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# Quantum state

In quantum physics, a quantum state is an oul' mathematical entity that provides a bleedin' probability distribution for the outcomes of each possible measurement on a system. Jesus, Mary and Joseph. Knowledge of the bleedin' quantum state together with the oul' rules for the oul' system's evolution in time exhausts all that can be predicted about the bleedin' system's behavior. Sure this is it. A mixture of quantum states is again a feckin' quantum state. Quantum states that cannot be written as a holy mixture of other states are called pure quantum states, while all other states are called mixed quantum states. A pure quantum state can be represented by a ray in an oul' Hilbert space over the oul' complex numbers,[1][2] while mixed states are represented by density matrices, which are positive semidefinite operators that act on Hilbert spaces.[3][4]

Pure states are also known as state vectors or wave functions, the bleedin' latter term applyin' particularly when they are represented as functions of position or momentum, would ye swally that? For example, when dealin' with the bleedin' energy spectrum of the feckin' electron in a hydrogen atom, the relevant state vectors are identified by the bleedin' principal quantum number n, the angular momentum quantum number , the feckin' magnetic quantum number m, and the bleedin' spin z-component sz. Sure this is it. For another example, if the spin of an electron is measured in any direction, e.g. with a Stern–Gerlach experiment, there are two possible results: up or down. Would ye swally this in a minute now? The Hilbert space for the electron's spin is therefore two-dimensional, constitutin' a qubit. A pure state here is represented by a bleedin' two-dimensional complex vector ${\displaystyle (\alpha ,\beta )}$, with a bleedin' length of one; that is, with

${\displaystyle |\alpha |^{2}+|\beta |^{2}=1,}$
where ${\displaystyle |\alpha |}$ and ${\displaystyle |\beta |}$ are the feckin' absolute values of ${\displaystyle \alpha }$ and ${\displaystyle \beta }$. Whisht now and eist liom. A mixed state, in this case, has the structure of a ${\displaystyle 2\times 2}$ matrix that is Hermitian and positive semi-definite, and has trace 1.[5] A more complicated case is given (in bra–ket notation) by the feckin' singlet state, which exemplifies quantum entanglement:
${\displaystyle \left|\psi \right\rangle ={\frac {1}{\sqrt {2}}}{\bigl (}\left|\uparrow \downarrow \right\rangle -\left|\downarrow \uparrow \right\rangle {\bigr )},}$
which involves superposition of joint spin states for two particles with spin 12. The singlet state satisfies the oul' property that if the particles' spins are measured along the bleedin' same direction then either the oul' spin of the bleedin' first particle is observed up and the feckin' spin of the second particle is observed down, or the bleedin' first one is observed down and the second one is observed up, both possibilities occurrin' with equal probability.

A mixed quantum state corresponds to a probabilistic mixture of pure states; however, different distributions of pure states can generate equivalent (i.e., physically indistinguishable) mixed states, to be sure. The Schrödinger–HJW theorem classifies the bleedin' multitude of ways to write a given mixed state as a convex combination of pure states.[6] Before a holy particular measurement is performed on a quantum system, the feckin' theory gives only a probability distribution for the bleedin' outcome, and the form that this distribution takes is completely determined by the feckin' quantum state and the feckin' linear operators describin' the measurement. In fairness now. Probability distributions for different measurements exhibit tradeoffs exemplified by the feckin' uncertainty principle: a feckin' state that implies an oul' narrow spread of possible outcomes for one experiment necessarily implies a wide spread of possible outcomes for another.

## Conceptual description

### Pure states

Probability densities for the feckin' electron of a bleedin' hydrogen atom in different quantum states.

In the bleedin' mathematical formulation of quantum mechanics, pure quantum states correspond to vectors in an oul' Hilbert space, while each observable quantity (such as the feckin' energy or momentum of a holy particle) is associated with a mathematical operator. The operator serves as a feckin' linear function which acts on the oul' states of the bleedin' system. I hope yiz are all ears now. The eigenvalues of the feckin' operator correspond to the oul' possible values of the feckin' observable. For example, it is possible to observe a feckin' particle with a holy momentum of 1 kg⋅m/s if and only if one of the bleedin' eigenvalues of the momentum operator is 1 kg⋅m/s, enda story. The correspondin' eigenvector (which physicists call an eigenstate) with eigenvalue 1 kg⋅m/s would be a holy quantum state with a definite, well-defined value of momentum of 1 kg⋅m/s, with no quantum uncertainty, fair play. If its momentum were measured, the oul' result is guaranteed to be 1 kg⋅m/s.

On the feckin' other hand, a system in a superposition of multiple different eigenstates does in general have quantum uncertainty for the oul' given observable. We can represent this linear combination of eigenstates as:

${\displaystyle |\Psi (t)\rangle =\sum _{n}C_{n}(t)|\Phi _{n}\rangle .}$

The coefficient which corresponds to a particular state in the bleedin' linear combination is a holy complex number, thus allowin' interference effects between states. Be the holy feck, this is a quare wan. The coefficients are time dependent, begorrah. How a bleedin' quantum state changes in time is governed by the feckin' time evolution operator. The symbols ${\displaystyle |}$ and ${\displaystyle \rangle }$[a] surroundin' the oul' ${\displaystyle \Psi }$ are part of bra–ket notation.

Statistical mixtures of states are a feckin' different type of linear combination. Bejaysus. A statistical mixture of states is an oul' statistical ensemble of independent systems, would ye believe it? Statistical mixtures represent the bleedin' degree of knowledge whilst the oul' uncertainty within quantum mechanics is fundamental, the hoor. Mathematically, a bleedin' statistical mixture is not a combination usin' complex coefficients, but rather a combination usin' real-valued, positive probabilities of different states ${\displaystyle \Phi _{n}}$. A number ${\displaystyle P_{n}}$ represents the bleedin' probability of a randomly selected system bein' in the bleedin' state ${\displaystyle \Phi _{n}}$. Unlike the bleedin' linear combination case each system is in a holy definite eigenstate.[7][8]

The expectation value ${\displaystyle {\langle A\rangle }_{\sigma }}$ of an observable A is a statistical mean of measured values of the bleedin' observable. It is this mean, and the bleedin' distribution of probabilities, that is predicted by physical theories.

There is no state which is simultaneously an eigenstate for all observables. Jaykers! For example, we cannot prepare a holy state such that both the position measurement Q(t) and the oul' momentum measurement P(t) (at the same time t) are known exactly; at least one of them will have a range of possible values.[b] This is the oul' content of the bleedin' Heisenberg uncertainty relation.

Moreover, in contrast to classical mechanics, it is unavoidable that performin' a feckin' measurement on the oul' system generally changes its state.[9][10][c] More precisely: After measurin' an observable A, the feckin' system will be in an eigenstate of A; thus the state has changed, unless the bleedin' system was already in that eigenstate. G'wan now. This expresses a bleedin' kind of logical consistency: If we measure A twice in the same run of the experiment, the oul' measurements bein' directly consecutive in time,[d] then they will produce the feckin' same results. G'wan now and listen to this wan. This has some strange consequences, however, as follows.

Consider two incompatible observables, A and B, where A corresponds to an oul' measurement earlier in time than B.[e] Suppose that the feckin' system is in an eigenstate of B at the feckin' experiment's beginnin', would ye believe it? If we measure only B, all runs of the bleedin' experiment will yield the oul' same result. If we measure first A and then B in the same run of the feckin' experiment, the bleedin' system will transfer to an eigenstate of A after the first measurement, and we will generally notice that the feckin' results of B are statistical. Thus: Quantum mechanical measurements influence one another, and the feckin' order in which they are performed is important.

Another feature of quantum states becomes relevant if we consider an oul' physical system that consists of multiple subsystems; for example, an experiment with two particles rather than one. Jasus. Quantum physics allows for certain states, called entangled states, that show certain statistical correlations between measurements on the feckin' two particles which cannot be explained by classical theory, would ye swally that? For details, see entanglement, game ball! These entangled states lead to experimentally testable properties (Bell's theorem) that allow us to distinguish between quantum theory and alternative classical (non-quantum) models.

### Schrödinger picture vs. Heisenberg picture

One can take the feckin' observables to be dependent on time, while the feckin' state σ was fixed once at the feckin' beginnin' of the feckin' experiment. Story? This approach is called the oul' Heisenberg picture, fair play. (This approach was taken in the oul' later part of the oul' discussion above, with time-varyin' observables P(t), Q(t).) One can, equivalently, treat the observables as fixed, while the feckin' state of the oul' system depends on time; that is known as the bleedin' Schrödinger picture, the cute hoor. (This approach was taken in the feckin' earlier part of the oul' discussion above, with a time-varyin' state ${\textstyle |\Psi (t)\rangle =\sum _{n}C_{n}(t)|\Phi _{n}\rangle }$.) Conceptually (and mathematically), the bleedin' two approaches are equivalent; choosin' one of them is a holy matter of convention.

Both viewpoints are used in quantum theory, like. While non-relativistic quantum mechanics is usually formulated in terms of the Schrödinger picture, the bleedin' Heisenberg picture is often preferred in a bleedin' relativistic context, that is, for quantum field theory. C'mere til I tell yiz. Compare with Dirac picture.[12]: 65

## Formalism in quantum physics

### Pure states as rays in a complex Hilbert space

Quantum physics is most commonly formulated in terms of linear algebra, as follows. Any given system is identified with some finite- or infinite-dimensional Hilbert space. The pure states correspond to vectors of norm 1. Thus the set of all pure states corresponds to the oul' unit sphere in the oul' Hilbert space, because the oul' unit sphere is defined as the feckin' set of all vectors with norm 1.

Multiplyin' a pure state by a scalar is physically inconsequential (as long as the oul' state is considered by itself). If a vector in a feckin' complex Hilbert space ${\displaystyle H}$ can be obtained from another vector by multiplyin' by some non-zero complex number, the oul' two vectors are said to correspond to the oul' same "ray" in ${\displaystyle H}$[1]: 50  and also to the bleedin' same point in the bleedin' projective Hilbert space of ${\displaystyle H}$.

