Bonds should only be allowed for first time, non-violent offenders. 164. Human Trafficking and Jurisdiction: We call upon the Texas Legislature to amend the Code of Criminal Procedure to allow victims of human trafficking to have convictions within the previous ten years for prostitution offenses set aside if the victims received these convictions as a direct result of being trafficked. The Texas Legislature shall pass legislation granting the Texas Attorney General full concurrent jurisdiction over multi-jurisdictional cases, to be limited specifically to those cases involving human trafficking. 165. Rule of Law Enforcement: We support rule of law and enforcement of laws, which maintain an ordered republic. We call for independent prosecutorial authority to prosecute crimes that maintain Table of Contents 28 order (such as sedition, riot, official oppression, election integrity, etc.) to be delegated to a statewide officer such as the Attorney General. We call on the Legislature to ensure that election crimes will be promptly prosecuted, even in counties with progressive district attorneys. Law Enforcement 166. Abortion Homicide Exemption: The physician homicide exemption of the Texas Penal Code Section 19.06 should be modified to apply only to non-elective abortions such as required to save the life of the mother. 167. Capital Punishment: Properly applied capital punishment is legitimate and should be reasonably swift, while respecting all due process. 168. Police, Firefighters, and Other First Responders: We support proper funding for robust training programs that provide first responders with intensive and comprehensive physical and academic training in the classroom and on the ground. We ask that the Texas Legislature properly fund preventative and responsive mental health care for the same. We recommend the Legislature increase funding for the Texas Rural Volunteer Fire Department Assistance Program. 169. No-Knock Raids: We call upon the Texas Legislature to improve no-knock warrant procedures to protect law enforcement and the community. 170. Political Policing: We believe that laws should be enforced uniformly, that punishment should meet the crime, and that law enforcement should never be used to target individuals for political purposes. We oppose the targeting of police officers by progressive district attorneys. We support automatic and prompt expunction of law enforcement officials’ records who are found not guilty in a court of law for job-related actions. The district attorney shall provide a letter of denial to prosecute within thirty (30) days. Family Law 171. Equal Parenting: We support legislation providing for equal and consistent parenting (possession and access) for every child, when both parents, one biological man and one biological woman, are fit, willing, and able, as is in the best interest of the child. 172. Child Protective Services: We support reforming or replacing Child Protective Services, and we ask for legislation that would support due process in family court proceedings, oversight of the system, and, if requested by either party, a jury determining the outcome of any case. The Texas Department of Family and Protective Services or any agency, company, or occupation charged with protecting abused or neglected children and its employees will not enjoy any immunity when violating state, local, or federal laws, guidelines, or constitutional rights and protections. 173. Protect Our Children: We support increasing the minimum sentencing for sexual assault on a minor (less than 14 years old) to 25 years and removing the option for parole for such offenders. We support: a) Notifying the victim or victim representative sixty (60) days in advance of any changes to scheduled court proceeding, filing, or continuances. b) Amending Texas Code of Procedures Article 56A.452 to comport with this plank. State Affairs Heritage Preservation 174. Alamo: The Alamo should be remembered and not “reimagined.” Texas’ authority over the Alamo shall not be infringed by any organization, including local governments, the federal government, the United Nations, or UNESCO. Decision-making authority must remain with Texas, and custodians must affirm the significance of the 1836 battle, maintain transparency, and protect all existing monuments, Table of Contents 29 especially the Cenotaph, which shall not be moved. Plaza and Plaza de Valero, the areas located in front of the Alamo Church, shall be dedicated to the public use as an open space. 175. Heritage: We call on governmental entities to protect all symbols of our American and Texan heritage. This includes opposing the removal of the Ten Commandments and other religious symbols, supporting the Pledge Protection Act, establishing penalties for flag desecration, preserving Texas history and sites, and restoring plaques honoring the Confederate widows’ pension fund. We support March 2nd Texas Independence Day being an official state holiday. We support having all schools, state agencies, and public offices celebrate and honor Constitution Day on September 17th of each year or the preceding Friday or following Monday closest thereto. 176. Historical Monuments: We believe that all Texas historical war memorials, including Confederate monuments, shall be protected from future removal or defacement, and we believe that those monuments that have been removed should be restored to their historical locations. We support the continuing allocation of funds that are necessary to preserve the USS Texas as a permanent monument to the ship, her crew of two world wars, and the history of the State of Texas. 177. Military Base Names: Publicly honor the southern heroes and rescind all name changes of our military bases. 178. Honor Our Flags: We appreciate and honor our flags and what they represent, and we strongly advocate for all public schools to display only the United States and Texas flags in every classroom and to begin each school day with the pledges to both. We call for the requirement of raising American and Texas flags on publicly-funded student campuses and the prohibition of other flags (other than United States military flags) to be hung on publicly funded buildings on student campuses. All Texas students attending publicly funded campuses will see American and Texas flags throughout university campuses. Individual Rights and Freedoms 179. Identity Theft and Data Privacy: The Texas Legislature shall expand existing privacy laws, and laws protecting against identity theft by limiting the ways in which internet providers, schools, federal and state government entities, and others may access, collect, store, and use Texans’ electronic communications, documents, metadata, and protected information. 180. Religious Freedom for Business: We support the removal of laws and regulations that are used to force business owners and employees to violate their conscience, sincerely held beliefs, or core values. Properly defining public accommodation as understood in the Civil Rights Act of 1964 requires that we: a) Prohibit any change to the legal definition by any federal, state, or local law to expand government control to restrict any 1st Amendment rights. b) Proscribe any law that requires any private business or individual to create or provide a custom product or service, any kind of expressive work, enter into a contract, or be coerced into any speech that is not their own. c) Prohibit businesses from professing, espousing, or adopting any views on sex, sexuality, gender, or gender identity, other than to guarantee that views and positions on these matters are not used as a basis for denial of access to available public accommodations. 181. Freedom of Speech and Religious Practice: As America is “one nation under God,” founded on Judeo-Christian principles, we affirm the constitutional right of all individuals to worship as they choose. We strongly believe in Religious Freedom and Freedom of Speech. Therefore, we demand: a) The repeal of the Johnson Amendment, which assaults the free speech of pastors and religious organizations. b) Protection of the 1st Amendment rights of any citizen to practice their religion and exercise their right to free speech in the public square, as well as in religious organization affiliations. Table of Contents 30 c) Texas judges and legislators uphold and defend our God-given unalienable rights of religious liberty and freedom of speech to protect against the intimidation and prevention of Christians and other people of faith from exercising these rights. d) Acknowledgment that the Church is a God-ordained institution with a sphere of authority separate from that of civil government, and thus the Church is not to be regulated, controlled, or taxed by any level of civil government. Nor shall services or other church functions ever again be shut down or suspended by overreaching civil authorities under any pretext whatsoever. e) Texas Legislation adopt language to oppose abuse of emergency powers to seize or block bank accounts or to criminalize political speech as a result of protected speech under the 1st Amendment. 182. Gender Identity and Government Policy: We support legislation in the State of Texas that ensures: a) That all public and private restrooms, changing facilities, showers, etc., be segregated based on biological sex. b) Opposition to any attempt to criminalize or penalize anyone for the wrong use of pronouns. c) All government agencies guarantee that views and positions on these matters are not used as a basis to deny access to public accommodations, as defined by the Civil Rights Act of 1964, nor to deny employment, or discriminate in employment decisions, solely on the basis of a person’s views on these matters. d) Mandating adherence to sex identification on all official documents that is in alignment with biological sex. Male is defined as the sex having the capacity to produce small gametes; female is defined as having the capacity to produce large gametes. Family and Gender Issues 183. Human Sexuality: We affirm God’s biblical design for marriage and family between one biological man and one biological woman, which has proven to be the foundation for all great nations in Western Civilization. We oppose homosexual marriage, regardless of state of origin. We urge the Texas Legislature to pass religious liberty protections for individuals, businesses, and government officials who believe marriage is between one man and one woman. We oppose the granting of special legal entitlements or creation of special status for sexual behavior or identity, regardless of state of origin. We oppose any criminal or civil penalties against those who oppose non-traditional sexual behavior out of faith, conviction, or belief in traditional values. 184. Protect Minors Until Age of Consent: A law shall be enacted to protect the rights of the individual until the age of consent is reached, to include: a) Prohibiting social transitioning or other treatments. b) Protecting against predatory sexual behaviors including, but not limited to, public “Drag Queen Story Hour“ sessions. c) Prohibiting the desensitization of children to sexual topics by inappropriately exposing them to, or normalizing of, sexual behavior. d) Requiring the disclosure of the above offenses to parents or guardians. 185. Definition of Marriage and Family: We support the definition of marriage as a God-ordained, legal, and moral covenant only between one biological man and one biological woman. Further, we support a traditional definition of family with only one biological man in the role of father and one biological woman in the role of mother. We are opposed to same-sex parenting, intentionally subjecting a child to the loss of their biological father or mother, and other non-traditional definitions of family. 186. State Authority Over Marriage: We support withholding jurisdiction from the federal courts and nullifying federal Executive Branch rules, orders, regulations, or licensing requirements in cases involving family law, and especially any changes in the definition of marriage. 187. Nullify Unconstitutional Ruling: We believe the Obergefell v. Hodges decision, overturning the Texas law prohibiting same-sex marriage in Texas, has no basis in the Constitution and should be nullified. Table of Contents 31 188. Spousal Benefits: We shall not recognize or grant to any unmarried person the legal rights or status of a spouse, as defined in Principle number 6 of this Platform, including granting benefits by political subdivisions. 189. No-Fault Divorce: The Texas Family Code shall be completely rewritten with regards to No-Fault Divorce and Child Custody. Suits related to these topics shall be delineated in such a way as to remove the need for any but the most minimal judicial interaction, and promote the maintenance of the traditional family via required intervention or counseling prior to any decree of divorce. We urge the Legislature to rescind unilateral no-fault divorce laws, to support covenant marriage, and to pass legislation extending the period of time in which a divorce may occur to six months after the date of filing for divorce. 190. Adoption: We encourage the Texas Legislature to remove as many barriers to adoption as possible and to make the process less intrusive while protecting children’s safety and best interests. We urge the Texas Legislature to adopt the following steps to promote adoption: a) Expand community-based care: Increase partnerships with local private and nonprofit charitable organizations to create a safer and more responsive system. b) Require a guilty verdict before placing individuals in the Central Registry: Currently an unsubstantiated accusation of abuse or neglect can lead to Texans’ being listed in a government database and cause them to lose jobs and suffer other penalties without ever being found guilty by a court. In some cases, even those found innocent remain listed on the database. No Texan shall be deprived of liberty or their right to earn a living without a fair hearing. c) Expand service options: Improve care for families of children at risk of entering foster care by providing more choice and flexibility for family preservation services outside of the current state-contracted services. d) End hidden foster care: Many families are threatened and coerced into giving up custody of their children without ever going to court. Texas shall protect parents’ rights and end coercive agreements by limiting CPS’s ability to separate families without court oversight. e) Repeal anonymous reporting: False reporting of families to CPS can lead to great harm and a large waste of resources. False reporting can be prevented while increasing the accuracy of reports to CPS by eliminating the option of anonymous reporting in favor of confidentiality. Professionals who must report suspicious activity shall be trained on alternatives to filing a CPS report and permitted to refer struggling families to community service providers. f) Make the adoption process more affordable, streamlined, and accessible. g) Provide more support to adoptive children and biological and adoptive families, such as access to affordable mental health care. h) Work to destigmatize adoption for both birth mothers and potential adoptive families. i) Increase funding to foster care background checks and wellness checks, including interviewing children in foster care. We urge communities and people of faith to promote adoption and, for those not called to adopt, to offer assistance to families that can. We believe that, in the best interests of the family and child, the State of Texas should allow children to be adopted only by married or single heterosexuals. 191. Child Rights: We call on the Texas Legislature to pass legislation to protect privacy in public schools and government buildings by ensuring that multi-use facilities, including showers, changing rooms, and bathrooms, are designated for and used only by persons based on the person’s biological sex. 192. Detransitioners: We support detransitioners and desisters in their pursuit of establishing a healthy lifestyle after being harmed by the gender identity industry, including: a) The extension of statute of limitations for medical malpractice suits from two years to ten years, or until the individual is 26 years old, whichever is longer, for persons who underwent social or medical “transition” as minors and have suffered mental or physical damage as a result. Table of Contents 32 b) Requiring health insurance companies that fund social or medical “transition” procedures or therapy to also fully fund the expenses incurred due to deleterious side effects and treatment necessary to detransition or desist safely. 193. Keep Prisons and Jails Single Sex: The Texas Legislature should mandate that juveniles and adults detained in jails, detention centers, or prisons be housed according to biological sex. Any search of an inmate or detainee which involves physical contact must be performed by an officer whose biological sex matches the biological sex of the inmate or detainee. Pro-Life Issues 194. Pro-Life: Abortion is not healthcare, it is homicide. Until the abolition of abortion is achieved, we support laws that restrict and regulate abortion, including but not limited to: a) Parental and informed consent, including the elimination of judicial bypass. b) Prohibition of licensing, liability, and malpractice insurance for abortionists and abortion facilities. c) Prohibition of financial kickbacks for abortion referrals. d) Prohibition of late-term abortions. e) Prohibition of abortions after the time an unborn child’s heartbeat is detected. f) Prohibition of the manufacturing, importation, sale, dispensing and use of abortifacients. The state may enforce criminal penalties, while private individuals may enforce civil penalties against companies and suppliers of online sales and delivery of any form of abortifacients. g) Elimination of causes of action for “wrongful birth.” h) Health insurance coverage for abortion services and abortifacients, which under Texas law shall be considered supplemental coverage and billed to the beneficiary. i) Criminal penalties be attached to any entity convicted of selling body parts of aborted children or, excluding the mother, of conducting an illegal abortion. j) Extending the private cause of action used in the Texas Heartbeat Act to all pro-life laws and policies in Texas. k) A Legislative study on the effects of abortion numbers post-Roe, and to discuss a better solution than the current failing legislation in effect. l) Supporting the right of Texas municipalities to protect mothers and their preborn children in their communities by passing enforceable city ordinances that further ban abortions within their city limits, closing loopholes in state abortion laws. 195. Abolish Abortion: We urge the Texas Legislature to enact legislation to abolish abortion by immediately securing the right to life and equal protection of the laws to all preborn children from the moment of fertilization and to oppose legislation that discriminates against any preborn children and violates the United States Constitution by denying such persons equal protection of the laws, and to adopt effective tools to ensure the enforcement of our laws to protect life when doctors or district attorneys fail to do so. 196. Inviolability of Life and Fundamental Right to Life: All innocent human life must be respected and safeguarded from fertilization to natural death. Therefore, the unborn, the aged, and the physically or mentally challenged have a fundamental individual right to life which cannot be infringed. We respect the uniqueness of human life and oppose practices that corrupt human DNA, mix human and animal DNA, or other transhumanist initiatives that do not respect the sanctity and uniqueness of human life. All humans are endowed by their Creator with sovereign rights of ownership of their person and DNA, regardless of any DNA modification, and claims to the contrary are invalid. Land Use 197. Property Annexation and De-Annexation: Homeowners and landowners in an area proposed for annexation shall have the right to vote to approve or reject the annexation, including infill development zones (IDZ), regardless of the population of the county. No annexation can occur within 45 days of any Table of Contents 33 election. In any city with a population of over 250,000, any neighborhood can elect to de-annex and reincorporate as a new city – for any reason – provided they receive at least 50% plus one vote of those voting in a regular election and living within the defined de-annexation boundaries. 198. HOA Governance: We support legislation restricting the power of HOAs: a) Informed consent: buyers must sign stating they read every page of the document, from page one to the end of the document. b) No authority to foreclose. c) No authority to restrict the flying of the United States flag. d) No restrictions on rainwater collection. e) No changes to terms without 85% consent of the home owners and property owners. f) Developers must turnover HOA once 70% of the original lots or land are purchased. g) Be obligated to abide by open meeting rules. h) Practice financial transparency. i) Only owners who live or own adjacent or within two lots can file complaints. j) Abolish any and all extra-judicially enforced policies. 199. Property Rights: Property ownership and free enterprise, the foundation of our collective wealth, must not be abridged or denied by government. We support legislation to protect these bedrock rights. Areas of concern are: annexation eminent domain (including foreign entities) property forfeiture natural resources and conservation easements extraterritorial jurisdiction seizure for public or private development nationalization of lands preservation of our 4th Amendment right to privacy groundwater and mineral rights Property owners shall be notified of their rights regarding condemnation, annexation, or easement, and the condemner shall be required to petition a court of jurisdiction to show public necessity. Taking of property shall result in immediate compensation of fair market value to the owner. These issues shall be administered by elected officials accountable to voters. We strongly encourage amendments to the existing Texas laws to further protect the homeowner’s property rights against squatters with minimal legal intervention. No state, county, municipality, HOA, or other entity can restrict a landowner or renter from harvesting rainwater on their property or the property they rent. 200. Eminent Domain: The use of eminent domain must exclude the seizure of private property for private economic development or increased tax revenue, in addition to compensation for damages and the value of the taking, and eminent domain laws should be restructured to include negotiated annual payments in perpetuity for the easement across the property that is affected payable from the company or corporation acquiring the easement. These provisions should also include toll roads and high-speed rail lines. Landowners who prevail in eminent domain cases disputing the taking of their property or the property value shall be reimbursed for all attorney’s fees, expert witness fees, appraisal fees, and costs. 201. No Part of Texas to be Owned by Foreign Interests: a) The State of Texas shall prohibit the sale of all real estate interests within its borders to all except American citizens (defined by current statute) or United States owned and operated corporations. The sole exception to this rule is that a lawful immigrant may own one single family residential property and/or small business. b) All land in Texas that is currently owned by foreign individuals, corporations, or interests with ties to such, will be required to sell their property and vacate immediately. Land that is within 50 miles of a military base, and owned by interests related to all foreign governments, or any other foreign state considered hostile to the United States, will be seized immediately under Eminent Domain. Our security is not for sale. c) Farmland in active agriculture service cannot be purchased and allowed to become inactive. The current trend of the rich to purchase land for the purpose of decommissioning it to restrict the amount of food available to the public must be stopped. The penalty for this type of treason shall include the loss of the property. Table of Contents 34 State Governance 202. Campaign Contribution Limits: We urge immediate repeal of all limits on campaign contributions by American citizens to the candidates or causes of their choice. However, the State of Texas should pass legislation for non-federal elections which bans campaign contributions and expenditures that originate from outside the State of Texas, including those by individuals, organizations, and political action committees, and including those on any questions, propositions, and amendments on any ballots. 203. Texas Independence: The Texas Legislature should pass a bill in its next session requiring a referendum in the next General Election for the people of Texas to determine whether or not the State of Texas should reassert its status as an independent nation. This referendum should be a legislative priority. 204. Marijuana Remains Illegal: We oppose the legalization of recreational marijuana and we support offering opportunities for drug treatment before penalties for its illegal possession or use. 205. Gun-Free Zones in Texas: There shall be no gun-free zones in Texas. Any public access property owner who demands law-abiding citizens disarm themselves shall assume liability for injuries they incur while on the “posted premises.” 206. Daylight Savings Time: Texas should no longer participate in Daylight Savings Time. 207. City of Austin Management: We support the revocation of home rule status for the city of Austin, and transfer of its management to the State of Texas. 208. Digital Passport: The Texas Legislature should ban all digital passports and digital identification. Government and Election Integrity Government Operations 209. Sexual Harassment: We believe sexual harassment by elected and appointed officials should not be tolerated. 210. Government Authority: We believe any government authority that has the ability to levy a tax or fee or to appraise property on the people should be accountable to those who pay the taxes via the electoral process, from the local level to the federal level. 211. Federal Land Disposition: All Federal lands shall be turned over to their respective states, except for land specifically authorized in the Constitution (military bases, federal buildings, post offices). 212. Tax-Funded Lobbying: We oppose using tax dollars to hire lobbyists or paying tax dollars to associations that lobby the Legislature, in particular by public schools. 213. Texas Speaker of the House, House Committees, and Legislative Quorum: In the Texas Legislature: a) We oppose the use of pledge cards and call for Republican members to caucus after each November General Election to determine, by secure secret ballot, their candidate for Speaker and Speaker Pro Tempore. We also call for Republican members to vote as a unified body for their selected Speaker and Speaker Pro Tempore candidate when the Legislature convenes in regular session, provided that the individual selected in caucus for speaker publicly pledges to comply with the entirety of this plank. b) Texas House standing committees shall advance the conservative grassroots agenda, not those of special interests and lobbyists. The chairman and a majority of members of key committees shall support the conservative agenda. All committees shall be chaired by Republicans when in the majority. Table of Contents 35 c) Any legislator who purposely refuses to attend a legally scheduled session for the purpose of denying a quorum shall lose any chairmanship, vice-chairmanship, and committee membership to which he/she may have been appointed. The individual shall lose any salary for that absence period. Additionally, the individual shall pay back any earnings and cost of living allowances earned during that period. 214. Caucus Priorities: We urge the Texas Senate Republican Caucus and the Texas House Republican Caucus to adopt and publish a list of legislative priorities before convening each regular session of the legislature. These priorities shall be in line with those of the Republican Party of Texas. 215. Timely Legislative Action: We call for changes to the House Rules and legislation that will improve timely legislative action in the Legislature including: a) Legislation to remove the constitutional provision that the Legislature cannot take action until sixty (60) days after convening. b) Requiring the House Calendars Committee to vote on a bill within fourteen days of receipt, or it is automatically put on the calendar. c) Automatic setting of a bill on the calendar within seven days if it is cosponsored by a simple majority plus one of House members. d) Votes on each bill individually that is not set on the calendar, rather than killing a slate of bills at once. e) Requiring assignment of all bills to a committee within seven days of the filing deadline or within two days of receipt from the opposite chamber. f) A petition process for legislators to petition the House Speaker to have the ability to move a bill out of committee by getting a prescribed number of signatures of the Representatives and having the bill sent to the floor with no changes within one week of presenting the petition that contains the prescribed number of signatures. 216. Unelected Bureaucrats: We oppose the appointment of unelected bureaucrats and we support defunding and abolishing the following departments or agencies: Alcohol, Tobacco, Firearms and Explosives (ATF) Centers for Disease Control (CDC) Housing and Urban Development (HUD) Commerce Federal Deposit Insurance Corporation (FDIC) Education Interior (specifically, the Bureau of Land Management) Energy National Food and Drug Administration (FDA) Health and Human Services (HHS) Occupational Safety and Health Administration (OSHA) Labor The Environmental Protection Agency (EPA) Labor Relations Board Transportation Security Administration (TSA) The Internal Revenue Service (IRS) and any other federal agency or department that is not authorized by the Constitution. In the interim, executive decisions by departments or agencies must be reviewed and approved by Congress before taking effect. We believe the people of Texas should elect their own Secretary of State and Texas Education Agency (TEA) Commissioner. 217. Sunset Commission: We support a majority citizen-led Sunset Commission. Elections 218. Bonds Create Bondage: State and local bond election ballots shall be required to include the amount of debt currently outstanding, current debt service payments, current per capita debt obligations, the amount of new debt being proposed, estimated debt service for the new debt, and estimated per capita burden being proposed. The bond issue must obtain a two-thirds (2/3) affirmative vote of at least 20% of registered voters in the voting jurisdiction. No public funds are to be spent Table of Contents 36 influencing a bond election. We oppose bundling of items on bond election ballots and “rolling polling“ for bond and tax rate increase elections. Any bond election, at any level of government in Texas, must state on the ballot, “This is a tax increase,” in bold print. 219. Pay-to-Play Endorsement Slates: We oppose “pay-to-play“ endorsement slates where a candidate has paid or promised to pay to be included. We oppose the use of the Republican brand and logos by private Political Action Committees (PAC) that endorse in the primaries. 220. In-Person Election Voting: In-person voting shall be conducted as a single period of time of no more than three days with no time gap between the first day of voting and Election Day. Election Day statutes must be used for the entire voting period. 221. Fair Elections Procedures: We support the right of eligible voters to cast a ballot in each election once, but we oppose illegal voting, illegal assistance, or allowing votes by ineligible persons. We support: a) Allowing, by open records request, a bit-by-bit forensic imaging of all electronic devices, including servers, at Central Counting immediately before and after each election. b) Vigorous enforcement of all election laws as written, and we oppose any laws, lawsuits, and judicial decisions that make voter fraud very difficult to deter, detect, or prosecute. c) Voter Photo ID. d) Prohibition of remote electronic and internet voting for public office and any ballot measure. e) That mail-in ballots must be requested and only granted to voters who cannot physically appear in-person, that the request must utilize the official application form only, and the ballots must arrive before 7 p.m. on Election Day (with the exception of APO addresses). Mail-in ballots should not be separated from their carrier envelopes until the polls close. f) Felony status for willful violations of the Texas Election Code. g) The constitutional authority of state legislatures to regulate voting, including disenfranchisement of convicted felons and the opposition to the federal takeover of our elections. h) Changes to the appropriate sections of Texas law that would deny or cancel a homestead exemption, driver license, and License to Carry if the addresses on those documents DO NOT match the address on the voter’s registration. i) Consolidating elections to Primary, Runoff, Special Called, and General Election days and locations. j) Sequentially numbered and signed paper ballots that contain anti-counterfeiting measures and are accounted through strong chain of custody procedures. k) An amendment to the Texas Constitution to restore authority to the Texas Attorney General to prosecute election crimes. l) The ability for civil lawsuits to be filed for election fraud or failure of officials to follow the Texas Election Code. m) Allowing trained poll watchers from anywhere in Texas with local party or candidate approval. n) Creating processes that will allow rapid adjudication of election law violation disputes as they occur and before violations can be successfully perpetrated. o) Direct prohibition on all types of preferential or ranked choice voting systems in ALL elections to include school board, municipal, county, statewide, and federal elections. p) Striking Chapter 43.007 Countywide Polling Place Program in the Texas Election Code and requiring PRECINCT ONLY voting for any in-person voting with the using of paper poll books only. q) The use of precinct-level, county-level, and state-level vote count verification processes and statistically based randomized audits. r) Expanding write-once data memory cards from Central Counting Stations into precinct-level ballot counters (SB 1661, 88th Regular Session). Table of Contents 37 s) Having judges and DPS officers as election marshals that are trained in election law and making them available during elections to hear and resolve election-related cases, to include candidate eligibility. t) Requiring persons who utilize the Reasonable Impediment Declaration to provide their thumbprints on the form for purposes of later verification. u) Requiring the Texas Election Code to limit the number of voters that a voter assistant can assist to no more than three voters per election. This stipulation includes the entire duration of the voting period. v) Requiring all State and Local Candidates to post a statement of their criminal history regarding any class A misdemeanors and felony offenses on their websites while they are running for public office. 222. Voter Registration: We support restoring integrity to the voter registration rolls and reducing voter fraud by: a) Repealing all motor voter laws. b) Requiring voters to re-register if they have not voted in a five-year period. c) Requiring photo ID proof of citizenship with the voter registration application for all registrants. d) Retaining the 30-day registration deadline. e) Conducting periodic checks on the voter rolls to ensure all currently registered voters are eligible, with penalties for election officials not performing the checks. f) Giving the Secretary of State enforcement authority to ensure county registrar compliance with Secretary of State directives. g) Revising Title 19 funding to avoid incentivizing retention of ineligible voters. h) Promoting a collaboration between states to ensure accuracy in Texas voter rolls without an interstate cross-check system. i) Having all Texas counties integrate into the Texas Secretary of State voter roll platform. No Texas county shall use a third party to manage, in any form, that county’s voter roll. 223. Campaigning at Poll Sites: We condemn the illegal actions of any government entity that does not uphold Texas Election Law on free speech at polling sites outside of the existing boundaries. The right to campaign, including the display of signage, at an appropriate distance (100 feet) from the polling place shall not be infringed. 224. Voting Rights: We support equal suffrage for all United States citizens of voting age. We oppose any identification of citizens by race, race, origin, creed, sexuality, or lifestyle choices, and oppose use of any such identification for purposes of creating voting districts. We urge that the Voting Rights Act of 1965, codified and updated in 1973, be repealed and not reauthorized. 225. Closed Primary: We support protecting the integrity of the Republican Primary Election by requiring a closed primary system in Texas. While we welcome people to join the Republican Party who support limited government and traditional values, we oppose campaigns to get liberal Democrats to cross over and move the Republican Party to the left in the Primary. 226. Redistricting: We support drawing districts based on eligible voters, not pure population. We believe districts should be geographically compact when possible. We oppose any redistricting map that is unfair to conservative candidates in the Primary or the General Election. 227. Republican Party Operations: We oppose all legislative actions that limit the Republican Party of Texas (RPT), the County Executive Committees (CEC), and subcommittees over the ability to adopt bylaws on calling meetings, filling vacancies, or administrating party accounts and contracts. Moreover, we support removing the Republican Party of Texas from the Election Code, which allows for closed primaries and for making and enforcing its own rules. We support the current framework in which the State Republican Executive Committee (SREC) — Chairman, Vice Chairman, and members from the Senatorial Districts — are elected by the delegates at the State Convention. Table of Contents 38 228. Hand-Counting Procedures: We support the rights of counties that are willing and able to competently and efficiently implement voting procedures that do not require the use of machines, and support implementing hand-counting procedures that are more efficient, verifiable, secure, fully auditable, and transparent, with video of each ballot and each counting station, through appropriate changes to Chapter 65 of the Texas Election Code. National Defense and Foreign Affairs Veterans Affairs 229. Preservation of Military Honor and Integrity: We believe that the honor and integrity of our active duty and military veterans must be protected. False accusations of Stolen Valor are harmful and unjust, damaging the reputations of those who have honorably served our country. To deter such malicious actions, we propose that making false accusations of Stolen Valor be classified as a criminal offense, with penalties equivalent to those for defamation. These changes will uphold the respect and dignity our military personnel and veterans deserve, ensuring that their sacrifices are honored and safeguarded. 230. Support of Our Armed Forces: The men and women who wear our country’s uniform, whether on active duty or in the Reserves, National Guard, or Texas State Guard, are the most important assets in our military arsenal. All current and prior military personnel and their families must have the benefits, healthcare, housing, education, and overall support they need. Injured military personnel deserve the best medical, mental health, and rehabilitative care our country has to offer. The Texas State Guard should receive the funding necessary for equipment, uniforms, healthcare, and pay needed to accomplish the mission of the Texas Guard motto, “Texans Serving Texas.” Department of Veteran Affairs monetary benefits shall match present national price index value in all programs. We support giving veterans true choice in healthcare by granting veterans insurance to see the doctors of their choosing, eliminating wait time deaths, suicides, misdiagnoses, and overall negligent healthcare. 231. Texas National Guard Benefits: The Legislature shall provide parity of benefits to the Texas National Guard, regardless of whether it is activated under state or federal orders. Parity would include, but not be limited to, the same benefits provided by the Hazlewood Act, and also the same benefits as, or equal benefits to, those from the VA that would be provided by federal deployment orders. 232. Eliminate Illegal Immigration Magnets: The State of Texas shall impose a 100% surcharge on remittances from an illegal alien to a recipient in a foreign nation. The state shall refuse to grant a birth certificate to any child born to illegal alien parents on Texas soil, shall require proof of legal residency for obtaining a Texas driver’s license or for enrolling in public school, and shall require proof of citizenship for obtaining voter registration and public benefits. Texas shall require all employers to screen new hires through the free E-Verify system to prevent the hiring of illegal aliens and of anyone not legally authorized to work in the United States, and to protect jobs for American workers. No tax dollars shall be provided for social or educational programs for illegal aliens. All existing laws providing for in-state tuition and non-emergency medical care shall be rescinded. All unverifiable foreign-issued identification cards shall be legally invalid in the United States. In addition, the State of Texas should ban the sale of land to illegal aliens and pursue litigation against developers who induce illegal immigration into Texas through marketing. 233. Military Readiness: We support a military force of sufficient strength and readiness to deter any threat to our national sovereignty or to the safety and freedom of our citizens. We oppose gender theory; diversity, equity, and inclusion (DEI) training and re-education; other social engineering topics; and the permitting of transgendered persons to serve in the military – elements that are poisoning our nation’s military effectiveness. We oppose expanding Selective Service to include women, as well as the lowering of standards for combat roles. Table of Contents 39 234. Cybersecurity: As foreign and domestic threats to cybersecurity evolve, the State of Texas must upgrade systems and system security, including conducting regular audits to identify and address vulnerabilities to meet these threats and share threat intelligence data among levels of government. The integrity of our state and local systems and infrastructure must be maintained. 235. Keeping America Safe: We recognize that many of the threats our nation faces are not from foreign nation states, but militant organizations like the Latin American-based drug cartels which engage in cyberattacks, drug smuggling, human trafficking, and other illegal activities that endanger our nation, our military, and our citizens. We support aggressive military and law enforcement actions to combat these organizations anywhere they pose a physical threat to American citizens. 236. Combating Illegal Immigration: We recognize illegal immigration as the greatest threat to American security and sovereignty. To combat the invasion at our border, we demand that both the federal government and the State of Texas enforce laws to their fullest extent, implement an entry/exit tracking system for all visa holders and biometric tracking of all inadmissible aliens, end Catch and Release by requiring the federal government to keep all suspected illegal aliens in custody pending final determination of immigration status, and devote all available resources to deportations. 237. Non-citizen Deportation Plank for Violent Protest or Riot: The State of Texas shall enforce strict deportation procedures for any non-citizen, including those on work or student visas, who are arrested for participating in a violent protest or riot. Upon arrest, these individuals shall be immediately processed for deportation to their country of origin. Those deported under this policy will be permanently classified as “inadmissible,” forever banning them from reentering the United States. Additionally, non-citizen students arrested for such activities will be expelled from their educational institutions without refunds and processed for deportation as described. 238. Putting American Citizens First in Immigration Policy: Our federal immigration system should be reformed to serve the interest of American citizens first. Accordingly, we support the implementation of a merit-based system that ensures the total number of new immigrants should be limited to a level that facilitates assimilation. In light of the millions of illegal aliens and unlawful expansion of mass immigration into the United States, a net migration moratorium should be enacted. We support the repeal of the H1-B visa program, ending the Diversity Visa Lottery, and preventing chain migration in the interest of protecting American jobs, technology, and national security. We support additional court resources to expedite final determination of status and deportation for both violent and non-violent illegal aliens and visa overstays. We call on Congress to end the Refugee Resettlement program and enter into Third Safe Country agreements for the purpose of lowering the number of asylees coming to the United States from around the globe. Border Security and Immigration 239. Border Security Funding: We expect both the Texas Legislature and the United States Congress to prioritize the allocation of funds to effectively secure the border through whatever means necessary, including, but not limited to barriers, a border wall, and/or fences everywhere along the border where they are feasible and useful, and appropriate numbers of personnel and technology over land, sea, and air shall be made available. Texas shall seek repayment from the Federal Government for expenses incurred in securing its international border. 240. State Self-Defense: We urge the Texas Legislature to invoke Article 1, Section 10, Clause 3, of the United States Constitution, also known as the “state self-defense clause,” which asserts that under an active invasion (as defined or declared by the Governor of the State or Texas Legislature), the sovereign state of Texas has the authority and duty to defend Texas citizens, by any and all appropriate measures, against “imminent danger” without delay. Texas shall take these measures as a sovereign state when it determines it is necessary to defend its territory from such assaults. We call for the Texas Legislature to create a Border Protection Unit with the authority to repel illegal crossings and deport apprehended illegal aliens. We must immediately equip the Texas Military Department with the necessary tools and authority to serve and protect our Texas territory and citizens. Table of Contents 40 241. Interstate Border Compact: We urge the Texas Legislature to establish effective interstate compact(s) with other states for the purpose of securing the Texas portion of the United States border and enforcing immigration laws. 242. End Sanctuary Cities, Require 287(g): The Texas Legislature should prioritize legislation to require all law enforcement entities within the state to participate in the United States Immigration and Nationality Act, Section 287(g) program, which allows local law enforcement officials to cooperate with federal immigration agents. State and federal funds shall be denied to any public or private entity, including but not limited to sanctuary cities that are not compliant with immigration laws. The State of Texas shall prosecute the responsible elected officials of sanctuary cities, counties, or states for obstruction of immigration laws. 243. Aid to Illegal Aliens: We call for prohibition of federal or state funding to any entity or organization that provides material aid or benefit to illegal aliens, and to revoke tax-exempt status and/or business licenses of any entity or organization found to have done so. Foreign Affairs 244. Defeating the Globalist Agenda: We recognize that we are living in a time of geopolitical upheaval and unprecedented complexity of threats to our liberties, constitutional governance, and national sovereignty. These threats emanate from “globalist” agents both within and outside our borders. The United States is a sovereign nation founded on the principles of freedom. We reject any assertion of authority over our nation or its citizens from foreign individuals or entities, such as the World Economic Forum, World Health Organization, and the United Nations. We stand firmly against the concept of a One World Government or The Great Reset. 245. Dependency on Foreign Nations: We call upon the United States to re-examine our trading relationships based on America’s economic and foreign policy interests in order to eliminate dependency on adversarial nations, such as China, for critical medical, technological, energy, and other vital resources. Critical infrastructure such as internet, cabling, electrical, power, and water facilities shall not be owned by companies with ownership in adversarial nations. 246. Western Pacific: We call upon the United States to move towards diplomatic recognition of Taiwan as an independent nation and renew our commitment to defend our security and vital economic interests in the Western Pacific region in the face of China’s military provocations, which threaten its neighbors and critical maritime trade routes. We call upon the United States to continue an effective containment strategy of North Korea. We hold that Taiwan, Japan, Australia, New Zealand, the Philippines, and South Korea are critical economic and strategic partners with the United States. 247. Europe: We value our historic and strategic alliance with our European partners. It is critical that European nations take increased responsibility for their individual and collective self-defense by fulfilling their financial and military obligations to the NATO Treaty, thereby ensuring the sustainability and effectiveness of the alliance. We call on the United States to reevaluate its role and participation in NATO, ensuring that our commitments are aligned with national interests and the fair distribution of defense responsibilities. If those countries do not promptly implement such changes, we call on the United States to create new alliances with like-minded, willing, and capable nations to replace NATO. Furthermore, we urge the United States to cease all military engagements, across all domains of warfare, in the Ukraine-Russia conflict and on the periphery of the Russian Federation’s borders, advocating instead for diplomatic solutions and national sovereignty. 248. Middle East, North Africa, and India: We recognize the critical importance of securing vital trade routes, including those in the Persian Gulf and Red Sea, for maintaining global supply chains and ensuring global security. We commend the implementation of the Abraham Accords and advocate for continued normalization of relations between Israel and Arab states, which is essential for regional stability and peace. We also emphasize the need for successful deterrence against hostile actions by Iran and its proxies to prevent further escalations in the region. We oppose the relocation of Gazans to United States. We also acknowledge the strategic partnership with India as a key ally in the region, enhancing our collective security and economic interests. Table of Contents 41 249. Israel. We strongly condemn the violence, harassment and hatred directed toward Jews and Israel worldwide. We respect Israel’s rights of sovereignty, self-determination, and self-defense, and therefore we support: a) The relocation of the United States Embassy in Israel to Jerusalem, Israel’s eternal and indivisible capital. b) Israel’s sovereignty over the Golan Heights. c) Israel’s right to exist, right to secure borders, and right to the land secured by practicing self-defense from aggressive enemies. d) Prohibition of a Palestinian state within the historical borders of Israel, as it would jeopardize Israel’s security, and it would force Israel to give up land that God gave to the Jewish people, as referenced in Genesis. e) Israel’s maintaining a qualitative military edge over any and all adversaries through continued support militarily, financially, and technologically. f) Prohibition of the anti-Semitic Boycott, Divestment, and Sanctions (BDS) movement as a form of warfare being waged upon Israel, on all levels, including and especially on college campuses, at the United Nations, and by anti-Western nongovernmental organizations. g) Israel’s right to recover their hostages, protect their citizens, and defend their borders. h) Opposition to any two-state solutions (Land for Peace). 250. Foreign Defense: We oppose any offensive foreign military action or any other involvement in a war that has not been declared by the United States Congress. Congress shall not abdicate war powers to the executive branch except when under imminent threat, and these powers are not to be used in a preemptive strike unless approved by Congress. The Texas National Guard should only be deployed to overseas combat zones under authorization of Congress through a declaration of war. 251. United Nations: The United Nations is a detriment to the sovereignty of the United States and other countries. We support the immediate withdrawal from the current United Nations and the removal of the United Nations from United States soil. We oppose participation in any United Nations entity, program, or initiative that would compromise American sovereignty to an external entity. 252. Unidentified Aerial Phenomena: We call on the United States Congress and the Executive Branch to uphold the principles of transparency and accountability by disclosing to the American people all pertinent information and knowledge held by United States government agencies and departments regarding the nature and origins of non-American Unidentified Aerial Phenomena (UAP). In line with the National Defense Authorization Act (NDAA) and existing protocols within the Department of Defense, it is imperative that any classified information relevant to UAPs be reviewed for declassification and public release, ensuring full transparency on these matters. These disclosures are essential for public trust, national security, and the integrity of our government institutions. Table of Contents 42 Resolutions 1. Resolution to Support Agenda 47 of President Trump: The Republican Party of Texas fully endorses President Trump’s Agenda 47 visionary positive solutions for the reconstruction of America, and will use these to recruit independent and traditional Democrat voters for a landslide victory in November 2024. 2. Resolution Against the Misuse of the Legal System for Political Purposes: We believe that the Department of Justice, some State Attorneys General, and a few local county district attorneys have engaged in a coordinated attack against former President Donald John Trump and his supporters, by falsely and maliciously indicting Republican leaders, lawyers, and past or present elected officials. We further believe that these attacks against President Trump and his supporters constitute a threat to the continued existence of this Constitutional Federal Republic. We fully understand that similar misuse of the legal system for political purposes constitutes some of the methods used by dictators to establish one-party statist control over many nations. We reject and condemn this misuse of the legal system for political purposes and urge the American people to reject it by overwhelmingly supporting the election of Donald John Trump as the 47th President of the United States of America. 3. 2023 Impeachment: We believe that the impeachment of Texas Attorney General Ken Paxton was unprofessional, petty, and a fiscally irresponsible abuse of constitutional power, in which members of the Texas House of Representatives attempted to circumvent the will of the voters of the 2022 Election by removing a political opponent with little to no evidence of wrongdoing, based on a non-transparent and secretive short-term investigation and an unprecedented quick impeachment proceeding that lasted less than a day. We believe that Attorney General Ken Paxton did not commit any impeachable offense, condemn the actions of the Texas House of Representatives for the impeachment, and call for the State to reimburse the Attorney General for lost wages and legal fees during his temporary removal from office. 4. Care for Women: Texas Republicans understand the lasting hurt that often results from a past abortion, we invite those who have past hurts from abortion to engage in loving conversations and healing counsel, and we invite post-abortive women into the pro-life conversation to spread awareness that abortion harms women. We maintain our belief in the sanctity of innocent human life, created in the image of God, which should be protected from fertilization until natural death. We are the party of LIFE. 5. Border Invasion: The Republican Party of Texas calls upon Texas law enforcement, the Texas judiciary, and the Governor to act to protect Texans from the border invasion and imminent danger, enforcing SB 4 (88th Regular Session) and other such measures as the Governor or Legislature sees fit to protect Texas, without paying heed to any stays or opinions of the federal judiciary or other branch of federal government to the contrary. 6. Urging Congress to Put Border Security First, Oppose Further Aid to Ukraine: The Republican Party of Texas calls for an immediate cessation of further taxpayer-funded financial, military, and humanitarian aid to Ukraine and urges Congress and the President to prioritize the allocation of resources towards securing the United States-Mexico border and deporting illegal aliens. 7. Special Legislative Session: The Republican Party of Texas calls upon Governor Abbott to call a special session in June 2024 to address election integrity and other issues. Specifically, the call should include the following: a) Legislation to implement independent, state-level prosecution of election integrity; to address abortion, human smuggling, sedition, and riot; and to respond to the Court of Criminal Appeal’s ruling in State v. Stephens. b) Legislation to require proof of citizenship for new voter registration and a thorough review of existing voter rolls to identify and remove illegal aliens, other non-citizens, those with unauthorized non-residential addresses, and long-time inactive voters from the voter rolls. Table of Contents 43 Index 2 25-Day Rule .......................................................... 26 A abortion.................................................................. 25 abolition of .................................................... 6, 32 alternatives ........................................................ 25 and human smuggling ....................................... 42 as health care ..................................................... 32 bans ................................................................... 32 care for women ................................................. 42 conscience clause .............................................. 25 home health care ............................................... 23 homicide exemption .......................................... 28 insurance ........................................................... 32 late term ............................................................ 32 Medicaid ........................................................... 23 pill ..................................................................... 25 Planned Parenthood .......................................... 25 providers ........................................................... 25 referrals ............................................................. 32 religious liberty provisions ............................... 20 trafficking .......................................................... 11 victims ............................................................... 42 addiction ................................................................ 24 adjudication ........................................................... 36 adoption ................................................................. 31 age of consent ........................................................ 30 Agenda 47 ............................................................. 42 agricultural ........................................................ 7, 10 Alamo .................................................................... 28 alcohol ..................................................................... 7 Alcohol, Tobacco, Firearms and Explosives (ATF) .......................................................................... 35 American Identity curriculum ......................................................... 17 American Institutions curriculum ......................................................... 20 ammunition, purchase requirements ....................... 6 abortion.................................................................. 25 annexation ............................................................. 33 appraise.................................................................. 34 property ............................................................. 34 artificial intelligence .............................................. 11 Attorney General ................................................... 28 Australia ................................................................ 40 authority ................................................................ 39 auto manufacturers .................................................. 7 B bail ......................................................................... 27 bailouts .................................................................. 14 ballots bond .................................................................. 14 bond elections ................................................... 35 fair procedures .................................................. 36 Internet .............................................................. 36 mail-in ............................................................... 36 rolling polling ................................................... 36 security .............................................................. 36 Speaker of the House ........................................ 34 belief ...................................................................... 24 belief in God ............................................................ 3 bilingual education curriculum ......................................................... 17 Bill of Rights, Texas .............................................. 21 Board of Pharmacy ................................................ 21 bond election ......................................................... 35 bond issues ............................................................ 