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IMR Press / FBL / Volume 27 / Issue 5 / DOI: 10.31083/j.fbl2705155
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Open Access Original Research
Atypical Extracellular Action Potentials from Posteromedial Hypothalamus in Anesthetized Humans
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1 Clinical Neurophysiology, Hospital Universitario La Princesa, 28006 Madrid, Spain
2 Fundación de Investigación Biomédica, Hospital Universitario La Princesa, 28006 Madrid, Spain
3 PhD Program in Neuroscience, Autonoma de Madrid University-Cajal Institute, 28029 Madrid, Spain
Front. Biosci. (Landmark Ed) 2022, 27(5), 155; https://doi.org/10.31083/j.fbl2705155
Submitted: 9 February 2022 | Revised: 16 April 2022 | Accepted: 24 April 2022 | Published: 12 May 2022
This is an open access article under the CC BY 4.0 license.
Abstract

Background: We obtained microelectrode recordings from four patients with intractable aggressivity who underwent surgery at posteromedial hypothalamus under general anaesthesia. We described two general types of extracellular action potentials (EAPs): typical/canonical and atypical. Methods: We analysed 337 units and 67 traces, which were characterized by the mean action potential (mAP). For the first phase, depolarization and repolarization, we computed amplitudes (V${}_{\text{FP}}$, V${}_{\text{Dep}}$ and V${}_{\text{Rep}}$) and durations (d${}_{\text{FP}}$, d${}_{\text{Dep}}$ and d${}_{\text{Rep}}$), maximum and minimum values of the first derivative (dV${}_{\text{max}}$, dV${}_{\text{min}}$), and amplitude and duration ratios. Results: Most of the canonical mAPs were positive (81.1%). EAPs with atypical mean action potentials (amAPs) were recorded in 42/337 cases. Only 35.6% of mAPs showed 2 phases. We identified the following types: N1P1N2 (38.3%), P1N1 (35.9%), amAP (12.5%), P1P2N1 (12.2%), N1P1 (4.7%), P1N1P2 (4.1%) and N1N2P1 (3.2%). We can define the properties of canonical forms as those units with (i) at least two opposite phases; (ii) $V_{\text{Dep}}\in{}[1.2,2.7]\times{}|{}V_{\text{Rep}}|{}$ and strongly related by this function $V_{\text{Rep}}=-0.56\left(\pm{}0.01\right)V_{\text{Dep}}-1.83(\pm{}0.79)$; (iii) a very strong relationship between dV${}_{\text{max}}$ and dV${}_{\text{min}}$, given by the equation ${dV}_{\text{min}}=-0.91\left(\pm{}0.03\right){dV}_{\text{max}}-0.37(\pm{}0.12)$, both of which were included in the depolarization phase; (iv) related with V${}_{\text{Dep}}$ by the equation ${dV}_{\text{max}}=0.08\left(\pm{}0.001\right)V_{\text{Dep}}-0.28(\pm{}0.14)$; and (v) $d_{\text{Dep}}\textasciitilde{}0.38d_{\text{Rep}}$. However, the first phase does not pertain to the same dynamic process responsible for depolarization and repolarization. Conclusions: Atypical units are described here for the first time and are true EAPs that differ strikingly from canonical forms. To date, they have been observed only in the hypothalamus, but future research is needed to assess their existence in other brain structures.

Keywords
axons
capacitive current
deep brain stimulation
gap junctions
intractable aggressiveness
microelectrode recordings
sorting spikes
Figures
Fig. 1.
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### Limits