### Bra–ket notation

Calculations in quantum mechanics make frequent use of linear operators, scalar products, dual spaces and Hermitian conjugation, the shitehawk. In order to make such calculations flow smoothly, and to make it unnecessary (in some contexts) to fully understand the underlyin' linear algebra, Paul Dirac invented a notation to describe quantum states, known as bra–ket notation, so it is. Although the bleedin' details of this are beyond the feckin' scope of this article, some consequences of this are:

• The expression used to denote a feckin' state vector (which corresponds to a pure quantum state) takes the form ${\displaystyle |\psi \rangle }$ (where the feckin' "${\displaystyle \psi }$" can be replaced by any other symbols, letters, numbers, or even words). This can be contrasted with the bleedin' usual mathematical notation, where vectors are usually lower-case latin letters, and it is clear from the bleedin' context that they are indeed vectors.
• Dirac defined two kinds of vector, bra and ket, dual to each other.[f]
• Each ket ${\displaystyle |\psi \rangle }$ is uniquely associated with a so-called bra, denoted ${\displaystyle \langle \psi |}$, which corresponds to the oul' same physical quantum state, fair play. Technically, the oul' bra is the bleedin' adjoint of the feckin' ket. Arra' would ye listen to this. It is an element of the oul' dual space, and related to the oul' ket by the bleedin' Riesz representation theorem. Be the holy feck, this is a quare wan. In an oul' finite-dimensional space with an oul' chosen basis, writin' ${\displaystyle |\psi \rangle }$ as an oul' column vector, ${\displaystyle \langle \psi |}$ is an oul' row vector; to obtain it just take the feckin' transpose and entry-wise complex conjugate of ${\displaystyle |\psi \rangle }$.
• Scalar products[g][h] (also called brackets) are written so as to look like an oul' bra and ket next to each other: ${\displaystyle \langle \psi _{1}|\psi _{2}\rangle }$. (The phrase "bra-ket" is supposed to resemble "bracket".)

### Spin

The angular momentum has the feckin' same dimension (M·L2·T−1) as the Planck constant and, at quantum scale, behaves as a holy discrete degree of freedom of a quantum system.[which?] Most particles possess a kind of intrinsic angular momentum that does not appear at all in classical mechanics and arises from Dirac's relativistic generalization of the theory. Mathematically it is described with spinors, fair play. In non-relativistic quantum mechanics the oul' group representations of the bleedin' Lie group SU(2) are used to describe this additional freedom. For an oul' given particle, the feckin' choice of representation (and hence the oul' range of possible values of the bleedin' spin observable) is specified by a non-negative number S that, in units of Planck's reduced constant ħ, is either an integer (0, 1, 2 ...) or a bleedin' half-integer (1/2, 3/2, 5/2 ...), would ye believe it? For a feckin' massive particle with spin S, its spin quantum number m always assumes one of the bleedin' 2S + 1 possible values in the feckin' set

${\displaystyle \{-S,-S+1,\ldots +S-1,+S\}}$

As a consequence, the bleedin' quantum state of a feckin' particle with spin is described by a bleedin' vector-valued wave function with values in C2S+1. Equivalently, it is represented by a bleedin' complex-valued function of four variables: one discrete quantum number variable (for the spin) is added to the oul' usual three continuous variables (for the position in space).

### Many-body states and particle statistics

The quantum state of a holy system of N particles, each potentially with spin, is described by a feckin' complex-valued function with four variables per particle, correspondin' to 3 spatial coordinates and spin, e.g.

${\displaystyle |\psi (\mathbf {r} _{1},\,m_{1};\;\dots ;\;\mathbf {r} _{N},\,m_{N})\rangle .}$

Here, the feckin' spin variables mν assume values from the feckin' set

${\displaystyle \{-S_{\nu },\,-S_{\nu }+1,\,\ldots ,\,+S_{\nu }-1,\,+S_{\nu }\}}$
where ${\displaystyle S_{\nu }}$ is the bleedin' spin of ν-th particle. Jaykers! ${\displaystyle S_{\nu }=0}$ for a feckin' particle that does not exhibit spin.

The treatment of identical particles is very different for bosons (particles with integer spin) versus fermions (particles with half-integer spin). The above N-particle function must either be symmetrized (in the bleedin' bosonic case) or anti-symmetrized (in the bleedin' fermionic case) with respect to the bleedin' particle numbers. If not all N particles are identical, but some of them are, then the feckin' function must be (anti)symmetrized separately over the variables correspondin' to each group of identical variables, accordin' to its statistics (bosonic or fermionic).

Electrons are fermions with S = 1/2, photons (quanta of light) are bosons with S = 1 (although in the vacuum they are massless and can't be described with Schrödinger mechanics).

When symmetrization or anti-symmetrization is unnecessary, N-particle spaces of states can be obtained simply by tensor products of one-particle spaces, to which we will return later.

### Basis states of one-particle systems

As with any Hilbert space, if a basis is chosen for the bleedin' Hilbert space of a system, then any ket can be expanded as a bleedin' linear combination of those basis elements. Symbolically, given basis kets ${\displaystyle |{k_{i}}\rangle }$, any ket ${\displaystyle |\psi \rangle }$ can be written

${\displaystyle |\psi \rangle =\sum _{i}c_{i}|{k_{i}}\rangle }$
where ci are complex numbers. In physical terms, this is described by sayin' that ${\displaystyle |\psi \rangle }$ has been expressed as a quantum superposition of the feckin' states ${\displaystyle |{k_{i}}\rangle }$, like. If the oul' basis kets are chosen to be orthonormal (as is often the feckin' case), then ${\displaystyle c_{i}=\langle {k_{i}}|\psi \rangle }$.

One property worth notin' is that the feckin' normalized states ${\displaystyle |\psi \rangle }$ are characterized by

${\displaystyle \langle \psi |\psi \rangle =1,}$
and for orthonormal basis this translates to
${\displaystyle \sum _{i}\left|c_{i}\right|^{2}=1.}$

Expansions of this sort play an important role in measurement in quantum mechanics. In particular, if the feckin' ${\displaystyle |{k_{i}}\rangle }$ are eigenstates (with eigenvalues ki) of an observable, and that observable is measured on the feckin' normalized state ${\displaystyle |\psi \rangle }$, then the feckin' probability that the result of the bleedin' measurement is ki is |ci|2. (The normalization condition above mandates that the feckin' total sum of probabilities is equal to one.)

A particularly important example is the bleedin' position basis, which is the oul' basis consistin' of eigenstates ${\displaystyle |\mathbf {r} \rangle }$ with eigenvalues ${\displaystyle \mathbf {r} }$ of the feckin' observable which corresponds to measurin' position.[i] If these eigenstates are nondegenerate (for example, if the oul' system is a holy single, spinless particle), then any ket ${\displaystyle |\psi \rangle }$ is associated with a complex-valued function of three-dimensional space

${\displaystyle \psi (\mathbf {r} )\equiv \langle \mathbf {r} |\psi \rangle .}$
[k] This function is called the wave function correspondin' to ${\displaystyle |\psi \rangle }$. Similarly to the feckin' discrete case above, the bleedin' probability density of the oul' particle bein' found at position ${\displaystyle \mathbf {r} }$ is ${\displaystyle |\psi (\mathbf {r} )|^{2}}$ and the feckin' normalized states have
${\displaystyle \int d^{3}\mathbf {r} \,|\psi (\mathbf {r} )|^{2}=1.}$
In terms of the bleedin' continuous set of position basis ${\displaystyle |\mathbf {r} \rangle }$, the bleedin' state ${\displaystyle |\psi \rangle }$ is:
${\displaystyle |\psi \rangle =\int d^{3}\mathbf {r} \,\psi (\mathbf {r} )|\mathbf {r} \rangle .}$

### Superposition of pure states

As mentioned above, quantum states may be superposed. Listen up now to this fierce wan. If ${\displaystyle |\alpha \rangle }$ and ${\displaystyle |\beta \rangle }$ are two kets correspondin' to quantum states, the ket

${\displaystyle c_{\alpha }|\alpha \rangle +c_{\beta }|\beta \rangle }$
is a different quantum state (possibly not normalized). Sufferin' Jaysus. Note that both the feckin' amplitudes and phases (arguments) of ${\displaystyle c_{\alpha }}$ and ${\displaystyle c_{\beta }}$ will influence the resultin' quantum state. In other words, for example, even though ${\displaystyle |\psi \rangle }$ and ${\displaystyle e^{i\theta }|\psi \rangle }$ (for real θ) correspond to the feckin' same physical quantum state, they are not interchangeable, since ${\displaystyle |\phi \rangle +|\psi \rangle }$ and ${\displaystyle |\phi \rangle +e^{i\theta }|\psi \rangle }$ will not correspond to the feckin' same physical state for all choices of ${\displaystyle |\phi \rangle }$. However, ${\displaystyle |\phi \rangle +|\psi \rangle }$ and ${\displaystyle e^{i\theta }(|\phi \rangle +|\psi \rangle )}$ will correspond to the feckin' same physical state. This is sometimes described by sayin' that "global" phase factors are unphysical, but "relative" phase factors are physical and important.

One practical example of superposition is the feckin' double-shlit experiment, in which superposition leads to quantum interference. Sufferin' Jaysus. The photon state is a bleedin' superposition of two different states, one correspondin' to the bleedin' photon travel through the left shlit, and the feckin' other correspondin' to travel through the oul' right shlit, like. The relative phase of those two states depends on the bleedin' difference of the feckin' distances from the two shlits, like. Dependin' on that phase, the interference is constructive at some locations and destructive in others, creatin' the feckin' interference pattern. Whisht now and eist liom. We may say that superposed states are in coherent superposition, by analogy with coherence in other wave phenomena.

Another example of the oul' importance of relative phase in quantum superposition is Rabi oscillations, where the bleedin' relative phase of two states varies in time due to the feckin' Schrödinger equation. Jaysis. The resultin' superposition ends up oscillatin' back and forth between two different states.

### Mixed states

A pure quantum state is a holy state which can be described by a feckin' single ket vector, as described above, you know yerself. A mixed quantum state is a feckin' statistical ensemble of pure states (see quantum statistical mechanics).

Mixed states arise in quantum mechanics in two different situations: first, when the oul' preparation of the bleedin' system is not fully known, and thus one must deal with an oul' statistical ensemble of possible preparations; and second, when one wants to describe an oul' physical system which is entangled with another, as its state can not be described by a holy pure state. In the first case, there could theoretically be another person who knows the feckin' full history of the bleedin' system, and therefore describe the feckin' same system as a bleedin' pure state; in this case, the feckin' density matrix is simply used to represent the bleedin' limited knowledge of an oul' quantum state, be the hokey! In the oul' second case, however, the feckin' existence of quantum entanglement theoretically prevents the existence of complete knowledge about the feckin' subsystem, and it's impossible for any person to describe the feckin' subsystem of an entangled pair as a pure state.

Mixed states inevitably arise from pure states when, for a composite quantum system ${\displaystyle H_{1}\otimes H_{2}}$ with an entangled state on it, the oul' part ${\displaystyle H_{2}}$ is inaccessible to the observer. The state of the bleedin' part ${\displaystyle H_{1}}$ is expressed then as the feckin' partial trace over ${\displaystyle H_{2}}$.