14 border .................................................................... 40 defense of by Texas........................................... 13 border invasion ...................................................... 42 border laws ............................................................ 13 border security ....................................................... 40 border wall ............................................................ 39 boycotts ................................................................... 8 bullion ................................................................... 15 Bureau of Land Management ................................ 35 C campaign contributions ......................................... 34 campus speech ....................................................... 20 capital punishment ................................................ 28 carbon dioxide, as non-polutant .............................. 9 casinos, gambling .................................................. 12 caucus .............................................................. 34, 35 Cenotaph ............................................................... 29 censorship .............................................................. 12 Census Bureau ......................................................... 5 Centers for Disease Control (CDC) ...................... 35 chairmanship ......................................................... 35 child custody ......................................................... 31 Child Protective Services ...................................... 28 child support welfare .............................................................. 23 China ..................................................................... 40 citizenship ............................................................. 22 civil asset forfeiture ............................................... 26 class action ............................................................ 27 climate change ......................................................... 9 climate justice........................................................ 10 cloning ................................................................... 25 closed primary ....................................................... 37 commerce .............................................................. 35 Commissioner of Education SBOE ................................................................ 18 Common Core national core curriculum ................................... 17 concurrent majority ................................................. 6 conflict of interest ................................................. 22 conscience ............................................................. 25 business rights ................................................... 29 conservative........................................................... 34 constitution issues related to ................................................. 13 Table of Contents 44 spending and taxation ....................................... 12 Constitution ............................................. 3, 4, 5, 6, 7 and Federal lands .............................................. 34 bureaucracy ....................................................... 35 state self-defense ............................................... 39 Texas ................................................................. 15 US ..................................................................... 17 Constitution Day ................................................... 29 Constitution of Texas ............................................... 3 Constitution of the United States ............................ 3 Constitution, and abortion ....................................... 6 Constitution, Equal Rights Amendment .................. 5 constitution, issues related to .................................. 4 Constitutional Carry ................................................ 4 Constitutional Convention US history ......................................................... 17 contact tracing ......................................................... 5 Corporation for Public Broadcasting..................... 15 corporations ............................................................. 8 counterfeiting ........................................................ 36 creator ...................................................................... 3 Critical Race Theory ................................... 8, 17, 20 currency, digital ....................................................... 8 cyber security ........................................................ 12 cybersecurity ......................................................... 39 D data, digital ............................................................ 12 debt ........................................................................ 35 Declaration of Independence ................................... 3 defined contribution plans ..................................... 12 definition of marriage ............................................ 30 DEI ........................................................................ 11 Department of Education, abolish ......................... 18 deportation ............................................................. 39 desisters ................................................................. 31 detransitioners ....................................................... 31 digital currency .................................................. 8, 15 disability welfare .............................................................. 23 disasters ................................................................. 12 disenfranchisement ................................................ 36 district attorneys .................................................... 28 Diversity, Equity, and Inclusion ............................ 11 divorce, no-fault .................................................... 31 doxing .................................................................... 11 drag queen story hour ............................................ 30 driver licenses ........................................................ 36 dysphoria treatment ........................................................... 24 E early voting ............................................................ 36 early warning, flooding and hurricanes ................. 10 easement ................................................................ 33 education ..................................................... 6, 15, 18 choice .................................................................. 3 funding .............................................................. 14 education on humanity of the preborn child preborn child ..................................................... 18 election hand-counting procedures ................................. 38 integrity ............................................................. 42 special legislative session ................................. 42 Election Day .......................................................... 36 elections bond .................................................................. 35 concurrent majority ............................................. 6 consolidating ..................................................... 36 crimes ................................................................ 28 election code ..................................................... 36 Election Day ..................................................... 36 fair election procedures..................................... 36 fraud .................................................................. 36 general............................................................... 34 in-person voting ................................................ 36 integrity ............................................................. 28 laws ................................................................... 36 open records ...................................................... 36 primary .............................................................. 37 Electoral College ..................................................... 5 electromagnetic pulse (EMP) .................................. 9 electronic devices .................................................. 36 eligible voters ........................................................ 36 embryonic stem cell research ................................ 25 emergency medical technicians ............................... 9 eminent domain ..................................................... 33 Employee Stock Ownership Plans (ESOPs) ........... 9 encryption .............................................................. 12 Endangered Species Act ........................................ 10 endorsement slates ................................................ 36 energy .............................................................. 10, 35 energy, and environment ......................................... 9 energy, green ......................................................... 11 energy, sources, renewable .................................... 10 enforcement ........................................................... 27 entitlements ........................................................... 24 Environmental Protection Agency ........................ 10 Environmental, Social, Governance (ESG) ............ 8 environmentalism .................................................... 9 equal access financial aid ...................................................... 20 equal parenting ...................................................... 28 ESG ......................................................................... 8 essential academic knowledge .............................. 16 curriculum ......................................................... 16 euthanasia .............................................................. 25 executive orders .................................................. 4, 5 F faith-based institutions ........................................................ 22 rehabilitation programs ..................................... 24 family .................................................................... 30 and CPS ............................................................ 28 law .................................................................... 28 parental rights in education ............................... 15 traditional ............................................................ 3 family, rights ........................................................... 5 family, sick and maternity leave.............................. 7 Fannie Mae ............................................................ 15 farms .................................................................. 7, 10 Table of Contents 45 federal agency ....................................................... 35 Federal Deposit Insurance Corporation (FDIC) .... 35 federal funds, related to Texas laws ...................... 13 federal land ............................................................ 34 Federal Reserve ..................................................... 15 Federalist Papers US history ......................................................... 17 felony, and voting .................................................. 36 fetal tissue .............................................................. 25 firearm, purchase requirements ............................... 6 firefighters ............................................................... 9 first responders ...................................................... 28 fit parent equal parenting .................................................. 28 flags desecration ........................................................ 29 display of........................................................... 29 flood ...................................................................... 10 flooding ................................................................. 10 flooding mitigation and early warning .................. 10 Food and Drug Administration (FDA) .................. 35 foreign military action ........................................... 41 forensic .................................................................. 36 foster care .............................................................. 31 Founders’ writings US history ......................................................... 17 Founding Documents curriculum ......................................................... 17 fraud ...................................................................... 27 election .............................................................. 36 Freddie Mac........................................................... 15 free trade .................................................................. 7 freedom.................................................................... 3 free-market curriculum ......................................................... 20 frivolous lawsuits .................................................. 27 funding .................................................................. 27 Title 19 .............................................................. 37 G gain-of-function ..................................................... 27 gambling ................................................................ 12 gambling, legalized ............................................... 12 gender ...................................................................... 8 gender identity ................................................. 19, 24 gender, biological .................................................... 8 gender, transition ..................................................... 8 general election ..................................................... 34 globalist institutions .............................................. 13 God image of .............................................................. 3 gold and silver as legal tender ............................... 15 gold and silver bullion ........................................... 15 government authority ............................................ 34 Governor.................................................. 4, 5, 22, 39 grassroots ............................................................... 34 Great Reset .............................................................. 8 grid, electric ....................................................... 9, 10 grooming of minors ............................................... 16 groundwater ........................................................... 10 gun free zones........................................................ 34 gun, purchase........................................................... 6 H handgun safety and proficiency training school security .................................................. 18 hate crime .............................................................. 26 Health and Human Services (HHS) ...................... 35 health savings account, Texas ............................... 12 health savings accounts ......................................... 22 high speed rail ....................................................... 11 HOAs..................................................................... 33 homeschool ........................................................... 20 homestead exemption ............................................ 36 homicide exemption .............................................. 28 homosexuality and gender issues .............................................. 24 and public funding ............................................ 20 behavior ............................................................ 24 criminal penalties related to opposition ............. 24 hormones ............................................................... 24 Housing and Urban Development (HUD)............. 35 human sexuality .................................................... 30 human trafficking .................................................. 27 hurricane mitigation and early warning ................ 10 I identity................................................................... 27 identity theft .......................................................... 29 illegal assistance .................................................... 36 illegal drug ............................................................ 24 illegal immigration .......................................... 38, 39 illegal voting.......................................................... 36 immigration ............................................................. 4 as border invasion ............................................. 42 immigration laws ................................................... 40 immigration system ............................................... 39 impeachment ................................................... 27, 42 income tax ....................................................... 13, 14 India....................................................................... 40 industries ........................................................... 8, 10 informed consent ................................................... 22 infrastructure ............................................. 10, 11, 39 inland waterways ................................................... 10 in-person voting ................................................................ 36 institutional care .................................................... 23 instructional excellence ......................................... 16 insurance ............................................................... 22 intelligence data .................................................... 39 Internal Revenue Service (IRS) ............................ 35 International Baccalaureate national core curriculum ................................... 17 internet................................................. 11, 12, 13, 36 Interstate Border Compact .................................... 40 invasion ................................................................. 39 J Japan ...................................................................... 40 judiciary................................................................... 4 jurors ..................................................................... 27 jury ........................................................................ 28 Table of Contents 46 L law enforcement .................................................... 28 laws election .............................................................. 36 laws, blue ................................................................. 7 laws, minimum wage............................................... 7 lawsuits .................................................................. 36 legal system, misuse .............................................. 42 legislative action, timely ....................................... 35 legislative priorities ............................................... 35 LGBTQ, identification .......................................... 24 License to Carry .................................................... 36 licensing ............................................................ 7, 24 liens fraudulent filings ............................................... 27 life innocent ............................................................... 3 sanctity of ............................................................ 3 life-affirming patient protection ............................ 25 limited government ............................................... 37 lockdowns................................................................ 5 LTC (License to Carry) school security .................................................. 18 M mail-in ballots ................................................................ 36 mandates ...................................................... 6, 11, 21 marijuana, recreational .......................................... 34 marriage officiation .......................................................... 26 traditional marriage ............................................. 3 marriage, definition of ........................................... 30 marriage, design of ................................................ 30 mass transit ...................................................... 11, 13 Medicaid .......................................................... 22, 23 medical care........................................................... 21 medical information .............................................. 21 medical supplies .................................................... 22 Medicare ................................................................ 23 mental health ......................................................... 23 mentally disabled................................................... 23 Middle East ........................................................... 40 military ............................................ 3, 20, 24, 34, 38 deployment of Texas Natinoal Guard ............... 41 foreign action .................................................... 41 military veterans .................................................... 38 minors .................................................................... 21 N National Core Curriculum, oppose curriculum ......................................................... 17 National Labor Relations Board ............................ 35 National Sexuality Education Standards national core curriculum, oppose ...................... 17 NATO .................................................................... 40 net-neutrality ......................................................... 11 New Zealand ......................................................... 40 no-fault divorce ..................................................... 31 no-knock raids ....................................................... 28 North Africa........................................................... 40 North Korea ........................................................... 40 Nuremberg Code ................................................... 21 O Obergefell v. Hodges ............................................. 30 obscenity exemption.............................................. 26 offshore ................................................................. 10 open meetings Texas ................................................................. 19 transparency in education ................................. 18 open records request .............................................. 36 oppression ............................................................. 28 opt-out ................................................................... 23 oversight ................................................................ 28 oversight of instructional materials curriculum ......................................................... 18 P pandemic ............................................................... 27 parent informed consent .............................................. 32 parent rights local control of health education....................... 19 parental consent .............................................................. 21 rights ..................................................... 15, 20, 21 rights enforcement ............................................ 16 rights in education............................................. 15 rights related to adult dependent children ......... 21 rights related to healthcare in schools ............... 16 parental, rights ......................................................... 5 parents ................................................................... 21 patent protection ...................................................... 8 patents ..................................................................... 8 Patient Protection and Affordable Care Act .......... 23 pay-to-play ............................................................ 36 PBS ........................................................................ 15 pedophilia, and public health ................................ 24 penalties ................................................................. 24 pensions ................................................................. 12 Permanent School Fund .................................. 13, 18 permitting, building ................................................. 8 Persian Gulf ........................................................... 40 Pharmacy Board .................................................... 22 Philippines ............................................................. 40 photo ID welfare .............................................................. 23 Photo ID ................................................................ 36 Planned Parenthood ............................................... 25 pledge cards ........................................................... 34 police ..................................................................... 28 political policing.................................................... 28 Poll Watchers......................................................... 36 pornography, and public health ............................. 24 power grid ............................................................... 9 power plants, nuclear ............................................ 10 prayer....................................................................... 6 preborn .............................................................. 6, 25 preborn child ......................................................... 18 prescription drugs .................................................. 22 President Trump .................................................... 42 Table of Contents 47 primary election..................................................... 37 prisons, housing by biological sex ........................ 32 privacy data .................................................................... 26 privacy, data........................................................... 12 private property ..................................................... 26 private sector ......................................................... 14 pronoun use gender identity .................................................. 19 pronouns ................................................................ 30 property ................................................................. 34 tax...................................................................... 14 property forfeiture ................................................. 33 prostitution ............................................................ 27 puberty ................................................................... 24 public employees ..................................................... 9 public funding embryonic stem call research ............................ 25 fetal tissue ......................................................... 25 human cloning................................................... 25 Planned Parenthood .......................................... 25 public funds ........................................................... 35 public official dereliction of duty ............................ 5 public-private partnerships .................................... 11 Q quarantine .............................................................. 21 quorum .................................................................. 35 R Rainy Day Fund ..... 22, See Economic Stabilization, Plank 63 ranches ................................................................... 10 readiness ................................................................ 38 recall of elected official ........................................... 4 recreational marijuana ........................................... 34 registration ............................................................. 36 regulations ............................................... 4, 7, 10, 21 religious ................................................................. 21 religious beliefs ..................................................... 25 religious freedom in schools .......................................................... 16 religious liberty ..................................................... 20 required minimum distribution requirement repeal .......................................................................... 13 research.................................................................. 27 gain-of-function ................................................ 27 residential care facility .......................................... 21 restrooms ............................................................... 30 right to keep and bear arms ..................................... 4 right to try .............................................................. 22 rights ...................................................................... 21 conscience ......................................................... 25 constitutional ..................................................... 21 elections ............................................................ 36 inalienable ........................................................... 3 judicial .............................................................. 27 medical ........................................................ 21, 22 medical freedom ............................................... 21 of parents..................................................... 15, 21 of students ......................................................... 16 parental ................................................. 15, 16, 20 parental, and adult dependent children ............. 21 parental, education ............................................ 15 protections ........................................................ 26 related to healthcare in schools ......................... 16 right to try ......................................................... 22 students and patriotic displays .......................... 17 Voting Rights Act of 1965 ................................ 37 rights, 2nd Amendment ........................................... 6 rights, age-related .................................................... 5 rights, citizen ........................................................... 5 rights, constitutional ................................................ 4 rights, cyber security ............................................. 12 rights, Equal Rights Amendment ............................ 5 rights, inalienable .................................................... 3 rights, labor ........................................................... 11 rights, local self-government ................................... 6 rights, National Right to Work Act.......................... 9 rights, of parents ...................................................... 5 rights, parental ......................................................... 5 rights, patent ............................................................ 8 rights, preborn ......................................................... 6 rights, second amendment ....................................... 6 rights, sovereign peoples’........................................ 5 rights, to secede ....................................................... 6 rights, water resources ........................................... 10 riot ......................................................................... 28 road construction ................................................... 13 roads, toll ............................................................... 11 Robin Hood ........................................................... 14 Robin Hood accounting......................................... 15 rolling polling ........................................................ 36 rolls voter registration ............................................... 37 rule of law ............................................................. 27 S safety ..................................................................... 38 safety inspection, state .......................................... 11 same-sex attraction ................................................ 24 same-sex marriage ................................................. 30 Sanctuary Cities .................................................... 40 SBOE..................................................................... 18 scholarships financial aid ...................................................... 20 school charters ....................................................... 18 school choice ......................................................... 15 school security ....................................................... 18 school-based mental health providers ................... 22 Secretary of State .................................................. 37 self-defense ........................................................... 39 sex identification on official documents ................. 30 sexual assault, penalty ........................................... 28 sexual harassment.................................................. 34 silver and gold as legal tender ............................... 15 social credit system ............................................... 15 social justice .................................................... 11, 20 social media ........................................................... 12 Social Security ........................................................ 9 South Korea ........................................................... 40 Table of Contents 48 sovereign ............................................................... 39 sovereignty ................................................ 3, 4, 5, 38 sovereignty, state ..................................................... 7 Speaker of the House............................................. 34 special interests ..................................................... 34 special purpose districts ........................................ 13 special status .......................................................... 24 spousal benefits ..................................................... 31 state bar, mandatory membership ............................ 7 State Board of Education................................. 13, 18 state funding education funding ............................................. 17 stem cell................................................................. 25 stem cell research .................................................. 25 subsidies ...................................................... 3, 10, 14 Sunset Commission ............................................... 35 Supplemental Nutrition Assistance Program (SNAP) ............................................................. 23 Supreme Court......................................................... 4 Supreme Court, Texas ........................................... 10 surgery ................................................................... 24 surveillance............................................................ 26 Swatting ................................................................. 11 T Taiwan ................................................................... 40 TASB ..................................................................... 18 tax34 burden ............................................................... 14 business franchise ............................................. 13 estate ................................................................. 13 Internet and phone ............................................ 13 inventory ........................................................... 13 on home sales .................................................... 13 property ............................................................. 14 relief .................................................................. 14 repeal ................................................................. 13 tax, gasoline ........................................................... 13 tax, increases ......................................................... 12 taxpayer funding .................................................... 24 taxpayer-funded lobby ........................................... 18 teacher and administrator certification, SBOE ...... 18 technology ............................................................. 39 Ten Commandments, display in public buildings . 29 tenure ..................................................................... 20 Texas Bar ................................................................. 7 Texas Child Mental Health Care Consortium ....... 22 Texas Election Code .................................. 14, 36, 37 Texas health savings accounts ............................... 12 Texas Medical Board (TMB) ................................ 21 Texas Military........................................................ 39 Texas Open Meetings Act transparency in education ................................. 19 threat ...................................................................... 39 tidal surge, effect and control ................................ 10 tideland boundaries ............................................... 10 Title 19 .................................................................. 37 tolls ........................................................................ 11 tolls, collection ...................................................... 11 toxic exposure ....................................................... 26 traditional marriage ................................................. 3 traditional values ............................................. 24, 37 training .................................................................. 28 transgender ............................................................ 24 transgender normalizing curriculum curriculum ......................................................... 19 transition, gender ..................................................... 8 Trump, President ................................................... 42 U UIL ........................................................................ 20 Ukraine .................................................................. 42 unborn ................................................................. 5, 8 unconstitutional ..................................................... 23 underage ................................................................ 24 unelected ............................................................... 35 UNESCO ............................................................... 28 unidentified aerial phenomena .............................. 41 unions ...................................................................... 9 unions, dues ............................................................. 9 United Nations ................................................ 28, 41 United States Constitution..................................... 17 utilities ..................................................................... 9 V vaccination ........................................................ 8, 21 vaccines fetal tissue ......................................................... 25 vehicle issues ................................................... 11, 13 vehicle, registration ............................................... 11 vehicles, commercial ............................................. 11 veterans ................................................................. 38 victims ................................................................... 27 voter fraud ............................................................. 36 voter registration ............................................. 37, 42 voting..................................................................... 36 concurrent majority ............................................. 6 early .................................................................. 36 W war memorials ....................................................... 29 warrant................................................................... 26 warrants ................................................................. 27 water districts .......................................................... 9 welfare ................................................................... 22 Western civilization curriculum ......................................................... 20 worship .................................................................. 29 writ of habeas corpus .............................................. 5 |
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What is the difference between "comply" and "obey" in the following sentences?
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Joey Hahm
B.S. in Biochemistry (college major), Georgia Institute of Technology (Graduated 2013)
·7y
I'm strictly copy pasta-ing so please don't report me for plagerism
“I read somewhere recently (Oswald Chambers, maybe?) that it's time to take the concept of obedience out of the mire and reclaim it as a worthy act …. I've been thinking about that idea off and on since.
Been asking myself, what does it mean to act obediently?
Here are my thoughts:
There is a pronounced difference between compliance and obedience; most of us seem to practice much compliance and very little obedience.
When we comply we may follow the rules, but we will do so from a place of duty, resentment or fear. Like the little
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I'm strictly copy pasta-ing so please don't report me for plagerism
“I read somewhere recently (Oswald Chambers, maybe?) that it's time to take the concept of obedience out of the mire and reclaim it as a worthy act …. I've been thinking about that idea off and on since.
Been asking myself, what does it mean to act obediently?
Here are my thoughts:
There is a pronounced difference between compliance and obedience; most of us seem to practice much compliance and very little obedience.
When we comply we may follow the rules, but we will do so from a place of duty, resentment or fear. Like the little girl who sat down when her father commanded it, but admitted later that “I was sitting down on the outside, but I was standing up on the inside!”
If we feel forced to do something we may comply, but we are resentful and resistant about it. When we feel strong-armed, coerced or guilted into following someone's dictates, we may go along with them; we may reluctantly do what's required just to “get them off our back” (we are passive-aggressive), or we may comply as a way to avoid punishment (we are fear-based), but one thing is for sure – when we comply, we are in victim consciousness.
Obedience, on the other hand, is an act of freedom: obedience is the spontaneous act of saying yes to something (or someone) that we believe in. To come into obedience means to align our will with something we respect and trust wholeheartedly to the point that we do not hesitate to obey its dictates. Obedience cannot be forced.
We choose freely to obey that which we have complete confidence in. We obey from love. Obedience is the act of a mind free from any vestiges of victim consciousness.
Obedience is not a sort of “rote following after,” nor is it done from a place of fear or recalcitrance. In obedience, we chose to believe in a particular path or set of principles and/or to follow a teacher who professes and exemplifies such principles, not from a “less than” status, but from a joyful recognition that there is something being offered that we very much want to embody in our own lives.
To be obedient means to set our sights on a vision or on a model of something we long for so much that we are willing to do whatever it takes, whatever that is, to achieve it.
Obedience is the natural and healthy response of someone who is humble enough to be teachable, open-minded enough to accept life's teachings in whatever form they come, and motivated enough to hear and obey the inner promptings of a higher self and Source.
Which are you? Are you more compliant, going through each day grumbling because you have to work a job you hate or grudgingly doing the bare minimum just to get by? Or are you obedient? – so attuned to and desirous of serving a higher good that your heart leaps to obey your innermost promptings? May it be so.
May you find great willingness to follow and obey truth wherever it appears in your life and may you find the ability to give yourself wholeheartedly to Source.
Blessings, Lynne”
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David Smith
Writer of short stories and other occasional stuff · Author has 3.1K answers and 2.3M answer views
·3y
Originally Answered: What's the difference between comply and obey? ·
I would say it has to do with what level of authority is being recognized. I will provide a couple of examples.
In most places in the USA, it is entirely legal to “film" (shoot video) of anything that takes place on, or can be seen from, public property. The wishes of anyone not to be filmed have no bearing. However, despite the fact that there is no legal obligation to do so, if I stop filming at someone's request, either as a courtesy because they do not wish to be filmed, or because they express concerns about security, then I am complying wIth their request.
In most places in the USA, includ
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I would say it has to do with what level of authority is being recognized. I will provide a couple of examples.
In most places in the USA, it is entirely legal to “film" (shoot video) of anything that takes place on, or can be seen from, public property. The wishes of anyone not to be filmed have no bearing. However, despite the fact that there is no legal obligation to do so, if I stop filming at someone's request, either as a courtesy because they do not wish to be filmed, or because they express concerns about security, then I am complying wIth their request.
In most places in the USA, including the state I live in, motor vehicle passengers in front seats are required by law to wear seatbelts. Despite the fact that I believe that seatbelts save lives, I do not agree with the notion that we need to have a law about it. However, it is a matter of law, and despite my personal opposition it is unlikely to be declared unconstitutional. Therefore I obey it.
We could say that I comply, without regard to whether or not there is a legal requirement, simply as a matter of courtesy or as a way of avoiding a hassle.
But we could say in this case that I also obey, because despite my objections I recognize that it is the law and therefore I submit to it.
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Neil Turner
English monoglot with phrase books · Author has 2.3K answers and 3.1M answer views
·3y
Originally Answered: What's the difference between comply and obey? ·
To comply is to follow some pre-defined rule(s). Usually one is not absolutely obliged to comply, but compliance may be a condition for some other activity: no bombing, no petting and wash your feet if you want to use the swimming pool; meet their food hygiene and packaging standards to sell your produce in the EU.
To obey is to execute an order. It may be one specific order - “Guards, seize him!” - or a general pledge - “I swear unconditional obedience to the Emperor and his appointed officers.” In some cases, for instance the command of an officer to a serving soldier, failure to obey may res
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To comply is to follow some pre-defined rule(s). Usually one is not absolutely obliged to comply, but compliance may be a condition for some other activity: no bombing, no petting and wash your feet if you want to use the swimming pool; meet their food hygiene and packaging standards to sell your produce in the EU.
To obey is to execute an order. It may be one specific order - “Guards, seize him!” - or a general pledge - “I swear unconditional obedience to the Emperor and his appointed officers.” In some cases, for instance the command of an officer to a serving soldier, failure to obey may result in penalties.
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David Greenspan
Word lover · Author has 262 answers and 3.2M answer views
·11y
You "comply" with rules, laws, and policies. You "obey" people and orders.
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·
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The words "comply" and "obey" both relate to following rules or requests, but they have slightly different connotations and usages. Here’s a breakdown:
Comply:
Definition: To act in accordance with a wish or command, often implying a formal or official context.
Usage: Typically used in situations involving regulations, standards, or requests that may require documentation or acknowledgment.
Example Sentence: "The company must comply with environmental regulations."
Obey:
Definition: To follow commands, rules, or instructions, often implying a more immediate, personal, or auth
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The words "comply" and "obey" both relate to following rules or requests, but they have slightly different connotations and usages. Here’s a breakdown:
Comply:
Definition: To act in accordance with a wish or command, often implying a formal or official context.
Usage: Typically used in situations involving regulations, standards, or requests that may require documentation or acknowledgment.
Example Sentence: "The company must comply with environmental regulations."
Obey:
Definition: To follow commands, rules, or instructions, often implying a more immediate, personal, or authoritative context.
Usage: Commonly used in everyday situations and can imply a moral or ethical obligation.
Example Sentence: "Children are expected to obey their parents."
Summary
Comply is often used in formal contexts and implies adherence to regulations or standards.
Obey is more personal and direct, often associated with commands from authority figures.
If you have specific sentences in mind, feel free to share them for a more tailored analysis!
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Al
Former Criminal Defense Lawyer, Washington State (1983–2016) · Author has 5.3K answers and 3M answer views
·3y
Originally Answered: What's the difference between comply and obey? ·
I use “to comply” to mean to follow a set of directives that generally apply to everyone in the same general situation under the same general set of circumstances. They are pre-established beforehand to cover such situations and are promulgated by a source with formal, generally recognized authority: an employer, a union rule, the board of health, the city council, a Federal law, etc.
I use “to obey” to mean to follow a directive, often one formed and given in the moment, that generally applies only to a certain individual or an informal group of such persons, generally by someone who has a per
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I use “to comply” to mean to follow a set of directives that generally apply to everyone in the same general situation under the same general set of circumstances. They are pre-established beforehand to cover such situations and are promulgated by a source with formal, generally recognized authority: an employer, a union rule, the board of health, the city council, a Federal law, etc.
I use “to obey” to mean to follow a directive, often one formed and given in the moment, that generally applies only to a certain individual or an informal group of such persons, generally by someone who has a personal relationship to the other: one’s parents, babysitter, teacher, etc,, or sometimes by someone who has the authority to give ad hoc direction: a police officer, firefighter, a physician.
Furthermore, the consequences of failure to comply tend to be a set penalty, or the loss of a benefit or privilege. Sometimes the rule sets forth conditions precedent to the grant of a privilege: one must be of age, have received a particular length of instruction, and have passed practical, knowledge, and vision tests to legally operate a motor vehicle. If one has not been granted a license and drives, anyway one faces a possible fine or jail term, and even if one has a license one can lose it by accumulating too many violations, or failure to pay those tickets.
The consequences of failure to obey can be severe, but are often I formal, and may range from disapproval, a scolding, or the loss of a privilege, up to loss of a job, loss of rank, discharge (from the military, or what a moral theologian would call A loss of grace with spiritual consequences.
The duty to obey tends to flow from moral obligation arising from a relationship: parent and child, teacher and pupil, lord and subject. There is usually a feeling of duty, and a corresponding subjective feeling of guilt from a failure to obey.
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Hal Mickelson
Former Corporate Attorney; AB, History, JD, Law · Author has 14.8K answers and 16M answer views
·3y
Originally Answered: What's the difference between comply and obey? ·
One complies with a rule, with a detailed set or rules or with a policy; one can also comply with instructions, directions or a military order. One obeys a rule, a detailed set of rules or a military order; we don’t talk about “obeying” a policy or instructions. There is a great deal of overlap between the two words; we’re more likely to use the word “comply” in an informal setting and where complicance is volutary; we’re more likely to use the word “obey” in a more structured setting where obedience is mandatory. You comply with a request for cooperation in your neighborhood; you obey the har
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One complies with a rule, with a detailed set or rules or with a policy; one can also comply with instructions, directions or a military order. One obeys a rule, a detailed set of rules or a military order; we don’t talk about “obeying” a policy or instructions. There is a great deal of overlap between the two words; we’re more likely to use the word “comply” in an informal setting and where complicance is volutary; we’re more likely to use the word “obey” in a more structured setting where obedience is mandatory. You comply with a request for cooperation in your neighborhood; you obey the harsh demands of a nasty boss. Your aunt complies with the rules of the cake-baking contest; the dog obeys its master.
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Mark Bradley
Economist
·9mo
What are the dumbest financial mistakes most Americans make?
Where do I start?
I’m a huge financial nerd, and have spent an embarrassing amount of time talking to people about their money habits.
Here are the biggest mistakes people are making and how to fix them:
Not having a separate high interest savings account: Having a separate account allows you to see the results of all your hard work and keep your money separate so you're less tempted to spend it. Plus with rates above 5.00%, the interest you can earn compared to most banks really adds up.
Here is a list of the top savings accounts available today. Deposit $5 before moving on because this is one
Continue Reading
Where do I start?
I’m a huge financial nerd, and have spent an embarrassing amount of time talking to people about their money habits.
Here are the biggest mistakes people are making and how to fix them:
Not having a separate high interest savings account: Having a separate account allows you to see the results of all your hard work and keep your money separate so you're less tempted to spend it. Plus with rates above 5.00%, the interest you can earn compared to most banks really adds up.
Here is a list of the top savings accounts available today. Deposit $5 before moving on because this is one of the biggest mistakes and easiest ones to fix.
Overpaying on car insurance: You’ve heard it a million times before, but the average American family still overspends by $417/year on car insurance. If you’ve been with the same insurer for years, chances are you are one of them.
Pull up Coverage.com, a free site that will compare prices for you, answer the questions on the page, and it will show you how much you could be saving.
That’s it. You’ll likely be saving a bunch of money. Here’s a link to give it a try.
Not using a debt relief program: People with $10K in credit card debt can get significant reductions by using a debt relief program. Don’t suffer alone, here’s a quick 2 minute quiz that will tell you if you qualify.
Not earning free money while investing: Most people think investing is complicated or for the rich. Not anymore. There are platforms that literally give you free money just to get started. Deposit as little as $25, and you could get up to $1000 in bonus funds. This is the siteI recommend to my friends.
How to get started
Hope this helps! Here are the links to get started:
Have a separate savings account
Stop overpaying for car insurance
Finally get out of debt
Start investing with a free bonus
Upvote ·
4.8K 4.8K
1K 1K
999 159
Chrysaor Jordan
speaks American English with native fluency · Author has 27.8K answers and 105.9M answer views
·11y
David Greenspan explained the nuance well.
One more detail: "comply" is a very formal word, while "obey" is less so.
And as User-12828854714828252077 said: they are synonyms.
Last word here: victrola.
Upvote ·
9 2
9 2
K Ramachandran Pillai
Sr. Office Superintendent at BITS Pilani, Hyderabad Campus (2006–present) · Author has 867 answers and 470K answer views
·3y
Originally Answered: What's the difference between comply and obey? ·
comply/ compliance is used in statutory an or legal matters. obey is like obey one’s teachers and parents; obey the orders etc.
Upvote ·
Promoted by Roundhouse Provisions
Matt Jennings
Former Youth Basketball Coach
·Updated 2y
What is the secret to staying relatively healthy at age 70?
Look at the legendary Chuck Norris’s advice since he is now a whopping 81 years old and yet has MORE energy than me. He found a key to healthy aging… and it was by doing the opposite of what most of people are told. Norris says he started learning about this revolutionary new method when he noticed most of the supplements he was taking did little or nothing to support his health. After extensive research, he discovered he could create dramatic changes to his health simply focusing on 3 things that sabotage our body as we age.
“This is the key to healthy aging,” says Norris. “I’m living proof.”
N
Continue Reading
Look at the legendary Chuck Norris’s advice since he is now a whopping 81 years old and yet has MORE energy than me. He found a key to healthy aging… and it was by doing the opposite of what most of people are told. Norris says he started learning about this revolutionary new method when he noticed most of the supplements he was taking did little or nothing to support his health. After extensive research, he discovered he could create dramatic changes to his health simply focusing on 3 things that sabotage our body as we age.
“This is the key to healthy aging,” says Norris. “I’m living proof.”
Now, Chuck Norris has put the entire method into a 15-minute video that explains the 3 “Internal Enemies” that can wreck our health as we age, and the simple ways to help combat them, using foods and herbs you may even have at home.
I’ve included the Chuck Norris video here so you can give it a shot.
Upvote ·
13K 13K
2.3K 2.3K
999 258
Chris Robida
Former 40 Years Teaching Elem., Middle Sch. & College (1973–2016) · Author has 4.8K answers and 2.5M answer views
·5y
Related
Which is correct, comply to or comply with?
I’ve never seen comply to, reply to, yes, but not comply to.
Comply with something is far more common.
,
Upvote ·
9 4
Tim Offenstein
Reading the Bible over and over again. · Author has 99 answers and 19.7K answer views
·1y
Related
What is the difference between "obey" and "submit" in the Bible?
According to a simple dictionary description, Submit = “to give over or yield to the power or authority of another”
Obey = “to comply with or follow the commands, restrictions, wishes, or instructions of”
So there’s not a lot of difference between the two terms, in the Bible or elsewhere. Here are a few passages that shine light on the terms Biblically speaking:
John 3:36, “Whoever believes in the Son has eternal life; whoever does not obey the Son shall not see life, but the wrath of God remains on him.
Romans 6:16, “Do you not know that if you present yourselves to anyone as obedient slaves, you
Continue Reading
According to a simple dictionary description, Submit = “to give over or yield to the power or authority of another”
Obey = “to comply with or follow the commands, restrictions, wishes, or instructions of”
So there’s not a lot of difference between the two terms, in the Bible or elsewhere. Here are a few passages that shine light on the terms Biblically speaking:
John 3:36, “Whoever believes in the Son has eternal life; whoever does not obey the Son shall not see life, but the wrath of God remains on him.
Romans 6:16, “Do you not know that if you present yourselves to anyone as obedient slaves, you are slaves of the one whom you obey, either of sin, which leads to death, or of obedience, which leads to righteousness?
Roman’s 8:7, “For the mind that is set on the flesh is hostile to God, for it does not submit to God's law; indeed, it cannot.”
James 4:7, “Submit yourselves therefore to God. Resist the devil, and he will flee from you. Draw near to God, and he will draw near to you.”
Hope this clarifies and answers your question.
Upvote ·
Linda Hagge
Former University Instructor in English and Linguistics at Iowa State University (1985–2010) · Author has 3.6K answers and 1.4M answer views
·3y
Related
What is the difference between “obey” and “abide”?
They are nothing alike. To obey means to do what someone else tells you. To abide means to accept something. Another meaning is “to stay in one place.”
I obey my parents.
I cannot abide racists.
Abide with me; fast falls the eventide.
Upvote ·
9 3
Just Me.
Mom/grandma; observationist; aspie; bilingual professional · Author has 101 answers and 28.9K answer views
·Mar 16
Related
Are "comply with" and "respect" synonyms?
No; they are not synonymous. There is a lot that can be said here, but I will keep it concise.
While one of the definitions of respect is to abide by or comply with, when you comply with, you typically do so because you must, you are “required” to obey with a rule, set of standards, laws, etc. (or else). In other words, there is usually a consequence if you don’t comply, and sometimes, people comply out of fear but never out of respect.
Respect comes from admiration, loyalty, belief, value, appreciation—even love in many cases.
You can comply with doing something and have absolutely no respect fo
Continue Reading
No; they are not synonymous. There is a lot that can be said here, but I will keep it concise.
While one of the definitions of respect is to abide by or comply with, when you comply with, you typically do so because you must, you are “required” to obey with a rule, set of standards, laws, etc. (or else). In other words, there is usually a consequence if you don’t comply, and sometimes, people comply out of fear but never out of respect.
Respect comes from admiration, loyalty, belief, value, appreciation—even love in many cases.
You can comply with doing something and have absolutely no respect for [it] nor those who impose, mandate, or require such a thing.
Upvote ·
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3778 | https://www.westfield.ma.edu/cmasi/gen_chem1/Solutions/Molarity/molarity.htm | genchem
Solutions and Molarity
The amount of material in a solution has to expressed in terms of concentration. You can see how confusing it would be if solutions were discussed in terms of moles.