#### Understand the intuitive concept of the limit of a function

• Find $$\mathop {\lim }\limits_{x \to 5} \;\frac{{\sqrt {{x^2} + 5x} - \sqrt {50} }}{{\sqrt {{x^2} - 10} - \sqrt {15} }}$$.
• $$\begin{array}{c}\mathop {\lim }\limits_{x \to 5} \;\frac{{\sqrt {{x^2} + 5x} - \sqrt {50} }}{{\sqrt {{x^2} - 10} - \sqrt {15} }} = \mathop {\lim }\limits_{x \to 5} \;\frac{{(\sqrt {{x^2} + 5x} - \sqrt {50} )(\sqrt {{x^2} + 5x} + \sqrt {50} )(\sqrt {{x^2} - 10} + \sqrt {15} )}}{{(\sqrt {{x^2} - 10} - \sqrt {15} )(\sqrt {{x^2} + 5x} + \sqrt {50} )(\sqrt {{x^2} - 10} + \sqrt {15} )}}\\ = \mathop {\lim }\limits_{x \to 5} \;{\frac{{[({x^2} + 5x) - 50](\sqrt {{x^2} - 10} + \sqrt {15} )}}{{[({x^2} - 10) - 15](\sqrt {{x^2} + 5x} + \sqrt {50} )}}^{^{}}}\\ = \mathop {\lim }\limits_{x \to 5} \;{\frac{{({x^2} + 5x - 50)(\sqrt {{x^2} - 10} + \sqrt {15} )}}{{({x^2} - 25)(\sqrt {{x^2} + 5x} + \sqrt {50} )}}^{^{}}}\\ = \mathop {\lim }\limits_{x \to 5} \;{\frac{{(x + 10)(x - 5)(\sqrt {{x^2} - 10} + \sqrt {15} )}}{{(x + 5)(x - 5)(\sqrt {{x^2} + 5x} + \sqrt {50} )}}^{^{}}}\\ = \mathop {\lim }\limits_{x \to 5} \;{\frac{{(x + 10)(\sqrt {{x^2} - 10} + \sqrt {15} )}}{{(x + 5)(\sqrt {{x^2} + 5x} + \sqrt {50} )}}^{^{}}}\\ = {\frac{{(5 + 10)(\sqrt {{5^2} - 10} + \sqrt {15} )}}{{(5 + 5)[\sqrt {{5^2} + 5(5)} + \sqrt {50} ]}}^{^{}}}\\ = {\frac{{30\sqrt {15} }}{{20\sqrt {50} }}^{^{}}}\\ = \frac{{3\sqrt {15} }}{{2 \times 5\sqrt 2 }} \times {\frac{{\sqrt 2 }}{{\sqrt 2 }}^{^{}}}\\ = {\underline{\underline {\frac{{3\sqrt {30} }}{{20}}}} ^{^{}}}\end{array}$$

#### Find the limit of a function

• [codemissing] lim θ
• Find $$\mathop {\lim }\limits_{x \to 0} \;\frac{{\sin (1 - {e^{2x}})}}{{1 - {e^x}}}$$.
• $$\begin{array}{c}\mathop {\lim }\limits_{x \to 0} \;\frac{{\sin (1 - {e^{2x}})}}{{1 - {e^x}}} = \mathop {\lim }\limits_{x \to 0} \;\frac{{\sin (1 - {e^{2x}})}}{{1 - {e^x}}} \cdot \frac{{1 + {e^x}}}{{1 + {e^x}}}\\ = \mathop {\lim }\limits_{x \to 0} \;\frac{{(1 + {e^x})\sin (1 - {e^{2x}})}}{{1 - {e^{2x}}}}\end{array}$$

Let $$t = 1 - {e^{2x}}$$, then $$x = \frac{1}{2}\ln (1 - t)$$.
When $$x \to 0$$, $$t \to 0$$.

$$\begin{array}{c}\mathop {\lim }\limits_{x \to 0} \;\frac{{\sin (1 - {e^{2x}})}}{{1 - {e^x}}} = \mathop {\lim }\limits_{t \to 0} \;\frac{{[1 + {e^{\frac{1}{2}\ln (1 - t)}}]\sin t}}{t}\\ = \mathop {\lim }\limits_{t \to 0} \;[1 + {e^{\frac{1}{2}\ln (1 - t)}}] \cdot \mathop {\lim }\limits_{t \to 0} \;\frac{{\sin t}}{t}\\ = [1 + {e^{\frac{1}{2}\ln (1 - 0)}}] \cdot 1\\ = \underline{\underline 2} \end{array}$$

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# What could cause a difference in coloration between the hemispheres of a gas giant?

Would it be possible for a gas giant to have differently colored hemispheres?

E.g. the northern hemisphere appearing blue when looked at, while the southern hemisphere would appear brown? (or any other combination of coloration such as red & green, etc.)