A mixed state cannot be described with a single ket vector, that's fierce now what? Instead, it is described by its associated density matrix (or density operator), usually denoted ρ. Note that density matrices can describe both mixed and pure states, treatin' them on the bleedin' same footin'. Sufferin' Jaysus. Moreover, a holy mixed quantum state on a given quantum system described by a Hilbert space ${\displaystyle H}$ can be always represented as the bleedin' partial trace of a holy pure quantum state (called a feckin' purification) on a holy larger bipartite system ${\displaystyle H\otimes K}$ for a feckin' sufficiently large Hilbert space ${\displaystyle K}$.

The density matrix describin' a mixed state is defined to be an operator of the oul' form

${\displaystyle \rho =\sum _{s}p_{s}|\psi _{s}\rangle \langle \psi _{s}|}$
where ${\displaystyle p_{s}}$ is the bleedin' fraction of the bleedin' ensemble in each pure state ${\displaystyle |\psi _{s}\rangle .}$ The density matrix can be thought of as a way of usin' the oul' one-particle formalism to describe the feckin' behavior of many similar particles by givin' a holy probability distribution (or ensemble) of states that these particles can be found in.

A simple criterion for checkin' whether a bleedin' density matrix is describin' an oul' pure or mixed state is that the oul' trace of ρ2 is equal to 1 if the state is pure, and less than 1 if the feckin' state is mixed.[l][14] Another, equivalent, criterion is that the oul' von Neumann entropy is 0 for a holy pure state, and strictly positive for a feckin' mixed state.

The rules for measurement in quantum mechanics are particularly simple to state in terms of density matrices. Listen up now to this fierce wan. For example, the ensemble average (expectation value) of a measurement correspondin' to an observable A is given by

${\displaystyle \langle A\rangle =\sum _{s}p_{s}\langle \psi _{s}|A|\psi _{s}\rangle =\sum _{s}\sum _{i}p_{s}a_{i}|\langle \alpha _{i}|\psi _{s}\rangle |^{2}=\operatorname {tr} (\rho A)}$
where ${\displaystyle |\alpha _{i}\rangle }$ and ${\displaystyle a_{i}}$ are eigenkets and eigenvalues, respectively, for the feckin' operator A, and "tr" denotes trace, the shitehawk. It is important to note that two types of averagin' are occurrin', one bein' a weighted quantum superposition over the bleedin' basis kets ${\displaystyle |\psi _{s}\rangle }$ of the pure states, and the oul' other bein' a statistical (said incoherent) average with the feckin' probabilities ps of those states.

Accordin' to Eugene Wigner,[15] the concept of mixture was put forward by Lev Landau.[16][13]: 38–41

## Mathematical generalizations

States can be formulated in terms of observables, rather than as vectors in a vector space. Jesus, Mary and Joseph. These are positive normalized linear functionals on a feckin' C*-algebra, or sometimes other classes of algebras of observables. See State on a feckin' C*-algebra and Gelfand–Naimark–Segal construction for more details.

## Notes

1. ^ Sometimes written ">"; see angle brackets.
2. ^ To avoid misunderstandings: Here we mean that Q(t) and P(t) are measured in the oul' same state, but not in the same run of the oul' experiment.
3. ^
4. ^ i.e. Be the holy feck, this is a quare wan. separated by a zero delay. Story? One can think of it as stoppin' the bleedin' time, then makin' the oul' two measurements one after the other, then resumin' the bleedin' time. G'wan now. Thus, the bleedin' measurements occurred at the oul' same time, but it is still possible to tell which was first.
5. ^ For concreteness' sake, suppose that A = Q(t1) and B = P(t2) in the above example, with t2 > t1 > 0.
6. ^ Dirac (1958),[11] p. 20: "The bra vectors, as they have been here introduced, are quite a different kind of vector from the feckin' kets, and so far there is no connexion between them except for the existence of a feckin' scalar product of a feckin' bra and a feckin' ket."
7. ^ Dirac (1958),[11] p. 19: "A scalar product B|A now appears as an oul' complete bracket expression."
8. ^ Gottfried (2013),[12] p. 31: "to define the feckin' scalar products as bein' between bras and kets."
9. ^ Note that a state ${\displaystyle |\psi \rangle }$ is an oul' superposition of different basis states ${\displaystyle |\mathbf {r} \rangle }$, so ${\displaystyle |\psi \rangle }$ and ${\displaystyle |\mathbf {r} \rangle }$ are elements of the oul' same Hilbert space. A particle in state ${\displaystyle |\mathbf {r} \rangle }$ is located precisely at position ${\displaystyle \mathbf {r} =(x,y,z)}$, while a feckin' particle in state ${\displaystyle |\psi \rangle }$ can be found at different positions with correspondin' probabilities.
10. ^ Landau (1965),[13] p. G'wan now and listen to this wan. 17: "∫ Ψf Ψf* dq = δ(f′ − f)" (the left side corresponds to f|f′⟩), "δ(f′ − f) df′ = 1".
11. ^ In the feckin' continuous case, the oul' basis kets ${\displaystyle |\mathbf {r} \rangle }$ are not unit kets (unlike the feckin' state ${\displaystyle |\psi \rangle }$): They are normalized accordin' to ${\textstyle \int d^{3}\mathbf {r} '\,\langle \mathbf {r} |\mathbf {r} '\rangle =1,}$[j] i.e., ${\displaystyle \langle \mathbf {r} |\mathbf {r} '\rangle =\delta (\mathbf {r} '-\mathbf {r} )}$ (a Dirac delta function), which means that ${\displaystyle \langle \mathbf {r} |\mathbf {r} \rangle =\infty .}$
12. ^ Note that this criterion works when the density matrix is normalized so that the bleedin' trace of ρ is 1, as it is for the oul' standard definition given in this section. Whisht now and eist liom. Occasionally a feckin' density matrix will be normalized differently, in which case the oul' criterion is ${\displaystyle \operatorname {Tr} (\rho ^{2})=(\operatorname {Tr} \rho )^{2}}$

## References

1. ^ a b Weinberg, S. (2002), The Quantum Theory of Fields, vol. I, Cambridge University Press, ISBN 978-0-521-55001-7
2. ^ Griffiths, David J. (2004), Introduction to Quantum Mechanics (2nd ed.), Prentice Hall, ISBN 978-0-13-111892-8
3. ^ Holevo, Alexander S. (2001). Whisht now and eist liom. Statistical Structure of Quantum Theory. Lecture Notes in Physics, fair play. Springer. ISBN 3-540-42082-7. OCLC 318268606.
4. ^ Peres, Asher (1995), the hoor. Quantum Theory: Concepts and Methods. Listen up now to this fierce wan. Kluwer Academic Publishers. Sure this is it. ISBN 0-7923-2549-4.
5. ^ Rieffel, Eleanor G.; Polak, Wolfgang H. Whisht now and eist liom. (2011-03-04). Quantum Computin': A Gentle Introduction. G'wan now. MIT Press. ISBN 978-0-262-01506-6.
6. ^ Kirkpatrick, K. Sufferin' Jaysus listen to this. A, game ball! (February 2006). "The Schrödinger-HJW Theorem", fair play. Foundations of Physics Letters, to be sure. 19 (1): 95–102, enda story. arXiv:quant-ph/0305068. Bibcode:2006FoPhL..19...95K. Whisht now and eist liom. doi:10.1007/s10702-006-1852-1. Bejaysus here's a quare one right here now. ISSN 0894-9875. S2CID 15995449.
7. ^ "Statistical Mixture of States", would ye believe it? Archived from the original on September 23, 2019, you know yerself. Retrieved November 9, 2021.
8. ^ "The Density Matrix". Archived from the original on January 15, 2012. Retrieved January 24, 2012.
9. ^ Heisenberg, W. (1927). Holy blatherin' Joseph, listen to this. Über den anschaulichen Inhalt der quantentheoretischen Kinematik und Mechanik, Z, would ye swally that? Phys. 43: 172–198. Be the holy feck, this is a quare wan. Translation as 'The actual content of quantum theoretical kinematics and mechanics'. Here's a quare one. Also translated as 'The physical content of quantum kinematics and mechanics' at pp. 62–84 by editors John Wheeler and Wojciech Zurek, in Quantum Theory and Measurement (1983), Princeton University Press, Princeton NJ.
10. ^ Bohr, N. (1927/1928). The quantum postulate and the feckin' recent development of atomic theory, Nature Supplement April 14 1928, 121: 580–590.
11. ^ a b c Dirac, P.A.M. (1958). The Principles of Quantum Mechanics, 4th edition, Oxford University Press, Oxford UK.
12. ^ a b Gottfried, Kurt; Yan, Tung-Mow (2003). Stop the lights! Quantum Mechanics: Fundamentals (2nd, illustrated ed.), would ye swally that? Springer. Sufferin' Jaysus listen to this. ISBN 9780387955766.
13. ^ a b Lev Landau; Evgeny Lifshitz (1965), to be sure. Quantum Mechanics — Non-Relativistic Theory (PDF). Course of Theoretical Physics, fair play. Vol. 3 (2nd ed.). London: Pergamon Press.
14. ^
15. ^ Eugene Wigner (1962). "Remarks on the feckin' mind-body question" (PDF). In fairness now. In I.J. Good (ed.). The Scientist Speculates. C'mere til I tell ya now. London: Heinemann. Bejaysus this is a quare tale altogether. pp. 284–302. Footnote 13 on p.180
16. ^ Lev Landau (1927). Would ye swally this in a minute now?"Das Dämpfungsproblem in der Wellenmechanik (The Dampin' Problem in Wave Mechanics)". Zeitschrift für Physik. Here's another quare one. 45 (5–6): 430–441. C'mere til I tell ya now. Bibcode:1927ZPhy...45..430L, Lord bless us and save us. doi:10.1007/bf01343064, game ball! S2CID 125732617. English translation reprinted in: D. Ter Haar, ed. Chrisht Almighty. (1965). Would ye swally this in a minute now?Collected papers of L.D. Landau. Here's a quare one for ye. Oxford: Pergamon Press. p.8–18


text:
# 1.The core diameter of multimode step index fibre is 60 micrometer. The difference in refractive index is 0.013. The core refractive index is 1.46. Determine the number of guided modes when the operating wavelength is 0.75 micrometer.

KIPTANUI DOMINIC
25 Points
11 months ago
n1 = 1.473
n2 = 1.46.
Core Diameter = 60 μm.
Wavelength = 0·75 μm

NA=  $\sqrt{1.473^{2}}-{1.46^{2}}$

N·A= 0.195

No. of modes =      $\frac{1}{2}( \frac{pie * D}{Wavelenght} *N.A)^{2}$

No. of modes =         $\frac{1}{2}( \frac{\frac{22}{7}*D((0.75X10^{-6} ))}{(0.75X10^{-6})) } *0.195)^{2}$

No. of Modes = 1200

format: plain


text:
# Sir Michael Atiyah has died

Over the weekend we heard the sad news that mathematician Sir Michael Atiyah has passed away on 11th January. One of the few mathematicians to have been awarded both a Fields medal and an Abel prize, Atiyah leaves behind an extensive mathematical legacy and will be missed by many.