If beaker A was labeled
"1 mole of sugar"
because 1 mole of sugar was placed in 2 L of water there would be no way of knowing how much sugar was left after some of the solution was removed.
So, it would be better if the jar was labeled "342 g of sugar in 2 L of solution". Then we would be able to tell how much sugar was left after some of the solution was removed.
Solutions are always described in terms of amount of solute dissolved in amount of solvent or in terms of amount of solute dissolved in amount of solution.
Molarity
There are a variety of ways to express concentration.
Commercial products are often labeled
% vol. (hydrogen peroxide, wine)
or
% mass (hydrochloric acid)
Polymer chemists use
g/100 mL
when performing viscosity experiments to determine the molar mass of their polymers.
Typically, chemists use molarity. Molarity is abbreviated (M) and it is defined as follows:
Believe it or not, molarity was invented for convenience.
Let's imagine, for a moment, a world in which concentration was measured in grams per liter of solution.
You are instructed to make 58 g NaCl by mixing together HCl and NaOH solutions.
In the lab you find two bottles. One bottle is labeled 36.5 g HCl per L of solution, and the other is labeled 20.0 g of NaOH per L of solution.
So, how much of each solution do you need to mix together?
The balanced equation is the only way to relate NaCl, HCl, and NaOH; therefore, the amounts of HCl, NaCl, and NaOH must be related using moles. That means all the concentrations must be converted to moles each time the solutions are used.
Now, let's imagine, for a moment, a world in which concentration was measured in grams per liter of solvent.
Now the problem is the volume... If you add a liter of water to 20.0 g of NaOH what is the volume of the solution? I do not know and neither would you because the NaOH will change the volume. So, there would be more than 1 L of solution
Molarity eliminates both of these problems.
Solution preparation is simplified because solvent is added until the proper volume of solution is obtained.
Gram to mole conversions are eliminated once the solution is prepared.
Molarity Problems
First kind: Determine the concentration of a solution.
What is the concentration (molarity) of a solution made by dissolving 0.20 mol ammonium nitrate (NH 4 NO 3) in enough water to make 500.0 mL of an NH 4 NO 3 solution.
Since molarity is just moles per liter if you know how many moles are present in a certain number of liters then you know the molarity.
Second kind: Determine the number of moles in a certain volume of a solution of known concentration.
How many moles of albumin are in 3.00 x 10 2 mL of a 0.015 M solution?
Essentially the problem is a conversion from mL to mol.
Molarity relates moles to L (or mL); so, you could use molarity as a conversion factor!
Use the frame advance button to step through the QuickTime movie.
Third kind: making a solution.
Make 250 mL of a 2.5 M aqueous CH 3 CH(OH)CH 3 (isopropanol) solution.
You need to now how much isopropanol to mix with the water. So, you need to determine the number of moles of isopropanol that you need, which will allow you to calculate the number of grams of isopropanol are needed.
Start with 250 mL because that is how much you want to make, and convert to L because that is what molarity uses.
Use the frame advance button to step through the QuickTime movie.
Since isopropanol is a liquid you could convert that number to mL and measure 48 mL isopropanol. Does that mean you need to add 202 mL H 2 O? NO, you must use a 250 mL volumetric flask and dilute to the line. Actually, you will add more than 202 mL water.
DilutionsM(who)V(what) = huh?
(I hate M 1 V 1 = M 2 V 2 because most people memorize this equation,
but do not know when to use it!)
10.0 mL of a 0.20 M NaCl solution is diluted to 100.0 mL. What is the concentration of the final solution?
Instead of using the "MV abomination" why not analyze the problem. To find the concentration of the final solution you need to know two things...
of moles
volume of solution
You know the final volume is 100.0 mL (that is what the problem says).
Now you need to know the number of moles of NaCl present in the solution. How can you find the number of moles?
All the NaCl that is in the new solution came from the old solution. So, find the number of moles of of NaCl that are in the original 10.0 mL solution.
So, molarity is just another conversion factor, like molar mass, or density.
return to GenChem Home Page |
3779 | https://flexbooks.ck12.org/cbook/ck-12-middle-school-math-concepts-grade-8/section/2.1/related/lesson/decimal-addition-msm7/ | Decimal Addition | CK-12 Foundation
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Back To Add and Subtract DecimalsBack
2.1
Decimal Addition
Written by:Jen Kershaw, M.ed |Kaitlyn Spong
Fact-checked by:The CK-12 Editorial Team
Last Modified: Sep 01, 2025
[Figure 1]
Lindsey made $415.85 babysitting over the summer. She went to the bank to deposit her money into a savings account. At the bank she learned that she can expect to earn 2% interest on her money in one year. She does the calculations and finds that 2% of 415.85 is 8.317. Assuming Lindsey doesn’t add any more money to her savings account, how can she figure out how much money will be in her account after one year?
In this concept, you will learn how to add decimals with and without rounding.
Adding Decimals
The decimal system lets us represent numbers that are less than 1 and other numbers that exist between whole numbers. In a decimal number, the decimal point divides the whole part of the number from the fractional part of the number.
To add decimal numbers:
Arrange the numbers vertically so that the decimal points line up.
Add zeros for any missing spaces so that each number has the same number of digits to the right of the decimal point.
Add from right to left, carrying just as if you were adding whole numbers.
Insert a decimal point into your answer so that it lines up with the decimal points of the original numbers.
Here is an example.
Find the sum of 1.87 and 2.3.
First, arrange the numbers vertically so that the decimal points line up.
1.87+2.3 _
Next, add a zero after the 3 in the second number so that both numbers have two digits to the right of the decimal point.
1.87+2.30 _
Now, add from right to left. Start by adding 7 and 0 to make 7. Then 8 and 3 to make 11. At this point you will carry the 1 and add it along with the 1 and the 2 to make 4. Insert a decimal point into your answer directly under the decimal points in the original numbers.
1.87+2.30 _ 4.17
The answer is 1.87+2.3=4.17.
Here is another example.
4.5+2.137=_
First, arrange the numbers vertically so that the decimal points line up.
4.5+2.137 _
Next, add two zeros after the 5 in the first number so that both numbers have three digits to the right of the decimal point.
4.500+2.137 _
Now, add from right to left. Start by adding 0 and 7 to make 7. Then 0 and 3 to make 3. Then 5 and 1 to make 6.Finally add 4 and 2 to make 6. Insert a decimal point into your answer directly under the decimal points in the original numbers.
4.500+2.137 _ 6.637
The answer is 4.5+2.137=6.637.
Sometimes you will want to round numbers before adding them. Recall that there are four steps for rounding numbers.
Identify the place you want to round to and find that digit within your number.
Look at the digit to the right of the place you want to round to.
If the digit to the right is between 5 and 9, increase the digit in the place you want to round to by 1. If the digit to the right is between 0 and 4, keep the digit in the place you want to round to the same.NOTE: If the digit in the place you want to round to is a 9 and you need to increase it by 1, change it to zero and increase the digit to its left by 1.
Replace all digits after the place you are rounding to with zeros. If these digits occur to the right of the decimal point, it is not necessary to write the zeros.
Here is an example.
Round each number to the nearest hundredth and then find the sum.
9.199,11.353,10.648
First, round each number to the nearest hundredth. Remember that the hundredths place is the second digit to the right of the decimal point. Look at each digit in the hundredths place and then the digit to the right of that to decide if you will need to round up the digit in the hundredths place or keep it the same.
9.199 rounds to 9.20
11.353 rounds to 11.35
10.648 rounds to 10.65
Now, find the sum of the three rounded numbers. Start by arranging the numbers vertically so that the decimal points line up.
9.20 11.35+10.65 _
Next, add from right to left. Start by adding 0, 5, and 5 to make 10. Carry the 1 and add it with the 2, 3, and 6 to make 12. Once again carry the 1 and add it with the 9 and 1 to make 11. Carry the 1 one more time and add it with the 1 and 1 to make 3. Insert a decimal point into your answer directly under the decimal points in the original numbers.
9.20 11.35+10.65 _ 31.20
The answer is 9.20+11.35+10.65=31.20.
Examples
Example 1
Earlier, you were given a problem about Lindsey and her savings account.
She deposited $415.85 into the account and expects to earn $8.317 in interest in one year. She wants to figure out how much money she will have in her account after one year.
First, round both numbers to the nearest cent, which is the nearest hundredth. $415.85 is already to the nearest hundredth, but $8.317 needs to be rounded.
8.317 rounds to 8.32
Now, find the sum of the two rounded numbers. Start by arranging the numbers vertically so that the decimal points line up.
415.85+8.32 _
Next, add from right to left. Start by adding 5 and 2 to make 7. Then add 8 and 3 to make 11. Carry the 1 and add it to 5 and 8 to make 14. Carry the 1 and add it to 41 to make 42. Insert a decimal point into your answer directly under the decimal points in the original numbers.
415.85+8.32 _ 424.17
The answer is $415.85+$8.32=$424.17, so Lindsey can expect to have $424.17 in her account after one year.
Example 2
Round each value to the nearest tenth and then find the sum.
4.56,3.47,2.04,6.21
First, round each number to the nearest tenth. Remember that the tenths place is the first digit to the right of the decimal point. Look at each digit in the tenths place and then the digit to the right of that to decide if you will need to round up the digit in the tenths place or keep it the same.
4.56 rounds to 4.6
3.47 rounds to 3.5
2.04 rounds to 2.0
6.21 rounds to 6.2
Now, find the sum of the four rounded numbers. Start by arranging the numbers vertically so that the decimal points line up.
4.6 3.5 2.0+6.2 _
Next, add from right to left. Start by adding 6, 5, 0, and 2 to make 13. Carry the 1 and add it with the 4, 3, 2, and 6 to make 16. Insert a decimal point into your answer directly under the decimal points in the original numbers.
4.6 3.5 2.0+6.2 _ 16.3
The answer is 4.6+3.5+2.0+6.2=16.3.
Example 3
4.56+1.2+37.89=_
First, arrange the numbers vertically so that the decimal points line up.
4.56 1.2+37.89 _
Next, add one zero after the 2 in the second number so that all numbers have two digits to the right of the decimal point.
4.56 1.20+37.89 _
Now, add from right to left. Start by adding 6, 0 and 9 to make 15. Carry the 1 and add it with the 5, 2, and 8 to make 16. Once again carry the 1 and add it with the 4, 1, and 7 to make 13. Carry the 1 one more time and add it with the 3 to make 4. Insert a decimal point into your answer directly under the decimal points in the original numbers.
4.56 1.20+37.89 _ 43.65
The answer is 4.56+1.20+37.89=43.65.
Example 4
14.2+56.78+123.4=_
First, arrange the numbers vertically so that the decimal points line up.
14.2 56.78+123.4 _
Next, add one zero after the 2 in the first number and one zero after the 4 in the third number so that all numbers have two digits to the right of the decimal point.
14.20 56.78+123.40 _
Now, add from right to left. Start by adding 0, 8, and 0 to make 8. Then add 2, 7, and 4 to make 13. Carry the 1 and add it with the 4, 6, and 3 to make 14. Once again carry the 1 and add it with the 1, 5, and 12 to make 19. Insert a decimal point into your answer directly under the decimal points in the original numbers.
14.20 56.78+123.40 _ 194.38
The answer is 14.20+56.78+123.40=194.38.
Example 5
189.34+123.5+7.2=_
First, arrange the numbers vertically so that the decimal points line up.
189.34 123.5+7.2 _
Next, add one zero after the 5 in the second number and one zero after the 2 in the third number so that all numbers have two digits to the right of the decimal point.
189.34 123.50+7.20 _
Now, add from right to left. Start by adding 4, 0, and 0 to make 4. Then add 3, 5, and 2 to make 10. Carry the 1 and add it with the 9, 3, and 7 to make 20. Carry the 2 and add it with the 8 and 2 to make 12. Carry the 1 and add it with the 1 and 1 to make 3. Insert a decimal point into your answer directly under the decimal points in the original numbers.
189.34 123.50+7.20 _ 320.04
The answer is 189.34+123.50+7.20=320.04.
Review
Find each sum.
29.451+711.36
0.956+0.9885
13.69+42.23
58.744+12.06+8.8
67.90+1.2+3.456
Round to the nearest tenth, then find the sum.
88.95+23.451
690.026+831.163
3.27+7.818
0.99+0.909+0.048
5.67+3.45+1.23
Round to the nearest hundredth, then find the sum.
511.51109+99.0986
0.744+7.005+0.7071
32.368+303.409
1.0262+1.0242
1.309+3.4590
Review (Answers)
Click HERE to see the answer key or go to the Table of Contents and click on the Answer Key under the 'Other Versions' option.
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3780 | https://www.youtube.com/watch?v=xjnlymFLSdM | Counting Numbers | Numbers 1-10 lesson for children
The Singing Walrus - English Songs For Kids
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"Counting Numbers 1-10" is a part of a fun series of educational videos for kids.
The first scene slowly introduces the numbers 1-10 by counting little chicks on the screen, first counting from 1-10, then counting backwards from 10-1.
In the second scene, children repeat counting numbers 1-10 once more, this time with colorful balloons.
Lastly, the video finishes with a fun number game where the children have to count the chicks on the screen!
This video is the perfect supplement to our popular numbers song "Counting from 1-10". It can be used as an introductory video before showing the numbers song.
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Laaaaaa! Hi kids! Today we're going to learn the numbers 1-10. Do you know how to count from 1-10? Let's take a look. One [kids] One Two [kids] Two Three [kids] Three Four [kids] Four Five [kids] Five Six [kids] Six Seven [kids] Seven Eight [kids] Eight Nine [kids] Nine Ten [kids] Ten Okay Now let's count from 10-1. Ten [kids] Ten Nine [kids] Nine Eight [kids] Eight Seven [kids] Seven Six [kids] Six Five [kids] Five Four [kids] Four Three [kids] Three Two [kids] Two One [kids] One Fantastic! Let's try it one more time. One [kids] One Two [kids] Two Three [kids] Three Four [kids] Four Five [kids] Five Six [kids] Six Seven [kids] Seven Eight [kids] Eight Nine [kids] Nine Ten [kids] Ten Great! Now, let's count from 10-1. Ten [kids] Ten Nine [kids] Nine Eight [kids] Eight Seven [kids] Seven Six [kids] Six Five [kids] Five Four [kids] Four Three [kids] Three Two [kids] Two One [kids] One Outstanding! Now we're going to see if you can count the number of chicks on the screen. Tell me how many there are. [kids] Four [kids] One [kids] Nine [kids] Five [kids] Eight [kids] Three [kids] Six [kids] Ten [kids] Two [kids] Seven Great job kids! Click here to watch our fun "Counting from "1-10" music video. You can sing along and practice. Don't forget to click 'Like', and don't forget to subscribe to our Youtube channel. Thanks for watching! |
3781 | https://artofproblemsolving.com/wiki/index.php/Logarithm?srsltid=AfmBOoqwHsYuX_2qfWfQEC02UM-Ygf-9X5ULdrqRL1GuVPMQYE7juuR7 | Art of Problem Solving
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Logarithm
Logarithms are right inverses of powers.
Contents
1 Definition
2 Applications
2.1 Discrete Logarithm
3 Problems
4 Natural Logarithm
5 Problems
5.1 Introductory
5.2 Intermediate
6 Videos
Definition
Logarithms, the right inverses of powers can be written as for some and , where .
Applications
Some of the real powerful uses of logarithms come down to never having to deal with massive numbers. For example, would be a pain to have to calculate any time you wanted to use it (say in a comparison of large numbers). Its natural logarithm though (partly due to left to right parenthesized exponentiation) is only 7 digits before the decimal point. Comparing the logs of the numbers to a given precision can allow easier comparison than computing and comparing the numbers themselves. Logs also allow (with repetition) to turn left to right exponentiation into power towers (especially useful for tetration (exponentiation repetition with the same exponent)). For example,
Therefore by: and identities 1 and 2 above (2 being used twice), we get:
such that:
Discrete Logarithm
An only partially related value is the discrete logarithm, used in cryptography via modular arithmetic. It's the lowest value such that for given being integers (as well as the unknowns being integers).
It's related to the usual logarithm by the fact that if isn't an integer power of then is a lower bound on .
Problems
Evaluate:
Evaluate:
Simplify the following expression:
where N=(100!)3.
Natural Logarithm
The natural logarithm is the logarithm with base e. It is usually denoted , an abbreviation of the French logarithme normal, so that However, in higher mathematics such as complex analysis, the base 10 logarithm is typically disposed with entirely, the symbol is taken to mean the logarithm base and the symbol is not used at all. (This is an example of conflicting mathematical conventions.)
can also be defined as the area under the curve between 1 and a, or .
All logarithms are undefined in nonpositive reals, as they are complex. From the identity , we have . Additionally, for positive real .
Problems
Introductory
What is the value of for which ?
Source
Positive integers and satisfy the condition Find the sum of all possible values of .
Source
Intermediate
The sequence is geometric with and common ratio where and are positive integers. Given that find the number of possible ordered pairs
Source
The solutions to the system of equations
are and . Find . Source
Videos
Five-minute Intro to Logarithms
Introduction to Logarithms
Logarithms Properties
Khan Academy
Two-minute Intro to Logarithms
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3782 | https://pubmed.ncbi.nlm.nih.gov/2190330/ | Initial evaluation of the patient with blunt abdominal trauma - PubMed
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. 1990 Jun;70(3):495-515.
doi: 10.1016/s0039-6109(16)45126-1.
Initial evaluation of the patient with blunt abdominal trauma
O J McAnena1,E E Moore,J A Marx
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Affiliation
1 Denver General Hospital, Colorado.
PMID: 2190330
DOI: 10.1016/s0039-6109(16)45126-1
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Initial evaluation of the patient with blunt abdominal trauma
O J McAnena et al. Surg Clin North Am.1990 Jun.
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. 1990 Jun;70(3):495-515.
doi: 10.1016/s0039-6109(16)45126-1.
Authors
O J McAnena1,E E Moore,J A Marx
Affiliation
1 Denver General Hospital, Colorado.
PMID: 2190330
DOI: 10.1016/s0039-6109(16)45126-1
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Abstract
Unrecognized abdominal injury remains a distressingly frequent cause of preventable death following blunt trauma. Peritoneal signs are often subtle, overshadowed by pain from associated injury, and masked by head trauma or intoxicants. The initial management of the patient with blunt abdominal trauma should parallel the primary survey of airway, breathing, and circulation. Diagnostic peritoneal lavage remains the cornerstone of triage in patients with life-threatening blunt abdominal trauma. The only absolute contraindication to the procedure is an existing indication for laparotomy. Computed tomography is useful as a complementary diagnostic tool in selected patients, and it is the critical test for guiding nonoperative management of known intraperitoneal trauma. Routine ancillary tests for potentially occult injuries include nasogastric-tube placement for ruptures of the left diaphragm, Gastrografin contrast study for duodenum perforation, and pyelography for urologic injury. Ultrasonography may become a valuable tool in the initial assessment of the injured abdomen. Ultimately, the most important principle in the management of blunt abdominal trauma is repeat physical examination by an experienced surgeon.
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3783 | https://www.zoology.ubc.ca/~krebs/downloads/krebs_chapter_08_2017.pdf | CHAPTER 8 SAMPLING DESIGNS - RANDOM SAMPLING, ADAPTIVE AND SYSTEMATIC SAMPLING (Version 4, 14 March 2013) Page 8.1 SIMPLE RANDOM SAMPLING ....................................................................... 325 8.1.1 Estimation of Parameters ................................................................... 328 8.1.2 Estimation of a Ratio ........................................................................... 334 8.1.3 Proportions and Percentages............................................................ 337 8.2 STRATIFIED RANDOM SAMPLING ................................................................ 341 8.2.1 Estimates of Parameters ..................................................................... 343 8.22 Allocation of Sample Size .................................................................... 348 8.2.2.1 Proportional Allocation ................................................................ 348 8.2.2.2 Optimal Allocation ........................................................................ 350 8.2.3 Construction of Strata ......................................................................... 356 8.2.4 Proportions and Percentages............................................................. 359 8.3 ADAPTIVE SAMPLING .................................................................................... 360 8.3.1 Adaptive cluster sampling .................................................................. 361 8.3.2 Stratified Adaptive Cluster Sampling ................................................ 365 8.4 SYSTEMATIC SAMPLING ............................................................................... 365 8.5 MULTISTAGE SAMPLING ............................................................................... 369 8.5.1 Sampling Units of Equal Size ............................................................. 370 8.5.2 Sampling Units of Unequal Size ......................................................... 373 8.6 SUMMARY ....................................................................................................... 375 SELECTED REFERENCES ................................................................................... 377 QUESTIONS AND PROBLEMS ............................................................................. 378 Ecologists sample whenever they cannot do a complete enumeration of the population. Very few plant and animal populations can be completely enumerated, and so most of our ecological information comes from samples. Good sampling methods are critically Chapter 8 Page 325 important in ecology because we want to have our samples be representative of the population under study. How do we sample representatively? This chapter will attempt to answer this question by summarizing the most common sampling designs that statisticians have developed over the last 80 years. Sampling is a practical business and there are two parts of its practicality. First, the gear used to gather the samples must be designed to work well under field conditions. In all areas of ecology there has been tremendous progress in the past 40 years to improve sampling techniques. I will not describe these improvements in this book - they are the subject of many more detailed handbooks. So if you need to know what plankton sampler is best for oligotrophic lakes, or what light trap is best for nocturnal moths, you should consult the specialist literature in your subject area. Second, the method of placement and the number of samples must be decided, and this is what statisticians call sampling design. Should samples be placed randomly or systematically? Should different habitats be sampled separately or all together? These are the general statistical questions I will address in this and the next chapter. I will develop a series of guidelines that will be useful in sampling plankton with nets, moths with light traps, and trees with distance methods. The methods discussed here are addressed in more detail by Cochran (1977), Jessen (1978), and Thompson (1992). 8.1 SIMPLE RANDOM SAMPLING The most convenient starting point for discussing sampling designs is simple random sampling. Like many statistical concepts, "random sampling" is easier to explain on paper than it is to apply in the field. Some background is essential before we can discuss random sampling. First, you must specify very clearly what the statistical population is that you are trying to study. The statistical population may or may not be a biological population, and the two ideas should not be confused. In many cases the statistical population is clearly specified: the white-tailed deer population of the Savannah River Ecological Area, or the whitefish population of Brooks Lake, or the black oaks of Warren Dunes State Park. But in other cases the statistical population has poorly defined boundaries: the mice that will enter live traps, the haddock population of George’s Bank, the aerial aphid population over southern England, the seed bank of Erigeron canadensis. Part of this vagueness in ecology depends on Chapter 8 Page 326 spatial scale and has no easy resolution. Part of this vagueness also flows from the fact that biological populations also change over time (Van Valen 1982). One strategy for dealing with this vagueness is to define the statistical population very sharply on a local scale, and then draw statistical inferences about it. But the biological population of interest is usually much larger than a local population, and one must then extrapolate to draw some general conclusions. You must think carefully about this problem. If you wish to draw statistical inferences about a widespread biological population, you should sample the widespread population. Only this way can you avoid extrapolations of unknown validity. The statistical population you wish to study is a function of the question you are asking. This problem of defining the statistical population and then relating it to the biological population of interest is enormous in field ecology, and almost no one discusses it. It is the first thing you should think about when designing your sampling scheme. Second, you must decide what the sampling unit, is in your population. The sampling unit could be simple, like an individual oak tree or an individual deer, or it can be more complex like a plankton sample, or a 4 m2 quadrat, or a branch of an apple tree. The sample units must potentially cover the whole of the population and they must not overlap. In most areas of ecology there is considerable practical experience available to help decide what sampling unit to use. The third step is to select a sample, and a variety of sampling plans can be adopted. The aim of the sampling plan is to maximize efficiency - to provide the best statistical estimates with the smallest possible confidence limits at the lowest cost. To achieve this aim we need some help from theoretical statistics so that we can estimate the precision and the cost of the particular sampling design we adopt. Statisticians always assume that a sample is taken according to the principles of probability sampling, as follows: 1. Define a set of distinct samples S1, S2, S3 ... in which certain specific sampling units are assigned to S1, some to S2, and so on. 2. Each possible sample is assigned a probability of selection. Chapter 8 Page 327 3. Select one of the Si samples by the appropriate probability and a random number table. If you collect a sample according to these principles of probability sampling, a statistician can determine the appropriate sampling theory to apply to the data you gather. Some types of probability sampling are more convenient than others, and simple random sampling is one. Simple random sampling is defined as follows: 1. A statistical population is defined that consists of N sampling units; 2. n units are selected from the possible samples in such a way that every unit has an equal chance of being chosen. The usual way of achieving simple random sampling is that each possible sample unit is numbered from 1 to N. A series of random numbers between 1 and N is then drawn either from a table of random numbers or from a set of numbers in a hat. The sample units which happen to have these random numbers are measured and constitute the sample to be analyzed. It is hard in ecology to follow these simple rules of random sampling if you wish the statistical population to be much larger than the local area you are actually studying. Usually, once a number is drawn in simple random sampling it is not replaced, so we have sampling without replacement. Thus, if you are using a table of random numbers and you get the same number twice, you ignore it the second time. It is possible to replace each unit after measuring so that you can sample with replacement but this is less often used in ecology. Sampling without replacement is more precise than sampling with replacement (Caughley 1977). Simple random sampling is sometimes confused with other types of sampling that are not based on probability sampling. Examples abound in ecology: 1. Accessibility sampling: the sample is restricted to those units that are readily accessible. Samples of forest stands may be taken only along roads, or deer may be counted only along trails. Chapter 8 Page 328 2. Haphazard sampling: the sample is selected haphazardly. A bottom sample may be collected whenever the investigator is ready, or ten dead fish may be picked up for chemical analysis from a large fish kill. 3. Judgmental sampling: the investigator selects on the basis of his or her experience a series of "typical" sample units. A botanist may select 'climax' stands of grassland to measure. 4. Volunteer sampling: the sample is self-selected by volunteers who will complete some questionnaire or be used in some physiological test. Hunters may complete survey forms to obtain data on kill statistics. The important point to remember is that all of these methods of sampling may give the correct results under the right conditions. Statisticians, however, reject all of these types of sampling because they can not be evaluated by the theorems of probability theory. Thus the universal recommendation: random sample! But in the real world it is not always possible to use random sampling, and the ecologist is often forced to use non-probability sampling if he or she wishes to get any information at all. In some cases it is possible to compare the results obtained with these methods to those obtained with simple random sampling (or with known parameters) so that you could decide empirically if the results were representative and accurate. But remember that you are always on shaky ground if you must use non-probability sampling, so that the means and standard deviations you calculate may not be close to the true values. Whenever possible use some conventional form of random sampling. 8.1.1 Estimation of Parameters In simple random sampling, one or more characteristics are measured on each experimental unit. For example, in quadrat sampling you might count the number of individuals of Solidago spp. and the number of individuals of Aster spp. In sampling deer, you might record for each individual its sex, age, weight, and fat index. In sampling starling nests you might count the number of eggs and measure their length. Chapter 8 Page 329 In all these cases and in many more, ecological interest is focused on four characteristics of the population1: 1. Total = X ; for example, the total number of Solidago individuals in the entire 100 ha study field. 2. Mean = x ; for example, the average number of Solidago per m2. 3. Ratio of two totals = R = x y ; for example, the number of Solidago per Aster in the study area. 4. Proportion of units in some defined class; for example, the proportion of male deer in the population. We have seen numerous examples of characteristics of these types in the previous seven chapters, and their estimation is covered in most introductory statistics books. We cover them again here briefly because we need to add to them an idea that is not usually considered in introductory books - the finite population correction. For any statistical population consisting of N units, we define the finite population correction (fpc) as: fpc 1-N n n N N − = = (8.1) where: fpc Finite population correction Total population size Sample size N n = = = and the fraction of the population sampled (n/N) is sometimes referred to as f. In a very large population the finite population correction will be 1.0, and when the whole population is measured, the fpc will be 0.0. Ecologists working with abundance data such as quadrat counts of plant density immediately run into a statistical problem at this point. Standard statistical procedures 1 We may also be interested in the variance of the population, and the same general procedures apply. Chapter 8 Page 330 deal with normally distributed raw data in which the standard deviation is independent of the mean. But ecological abundance data typically show a positive skew (e.g. Figure 4.6 page 000) with the standard deviation increasing with the mean. These complications violate the assumptions of parametric statistics, and the simplest way of correcting ecological abundance data is to transform all counts by a log transformation: ( ) log Y X = (8.2) where Y = transformed data X = original data All analysis is now done on these Y-values which tend to satisfy the assumptions of parametric statistics1. In the analyses that follow with abundance data we effectively replace all the observed X values with their Y transformed counterpart and use the conventional statistical formulas found in most textbooks that define the variable of interest as X. Cochran (1977) demonstrates that unbiased estimates of the population mean and total for normally distributed data are given by the following formulas. For the mean: i x x n = ∑ (8.3) where: = Population mean = Observed value of in sample = Sample size i x x x i n For the variance2 of the measurements we have the usual formula: ( ) 2 2 1 i x x s n − = − ∑ (8.4) 1 This transformation will be discussed in detail in Chapter 15, page 000. 2 If the data are compiled in a frequency distribution, the appropriate formulas are supplied in Appendix 1, page 000. Chapter 8 Page 331 and the standard error of the population mean x is given by: ( ) 2 1 x s s f n = − (8.5) where: 2 Standard error of the mean Variance of the measurements as defined above (8.4) Sample size Sampling fraction / x s x s n f n N = = = = = These formulas are similar to those you have always used, except for the introduction of the finite population correction. Note that when the sampling fraction (n/N) is low, the finite population correction is nearly 1.0 and so the size of the population has no effect on the size of the standard error. For example, if you take a sample of 500 measurements from two populations with the same variance ( 2 s = 484.0) and population A is small (N = 10,000) and population B is a thousand times larger (N = 10,000,000), the standard errors of the mean differ by only 2.5% because of the finite population correction. For this reason, the finite population correction is usually ignored whenever the sampling fraction (n/N) is less than 5% (Cochran 1977). Estimates of the population total are closely related to these formulas for the population mean. For the population total: ˆ X N x = (8.6) where: ˆ Estimated population total Total size of population Mean value of population X N x = = = The standard error of this estimate is given by: X x s N s = (8.7) where: = Standard error of the population total = Total size of the population = Standard error of the mean from equation (8.4) X x s N s Confidence limits for both the population mean and the population total are usually derived by the normal approximation: Chapter 8 Page 332 x x t s α ± (8.8) where: ( ) = Student's value for - 1 degrees of freedom for the 1-level of confidence t t n α α This formula is used all the time in statistics books with the warning not to use it except with "large" sample sizes. Cochran (1977, p.41) gives a rule-of-thumb that is useful for data that has a strong positive skew (like the negative binomial curves in Figure 4.8, page 000). First, define Fisher’s measure of skewness: ( ) 1 3 3 Fisher's measure of skewness 1 g x x ns = = − ∑ (8.9) Cochran’s rule is that you have a large enough sample to use the normal approximation (equation 8.8) if – 2 1 25 n g > (8.10) where: = Sample size = Standard deviation n s Sokal and Rohlf (1995, p. 116) show how to calculate g1 from sample data and many statistical packages provide computer programs to do these calculations. Box 8.1 (page 000) illustrates the use of these formulae for simple random sampling. Box 8.1 ESTIMATION OF POPULATION MEAN AND POPULATION TOTAL FROM SIMPLE RANDOM SAMPLING OF A FINITE POPULATION A biologist obtained body weights of male reindeer calves from a herd during the seasonal roundup. He obtained weights on 315 calves out of a total of 1262 in the herd, checked the assumption of a normal distribution, and summarized the data: Body weight class (kg) Midpoint, x Observed frequency, fx 29.5-34.5 32 4 34.5-39.5 37 13 39.5-44.5 42 20 44.5-49.5 47 49 49.5-54.5 52 61 Chapter 8 Page 333 54.5-59.5 57 72 59.5-64.5 62 57 64.5-69.5 67 25 69.5-74.5 72 12 74.5-79.5 77 2 The observed mean is, from the grouped version of equation (8.3), (4)(32) + (13)(37) + (20)(42) + 315 17,255 = = 54.78 kg 315 x f x x n = = ∑ The observed variance is, from the grouped version of equation (8.4), ( ) 2 2 2 2 / - 1 969,685 - (17,255) / 315 78.008 314 x x f x f x n s n − = = = ∑ ∑ The standard error of the mean weight, from equation (8.5): 2 1 78.008 315 1- 0.4311 315 1262 x s s f n = − = = From equation (8.8) we calculate g1 = ( ) ( ) 3 3 1/ ns x x − ∑ = -0.164. Hence, applying equation (8.10) 2 1 2 25 315 25( 0.164) 0.67 n g > >> − = so that we have a “large” sample according to Cochran’s rule of thumb. Hence we can compute normal confidence limits from equation (8.8): for α = .05 the t value for 315 d.f. is 1.97 x x t s α ± 54.78 1.97(0.431) ± or 95% confidence limits of 53.93 to 55.63 kg. To calculate the population total biomass, we have from equation (8.6): Chapter 8 Page 334 ˆ 1262(54.78) 69,129.6 kg X N x = = = and the standard error of this total biomass estimate for the herd is, from equation (8.7), ˆ 1262(0.4311) 544.05 x X s Ns = = = and normal 95% confidence limits can be calculated as above to give for the population total: 69,130 1072 kg ± It is important to remember that if you have carried out a log transform on abundance data, all of these estimates are in the log-scale. You may wish to transform them back to the original scale of measurement, and this procedure is not immediately obvious (i.e. do not simply take anti-logs) and is explained in Chapter 15, Section 15.1.1 (page 000). 8.1.2 Estimation of a Ratio Ratios are not as commonly used in ecological work as they are in taxonomy, but sometimes ecologists wish to estimate from a simple random sample a ratio of two variables, both of which vary from sampling unit to sampling unit. For example, wildlife managers may wish to estimate the wolf/moose ratio for several game management zones, or behavioral ecologists may wish to measure the ratio of breeding females to breeding males in a bird population. Ratios are peculiar statistical variables with strange properties that few biologists appreciate (Atchley et al. 1976). Ratios of two variables are not just like ordinary measurements and to estimate means, standard errors, and confidence intervals for ecological ratios, you should use the following formulae from Cochran (1977): For the mean ratio: ˆ x R y = (8.11) where: Chapter 8 Page 335 ˆ Estimated mean ratio of to Observed mean value of Observed mean value of R x y x x y y = = = The standard error of this estimated ratio is: 2 2 2 ˆ ˆ ˆ 2 1 1 R x R xy R y f s n n y − + − = − ∑ ∑ ∑ (8.12) where: ˆ Estimated standard error of the ratio Sampling fraction / Sample size Observed mean of measurement (denominator of ratio) R s R f n N n y Y = = = = = and the summation terms are the usual ones defined in Appendix 1 (page 000). The estimation of confidence intervals for ratios from the usual normal approximation (eq. 8.8) is not valid unless sample size is large as defined above (page 332) (Sukhatme and Sukhatme 1970). Ratio variables are often skewed to the right and not normally distributed, particularly when the coefficient of variation of the denominator is relatively high (Atchley et al. 1976). The message is that you should treat the computed confidence intervals of a ratio as only an approximation unless sample size is large. Box 8.2 illustrates the use of these formulas for calculating a ratio estimate. Box 8.2 ESTIMATION OF A RATIO OF TWO VARIABLES FROM SIMPLE RANDOM SAMPLING OF A FINITE POPULATION Wildlife ecologists interested in measuring the impact of wolf predation on moose populations in British Columbia obtained estimates by aerial counting of the population size of wolves and moose and 11 subregions which constituted 45% of the total game management zone. Subregion No. of wolves No. of moose Wolves/moose A 8 190 0.0421 B 15 370 0.0405 C 9 460 0.0196 D 27 725 0.0372 E 14 265 0.0528 F 3 87 0.0345 G 12 410 0.0293 Chapter 8 Page 336 H 19 675 0.0281 I 7 290 0.0241 J 10 370 0.0270 K 16 510 0.0314 x = 0.03333 SE = 0.00284 95% CL = 0.02701 to 0.03965 The mean numbers of wolves and moose are 8 + 15 + 9 + 27 + 140 12.727 wolves 11 11 190 + 370 + 460 + 4352 395.64 moose 11 11 x x n y y n = = = = = = = = ∑ ∑ The mean ratio of wolves to moose is estimated from equation (8.11): 12.727 ˆ 0.03217 wolves/moose 395.64 x R y = = = The standard error of this estimate (equation 8.12) requires three sums of the data: 2 2 2 2 2 2 2 2 8 + 15 + 9 + 2214 y 190 + 370 + 460 + 2,092,844 (8)(190) + (15)(370) + (9)(460) + 66,391 x xy = = = = = = ∑ ∑ ∑ From equation (8.12): 2 2 2 ˆ 2 ˆ ˆ 2 1 1 1 - 0.45 2214 - 2(0.032)(66,391) + (0.032 )(2,092,844) 11 - 1 11(395.64) 0.7416 108.31 0.00186 1312.19 10 R x R xy R y f s n n y − + − = − = = = ∑ ∑ ∑ the 95% confidence limits for this ratio estimate are thus (tα for 10 d.f. for α = .05 is 2.228): ˆ or 0.03217 2.228(0.00186) R R t s α ± ± Chapter 8 Page 337 or 0.02803 to 0.03631 wolves per moose. 8.1.3 Proportions and Percentages The use of proportions and percentages is common in ecological work. Estimates of the sex ratio in a population, the percentage of successful nests, the incidence of disease, and a variety of other measures are all examples of proportions. In all these cases we assume there are 2 classes in the population, and all individuals fall into one class or the other. We may be interested in either the number or the proportion of type X individuals from a simple random sample: Population Sample No. of total individuals N n No. of individuals of type X A a Proportion of type X individuals A P N = ˆ a p n = In statistical work the binomial distribution is usually applied to samples of this type, but when the population is finite the more proper distribution to use is the hypergeometric distribution1 (Cochran 1977). Fortunately, the binomial distribution is an adequate approximation for the hypergeometric except when sample size is very small. For proportions, the sample estimate of the proportion P is simply: ˆ a p n = (8.13) where: ˆ Proportion of type individuals Number of type individuals in sample Sample size p X a X n = = = The standard error of the estimated proportion ˆ p is from Cochran (1977), ˆ ˆ ˆ 1 1 P pq s f n = − − (8.14) 1 Zar (1996, pg. 520) has a good brief description of the hypergeometric distribution. Chapter 8 Page 338 where: ˆ = Standard error of the estimated population Sampling fraction / ˆ Estimated proportion of types ˆ ˆ 1 - Sample size P s p f n N p X q p n = = = = = For example, if in a population of 3500 deer, you observe a sample of 850 of which 400 are males: ˆ ˆ 400/850 0.4706 850 (0.4706)(1 - 0.4706) 1 0.0149 3500 849 P p s = = = − = To obtain confidence limits for the proportion of x-types in the population several methods are available (as we have already seen in Chapter 2, page 000). Confidence limits can be read directly from graphs such as Figure 2.2 (page 000), or obtained more accurately from tables such as Burnstein (1971) or from Program EXTRAS (Appendix 2, page 000). For small sample sizes the exact confidence limits can be read from tables of the hypergeometric distribution in Lieberman and Owen (1961). If sample size is large, confidence limits can be approximated from the normal distribution. Table 8.1 lists sample sizes that qualify as "large". The normal approximation to the binomial gives confidence limits of: ˆ ˆ 1 ˆ 1 1 2 pq p z f n n α ± − + − or ˆ 1 ˆ 2 P p z s n α ± + (8.15) where: Chapter 8 Page 339 ˆ ˆ = Estimated proportion of types = Standard normal deviate (1.96 for 95% confidence intervals, 2.576 for 99% confidence intervals) Standard error of the estimated proportion (equat P p X z s α = ion 8.13) Sampling fraction / ˆ ˆ 1 - proportion of types in sample Sample size f n N q p Y n = = = = = TABLE 8.1 SAMPLE SIZES NEEDED TO USE THE NORMAL APPROXIMATION (EQUATION 8.15) FOR CALCULATING CONFIDENCE INTERVALS FOR PROPORTIONSa Proportion, p Number of individuals in the smaller class, np Total sample size, N 0.5 15 30 0.4 20 50 0.3 24 80 0.2 40 200 0.1 60 600 0.05 70 1400 a For a given value of p do not use the normal approximation unless you have a sample size this large or larger. Source: Cochran, 1977. The fraction (1/2n) is a correction for continuity, which attempts to correct partly for the fact that individuals come in units of one, so it is possible, for example, to observe 216 male deer or 217, but not 216.5. Without this correction the normal approximation usually gives a confidence belt that is too narrow. For the example above, the 95% confidence interval would be: 1 0.4706 1.96(0.0149) 2(850) or 0.4706 0.0298 (0.44 to 0.50 males) ± + ± Note that the correction for continuity in this case is very small and if ignored would not change the confidence limits except in the fourth decimal place. Chapter 8 Page 340 Not all biological attributes come as two classes like males and females of course, and we may wish to estimate the proportion of organisms in three or four classes (instar I, II, III and IV in insects for example). These data can be treated most simply by collapsing them into two-classes, instar II vs. all other instars for example, and using the methods described above. A better approach is described in Section 13.4.1 for multinomial data. One practical illustration of the problem of estimating proportions comes from studies of disease incidence (Ossiander and Wedemeyer 1973). In many hatchery populations of fish, samples need to be taken periodically and analyzed for disease. Because of the cost and time associated with disease analysis, individual fish are not always the sampling unit. Instead, groups of 5, 10 or more fish may be pooled and the resulting pool analyzed for disease. One diseased fish in a group of 10 will cause that whole group to be assessed as disease-positive. Worlund and Taylor (1983) developed a method for estimating disease incidence in populations when samples are pooled. The sampling problem is acute here because disease incidence will often be only 1-2%, and at low incidences of disease, larger group sizes are more efficient in estimating the proportion diseased. Table 8.2 gives the confidence intervals expected for various sizes of groups and number of groups when the expected disease incidence varies from 1-10%. For group size = 1, these limits are the same as those derived above (equation 8.14). But Table 8.2 shows clearly that, at low incidence, larger group sizes are much more precise than smaller group sizes. Worlund and Taylor (1983) provide more details on optimal sampling design for such disease studies. One problem with disease studies is that diseased animals might be much easier to catch than healthy animals, and one must be particularly concerned with obtaining a random sample of the population. TABLE 8.2 WIDTH OF 90% CONFIDENCE INTERVALS FOR DISEASE INCIDENCEa Percent disease incidence No. 1.0% 2.0% 5.0% 10% of Group size, k Group size, k Group size,k Group size, k Chapter 8 Page 341 groups , n 1 5 10 159 1 5 10 79 1 5 10 31 1 5 10 15 12 4.7 2.1 1.5 - 6.6 3.0 2.2 - 10.3 4.9 3.7 - 14.2 7.1 - - 20 3.6 1.6 1.2 - 5.1 2.3 1.7 - 8.0 3.8 2.8 - 11.0 5.5 4.5 - 30 3.0 1.3 1.0 0.4 4.2 1.9 1.4 0.7 6.5 3.1 2.3 1.8 9.0 4.5 3.7 3.5 60 2.1 1.0 0.7 0.3 3.0 1.4 1.0 0.5 4.6 2.2 1.6 1.3 6.4 3.2 2.6 2.5 90 1.7 0.8 0.6 0.2 2.4 1.1 0.8 0.4 3.8 1.8 1.3 1.0 5.2 2.6 2.1 2.0 120 1.5 0.7 0.5 0.2 2.1 1.0 0.7 0.4 3.3 1.5 1.2 0.8 4.5 2.2 1.8 1.8 150 1.3 0.6 0.4 0.2 1.9 0.8 0.6 0.3 2.9 1.4 1.0 0.8 4.0 2.0 1.6 1.6 250 1.0 0.5 0.3 0.1 1.4 0.7 0.5 0.2 2.3 1.1 0.8 0.6 3.1 1.6 1.3 1.2 350 0.9 0.4 0.3 0.1 1.2 0.6 0.4 0.2 1.9 0.9 0.7 0.5 2.6 1.3 1.1 1.0 450 0.8 0.3 0.2 0.1 1.1 0.5 0.4 0.2 1.7 0.8 0.6 0.5 2.3 1.2 1.0 0.9 a A number of groups (n) of group size k are tested for disease. If one individual in a group has a disease, the whole group is diagnosed as disease positive. The number in the table should be read as “ % d ± ,” that is, as one-half of the width of the confidence interval. Source: Worlund and Taylor, 1983. 