A tribute to former President of the Royal Society Sir Michael Atiyah OM FRS (1929 – 2019), on the Royal Society website

Mathematician Sir Michael Atiyah dies aged 89, at BBC News

Michael Atiyah, Mathematician in Newton’s Footsteps, Dies at 89 at the New York Times

# Carnival of Mathematics 165

The next issue of the Carnival of Mathematics, rounding up blog posts from the month of December, is now online at Ganit Charcha. The Carnival rounds up maths blog posts from all over the internet, including some from our own Aperiodical. See our Carnival of Mathematics page for more information.

# Mathematical Obituaries, December 2018

As 2018 drew to a close, we sadly lost several pioneering figures in mathematics. UPDATE: We have sadly also lost Rich Haberman, added 14/1/19.

# Happy Thirdsday!

Today is the third of January, and the third day of the year – and since this year it also falls on a Thursday, making for excellent pun opportunities, a group of mathematicians including Jim Propp, Evelyn Lamb, Zoe GriffithsBen Orlin, Matt Parker and several others have chosen to use today to celebrate the number $\frac{1}{3}$ (and in America, you’d even write the date as 1/3). Today is officially Thirdsday! Celebrate by:

I personally will be sketching the middle third Cantor set, as it’s my favourite fractal.

# Not Mentioned on The Aperiodical, 2018

We’ve had a bit of a break over the holidays, but mathematical news stops for no mince pie. From new prime numbers to mathematical doodling challenges, here’s a round-up of some of the facts/stories that we’ve seen while trying not to do any work.

# Particularly mathematical New Years Honours 2019

It’s that time of year when we take a look at the UK Government’s New Years Honours list for any particularly mathematical entries. Here is the selection for this year – any more, let us know in the comments and we’ll add to the list.  Tim Harford, journalist and presenter of BBC Radio programme More… Read more »

# Aperiodvent, Day 24: The Equation for a Merry Xmas

Merry Christmas, Aperi-readers. Please enjoy this pleasing algebraic manipulation, and we’ll see you when we’ve finished eating all the mince pies! This post is part of the Aperiodical’s 2018 Aperiodvent Calendar.


text:
# How to add a comma between author and year

I use the authoryear style in biblatex, and when I do for instance \cite{bauman2000}, I get:

Bauman 2000

But I would like to add a comma between the author and the year, in order to have

Bauman, 2000

Is there a simple command to get this change ?

• Welcome to TeX.SX! – egreg Sep 19 '13 at 10:20
• Welcome to this site. I hope you will have a wonderful time here. – Masroor Sep 19 '13 at 10:38

The space between the name and the year is controlled by the macro \nameyeardelim. So you have to redefine it:

\renewcommand*{\nameyeardelim}{\addcomma\space}

Here a small example:

\documentclass[]{article}
\usepackage[style=authoryear]{biblatex}
\begin{document}
\cite{companion}
\end{document}

The default definition can be found in the file biblatex.def. There are other useful definitions are also done.

• Thanks it works !! Is there a way to flag this question as "resolved" ? – Leo Sep 19 '13 at 11:47
• @Leo: Yes, there is a checkmark under the arrows. I am glad that it helps. – Marco Daniel Sep 19 '13 at 11:57
• The number of small changes I've had to add to biblatex in order to make it produce standard citations is amazing.. Makes me wonder if this is the best option anyway. – Doggie52 Mar 10 '15 at 13:52
• After adding the above line in my code, I got "... command already defined ..." error. I then edited corresponding line in the default definition file biblatex.def (in my system it is located in /usr/share/texlive/texmf-dist/tex/latex/biblatex/biblatex.def). Then it worked. Editing this file may require permission. – Sisay Chala Nov 4 '15 at 14:53
• In more recent versions of biblatex nameyeardelim is a context-sensitive delimiter, so \DeclareDelimFormat{nameyeardelim}{\addcomma\space} would be the preferred way to do the same thing as in this answer. – moewe 9 hours ago


text:
# Coxeter Groups¶

sage.combinat.root_system.coxeter_group.CoxeterGroup(data, implementation='reflection', base_ring=None, index_set=None)

Return an implementation of the Coxeter group given by data.

INPUT:

• data – a Cartan type (or coercible into; see CartanType) or a Coxeter matrix or graph
• implementation – (default: 'reflection') can be one of the following:
• 'permutation' - as a permutation representation
• 'matrix' - as a Weyl group (as a matrix group acting on the root space); if this is not implemented, this uses the “reflection” implementation
• 'coxeter3' - using the coxeter3 package
• 'reflection' - as elements in the reflection representation; see CoxeterMatrixGroup
• base_ring – (optional) the base ring for the 'reflection' implementation
• index_set – (optional) the index set for the 'reflection' implementation

EXAMPLES:

Now assume that data represents a Cartan type. If implementation is not specified, the reflection representation is returned:

sage: W = CoxeterGroup(["A",2])
sage: W
Finite Coxeter group over Integer Ring with Coxeter matrix:
[1 3]
[3 1]

sage: W = CoxeterGroup(["A",3,1]); W
Coxeter group over Integer Ring with Coxeter matrix:
[1 3 2 3]
[3 1 3 2]
[2 3 1 3]
[3 2 3 1]

sage: W = CoxeterGroup(['H',3]); W
Finite Coxeter group over Number Field in a with defining polynomial x^2 - 5 with a = 2.236067977499790? with Coxeter matrix:
[1 3 2]
[3 1 5]
[2 5 1]

We now use the implementation option:

sage: W = CoxeterGroup(["A",2], implementation = "permutation") # optional - gap3
sage: W                                                         # optional - gap3
Permutation Group with generators [(1,4)(2,3)(5,6), (1,3)(2,5)(4,6)]
sage: W.category()                       # optional - gap3
Join of Category of finite enumerated permutation groups
and Category of finite weyl groups
and Category of well generated finite irreducible complex reflection groups

sage: W = CoxeterGroup(["A",2], implementation="matrix")
sage: W
Weyl Group of type ['A', 2] (as a matrix group acting on the ambient space)

sage: W = CoxeterGroup(["H",3], implementation="matrix")
sage: W
Finite Coxeter group over Number Field in a with defining polynomial x^2 - 5 with a = 2.236067977499790? with Coxeter matrix:
[1 3 2]
[3 1 5]
[2 5 1]

sage: W = CoxeterGroup(["H",3], implementation="reflection")
sage: W
Finite Coxeter group over Number Field in a with defining polynomial x^2 - 5 with a = 2.236067977499790? with Coxeter matrix:
[1 3 2]
[3 1 5]
[2 5 1]

sage: W = CoxeterGroup(["A",4,1], implementation="permutation")
Traceback (most recent call last):
...
ValueError: the type must be finite

sage: W = CoxeterGroup(["A",4], implementation="chevie"); W     # optional - gap3
Irreducible real reflection group of rank 4 and type A4

We use the different options for the “reflection” implementation:

sage: W = CoxeterGroup(["H",3], implementation="reflection", base_ring=RR)
sage: W
Finite Coxeter group over Real Field with 53 bits of precision with Coxeter matrix:
[1 3 2]
[3 1 5]
[2 5 1]
sage: W = CoxeterGroup([[1,10],[10,1]], implementation="reflection", index_set=['a','b'], base_ring=SR)
sage: W
Finite Coxeter group over Symbolic Ring with Coxeter matrix:
[ 1 10]
[10  1]


text:
## African Diaspora Journal of Mathematics

### An Extension of Fuglede-Putnam Theorem for $w$-Hyponormal Operators

M. H. M. Rashid

#### Abstract

In this paper, we prove the following: assume that either (i) $T^*$ is $w$-hyponormal and $S$ is $w$-hyponormal such that $\ker(T^*)\subset\ker(T)$ and $\ker(S)\subset\ker(S^*)$ or (ii) $T^*$ is $p$-hyponormal or $\log$-hyponormal and $S$ is $w$-hyponormal such that $\ker(S)\subset\ker(S^*)$ or (iii) $T^*$ is an injective $w$-hyponormal and $S$ is a dominant holds. Then the pair $(T,S)$ satisfy Fuglede-Putnam theorem. Also, other related results are given.

#### Article information

Source
Afr. Diaspora J. Math. (N.S.), Volume 14, Number 1 (2012), 106-118.

Dates
First available in Project Euclid: 18 July 2013

Mathematical Reviews number (MathSciNet)
MR3080400

Zentralblatt MATH identifier
06227069

Subjects
Primary: 47B20; 47A10; 47A11

#### Citation

Rashid, M. H. M. An Extension of Fuglede-Putnam Theorem for $w$-Hyponormal Operators. Afr. Diaspora J. Math. (N.S.) 14 (2012), no. 1, 106--118. https://projecteuclid.org/euclid.adjm/1374153559

#### References

• A. Aluthge, On $p$-hyponormal operators for $0 < p < 1,$ Integral Equations Operator Theory 13(1990), 307–315.
• A. Aluthge, D. Wang, $w$-hyponormal operators, Integral Equations Operator Theory 36(2000), 1–10.
• T. Ando, Operators with a norm condition, Acta. Sci. Math. 33 (No.4)(1972), 359–365.
• A. Bachir and F. Lombaria, Fuglede-Putnnam theorem for $w$-hyponormal operators, Math. Ineq. Appl. 12 (2012), 777–786.
• S.K. Berberian, Extensions of a theorem of Fuglede and Putnam, Proc. Amer. Math. Soc. 71 (1978), 113-114.
• M. Ch$\bar{o},$ T. Huruya and Y. O. Kim, A note on $w$-hyponormal operators, J. Ineq. Appl. 7(2002), 1–10.
• R. G. Douglas, On majorization, factorization, and range inclusion of operators on Hilbert space, Proc. Amer. Math. Soc. 17 (1966), 413–415.
• B. P. Duggal, On generalised Putnam-Fuglede theorems, Mh. Math. 107 (1989), 309-332.
• B. P. Duggal, A remark on generalised Putnam-Fuglede theorems, Proc. Amer. Math. Soc. 129 (2001), 83-87.
• M. Fujii, D. Jung, S.-H. Lee, M.-Y. Lee and R. Nakamoto, Some classses of operators related to paranormal and log-hyponormal operators, Math. Japon. 51 (No.3) (2000), 395–402.
• T. Furuta, On the Class of Paranormal operators, Proc. Jaban. Acad. 43(1967), 594-598.
• T. Furuta, M. Ito and T. Yamazaki, A subclass of paranormal operators including class of $log$-hyponormal and several related classes, Sci. Math. 1(1998), 389–403.
• F. Hansen, An equality, Math. Ann. 246 (1980), 249–250.
• M. Ito and T. Yamazaki, Relations betweens two equalities $(B^{\frac{r}{2}}A^pB^{\frac{r}{2}})^{\frac{r}{r+p}}\geq B^r$ and $A^p\geq (A^{\frac{p}{2}}B^rA^{\frac{p}{2}})^{\frac{p}{r+p}}$ and their applications, Integral Equations Operator Theory 44(2002), 442–450.
• M. Ito, Some classes of operators associated with generalized Aluthge transformation, Sut J. Math. 35 (No.1)(1999), 149–165.
• I. H. Jeon, J.I. Lee and A. Uchiyama, On quasisimilarity for $\log$-hyponormal operator, Glasg. Math. J. 46(2004), 169–176.
• I. H. Jeon and B. P. Duggal, $p$-Hyponormal operators and quasisimilarity, Integral Equations and Operator Theory 49(No.3) (2004), 397–403.
• I.B. Jung, E. Ko and C. Pearcu, Aluthge transforms of operators, Integral Equations Operator Theory 37 (2000), 437–448.
• I.H. Kim, The Fuglede-Putnam theorem for $(p,k)$-quasihyponormal operators, J. Ineq. Appl. (2006), Article ID 47481, 1–7.
• M. H. Mortad, Yet More Versions of the Fuglede-Putnam Theorem, Glasg. Math. J. 51 (No.3)(2009), 473–480.
• M. H. Mortad, An All-Unbounded-Operator Version of the Fuglede-Putnam Theorem, Complex Anal. Oper. Theory 6 (No.6) (2012), 1269–1273.
• T. Okuyama and K. Watanabe, The Fuglede-Putnam Theorem and a Generalization of Barría's Lemma, Proc. Amer. Math. Soc. 126 (No.9) (1998), 2631–2634.
• G.K. Pedersen, Some operator monotone functions, Proc. Amer. Math. Soc. 36(1972), 309–310.
• M. Radjabalipour, An extension of Putnam-Fuglede theorem for hyponormal operators, Math. Z. 194(1987), 117-120.
• M.H.M. Rashid, M. S. M. Noorani and A. S. Saari, On the Spectra of Some Non-Normal Operators, Bulletin of the Malaysian Mathematical Sciences Society 31(No.2)(2008), 135–143.
• M.H.M. Rashid and H.Zguitti, Weyl type theorems and class $A(s,t)$ operators, Math. Ineq. Appl. 14 (No.3) (2011), 581-594.
• M. Rosenblum, On a Theorem of Fuglede and Putnam, J. Lond. Math. Soc. 33(1958), 376–377.
• J. G. Stampfli, B. L. Wadhwa, An asymmetric Putnam-Fuglede theorem for dominant operators, Indiana Univ. Math. 25 (No.4)(1976), 359–365.
• J. Stochel, An Asymmetric Putnam-Fuglede Theorem for Unbounded Operators, Proc. Amer. Math. Soc. 129 (No.8) (2001), 2261–2271.
• K. Takahashi, On the converse of Putnam-Fuglede theorem, Acta Sci. Math. (Szeged) 43(1981), 123–125.
• A. Uchiyama and K. Tanahashi, Fuglede-Putnam theorem for $p$-hyponormal or $\log$-hyponormal operators, Glasg. Math. J. 44 (2002), 397-410.
• A. Uchiyama and K. Tanahashi, On the Riesz idempotent of class $A$ operators, Math. Ineq. Appl. 5 (No.2) (2002), 291–298.
• L. R. Williams, Quasi-similarity and hyponormal operators, Integral Equations Operator Theory 5 (1981), 678-686.


text:
#### Archived

This topic is now archived and is closed to further replies.

# Opening Files: very weird error.

This topic is 5630 days old which is more than the 365 day threshold we allow for new replies. Please post a new topic.

## Recommended Posts

Hi! I have modified the function included in TOTWGPG to load bitmap files and today I discovered a very weird erro while I was using this function in my app. Here´s the section that´s giving me a headache: int LoadBitmapFile(BITMAP_FILE_PTR Bitmap,char *lpFileName,DDPIXELFORMAT &ddPixel) { int iFileHandle, iCounter; UCHAR *lpTempBuffer = NULL; OFSTRUCT FileData; memset(&FileData,0,sizeof(OFSTRUCT)); if(-1==(iFileHandle = OpenFile(lpFileName,&FileData,OF_READ))) { DWORD d = GetLastError(); return BMP_FAILED; } // More code. } // End LoadBitmapFile(). One function call might look like this: LoadBitmapFile(&Bitmap,"cdabc.bmp",ddPixel); The weird thing is that if I write for the name "cdab.bmp" (assuming there´s such a BMP, of course), the iFileHandle var inside LoadBitmapFile gets a proper value (different from -1). But if I pass "cdabc.bmp" (or "bitmap24.bmp", a longer name), iFileHandle gets the value -1. GetLastError() returned the error code 111 which, according to MSDN, is 111 The file name is too long. ERROR_BUFFER_OVERFLOW Does somebody have a clue of what´s going on in here? If you need more pieces of the code just ask me. Thanks in advance!!

##### Share on other sites
Try sticking with dos 8.3 format. Or write a new bmp loader.

##### Share on other sites
quote:
Original post by goldsword
Try sticking with dos 8.3 format.

The book has examples with files with more characters than 8.

quote:
Original post by goldsword
Or write a new bmp loader.

I don´t want to write another one, I´d like to know what´s going on here.

Anybody can help me? Thanks!

##### Share on other sites
It seems like every day 10 people have trouble with the code in this book. I wouldn''t trust it too much. You might want to look into CreateFile, fopen, or fstream to open the file. Maybe they''ll solve the problem.

How To Ask Questions The Smart Way | Google can help with your question | Search MSDN for help with standard C or Windows functions

##### Share on other sites
No, I think the code is really trustable. I´m pretty sure it´s me who´s making a silly mistake. Anyway, I´ll try to make it work in another way. Is it possible that some other function in the program is causing this problem? Thanks!

##### Share on other sites
From MSDN: "The OFSTRUCT structure contains a pathname string member whose length is limited to OFS_MAXPATHNAME characters. OFS_MAXPATHNAME is currently defined to be 128. Because of this, you cannot use the OpenFile function to open a file whose path length exceeds 128 characters. The CreateFile function does not have such a path length limitation."

##### Share on other sites
quote:
Original post by dalleboy
From MSDN: "The OFSTRUCT structure contains a pathname string member whose length is limited to OFS_MAXPATHNAME characters. OFS_MAXPATHNAME is currently defined to be 128. Because of this, you cannot use the OpenFile function to open a file whose path length exceeds 128 characters. The CreateFile function does not have such a path length limitation."

I don´t think "cdabc.bmp" exceeds 128 characters. The problem must be somewhere else.

##### Share on other sites
I can't tell you why it's not working, but I can tell you to switch over to CreateFile. MSDN says OpenFile is there to support 16-bit versions of Windows, which not a lot of people use anymore

[edited by - Zipster on August 19, 2002 3:58:27 AM]

##### Share on other sites
quote:
Original post by Ignacio Liverotti
I don''t think "cdabc.bmp" exceeds 128 characters. The problem must be somewhere else.

Read my post again, MSDN said: "you cannot use the OpenFile function to open a file whose path length exceeds 128 characters". And with path I think they mean the full path to the file including the drive, subdirectories and the file itself.

##### Share on other sites
Thanks for your posts! I´ll rewrite the function using CreateFile.

dalleboy: I think my path exceeded the 128 characters, because it is c:\archivos de programa\microsoft visual studio\... a lot more. When I read your first post I tought you referred to the file´s name.

Thanks for your answers again! During my next section (I have a dialup connection) I´ll tell you if CreateFile did the trick.

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Ok, I tried to rewrite the function using the Win32 File I/O functions, but there´s a little tiny thing I don´t understand about one of them: ReadFile(). Its 4th parameter:

MSDN says:

Pointer to the number of bytes read. ReadFile sets this value to zero before doing any work or error checking. If this parameter is zero when ReadFile returns TRUE on a named pipe, the other end of the message-mode pipe called the WriteFile function with nNumberOfBytesToWrite set to zero.
Windows 95 and Windows 98: This parameter cannot be NULL.

I read the explanation many times but I´m not sure about what to pass in this parameter. Can you help me with this one? Thanks a lot!

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quote:
Original post by Martee

Do I have to use that pointer later? Is it useful?

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Well in Windows 95 & 98 you need to pass a dummy variable. Regardless of whether you use it or not, a variable must be passed on those Windows versions. It could be useful later when debugging your code however.

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You pass a reference to a DWORD and the function stores the number of actual number of bytes read into your buffer. So a check on the function working correctly might look something like this

//error handling code here.

Hope this helps fatty
Aface

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Thanks for all the replies! The functions is working fine now.


text:
# Cylinder

Cylinder (right circular cylinder) is a solid of revolution generated by rotating a rectangle or square about one of its sides. Surface area of a right circular cylinder consists of 2 circles and one rectangle (or square). A solid circular cylinder can be seen as the limiting case of a n-gonal prism where n approaches infinity.

## Surface Area

The surface area of a right circular cylinder consists of the area of top and bottom base (which are equal) and side:

$$S=2\pi r^{2} + 2\pi rh$$

where,

π— Pi also referred to as Archimedes' constant is a mathematical constant, that is defined as the ratio of a circle's circumference to its diameter. It is approximately equal to 3.14159265359;
$$V=\pi r^{2} h$$


text:
# Finding a kernel for d-Bounded degree deletion

In $$d$$ Bounded degree deletion problem, we are given an undirected graph $$G$$ and a positive integer $$k$$, and the task is to find at most $$k$$ such vertices whose removal decreases the the maximum vertex degree of the graph to at most $$d$$.

The question is to how to find a polynomial kernel (in $$k$$ and $$d$$) for this problem.

I seem to be able to get the only reduction rule that if any vertex has degree $$> k+d$$, it has to be there in the deletion set (if the answer to instance is yes). Because if it isn't, then at least $$k+1$$ of its neighbors have to be in deletion set. I can't seem to move beyond this point.

The exercise is from this book (exercise $$2.9$$).

I am also aware that we can remove edges between vertices with degree $$< d$$, and find solution in the modified graph (hint from the book). But I am not sure how it will be useful, in getting a bound over number of vertices/edges in $$k$$ and $$d$$.

I would appreciate only hints if possible (something maybe beyond the book hints).

PS: for $$d=0$$ this reduces to vertex cover problem.