8.2 STRATIFIED RANDOM SAMPLING One of the most powerful tools you can use in sampling design is to stratify your population. Ecologists do this all the time intuitively. Figure 8.1 gives a simple example. Population density is one of the most common bases of stratification in ecological work. When an ecologist recognizes good and poor habitats, he or she is implicitly stratifying the study area. In stratified sampling the statistical population of N units is divided into subpopulations which do not overlap and which together comprise the entire population. Thus: N = N1 + N2 + N3 + ..... NL where L = total number of subpopulations Chapter 8 Page 342 Stratum C Stratum B Stratum A Figure 8.1 The idea of stratification in estimating the size of a plant or animal population. Stratification is made on the basis of population density. Stratum A has about ten times the density of Stratum C. The subpopulations are called strata by statisticians. Clearly if there is only one stratum, we are back to the kind of sampling we have discussed earlier in this chapter. To obtain the full benefit from stratification you must know the sizes of all the strata (N1, N2, ...). In many ecological examples, stratification is done on the basis of geographical area, and the sizes of the strata are easily found in m2 or km2, for example. There is no need for the strata to be of the same size. Once you have determined what the strata are, you sample each stratum separately. The sample sizes for each stratum are denoted by subscripts: n1 = sample size in stratum 1 n2 = sample size in stratum 2 and so on. If within each stratum you sample using the principles of simple random sampling outlined above (page 327), the whole procedure is called stratified random sampling. It is not necessary to sample each stratum randomly, and you could, for example, sample systematically within a stratum. But the problems outlined above would then mean that it would be difficult to estimate how reliable such sampling is. So it is recommended to sample randomly within each stratum. Chapter 8 Page 343 Ecologists have many different reasons for wishing to stratify their sampling. Four general reasons are common (Cochran 1977): 1. Estimates of means and confidence intervals may be required separately for each subpopulation. 2. Sampling problems may differ greatly in different areas. Animals may be easier or harder to count in some habitats than they are in others. Offshore samples may require larger boats and be more expensive to get than nearshore samples. 3. Stratification may result in a gain in precision in the estimates of the parameters of the whole population. Confidence intervals can be narrowed appreciably when strata are chosen well. 4. Administrative convenience may require stratification if different field laboratories are doing different parts of the sampling. Point 3 is perhaps the most critical one on this list, and I will now discuss how estimates are made from stratified sampling and illustrate the gains one can achieve. 8.2.1 Estimates of Parameters For each of the subpopulations (N1, N2, ...) all of the principles and procedures of estimation outlined above can be used. Thus, for example, the mean for stratum 1 can be estimated from equation (8.2) and the variance from equation (8.3). New formulas are however required to estimate the mean for the whole population N. It will be convenient, before I present these formulae, to outline one example of stratified sampling so that the equations can be related more easily to the ecological framework. Table 8.3 gives information on a stratified random sample taken on a caribou herd in central Alaska by Siniff and Skoog (1964). They used as their sampling unit a quadrat of 4 sq. miles, and they stratified the whole study zone into six strata, based on a pilot survey of caribou densities in different regions. Table 8.3 shows that the 699 total sampling units were divided very unequally into the six strata, so that the largest stratum (A) was 22 times the size of the smallest stratum (D). Chapter 8 Page 344 TABLE 8.3 STRATIFIED RANDOM SAMPLING OF THE NELCHINA CARIBOU HERD IN ALASKA BY SINIFF AND SKOOG (1964)a Stratum Stratum size, Nh Stratum weight, Wh Sample size, nh Mean no. of caribou counted per sampling unit, h x Variance of Caribou counts, 2 h s A 400 0.572 98 24.1 5575 B 30 0.043 10 25.6 4064 C 61 0.087 37 267.6 347,556 D 18 0.026 6 179.0 22,798 E 70 0.100 39 293.7 123,578 F 120 0.172 21 33.2 9795 Total 699 1.000 211 a Six strata were delimited in preliminary surveys based on the relative caribou density. Each sampling unit was 4 square miles. A random sample was selected in each stratum and counts were made from airplanes. Source: Siniff and Skoog, 1964. We define the following notation for use with stratified sampling: Stratum weight = = h h N W N (8.16) where: Nh = Size of stratum h (number of possible sample units in stratum h) N = Size of entire statistical population The stratum weights are proportions and must add up to 1.0 (Table 8.3). Note that the Nh must be expressed in "sample units". If the sample unit is 0.25 m2, the sizes of the strata must be expressed in units of 0.25 m2 (not as hectares, or km2). Simple random sampling is now applied to each stratum separately and the means and variances calculated for each stratum from equations (8.2) and (8.3). We will defer until the next section a discussion on how to decide sample size in each stratum. Table 8.3 gives sample data for a caribou population. Chapter 8 Page 345 The overall mean per sampling unit for the entire population is estimated as follows (Cochran 1977): 1 ST = L h h h N x x N = ∑ (8.17) where: ST = Stratified population mean per sampling unit = Size of stratum = Stratum number (1, 2, 3, , ) = Observed mean for stratum = Total population size = h h h x N h h L x h N N ∑ Note that ST x is a weighted mean in which the stratum sizes are used as weights. For the data in Table 8.3, we have: ST (400)(24.1) + (30)(25.6) + (61)(267.6 + = 699 = 77.96 caribou/sample unit x Given the density of caribou per sampling unit, we can calculate the size of the entire caribou population from the equation: ST ST ˆ = X N x (8.18) where: ST ST ˆ Population total Number of sample units in entire population x Stratified mean per sampling unit (equation 8.17) X N = = = For the caribou example: ST ˆ = 699(77.96) = 54,497 caribou X so the entire caribou herd is estimated to be around 55 thousand animals at the time of the study. The variance of the stratified mean is given by Cochran (1977, page 92) as: Chapter 8 Page 346 ( ) ( ) 2 2 ST 1 Variance of 1 L h h h h h w s x f n = = − ∑ (8.19) where: 2 / Stratum weight (equation 8.16) Observed variance of stratum (equation 8.4) Sample size in stratum Sampling fraction in stratum / h h h h h h W s h n h f h n N = = = = = The last term in this summation is the finite population correction and it can be ignored if you are sampling less that 5% of the sample units in each stratum. Note that the variance of the stratified means depends only on the size of the variances within each stratum. If you could divide a highly variable population into homogeneous strata such that all measurements within a stratum were equal, the variance of the stratified mean would be zero, which means that the stratified mean would be without any error! In practice of course you cannot achieve this but the general principle still pertains: pick homogeneous strata and you gain precision. For the caribou data in Table 8.3 we have: ( ) ( ) 2 ST 2 (0.572) (5575) 98 Variance of = 1 - 98 400 (0.043) 4064 10 + 1 - + 10 30 = 69.803 x The standard error of the stratified mean is the square root of its variance: ( ) ST ST Standard error of = Variance of = 69.803 = 8.355 x x Note that the variance of the stratified mean cannot be calculated unless there are at least 2 samples in each stratum. The variance of the population total is simply: ( ) ( ) ( ) 2 ST ST ˆ Variance of = variance of X N x (8.20) For the caribou the variance of the total population estimate is: ( ) 2 ST ˆ Variance of = 699 (69.803) 34,105,734 X = Chapter 8 Page 347 and the standard error of the total is the square root of this variance, or 5840. The confidence limits for the stratified mean and the stratified population total are obtained in the usual way: ST ST (standard error of ) x t x α ± (8.21) ST ST ˆ ˆ (standard error of ) X t X α ± (8.22) The only problem is what value of Student's t to use. The appropriate number of degrees of freedom lies somewhere between the lowest of the values (nh - 1) and the total sample size ( 1 h n − ∑ ). Cochran (1977, p.95) recommends calculating an effective number of degrees of freedom from the approximate formula: ( ) 2 2 1 2 4 1 d.f. 1 L h h h L h h h h g s g s n − − ≈ − ∑ ∑ (8.23) where: ( ) 2 d.f. Effective number of degrees of freedom for the confidence limits in equations (8.21) and (8.22) / Observed variance in stratum Sample size in stratum S h h h h h h h h g N N n n s h n h N = = − = = = ize of stratum h For example, from the data in Table 8.3 we obtain Stratum gh A 1232.65 B 60.00 C 39.57 D 36.00 E 55.64 F 565.71 Chapter 8 Page 348 and from equation (8.22): 2 12 (34,106,392) d.f. = 134.3 8.6614 X 10 ≈ and thus for 95% confidence intervals for this example tα = 1.98. Thus for the population mean from equation (8.20) the 95% confidence limits are: 77.96 1.98(8.35) ± or from 61.4 to 94.5 caribou per 4 sq. miles. For the population total from equation (8.21) the 95% confidence limits are: 54,497 1.98(5840) ± or from 42,933 to 66,060 caribou in the entire herd. 8.22 Allocation of Sample Size In planning a stratified sampling program you need to decide how many sample units you should measure in each stratum. Two alternate strategies are available for allocating samples to strata - proportional allocation or optimal allocation. 8.2.2.1 Proportional Allocation The simplest approach to stratified sampling is to allocate samples to strata on the basis of a constant sampling fraction in each stratum. For example, you might decide to sample 10% of all the sample units in each stratum. In the terminology defined above: h h n N n N = (8.24) For example, in the caribou population of Table 8.3, if you wished to sample 10% of the units, you would count 40 units in stratum A, 3 in stratum B, 6 in stratum C, 2 in stratum D, 7 in E and 12 in F. Note that you should always constrain this rule so that at least 2 units are sampled in each stratum so that variances can be estimated. Equation (8.23) tells us what fraction of samples to assign to each stratum but we still do not know how many samples we need to take in total (n). In some situations this Chapter 8 Page 349 is fixed and beyond control. But if you are able to plan ahead you can determine the sample size you require as follows. • Decide on the absolute size of the confidence interval you require in the final estimate. For example, in the caribou case you may wish to know the density to ±10 caribou/4 sq. miles with 95% confidence. • Calculate the estimated total number of samples needed for an infinite population from the approximate formula (Cochran 1977 p. 104): 2 2 4 h h W s n d ≈ ∑ (8.25) where: 2 Total sample size required (for large population) Stratum weight Observed variance of stratum Desired absolute precision of stratified mean (width of confidence interval h h n W s h d = = = = is ) d ± This formula is used when 95% confidence intervals are specified in d. If 99% confidence intervals are specified, replace 4 in equation (8.24) with 7.08, and for 90% confidence intervals, use 2.79 instead of 4. For a finite population correct this estimated n by equation (7.6), page 000: = 1 n n nN + where: = Total sample size needed in finite population of size n N For the caribou data in Table 8.3, if an absolute precision of ± 10 caribou / 4 square miles is needed: [ ] 2 4 (0.572)(5575) + (0.043)(4064) + 1 = 1933.6 10 n ≈ Note that this recommended sample size is more than the total sample units available! For a finite population of 699 sample units: Chapter 8 Page 350 1933.6 513.4 sample units 1933.6 1 + 699 n∗≈ = These 514 sample units would then be distributed to the six strata in proportion to the stratum weight. Thus, for example, stratum A would be given (0.572)(514) or 294 sample units. Note again the message that if you wish to have high precision in your estimates, you will have to take a large sample size. 8.2.2.2 Optimal Allocation When deciding on sample sizes to be obtained in each stratum, you will find that proportional allocation is the simplest procedure. But it is not the most efficient, and if you have prior information on the sampling methods, more powerful allocation plans can be specified. In particular, you can minimize the cost of sampling with the following general approach developed by Cochran (1977). Assume that you can specify the cost of sampling according to a simple cost function: O h h C c c n = + ∑ (8.26) where: O Total cost of sampling Overhead cost Cost of taking one sample in stratum Number of samples taken in stratum h h C c c h n h = = = = Of course the cost of taking one sample might be equal in all strata but this is not always true. Costs can be expressed in money or in time units. Economists have developed much more complex cost models but we shall stick to this simple model here. Cochran (1977 p. 95) demonstrates that, given the cost function above, the standard error of the stratified mean is at a minimum when: is proportional to h h h h N s n c Chapter 8 Page 351 This means that we should apportion samples among the strata by the ratio: / / h h h h h h h N s c n n N s c = ∑ (8.27) This formula leads to three useful rules-of-thumb in stratified sampling: in a given stratum, take a larger sample if 1. The stratum is larger 2. The stratum is more variable internally 3. Sampling is cheaper in the stratum Once we have done this we can now go in one of two ways: (1) minimize the standard error of the stratified mean for a fixed total cost. If the cost is fixed, the total sample size is dictated by: ( ) ( ) ( ) O / h h h h h h C c N s c n N s c − = ∑ ∑ (8.28) where: O Total sample size to be used in stratified sampling for all strata combined Total cost (fixed in advance) Overhead cost Size of stratum standard deviation of stratum cost to h h h n C c N h s h c = = = = = = take one sample in stratum h Box 8.3 illustrates the use of these equations for optimal allocation. Box 8.3 OPTIMAL AND PROPORTIONAL ALLOCATION IN STRATIFIED RANDOM SAMPLING Russell (1972) sampled a clam population using stratified random sampling and obtained the following data: Stratum Size of stratum, Nh Stratum weight, Wh Sample size, nh Mean (bushels), xh Variance, 2 h s A 5703.9 0.4281 4 0.44 0.068 B 1270.0 0.0953 6 1.17 0.042 C 1286.4 0.0965 3 3.92 2.146 Chapter 8 Page 352 D 5063.9 0.3800 5 1.80 0.794 N = 13,324.2 1.0000 18 Stratum weights are calculated as in equation (8.16). I use these data to illustrate hypothetically how to design proportional and optimal allocation sampling plans. Proportional Allocation If you were planning this sampling program based on proportional allocation, you would allocate the samples in proportion to stratum weight (equation 8.24): Stratum Fraction of samples to be allocated to this stratum A 0.43 B 0.10 C 0.10 D 0.38 Thus, if sampling was constrained to take only 18 samples (as in the actual data), you would allocate these as 7, 2, 2, and 7 to the four strata. Note that proportional allocation can never be exact in the real world because you must always have two samples in each stratum and you must round off the sample sizes. If you wish to specify a level of precision to be attained by proportional allocation, you proceed as follows. For example, assume you desire an absolute precision of the stratified mean of d = ± 0.1 bushels at 95% confidence. From equation (8.25): [ ] 2 2 2 4 4 (0.4281)(0.068) (0.0953)(0.042) + (0.1) 217 samples h h W s n d + ≈ = ≈ ∑ (assuming the sampling fraction is negligible in all strata). These 217 samples would be distributed to the four strata according to the fractions given above - 43% to stratum A, 10% to stratum B, etc. Optimal Allocation In this example proportional allocation is very inefficient because the variances are very different in the four strata, as well as the means. Optimal allocation is thus to be preferred. To illustrate the calculations, we consider a hypothetical case in which the cost per sample varies in the different strata. Assume that the overhead cost in equation (8.25) is $100 and the coasts per sample are c1 = $10 c2 = $20 c3 = $30 c4 = $40 Apply equation (8.27) to determine the fraction of samples in each stratum: Chapter 8 Page 353 / / h h h h h h h N s c n n N s c = ∑ These fractions are calculated as follows: Stratum Nh sh h c / h h N s c Estimated fraction, nh/n A 5703.9 0.2608 3.162 470.35 0.2966 B 1270.0 0.2049 4.472 58.20 0.0367 C 1286.4 1.4649 5.477 344.06 0.2169 D 5063.9 0.8911 6.325 713.45 0.4498 Total = 1586.06 1.0000 We can now proceed to calculate the total sample size needed for optimal allocation under two possible assumptions: Minimize the Standard Error of the Stratified Mean In this case cost is fixed. Assume for this example that $200 is available. Then, from equation (8.28), ( ) ( ) O / / (2000 - 100)(1586.06) 70.9 (rounded to 71 samples) 44,729.745 h h h h h h C c N s c n N s c − = = = ∑ ∑ Note that only the denominator needs to be calculated, since we have already computed the numerator sum. We allocate these 71 samples according to the fractions just established: Stratum Fraction of samples Total no. samples allocation of 68 total A 0.2966 21.1 (21) B 0.0367 2.6 (3) C 0.2169 15.4 (15) D 0.4498 31.9 (32) Minimize the Total cost for a Specified Standard Error In this case you must decide in advance what level of precision you require. In this hypothetical calculation, use the same value as above, d = ± 0.1 bushels (95% confidence limit). In this case the desired variance (V) of the stratified mean is Chapter 8 Page 354 2 2 0.1 = = =0.0025 2 d V t Applying formula (8.29): ( )( ) 2 / = (1/ )( ) h h h h h h h h W s c W s c n V N W s + ∑ ∑ ∑ We need to compute three sums: 2 (0.4281)(0.2608)(3.162) + (0.0953)(0.2049)(4.472)+ 3.3549 (0.4281)(0.068) (0.0953)(0.2049) / + + 3.162 4.472 0.1191 (0.4281)(0.06 h h h h h h h h W s c W s c W s = = = = = ∑ ∑ ∑ 8) + (0.0953)(0.042) + 0.5419 = Thus: (3.3549)(0.1191) = = 157.3 (rounded to 157 samples) 0.0025 (0.5419 /13,324.2) n + We allocate these 157 samples according to the fractions established for optimal allocation Stratum Fraction of samples Total no. of samples allocated of 157 total A 0.2966 46.6 (47) B 0.0367 5.8 (6) C 0.2169 34.1 (34) D 0.4498 70.6 (71) Note that in this hypothetical example, many fewer samples are required under optimal allocation (n = 157) than under proportional allocation (n = 217) to achieve the same confidence level ( d = ± 0.1 bushels). Program SAMPLE (Appendix 2, page 000) does these calculations. (2) minimize the total cost for a specified value of the standard error of the stratified mean. If you specify in advance the level of precision you need in the stratified mean, you can estimate the total sample size by the formula: ( )( ) ( ) 2 / (1/ ) h h h h h h h h W s c W s c n V N W s = + ∑ ∑ ∑ (8.29) Chapter 8 Page 355 where: Total sample size to be used in stratified sampling Stratum weight Standard deviation in stratum Cost to take one sample in stratum Total number of sample units in entire populat h h h n W s h c h N = = = = = ( ) 2 ion Desired variance of the stratified mean / Desired absolute width of the confidence interval for 1-V d t d α α = = = Student's value for 1- confidence limits ( 2 for 95% confidence limits, 2.66 for 99% confidence limits, 1.67 for 90% of confidence limits) t t t t t α α = ≈ ≈ ≈ Box 8.3 illustrates the application of these formulas. If you do not know anything about the cost of sampling, you can estimate the sample sizes required for optimal allocation from the two formulae: 1. To estimate the total sample size needed (n): ( ) ( ) 2 2 (1/ ) h h h h W s n V N W s = + ∑ ∑ (8.30) where : Desired variance of the stratified mean V = and the other terms are defined above. 2. To estimate the sample size in each stratum: h h h h h N s n n N s = ∑ (8.31) where: Total sample size estimated in equation (8.29) n = and the other terms are defined above. These two formulae are just variations of the ones given above in which sampling costs are presumed to be equal in all strata. Proportional allocation can be applied to any ecological situation. Optimal allocation should always be preferred, if you have the necessary background Chapter 8 Page 356 information to estimate the costs and the relative variability of the different strata. A pilot survey can give much of this information and help to fine tune the stratification. Stratified random sampling is almost always more precise than simple random sampling. If used intelligently, stratification can result in a large gain in precision, that is, in a smaller confidence interval for the same amount of work (Cochran 1977). The critical factor is always to chose strata that are relatively homogeneous. Cochran (1977 p. 98) has shown that with optimal allocation, the theoretical expectation is that: S.E.(optimal) ≤ S.E.(proportional) ≤ S.E.(random) where: S.E.(optimal) = Standard error of the stratified mean obtained with optimal allocation of sample sizes S.E.(proportional) = Standard error of the stratified mean obtained with proportional allocation S.E.(random) = Standard error of the mean obtained for the whole population using simple random sampling Thus comes the simple recommendation: always stratify your samples! Unless you are perverse or very unlucky and choose strata that are very heterogeneous, you will always gain by using stratified sampling. 8.2.3 Construction of Strata How many strata should you use, if you are going to use stratified random sampling? The answer to this simple question is not easy. It is clear in the real world that a point of diminishing returns is quickly reached, so that the number of strata should normally not exceed 6 (Cochran 1977, p. 134). Often even fewer strata are desirable (Iachan 1985), but this will depend on the strength of the gradient. Note that in some cases estimates of means are needed for different geographical regions and a larger number of strata can be used. Duck populations in Canada and the USA are estimated using stratified sampling with 49 strata (Johnson and Grier 1988) in order to have regional estimates of production. But in general you should not expect to gain much in precision by increasing the number of strata beyond about 6. Chapter 8 Page 357 Given that you wish to set up 2-6 strata, how can you best decide on the boundaries of the strata? Stratification may be decided a priori from your ecological knowledge of the sampling situation in different microhabitats. If this is the case, you do not need any statistical help. But sometimes you may wish to stratify on the basis of the variable being measured (x) or some auxiliary variable (y) that is correlated with x. For example, you may be measuring population density of clams (x) and you may use water depth (y) as a stratification variable. Several rules are available for deciding boundaries to strata (Iachan 1985) and only one is presented here, the cum f rule. This is defined as: cumulative square-root of frequency of quadrats cum f = This rule is applied as follows: 1. Tabulate the available data in a frequency distribution based on the stratification variable. Table 8.4 gives some data for illustration. 2. Calculate the square root of the observed frequency and accumulate these square roots down the table. 3. Obtain the upper stratum boundaries for L strata from the equally spaced points: Maximum cumulative Boundary of stratum = f i i L (8.32) For example, in Table 8.4 if you wished to use five strata the upper boundaries of strata 1 and 2 would be: 41.624 Boundary of stratum 1 = (1) = 8.32 5 41.624 Boundary of stratum 2 (2) 16.65 5 = = These boundaries are in units of cum f . In this example, 8.32 is between depths 20 and 21, and the boundary 20.5 meters can be used to separate samples belonging to stratum 1 from those in stratum 2. Similarly, the lower boundary of the second stratum is 16.65 cum f units which falls between depths 25 and 26 meters in Table 8.4. Chapter 8 Page 358 Using the cum f rule, you can stratify your samples after they are collected, an important practical advantage in sampling. You need to have measurements on a stratification variable (like depth in this example) in order to use the cum f rule. TABLE 8.4 DATA ON THE ABUNDANCE OF SURF CLAMS OFF THE COAST OF NEW JERSEY IN 1981 ARRANGED IN ORDER BY DEPTH OF SAMPLESa Class Depth, y (m) No,. of samples, f f cum f Observed no. of clams, x 1 14 4 2.00000 2.000 34, 128, 13, 0 2 15 1 1.00000 3.000 27 3 18 2 1.41421 4.414 361, 4 Stratum 1 4 19 3 1.73205 6.146 0, 5, 363 5 20 4 2.00000 8.146 176, 32, 122, 41 6 21 1 1.00000 9.146 21 7 22 2 1.41421 10.560 0, 0 8 23 5 2.23607 12.796 9, 112, 255, 3, 65 Stratum 2 9 24 4 2.00000 14.796 122, 102, 0, 7 10 25 2 1.41421 16.210 18, 1 11 26 2 1.41421 17.625 14, 9 12 27 1 1.00000 18.625 3 13 28 2 1.41421 20.039 8, 30 Stratum 3 14 29 3 1.73205 21.771 35, 25, 46 15 30 1 1.00000 22.771 15 16 32 1 1.00000 23.771 11 17 33 4 2.00000 25.771 9, 0, 4, 19 18 34 2 1.41421 27.185 11, 7 19 35 3 1.73205 28.917 2, 10, 97 Stratum 4 20 36 2 1.41421 30.332 0, 10 21 37 3 1.73205 32.064 2, 1, 10 22 38 2 1.41421 33.478 4, 13 23 40 3 1.73205 35.210 0, 1, 2 24 41 4 2.00000 37.210 0, 2, 2, 15 25 42 1 1.00000 38.210 13 Stratum 5 Chapter 8 Page 359 26 45 2 1.41421 39.624 0, 0 27 49 1 1.00000 40.624 0 28 52 1 1.00000 41.624 0 a Stratification is carried out on the basis of the auxiliary variable depth in order to increase the precision of the estimate of clam abundance for this region. Source: Iachan, 1985. 8.2.4 Proportions and Percentages Stratified random sampling can also be applied to the estimation of a proportion like the sex ratio of a population. Again the rule-of-thumb is to construct strata that are relatively homogeneous, if you are to achieve the maximum benefit from stratification. Since the general procedures for proportions are similar to those outlined above for continuous and discrete variables, I will just present the formulae here that are specific for proportions. Cochran (1977 p. 106) summarizes these and gives more details. We estimate the proportion of x-types in each of the strata from equation (8.12) (page 000). Then, we have for the stratified mean proportion: ST ˆ ˆ = h h N p p N ∑ (8.33) where: ST ˆ Stratified mean proportion Size of stratum ˆ Estimated proportion for stratum (from equation 8.13) Total population size (total number of sample units) h h p N h p h N = = = = The standard error of this stratified mean proportion is: ( ) ( ) 2 ST ˆ ˆ 1 ˆ S.E. = 1 1 h h h h h n h N N n p q p N N n − − − ∑ (8.34) where: ( ) ST ˆ S.E. Standard error of the stratified mean proportion ˆ ˆ 1 - Sample size in stratum h h h p q p n h = = = and all other terms are as defined above. Chapter 8 Page 360 Confidence limits for the stratified mean proportion are obtained using the t-distribution as outlined above for equation 8.20 (page 347). Optimal allocation can be achieved when designing a stratified sampling plan for proportions using all of the equations given above (8.26-8.30) and replacing the estimated standard deviation by: ˆ ˆ 1 h h h h p q s n = − (8.35) where: Standard deviation of the proportion in stratum ˆ Fraction of types in stratum ˆ ˆ 1- Sample size in stratum h h h h h s p h p x h q p n h = = = = Program SAMPLE in Appendix 2 (page 000) does all these calculations for stratified random sampling, and will compute proportional and optimal allocations from specified input to assist in planning a stratified sampling program. 8.3 Adaptive Sampling Most of the methods discussed in sampling theory are limited to sampling designs in which the selection of the samples can be done before the survey, so that none of the decisions about sampling depend in any way on what is observed as one gathers the data. A new method of sampling that makes use of the data gathered is called adaptive sampling. For example, in doing a survey of a rare plant, a botanist may feel inclined to sample more intensively in an area where one individual is located to see if others occur in a clump. The primary purpose of adaptive sampling designs is to take advantage of spatial pattern in the population to obtain more precise measures of population abundance. In many situations adaptive sampling is much more efficient for a given amount of effort than the conventional random sampling designs discussed above. Thompson (1992) presents a summary of these methods. Chapter 8 Page 361 8.3.1 Adaptive cluster sampling When organisms are rare and highly clustered in their geographical distribution, many randomly selected quadrats will contain no animals or plants. In these cases it may be useful to consider sampling clusters in a non-random way. Adaptive cluster sampling begins in the usual way with an initial sample of quadrats selected by simple random sampling with replacement, or simple random sampling without replacement. When one of the selected quadrats contains the organism of interest, additional quadrats in the vicinity of the original quadrat are added to the sample. Adaptive cluster sampling is ideally suited to populations which are highly clumped. Figure 8.2 illustrates a hypothetical example. Figure 8.2 A study area with 400 possible quadrats from which a random sample of n = 10 quadrats (shaded) has been selected using simple random sampling without replacement. Of the 10 quadrats, 7 contain no organisms and 3 are occupied by one individual. This hypothetical population of 60 plants is highly clumped. To use adaptive cluster sampling we must first make some definitions of the sampling universe: Chapter 8 Page 362 condition of selection of a quadrat: a quadrat is selected if it contains at least y organisms (often y = 1) neighborhood of quadrat x: all quadrats having one side in common with quadrat x edge quadrats: quadrats that do not satisfy the condition of selection but are next to quadrats that do satisfy the condition (i.e. empty quadrats) network: a group of quadrats such that the random selection of any one of the quadrats would lead to all of them being included in the sample. These definitions are shown more clearly in Figure 8.3, which is identical to Figure 8.2 except that the networks and their edge quadrats are all shown as shaded. Figure 8.3 The same study area shown in Figure 8.2 with 400 possible quadrats from which a random sample of n = 10 quadrats has been selected. All the clusters and edge quadrats are shaded. The observer would count plants in all of the 37 shaded quadrats. It is clear that we cannot simply calculate the mean of the 37 quadrats counted in this example to get an unbiased estimate of mean abundance. To estimate the mean abundance from adaptive cluster sampling without bias we proceed as follows (Thompson 1992): Chapter 8 Page 363 (1) Calculate the average abundance of each of the networks: k k i i y w m = ∑ (8.36) where wi = Average abundance of the i-th network yj = Abundance of the organism in each of the k-quadrats in the i-th network mi = Number of quadrats in the i-th network (2) From these values we obtain an estimator of the mean abundance as follows: i i w x n = ∑ (8.37) where x = Unbiased estimate of mean abundance from adaptive cluster sampling n = Number of initial sampling units selected via random sampling If the initial sample is selected with replacement, the variance of this mean is given by: ( ) ( ) ( ) 2 1 ˆ var x 1 n i i w x n n = − = − ∑ (8.38) where ( ) ˆ var x = estimated variance of mean abundance for sampling with replacement and all other terms are defined above. If the initial sample is selected without replacement, the variance of the mean is given by: ( ) ( ) ( ) ( ) 2 1 ˆ var x 1 n i i N n w x N n n = − − = − ∑ (8.39) where N = total number of possible sample quadrats in the sampling universe We can illustrate these calculations with the simple example shown in Figure 8.3. From the initial random sample of n = 10 quadrats, three quadrats intersected networks in the lower and right side of the study area. Two of these networks each have 2 plants Chapter 8 Page 364 in them and one network has 5 plants. From these data we obtain from equation (8.36): 2 2 5 0 0 7 8 15 1 1 0.08690 plants per quadrat 10 i i w x n + + + + + = = = ∑ Since we were sampling without replacement we use equation (8.39) to estimate the variance of this mean: ( ) ( ) ( ) ( ) ( ) ( )( )( ) 2 1 2 2 ˆ var x 1 2 2 400 10 0.0869 0.0869 7 8 400 10 10 1 0.0019470 n i i N n w x N n n = − − = − − − + − + = − = ∑ We can obtain confidence limits from these estimates in the usual way: ( ) ˆ var x t x α ± For this example with n = 10, for 95% confidence limits tα = 2.262 and the confidence limits become: ( )( ) 0.0869 2.262 0.0019470 0.0869 0.0983 ± = ± or from 0.0 to 0.185 plants per quadrat. The confidence limits extend below 0.0 but since this is biologically impossible, the lower limit is set to 0. The wide confidence limits reflect the small sample size in this hypothetical example. When should one consider using adaptive sampling? Much depends on the abundance and the spatial pattern of the animals or the plants being studied. In general the more clustered the population and the rarer the organism, the more efficient it will be to use adaptive cluster sampling. Thompson (1992) shows, for example, from the data in Figure 8.2 that adaptive sampling is about 12% more efficient than simple random sampling for n = 10 quadrats and nearly 50% more Chapter 8 Page 365 efficient when n = 30 quadrats. In any particular situation it may well pay to conduct a pilot experiment with simple random sampling and adaptive cluster sampling to determine the size of the resulting variances. 8.3.2 Stratified Adaptive Cluster Sampling The general principle of adaptive sampling can also be applied to situations that are well enough studied to utilize stratified sampling. In stratified adaptive sampling random samples are taken from each stratum in the usual way with the added condition that whenever a sample quadrat satisfies some initial conditions (e.g. an animal is present), additional quadrats from the neighborhood of that quadrat are added to the sample. This type of sampling design would allow one to take advantage of the fact that a population may be well stratified but clustered in each stratum in an unknown pattern. Large gains in efficiency are possible if the organisms are clustered within each stratum. Thompson (1992, Chap. 26) discusses the details of the estimation problem for stratified adaptive cluster sampling. The conventional stratified sampling estimators cannot be used for this adaptive design since the neighborhood samples are not selected randomly. 8.4 SYSTEMATIC SAMPLING Ecologists often use systematic sampling in the field. For example, mouse traps may be placed on a line or a square grid at 50 m intervals. Or the point-quarter distance method might be applied along a compass line with 100 m between points. There are many reasons why systematic sampling is used in practice, but the usual reasons are simplicity of application in the field, and the desire to sample evenly across a whole habitat. The most common type of systematic sampling used in ecology is the centric systematic area-sample illustrated in Figure 8.4. The study area is subdivided into equal squares and a sampling unit is taken from the center of each square. The samples along the outer edge are thus half the distance to the boundary as they are to the nearest sample (Fig. 8.4). Note that once the number of samples has been specified, there is only one centric sample for any area - all others would be eccentric samples (Milne 1959). Chapter 8 Page 366 Figure 8.4 Example of a study area subdivided into 20 equal-size squares with one sample taken at the center of each square. This is a centric systematic area-sample. Statisticians have usually condemned systematic sampling in favor of random sampling and have cataloged all the pitfalls that may accompany systematic sampling (Cochran 1977). The most relevant problem for an ecologist is the possible existence of periodic variation in the system under analysis. Figure 8.5 illustrates a hypothetical example in which an environmental variable (soil water content, for example) varies in a sine-wave over the study plot. If you are unlucky and happen to sample at the same periodicity as the sine wave, you can obtain a biased estimate of the mean and the variance (A in Fig. 8.5). Observed Value of X Distance along transect line A A A B B B B B A A A B Chapter 8 Page 367 Figure 8.5 Hypothetical illustration of periodic variation in an ecological variable and the effects of using systematic sampling on estimating the mean of this variable. If you are unlucky and sample at A, you always get the same measurement and obtain a highly biased estimate of the mean. If you are lucky and sample at B, you get exactly the same mean and variance as if you had used random sampling. The important question is whether such periodic variation exists in the ecological world. But what is the likelihood that these problems like periodic variation will occur in actual field data? Milne (1959) attempted to answer this question by looking at systematic samples taken on biological populations that had been completely enumerated (so that the true mean and variance were known). He analyzed data from 50 populations and found that, in practice, there was no error introduced by assuming that a centric systematic sample is a simple random sample, and using all the appropriate formulae from random sampling theory. Periodic variation like that in Figure 8.5 does not seem to occur in ecological systems. Rather, most ecological patterns are highly clumped and irregular, so that in practice the statistician's worry about periodic influences (Fig. 8.5) seems to be a misplaced concern (Milne 1959). The practical recommendation is thus: you can use systematic sampling but watch for possible periodic trends. Caughley (1977) discusses the problems of using systematic sampling in aerial surveys. He simulated a computer population of kangaroos, using some observed aerial counts, and then sampled this computer population with several sampling designs, as outlined in Chapter 4 (pp. 000-000). Table 8.5 summarizes the results based on 20,000 replicate estimates done by a computer on the hypothetical kangaroo population. All sampling designs provided equally good estimates of the mean kangaroo density and all means were unbiased. But the standard error estimated from systematic sampling was underestimated, compared with the true value. This bias would reduce the size of the confidence belt in systematic samples, so that confidence limits based on systematic sampling would not be valid because they would be too narrow. The results of Caughley (1977) should not to be generalized to all aerial surveys but they inject a note of warning into the planning of aerial counts if systematic sampling is used. Chapter 8 Page 368 TABLE 8.5 SIMULATED COMPUTER SAMPLING OF AERIAL TRANSECTS OF EQUAL LENGTH FOR A KANGAROO POPULATION IN NEW SOUTH WALESa Sampling system Method of analysis PPSb with replacement Random with replacement Random without replacement Systematic Coefficient of variation 2% sampling rate (n = 10 transects) 9 9 9 9 20% sampling rate (n = 100 transects) 3 3 2 3 Bias in standard error (%) 2% sampling rate (n = 10 transects) 20% sampling rate (n = 100 transects) 0 0 0 -23 a Data from actual transects were used to set up the computer population, which was then sampled 20,000 times at two different levels of sampling. The percentage coefficient of variation of the population estimates and the relative bias of the calculated standard errors of the population estimates were compared for random and systematic sampling. b PPS, Probability-proportional-to-size sampling, discussed previously in Chapter 4. Source: Caughley, 1977b. There is probably no issue on which field ecologists and statisticians differ more than on the use of random vs. systematic sampling in field studies. If gradients across the study area are important to recognize, systematic sampling like that shown in Figure 8.4 will be more useful than random sampling to an ecologist. This decision will be strongly affected by the exact ecological questions being studied. Some combination of systematic and random sampling may be useful in practice and the most important message for a field ecologist is to avoid haphazard or judgmental sampling. The general conclusion with regard to ecological variables is that systematic sampling can often be applied, and the resulting data treated as random sampling data, without bias. But there will always be a worry that periodic effects may influence the estimates, so that if you have a choice of taking a random sample or a systematic Chapter 8 Page 369 one, always choose random sampling. But if the cost and inconvenience of randomization is too great, you may lose little by sampling in a systematic way. 8.5 MULTISTAGE SAMPLING Ecologists often subsample. For example, a plankton sample may be collected by passing 100 liters of water through a net. This sample may contain thousands of individual copepods or cladocerans and to avoid counting the whole sample, a limnologist will count 1/100 or 1/1000 of the sample. Statisticians describe subsampling in two ways. We can view the sampling unit in this case to be the 100 liter sample of plankton and recognize that this sample unit can be divided into many smaller samples, called subsamples or elements. Figure 8.6 shows schematically how subsampling can be viewed. The technique of subsampling has also been called two-stage sampling because the sample is taken in two steps: 1. Select a sample of units (called the primary units) 2. Select a sample of elements within each unit. Figure 8.6 Schematic illustration of two-stage sampling. In this example seven primary sampling units occur in the study area, and they contain different number of elements (from 4 to 49). For example, the 7 primary units could be 7 lakes of varying size and the elements could be benthic areas of 10 cm2. Alternatively they could be 7 woodlands varying in size from 4 to 49 ha. (• ) Sample elements selected for counting. Chapter 8 Page 370 Many examples can be cited: Type of data Primary sample unit Subsamples or Elements Aphid infestation Sycamore tree Leaves within a tree DDT contamination Clutch of eggs Individual eggs Plankton 100-liter sample 1 ml subsample Pollen profiles in peat 1 cm3 peat at given depth Microscope slides of pollen grains Fish population in streams Entire stream Habitat section of stream If every primary sampling unit contains the same number of elements, subsampling is relatively easy and straightforward (Cochran, 1977, Chapter 10). But in most ecological situations, the primary sample units are of unequal size, and sampling is more complex. For example, different sycamore trees will have different numbers of leaves. It is important in ecological multistage sampling that the elements are of equal size– only the primary sampling units can vary in size. For example, if you are surveying different size woodlands, you should use a constant quadrat size in all of the woodlands. Clearly subsampling could be done at several levels and thus two-stage sampling can be generalized to 3-stage sampling, and the general term multistage sampling is used to describe any design in which there are two or more levels of sample selection (Hankin 1984). 8.5.1 Sampling Units of Equal Size Consider first the simplest case of multistage sampling in which n primary sample units are picked and m subsamples are taken in each unit. For example, you might take 20 plankton samples (each of 100 liters of lake water) and from each of these 20 samples count four subsamples (elements) of 1 ml each. We adopt this notation: xij = measured value for the j element in primary unit i xi = mean value per element in primary unit I = 1 m ij j x m = ∑ The mean of the total sample is given by: Chapter 8 Page 371 1 = n i i x x n = = ∑ (8.40) The standard error of this mean is (from Cochran 1977 p. 277): ( ) 1 2 2 2 1 1 2 1 1 S.E. = f f f x s s n mn = − − + (8.41) where: 1 2 = Sampling fraction in first stage = Number of primary units sampled/Total number of primary units = Sampling fraction in second stage = Number of elements sampled/Number of elements per unit f f n = Number of primary units sampled = Number of elements sampled per unit m ( ) 2 n 2 1 = / 1 = Variance among primary unit means i s x x n = − − ∑ (8.42) ( ) 2 2 2 = / ( 1) = Variance among elements within primary units n m ij i s x x n m − − ∑∑ (8.43) If the sampling fractions are small, the finite population corrections (f1, f2) can be omitted. Note that the standard error can be easily decomposed into 2 pieces, the first piece due to variation among primary sampling units, and the second piece due to variation within the units (among the subsamples). Box 8.4 illustrates the use of these formulae in subsampling. Program SAMPLE (Appendix 6, page 000) will do these calculations. Box 8.4 MULTISTAGE SAMPLING: SUBSAMPLING WITH PRIMARY UNITS OF EQUAL SIZE A limnologist estimated the abundance of the cladoceran Daphnia magna by filtering 1000 liters of lake water (= sampling unit) and subdividing it into 100 subsamples (= elements), of which 3 were randomly chosen for counting. One day when he sampled he got these results (number of Daphnia counted): Subsample Sample 9.1 Sample 9.2 Sample 9.3 Sample 9.4 1 46 33 27 39 2 30 21 14 31 Chapter 8 Page 372 3 42 56 65 45 Mean 39.33 36.67 35.33 38.33 In this example, n = 4 primary units sampled and N is very large (> 105) so the sampling fraction in the first stage (f1 ) is effectively 0.0. In the second stage m = 3 of a total of M = 100 possible elements, so the sampling fraction f2 is 0.03. The mean number of Daphnia per 10 liters is given by equation (8.40): 1 39.33 36.67 35.33 38.33 + + + 37.42 4 4 4 4 n i i x x n Daphnia = = = = = ∑ Note that this is the same as the mean that would be estimated if the entire data set were treated as a simple random sample. This would not be the case if the number of subsamples varied from primary sample unit to unit. The standard error of the estimated mean is from equation (8.41): ( ) ( ) 1 2 2 2 1 1 2 = 1 1 S.E.
f f f x s s n mn − − = + First, calculate 2 2 1 2 and s s : ( ) ( ) 2 2 2 2 1 2 2 2 2 2 (39.33 - 37.42) (36.67 - 37.42) + + 1 3 3 3.1368 (variance among primary unit means) 46 39.33 (30 39.33) ( 1) 4(2) 4(2) 284.333 (variance among s i ij i x x s n x x s n m = − = = − = − − − = = + + − = ∑ ∑∑ ubsamples) It is clear from the original data that there is much more variation among the subsamples than there is among the primary samples. Since f1 is nearly zero, the second term disappears and ( ) = 1 S.E. = (3.1368) = 0.8856 4 x The 95% confidence interval would be (tα = 2.20 for 11 d.f.): ( ) = S.E.