• Hint (not sure this is on the right track, but it's what I'd try): [EDIT: In the reduced problem in which the maximum degree is $k+d$:] Think about the maximum possible amount of "progress" any single deletion could make towards the goal. If you have a YES-instance, then you know that $k$ times that amount of progress (which is an upper bound on the most progress that $k$ deletions could make) must be sufficient to solve the problem. I think this should yield an upper bound on the vertex count. Oct 18, 2020 at 16:58
• Thanks j_random_hacker , for anyone looking have added answer to my question. Oct 21, 2020 at 15:57

Reduction Rule 1. Let $$V$$ be the set of vertices which are isolated. Convert the instance from $$I = (G,k,d)$$ to $$I^{'} = (G -V, k,d)$$. If $$I^{'}$$ is a yes instance, then so is $$I$$, because adding back the isolated vertices do not add on to the degree of other vertices. And isolated vertices have already have degree 0 ($$\le d$$ as $$d \ge 0$$). And if $$I$$ is a yes instance so is $$I^{'}$$, as $$G-V$$ is a sub-graph of $$G$$, and $$k$$ and $$d$$ remain same across the two instances.

Reduction Rule 2. If a vertex $$v$$ has degree $$\ge k+d+1$$ then it has to be kept in the deletion set. Otherwise $$k+1$$ of its neighbours have to be kept in the deletion set. Which cannot be done, as maximum size of deletion set is $$k$$. Thus after this reduction all vertices will have degree $$\le k+d$$.

Reduction Rule 3 If there are two vertices $$v$$ and $$w$$, such that degree of both of them is $$\le d$$. Then the edge $$vw$$ can be removed converting the instance from $$I = (G, k,d)$$ to $$I^{'} = (G-\{vw\}, k,d)$$. If $$I$$ is a yes instance so is $$I^{'}$$ as $$G-\{vw\}$$ is a sub-graph of $$G$$, with $$k$$ and $$d$$ remaining same across the instances. And if $$I^{'}$$ is a yes instance so is $$I$$, as adding back $$vw$$ can make the degree of $$v$$ and $$w$$ at most $$d$$ (as they initially had degree $$\le d$$).

Rule $$1$$ might be applied after applying Rule $$3$$ as well.

Now considering a graph where vertices with degree $$\le d$$ don't share an edge, no vertex is isolated and and all vertices have degree $$\le k+d$$ (ie. none of the above rules applied). Let $$A$$ be the set of vertices with degree $$\le d$$ and let $$B$$ be the set of vertices with degree $$> d$$. We can argue that $$|B| \le k(k+d) +k$$, otherwise the output is no. Because if $$|B| > k(k+d) +k$$, then as most $$k$$ elements will be chose from $$B$$ for the deletion set, and $$> k(k+d)$$ elements will be left (in $$B$$). And as each of them has degree $$> d$$, they each have at least one edge going into the deletion set. But the deletion set can only accommodate $$k(k+d)$$ edges. Also as each vertex in $$A$$ has at least one neighbor (and all of them in $$B$$), the maximum number of vertices in $$A$$, will be bounded as $$|A| \le |B|(k+d) \le (k(k+d)+k)(k+d) = k(k+d)(k+d+1)$$. So $$|A| + |B| \le k(k+d) + k(k+d)(k+d+1) = k(k+d)(k+d+2)$$.

• the above bound of |B| isn't clear to me, can you please explain it? I'm not sure why the fact that each vertex in B has degree > d, means they each have at least one edge going into the deletion set... thanks Apr 13, 2021 at 13:20
• Let's suppose we have a vertex $v$ with degree $>d$, with no edge going into the deletion set. Then by definition, this is not a deletion set leading to a contradiction. As after removing vertices in the deletion set each vertex left should have bounded degree of $<=d$. Apr 13, 2021 at 13:58
• @sashas any idea how to it on O(k)? (I think Expansion Lemma should be in used here) Jul 14, 2021 at 20:10


text:
# Appendicies in LateX

1. Apr 30, 2009

### philip041

I have written

\begin{appendix}
\section{name1}
blah blah
\section{name2}
\subsection{name3}
\end{appendix}

It calls them 'A name1' etc, I want it to write 'Appendix A: name1'

How do I get it to do this?

My class environment is article which I don't want to change because that will change other things like the numbering.

help!

2. Apr 30, 2009

### Dr Transport

You have to go into the style file and hack it to ensure that you get Appendix A etc..as opposed to A, B etc......

3. May 1, 2009

how?

4. May 1, 2009

### cristo

Staff Emeritus
Locate the article.cls file, find the line that defines the appendix layout, and amend it to what you want it to look like.

5. May 4, 2009

### turin

Noooo! First, COPY article.cls to a file with a different name and a .cls extension. Then, do the hacking.

LaTeX How to type something in $\LaTeX$ in the title itself? Jan 13, 2018


text:
How do I put the top of an image at the top of a page in LaTeX?

1. Nov 23, 2012

PhizKid

So I just have one line for the image:

Code (Text):
\includegraphics[width=10mm]{C:/Users/PhizKid/Desktop/image1.png}\\
And then some text to explain the image below it.

But when I try to print it out, approximately top half of the image is cut off and it runs off the top of the page.

How do I move the image down so that the top of the image is at the top of the page, but below the page number at the top of the page (where the text usually starts if you begin to type in a new page)?

I tried using vspace{50mm} and \\[50\baselineskip] before the image code, but the image doesn't move up or down at all.