37.42 2.20(0.8856) x t x α = ± ± Chapter 8 Page 373 or from 35.5 to 39.4 Daphnia per 10 liters of lake water. If you wish to express these values in terms of the original sampling unit of 1000 liters, multiply them by 100 Program SAMPLE (Appendix 6) will do these calculations. 8.5.2 Sampling Units of Unequal Size If the sampling units are of varying size, calculations of the estimated means and variances are more complex. I will not attempt to summarize the specific details in this book because they are too complex to be condensed simply (cf. Cochran 1977, Chapter 11). There are two basic choices that you must make in selecting a multistage sampling design model - whether to chose the primary sampling units with equal probability or with probability proportional to size (PPS). For example, in Figure 8.6 we could choose two of the 7 primary sampling units at random, assigning probability of 1 in 7 to each. Alternatively, we could note that there are 123 elements in the study area in Figure 8.6, and that the largest unit has 49 elements, so its probability of selection should be 49/123 or 0.40, while the smallest unit has only 4 elements, so its probability of selection should be 0.03. It is usually more efficient to sample a population with some type of PPS sampling design (Cochran 1977). But the problem is that, before you can use PPS sampling, you must know (or have a good estimate of) all the numbers of elements in each of the primary units in the population (the equivalent to the information in Figure 8.6). If you do not know this, (as is often the case in ecology), you must revert to simple random sampling or stratified sampling, or do a pilot study to get the required information. Cochran (1977) shows that often there is little loss in precision by making a rough estimate of the size of each primary unit, and using probability-proportional-to-estimated-size (PPES) sampling. Cochran (1977, Chapter 11) has a clear discussion of the various methods of estimation that are applied to multistage sampling designs. Hankin (1984) discusses the application of multistage sampling designs to fisheries research, and notes the need for a computer to calculate estimates from the more complex models (Chao 1982). We have already applied a relatively simple form of PPS sampling to aerial Chapter 8 Page 374 census (pp. 000-000). Ecologists with more complex multistage sampling designs should consult a statistician. Program SAMPLE (Appendix 6, page 000) will do the calculations in Cochran (1977, Chapter 11) for unequal size sampling units for equal probability, PPS, or PPES sampling. With multistage sampling you must choose the sample sizes to be taken at each stage of sampling. How many sycamore trees should you choose as primary sampling units? How many leaves should you sample from each tree? The usual recommendation is to sample the same fraction of elements in each sampling unit, since this will normally achieve a near-optimal estimate of the mean (Cochran 1977 p. 323). To choose the number of primary units to sample in comparison to the number of elements to sample within units, you need to know the relative variances of the two levels. For example, the total aphid population per sycamore tree may not be very variable from tree to tree, but there may be great variability from leaf to leaf in aphid numbers. If this is the case, you should sample relatively few trees and sample more leaves per tree. Cochran (1977) should be consulted for the detailed formulae, which depend somewhat on the relative costs of sampling more units compared with sampling more elements. Schweigert and Sibert (1983) discuss the sample size problem in multistage sampling of commercial fisheries. One useful rule-of-thumb is to sample an average of m elements per primary sampling unit where: 2 2 2 U s m s ≈ (8.44) where: ( ) ( ) [ ] ( ) 2 2 2 2 2 2 1 2 Optimal number of elements to sample per primary unit Variance among elements within primary units / 1 as defined in equation (8.43) / = component of va n m ij i U m s x x n m s s s M = = = − − = − ∑∑ ( ) [ ] 2 1 2 1 riance among unit means Variance among primary units / 1 as defined in equation (8.41) n i i s x x n = = = = − − ∑ Chapter 8 Page 375 For example, from the data in Box 8.4, 2 1 s = 3.14 and 2 2 s = 284.3 and M = 100 possible subsamples per primary unit: thus - 2 284.3 = 3.14 - = 0.29 100 284.3 = 31.3 elements 0.29 U s m ≈ There are 100 possible subsamples to be counted, and this result suggests you should count 31 of the 100. This result reflects the large variance between subsample counts in the data of Box 8.4. Once you have an estimate of the optimal number of subsamples (m in equation 8.44) you can determine the sample size of the primary units (n) from knowing what standard error you desire in the total mean x = (equation 8.39) from the approximate formula: 2 2 2 2 2 1 2 1 1 1 1 S.E. = s x s s s n M mn N = − + − (8.45) where: ( ) 2 1 2 2 = S.E. x Desired standard error of mean Sample size of primary units needed Variance among primary units (equation 8.41) Variance among elements (equation 8.38) Total number of element n s s M = = = = = s per primary unit Optimal number of elements to subsample (equation 8.43) m = This equation can be solved for n if all the other parameters have been estimated in a pilot survey or guessed from prior knowledge. 8.6 SUMMARY If you cannot count or measure the entire population, you must sample. Several types of sampling designs can be used in ecology. The more complex the design, the more efficient it is, but to use complex designs correctly you must already know a great deal about your population. Chapter 8 Page 376 Simple random sampling is the easiest and most common sampling design. Each possible sample unit must have an equal chance of being selected to obtain a random sample. All the formulas of statistics are based on random sampling, and probability theory is the foundation of statistics. Thus you should always sample randomly when you have a choice. In some cases the statistical population is finite in size and the idea of a finite population correction must be added into formulae for variances and standard errors. These formulas are reviewed for measurements, ratios, and proportions. Often a statistical population can be subdivided into homogeneous subpopulations, and random sampling can be applied to each subpopulation separately. This is stratified random sampling, and represents the single most powerful sampling design that ecologists can adopt in the field with relative ease. Stratified sampling is almost always more precise than simple random sampling, and every ecologist should use it whenever possible. Sample size allocation in stratified sampling can be determined using proportional or optimal allocation. To use optimal allocation you need to have rough estimates of the variances in each of the strata and the cost of sampling each strata. Optimal allocation is more precise than proportional allocation, and is to be preferred. Some simple rules are presented to allow you to estimate the optimal number of strata you should define in setting up a program of stratified random sampling. If organisms are rare and patchily distributed, you should consider using adaptive cluster sampling to estimate abundance. When a randomly placed quadrat contains a rare species, adaptive sampling adds quadrats in the vicinity of the original quadrat to sample the potential cluster. This additional non-random sampling requires special formulas to estimate abundance without bias. Systematic sampling is easier to apply in the field than random sampling, but may produce biased estimates of means and confidence limits if there are periodicities in the data. In field ecology this is usually not the case, and systematic samples seem to be the equivalent of random samples in many field situations. If a gradient exists in the Chapter 8 Page 377 ecological community, systematic sampling will be better than random sampling for describing it. More complex multistage sampling designs involve sampling in two or more stages, often called subsampling. If all the sample units are equal in size, calculations are simple. But in many ecological situations the sampling units are not of equal size, and the sampling design can become very complex so you should consult a statistician. Multistage sampling requires considerable background information and unless this is available, ecologists are usually better off using stratified random sampling. It is always useful to talk to a professional statistician about your sampling design before you begin a large research project. Many sampling designs are available, and pitfalls abound. SELECTED REFERENCES Abrahamson, I.L., Nelson, C.R., and Affleck, D.L.R. 2011. Assessing the performance of sampling designs for measuring the abundance of understory plants. Ecological Applications 21(2): 452-464. Anganuzzi, A.A. & Buckland, S.T. 1993. Post-stratification as a bias reduction technique. Journal of Wildlife Management 57: 827-834. Cochran, W.G. 1977. Sampling Techniques, 3rd ed. John Wiley and Sons, New York. Legendre, P., Dale, M.R.T., Fortin, M.J., Gurevitch, J., Hohn, M., and Myers, D. 2002. The consequences of spatial structure for the design and analysis of ecological field surveys. Ecography 25(5): 601-615. Milne, A. 1959. The centric systematic area-sample treated as a random sample. Biometrics 15: 270-297. Morin, A. 1985. Variability of density estimates and the optimization of sampling programs for stream benthos. Canadian Journal of Fisheries and Aquatic Sciences 42: 1530-1534.Peterson, C.H., McDonald, L.L., Green, R.H., and Erickson, W.P. 2001. Sampling design begets conclusions: the statistical basis for detection of injury to and recovery of shoreline communities after the 'Exxon Valdez¹ oil spill. Marine Ecology Progress Series 210: 255-283. Thompson, S.K. 2012. Sampling. 3rd ed. John Wiley and Sons, Holboken, New Jersey. Chapter 8 Page 378 Underwood, A.J., and Chapman, M.G. 2003. Power, precaution, Type II error and sampling design in assessment of environmental impacts. Journal of Experimental Marine Biology and Ecology 296(1): 49-70. QUESTIONS AND PROBLEMS 8.1. Reverse the ratio calculations for the wolf-moose data in Box 8.2 (page 000) and calculate the estimated ratio of moose to wolves for these data, along with the 95% confidence interval. Are these limits the reciprocal of those calculated in Box 8.2? Why or why not? (a) How would these estimates change if the population was considered infinite instead of finite? 8.2. Assume that the total volume of the lake in the example in Box 8.4 (page 000) is 1 million liters (N). Calculate the confidence limits that occur under this assumption and compare them with those in Box 8.4 which assumes N is infinite. 8.3. In the wood lemming (Myopus schisticolor) in Scandinavia there are two kinds of females - normal females that produce equal numbers of males and females, and special females that produce only female offspring. In a spruce forest with an estimated total population of 72 females, a geneticist found in a sample of 41 females, 15 individuals were female-only types. What is the estimate of the fraction of normal females in this population? What are the 95% confidence limits? 8.4. Hoisaeter and Matthiesen (1979) report the following data for the estimation of seaweed (Ulva) biomass for a reef flat in the Philippines: (quadrat size 0.25 m2) Stratum Area (m2) Sample size Mean (g) Variance I (near shore) 2175 9 0.5889 0.1661 II 3996 14 19.3857 179.1121 III 1590 7 2.1429 3.7962 IV 1039 6 0.2000 0.1120 Estimate the total Ulva biomass for the study zone, along with its 95% confidence limits. Calculate proportional and optimal allocations of samples for these data, assuming the cost of sampling is equal in all strata, and you require a confidence belt of ±25% of the mean. 8.5. Tabulate the observed no. of clams (x) in column 6 of Table 8.4 (page 000) in a cumulative frequency distribution. Estimate the optimal strata boundaries for this variable, based on 3 strata, using the cum f procedure. How do the results of Chapter 8 Page 379 this stratification differ from those stratified on the depth variable (as in Table 8.4)? 8.6. A plant ecologist subsampled 0.25 m2 areas within 9 different deer-exclosures of 16 m2. She subsampled 2 areas within each exclosure, and randomly selected 9 of 18 exclosures that had been established in the study area. She got these results for herb biomass (g dry weight per 0.25 m2): Exclosure no. Subsample 3 5 8 9 12 13 15 16 18 A 2 5 32 23 19 16 23 25 13 B 26 3 6 9 8 7 9 3 9 Estimate the mean biomass per 0.25 m2 for these exclosures, along with 95% confidence limits. How would the confidence limits for the mean be affected if you assumed all these estimates were replicates from simple random sampling with n = 18? What recommendation would you give regarding the optimal number of subsamples for these data? 8.7. Describe an ecological sampling situation in which you would not recommend using stratified random sampling. In what situation would you not recommend using adaptive cluster sampling? 8.8. How does multistage sampling differ from stratified random sampling? 8.9. Use the marked 25 x 25 grid on Figure 6.2 of the redwood seedlings (page 000) to set out an adaptive sampling program to estimate density of these seedlings. From a random number table select 15 of the possible 625 plots and apply adaptive cluster sampling to estimate density. Compare your results with simple random sampling of n = 15 quadrats. |
3784 | https://discuss.codecademy.com/t/python-challenge-number-permutation/636205 | Python Challenge - Number Permutation
Community%20FAQs%20on%20Codecademy%20Exercises1000×208 141 KB
This community-built FAQ covers the “Number Permutation” code challenge in Python. You can find that challenge here, or pick any challenge you like from our list.
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Does anybody know how to do this? I’ve been trying, but I can’t.
@aryabadri8979868753 found a solution, posted below
image726×415 26.2 KB
I did a brute force approach instead of recursion this time:
I iterated through all possible series of 5 numbers from 0 - 9 (with repetition), and counted only the relevant ones (meaning the ones where the numbers added up to the given number z, where there were no zeros that weren’t at the beginning).
I used an integer (meaning its digits) as each permutation, so that I could iterate through all the permutations.
`def checkPermutation(n, sum_of_digits = None, max_exp_of_10 = 4):
get sum of digits of n, check if equals sum_of_digits
can have leading 0s, but no 0s after a non-zero
previous_was_nonzero = False
total = 0 # will store sum of digits so far
checker = 10 max_exp_of_10
while checker >= 1:
digit = n // checker
if digit == 0:
if previous_was_nonzero:
if zero comes after nonzero, return False
return False
else:
total += (digit % 10)
previous_was_nonzero = True
n = n % checker
checker = checker // 10
if sum_of_digits is None:
return total
return (total == sum_of_digits)
def makeNumber(z):
if (z < 1) or (z > 45):
return 0
count = 0
for n in range(1, 100000):
if checkPermutation(n, z):
count += 1
return count
print(makeNumber(21))def makeNumber(z):
lst = # temp lst to create tuple in, inner list.
cnt = 0 # store the count of permitations
for idx in range(5): # Limit to 5 digit in tuple
lst = [tuple(i,) + (x,) for i in lst for x in range(1, z+1) if x <= 9]
create permutations in a lst
for i in lst: # loop trough list and put a condition to get the count of the sum that equals z
if sum(i) == z:
cnt += 1
return cnt
print(makeNumber(4))`
A small summary of my code:
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3785 | https://artowen.su.domains/courses/200/lec02.pdf | Stat 200: Introduction to Statistical Inference Autumn 2018/19 Lecture 2: More probability review Lecturer: Art B. Owen September 27 Disclaimer: These notes have not been subjected to the usual scrutiny reserved for formal publications. They are meant as a memory aid for students who took stat 200 at Stanford University. They may be distributed outside this class only with the permission of the instructor. Also, Stanford University holds the copyright.
Abstract These notes are mnemonics about what was covered in class. They don’t replace being present or reading the book. Reading ahead in the book is very effective.
This is a continuation of probability review. It should be reminding you of things you learned in probability. It is not practical do all of probability in week one of a statistics course.
2.1 Moments We begin with expected values. For a discrete random variable (RV) X, it’s expectation is E(X) = X i p(xi)xi, if X i p(xi)|xi| < ∞.
If X is a continuous RV then E(X) = Z ∞ −∞ f(x)x dx if Z ∞ −∞ f(x)|x| dx < ∞.
The ’if clause’ is important to make the quantity a well-defined finite value. We can think of E(X) as one kind of ‘typical’ outcome for X. As sketched in class the PDF of a continuous X would ‘balance’ at E(X).
For random variables X and Y and constants a, b and c, E(aX + bY + c) = aE(X) + bE(Y ) + c.
In this and other identities we assume that the expecations involved all exist without always saying that.
The expected value of X is also called the first moment. The second moment is E(X2), the third is E(X3) and so on. When E(X) = µ we define the k’th centered moment as E((X −µ)k). The variance of X is Var(X) = E((X −E(X))2) so it is the second centered moment. It quantifies how spread out f is, that is, how far from E(X) that X may get. The standard deviation of X is p Var(X). A little algebra gives Var(X) = E(X2) −E(X)2.
A litle more algebra would give Var(X + Y ) = Var(X) + Var(Y ), if X and Y are independent.
c ⃝Stanford University 2018 2-2 Lecture 2: More probability review For random variables X and Y , thier covariance is Cov(X, Y ) = E((X −E(X))(Y −E(Y ))) = · · · = E(XY ) −E(X)E(Y ), where I use · · · to replace algebra that I know you can do especially when running through the algebra adds no statistical idea. [If you are not sure why one works, then plug and chug to see it work out. There’s no shame in that. Besides, nobody will know. Also you might find a typo that way.] A positive covariance is one way to quantify that X and Y tend to increase together. A negative value quantifies a tendency for increases in one to accompany decreases in the other.
One of the most famous and important statistical facts is that a nonzero Cov(X, Y ) does not necessarily mean that if you somehow caused a change in X it would bring the corresponding change in Y either at all or on average.
We can standardize X by subtracting its mean and dividing by its standard deviation, getting (X − E(X))/ p Var(X). The standardized random variable has mean 0 and variance 1 (work it out if you doubt).
The covariance between two standardized variables is their correlation Corr(X, Y ) ≡Cov X −E(X)) p Var(X) , Y −E(Y )) p Var(Y ) .
Here ≡means not just equal, but equal by definition of what is on the left hand side. The correlation is not well defined if either Var(X) = 0 or Var(Y ) = 0.
2.2 Some distributions Here we look at a few important distributions but not all of them. Appendix A of Rice has many of the most important ones. So does Chapter 6 (distributions related to the normal distribution).
A Bernoulli random variable X takes only two values 0 and 1. These are sometimes called ‘failure’ and ‘success’ respectively. Examples for X = 1: you got the basketball through the hoop, the coin came up heads, you received lighting from the sky. We say X ∼Bern(p) when the Bernoulli random variable has Pr(X = 1) = p. Of course 0 ⩽p ⩽1. Bernoulli random variables are commonly ‘indicator’ variables that indicate that some event occured. When the event is A we might use notation X = 1A ≡ ( 1, A occurs 0, else.
This lets us turn events into numbers or turn logic into algebra, for instance Pr(A) = E(1A), and Pr(A ∪B) = E(1A∪B) = E(1A + 1B −1A∩B) = Pr(A) + Pr(B) −Pr(A ∩B).
It is easy to see that E(X) = p. Also X = X2 = X3 = X4 · · · so moments of Bernoulli variables are easy to work out. They all equal p.
Let X1, X2, . . . , Xn be independent Bern(p) random variables. The Binomial distribution Bin(n, p) is formed as the distribution of Y = n X i=1 Xi.
Perhaps this is the number of baskets you get in n tries. A direct combinatorial argument gives Pr(Y = y) = ( n y py(1 −p)n−y, y = 0, 1, 2, . . . , n 0, else.
Lecture 2: More probability review 2-3 We can find by manipulating sums that E(Y ) = n X y=0 n y py(1 −p)n−y × y = np.
Begin by removing y = 0 from the sum and expanding n y = n!/(y!(n −y)!). It is easier to notice that E(Y ) = n X i=1 E(Xi) = n X i=1 p = np, though you should work it both ways if the methods are new to you. Similarly from independence Var(Y ) = nVar(X1) = np(1 −p), or you can do it by manipulating the sum.
It is important to have Pr(Xi = 1) be the same for all i = 1, . . . , n. Similarly the Xi should be independent.
Here we skipped: geometric distribution (number of tosses to first basket) negative binomial (number of tosses to attain some number of baskets) and the hypergeometric (all about red and black balls in an urn).
Read about those in Rice.
The Poisson distribution with parameter λ ⩾0, written X ∼Poi(λ) has Pr(X = x) = e−λλx x!
, x = 0, 1, 2, · · · 0, else.
If we had asked for a distribution with Pr(X = x) = cλx/x! for some constant c (commonly written Pr(X = x) ∝λx/x! reading ∝as ‘proportional to’) then we would have needed the constant c to satisfy 1 c = ∞ X x=0 λx x! .
We remember from calculus that the right hand side is exp(λ) and so we find that c = exp(−λ).
Now, working in slow motion, and knowing ahead of time what order to take the steps E(X) = ∞ X x=0 e−λλx x!
× x = ∞ X x=1 e−λλx x!
× x = ∞ X x=1 e−λλx (x −1)!
= ∞ X r=0 e−λλr+1 r!
= λ × ∞ X r=0 e−λλr r!
= λ.
Be sure you understand why each step is ok. When working out expected values you might have to try a few things to find the right sequence. In these problems it is very common that the fact that P∞ x=0 Pr(X = x) = 1 2-4 Lecture 2: More probability review comes in handy. So you try to wrangle the expression to include that. Here the sum came through for the exact same λ we started with. In other settings you see the pattern for a different parameter value than you started with.
We saw in class that lim n→∞Pr(Bin(n, λ/n) = x) = Pr(Poi(λ) = x) for x = 0, 1, . . . . This means that the Poisson random variable can be obtained in a setting where some very low probability Bernoulli event is tracked a large number n of times. We really needed to use Bin(n, p) with p = λ/n. If we used p = λ/nr for any r < 1 then we would have a limit where E(X) = np = λ × n1−r →∞ which is not an interesting or useful limit; it is not even a distribution. If instead r < 1 then the limit is Poi(0) which is the same as Bern(0), i.e., a random variable that is always 0, once again not very useful.
2.3 Inequalities We proved Markov’s inequality. If Pr(X ⩾0) = 1 then Pr(X ⩾t) ⩽E(X) t .
The proof in class used indicator variables starting with X = X × 1X<t + X × 1X⩾t which holds because 1X<t + 1X⩾t = 1. Then X × 1X 0 Pr(|X −µ| ⩾t) ⩽σ2/t2.
(Use Markov’s on Y = (X −µ)2). If t = kσ then Pr(|X −µ| ⩾kσ) ⩽1/k2.
2.4 Tower property and variance decomposition Let X and Y be random variables. Read how Rice defines E(Y | X) and Var(Y | X) (around page 150).
Then E(Y ) = E(E(Y | X)) (2.1) is called the tower property of expectation. Also Var(Y ) = E(Var(Y | X)) + Var(E(Y | X)) (2.2) is used a lot in statistics. An important consequence is that E(Y | X) can never have higher variance than Y has.
Suppose we had to guess at Y . If we guess m then our mean squared error is E((Y −m)2) = · · · = Var(Y ) + (m −E(Y ))2.
Lecture 2: More probability review 2-5 [You should be able to fill in the dots.] Our most accurate guess is m = E(Y ).
If we know X = x then our best guess for Y is E(Y | X = x). Now E(Y | X) is a random variable taking the value E(Y | X = x) when X = x. So somebody who knows X (whatever it turns out to be) would guess E(Y | X). Their expected squared error is E (E(Y | X) −Y )2 = E (E(Y | X) −Y ) 2 + Var E(Y | X) −Y .
The first term is 02 = 0 by the tower property (2.1). The second term is no larger than Var(Y ) by (2.2). As a consequence E(Y | X) is a better guess than E(Y ) or at least as good. Knowing X can never make you a worse guesser about Y .
2.5 Moment generating function The moment generating function of X, denoted M(t) or sometimes MX(t) is E(etX), a function of t ∈R.
Often it is infinite for some t but finite for other t. It is most useful when M(t) exists for all t ∈(−h, h), that is for |t| < h for some h > 0. In that case the MGF uniquely determines the distribution of X. That is, if random variables X and Y with CDFs FX and FY have MGFs MX(t) = MY (t) for all |t| < h > 0 then FX = FY . Same MGF means same distribution.
The characteristic function φ(t) = E(eitX) = M(it) always exists and it characterizes the distribution of X.
It is useful when the MGF does not exist but it requires wrangling complex numbers.
We use the MGF to generate moments. As we saw in class M ′(0) = E(X).
Similarly M ′′(0) = E(X2) and for integers r ⩾1 taking the r’th derivative we get M (r)(0) = E(Xr). For continuous random variables these findings require us to be able to interchange differentation and integration.
Our main use of MGFs is to find the distribution of the sum of two independent random variables with known distributions.
We saw that a Gamma random variable with shape α has MGF M(t) = (1 −t)−α for all |t| < h = 1.
From the definitions of MGF we see that for independent X and Y , MX+Y (t) = MX(t) × MY (t). If they’re Gam(α) then MX+Y (t) = (1 −t)−2α which we recognize as the MGF of the Gam(2α) distribution. If we sum n independent Gam(α) random variables we get a Gam(nα) random variable.
The Gamma distribution has two parameters. In addition to the shape α there is a rate λ > 0. If X ∼ Gam(α) and Y = X/λ then Y ∼Gam(α, λ). Now MY (t) = (1 −t/λ)−α for |t| < λ.
For us the most important Gamma distributions will be χ2. The χ2 (n) distribution is Gam(n/2, 1/2).
2.6 Convergence You should know Rice chapter 5 on convergence of distributions. Here are the key ideas.
First, the law of large numbers. Let Xi be IID with mean µ and let ¯ Xn ≡(1/n) Pn i=1 Xi be their average.
The law of large numbers states that for any ϵ > 0 lim n→∞Pr(| ¯ Xn −µ| > ϵ) = 0.
2-6 Lecture 2: More probability review In this sense averages settle down to their corresponding expecations.
Rice proves this via Chebychev assuming also Var(Xi) < ∞, but it holds without that assumption.
This is not the only law of large numbers, but it is the one that we will use.
The random variables ¯ Xn “converge in probability” to the value µ.
Sometimes a sequence of random variables does not converge to a constant but has instead a limit that is also a random variable. Let Xn be a sequence of random variable with CDFs Fn and let X be a random variable with CDF F. If limn→∞Fn(x) = F(x) holds at all x where F is continuous then we say that the sequence Xn converges in distribution to X.
To understand the exception ‘where F is continuous’ think about Fn(x) = 1x⩾1/n and F(x) = 1x⩾0. What kind of random variable is Xn, what kind is X and do we think these Xn are converging to X?
Now let Xn have MGF Mn, let X have MGF M and suppose that Mn(t) →M(t) as n →∞holds for all |t| < h where h > 0. Then Xn converges in distribution to X. This follows from the continuity theorem in Rice.
The central limit theorem is about averages having nearly the normal distribution. See Rice p 184. Let ϕ(x) = 1 √ 2π e−1 2 x2 be the N(0, 1) PDF and Φ(x) = Z x −∞ ϕ(z) dz be the corresponding CDF. Then for IID variables Xi with mean µ and variance σ2 > 0 lim n→∞Pr ¯ Xn −µ σ/√n = Φ(x) for all x ∈R. This means that Zn ≡( ¯ X −µ)/(σ/√n) converges in distribution to N(0, 1). We can tell from the formula that Zn has mean 0 and variance 1. The CLT adds the normality (in the limit).
Rice does the CLT a little differently. Read his treatment carefully to be sure it is the same. He proves it via the MGF.
We can rearrange the CLT result to find that ¯ Xn has distribution close to N(µ, σ2/n). |
3786 | http://econdse.org/wp-content/uploads/2016/04/linear_difference_eq-LectureNotes-Tirelli.pdf | Linear Difference Equations Mario Tirelli Still a preliminary version March 2, 2014 Contents 1.
Introduction 1 2.
Linear difference equations 2 2.1.
Equations of first order with a single variable 2 2.2.
Equations of first order with m > 1 variables (systems of equations) 6 2.3.
Higher order difference equations 13 2.4.
Solution stability 16 2.5.
Stochastic difference equations 19 1. Introduction Dynamic economic models are a useful tool to study economic dynamics and get a better un-derstanding of relevant phenomena such as growth and business cycle. Equilibrium conditions are normally identified by a system of difference equations and a set of boundary conditions (describing limit values of some variables).
Thus, studying equilibrium properties requires studying the properties of a system of difference equations. In many interesting cases such dif-ference equations are nonlinear; as you will see, even the textbook, deterministic, neoclassical, growth model is represented by a system of nonlinear equations. Although nonlinearities make the study of dynamic economic systems difficult, we can still hope to derive useful characteri-zations and results. This is achieved either locally, in a neighborhood of an equilibrium point, or globally for log-linearized systems. In both cases, non-linear systems are studied using the theory of linear difference equations. In these notes we shall summarize some useful results in this theory and apply them to deal with dynamic economic systems.
In these class notes I present some useful material on how to solve linear difference equations and study solution stability. These notes are incomplete; many important questions are left unexplained; but students can find additional, sharp and clear material in Elaydi (2005) ”An Introduction to Difference Equations”, Springer-Verlag. A general, introductory, reference is Simon & Blume, ”Mathematics for Economists”, Norton. For issues on linear algebra, my favorite textbook is Serge Lang ”Linear Algebra”, Springer-Verlag.
1 2. Linear difference equations 2.1. Equations of first order with a single variable. Let us start with equations in one variable, (1) xt + axt−1 = bt This is a first-order difference equation because only one lag of x appears. In this equation, a is a time-independent coefficient and bt is the forcing term. When bt = 0, the difference equation is said to be homogeneous, and otherwise non-homogeneous. When the forcing term is a constant (bt = b for all t), the difference equation 1 is non-homogeneous and autonomous (or time-invariant). Finally, when bt is time-dependent the equation is said to be non-autonomous; this is a more general formulation, allowing, for example, to capture seasonality, deterministic shocks or perturbations. The most general form of linear difference equation is one in which also the coefficient a is time-varying.
2.1.1. Autonomous equations. Let us start with a non-homogeneous, autonomous difference equation, with bt being equal to a time-invariant scalar b. A first approach is to solve 1 by successive iteration. Suppose the initial value x0 is given, then x1 = b −ax0 x2 = b −a(b −ax0) = b −ab + a2x0 x3 = b −a(b −ab + a2x0) = b −ab + a2b −a3x0 ................
xt = b(1 + (−a) + (−a)2, .., +(−a)t−1) + (−a)tx0 = 1 −(−a)t 1 −(−a) b + (−a)tx0, if a ̸= −1 Rearranging, gathering terms in (−a)t, (2) xt = (−a)t [ x0 − b 1 + a ] + b 1 + a A general method, analogous to the one used for differential equations, is based on the Superposition Principle (see theorem [SP] below): the solution of a linear difference equation is the sum of the solution of its homogeneous part, the complementary solution, and the particular solution. A particular solution is any solution to the non-homogeneous difference equation, xt = xco t + xpa t For autonomous equations, a (very convenient) particular solution is the steady-state solution, xpa t = x∗, which is constant over time. Consider the homogeneous equation, xt + axt−1=0.
Recalling differential equations, one may guess a solution to this equation to be, xt = ckt, 2 where c is analogous to a constant of integration and k = −a.1 To see the latest, substitute the guessed solution in the equation, ckt + ackt−1 = 0; simplifying, ckt−1(k + a) = 0, which is satisfied if and only if k = −a. To summarize, the complementary solution is, xco t = c(−a)t As a particular solution take the steady-state x∗; substituting xt = x∗, x∗+ ax∗= b, hence xpa t = b 1 + a Therefore the general solution is, (3) xt = xco t + xpa t = c(−a)t + b 1 + a This solution is identical to 2, the one found by iterated substitutions, except for c that is still to be determined. Take t = 0, then x0 = c(−a)0 + b/(1 + a), implying c = x0 −b/(1 + a), if a ̸= −1.
It remains to consider the case in which a = −1, implying xt = xt−1 + b. Incidentally, the method used to find the general solution, and precisely the complementary solution, is called undetermined coefficients. This is based on guessing a functional form for xt and verify that this solves the equation (for some coefficients). Guess, xt = αt, a trend line with initial value at α to be determined. Substituting, α(t+1)−αt−b = 0, satisfied for α = b. Hence, if a = −1, (4) xt = bt + x0 You can check that any other guess on functional forms would not work.
We conclude this section establishing the Superposition Principle (SP).
Theorem 2.1 (SP): Any solution xt of the equation xt + axt−1 = bt can be written as xt = xco t + xpa t with xco t = c(−a)t, for a particular solution xpa t .
Proof. Let zt := xt −xpa t , where xt solves 1.
zt+1 = xt+1 −xpa t+1 = −axt + bt −(−axpa t + bt) = −a (xt −xpa t ) = −azt−1 That is, zt = c(−a)t solves the the homogeneous equation. Therefore, buy definition of zt, xt = zt + xpa t = c(−a)t + xpa t as desired.
□ 1Recall that for homogeneous differential equations of the type, x′(t)+ax(t) = 0, the complementary solution is xco(t) = ce(−a)t. This can be derived, first, by integrating x′ x + a = 0 over t, which yields log[x(t)] = B −at, with B defined by collecting the two constants of integration. Then, by taking the exponential of both sides and letting c := eB. In the difference equation, the instantaneous rate of change e(−a)t is replaced by (−a)t.
3 The Superposition Principle is a general result that, as you can see from its proof, extends to non-autonomous equations and to systems of linear equations.
2.1.2. Non-autonomous equations, lags and leads. Consider non-autonomous equations, assum-ing a time-varying term bt.2 In general, the solutions of these equations will take the functional form of bt.
If bt is an exponential or it is a polynomial of order p, then the solution will, respectively, have the form of an exponential or of a p-order polynomial in t.
A particularly useful method to solve equations is via the introduction of lag and lead oper-ators. In general, an operator is a linear function between vector spaces. L and F, respectively, are lag and lead operators if for every sequence (xt) xt−1 = Lxt = F−1xt xt+1 = Fxt = L−1xt.
Observe that we can repeatedly apply the operator to shift a variable over time, L2xt = L (Lxt) = Lxt−1 = xt−2 and, more generally, xt−p = Lpxt = F−pxt, xt+p = Fpxt = L−pxt.
Consider equation 1 and relabel the coefficient α := −a, so that (5) xt = αxt−1 + bt The solution to the homogeneous part is as above, xco t = αtx0.
To find the particular solution, we have to distinguish whether or not α is less than one in modulus.
|α| < 1 (1 −αL)xt = bt xpa t = 1 1 −αLbt Since |α| < 1 we know that the fraction is the infinite sum of a geometric sequence of common ratio αL, (1 −αL)−1 = ( 1 + αL + (αL)2 + (αL)3 + ......
) xco t = ( 1 + αL + (αL)2 + (αL)3 + ......
) bt = bt + αbt−1 + α2bt−2 + α3bt−3 + .........
Hence, xpa t = t ∑ s=−∞ αt−sbs.
2Non-autonomous equations may also be of the form xt = αt−1xt−1 + b, with time-varying coefficients.
4 In particular, if x0 is a given initial state (i.e. x−t = 0 for all t ≥0) the latest reduces to a finite sum, xpa t = t ∑ s=0 αt−sbs, All these infinite sums will not be well defined for |α| ≥1.
|α| > 1 In this case it suffices to solve the equation forward, i.e. use the lead operators. Start with xt+1 = αxt + bt+1.
xt+1 −αxt = bt+1 (F −α) xt = bt+1 ( 1 −F α ) xt = −1 αbt+1 with γ := α−1 is less than one in modulus.3 xpa t = − 1 1 −γFγbt+1 = − ( 1 + γF + (γF)2 + (γF)3 + ......
) γbt+1 = −γ ( bt+1 + γbt+2 + γ2bt+3 + .... + γsbt+1+s + ....
) xpa t = −1 α ∞ ∑ s=0 ( 1 α )s bt+1+s To summarize, whenever x0 is the initial state, the particular solution is (6) xpa t = ∑t s=0 αt−sbs if |α| < 1 −1 α ∑∞ s=0 ( 1 α )s bt+1+s if |α| > 1 You can check that if bt = b at all dates t, we obtain the solution to the autonomous system in 3 and 4 (given that α := −a).
Remark 2.1 (Variable coefficients). So far we have assumed that the coefficients a or α are constant over time. A non-autonomous equation may also be so because coefficients vary over time.
xt+1 = αt+1xt + bt+1 3Notice that we could have also proceeded, starting from xt = (1 −αL)−1bt as above, and observing that, 1 1 −αL = (αL)−1 (αL)−1(1 −αL) = −(αL)−1 1 −(αL)−1 = −F α × 1 1 −F α = −γF 1 −γF with γ := α−1 is less than one in modulus.
xpa t = − 1 1 −γFγFbt = −γ ( 1 + γF + (γF)2 + (γF)3 + ......
) bt+1 5 You can check, by iterated substitutions (backwards if |α| < 1, and forward if |α| > 1), that this extension leads to the following particular solutions.
(7) xpa t = ∑t−1 k=1 (∏k−1 s=0 αt−s ) bt−k + bt, if |α| < 1; −∑∞ k=0 (∏k s=0 ( 1 αt+1+s )) bt+1+k, if |α| > 1.
The complementary solution is, xco t = ( t ∏ s=0 αs+1 ) x0 Example 2.1. Consider the individual, sequential, budget constraint, (1 + rt)at = ct −et + at+1 where the gross return on assets a, (1 + rt) > 1 for all t (i.e. there is positive interest rate in every date) and (et)t is the lifetime endowment profile (e.g. exogenous labor income). Solve this difference equation forward, by iterated substitutions, to get the intertemporal budget constraint, given the limiting condition (transversality+ no-Ponzi) lims→∞ at+1+s ∏s−1 k=0(1+rt+k) = 0. Alternatively, compute the particular solution applying the above rule, and solve the for the complementary forward, by iterated substitutions.
2.2. Equations of first order with m > 1 variables (systems of equations). Consider the following difference equation in m variables, indeed a system of difference equations, x1t x2t .
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xmt = a11 a12 . . .
a1m a21 ...
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a2m . . .
amm x1t−1 x2t−1 .
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xmt−1 + b1t b2t .
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bmt In vector notation, (8) xt = Axt−1 + bt where A is a square m−matrix of coefficients aij.
First, notice that if A were a diagonal matrix, then the system would consist in m indepen-dent variables, with equations of the type xit = aiixit−1 + bit In this lucky case a solution to the system is just the one in 6 for each single equation.
A second case is when we can transform 8 into a system whose coefficient matrix is indeed diagonal, or closed to diagonal, solve the new system and map back the solutions to attain solutions of the original system. We shall illustrate this methodology next, starting from the simplest case of an autonomous system.
6 2.2.1. Autonomous systems. Let bt = b and, without loss of generality, assume m = 2.
First derive the eigenvalues of A. The eigenvalues λ of A are the m roots of the following polynomial (or characteristic) equation p(λ) = det (A −λIm) = det ( a11 −λ a12 a21 a22 −λ ) = (a11 −λ)(a22 −λ) −a12a21 p(λ) = 0 if and only if (a11 −λ)(a22 −λ)−a12a21 = 0. Simplifying, λ2 −λ(a11 +a22)+(a11a22 − a12a21) = 0 or λ2 −λtr + det = 0 with tr denoting the trace of A, and det its determinant. Notice that the eigenvalues of A are the roots of the polynomial p(·) = 0; hence,4 (9) λ1, λ2 = tr ± √ tr2 −4det 2 The discriminant is ∆:= (tr)2 −4det. If ∆> 0 we have distinct, real roots; if ∆< 0 we have distinct, complex roots; if ∆= 0 roots are coincident, both equal to tr/2.
Second, derive the eigenvectors of A. The ith eigenvector of A (associated to eigenvalue λi) is the vector vi in Rm, such that (10) (A −λiIm) vi = 0 Observe that the latest is homogeneous of degree 0 in vi (i.e. if vi solves 10, cvi does also solve 10 for any scalar c ̸= 0). Thus, we can normalize vi, for example, by taking one of its components equal to unity (i.e. c = 1/vm if vm ̸= 0), vi = ( νi 1 ) For the case m > 2, νi is a vector of m −1 components. Expression 10 is a system in m = 2 equations and one unknown, the scalar νi. To find a solution, we write the system as, ( a11 a12 a21 a22 ) ( νi 1 ) − ( νi 1 ) λi = 0 Consisting in two equations, a11νi + a12 −νiλi = 0 a21νi + a22 −λi = 0 that, solved by substitution, yield, a12 λi −a11 = νi = λi −a22 a21 4Consider a matrix A with m eigenvalues (λi). Two useful properties of eigenvalues are: i) ∏m i=1 λi = det(A) ii) ∑m i=1 λi = tr(A).
7 To check this is a viable solution, recall that λi is an eigenvalue; that is λi is such that p(λi) = 0, implying a21a12 −(λi −a22)(λi −a11) = 0. Therefore, we conclude the ith eigenvector is, ( νi 1 ) = ( λi−a22 a21 1 ) We start by deriving a particular solution for the autonomous system, with a 2−dimensional square matrix A.
Particular solution The particular solution is the steady-state, if it exists. Let xt = x∗for all t. Then, the system reduces to x∗= Ax∗+ b ⇐ ⇒(I2 −A)x∗= b If (I2 −A) is invertible, xpa t := x∗= (I2 −A)−1b (I2 −A)−1 = 1 p(1) ( 1 −a22 a12 a21 1 −a11 ) which is well defined as long as p(1) = det(I −A) ̸= 0 or, equivalently, if and only if 1−tr(A)+ det(A) ̸= 0.5 Complementary and general solutions Consider the homogeneous difference equation, xt = Axt−1. Let the matrix of eigenvalues and eigenvectors of A be, Λ := ( λ1 0 0 λ2 ) , P := [v1, v2] = ( λ1−a22 a21 λ2−a22 a21 1 1 ) Case 1: distinct eigenvalues.
Observe that P is of full rank (hence invertible) provided eigenvalues are distinct, λ1 ̸= λ2.
Hence, by definition 10 (see also the following Remark), AP = PΛ ⇐ ⇒A = PΛP −1 ⇐ ⇒P −1AP = Λ Distinct eigenvalues allow for a diagonalization of A (i.e to transform A into a diagonal matrix Λ). Substituting into the homogeneous system, xt = PΛP −1xt−1 or P −1xt = ΛP −1xt−1 We can now perform a change of variables, zt := P −1xt, zt = Λzt−1 The strategy is to derive a solution for zt to latter recover the solution of the original variable xt. The solution is easily determined since the matrix Λ is diagonal: given initial conditions, z0 := P −1x0, zt = ( z10λt 1 z20λt 2 ) 5You should verify that det(I −A) = 1 −tr(A) −det(A), which is the characteristic equation of A, p(λ) evaluated at 1.
8 Therefore, (11) xco t = Pzt = ( λ1−a22 a21 λ2−a22 a21 1 1 ) ( z10λt 1 z20λt 2 ) More generally, xit = x∗ i + m ∑ j=1 νijzj0λt j, for all i = 1, .., m −1 (12) xmt = x∗ m + m ∑ j=1 zj0λt j In matrix form, the general solution is (13) xt = x∗+ PΛtz0, z0 := P −1x0 Remark 2.2 (Matrix diagonalization). In general, an m × m matrix is diagonalizable if and only if it has m linearly independent eigenvectors (these vectors form a m−dimensional, com-plete, basis). To see why, observe that, by definition, P := [v1, v2, ..., vm], with vi being the ith−column of P. Since A is diagonalizable, P −1AP = Λ; implying that P is nonsingular, i.e. its columns (and hence the eigenvectors (vi) of A) are linearly independent. Conversely, if there are k linearly independent eigenvectors of a m−square matrix A, P is invertible, hence A is diagonalizable.
Eigenvectors are not all linearly independent when some eigenvalue λi that is repeated, with an algebraic multiplicity ri > 1 and it has associated eigenvectors who form a basis of dimension mi < ri (i.e. has geometric multiplicity mi less than the algebraic multiplicity ri). We are going to cover this in case 2 below.
Case 2: repeated eigenvalues.
Suppose A is not already a diagonal matrix; then, it is not diagonalizable whenever it has repeated eigenvalues; indeed, in this case not all eigenvectors are linearly independent. For example, matrices such as, ( 4 1 −1 2 ) and 4 2 −4 1 4 −3 1 1 0 are not diagonalizable. The first has both eigenvalues equal to 3 (its discriminant is zero). The second matrix has two repeated eigenvalues equal to 3 and one equal to 2. In all these cases, of multiple eigenvalues, some eigenvectors [v1, .., vm] are linearly dependent; hence, P fails to be invertible and A to be diagonalizable. Yet we can still transform A so as to simplify our system of difference equations, appealing to the Jordan canonical form. This transformation is attained as follows: (1) form the characteristic polynomial p(λ) and derive the eigenvalues (λi) of A; determine the ”algebraic multiplicity” of λi, which is by convention 1 when all eigenvalues are distinct; 9 (2) compute the associated eigenvectors: for each distinct λi, find the normalized m−vector vi that solves equation 10 above; instead, when λi has algebraic multiplicity ri > 1, compute the generalized eigenvectors of the repeated roots, as we are going to explain below; the number of mi linearly independent eigenvectors defines the geometric multi-plicity of the associated eigenvalue (for a distinct eigenvalue, both multiplicity indexes are one); (3) next, form P with the derived (linearly independent) eigenvectors to attain the Jordan canonical form of A, J := P −1AP; (4) finally, proceed as above with a change of variables in the original system of difference equations.
We now provide a definition of generalized eigenvectors. Suppose that an eigenvalue λi of A has an algebraic multiplicity ri > 1. The associated generalized eigenvector vi is a nonzero vector such that (A −λiI)vi ̸= 0 but (A −λiI)hvi = 0, for some h > 0. Further, suppose that some eigenvalue of A, say λi is repeated ri > 1 times. The (standard) ri eigenvectors associated to λi are linearly dependent (identical if normalized), that is to say that there is a unique linearly independent eigenvector, vi corresponding to λi, satisfying (A −λiI)vi = 0.
Next, we form a sequence of ri −1 generalized eigenvectors (v2, ..., vri) corresponding to λi, defined as vectors solving the following difference equations, (A −λiIm)vi = vi−1, i = 2, ..., ri ⇐ ⇒ (A −λiIm)v2 = v1 (A −λiIm)2v3 = v1 ........................
(A −λiIm)ri−1vri = v1 Clearly, (A −λiIm)vi ̸= 0. Moreover, pre-multiplying (A −λiIm)ri−1vri = v1 by (A −λiIm) we can derive a more general formulation of a the j −th generalized eigenvector associated to an eigenvalue λi with multiplicity ri, (A −λiIm)rivri = 0 This computation is reiterated stepwise for each eigenvalue i with multiplicity ri > 1.
To better illustrate computations, consider the following example.
Example 2.2. Let A be the 3 × 3 matrix above, with an eigenvalue of 2 and two eigenvalues of 3 (i.e. an eigenvalue with algebraic multiplicity of 2).
λ1 = 2, v1 solves (A −λ1I3)v1 = 0 λ2 = 3 =: λ, v2 solves (A −λI3)v2 = 0 λ3 = λ, v3 solves (A −λI3)v3 = v2 ⇐ ⇒(A −λI3)2v3 = 0 Carrying out computations, P := [v1, v2, v3] = 1 2 2 1 1 2 1 1 1 10 P is nonsingular (with a determinant equal to 1); hence, the Jordan matrix is well defined: J := P −1AP = 2 0 0 0 3 1 0 0 3 = ( J1 0 0 J2 ) Notice that J is almost (but not really) diagonal, with J1 = λ1 = 2 and J2 that is almost diagonal of the form, J2 = ( λ 1 0 λ ) = λI2 + ( 0 1 0 0 ) Then, A = PJP −1 which, substituted into the homogeneous difference equation system, gives xt = PJP −1xt−1 or P −1xt = JP −1xt−1 We can now perform a change of variables, zt := P −1xt, zt = Jzt−1 Notice, that everything is identical to the case in which A were diagonalizable except that the Jordan matrix J replaces the diagonal matrix of eigenvalues Λ. Yet, since J is not diagonal, something changes in the solution. We illustrate so in the context of the latest example.
zt = Jzt−1 ⇐ ⇒ z1t = 2z1t−1 z2t = 3z2t−1 + z3t−1 z3t = 3z3t−1 The first and third equations have a solution of the form zit = zi0λt i. The second equation is of the form, xt = αxt−1 + bt with |αj| > 1, bt := x0αt−1. In the present context, xt = z3t, α = λ = 3, x0 = z30. Looking back to the particular solution in 6, one sees that it does not work here, because bt is exponential (the particular solution in 6 would not be well defined). Instead, one can find an alternative particular using undetermined coefficients. For an exponential term bt one should guess an exponential functional form for the variable. Hence, let z2t = kλt, for an unknown (undetermined) coefficient k, and substitute into the equation to check whether or not the guess is correct: kλt = αkλt−1 + z30λt−1; dividing through by λt−1, the equation is satisfied for kλ = 3k + z30, or k = z30/(λ −3). The latest is not well defined since λ = 3. So, guess an alternative functional form, z2t = ktλt and verify: ktλt = λ ( k(t −1)λt−1) + z30λt−1 simplifying, k = z30/λ, with λ = 3. So, zpa 2t = z30tλt−1 By the Superimposition Principle, the general solution is, z2t = z20λt + z30tλt−1 = (z20λ + z30) λt−1 11 where the complementary solution is the solution of the homogeneous equation, z2t = 3z2t−1.
Finally, it is easy to derive a solution for the original system in x.
To conclude, the general solution of an autonomous system is analogous to 13, with the novelty that Λ is replaced by J and the matrix P contains generalized eigenvectors: (14) xt = x∗+ PJtz0, z0 := P −1x0 Finally, we point out that the Jordan form, which we derived in the above example, posses the following general representation: form g distinct eigenvalues, J = diag (J1, ..., Jg) , with a typical (Jordan) block, Ji = λi 1 0 . . .
0 0 ...
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1 0 . . .
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0 λi = λiI + 0 1 0 . . .
0 0 ...
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1 0 . . .
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0 0 Each block Ji is a ri × ri matrix, where ri is the algebraic multiplicity of λi and ∑g i=1 ri = m (i.e. J, as a m−square matrix, conforms to A). In our example above, r1 = 1 and r2 = 2.
Clearly, for ordinary matrices, with distinct eigenvalues, g = m, there are as many 1 × 1 Jordan blocks as the eigenvalues, and J = Λ.
2.2.2. Non-autonomous systems. Let us go back to the general case, with bt that is time-varying. The complementary solution is identical to the one just derive for the autonomous system. It remains to derive a particular solution and apply the Superposition Principle. We can do so using the method of undetermined coefficients. To guess a solution, we use the form of the solution found for single difference equations in section 2.1.2: we guess, (15) xpa t = t ∑ s=0 (A)t−s bs Let Rs,t := ∏t−1 i=s Ai, so that xpa t = ∑t s=0 Rs,tbs; and observe that Rs,t = ARs,t−1, Rt,t = A0 = I for all t.
xpa t+1 = t+1 ∑ s=0 Rs,t+1bs = t ∑ s=0 Rs,t+1bs + Rt+1,t+1bt+1 = A t ∑ s=0 Rs,tbs + bt+1 = Axpa t + bt+1 12 Although this is fully correct, it may not be computationally clever. In fact, we may have a much easier problem to solve if, rather than considering the original system, we pick the diagonalized one. For simplicity, consider the case in which A has distinct eigenvalues, in the diagonal matrix Λ, and thus independent eigenvectors in P. The transformed system is, zt = Λzt−1 + ct where ct := P −1bt.6 Therefore, for m = 2, we attained two independent, difference equations of the first order, zit = λizit−1 + cit, i = 1, 2 whose solutions are in 6. Alternatively, we can also express the particular solution as 15, zpa t = t ∑ s=0 (Λ)t−s cs with the advantage that Λn = diag[λn 1, λn 2] is easy to compute.7 Finally, we can retrieve the solution of the original system by using xt = Pzt.
An extension to the case in which A has repeated eigenvalues and to those in which it has time-variable coefficients should be straightforward.
2.3. Higher order difference equations. The order p of a difference equation refers to the higher order lag with which a variable is represented. A difference equation of order p in one variable take the form, (16) xt −α1xt−1 −α2xt−2 −...... −αpxt−p = bt The strategy to solve higher order, linear, difference equations is to transform them into systems with more variables of order one and exploit the techniques illustrated in section 2.2.
For expositional simplicity, we illustrate the procedure for the case of p = 2. We can rewrite 16 as xt = α1xt−1 + α2xt−2 + bt xt−1 = xt−1 or ( xt xt−1 ) = ( α1 α2 1 0 ) ( xt−1 xt−2 ) + ( bt 0 ) Looking at the system in vector notation, we can write (17) xt = Axt−1 + bt, which is of first order. We can solve this system applying the techniques learned above for equations of first order with m variables.
Complementary solution The eigenvalues (λ1, λ2) of A can be derived applying formula 9, with tr = α1, det = −α2 6You can check that for m = 2, c1t = (b1t −ν2b2)/(ν1 −ν2) and c2t = (−b1t + ν1b2t)/(ν1 −ν2).
7Here however one should be careful to distinguish cases in which the eigenvalues are bigger or less than one in absolute value.
13 Hence, λ1, λ2 = α1 2 ± 1 2 √ α2 1 + 4α2 The associated eigenvectors are vi = ( λi 1 ) , i = 1, 2.
Using the same notation as above, P := (v1, v2) = ( λ1−a22 a21 λ2−a22 a21 1 1 ) = ( λ1 λ2 1 1 ) which, again, is invertible if λ1 ̸= λ2. Finally, use equation 11, ( xt xt−1 ) = Pzt = ( λ1 λ2 1 1 ) ( z10λt 1 z20λt 2 ) with (18) z0 := P −1x0 = 1 λ1 −λ2 ( 1 −λ2 −1 λ1 ) ( x0 x−1 ) which, assuming x−1 = 0, yields, z10 = x0/(λ1 −λ2) and z20 = −x0/(λ1 −λ2) Extracting the solution for xt, (19) xco t = λ1z10λt 1 + λ2z20λt 2 This, for equations of order p, generalizes to, xco t = p ∑ i=1 λizi0λt i Particular solution Here we can simply refer to the particular solution of a non-autonomous system 15 derived above. An alternative procedure, specific to single difference equations of order p is to derive the particular solution exploiting the Fundamental Theorem of Algebra (FTA); this solution method is known as factorization. FTA says that any polynomial of order p can be factor into exactly p terms, each of which contains one characteristic root. That is, the polynomial in z, zp + a1zp−1 + ... + ap−1z + ap = 0 can equivalently be written (or factorized) as, (γ1 −z)(γ2 −z) · · · (γp −z) = 0, where (γi) are the (possibly non-distinct) characteristic roots of the polynomial.
Again, for simplicity, consider the p = 2 version of the difference equation 16, xt −α1xt−1 −α2xt−2 = bt Using the lag operator, (20) (1 −α1L −α2L2)xt = bt 14 The left hand side of this equation is, (1 −α1L −α2L2)xt equivalent to (1 + a1L + a2L2)xt with ai := −αi for all i. Now we can apply the FTA and factor the polynomial in L, as p(L) = (1 −λ1L)(1 −λ2L), where (λ1, λ2) are its characteristic roots. This is not immediately obvious. Indeed, it is key to notice the difference between 2.3 and 2.3. The form of the polynomial 2.3 is analogous to the one obtained by dividing 2.3 through by zp, 1 + a1z−1 + ... + ap−1z1−p + apz−p = 0 This, can factorized as, (1 −γ1z−1)(1 −γ2z−1) · · · (1 −γpz−1) = 0, and the equivalence between the two forms can easily be checked for the case of p = 2. By analogy, treating L as z−1, we have explained the form of 2.3.
Since, the characteristic roots, by definition, satisfy p(λi) = 0 for all i. So, (1 −λ1L)(1 −λ2L)xt = bt Therefore, xpa t = bt (1 −λ1L)(1 −λ2L) with λ1, λ2 = α1 2 ± 1 2 √ α2 1 + 4α2 Suppose that either one of the roots, say λ2 is less than one in modulus, then a way to solve the equation is as follows, (1 −λ1L)xpa t = bt 1 −λ2L yielding, xt = λ1xt−1 + ct which is a simple equation of first order, with the last term, ct := bt 1 −λ2L = t ∑ s=−∞ λt−s 2 bs The extension to equations of order p yields, xpa t = bt (1 −λ1L)(1 −λ2L) · · · (1 −λpL) 15 To determine λ′s. Go back to the original system representation 17 of a generic difference equation of order p. (λi) are the p eigenvalues of matrix A = a1 a2 . . .
ap 1 0 . . .
0 .
.
.
...
...
.
.
.
0 . . .
1 0 Finally, a general solution can simply be found using, xt = xco t + xpa t .
2.4. Solution stability. One of the main objectives in the study of a dynamical system is to analyze the behavior of its solutions near a steady-state. This is the concern of stability theory. In this subsection, we introduce the basic concepts and results of stability theory for linear difference equations. These will latter be extended to analyze local stability of non-linear equations.
2.4.1. First order equations. Let us start with a first order difference equation as 5, xt = αxt−1 + b A steady-state x∗is such that xt = x∗at all t.
Definition 1 (Stability). The steady-state x∗of 5 is stable if given ϵ > 0 there exists δ > 0 such that |x0 −x∗| < δ = ⇒|xt −x∗| < ϵ, for all t > 0.
If x∗is not stable, then it is called unstable.
A stable solution is represented in figure 1,8 where n = t, the initial period is labeled as n0, the horizontal axis crosses the vertical at the steady-state, the δ-neighborhood contains the initial value of xt, there exists an ϵ neighborhood containing the values of xt for all t = n ≥n0 n x1 n0 x 2 Figure 1. x∗is stable 8This figure is taken from Elaydi (2005), p.177.
16 Definition 2 (Asymptotic stability). The steady-state x∗of 5 is asymptotically stable (or attracting) if there exists a η > 0 such that |x0 −x∗| < η = ⇒ lim t→∞xt = x∗.
If η = ∞, x∗is globally asymptotically stable (x∗is a global attractor).
Loosely speaking, a x∗is asymptotically stable if every solution xt that starts (at t = 0) near x∗converges to x∗. It is globally asymptotically stable if just about every solution xt tends to x∗over time. x∗is stable if every solution xt starting (at t = 0) close to x∗(i.e. in ball of ray δ centered in x∗), does also remain close to it over time (i.e. it will stay in an ϵ−ball as t goes to infinity). Notice that asymptotic stability is a stronger requirement; it implies stability, but it is not implied by it.
Theorem 2.2 (Stability 1): A steady-state x∗is asymptotically stable if |α| < 1 and asymptotically unstable if |α| > 1. If |α| = 1 the steady-state x∗is stable.
The proof of the theorem goes as follows. First, it is clear that the system has no dynamics if α = 0, since then xt = x∗= b. Next, suppose |α| < 1, but α ̸= 0. Notice that, however distant from x∗is an initial state x0, the system will converge to the steady state: a solution to equation 5 is xt = αtx0+x∗, hence as t goes to infinity xt approaches x∗(because αt →0). The sign of α determines the transitional dynamics. If negative, the sequence will oscillate around x∗until convergence (in fact, when t is even -odd-, at is positive -negative-). Conversely, when |α| > 1, but α ̸= 0, the dynamics is explosive; and transitional dynamics is either oscillatory or not, depending on whether a is negative or positive. It remains to consider the case in which |α| = 1. If α = 1 we have xt = x0, hence constant. If α = −1, xt = ±x0, respectively, in even and odd times t, hence a stable oscillatory dynamics.
A diagrammatic analysis is also instructive. In a Cartesian diagram (xt, xt+1) one can graph equation 5 as time passes. The intersection of 5 with the 45◦−line is the steady state x∗. Then, take as initial state x0 any point to the right, or to the left, of x∗and draw the phase diagram as follows. From x0 on the horizontal axis recover x1(x0) from the graph of equation 5 and placed it on the vertical axis; then, use the 45◦−line to map x1(x0) into the horizontal axis.
Reiterate: given the new state x1(x0), find x2(x1(x0)) from the graph of equation 5 and placed it on the vertical axis; then, use the 45◦−line to map x2(x1(x0)) into the horizontal axis. Each of these steps either brigs the value of the sequence closer to x∗(asymptotical stable case) or it pushes it away from it (asymptotically unstable case). You can check that x∗is asymptotically stable if the slope of the phase diagram of equation 5, at x∗, is less than one (crossing the 45o-line from below with a slope less than 1 or from above with a grater than −1).
2.4.2. Systems of equations. We deal directly with first-order systems because, as we have learned above, this does also allow to study higher order difference equations. Consider the autonomous system 8 in m variables and equations, xt+1 = Axt + b 17 Again, we aim at studying the stability the solution of 8, which we recall being 12 xit = x∗ i + m ∑ j=1 vijzj0λt j, i = 1, .., m, where (λi) are the eigenvalues of A and vij are elements of the matrix P, whose columns are formed by the eigenvectors of A, and (z0) are initial conditions. Notice that what drives the dynamics, essentially, are the eigenvalues. If they are all less than one in absolute values, then the system solution will tend to x∗. Instead, if even one of the eigenvalues is larger than one in absolute value, the system solution will be asymptotically unstable: if |λi| > 1 for some i, and zi0 ̸= 0, some variable of the system will tend to diverge from its steady-state value (even if the ith−row of P, vi, is zero, xmt will be asymptotically unstable).
We summarize these considerations in the following, Theorem 2.3 (Stability 2): Let x∗be a steady-state of the system on Rm.
(1) x∗is asymptotically stable if all the eigenvalues (λi) of A are strictly less than one in modulus. The solution is said to be a sink.
(2) x∗is asymptotically unstable if at least one of the eigenvalues of A has modulus grater than one. The solution is said to define a source or a saddle, respectively, depending if |λi| > 1 for all i or for some (but not all) i.
(3) x∗is stable if all the eigenvalues of A have modulus weakly less than one, and some has modulus equal to one.
In the special case in which the characteristic polynomial is of order 2 (i.e. m = 2 as in our growth model), then there is a simple stability test one can carry out. Recall the characteristic polynomial of A is p(λ) = λ2 −trλ + det; its zeros, λ1, λ2, are the eigenvalues of A. This can be also seen a 2nd-order polynomial in λ that can be factor using its roots (corresponding to the eigenvalues of A), p(λ) = (λ −λ1)(λ −λ2) so that, by definition λ = λi satisfies p(λi) = 0 for all i = 1, 2.
Next, for stability analysis, it is interesting to evaluate p(λ) at 1. If λ = 1 solves 2.4.2, p(1) = (1 −λ1)(1 −λ2) = 0, i.e. if det = tr −1, then the system has either one of the eigenvalues equal to unity. Similarly, when p(−1) = (−1 −λ1)(−1 −λ2) = 0 ⇐ ⇒(1 + λ1)(1 + λ2) = 0 i.e. det = −(tr + 1), then either λ1 or λ2 are equal to −1. This is also useful to verify if the eigenvalues are both ling above or below one. When p(1) > 0, (1 −λ1)(1 −λ2) > 0, occurring when either λ1, λ2 ≥1 or λ1, λ2 ≥1, both with strict inequality if eigenvalues are repeated (λ1 = λ2). In contrast, when p(1) < 0, (1 −λ1)(1 −λ2) < 0, the eigenvalues lie on opposite sides of unity. We summarize the latest results in the following.
18 Remark 2.3 (Testing stability). Let p(z) := (z −λ1)(z −λ2).
p(1) = 0 iffλ1 or λ2 are equal to 1 iffdet = tr −1.
p(−1) = 0 iffλ1 or λ2 are equal to −1 iffdet = −(tr + 1).
p(1) > 0(< 0) iffλ1, λ2 lie on the same (opposite) side of 1, iffλ1, λ2 > 1 or λ1, λ2 < 1 (λi < 1, λ−i > 1, i = 1, 2), det > tr −1 (det < tr −1).
p(−1) > 0(< 0) iffλ1 and λ2 lie on the same (opposite) side of −1, iffdet > −(tr + 1) (det < −(tr + 1)).
Notice that if one knows that either p(1) < 0 or p(−1) < 0, it means that |λi| < 1 for some (at least one) i; i.e. there is at least a stable eigenvalue, hence the solution is either a saddle or a sink. Instead, if one knows that p(1) > 0 or p(−1) > 0 and |λi| < 1 for some i, also the other eigenvalue λ−i satisfies |λ−i| < 1 and the solution is a sink.
2.5. Stochastic difference equations.
2.5.1. Definitions. Stochastic models are those in which some variable is uncertain, i.e. it is regarded as random. Suppose x is such variable, x = (x0, x1, ....., xt, ....) it is an infinite sequence of random variables, a stochastic process. At each date t the t−element of this process, xt, is a random variable whose realization is on some domain Xt. A realization of x is an infinite sequence of realizations, each for every t−element. A finite, t, realization of x, xt = (x0, x1, .., xt) is said to be a history up to t, or a time-series.
A stochastic process is said to be strictly stationary if for all k ≥0 the sequences (x0, x1, ....., xt, ....) and (xk, xk+1, ....., xk+t, ....) have the same distribution. A stochastic process is said to be mean stationary if the mean of its elements is time invariant: E(xt) = E(xt+k) for all k ≥0. A sto-chastic process is said to be covariance stationary if the covariance of its elements depends only on theirs distance: Cov(xt, xt+k) = Cov(xT , xT+k) for all t, T, k ≥0. A stochastic process is said to be independently and identically distributed (iid) if each and every one of its elements, xt, has the same marginal distribution. This is different concept from stationarity that also place restrictions on the joint distributions; iid implies stationarity if the process is also serially independent, i.e. if every collection of sequence coordinate (or history of random variables) is totally independent.
A stochastic process has the Markov property if for all s ≥1, Prob(xt+1|xt, xt−1, ..., xt−s) = Prob(xt+1|xt) We assume that agents have the same information (degree of uncertainty); that is, they all know the probability space governing the system (space of possible events or state space, prob-ability distributions of stochastic processes, realizations of past events). At each time t agents, having observed the current and past realizations of each variable xt, form their expectations on xt+1 rationally, i.e. by computing the conditional expectation, txt+1 := Et(xt+1|Ωt), where Ωt represents the realizations of current and past variables. If the process is (at least) mean sta-tionary, txt+1 := E(xt+1|Ωt). If it is also Markovian, one takes Ωt = xt, i.e. txt+1 := E(xt+1|xt).
Law of Iterated Expectations: If Ωt ⊆Ωt+1, E (E (xt|Ωt+1) |Ωt) = E (xt|Ωt) 19 2.5.2. Solving stochastic difference equations. Solutions can basically be found applying the techniques of section 2.2 on deterministic difference equations. We clarify this with a simple example. For simplicity assume that stochastic processes are mean stationary, and let Et (·) := E (·|Ωt).
Example 2.3. Consider the following first order equation, Etxt+1 = αxt + Etut+1, |α| > 1.
Apply the operator L.
LEtxt+1 = αLxt + LEtut+1 and find, Etxt = αxt−1 + Etut Observe that Etxt = xt. Hence, xt = αxt−1 + Etut Find the general solution applying 6, xt = αtx0 −1 α ∞ ∑ s=t+1 ( 1 α )s−t Etus 20 |
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View definitions for relinquish
relinquish
verb as in give up, let go
Strongest matches
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Finally McIlroy landed a hot putt on the 14th, earning a birdie which put Europe into a lead they would not relinquish.
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It was the moment England needed and, after Zoe Harrison added the extras, they never relinquished their lead.
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They failed to win the opening hole in any of the first dozen matches as Europe romped to an early lead they wouldn't relinquish.
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North Korea declared itself a nuclear power in 2022 and vowed to never relinquish its weapons.
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"Both parties over decades have not wanted to relinquish the very powerful licencing system."
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What is the factoring the difference of two squares?
Common Core State Standards
How to factor the difference of two squares
Factor the difference of two squares examples
Example 1: x² variable has a coefficient of 1 Example 2: y² variable has a coefficient of 1 Example 3: x² variable has a coefficient greater than 1 Example 4: coefficient of both terms is greater than 1 Example 5: exponents greater than 2 Example 6: exponents greater than 2
Teaching tips for factoring the difference of squares
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Practice factoring the difference of two squares questions
Factoring the difference of two squares FAQs
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Math resources Algebra Factoring
Factoring the difference of two squares
Factoring the difference of two squares
Here you will learn how to factor the difference of two square terms to problem solve. You have already learned other techniques for factoring, now, you can add to your factoring strategies, factoring the difference of perfect square terms.
Students first learn how to factor in 7 th grade math where they learn how to factor the greatest common factor out of an expression. They extend that knowledge of factoring in an algebra 1 class.
What is factoring the difference of two squares?
The Difference of two squares is an algebraic expression where the first expression and the second expression are perfect square terms with the second square term being subtracted from the first. The difference of two squares is always in the form of:
Algebraic expressions that are the difference of squares are factorable.
They factor to be:
Let’s take a look algebraically and geometrically why the expression factors in this way.
If the expression were to be written in standard form of a quadratic, it would be written like this:
a^2-b^2=a^2+0 a b-b^2
You can use the trial and error method of factoring to factor the quadratic expression:
a^2+0 a b-b^2
The first term factors to a \times a and the last term factors to (- \, b) \times b
(a-b)(a+b)
The inside terms multiply to - \, ab and the outside terms multiply to be + \, ab
When you add ab and - \, ab, it equals 0ab or just 0 (they cancel each other out). This means that when you factor, a^2-b^2 it factors to be (a-b)(a+b). This works all the time with the difference of two perfect squares.
[FREE] Factoring The Difference Of Two Squares Worksheet (High School)
Use this worksheet to check your high school students’ understanding of factoring the difference of two squares. 15 questions with answers to identify areas of strength and support!
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[FREE] Factoring The Difference Of Two Squares Worksheet (High School)
Use this worksheet to check your high school students’ understanding of factoring the difference of two squares. 15 questions with answers to identify areas of strength and support!
DOWNLOAD FREE
Now, let’s look at why this works geometrically using squares.
Take a square with side length a. This has an area of a^{2}.
Let b=\cfrac{1}{2} \, a. Draw a square of side length b from a vertex (you may notice that a vertex of square b is the centre of square a ). Square b has an area of b^{2}.
Remove the square with side length b from the diagram. The remaining area is now a^{2}-b^{2} as the area of b^2 has been subtracted from the area of a^2.
The width of the top of the L -shape is a-b as the length of b has been subtracted from the side length of the square a.
This is the same as the furthest right vertical of the L -shape, however as b=\cfrac{1}{2} \, a, \, a-b=a-\cfrac{1}{2} \, a=\cfrac{1}{2} \, a=b. So, the furthest right vertical side has length b.
If you then split the L -shape into a rectangle and a square, you can then move the square to make a longer rectangle. The area has not changed as nothing has been removed here:
The rectangle now has a width of a-b, a height of a+b and an area of a^{2}-b^{2} which if substituted into the standard formula for the area of a rectangle is equivalent.
A=L\times{W}=(a+b)\times(a-b)=a^{2}-b^{2}
This geometrical representation with squares further proves that:
a^2-b^2=(a+b)(a-b)
What is the factoring the difference of two squares?
Common Core State Standards
How does this relate to high school math?
High School Algebra – Seeing Structure in Expressions: (HSA-SSE.A.2)Use the structure of an expression to identify ways to rewrite it. For example, see x^4 - y^4 as (x^2)^2 - (y^2)^2, thus recognizing it as a difference of squares that can be factored as (x^2 - y^2)(x^2 + y^2).
High School Algebra – Seeing Structure in Expressions: (HSA-SSE.B.3a)Choose and produce an equivalent form of an expression to reveal and explain properties of the quantity represented by the expression.★ a. Factor a quadratic expression to reveal the zeros of the function it defines.
How to factor the difference of two squares
In order to factor an algebraic expression using the difference of two squares:
Write down two sets of parentheses.
Square root the first term and write it on the left hand side of both parentheses.
Square root the last term and write it on the right hand side of both parentheses.
Put a \textbf{“ +”} in the middle of one bracket and a \textbf{“- “} in the middle of the other (the order doesn’t matter).
Factor the difference of two squares examples
Example 1: x² variable has a coefficient of 1
Factor the expression x^{2}-9.
Write down two sets of parentheses.
(\qquad)(\qquad)
2Square root the first term and write it on the left hand side of both parentheses.
x^2 is the first term and \sqrt{x^2}=x
Each set of parentheses starts with x.
(x\quad)(x\quad)
3Square root the last term and write it on the right hand side of both parentheses.
9 is the second term and \sqrt{9}=3
Each set of parentheses ends with 3.
(x\quad{3})(x\quad{3})
4Put a \textbf{“ +”} in the middle of one bracket and a \textbf{“- “} in the middle of the other (the order doesn’t matter).
(x+3)(x-3)
Remember, you can always check to see if your answer is correct by multiplying the factored expression back out.
x^{2}+3x-3x-9=x^{2}-9
x^{2}-9 factors to be (x+3)(x-3) .
Note: The order of the “+” and “-” does not matter.
Example 2: y² variable has a coefficient of 1
Factor the expression 64-y^{2} .
Write down two sets of parentheses.
(\qquad)(\qquad)
Square root the first term and write it on the left hand side of both parentheses.
The first term is 64 and \sqrt{64}=8 .
Each set of parentheses starts with 8.
({8}\quad)({8}\quad)
Square root the last term and write it on the right hand side of both parentheses.
The second term is y^2 and \sqrt{y^2}=y .
Each set of parentheses ends with y.
({8}\quad{y})({8}\quad{y})
Put a \textbf{“ +”} in the middle of one bracket and a \textbf{“- “} in the middle of the other (the order doesn’t matter).
(8-y)(8+y)
Check:
64+8y-8y-y^{2}=64-y^{2}
64-y^{2} factors to be (8-y)(8+y) .
Note: The order of the “+” and “-” does not matter.
Example 3: x² variable has a coefficient greater than 1
Factor the expression 25x^{2}-16 .
Write down two sets of parentheses.
(\qquad)(\qquad)
Square root the first term and write it on the left hand side of both parentheses.
The first term is 25x^2 and \sqrt{25x^{2}}=5x
Each set of parentheses starts with 5x.
(5x\quad)(5x\quad)
Square root the last term and write it on the right hand side of both parentheses.
The second term is 16 and \sqrt{16}=4
Each set of parentheses ends with 4.
(5x\quad{4})(5x\quad{4})
Put a \textbf{“ +”} in the middle of one bracket and a \textbf{“- “} in the middle of the other (the order doesn’t matter).
(5 x+4)(5 x-4)
Check:
25x^{2}+20x-20x-16=25x^{2}-16
25x^{2}-16 factors to be (5x+4)(5x-4) .
Example 4: coefficient of both terms is greater than 1
Factor the expression fully.
4x^{2}-81y^{2}
Write down two sets of parentheses.
(\qquad)(\qquad)
Square root the first term and write it on the left hand side of both parentheses.
The first term is 4x^2 and \sqrt{4x^{2}}=2x
Each set of parentheses starts with 2x.
(2x\quad)(2x\quad)
Square root the last term and write it on the right hand side of both parentheses.
The second term is 81 y^2 and \sqrt{81 y^2}=9 y
Each set of parentheses ends with 9y.
(2x\quad{9y})(2x\quad{9y})
Put a \textbf{“ +”} in the middle of one bracket and a \textbf{“- “} in the middle of the other (the order doesn’t matter).
(2x+9y)(2x-9y)
Check:
4x^{2}+18xy-18xy-81y^{2}=4x^{2}-81y^{2}
4x^{2}-81y^{2} factors to be (2x+9y)(2x-9y) .
Example 5: exponents greater than 2
Factor the expression completely.
x^{4}-100
Write down two sets of parentheses.
(\qquad)(\qquad)
Square root the first term and write it on the left hand side of both parentheses.
The first term is x^4 and \sqrt{x^4}=x^2
Each set of parentheses starts with x^2.
\left(x^2\quad\right)\left(x^2\quad\right)
Square root the last term and write it on the right hand side of both parentheses.
The second term is 100 and \sqrt{100}=10
Each set of parentheses ends with 10.
\left(x^{2}\quad{10}\right)\left(x^{2}\quad{10}\right)
Put a \textbf{“ +”} in the middle of one bracket and a \textbf{“- “} in the middle of the other (the order doesn’t matter).
\left(x^2+10\right)\left(x^2-10\right)
Check:
x^{4}+10x^{2}-10x^{2}-100=x^{4}-100
x^{4}-100 factors to be \left(x^{2}-10\right)\left(x^{2}+10\right) .
Example 6: exponents greater than 2
Factor the expression completely.
2m^6-72
Write down two sets of parentheses.
As both of the terms in the expression are a multiple of 2, factor this first.
2(m^{6}-36)
When writing the two pairs of parentheses, place the factor of 2 infront.
2(\qquad)(\qquad)
The remaining expression to factor is m^{6}-36.
Square root the first term and write it on the left hand side of both parentheses.
The first term is m^6 and \sqrt{m^6}=m^3
Each set of parentheses starts with m^3.
\left(m^3\right)\left(m^3\right)
Square root the last term and write it on the right hand side of both parentheses.
The second term is 36 and \sqrt{36}=6
Each set of parentheses ends with 6.
\left(m^3\quad{6}\right)\left(m^3\quad{6}\right)
Put a \textbf{“ +”} in the middle of one bracket and a \textbf{“- “} in the middle of the other (the order doesn’t matter).
\left(m^3-6\right)\left(m^3+6\right)
Check:
2\left(m^{6}-6m^{3}+6m^{3}-36\right)=2\left(m^{6}-36\right)=2m^{6}-72
2m^6-72 factors fully to be 2\left(m^3-6\right)\left(m^3+6\right) .
Teaching tips for factoring the difference of squares
Use visual tools such as algebra tiles or digital algebra tiles so students can connect the area model with arrays to factoring.
Although worksheets have their place, having students work in collaborative groups to problem solve is essential to the math classroom.
Easy mistakes to make
Forgetting that in the parentheses the signs are differentThere must be a + in one bracket, and a – in the other, the order doesn’t matter for example (\quad+\quad)(\quad-\quad) or (\quad-\quad)(\quad+\quad) .
Not finding the square root when factoring the difference of squares.For example, factoring 16x^{2}-4 to be (8x-2)(8x+2). The correct factored expression should be (4x-2)(4x+2) as \sqrt{16x^{2}}=4x.
Related factoring lessons
Factoring
How to factor quadratic equations
Factor by grouping
Practice factoring the difference of two squares questions
Factor the expression x^{2}-25 .
(x-5)(x-5)
(x+5)(x-5)
(x+1)(x-25)
(x-1)(x+25)
The expression x^2-25 is the difference of two squares. To factor it, take the square root of both the first and the second expressions and place the square roots in two sets of parentheses. The middle sign of one set of parentheses is “+” and the second one is “-”.
The first term in each pair of parentheses is \sqrt{x^2}=x
The second term in each pair of parentheses is \sqrt{25}=5
(x+5)(x-5) is the factored expression.
Factor the expression y^{2}-81.
(y+9)(y-9)
(y-9)(y-9)
(y+1)(y-81)
(y-1)(y+81)
The expression y^2-81 is the difference of two squares. To factor it, take the square root of both the first and the second expressions and place the square roots in two sets of parentheses. The middle sign of one set of parentheses is “+” and the second one is “-”.
The first term in each pair of parentheses is \sqrt{y^2}=y
The second term in each pair of parentheses is \sqrt{81}=9
(y+9)(y-9) is the factored expression.
Factor the expression 49-y^{2} .
(7-y)(7-y)
(49+y)(1-y)
(7+y)(7-y)
(1+y)(49-y)
The expression 49-y^2 is the difference of two squares. To factor it, take the square root of both the first and the second expressions and place the square roots in two sets of parentheses. The middle sign of one set of parentheses is “+” and the second one is “-”.
The first term in each pair of parentheses is \sqrt{49}=7
The second term in each pair of parentheses is \sqrt{y^{2}}=y
(7+y)(7-y) is the factored expression.
Factor the expression 4-x^4 .
(4+x)(1-x)
\left(x^2+2\right)\left(x^2-2\right)
\left(1+x^2\right)\left(4-x^2\right)
\left(2+x^2\right)\left(2-x^2\right)
The expression 4-x^4 is the difference of two squares. To factor it, take the square root of both the first and the second expressions and place the square roots in two sets of parentheses. The middle sign of one set of parentheses is “+” and the second one is “-”.
The first term in each pair of parentheses is \sqrt{4}=2
The second term in each pair of parentheses is \sqrt{x^{4}}=x^{2}
(2+x^{2})(2-x^{2}) is the factored expression.
Factor the expression 9x^{2}-100 .
(3x+10)(x-10)
(3x-10)(3x-10)
(9x+100)(x-1)
(9x+1)(x-100)
The expression 9x^{2}-100 is the difference of two squares. To factor it, take the square root of both the first and the second expressions and place the square roots in two sets of parentheses. The middle sign of one set of parentheses is “+” and the second one is “-”.
The first term in each pair of parentheses is \sqrt{9x^{2}}=3x
The second term in each pair of parentheses is \sqrt{100}=10
(3x+10)(3x-10) is the factored expression.
Factor the expression fully 363-27y^{4} .