Last edited: Nov 23, 2012


text: simple linear regression intercept: -48.996 In other words, it is an observation whose dependent-variable value is unusual given its values on the predictor variables. simple linear regression slope: 0.647 In this case, the data is linear and is compatible with the Linear Regression Algorithm. However, as noted in Section 9.1, the predicted responses, estimated slope coefficients, and hypothesis test results are not affected by the inclusion of the outlier. Then, I predict on both the datasets. Standardized residuals (sometimes referred to as "internally studentized residuals") are defined for each observation, i = 1, ..., n as an ordinary residual divided by an estimate of its standard deviation: $r_{i}=\frac{e_{i}}{s(e_{i})}=\frac{e_{i}}{\sqrt{MSE(1-h_{ii})}}$. While there is no solid mathematical definition, there are guidelines and statistical tests you can use to find outlier candidates. Code below helps you to plot and find potential outliers. RANSAC intercept: 0.004 Firstly, let’s plot Cook’s distance and the QQ Plot: Cook’s Distance. Copyright © 2018 The Pennsylvania State University In … How do outliers influence the least squares line in this plot? Real Life Examples of Various Distributions. In this post, I’ll explain what outliers are and why they are problematic, and present various methods for finding them. RANSAC slope: 1.0 Ordinary regression can be impacted by outliers in two ways: First, an extreme outlier in the y-direction at x-values near $\bar x$ can affect the fit in that area in the same way an outlier can affect a mean. Notwithstanding, a few newcomers will in general spotlight a lot on hypothesis and insufficient on commonsense application. After a regression line has been computed for a group of data, a point which lies far from the line (and thus has a large residual value) is known as an outlier.Such points may represent erroneous data, or may indicate a poorly fitting regression line. Finding outliers depends on subject-area knowledge and an understanding of the data collection process. In order to distinguish the effect clearly, I manually introduce extreme values to the original cars dataset. Therefore, the outlier in this case is not deemed influential (except with respect to MSE). Treating or altering the outlier/extreme values in genuine observations is not a standard operating procedure. Types of Outliers in Linear Regression. It doesn't have anything to do with what the response variable (Y) is; we just look at these points because they potentially have a significant impact on coefficient estimates and standard errors. The Hertzsprung–Russell diagram of stars plotted by luminosity and color. Previously in Lesson 4 we mentioned two measures that we use to help identify outliers. Instead, treat them simply as red warning flags to investigate the data points further. The approach is similar to that used in Example 1. RANSAC slope: 0.997 Outliers can be problematic because they can effect the results of an analysis. least trimmed squares intercept: -0.004, ideal slope: 1 Outliers: In linear regression, an outlier is an observation with large residual. least trimmed squares slope: 1.0 simple linear regression intercept: -1.503 Scatter Plot. Did you notice that the mean square error MSE is substantially inflated from 6.72 to 22.19 by the presence of the outlier? kanban. Let's try doing that to our Example #2 data set. Some statistical software flags any observation with a standardized residual that is larger than 2 (in absolute value). But, why should we? Outliers that fall horizontally away from the center of the cloud are called leverage points. All data sets have at least one outlier. All three of the other methods worked well, and LTS and Theil-Sen gave the best results for this specific data set and outlier type. For instance, the slope of a simple linear regression may significantly varies with just one outlier, whereas non-parametric tests such as the Wilcoxon test are usually robust to outliers. Data analytics is the study of dissecting crude data so as to make decisions about that data. Example 1. With an outlier free slope of 1 and intercept of 0, these are the results: ideal slope: 1 One easy way to learn the answer to this question is to analyze a data set twice—once with and once without the outlier—and to observe differences in the results. An alternative is to use studentized residuals. The good thing about standardized residuals is that they quantify how large the residuals are in standard deviation units, and therefore can be easily used to identify outliers: Using a cutoff of 2 may be a little conservative, but perhaps it is better to be safe than sorry. simple linear regression slope: 0.639 The output of the analysis is given in Figure 7. In the present worldwide commercial center, it isn't sufficient to assemble data and do the math; you should realize how to apply that data to genuine situations such that will affect conduct. To do so, click the Analyze tab, then Descriptive Statistics, then Explore: In the new window that pops up, drag the variable income into the box labelled Dependent List. machine learning projects for final year In case you will succeed, you have to begin building machine learning projects in the near future. For example, consider again the (contrived) data set containing n = 4 data points (x, y): The column labeled "FITS1" contains the predicted responses, the column labeled "RESI1" contains the ordinary residuals, the column labeled "HI1" contains the leverages hii, and the column labeled "SRES1" contains the standardized residuals. Let's take another look at the following data set (influence2.txt). Outliers in regression are observations that fall far from the “cloud” of points. How distant are the outliers from other observations? As you can see, the first residual (-0.2) is obtained by subtracting 2.2 from 2; the second residual (0.6) is obtained by subtracting 4.4 from 5; and so on. Density plot: To see the distribution of the predictor variable. Or, you can use the Q-Q plot to identify an outlier, which is introduced here. An outlier is an observation that lies abnormally far away from other values in a dataset. intervals. Obviously, the red-dashed line is not a good fit for these data. simple linear regression intercept: 8.915 From our regression model example, we can start investigating outliers observation by using Q-Q plot. replace a point in the sample with another point from the original pool of data, perform simple linear regression, and calculate the score; if it improved, keep the newpoint; repeat a bunch of times. Outlier Treatment. I'm taking sample data with a few different types of outliers, and calculating the slope and intercept using the following methods: Overall, simple linear regression resulted in noticeable errors for all three outlier types. Here, we see that the standardized residual for a given data point depends not only on the ordinary residual, but also the size of the mean square error (MSE) and the leverage hii. keep the sample from step 1 that gave you the best score. least trimmed squares slope: 1.0 Who Was the Best Running Back in Modern NFL History? Privacy and Legal Statements Projects assist you with improving your applied ML skills rapidly while allowing you to investigate an intriguing point. And the red dash line tells us how the regression model would look if we were to include the outliers. We can see that a plot of Cook’s distance shows clear outliers, and the QQ plot demonstrates the same (with a significant number of our observations not lying on the regression line). Figure $$\PageIndex{1}$$: Six plots, each with a least squares line and residual plot. ideal intercept: 0 One way to determine if outliers are present is to create a box plot for the dataset. Read article. There are six plots shown in Figure 8.3.3 along with the least squares line and residual plots. Theil-Sen estimator slope: 1.0 Types of outliers . This method is highly resistant to leverage points and is robust to outliers in the response. Example 2: Find any outliers or influencers for the data in Example 1 of Method of Least Squares for Multiple Regression. How to Handle Outliers in Regression Problems -- In this article, we discuss a general framework to drastically reduce the influence of outliers in most contexts. The column labeled "FITS1" contains the predicted responses, while the column labeled "RESI1" contains the ordinary residuals. In the program you will initially gain proficiency with the specialized skills, including R and Python dialects most usually utilized in data analytics programming and usage; Python Training in Chennai at that point center around the commonsense application, in view of genuine business issues in a scope of industry segments, for example, wellbeing, promoting and account. Outliers in regression are observations that fall far from the “cloud” of points. Linear Regression is without a doubt one of the most widely used machine algorithms because of the simple mathematics behind it and the ease with … Theil-Sen intercept: -0.004 How Do I Change A DataGrid Cell's Background Color Based On Its Value With MVVM In C#? Theil-Sen estimator slope: 1.0 These points are especially important because they can have a strong inﬂuence on the least squares line. In these cases, the outliers influenced the slope of the least squares lines. This method finds a line (plane or hyperplane) that minimizes a robust estimate of the scale (from which the method gets the S in its name) of the residuals. Your best option to use regression to find outliers is to use robust regression. If we regress y on x using the data set without the outlier, we obtain: And if we regress y on x using the full data set with the outlier, we obtain: What aspect of the regression analysis changes substantially because of the existence of the outlier? ‹ 9.2 - Using Leverages to Help Identify Extreme X Values, Lesson 1: Statistical Inference Foundations, Lesson 2: Simple Linear Regression (SLR) Model, Lesson 4: SLR Assumptions, Estimation & Prediction, Lesson 5: Multiple Linear Regression (MLR) Model & Evaluation, Lesson 6: MLR Assumptions, Estimation & Prediction, 9.1 - Distinction Between Outliers and High Leverage Observations, 9.2 - Using Leverages to Help Identify Extreme X Values, 9.3 - Identifying Outliers (Unusual Y Values), 9.5 - Identifying Influential Data Points, 9.6 - Further Examples with Influential Points, 9.7 - A Strategy for Dealing with Problematic Data Points, Lesson 12: Logistic, Poisson & Nonlinear Regression, Website for Applied Regression Modeling, 2nd edition, An observation with a standardized residual that is larger than 3 (in absolute value) is deemed by some to be an. In our previous look at this data set, we considered the red data point an outlier, because it does not follow the general trend of the rest of the data. In addition to the regression output being displayed in the output window, leverage values will be saved as an additional variable in your data set. It applies to problems such as clustering (finding centroids,) regression, measuring correlation or R-Squared, and many more. In Linear Fit, the outliers can be shown in results table by checking on the Outliers checkbox in the Linear Fit dialog. least trimmed squares intercept: -0.003, ideal slope: 1 ideal intercept: 0 Identifying outliers In Chapter 5, we will discuss how outliers can affect the results of a linear regression model and how we can deal with them. Learn how to quickly conduct a least-squares linear regression using DataGraph. You may also calculate the leverages using the SPSS menus: From the Analyze menu, select Regression, and then Linear. Furthermore, you can include projects into your portfolio, making it simpler to get a vocation, discover cool profession openings, and Final Year Project Centers in Chennai even arrange a more significant compensation. Theil-Sen estimator slope: 0.999 For now, it is enough to simply identify them and note how the relationship between two variables may change as a result of removing outliers. To better understand the implications of outliers better, I am going to compare the fit of a simple linear regression model on cars dataset with and without outliers. There are three ways we can find and evaluate outlier points: 1) Leverage points These are points with outlying predictor values (the X's). As you know, the major problem with ordinary residuals is that their magnitude depends on the units of measurement, thereby making it difficult to use the residuals as a way of detecting unusual y values. Any points that are outside these two lines are outliers. In (5), data with no clear trend were assigned a line with a large trend simply due to one outlier (!). So, it’s difficult to use residuals to determine whether an observation is an outlier, or to assess whether the variance is constant. randomly sample 60% of the points, perform simple linear regression on them, and repeat 20 times, keep the sample from step 1 that gave you the best score, replace a point in the sample with another point from the original pool of data, perform simple linear regression, and calculate the score; if it improved, keep the newpoint; repeat a bunch of times, 20% of points are all way-off in the same direction, 20% of points have large, random errors added to them, 1 point is massively off; error is 50x the total scale of the data. These points are especially important because they can have a strong influence on the least squares line. Figure 7 – Test for outliers and influencers for data in Example 2. We will use the following dataset in Excel to illustrate two methods for finding outliers: Method 1: Use the interquartile range However, this time, we add a little more detail. Let us see how to make each one of them. Theil-Sen intercept: -0.001 RANSAC intercept: -0.111 RANSAC slope: 1.03 QQ Plot. RANSAC intercept: -2.132 The development of artificial intelligence (AI) has propelled more programming architects, information scientists, and different experts to investigate the plausibility of a vocation in machine learning. Data analytics advances and procedures are generally utilized in business ventures to empower associations to settle on progressively Python Training in Chennai educated business choices. Types of outliers. Then click Statistics and make sure the box next to Percentiles is checked. The value of MSE is 0.40. least trimmed squares intercept: 0.001, I used python for all of this, and you can see and work with the code here. Let’s see how we can use a robust regression to mitigate for these outliers. The key here is not to take the cutoffs of either 2 or 3 too literally. Let's see what the standardized residual of the red data point suggests: Indeed, its standardized residual (3.68) leads this software to flag the data point as being an observation with a "Large residual.". Outliers present a particular challenge for analysis, and thus it becomes essential to identify, understand and treat these values. ideal intercept: 0 As you know, ordinary residuals are defined for each observation, i = 1, ..., n as the difference between the observed and predicted responses: For example, consider the following very small (contrived) data set containing n = 4 data points (x, y). least trimmed squares slope: 1.0 We can eliminate the units of measurement by dividing the residuals by an estimate of their standard deviation, thereby obtaining what are known as standardized residuals. An unusual value is a value which is well outside the usual norm. Background. simple linear regression slope: 0.97 Outliers in data can distort predictions and affect the accuracy, if you don’t detect and handle them appropriately especially in regression models. Why outliers detection is important? Nevada (28th observation) and Rhode Island (39th observation) are states that detected as potential outliers. This is a wonderful article, Given so much info in it, These type of articles keeps the users interest in the website, and keep on sharing more ... good luck. You can also exculde the outliers from an analysis routine. However, as noted in Section 9.1, the predicted responses, estimated slope coefficients, and hypothesis test results are not affected by the inclusion of the outlier. Having outliers in your predictor can drastically affect the predictions as they can easily affect the direction/slope of the line of best fit. Contact the Department of Statistics Online Programs. In this section, we identify criteria for determining which outliers are important and inﬂuential. Robust Regression and Outlier Detection is a book on robust statistics, particularly focusing on the breakdown point of methods for robust regression. With the outliers the line is pulled up and away from some of the observations in the larger group. An outlier may indicate a sample peculiarity or may indicate a data entry error or other problem. In order to determine if a point is inﬂuential, visualize the regression line with and without the point. Example 8.3.2. Recalling that MSE appears in all of our confidence and prediction interval formulas, the inflated size of MSE would thereby cause a detrimental increase in the width of all of our confidence and prediction  intervals. We sure spend an awful lot of time worrying about outliers. We need to find and graph the lines that are two standard deviations below and above the regression line. It was written by Peter Rousseeuw and Annick M. Leroy, and published in 1987 by Wiley. Ideally, a close to normal distribution (a bell shaped curve), without being skewed to the left or right is preferred. Therefore, the first standardized residual (-0.57735) is obtained by: r1 = −0.2 √0.4(1−0.7) =−0.57735 r 1 = − 0.2 0.4 ( 1 − 0.7) = − 0.57735. and the second standardized residual is obtained by: r2 = 0.6 √0.4(1−0.3) =1.13389 r 2 = 0.6 0.4 ( 1 − 0.3) = 1.13389. and so on. Therefore, the outlier in this case is not deemed influential (except with respect to MSE). What impact does their existence have on our regression analyses? They are: We briefly review these measures here. Those are the four stars with the lower temperatures. Therefore, the first standardized residual (-0.57735) is obtained by: $r_{1}=\frac{-0.2}{\sqrt{0.4(1-0.7)}}=-0.57735$. What Are the Most Common Scores in Baseball? and the second standardized residual is obtained by: $r_{2}=\frac{0.6}{\sqrt{0.4(1-0.3)}}=1.13389$. Without the outliers the regression line would be steeper, and lie closer to the larger group of observations. Outliers and Influential Observations. Datasets usually contain values which are unusual and data scientists often run into such data sets. High leverage points that actually inﬂuence the slope of the regression line are called inﬂuential points. So lets see how to detect and remove outliers from your data inPython using 1.5 IQR rule. The solid blue line shows us how the regression model would look if we were to ignore the outliers. The unusual values which do not follow the norm are called an outlier. Without the outlier there is no evident relationship between x and y. However, this method was also found to be inefficient. 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Will succeed, you can also exculde the outliers error or other problem observations in response. And the QQ plot: to see the distribution of the outlier } \ ) Six. Copyright © 2018 the Pennsylvania State University Privacy and Legal Statements Contact the Department of Online. 'S try doing that to our Example # 2 data set were to include the outliers the line best! Given in Figure 8.3.3 along with the outliers the line is pulled up away... Red dash line tells us how the regression model would look if we were to ignore the outliers Online.! Contact the Department of Statistics Online Programs find potential outliers or R-Squared, and published in by... The mean square error MSE is substantially inflated from 6.72 to 22.19 by the presence of the predictor variable is! At the following data set and Rhode Island ( 39th observation ) are states that detected as potential how to find outliers in linear regression... 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text:
## Any idea how to get something like this to work?

Question

I’m not sure, maybe it’s late but can someone help me figure out why this is not showing a link and instead just says

Restricted access. Role required. Make sure you read the #rules on theLink Textserver.