(11-3y)(11+3y)
\left(121+y^2\right)\left(1-9y^2\right)
\left(11+9 y^2\right)\left(11-9 y^2\right)
3\left(11+3 y^2\right)\left(11-3 y^2\right)
The expression 363-27y^{4} is the difference of two squares. To factor it, first take out a common factor of 3 from the two terms in the expression, then take the square root of both the first and the second expressions and place the square roots in two sets of parentheses. The middle sign of one set of parentheses is “+” and the second one is “-”.
Take out the common factor of 3\text{:}
3\left(121-9y^{4}\right)
The first term in each pair of parentheses is \sqrt{121}=11
The second term in each pair of parentheses is \sqrt{9y^{4}}=3y^{2}
3(11+3y^{2})(11-3y^{2}) is the factored expression.
Factor the expression 81x^{2}-16y^{2} fully.
(81x+y)(x-4y)
(x+4y)(81x-y)
(9x+4y)(9x-4y)
(9x-4y)(9x-4y)
The expression 81x^{2}-16y^{2} is the difference of two squares. To factor it, take the square root of both the first and the second expressions and place the square roots in two sets of parentheses. The middle sign of one set of parentheses is “+” and the second one is “-”.
The first term in each pair of parentheses is \sqrt{81x^{2}}=9x
The second term in each pair of parentheses is \sqrt{16y^{2}}=4y
(9x+4y)(9x-4y) is the factored expression.
Factoring the difference of two squares FAQs
What factoring (factorising) strategies can you use to factor polynomials?
When factoring polynomials, you can apply the same strategies used for factoring trinomials, like quadratics. Many times when factoring polynomials, you will have to apply more than one factoring strategy.
Can you factor the sum of squares?
Binomials that are the sum of squares cannot be factored. They are considered to be prime expressions. For example, if asked to factor x^2+y^2, you would leave that binomial as the final answer.
Can fractions be in a perfect square binomial expression?
Yes, some fractions are considered perfect squares, for example, \cfrac{1}{4} \, x^2-1 is the difference of two perfect squares, which can be factored.
The next lessons are
Quadratic graphs
Quadratic equations
Radicals
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Introduction
What is the factoring the difference of two squares?
Common Core State Standards
How to factor the difference of two squares
Factor the difference of two squares examples
↓
Example 1: x² variable has a coefficient of 1 Example 2: y² variable has a coefficient of 1 Example 3: x² variable has a coefficient greater than 1 Example 4: coefficient of both terms is greater than 1 Example 5: exponents greater than 2 Example 6: exponents greater than 2
Teaching tips for factoring the difference of squares
Easy mistakes to make
Related factoring Lessons
Practice factoring the difference of two squares questions
Factoring the difference of two squares FAQs
Next lessons
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3789 | https://courses.lumenlearning.com/uvu-combinedalgebra/chapter/4-2-transformations-of-the-quadratic-function-latexfxx2-latex/ | Chapter 4: Quadratic Functions
4.2: Transformations of the Quadratic Function
Learning Outcomes
For the quadratic function [latex]f(x)=x^2[/latex],
Perform vertical and horizontal shifts
Perform vertical compressions and stretches
Perform reflections across the [latex]x[/latex]-axis
Explain the vertex form of a quadratic function
Write the equation of a transformed quadratic function using the vertex form
Determine the equation of a quadratic function given its vertex and a point on the graph.
Vertical Shifts
If we shift the graph of the function [latex]f(x)=x^2[/latex] up 5 units, all of the points on the graph increase their [latex]y[/latex]-coordinates by 5, but their [latex]x[/latex]-coordinates remain the same. Therefore, the equation of the function [latex]f(x)=x^2[/latex] after it has been shifted up 5 units transforms to [latex]f(x)=x^2+5[/latex]. Table 1 shows the changes to specific values of this function, which are replicated on the graph in figure 1.
| | | | |
--- --- |
| [latex]x[/latex] | [latex]x^2[/latex] | [latex]x^2+5[/latex] | Figure 1. Shifting the graph of [latex]f(x)=x^2[/latex] up 5 units. Every point latex[/latex] moves to (latex[/latex]. |
| -3 | 9 | 14 |
| -2 | 4 | 9 |
| -1 | 1 | 6 |
| 0 | 0 | 5 |
| 1 | 1 | 6 |
| 2 | 4 | 9 |
| 3 | 9 | 14 |
| Table 1. [latex]f(x)=x^2[/latex] is transformed to [latex]f(x)=x^2+5[/latex] |
If we shift the graph of the function [latex]f(x)=x^2[/latex] down 3 units, all of the points on the graph decrease their [latex]y[/latex]-coordinates by 3, but their [latex]x[/latex]-coordinates remain the same. Therefore, the equation of the function [latex]f(x)=x^2[/latex] after it has been shifted down 3 units transforms to [latex]f(x)=x^2-3[/latex]. Table 2 shows the changes to specific values of this function, which are replicated on the graph in figure 2.
| | | | |
--- --- |
| [latex]x[/latex] | [latex]x^2[/latex] | [latex]x^2-3[/latex] | Figure 2. Shifting the graph of [latex]f(x)=x^2[/latex] down 3 units. Every point latex[/latex] moves to latex[/latex]. |
| -3 | 9 | 6 |
| -2 | 4 | 1 |
| -1 | 1 | -2 |
| 0 | 0 | -3 |
| 1 | 1 | -2 |
| 2 | 4 | 1 |
| 3 | 9 | 6 |
| Table 2. [latex]f(x)=x^2[/latex] is transformed to [latex]f(x)=x^2-3[/latex] |
Vertical shifts
We can represent a vertical shift of the graph of [latex]f(x)=x^2[/latex] by adding or subtracting a constant, [latex]k[/latex], to the function:
[latex]f(x)=x^2 + k[/latex]
If [latex]k>0[/latex], the graph shifts upwards and if [latex]k<0[/latex] the graph shifts downwards.
Change the value of [latex]k[/latex] in the interactive desmos graph in figure 3 by moving the vertex up or down.
Figure 3. Interactive transformation from [latex]f(x)=x^2[/latex] to [latex]f(x)=x^2+k[/latex]
Every point latex[/latex] moves to latex[/latex].
Example 1
If the function [latex]f(x)=x^2[/latex] is shifted down 7 units, what is the equation of the transformed function?
Solution
When a function is shifted down by 7 units, all of the function values are decreased by 7.
[latex]f(x)=x^2[/latex] is transformed to [latex]f(x)=x^2-7[/latex]
The equation for the graph of [latex]f(x)=x^2[/latex] that has been shifted up 4 units is
[latex]f(x)=x^2+4[/latex]
The equation for the graph of [latex]f(x)=x^2[/latex] that has been shifted down 4 units is
[latex]f(x)=x^2-4[/latex]
Try It 1
If the function [latex]f(x)=x^2[/latex] is shifted up 4 units, what is the equation of the transformed function?
Show Answer
[latex]f(x)=x^2+4[/latex]
Horizontal Shifts
If we shift the graph of the function [latex]f(x)=x^2[/latex] right 4 units, all of the points on the graph increase their [latex]x[/latex]-coordinates by 4, but their [latex]y[/latex]-coordinates remain the same. The vertex (0, 0) in the original graph is moved to (4, 0) (figure 3). Any point latex[/latex] on the original graph is moved to latex[/latex].
But what happens to the original function [latex]f(x)=x^2[/latex]? An automatic assumption may be that since [latex]x[/latex] moves to [latex]x+4[/latex] that the function will become [latex]f(x)=(x+4)^2[/latex]. But that is NOT the case. Remember that the new vertex is (4, 0) and if we substitute [latex]x=4[/latex] into the function [latex]f(x)=(x+4)^2[/latex] we get [latex]f(4)=(4+4)^2=64\ne0[/latex]!! The way to get a function value of zero is for the transformed function to be [latex]f(x)=(x-4)^2[/latex]. Then [latex]f(4)=(4-4)^2=0[/latex]. So the function [latex]f(x)=x^2[/latex] transforms to [latex]f(x)=(x-4)^2[/latex] after being shifted 4 units to the right. The reason is that when we move the function 4 units to the right, the [latex]x[/latex]-value increases by 4 and to keep the corresponding [latex]y[/latex]-coordinate the same in the transformed function, the [latex]x[/latex]-coordinate of the transformed function needs to subtract 4 to get back to the original [latex]x[/latex] that is associated with the original [latex]y[/latex]-value. Table 3 shows the changes to specific values of this function, and the graph is shown in figure 4.
| | | | |
--- --- |
| [latex]x[/latex] | [latex]x-4[/latex] | latex^2[/latex] | Figure 4. Shift the graph right 4 units. |
| 1 | -3 | 9 |
| 2 | -2 | 4 |
| 3 | -1 | 1 |
| 4 | 0 | 0 |
| 5 | 1 | 1 |
| 6 | 2 | 4 |
| 7 | 3 | 9 |
| Table 3. Shifting the graph right by 4 units transforms [latex]f(x)=x^2[/latex] into [latex]f(x)=(x-4)^2[/latex] |
On the other hand, if we shift the graph of the function [latex]f(x)=x^2[/latex] left by 7 units, all of the points on the graph decrease their [latex]x[/latex]-coordinates by 7, but their [latex]y[/latex]-coordinates remain the same. So any point latex[/latex] on the original graph moves to latex=(x+7)^2[/latex]. Table 4 shows the changes to specific values of this function, and the graph is shown in figure 5.
| | | | |
--- --- |
| [latex]x[/latex] | [latex]x+7[/latex] | latex^2[/latex] | Figure 5. Shifting [latex]f(x)=x^2[/latex] left by 7 units to get [latex]f(x)=(x+7)^2[/latex] |
| -10 | -3 | 9 |
| -9 | -2 | 4 |
| -8 | -1 | 1 |
| -7 | 0 | 0 |
| -6 | 1 | 1 |
| -5 | 2 | 4 |
| -4 | 3 | 9 |
| Table 4. Shifting the graph left by 7 units transforms [latex]f(x)=x^2[/latex] into [latex]f(x)=(x+7)^2[/latex] |
horizontal shifts
We can represent a horizontal shift of the graph of [latex]f(x)=x^2[/latex] by adding or subtracting a constant, [latex]h[/latex], to the variable [latex]x[/latex], before squaring.
[latex]f(x)=(x-h)^2[/latex]
If [latex]h>0[/latex] the graph shifts toward the right and if [latex]h<0[/latex] the graph shifts to the left.
Change the value of [latex]h[/latex] in the interactive desmos graph in figure 6 by moving the vertex left or right.
Figure 6. Interactive transformation from [latex]f(x)=x^2[/latex] to [latex]f(x)=(x-h)^2[/latex]
tip for success
Remember that the negative sign inside the argument of the vertex form of a parabola (in the parentheses with the variable [latex]x[/latex] ) is part of the formula [latex]f(x)=(x-h)^2 +k[/latex].
If [latex]h>0[/latex], we have [latex]f(x)=(x-h)^2+k=(x-\text{positive number})^2 +k[/latex]. You’ll see the negative sign, but the graph will shift right.
If [latex]h<0[/latex], we have [latex]f(x)=(x-h)^2+k=(x-\text{negative number})^2 +k \rightarrow f(x)=(x+\text{positive number})^2+k[/latex], since subtracting a negative number is equivalent to adding a positive number. You’ll see the positive sign, but the graph will shift left.
Example 2
If the function [latex]f(x)=x^2[/latex] is shifted left by 5 units, what is the equation of the transformed function?
Solution
When a function is shifted left by 5 units, all of the [latex]x[/latex]-values are decreased by 5. So we have to add 5 to the [latex]x[/latex]-values in the transformed function to keep the correct function values.
[latex]f(x)=x^2[/latex] is transformed to [latex]f(x)=(x+5)^2[/latex]
The equation for the graph of [latex]f(x)=x^2[/latex] that has been shifted right 2 units is
[latex]f(x)=(x-2)^2[/latex]
The equation for the graph of [latex]f(x)=^2[/latex] that has been shifted left 2 units is
[latex]f(x)=(x+2)^2[/latex]
Try It 2
If the function [latex]f(x)=x^2[/latex] is shifted right by 8 units, what is the equation of the transformed function?
Show Answer
[latex]f(x)=(x-8)^2[/latex]
Stretching Up and Compressing Down
If we vertically stretch the graph of the function [latex]f(x)=x^2[/latex] by a factor of 2, all of the [latex]y[/latex]-coordinates of the points on the graph are multiplied by 2, but their [latex]x[/latex]-coordinates remain the same. Since the vertex of [latex]f(x)=x^2[/latex] is (0, 0), it transforms to (0, 0). That is, [latex]2\times0^2=2\times0=0[/latex]. The vertex, since it is an [latex]x[/latex]-intercept, stays at the same location (0, 0). In other words, the vertex is the anchor point in the stretching process. The equation of the function after the graph is stretched by a factor of 2 is [latex]f(x)=2x^2[/latex]. The reason for multiplying [latex]x^2[/latex] by 2 is that each [latex]y[/latex]-coordinate is doubled, and since[latex]y=x^2[/latex], [latex]x^2[/latex] is doubled. Table 5 shows this change and the graph is shown in figure 7.
| | | | |
--- --- |
| [latex]x[/latex] | [latex]x^2[/latex] | [latex]2x^2[/latex] | Figure 7. [latex]f(x)=x^2[/latex] transformed to [latex]f(x)=2x^2[/latex]. Each [latex]y[/latex]-value is doubled. |
| -3 | 9 | 18 |
| -2 | 4 | 8 |
| -1 | 1 | 2 |
| 0 | 0 | 0 |
| 1 | 1 | 2 |
| 2 | 4 | 8 |
| 3 | 9 | 18 |
| Table 5. Stretching the graph vertically by a factor of 2 transforms [latex]f(x)=x^2[/latex] into [latex]f(x)=2x^2[/latex] |
On the other hand, if our factor is [latex]\dfrac{1}{2}[/latex] and we vertically compress the graph of the function [latex]f(x)=x^2[/latex] all of the [latex]y[/latex]-coordinates of the points on the graph are halved, but their [latex]x[/latex]-coordinates remain the same. This means the [latex]y[/latex]-coordinates are multiplied by [latex]\dfrac{1}{2}[/latex], or divided by 2. The [latex]x[/latex]-intercept (0, 0) of [latex]f(x)=x^2[/latex] is the only point that doesn’t move. In other words, the [latex]x[/latex]-intercept is the anchor point in the compressing process. The equation of the function after being compressed is [latex]f(x)=\dfrac{1}{2}x^2[/latex]. The reason for multiplying [latex]x^2[/latex] by [latex]\dfrac{1}{2}[/latex] is that each [latex]y[/latex]-coordinate becomes half of the original value when it is divided by 2. Table 6 shows this change and the graph is shown in figure 8.
| | | | |
--- --- |
| [latex]x[/latex] | [latex]x^2[/latex] | [latex]\frac{1}{2}x^2[/latex] | Figure 8. Compressing the graph when the factor is [latex]\frac{1}{2}[/latex]. Each [latex]y[/latex]-value is divided by 2. |
| -3 | 9 | 4.5 |
| -2 | 4 | 2 |
| -1 | 1 | 0.5 |
| 0 | 0 | 0 |
| 1 | 1 | 0.5 |
| 2 | 4 | 2 |
| 3 | 9 | 4.5 |
| Table 6. When the factor is [latex]\dfrac{1}{2}[/latex] the graph is compressed and transforms [latex]f(x)=x^2[/latex] into [latex]f(x)=\dfrac{1}{2}x^2[/latex] |
Change the value of [latex]a[/latex] in the interactive desmos graph in figure 9 by moving the point latex[/latex] vertically.
Figure 9. Interactive transformation from [latex]f(x)=x^2[/latex] to [latex]f(x)=ax^2[/latex]
vertical stretching and compressing
A stretch or compression of the graph of [latex]f(x)=x^2[/latex] can be represented by multiplying the [latex]x^2[/latex] by a constant, [latex]a>0[/latex].
[latex]f(x)=ax^2[/latex]
The magnitude of [latex]a[/latex] indicates the stretch/compression of the graph. If [latex]a>1[/latex], the graph is stretched vertically by a factor of [latex]a.[/latex] If [latex]0### Example 3
If the function [latex]f(x)=x^2[/latex] is compressed to one-quarter of its original height, what is the equation of the transformed function? What happens to the points (0, 0) and (1, 1) on the graph of the original function?
Solution
When a function is compressed to one-quarter of its original height, [latex]a=\dfrac{1}{4}[/latex] in the transformed equation [latex]f(x)=ax^2[/latex]. So [latex]f(x)=x^2[/latex] is transformed to [latex]f(x)=\dfrac{1}{4}x^2[/latex].
The point (0, 0) stays the same and the point (1, 1) moves to [latex]\left (1,\dfrac{1}{4}\right )[/latex] since the [latex]y[/latex]-values are divided by 4.
Try It 3
If the function [latex]f(x)=x^2[/latex] is stretched by a factor of 9, what is the equation of the transformed function? What happens to the points (0, 0) and (1, 1) on the graph of the original function?
Show Answer
[latex]f(x)=x^2[/latex] is transformed to [latex]f(x)=9x^2[/latex].
(0, 0) stays the same.
(1, 1) moves to (1, 9).
Reflection across the [latex]x[/latex]-axis
When the graph of the function [latex]f(x)=x^2[/latex] is reflected across the [latex]x[/latex]-axis, the [latex]y[/latex]-coordinates of all of the points on the graph change their signs, from positive to negative values or from negative to positive values, while the [latex]x[/latex]-coordinates remain the same. The equation of the function after [latex]f(x)=x^2[/latex] is reflected across the [latex]x[/latex]-axis is [latex]f(x)=-x^2[/latex]. The graph changes from opening upwards to opening downwards. Table 7 shows the effect of such a reflection on the functions values and the graph is shown in figure 10.
| | | | |
--- --- |
| [latex]x[/latex] | [latex]x^2[/latex] | [latex]-x^2[/latex] | Figure 10. Reflection across [latex]x[/latex]=axis. latex[/latex] transforms to latex[/latex]. |
| -3 | 9 | -9 |
| -2 | 4 | -4 |
| -1 | 1 | -1 |
| 0 | 0 | 0 |
| 1 | 1 | -1 |
| 2 | 4 | -4 |
| 3 | 9 | -9 |
| Table 7. Reflecting the graph of [latex]f(x)=x^2[/latex] across the [latex]x[/latex]-axis transforms [latex]f(x)=x^2[/latex] into [latex]f(x)=-x^2[/latex] |
Vertex Form of a Quadratic Function
All of the transformations we have applied to the function [latex]f(x)=x^2[/latex] can be combined. The result is the function [latex]f(x)=a(x-h)^2+k[/latex], where [latex]a[/latex] represents vertical stretching ([latex]a>1[/latex]) and compression ([latex]00[/latex] and left when [latex]h<0[/latex]; and [latex]k[/latex] represents a vertical shift up when [latex]k>0[/latex] and down when [latex]k<0[/latex]. In addition, the vertex will move from (0, 0) to latex[/latex].
VERTEX form of a quadratic equation
The vertex form of a quadratic functionis
[latex]f\left(x\right)=a{\left(x-h\right)}^{2}+k[/latex]
where [latex]\left(h, k\right)[/latex] is the vertex, and [latex]a[/latex] represents stretching ([latex]a>1[/latex]), compression ([latex]0If we were to multiply and simplify the vertex form of a quadratic function, the quadratic function would turn into standard form: [latex]f(x)=ax^2+bx+c[/latex], where [latex]a, b, c[/latex] are real numbers. the value of [latex]a[/latex] in standard form is the same value of [latex]a[/latex] in vertex form.
Move the red dots in figure 11 to change the values of [latex]a, h[/latex], and [latex]k[/latex], and watch the transformation unfold.
Figure 11. Interactive transformation from [latex]f(x)=x^2[/latex] to [latex]f(x)=a(x-h)^2+k[/latex]
Example 4
The function [latex]f(x)=x^2[/latex] is transformed by shifting the graph to the right by 4 units then down by 3 units. Write the vertex form of the function that represents this transformation.
Solution
Moving the graph right by 4 units means that [latex]h=4[/latex].
Moving the graph down by 3 units means tha [latex]k=-3[/latex]
Vertex form is [latex]f(x)=a(x-h)^2+k[/latex] so the transformed function is [latex]f(x)=(x-4)^2-3[/latex].
Example 5
The function [latex]f(x)=x^2[/latex] is transformed by reflecting the graph across the [latex]x[/latex]-axis, stretching the graph by a factor of 4, shifting the graph to the left by 1 unit then up by 7 units. Write the vertex form of the function that represents this transformation.
Solution
Reflecting the graph across the [latex]x[/latex]-axis and stretching the graph by a factor of 4 means that [latex]a=-4[/latex].
Moving the graph left by 1 unit means that [latex]h=-1[/latex].
Moving the graph up by 7 units means that [latex]k=7[/latex]
Vertex form is [latex]f(x)=a(x-h)^2+k[/latex] so the transformed function is [latex]f(x)=-4(x+1)^2+7[/latex].
Try It 4
The function [latex]f(x)=x^2[/latex] is transformed by reflecting the graph across the [latex]x[/latex]-axis and shifting the graph to the left by 3 units. Write the vertex form of the function that represents this transformation.
Show Answer
[latex]f(x)=-(x+3)^2[/latex]
Try It 5
The function [latex]f(x)=x^2[/latex] is transformed by compressing the graph to one-half of its original height, shifting the graph to the right by 5 units then down 4 units. Write the vertex form of the function that represents this transformation.
Show Answer
[latex]f(x)=\dfrac{1}{2}(x-5)^2-4[/latex]
Determining the Equation of a Quadratic Function
The vertex form of a quadratic function is [latex]f(x)=a(x-h)^2+k[/latex], where [latex]a, h, k[/latex] are real numbers and [latex]a\ne 0[/latex]. So, if we know the values of [latex]a, h, k[/latex], we can write the corresponding function. For example, if we are told that [latex]h=2,\;k=5,\;a=3[/latex]. Then the function is [latex]f(x)=3(x-2)^2+5[/latex]. Figure 12 shows the graph.
Figure 12. Graph of [latex]f(x)=3(x-2)^2+5[/latex]
What if we were told instead that the vertex is (2, 5) and the graph passes through the point (1, 8)? Can we find the equation of the function knowing this information?
First off, we know that by definition the vertex gives us the values of [latex]h=2[/latex] and [latex]k=5[/latex]. This means that the function is [latex]f(x)=a(x-2)^2+5[/latex]. Now all we need is the value of [latex]a[/latex].Since the point (1, 8) lies on the graph, it must be a solution of the equation that represents the function, [latex]y=a(x-2)^2+5[/latex]. Substituting [latex]x=1,\;y=8[/latex] leaves [latex]a[/latex] as the only unknown so we can solve for [latex]a[/latex]:
Now we know [latex]h=2,\;k=5,\;a=3[/latex] so [latex]f(x)=3(x-2)^2+5[/latex].
Example 6
Determine the function whose graph has a vertex at (–4, 1) and that passes through the point (0, 2).
Solution
Since we know the vertex, we know [latex]h=-4,\;k=1[/latex].
Therefore, [latex]f(x)=a(x+4)^2+1[/latex].
To find [latex]a[/latex], we substitute the point (0, 2) for latex[/latex]:
Consequently, [latex]f(x)=\dfrac{1}{16}\left (x+4\right )^2+1[/latex].
Try It 6
Determine the function whose graph has a vertex at (6, –2) and that passes through the point (1, –5).
Show Answer
Candela Citations
CC licensed content, Original
Transformations of the Quadratic Function . Authored by: Leo Chang and Hazel McKenna. Provided by: Utah Valley University. License: CC BY: Attribution
Figure 3. Interactive transformation from [latex]f(x)=x^2[/latex] to [latex]f(x)=x^2+b[/latex]. Authored by: Hazel McKenna. Provided by: Utah Valley University. Located at: License: CC BY: Attribution
Figure 6. Interactive transformation from [latex]f(x)=x^2[/latex] to [latex]f(x)=(x-h)^2[/latex]. Authored by: Hazel McKenna. Provided by: Utah Valley University. Located at: License: CC BY: Attribution
Figure 9. Interactive transformation from [latex]f(x)=x^2[/latex] to [latex]f(x)=ax^2[/latex]. Authored by: Hazel McKenna. Provided by: Utah Valley University. Located at: License: CC BY: Attribution
Figure 11. Interactive transformation from [latex]f(x)=x^2[/latex] to [latex]f(x)=a(x-h)^2+k[/latex]. Authored by: Hazel McKenna. Provided by: Utah Valley University. Located at: License: CC BY: Attribution
Determining the Equation of a Quadratic Function. Authored by: Hazel McKenna. Provided by: Utah Valley University. License: CC BY: Attribution
Adaptation and Revision. Authored by: Leo Chang and Hazel McKenna. Provided by: Utah Valley University. License: CC BY: Attribution
Examples 1 - 6 ; Try It: hjm387, hjm484, hjm713, hjm212, hjm992, hjm383. Authored by: Hazel McKenna. Provided by: Utah Valley University. License: CC BY: Attribution
All graphs created using desmos graphing calculator. Authored by: Hazel McKenna and Leo Chang. Provided by: Utah Valley University. License: CC BY: Attribution
CC licensed content, Shared previously
College Algebra. Authored by: Abramson, Jay et al.. Provided by: OpenStax. Located at: License: CC BY: Attribution. License Terms: Download for free at
Licenses and Attributions
CC licensed content, Original
Transformations of the Quadratic Function . Authored by: Leo Chang and Hazel McKenna. Provided by: Utah Valley University. License: CC BY: Attribution
Figure 3. Interactive transformation from [latex]f(x)=x^2[/latex] to [latex]f(x)=x^2+b[/latex]. Authored by: Hazel McKenna. Provided by: Utah Valley University. Located at: License: CC BY: Attribution
Figure 6. Interactive transformation from [latex]f(x)=x^2[/latex] to [latex]f(x)=(x-h)^2[/latex]. Authored by: Hazel McKenna. Provided by: Utah Valley University. Located at: License: CC BY: Attribution
Figure 9. Interactive transformation from [latex]f(x)=x^2[/latex] to [latex]f(x)=ax^2[/latex]. Authored by: Hazel McKenna. Provided by: Utah Valley University. Located at: License: CC BY: Attribution
Figure 11. Interactive transformation from [latex]f(x)=x^2[/latex] to [latex]f(x)=a(x-h)^2+k[/latex]. Authored by: Hazel McKenna. Provided by: Utah Valley University. Located at: License: CC BY: Attribution
Determining the Equation of a Quadratic Function. Authored by: Hazel McKenna. Provided by: Utah Valley University. License: CC BY: Attribution
Adaptation and Revision. Authored by: Leo Chang and Hazel McKenna. Provided by: Utah Valley University. License: CC BY: Attribution
Examples 1 - 6 ; Try It: hjm387, hjm484, hjm713, hjm212, hjm992, hjm383. Authored by: Hazel McKenna. Provided by: Utah Valley University. License: CC BY: Attribution
All graphs created using desmos graphing calculator. Authored by: Hazel McKenna and Leo Chang. Provided by: Utah Valley University. License: CC BY: Attribution
CC licensed content, Shared previously
College Algebra. Authored by: Abramson, Jay et al.. Provided by: OpenStax. Located at: License: CC BY: Attribution. License Terms: Download for free at
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3790 | https://math.stackexchange.com/questions/1107474/when-does-a-triangles-circumcenter-lie-on-its-incircle | Skip to main content
When does a triangle's circumcenter lie on its incircle?
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When does a triangle's circumcenter to lie on its incircle? The isosceles right triangle is an example of such a triangle ...
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edited Feb 2, 2015 at 4:42
Zubin Mukerjee
asked Jan 17, 2015 at 4:32
Zubin MukerjeeZubin Mukerjee
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Yes: Consider the isosceles right triangle. The circumcenter is the midpoint of the hypotenuse; that's also the point where the incircle touches that side.
– Blue
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@Zubin Mukerjee you can add a condition that the triangle is not isosceles then question may become interesting.
– Satvik Mashkaria
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Yes: Consider the isosceles right triangle. The circumcenter is the midpoint of the hypotenuse; that's also the point where the incircle touches that side.
Moreover, consider making the vertex angle of that triangle smaller and smaller (while remaining isosceles). The circumcenter moves to the interior of the incircle until ---when the triangle becomes equilateral--- it coincides with the incenter. For a sufficiently-small vertex angle, the circumcenter will be diametrically opposite the incircle's point of tangency with the base.
More generally, we can reference the formula for the distance from circumcenter (K) to incenter (J):
|JK¯¯¯¯¯¯¯¯|2=R(R−2r)
where R and r are, respectively, the circumradius and inradius. For K to lie on the incircle, |JK¯¯¯¯¯¯¯¯|=r, so that
R2−2Rr=r2(⋆)
Given that
r=4RsinA2sinB2sinC2
(⋆) is equivalent to
1−8sinA2sinB2sinC2−16sin2A2sin2B2sin2C2=0
which becomes
sinA2sinB2sinC2=14(2–√−1)(⋆⋆)
(We discard the solution −(2–√+1)/4, since all the sines are necessarily positive.)
We can also write (⋆⋆) as
cosA+cosB+cosC=2–√
(whereby the isosceles right triangle solution is evident); the Law of Cosines, and a bit of clean-up, then provides this form of the relation:
(−a+b+c)(a−b+c)(a+b−c)=2abc(2–√−1)
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edited Jan 17, 2015 at 6:27
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answered Jan 17, 2015 at 4:43
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Incidentally, the "sufficiently-small" vertex angle I mentioned is about 34.0625∘.
– Blue
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I added a picture to your answer. Hope you don't mind.
– user856
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I thought that vertex angle was ≈34°03′45′′ :-). The apparent 1/16 in the fractional part is a misdirection; it is not a rational-degree angle.
– Oscar Lanzi
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Try to prove that the ratio between tha greatest and smaller side of triangle is less than 2
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answered Feb 24, 2017 at 8:31
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Can you further elaborate
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3791 | https://codanics.com/population-mean-vs-sample-mean/ | Published Time: 2023-11-25T16:50:42+00:00
Population Mean vs. Sample Mean - Codanics
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Population Mean vs. Sample Mean
November 25, 2023 November 26, 2023Dr. Aammar Tufail
Population Mean vs. Sample Mean: Navigating the World of Averages 🌐📊
Welcome to the Intriguing World of Means!
In the realm of statistics and data science, two heroes emerge when we talk about averages: the Population Mean and the Sample Mean. While they might sound similar, they play distinct roles in how we interpret data. Understanding the difference between these two is not just a statistical necessity; it’s the key to accurate data interpretation and analysis. Let’s explore these concepts in detail. 🚀
Population Mean: The Big Picture 🌍
The Population Mean (μ) is like an aerial view of a landscape, providing a comprehensive picture. It represents the average of all values in a population – the entire group you’re interested in studying.
Definition: The Population Mean is the sum of all the values in the population divided by the number of values in that population.
Equation: [ \mu = \frac{\sum_{i=1}^{N} X_i}{N} ]
Here, ( \mu ) is the population mean, ( X_i ) represents each value in the population, and ( N ) is the total number of values.
Use Case: Ideal for smaller, well-defined populations or when you have access to complete data.
Example: The average age of all the students in a small school.
Sample Mean: The Practical Approach 📈
In contrast, the Sample Mean ( ( \bar{x} ) ) is like zooming in on a specific part of the landscape. It deals with a subset of the population, providing an estimate of the population mean.
Definition: The Sample Mean is the average of all values in a sample, which is a representative segment of the population.
Equation:[ \bar{x} = \frac{\sum_{i=1}^{n} x_i}{n} ]
Here, ( \bar{x} ) is the population mean, ( x_i ) represents each value in the population, and ( n ) is the total number of values.
Use Case: Used when it’s impractical or impossible to examine the entire population.
Example: The average age of 50 randomly selected students from the same school.
Key Differences: Understanding the Nuances 🚧
Scope and Accessibility: The Population Mean encompasses the whole group, while the Sample Mean focuses on a part of that group.
Purpose and Application: The Population Mean provides the true average, while the Sample Mean is a practical estimation, often used in inferential statistics to draw conclusions about the population.
Representation: They are represented differently in equations, with μ for the population mean and ( \bar{x} ) for the sample mean.
Why Both are Important? 🌟
Feasibility and Practicality: In many real-world scenarios, especially with large populations, it’s only feasible to calculate the sample mean.
Inference and Prediction: The sample mean allows statisticians and data scientists to make predictions about the population mean, which is especially useful in surveys and experiments.
Conclusion: Balancing the Big Picture with Practical Insights 🎯
Understanding the difference between the population mean and the sample mean is crucial in statistics. It’s about balancing the comprehensive view provided by the population mean with the practical insights offered by the sample mean. Whether you are a researcher, a data scientist, or a student, recognizing when to use each will enhance the accuracy and reliability of your data analysis.
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7 Comments.
Muhammad_Faizansays: June 5, 2024 at 1:57 pm I learned Population Mean and Sample Mean. When to use them and what are their results. Reply
saima Shahzadisays: December 20, 2023 at 4:32 pm done Reply
Shahid Umarsays: December 3, 2023 at 10:25 pm In the context of mean, I learned the clear picture of mean in two types mentioned in this blog (i,e: population mean and sampling mean) Reply
Javed Alisays: November 28, 2023 at 4:36 pm AOA, This blog on “Population Mean vs. Sample Mean” provides a comprehensive explanation of the differences between these two concepts. It highlights the importance of understanding when to use each means and how they play distinct roles in data interpretation and analysis. This blog effectively presents the definitions, equations, and key differences between the population mean and the sample mean. It is an informative and well-written blog that enhances knowledge in the field of statistics. ALLAH PAK ap ko dono jahan ki bhalian aata kry AAMEEN. Reply
Nimra Ishaqsays: November 26, 2023 at 9:38 pm Fantastic blog. Reply
Adeel Hamidsays: November 26, 2023 at 8:33 pm Its very informative Reply
Ansa Anjumsays: November 26, 2023 at 7:54 pm It was interesting to know thier difference and thier different formulas before reading this blog i was unaware of this difference …Thanks alot giving us great knowledge Reply
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Finding an integrating factor (Differential Equations, exact equations)
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$\begingroup$
The given formula is:
$4\left(\frac{x^3}{y^2} + \frac{3}{y} \right)dx + 3 \left(\frac{x}{y^2} + 4y\right)dy = 0$
This gives $M_y = - \frac{8x^3}{y^3} - \frac{12}{y^2}$ and $N_x = \frac{3}{y^2}$. So clearly as given this equation isn't exact.
The problem text states that I am to "Find an integrating factor and solve the given equation." However, I cannot find μ using any technique that I know or that I could come up with.
The only ways that the book up to this point has discussed coming up with integrating factors are in the form of $\frac{d\mu}{dx} = \frac{M_y - N_x}{N}\mu$ and $\frac{d\mu}{dy} = \frac{N_x - M_y}{M}\mu$ and the one oddball one where μ is a function of xy: $\frac{d\mu}{dxy} = \frac{N_x - M_y}{xM - yN}\mu$. Working these and also deriving and working μ(y/x) and μ(x/y).
I tried the final three because on the next problem there's a hint that pulls back to a different problem that was to provide a proof of a case where μ is a function of xy. And as this is an even-numbered problem, it's trickier than the odd numbered ones.
In the interest of showing my work. Here are the cases:
$\frac{d\mu}{dx} = \frac{- \frac{8x^3}{y^3} - \frac{15}{y^2}}{\frac{3x}{y^2} + 12y}\mu = Not\, promising...$
$\frac{d\mu}{dy} = \frac{\frac{8x^3}{y^3} + \frac{15}{y^2}}{\frac{4x^3}{y^2} + \frac{12}{y}}\mu = Not\, promising...$
$\frac{d\mu}{dxy} = \frac{- \frac{8x^3}{y^3} - \frac{15}{y^2}}{\frac{4x^4}{y^2} + \frac{15x}{y} + 12y^2}\mu = Really\,not\,promising...$
And for the division ones (are these right?):
$\frac{\partial\mu(\frac{x}{y})}{\partial x} = \frac{\frac{d\mu}{dx}}{y}$
$\frac{\partial\mu(\frac{x}{y})}{\partial y} = -\frac{x\frac{d\mu}{dx}}{y^2}$
And vica versa for the inverted case (are these right?):
$\frac{\partial\mu(\frac{y}{x})}{\partial x} = -\frac{y\frac{d\mu}{dx}}{x^2}$
$\frac{\partial\mu(\frac{y}{x})}{\partial y} = \frac{\frac{d\mu}{dx}}{x}$
And you can easily verify that nothing good comes of using these ideas.
So... I'm lost. I probably made a really basic mistake early on, that's causing the numbers to not line up correctly, but for the life of me I just cannot find it.
μ(y) case looks the closest to being good. But it's just not working out.
ordinary-differential-equations
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edited Sep 5, 2015 at 5:51
OmnipotentEntityOmnipotentEntity
asked Sep 5, 2015 at 5:47
OmnipotentEntityOmnipotentEntity
1,3121010 silver badges2727 bronze badges
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$\begingroup$ Please use LATEX for math expressions. $\endgroup$
Gummy bears
– Gummy bears
2015-09-05 05:48:54 +00:00
Commented Sep 5, 2015 at 5:48
1
$\begingroup$ Use '$$' signs at the starting and end of every math expression. $\endgroup$
Gummy bears
– Gummy bears
2015-09-05 05:51:00 +00:00
Commented Sep 5, 2015 at 5:51
1
$\begingroup$ Try multiplying the equation by $y^2$.. May help. $\endgroup$
Gummy bears
– Gummy bears
2015-09-05 06:02:00 +00:00
Commented Sep 5, 2015 at 6:02
1
$\begingroup$ That does indeed make the equations much nicer, but not seemingly any more solvable. :( $\endgroup$
OmnipotentEntity
– OmnipotentEntity
2015-09-05 06:06:16 +00:00
Commented Sep 5, 2015 at 6:06
1
$\begingroup$ Comment to the question (v2): Integrating factor should satisfy $(\frac{x^3}{3}+y)\frac{du}{dy}-(\frac{x}{4} +y^3) \frac{du}{dx}+\frac{3}{4}=0$... $\endgroup$
Qmechanic
– Qmechanic
2015-09-05 12:07:12 +00:00
Commented Sep 5, 2015 at 12:07
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Is the book you are referring to by any chance this one here? I've found that excerpt - after searching for longer than I care to admit ;) - and your description of the next problem (no. 31) referring back to another problem fits exactly!
Anyway, if you look closely at problem no. 30 (yours), you see that you have misplaced a parenthesis (already in your reddit post - yeah, I found that one, too) which makes this problem way harder than it really is.${}^{}$
The correct inexact differential is $$ \left(4\frac{x^3}{y^2}+\frac{3}{y}\right)\text{d}x + \left(3\frac{x}{y^2}+4y\right)\text{d}y $$ so you see that the factors of $3$ and $4$ should be inside the parentheses. Now if you multiply through by $\mu(y)=y^2$, you get $$ \left(4x^3+3y\right)\text{d}x + \left(3x+4y^3\right)\text{d}y $$ which gives you an exact differential, since $$ \partial_y\left(4x^3+3y\right)=3=\partial_x\left(3x+4y^3\right). $$
In the end, your intuition that the $\mu=\mu(y)$ looked the best was right, the factors were simply slightly off.
${}^{}$ To be fair, I have no idea how to find the integrating factor for the differential as it is posed in the current version (v2). While one immediately sees by comparison with the other problems in the linked PDF that this cannot be the intended procedure, it should be doable in general by using the method of characteristics to solve the semilinear partial differential equation for $\mu(x,y)$, but I've been running in circles on that path for a while now and am out of good ideas.
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answered Mar 26, 2017 at 12:00
Wojciech MorawiecWojciech Morawiec
15877 bronze badges
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$\begingroup$ Really digging up the old posts. :) Thanks for the effort and help. $\endgroup$
OmnipotentEntity
– OmnipotentEntity
2017-03-26 22:55:08 +00:00
Commented Mar 26, 2017 at 22:55
$\begingroup$ @OmnipotentEntity I was as surprised as you that I was able to help with such an old question. All I wanted was some insight into a similar problem I had with integrating factors, but as you can see by the footnote, I'm not quite there yet. Happy to help anyway! :) $\endgroup$
Wojciech Morawiec
– Wojciech Morawiec
2017-03-26 23:00:02 +00:00
Commented Mar 26, 2017 at 23:00
$\begingroup$ Obviously, the question is old enough that it's completely useless. But it is nice to know where the issue was. $\endgroup$
OmnipotentEntity
– OmnipotentEntity
2017-03-26 23:04:25 +00:00
Commented Mar 26, 2017 at 23:04
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3793 | https://www.quora.com/Why-is-IF5-a-liquid-while-IF7-is-a-gas | Why is IF5 a liquid while IF7 is a gas? - Quora
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Chemistry
Iodine Heptafluoride
Changes of States of Matt...
Intermolecular Bond
Inorganic Compounds
Gas (state of matter)
Properties of Matter
Iodine
Basic Physical Chemistry
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Why is IF5 a liquid while IF7 is a gas?
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James Semper
Chemistry Major · Author has 559 answers and 3.5M answer views
·8y
IF5 is slightly polar. Look at the structure on Google. It has a lone pair of electrons. IF7 does not have this lone pair and it is the same all around.
Polarity increases the attraction from one molecule to another and helps keep them as a liquid in this case.
So I believe your assumption is right. Though LD isn't the only thing at play since we have a slight dipole-dipole interaction.
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Why is gas so named when it is a liquid?