The code I’m using is:

\$result->error_message = esc_html__( 'Restricted access. Role required. Make sure you read the #rules on the' . 'Link Text' . 'server.', 'social-login' );

0
3 weeks 2021-07-14T03:31:08-05:00 0 Answers 0 views 0


text:
# Conservation of angular momentum of a bullet

1. Sep 1, 2013

### refrigerator

I was thinking of a couple basic mechanics problems lately. What if you have a rod sitting in space at rest, and you shoot a bullet at its center. Linear momentum is conserved, and the problem is quite trivial.

Now what if the bullet hit the rod slightly off of the rod's CG? I think linear momentum is conserved so the CG velocity of the rod is the same as with the problem above. However, I think the rod should also rotate.

So my question is, is angular momentum conserved here? Initially it seems the rod and bullet have purely linear motion, so no angular momentum, but after collision there clearly would be angular momentum.

I have thought about this but can't seem to figure out where the error in my reasoning is. Can somebody show me what's going on, or at least try to give me a fresh perspective?

Refrigerator

2. Sep 1, 2013

### D H

Staff Emeritus
Even a point mass can have angular momentum. Suppose you fire that bullet horizontally from a gun while standing upright. That bullet has non-zero angular momentum from the perspective of a reference frame with its origin at your feet.

When you are looking at angular momentum, you have to look at the total angular momentum of the system, not just that of spinning things like your rod.

3. Sep 2, 2013

### dauto

The bullet has orbital angular momentum even though it has no spin angular momentum

4. Sep 2, 2013

### Khashishi

Angular momentum is always conserved but it has a different value depending on where you put the origin. Even if the bullet is moving in a straight line, it still has angular momentum measured relative to a point that is displaced from the path of the bullet.

Usually you want to do this problem in the center of mass frame. If the bullet is hitting the end of the rod, the bullet flies some distance from the center of mass.


text:
# Solve this system of linear ODEs?

## Homework Statement

Solve this system of linear ODEs:

1) x''(t) = x + y
2) y''(t) = x + y

Just fyi, this is part of a much larger problem but I need to solve this system!

See above.

## The Attempt at a Solution

Okay so I think the most logical way to solve this would be to set x''(t) = y''(t).

Then
x''(t) = y''(t)
x' = y' + c1
x = y + c1t + c2

which implies

3) x'' = 2x - c1t - c2
4) y'' = 2y + c1t + c2

But I am not sure what to do from here. Apparently there should be 4 arbitrary constants in the final answer. But wouldn't solving x'' for x give you 2 and then solving y'' for y give you two more?

Thanks for any help.

Related Calculus and Beyond Homework Help News on Phys.org
cronxeh
Gold Member
Did you find the eigenvalues and eigenvectors? What is the general solution to this linear system of ODEs? Yes you will end up with 4 constants but thats the minor step

You could also write it as a vector equation
$$\vec x''(t) = \left(\begin{array}{cc} 1&1\\1&1\end{array}\right)\vec x(t)$$
and go on from there.


text:
Table 4

Algorithm assessment

Correct classificationsa 6,006 (91.4%) 5,539 (84.3%) 6,520 (99.2%)
Incorrect classificationsb 564 (8.6%) 1,031 (15.7%) 50 (0.8%)
Kappa statistic (κ0.89 0.79 0.99
Total number of classificationsc 6,570 6,570 6,570


text:
nLab Template page

This page is meant to provide a general example and template for new nLab-pages. You can look at its source code to see how the various parts are done. A minimal template is given first which can be copy-and-pasted into newly created pages. See HowTo for more details.

category: meta

Minimal Template Code

# Contents (or put a title here)
* this block creates the table of contents, leave as is
{: toc}

## Idea

(...)

## Definition

(...)

## Properties

(...)

## Examples

(...)

## References

(...)

A more detailed example follows. Check out the source code here to see how it’s coded:

Contents

Idea

It is an old observation that xyz. One notices that from the nPOV this is just an abc. This leads to the definition of a uvw. It is useful for doing klm and provides the basis for the more general theory of äöü.

Abstract

A uvw is effectively a uv together with a w. Its main property is encoded in Somebody’s Theorem which says that it consists of precisely three letters. The archetypical example of a uvw is $\mu \nu \omega$; details will be explained in the special examples paragraph.

Definition

As Jacques Distler said,

Definition

(uvw)

A uvw is a UVW in which all letters are lower case.

This may be summed up in the slogan:

A uvw is just what it looks like.

Properties

Lemma

Every uvw (Def. ) contains at least one letter.

By inspection.

Proposition

Every uvw contains strictly more than one letter.

Proof

Use the above lemma and continue counting:

(1)$1 + 1 = 2 \,.$

Theorem

Every uvw (Def. ) contains exactly three letters.

Proof

Along the lines of the above proposition, we use equation (1) and then conclude with

$2 + 1 = 3 \,.$

Notice that this is indeed independent of in which order we sum up the letters, in that the following diagram commutes:

$\array{ \mathbb{N}\times \mathbb{N} \times \mathbb{N} & \overset{Id \times + }{\longrightarrow} & \mathbb{N} \times \mathbb{N} \\ {}^{\mathllap{+ \times Id}} \big\downarrow && \big\downarrow^{\mathrlap{+}} \\ \mathbb{N} \times \mathbb{N} & \underset{+}{\longrightarrow} & \mathbb{N} } \,.$

Corollary

No uvw contains more than three letters.

Examples

Special cases

• First case

• Second case

• Third case

First person: I listed all of the special cases that I know above, but didn't Grothendieck study an important version too?

Second person: No, you're thinking of Lawvere. When I find the reference, I'll put it here.

Specific examples

For ease of reference, we will number the examples.

Example

The first example is obvious.

Example

The second example is a slight variation of Exp. .

Example

The third example is completely different from both Exp. .

References

The original definition appeared in section 3 of


text:
eta {ACTCD} R Documentation

## Ideal Response Patterns for all possible attribute profiles

### Description

This function is used to calculate ideal response patterns for all possible attribute profiles based on the DINA model (Junker & Sijtsma, 2001) or conjunctive-type cognitive diagnostic models.

### Usage

eta(K, J, Q)

### Arguments

 K The number of attributes. J The number of items. Q A required J \times K binary item-by-attribute association matrix (Q-matrix), where K is the number of attributes. The j^{th} row of the matrix is an indicator vector, 1 indicating attributes are required and 0 indicating attributes are not required to master item j.

### Value

A 2^K \times J binary matrix will be returned. Each row of ideal response patterns is corresponding to each of the 2^K possible attribute patterns, which can be obtained from alpha.

### References

Junker, B., & Sijtsma, K. (2001). Cognitive Assessment Models with Few Assumptions, and Connections with Nonparametric Item Response Theory. Applied Psychological Measurement, 25(3), 258-272.

alpha

### Examples

# Generating ideal response patterns
data(sim.Q)
K <- ncol(sim.Q)
J <- nrow(sim.Q)
IRP <- eta(K, J, sim.Q)

[Package ACTCD version 1.2-0 Index]


text:
# Write A Decimal That Is 1/10 Of 0.9

Write A Decimal That Is 1/10 Of 0.9. In the fraction $$\frac{9}{10}$$, the number 9. 0.9 = 0.91 step 2:

Write the given decimal in expanded form: Scroll down for detailed steps on how to convert 0.9 into a fraction. 0.9=9/10 0.9 is a decimal fraction.

### To Convert Fractions Into Decimals You Multiply The Denominator Until You.

So its 90/100 so in decimal form its. Steps to convert 0.9/10 to decimal steps: 0.9 1 = 0.9 x.

### Convert 0.9 10 As A Decimal?

Expand the denominator to be a power of 10. A decimal that is 1/10 of 0.9 is: 0.9=9/10 0.9 is a decimal fraction.

### Multiply Both Top And Bottom By 10 For Every Number After The Decimal Point:

Write the following numbers in the place value table : 3/5 is expanded to 6/10 by multiplying the numerator by 2 and denominator by 2:. 0.9 = 0.90 = 0.900 to proove that let's change the decimals to fractions:

### To Convert 1 9/10 To Decimal You Need Simply Divide Numerator 9 By Denominator 10:

First things first, if you don't know what a numerator and a denominator are in a fraction, we need to recap that: Write the given decimal in expanded form: 0.9 ÷ 10 = 0.09.

### Just Divide The Top Of The Fraction By The Bottom, And Read Off The Answer!

To rewrite a percent as a decimal dropped the percent sign and multiply by 1 100 and we will get 0.9 How to convert fraction to decimal method #1. Writing decimals in expanded form.

Read:   An Individual Center Of Gravity Can Usually Be Found


text:
# Set of strings $S$ such that no string from $S$ is a substring of another one

Let $X$ be a subset of $\{0,1\}^*$ with the following property: for every pair of distinct strings $x_1$, $x_2$ from $X$

$x_1$ is not a substring of $x_2$ and $x_2$ is not a substring of $x_1$.

How much can be the log-density of $X$,

i.e. $\lim \frac{ \log X_n}{n}$, where $X_n$ is the cardinality of all strings of length $n$ in $X$?

I'm going to use $\log_2$ instead of $\log$ throughout; this only results in constant factor change (and the largest possible answer becomes $1$, which is convenient).
Fedor Petrov's answer implies that $\lim \frac{\log_2 X_n}{n} < 1$ for any $X$. We can, however, construct a family $X$ with $\lim \frac{\log_2 X_n}{n} > 1 - \varepsilon$ for any $\varepsilon > 0$. To do that, choose $k \in \mathbb{N}$, and let $Y_k \subset \{0, 1\}^*$ consist of all strings without $k$ consecutive zeroes. Finally, let $X = \{0^k1s10^k | s \in Y_k\}$. $X$ is substring-free since $0^k$ can only occur as a prefix or as a suffix, and $x, y \in X$ with $x$ containing $y$ would imply that $x$ contains $0^k$ in the middle.
For this particular $X$, we have $X_n = \Theta(\alpha_k^n)$, where $\alpha_k$ is the largest root of $x^k = x^{k - 1} + \ldots + 1$. It is straightforward that $\lim_{k \to \infty} \alpha_k = 2$, hence for a given $\varepsilon > 0$ choosing $k$ such that $\alpha_k > 2^{1 - \varepsilon}$ does the job.
• I think you meant to write $\log 2$ in many places instead of $2$. – Noah Stephens-Davidowitz Aug 20 '18 at 21:07
Even if we forbid one substring $a$, the density becomes 0: if $m$ is a length of $a$, then there exist at most $(2^m-1)^{\lceil n/m\rceil}$ strings of length $n$ without substring $a$.
• Sorry, I wanted to ask about $\lim \frac{\log X_n}{n}$ – Alexey Milovanov Aug 13 '18 at 11:08