I take it you mean gasoline? Why is gasoline called “gas” when it’s really a liquid?
in 1862, John Cassell patented a petroleum distillate for use in petroleum lamps. He called it Cazeline. in 1864, the Germans spelled it Gazeline. About the same time it was being marketed in the U.S. as Gasoline. I suspect Gazeline and Gasoline were knockoff infringements on Cassell’s patents. Anyway, in 1876, Nicolaus Otto developed the “otto engine” that ran off Gazeline and became known as the first Gasoline engine
Gas (fuel) is an abbreviation for gasoline and has nothing to to with it’s state of matter.
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Richard Woods
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Why is gas a liquid?
Gas is not a liquid. The state of matter at ambient temperature are solid, liquid and gaseous.
Water is is a liquid at normal room temperature, a solid at below freezing and a gas above boiling.
If we stop thinking ‘heat’ and think thermal energy it gets easier. So all matter is made of atoms which are in motion. They relate to each other in molecules. Depending on their nature they will respond to the input of energy by becoming agitated (moving faster) and to a lack of energy by slowing down.
Depending on loads of other stuff they will become more dense at lower temperatures and less dense at h
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Gas is not a liquid. The state of matter at ambient temperature are solid, liquid and gaseous.
Water is is a liquid at normal room temperature, a solid at below freezing and a gas above boiling.
If we stop thinking ‘heat’ and think thermal energy it gets easier. So all matter is made of atoms which are in motion. They relate to each other in molecules. Depending on their nature they will respond to the input of energy by becoming agitated (moving faster) and to a lack of energy by slowing down.
Depending on loads of other stuff they will become more dense at lower temperatures and less dense at higher temperatures, depending on the molecules.
So heat is an input of energy and cooling is the withdrawal of energy. Water molecules become turgid and finally solid at freezing (it happens that in water this is super-sudden) and then become fluid and behave as liquid until they vapourise into a gas at boiling point.
And I recalled this from 4th year physics thanks to Stinks Brown but it drew me to find the following image and remind me of the proper words for the transformation - sublimate and desublimate so thank you.
The left end is 0 degrees Celsius and the right end 100 degrees Celsius. Water being a bit odd it is a liquid all the way between.
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Martin J Pitt
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How do you separate liquid from gas?
This is a common problem in chemical engineering (which is actually mostly about separation). Here are some examples:
When getting natural gas from underground to the refinery the pipeline usually contains liquid (from vapors which have condensed plus salt water). This collects into lumps of liquid called “slugs” which are pushed along the pipeline. Thus the first chemical engineering operation is to separate them from the gas. This is done by dividing the flow into several pipes, reducing the velocity, so that the liquid no longer fills the pipe and can be drawn off from the bottom. This is ca
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This is a common problem in chemical engineering (which is actually mostly about separation). Here are some examples:
When getting natural gas from underground to the refinery the pipeline usually contains liquid (from vapors which have condensed plus salt water). This collects into lumps of liquid called “slugs” which are pushed along the pipeline. Thus the first chemical engineering operation is to separate them from the gas. This is done by dividing the flow into several pipes, reducing the velocity, so that the liquid no longer fills the pipe and can be drawn off from the bottom. This is called a slug catcher.
The same basic process is used in many separators of other designs, i.e. to slow the flow of liquid and gas in a larger vessel, so that the liquid drops can fall by gravity.
We can also use a hydrocyclone where the liquid drops are thrown out to the wall and collect from the bottom of a cone. Here is a 1916 example!
We can use a sort of filter which works by droplets being trapped then joined together by a process called coalescence. The wet filter then traps more liquid and when the drops are large enough, they drop down to a collector. Here is a vapor (or gas) liquid separator. The larger vessel causes the big drops to fall down and collect in a pool. The gas/vapor going out the top goes through a metal mesh which acts as a coalescer,
To get gas out of liquid we can use similar principles, but often heat the liquid in order to remove some of the gas which is actually dissolved, as here to remove air from the water going into a boiler to make steam;
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The difference in physical states between IF₅ (iodine pentafluoride) and IF₇ (iodine heptafluoride) can be attributed to several factors, primarily molecular structure, intermolecular forces, and molecular weight.
Molecular Structure and Shape:
IF₅: Iodine pentafluoride has a square pyramidal molecular geometry. This shape allows for stronger dipole-dipole interactions due to the polar nature of the molecule. The presence of five fluorine atoms creates a significant dipole moment that leads to stronger intermolecular attractions.
IF₇: Iodine heptafluoride, on the other hand, has a pentagonal b
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The difference in physical states between IF₅ (iodine pentafluoride) and IF₇ (iodine heptafluoride) can be attributed to several factors, primarily molecular structure, intermolecular forces, and molecular weight.
Molecular Structure and Shape:
IF₅: Iodine pentafluoride has a square pyramidal molecular geometry. This shape allows for stronger dipole-dipole interactions due to the polar nature of the molecule. The presence of five fluorine atoms creates a significant dipole moment that leads to stronger intermolecular attractions.
IF₇: Iodine heptafluoride, on the other hand, has a pentagonal bipyramidal structure. While it is also polar, the larger number of fluorine atoms creates a more symmetrical distribution of charge, which reduces the overall dipole moment compared to IF₅.
Intermolecular Forces:
The stronger dipole-dipole interactions in IF₅ result in higher intermolecular forces, which contribute to its liquid state at room temperature.
In IF₇, the weaker dipole-dipole interactions and increased symmetry lead to lower intermolecular forces, allowing it to exist as a gas under similar conditions.
Molecular Weight:
IF₇ has a higher molecular weight than IF₅, but the effect of molecular weight is less significant than the differences in molecular structure and intermolecular forces. While higher molecular weight can lead to increased van der Waals forces, in this case, the strong dipole-dipole interactions in IF₅ dominate.
Conclusion:
In summary, IF₅ is a liquid due to its stronger dipole-dipole interactions stemming from its molecular shape and polarity, while IF₇ is a gas because its molecular structure leads to weaker intermolecular forces, despite its larger molecular weight.
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Antonello Frau
PhD in Chemistry&Materials Science and Engineering, University of Houston (Graduated 2009) · Author has 2.9K answers and 25.5M answer views
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Why is CCl₄ a liquid at room temperature but CH₄ a gas?
This is actually a very good question.
Here is carbon tetrachloride, CCl₄:
Now, here is methane, CH₄:
The smallest units both compounds are made of (CCl₄ and CH₄ molecules, that is) have the same molecular geometry - they are both tetrahedra. However, the first one is a liquid whereas the second one is a gas.
The difference in physical state between CCl₄ and CH₄ is due to the different atoms the central C atoms are bonded to: 4 chlorine atoms versus 4 hydrogen atoms. In other words, 4 Cl atoms have more electrons than 4 H atoms and this causes the intermolecular forces (Van der Waals force) among
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This is actually a very good question.
Here is carbon tetrachloride, CCl₄:
Now, here is methane, CH₄:
The smallest units both compounds are made of (CCl₄ and CH₄ molecules, that is) have the same molecular geometry - they are both tetrahedra. However, the first one is a liquid whereas the second one is a gas.
The difference in physical state between CCl₄ and CH₄ is due to the different atoms the central C atoms are bonded to: 4 chlorine atoms versus 4 hydrogen atoms. In other words, 4 Cl atoms have more electrons than 4 H atoms and this causes the intermolecular forces (Van der Waals force) among adjacent CCl₄ molecules to be more, statistics-wise.
Such weak forces that hold molecules closer and, eventually, determine the physical state of a compound in absence of stronger interactions (e.g. hydrogen bonding), scale up with the number of electrons because they arise from them. This is, for instance, the reason why elemental halogens (F₂, Cl₂, Br₂, and I₂) gradually change from gases (the first two ones) to a liquid (Br₂) and to solid (I₂).
Therefore, the more the van der Waals forces among molecules, the more energy is required to take them apart and bring a compound to, say, boil. This is why CCl₄ is a liquid at room temperature whilst CH₄ is a gas.
The reasoning about electronegativity differences (C-Cl bonds versus C-H bonds) would be plausible if the molecules were not symmetrical: in these examples, however, the molecular geometries of CCl₄ and CH₄ are such that neither has a nonzero dipole moment.
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James Semper
Chemistry Major · Author has 559 answers and 3.5M answer views
·10y
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Why is IF5 molecule polar?
The electronegativity of F is more than that of I so there are dipole moments on each of these bonds. There is one side of this molecule that has a lone pair of electrons. This lone pair pushes the bonds farther away. You can see that in this image because there isn't a perfect 90 degrees between the top bond and the bonds on the side. Each of these bonds has a negative dipole moment in the outward direction. So you can see that the top side of this molecule would be mostly negative while the side with the lone pair would be more positive. A simpler way of looking at is that not every side loo
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The electronegativity of F is more than that of I so there are dipole moments on each of these bonds. There is one side of this molecule that has a lone pair of electrons. This lone pair pushes the bonds farther away. You can see that in this image because there isn't a perfect 90 degrees between the top bond and the bonds on the side. Each of these bonds has a negative dipole moment in the outward direction. So you can see that the top side of this molecule would be mostly negative while the side with the lone pair would be more positive. A simpler way of looking at is that not every side looks the same so there is more than likely an overall dipole moment that makes this a polar molecule.
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A Learner
Author has 68 answers and 106.1K answer views
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Why is H2O a liquid while HF is a gas at room temperature?
It would be worth specifying that hydrogen fluoride boils at about 20∘C, so if by room temperature you mean 25∘C, then hydrogen fluoride will actually be a gas.
Impotant thing is that they have different number of lone pairs present on the more electronegative atom.
In water's case, you have two partial positive hydrogen atoms and two lone pairs of electrons on the oxygen atom, which means that each water molecule can form hydrogen bonds with four other water molecules.
Hydrogen fluoride has three lone pairs of electrons, but only one hydrogen atom. This means that on average you will have insuff
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It would be worth specifying that hydrogen fluoride boils at about 20∘C, so if by room temperature you mean 25∘C, then hydrogen fluoride will actually be a gas.
Impotant thing is that they have different number of lone pairs present on the more electronegative atom.
In water's case, you have two partial positive hydrogen atoms and two lone pairs of electrons on the oxygen atom, which means that each water molecule can form hydrogen bonds with four other water molecules.
Hydrogen fluoride has three lone pairs of electrons, but only one hydrogen atom. This means that on average you will have insufficient partial positive hydrogen atoms to allow for the majority of hydrogen fluoride molecules to hydrogen bond at a particular moment.
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Ross J. Wood
Former chemistry PhD student turned public servant · Author has 260 answers and 944.3K answer views
·7y
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Why is ethyl alcohol liquid while ethyl chloride is gas?
The boiling point of substances reflects the strength of the intermolecular forces present. This is because it’s these forces that must be broken to change a substance from the liquid phase to the gas phase. The stronger the intermolecular forces the higher the boiling point.
Hydrogen bonding is one of the strongest intermolecular forces. They form between a hydrogen atom that’s bonded to an oxygen, nitrogen, fluorine or chlorine, and a nucleophile (usually a lone pair of electrons). For a hydrogen bond to form, both these must be present.
Ethanol can both hydrogen bond donate and accept. It has
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The boiling point of substances reflects the strength of the intermolecular forces present. This is because it’s these forces that must be broken to change a substance from the liquid phase to the gas phase. The stronger the intermolecular forces the higher the boiling point.
Hydrogen bonding is one of the strongest intermolecular forces. They form between a hydrogen atom that’s bonded to an oxygen, nitrogen, fluorine or chlorine, and a nucleophile (usually a lone pair of electrons). For a hydrogen bond to form, both these must be present.
Ethanol can both hydrogen bond donate and accept. It has a hydrogen bonded to an oxygen, and an oxygen atom with lone pairs of electrons. This means that ethanol can hydrogen bond with itself, which increases the strength of interaction with other ethanol molecules, which increases its boiling point.
With ethyl chloride, even though it conatins a chlorine atom with lone pairs of electrons, it doesn’t contain an ‘acidic’ hydrogen. This means it could possibly hydrogen bond with a molecule of different type, but it can’t hydrogen bond with itself. This lowers the strength of its intermolecular forces, and therefore, lowers its boiling point. A similar effect is seen with propanone.
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Mike Jones
MAEd in Chemistry&Physics, Western Carolina University (Graduated 1974) · Upvoted by
Antonello Frau
, PhD Chemistry & Materials Science and Engineering, University of Houston (2009) · Author has 6.2K answers and 8.4M answer views
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Why is CCl₄ a liquid at room temperature but CH₄ a gas?
Methane is gas and tetrachloromethane is a liquid because of the difference in the magnitude of the London dispersion forces. London forces are one of the three van der Waals forces: Keesom forces (dipole-dipole attraction), Debye forces (induced attraction) and London forces (which are exhibited by all molecules); and the only ones exhibited by both CH4 and CCl4. Methane and CCl4 are symmetrical molecules and do not have a net dipole moment. Therefore, no Keesom forces or Debye forces, only London forces.
The strength of London forces is proportional to the polarizability of the molecule, whic
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Methane is gas and tetrachloromethane is a liquid because of the difference in the magnitude of the London dispersion forces. London forces are one of the three van der Waals forces: Keesom forces (dipole-dipole attraction), Debye forces (induced attraction) and London forces (which are exhibited by all molecules); and the only ones exhibited by both CH4 and CCl4. Methane and CCl4 are symmetrical molecules and do not have a net dipole moment. Therefore, no Keesom forces or Debye forces, only London forces.
The strength of London forces is proportional to the polarizability of the molecule, which it turn is determined by the total number of electrons and the area over which they are spread. CCl4 has more electrons (74, vs 10 for CH4) and is larger because Cl atoms are larger. Therefore CCl4 has stronger London forces, enough to make it a liquid at room temperature.
A common misconception is that London dispersion forces are very weak. In some cases London forces may be stronger than Keesom forces and Debye forces and second only to those of hydrogen bonding.
Another misconception about London forces is that they are somehow proportional to the molar mass. That is usually a coincidence because more massive molecules usually have more electrons and more surface area.
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Daniel James Berger
PhD in organic/organosilicon chemistry · Author has 4.6K answers and 11.3M answer views
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Why do we call gas gas when it's a liquid?
“Gas” is short for “gasoline.”
The etymology of “gasoline” seems to be that
It’s a substance related to illuminating gas.
It’s a liquid (“ol” for “oil”).
“ine” is just a generic sort of “chemistry suffix.”
Why does gasoline have the word "gas" in it, if it's never gaseous? (English Language & Usage stack exchange)
There’s a possibly new, possibly over-speculative etymology at the Oxford Dictionaries blog: The etymology of gasoline | OxfordWords blog in which “gasoline” is said to be a corruption of a trade name, cazeline — similar to the way we call all facial tissues “kleenex.”
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Jerome Zoeller
Former Research Scientist at Texas A&M University (TAMU) (1986–2001) · Author has 9.7K answers and 5.4M answer views
·8y
IF7 has a higher vapor pressure and a lower boiling point. There are algorithms that allow you to calculate the partial pressure of a compound from its molecular formula.
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Davie UD
B.Sc (First Class) in Chemistry, University of Colombo (Graduated 2021)
·4y
Related
Why is IF5 molecule polar?
First lets recall the requirments for a molecule to be non-polar, so that we can clearly identify whats polar and non-polar.
Should have a Primary molecular geometry ( linear -sp, trginal planar - sp2, tetrahedral, trigonal bi-pyramidal, ocathedral) or Square planar.
Should have same species of atoms connected to the central atom.
Then only the resultant dipole moment acting on the central atom would be ZERO - which is what non-polar actually means.
IF5 already fulfills the 2nd requirement.
Now we have to derive the shape of IF5 to see if it fulfills the 1st requirement. For this we have to use the
Continue Reading
First lets recall the requirments for a molecule to be non-polar, so that we can clearly identify whats polar and non-polar.
Should have a Primary molecular geometry ( linear -sp, trginal planar - sp2, tetrahedral, trigonal bi-pyramidal, ocathedral) or Square planar.
Should have same species of atoms connected to the central atom.
Then only the resultant dipole moment acting on the central atom would be ZERO - which is what non-polar actually means.
IF5 already fulfills the 2nd requirement.
Now we have to derive the shape of IF5 to see if it fulfills the 1st requirement. For this we have to use the VSEPR theory. In this method, we count the total number of valence electrons offered by both central atom and surrounding atoms for bond forming (valence electrons) and dreive its shape.
I = 7
5 x F = 7 x 5 = 35
Total = 42
No of electron Pairs = 42/2 = 21. Now deduct 5 pairs alocated for the five I-F bonds. Then filling out the lone pairs of the 5 F atoms ( 3 each), deduct 15 more lone pairs. Now you are only left with one 1 loone pair which is then placed on I.
So this results in 5 bond pairs, and 1 lone pair - Molecular Geometry (shape) is square pyramidal.
This is not one of the symmetric molecular shapes mentioned in the first requirement. Therefore, IF5 is definitely POLAR.
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3794 | https://pubmed.ncbi.nlm.nih.gov/25428714/ | Diagnostic utility of zinc protoporphyrin to detect iron deficiency in Kenyan pregnant women - PubMed
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. 2014 Nov 26:12:229.
doi: 10.1186/s12916-014-0229-8.
Diagnostic utility of zinc protoporphyrin to detect iron deficiency in Kenyan pregnant women
Martin N Mwangi,Sumi Maskey,Pauline E A Andang o,Noel K Shinali,Johanna M Roth,Laura Trijsburg,Alice M Mwangi,Han Zuilhof,Barend van Lagen,Huub Fj Savelkoul,Ayşe Y Demir,Hans Verhoef
PMID: 25428714
PMCID: PMC4276103
DOI: 10.1186/s12916-014-0229-8
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Randomized Controlled Trial
Diagnostic utility of zinc protoporphyrin to detect iron deficiency in Kenyan pregnant women
Martin N Mwangi et al. BMC Med.2014.
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. 2014 Nov 26:12:229.
doi: 10.1186/s12916-014-0229-8.
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Martin N Mwangi,Sumi Maskey,Pauline E A Andang o,Noel K Shinali,Johanna M Roth,Laura Trijsburg,Alice M Mwangi,Han Zuilhof,Barend van Lagen,Huub Fj Savelkoul,Ayşe Y Demir,Hans Verhoef
PMID: 25428714
PMCID: PMC4276103
DOI: 10.1186/s12916-014-0229-8
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Background: Iron-deficient erythropoiesis results in excess formation of zinc protoporphyrin (ZPP), which can be measured instantly and at low assay cost using portable haematofluorometers. ZPP is used as a screening marker of iron deficiency in individual pregnant women and children, but also to assess population iron status in combination with haemoglobin concentration. We examined associations between ZPP and disorders that are common in Africa. In addition, we assessed the diagnostic utility of ZPP (measured in whole blood and erythrocytes), alone or in combination with haemoglobin concentration, in detecting iron deficiency (plasma ferritin concentration <15 μg/L).
Methods: Single blood samples were collected from a population sample of 470 rural Kenyan women with singleton pregnancies, gestational age 13 to 23 weeks, and haemoglobin concentration ≥90 g/L. We used linear regression analysis to assess associations between ZPP and iron markers (including anaemia), factors known or suspected to be associated with iron status, inflammation markers (plasma concentrations of C-reactive protein and α 1-acid glycoprotein), infections (Plasmodium infection, HIV infection), and other disorders (α(+)-thalassaemia, plasma concentrations of total bilirubin, and lactate dehydrogenase). Subsequently, in those without inflammation, Plasmodium infection, or HIV infection, we used logistic discriminant analysis and examined receiver operating characteristics curves with corresponding area-under-the-curve to assess diagnostic performance of ZPP, alone and in combination with haemoglobin concentration.
Results: Individually, whole blood ZPP, erythrocyte ZPP, and erythrocyte protoporphyrin had limited ability to discriminate between women with and without iron deficiency. Combining each of these markers with haemoglobin concentration had no additional diagnostic value. Conventional cut off points for whole blood ZPP (>70 μmol/mol haem) resulted in gross overestimates of the prevalence of iron deficiency.
Conclusions: Erythrocyte ZPP has limited value to rule out iron deficiency when used for screening in conditions with a low prevalence (e.g., 10%). ZPP is of unreliable diagnostic utility when discriminating between pregnant women with and without iron deficiency. Based on these findings, guidelines on the use of ZPP to assess iron status in individuals or populations of pregnant women need review.
Trial registration:NCT01308112 (2 March 2011).
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Ability of erythrocyte protoporphyrin, either…
Figure 1
Ability of erythrocyte protoporphyrin, either alone or combined with haemoglobin concentration, to discriminate…
Figure 1
Ability of erythrocyte protoporphyrin, either alone or combined with haemoglobin concentration, to discriminate between pregnant women with and without iron deficiency.(Panel A) Receiver operating characteristics (ROC) curve for various blood markers, used alone, to discriminate between iron-deficient and iron-replete women. Area-under-the-curve (AUC; 95% CI): whole blood ZPP: (0.66; 0.57–0.74); erythrocyte ZPP: (0.73; 0.65–0.80); EP: (0.59; 0.50–0.68); haemoglobin concentration: (0.61; 0.52–0.70). (Panel B) Cumulative relative frequency distribution of erythrocyte ZPP, the best indicator when used as a single test (Panel A) to discriminate between iron-deficient and iron-replete women. The black circle in Panel A and the dotted black line in Panel B indicate the erythrocyte ZPP:haem ratio of 34 μmol/mol whereby the total diagnostic error is minimised at a prevalence of iron deficiency of 50%. (Panels C, E, andG) ROC curves for whole blood ZPP, erythrocyte ZPP, and EP, either alone or each in combination with haemoglobin concentration. AUC; 95% CI: combined whole blood ZPP with haemoglobin concentration: (0.64; 0.56–0.73); combined erythrocyte ZPP with haemoglobin concentration: (0.72; 0.64–0.80); combined EP with haemoglobin concentration: (0.64; 0.55–0.73). (Panel D) Bivariate scatterplot for whole blood ZPP and haemoglobin concentration, by iron status; (Panel F) Bivariate scatterplot for erythrocyte ZPP and haemoglobin concentration, by iron status; (Panel H) Bivariate scatterplot for EP and haemoglobin concentration, by iron status. Grey dashed lines in ROC curves indicate a ‘worst’ possible test, which has no discriminatory value and an area-under-the-curve (AUC) of 0.5. An ideal marker would have a curve that runs from the lower-left via the upper-left to the upper-right corner, yielding an AUC of 1.0.
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References
NCCLS . Erythrocyte Protoporphyrin Testing; Approved Guideline. NCCLS document C42-A. National Committee for Clinical Laboratory Standards: Wayne, PA; 1996.
WHO/CDC . Assessing the Iron Status of Populations. 2. Geneva, Switzerland: World Health Organization: Report of a joint World Health Organization/Centers for Disease Control and Prevention Technical Consultation on the assessment of iron status at the population level. Geneva, Switzerland, 6th–8th April, 2004; 2007.
Schifman RB, Thomasson JE, Evers JM. Red blood cell zinc protoporphyrin testing for iron deficiency anemia in pregnancy. Am J Obstet Gynecol. 1987;157:304–307. doi: 10.1016/S0002-9378(87)80157-6. - DOI - PubMed
Romslo I, Haram K, Sagen N, Augensen K. Iron requirement in normal pregnancy as assessed by serum ferritin, serum transferrin saturation and erythrocyte protoporphyrin determinations. Br J Obstet Gynaecol. 1983;90:101–107. doi: 10.1111/j.1471-0528.1983.tb08891.x. - DOI - PubMed
Harthoorn-Lasthuizen EJ, Lindemans J, Langenhuijsen MMAC. Erythrocyte zinc protoporphyrin testing in pregnancy. Acta Obstet Gynecol Scand. 2000;79:660–666. - PubMed
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3795 | https://brilliant.org/wiki/trigonometric-r-method/ | Trigonometric R method
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Aditya Virani,
Alexander Katz,
and
Jimin Khim
contributed
The trigonometric R method is a method of rewriting a weighted sum of sines and cosines as a single instance of sine (or cosine). This allows for easier analysis in many cases, as a single instance of a basic trigonometric function is often easier to work with than multiple are.
The R method is most often used to find the extrema (maximum and minimum) of combinations of trigonometric functions, since the extrema of a basic trigonometric function are easy to work with (both sine and cosine have a minimum of -1 and a maximum of 1).
Contents
Formal Statement and Proof
Using the R Method
Applications to Extrema
See Also
Formal Statement and Proof
The R method can be used to convert an expression of the form asinθ+bcosθ into a single instance of either sine or cosine:
The sine form:
For any a,b,θ∈R, the following corresponds:
asinθ+bcosθ=Rsin(θ+α),
where R=a2+b2 and tanα=ab, or α=arctanab.
The cosine form:
For any a,b,θ∈R, the following corresponds:
asinθ+bcosθ=Rcos(θ−α),
where R=a2+b2 and tanα=ba, or α=arctanba.
First, let's prove the sine form. Let a=Rcosα and b=Rsinα. Then the following equations correspond:
a2+b2tanαasinθ+bcosθ=R2(cos2α+sin2α)=R2=cosαsinα=ab=Rsinθcosα+Rcosθsinα=Rsin(θ+α). □
To prove the cosine form, let a=Rsinα and b=Rcosα. Then the following equations correspond:
a2+b2tanαasinθ+bcosθ=R2(sin2α+cos2α)=R2=cosαsinα=ba=Rcosθcosα+Rsinθsinα=Rcos(θ−α). □
Using the R Method
Given the condition R>0 and 0<θ<2π, what are the appropriate values of R and θ for the following equation:
sin8π+cos8π=Rsinθ?
Since R2=12+12=2 and arctan11=4π, we have
sin8π+cos8π=2sin(8π+4π)=2sin83π. □
Therefore, R=2 and θ=83π. □
Simplify sin6π+3cos6π using the R method.
(1) Using the sine form:
Since R2=12+(3)2=4 and arctan3=3π, we have
sin6π+3cos6π=2sin(6π+3π)=2sin2π=2. □
(2) Using the cosine form:
Since R2=12+(3)2=4 and arctan31=6π, we have
sin6π+3cos6π=2cos(6π−6π)=2cos0=2. □
20
25
30
35
If 3cosθ−sinθ=31 and 0<θ<2π, the value of 3sinθ+cosθ can be expressed as 3a. What is the value of a?
The correct answer is: 35
Solve for x:
3sinx−cosx=2,−π<x<π.
Since R2=(3)2+(−1)2=4 and arctan3−1=−6π, we have
3sinx−cosx=2sin(x−6π)⇒sin(x−6π)=2=22.
Let x−6π=θ. Given −π<x<π, we have −67π<θ<65π. In this interval the values of θ that satisfy sinθ=22 are 4π and 43π. Thus we have
x−6π=4πx−6π=43π⟹x=125π⟹x=1211π.
Therefore the answer is x=125π,1211π. □
When using the R method, there are numerous possible values of R and α, since sine and cosine are periodic functions. Adding 2nπ to α for any integer n would give the same answer. Also, adding (2n−1)π to α for any integer n and changing the sign of R would also be equivalent to the answer. However we most often use −2π<α<2π and R>0 for convenience.
Applications to Extrema
The R method is frequently used to find the maximum or minimum of equations in the form of asinθ+bcosθ. It is known that −1≤sinx≤1 and −1≤cosx≤1 for any real number x, so the maximum and minimum of asinθ+bcosθ are R and −R, respectively.
Find the sum of the maximum and minimum values of
3sin(x−3π)−4cos(x−3π)+2.
Since 32+(−4)2=52, we have R=5. This implies that
−5≤3sin(x−3π)−4cos(x−3π)≤5⇒−3≤3sin(x−3π)−4cos(x−3π)+2≤7.
Therefore the answer is −3+7=4. □
If the maximum value of 7cosx+6sinx is expressed as a, what is the value of a?
This problem is posed by Minimario M.
See Also
Cite as:
Trigonometric R method.
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3796 | https://www.youtube.com/watch?v=vQCkYm3v3aA | Distance and displacement introduction | One-dimensional motion | AP Physics 1 | Khan Academy
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So let's say we have a sheep and it is hungry. So that is my sheep, my best quick drawing of a sheep. And it is just following the grass wherever it finds good grass to eat. And so in pursuit of tasty grass, it first goes 10 kilometers to the east. 10 kilometers east. And then it takes a right turn and then it goes 5 kilometers south. Five kilometers south. And then it takes another right turn and it goes 10 kilometers west. 10 kilometers west. My question to you is how far has this sheep traveled? and pause the video and see if you can figure that out. How far has that sheep traveled? All right. Now, let's try to answer this together. And it's a little bit of a trick question because depending how you interpret the idea how far something has traveled, there could actually be two valid answers here. One answer is you could say, what is the total distance traveled? So, let me write this one down. This is an important concept. distance traveled and this would be the total length of its journey or the total length of its path. So the distance traveled in this situation would be 10 km + 5 km + 5 km plus 10 km plus 10 km which would give us 25 km. That would be 25 km. So some of y'all might have said that the sheep traveled 25 km. If that's what you came up with, what you calculated is the total distance traveled. Now, some of you might have said, "Wait, hold on a second. The sheep was here before and then it ends up right over here at the end of its journey." And so, its change in position is it would have moved a net 5 kilometers south from this point to this point would have been five kilometers south. It doesn't matter what it did along the way, what its path was. At the end of the day, it ended up 5 kilometers south of where it started out. And this would also be a valid way of saying how far it traveled. And this notion is known as displacement. So the displacement would be in this situation five kilometers south. So pause this video. What do you think is the definition or the difference between saying the distance something traveled versus its displacement? Well, as we mentioned, the distance something travels is the entire length of its journey. It's the the entire path. And this thing will never be negative. You could travel zero kilometers, but as long as you are you are your position keeps changing, this thing will only increase because it's the total length of your path. Your displacement is your change in position. And since it's a change in position, not only will we mention the five kilometers, we'll say in what direction. The position of the sheep has changed 5 kilometers to the south. So let me write this down. This is total length of path. Total length of path is the distance traveled while the displacement is the change in position. Change in position. |
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Author:Mr E. Bissoon
Equilateral Triangle
You can construct the equilateral triangle below (with perfect 60 angles) without using a protractor. In plain terms: In this worked example, the initial line is drawn using a ruler to an exact length. In this case the line is 6cm long. I now need to draw two small arcs using a compass set to the same distance (of 6cm). From one end of the line I will draw a small arc in the the top middle region above the line. I will then draw another small arc but this time from the other end of the line to the same [top middle] region above the line. This will create an intersect point from where I can draw two more lines to each end of the first line that I drew. This triangle has three perfect 60 internal angles and each side is of equal length to the others thus making it an equilateral triangle. You can bisect one of the angles to make a 30 angle. (Take a look at the dotted lines that represent the path of the compass. Note that I am not drawing the entire path of the dotted line. I am drawing the two small arcs only. Don't forget that if you are asked to construct a triangle in this way then ensure that you do not erase the arcs as this is evidence of your work.) Run the graphic below to watch how to construct this equilateral triangle.
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3798 | https://www.youtube.com/watch?v=uEAN8A3vYEU | Intro to cyclic groups and the group of units mod p
Andrew McCrady
5170 subscribers
4 likes
Description
340 views
Posted: 6 Sep 2020
This is a short lecture that introduces cyclic groups and primitive roots in the group of units mod p. This is for my online number theory course.
Transcript:
this is a short video about section 2.5 and stein's book on elementary number theory and we're going to talk about the first part the whole section is about the structure of the group z mod pz star remember that this is the group of units and zmod pz so the main result that we're going to do where of course you could take p to be a prime number but anyway that's what i was saying the main result we're going to look at in this section 2.5 what it all kind of culminates in is that this group is always cyclic now what i want to do is tell you what a cyclic group is so if you've had some modern algebra or abstract algebra you already know what this is but i'll say it a little bit more generally because it's more of a group theory thing than it is anything for necessarily just z mod p z or z my pz star so i'll say g is a group and i'll use like multiplication as my i'll i'll denote my operation as if it was looks like multiplication so i'll use exponents so let's say g is cyclic that means that there exists some element a in g such that the following happens for any for any other element g and g there should exist some natural number k such that a to the k gives you g so what i'm trying to get across is that every element of g is really just a power of a so the single element a in some sense generates every single other element of the group and you'd call such an a a generator so now let's do maybe two concrete examples before we go any further so don't look below don't look at primitive root yet i'm not talking about that yet i'm just trying to talk more about cyclic groups so what i'm going to do is i'm going to compare the following two groups z mod 8c star and i'm going to compare it to z mod 7z star and i'm going to try to look at which one of these is cyclic are they both cyclic are neither cyclic or what and you could do these by hand essentially all i'm asking you to do is list the elements in each group and just start computing powers of each element if you have a single element that will when you take powers of it you get every other element of the group then that means it's cyclic now i'm going to show you how to do that on sage just to kind of again demonstrate the novelty of sage why it's pretty cool so what we're going to do is let's start with z mod 8c so r is equal to the integers so i'm just naming it r integers eight so remember what that does so r is the ring z mod and z now i just want to look at the multiplicative group of units so i'm just gonna name that our underscore units and how do i make those so the way i'm going to make them is i'm going to make them as a list so i'm going to look at x for x in r if so i only want the following x's if the greatest common divisor of x with 8 is 1. so remember the group of units mod n are exactly the elements that are relatively prime to n so in this case i expect this to spit out one three five and seven those would be uh the things in z z that are relatively prime to n okay cool and so this is the multiplicative group z mod 8c star all right and so let's see i probably need to actually call it r underscore units and i expect it cool gives me one three five and seven so anytime you see r units right now that's the group you're thinking of one three five and seven the operation is you're multiplying things but you're always reducing everything mod eight so what i'm going to do now is i'm just going to compute the order of each element in this group and the way that i'll do that is with like a for loop say this is one way to do it so i could do it all at once again just the novelty of sage being able to do some of these computations all at one time it's something like 4x and r units so what this is going to do is i'm about to cycle through one three five and seven and do the following to each element of one three five and seven here's what i want to do i want to print x and then right next to it i want to print the order of x so x dot multiplicative underscore order leave those open so again what this will do this computes the order of each x in my z mod 8c star so let's take a look at what does this look like whoops that's not what i wanted how about now cool so the first column are the one three five and seven that is z mod ac star and the second column is the order of each element so you should read that one has order one three has order two five has order two seven has order two okay cool so if you think about it what does that say that says that's each element uh is really like its own inverse like 3 times 3 gives you back the identity it's how you should read the order of 3 is 2. so in particular it's not possible like i can't take powers of three and ever get five or ever get seven i can't take powers of five i never recover three or seven and i can't take powers of seven and never recover uh three or five so this is not cyclic so what this shows this shows that's zmod etsy star is not cyclic so what i'm going to do is a similar computation with let's compare it to z mod 7z star so i'll name it s now so let's say s is integers integers seven and i won't comment as many things this time since it's the same idea i'll say s underscore units that's a good name maybe and i'll say i'll use x again x for x and s if g c d and now i want x and seven to be one so what this does s units now i will comment this this is the group z mod 7z star i just put the little character to indicate the asterisk is up in the exponent i'm gonna do the same thing for x and s units i'm going to print x comma x dot multiplicative order and let's see what that does okay cool so by the way maybe what am i looking for maybe that's not maybe you can't tell like well how do i know what i'm looking for how do i know what's a good exponent like 6 looks like a pretty big exponent is that cool let me give you another way to try to find this one so i'll try to do this a little bit more explicitly so what i'm actually going to do is i'm going to go ahead and show you what the power of each element is by the way here um let's see what should this be maybe be good so this is which is equal to like as a set this would be one two three four five six so those are the numbers that are relatively prime to seven that are less than or equal to seven so those are the things i'm going to take powers of so now i'll come back down here what i'm going to do is i want to just show you what happens when you start taking powers of 2. what happens when you start taking powers of 3 etc and the question is do the powers of 2 give me 3 do they give me 4 did they give me 5 do they give me 6. the answer to that is yes then 2 is a generator so let's do that so what i might do you could do that by hand that might be a lot of work i'm going to do another for loop here so 4k in range say 1 to 7. so why do i do 7 if you look at that group it has six elements in it so euler's here told me the power of any element if i raise an element to the power of six then that should automatically get me back to one so the biggest exponent i need to consider is six so i put seven just because sage doesn't include that right endpoint so if you push enter again it should indent and here's what i want to do i'm going to print something like x and then comma i want to actually like show you the symbol i'm going to raise it to comma and then i want k comma i want to actually show you like the symbol that it's equal to i'm just going to plop that down as a string if that makes any sense to you cool if it doesn't don't worry about it and then i want to compute the actual power here so let's see what this looks like all right so you know the first set of blocks there you see one raised to the one through sixth power it gives you one so let's look at two two is the first interesting one when we start raising um two um to different powers here i see that it just keeps bouncing back between one two and four so two is not a generator um let's take a look at what happens when i start raising three to uh that power then what do i get i get just three two six four five one oh so check out three if you zoom in on that but the powers of three if you look what's going on in the equal sign to the right i recovered one three i guess maybe i should start i get three two six four five one that is all the elements that are in the group so that shows that three is a generator so what does this demonstrate so this tells us from this we see oops i got scoot over don't i so this shows that 3 generates the group so in that case this group z mod 7z star is cyclic great so sick looks a cool property because in some sense you really just need to tell me about like what's going on with the generator because everybody else in the group depends on just what the generator does okay so in the same video i'm going to tell you about something that's related and the relate on a similar idea is what's going to be called a primitive element and again i'm relaxing the condition that i need the modulus to be prime just for any n so just by definition a primitive element modulo n it should be an element of this multiplicative group that has order 5n so in the examples that we've done so far in our case we just saw that three had order six right because three if you look at three to the sixth is the first time that you ever get one so six is the smallest exponent and smallest natural number exponent you could raise 3 2 in order to get 1. so 6 is the order of 3. so 3 would be a primitive element whereas in our example that we did earlier with 8 here you know if i was to actually write this out you know what is this we said that this should be the group or the set one three five seven and the question is do you have an element that has order four right that's what phi of eight is it's 4. in this case no the order if you look at this list here the highest order of an element you get is 2. so this group does not have a primitive element so a big question is how do i know when something like z mod nz star is cyclic and how do i know when something like z mod nz star has a primitive element let me also show you in sage there's a way to check for primitive elements kind of similarly to how i was doing stuff for cyclic groups but maybe it'd be good to take a look and say how to do that and so uh let's see i mean you could also just see from well i'll just go ahead and do it why not this will be good so let's take a look at say let's say r let's rename it now let's say r and let me give myself some space okay let's say r is and let's relax the condition that your modulus is prime let's do nine say say r is integers oops all right cool and i'm going to do the same thing where our units is equal to x 4x and r if remember i want the units so those would be the numbers that are relatively prime to nine so gcd x9 is exactly one and what should those be let's actually call it just so you see it so that's what's what r units actually is so what i'm gonna do now is i'm going to check is there a primitive root in there remember what i'm asking is do any of those elements actually have order 6 and how do i know it's 6 well because i can count that that's how many elements are in the set or remember sage sage can do what is euler five nine and i know it should be six and you see six showed up right beneath my list there so do we have an element of order phi of nine which is six way we'll do this is for x and r units then i will say print x comma x dot is underscore primitive underscore root and those so is underscore primitive under square root is like some attribute that you can put on x just like how finding its multiplicative order was and what this spits out is if it's a primitive root it says true so how should you take that that says two and five are primitive roots and so two has order six and five has order six there's another way that you could see that there's one more thing if you look in the book it tells us one more way to talk about to find primitive roots in sage if you scroll down here but just be careful you notice it's a different function so print the prime and print primitive root p what that's going to do though is that's going to give you the smallest positive printer uh integer goodness primitive root modulo n it's only going to give you the smallest one for example if you were to do that here it would only give you two it would leave off five so maybe this is a good way to check what all the primitive roots are in that particular group |
3799 | https://artofproblemsolving.com/wiki/index.php/Circumradius?srsltid=AfmBOoo2w228oPFPooiVp4i938a2VnwGFi6P4XFZy8dSxlt3OrJJ8hFA | Art of Problem Solving
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Circumradius
The circumradius of a cyclicpolygon is the radius of the circumscribed circle of that polygon. For a triangle, it is the measure of the radius of the circle that circumscribes the triangle. Since every triangle is cyclic, every triangle has a circumscribed circle, or a circumcircle.
Contents
[hide]
1 Formula for a Triangle
2 Proof
3 Formula for Circumradius
4 Circumradius, bisector and altitude
5 Euler's Theorem for a Triangle
6 Proof
7 Right triangles
7.1 Theorem
8 Equilateral triangles
9 If all three sides are known
10 If you know just one side and its opposite angle
11 See also
Formula for a Triangle
Let and denote the triangle's three sides and let denote the area of the triangle. Then, the measure of the circumradius of the triangle is simply . This can be rewritten as .
Proof
We let , , , , and . We know that is a right angle because is the diameter. Also, because they both subtend arc . Therefore, by AA similarity, so we have or However, remember that . Substituting this in gives us and then simplifying to get and we are done.
Formula for Circumradius
Where is the circumradius, is the inradius, and , , and are the respective sides of the triangle and is the semiperimeter. Note that this is similar to the previously mentioned formula; the reason being that .
But, if you don't know the inradius, you can find the area of the triangle by Heron’s Formula:
Circumradius, bisector and altitude
Circumradius and altitude are isogonals with respect bisector and vertex of triangle.
Euler's Theorem for a Triangle
Let have circumcenter and incenter .Then
Proof
See
Right triangles
The hypotenuse of the triangle is the diameter of its circumcircle, and the circumcenter is its midpoint, so the circumradius is equal to half of the hypotenuse of the right triangle.
This results in a well-known theorem:
Theorem
The midpoint of the hypotenuse is equidistant from the vertices of the right triangle. The midpoint of the hypotenuse is the circumcenter of a right triangle.
Equilateral triangles
where is the length of a side of the triangle.
If all three sides are known
Which follows from the Heron's Formula and .
If you know just one side and its opposite angle
by the Law of Sines.
(Extended Law of Sines)
See also
Inradius
Semiperimeter
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