What processes would cause such a world to be formed?

• this might help you out: en.wikipedia.org/wiki/Sudarsky%27s_gas_giant_classification the colour is affected by temperature so I don't think so – Styxal Apr 12 '17 at 8:17
• look at the bands of jupiter: each band goes in an opposite direction to the surrounding bands, suppose you have the northern hemisphere spinning clockwise and the southern hemisphere spinning anticlockwise then you may get different gasses in them therefore different colours – Alex Robinson Apr 12 '17 at 8:19
• @Cursed1701, how can the norhern half of a planet spin in different direction with respect to the southern half? – L.Dutch Apr 12 '17 at 8:30
• @L.Dutch a writers best friend - Handwavium – Alex Robinson Apr 12 '17 at 8:30
• @Gstestso I've had a go at trying to make your question a little more structured/clearer (in regard to both: title & body) - please feel free to roll back to your previous version should I have misunderstood your intent :) – dot_Sp0T Apr 12 '17 at 9:49

The planet should be visible in one color from a long distance according to @Styxal's answer.

On the other hand, when you are close enough to see clouds in its atmosphere then you have some other options:

• Jupiter has the Great Red Spot and we don't know exactly what is it.
• [Sci-fi] Imagine a gas which reacts with light from the star (chemically or like a laser) and this reaction causes it to change color for a short amount of time. An observer will see an area with a different color and a gradient around this area due to winds.
• [Alien] A very, very big factory which pollutes the stratosphere. It would be a spot of different color with a gradient around it.
• [Alien] Messaging system. Aliens change the color and the observer from another star could see it and make a decision - send supplies or an army
• you've made me think here, and something like light pollution could maybe change the colour of one side – Styxal Apr 12 '17 at 9:45

Some others have pointed out Sudarsky’s colors, but ignored that this is not the whole story:

The discrepancy [that neither Jupiter nor Saturn actually look like Sudarsky’s type I color] can be partially accounted for by taking into account non-equilibrium condensates such as tholin or phosphorus, which are responsible for the coloured clouds in the Jovian atmosphere, and are not modelled in the calculations.

So, a “non-equilibrium process” produces a strong color, and is confined to one hemisphere either because the colors vanish before it crosses the equator, and the process is associated with the location.

Consider that spots like the Great Red Spot can move around but can't cross tbe equator. Suppose that's true for air mass in general — the convection cells are separate for northern and southern. So, if something like a Big Red Spot was spewing tholins into the air, it would tint the entire hemisphere. But the mixing of air across the equator is much smaller in magnitude, and the color has a limited lifespan (either it disappears as molecules break down, or it returns to its proper deeper layer). So, it doesn’t color the whole planet.

Note that Saturn has an unusual feature at one pole. Something that exists at one pole only would be perfect for producing the color: either dreging it up from a deep layer, or producing molecules through some energetic process.

Another possibility for the latter would be the presence of a star bright in UV located in the north (or south) sky. This causes novel molecules to be produced where the star is visible. The slow production is balanced with the slow disintegration and slower cross-hemisphere mixing of the material.

A more exotic solution would be related to hyperfine splitting or shifting. Some chemical process happens slightly differently depending on whether it’s in a north or south magnetic field; this slight change gets magnified through shifting equilibrium etc.

Well, that seems difficult since the magnetic field alone won’t break the symmetry. But you don’t need anything, if the breaking is spontaneous. You have groups of interacting molecules that can settle into different possible equilibrium states. It happens to be in different states in north and south hemispheres, thanks to low mixing between them. It might be seasonal, with different temperature making the difference; so the summer and winter hemispheres will have characteristic colors.

If you have a planet which is tidally locked with its star (a configuration which would be slow to achieve especially if there are other planets, but not impossible), you can get massive temperature disparities. The problem here is that convection streams in the atmosphere would distribute the heat, minimising the temperature difference.

A $~200K$ temperature difference isn't extreme however. In this case the dark side would allow the chemical interactions to form ammonia clouds, at the rim in the twilight you would see water vapour clouds.

Gas giants are known for their long lasting storm patterns. It is possible that the kinds of storms of the light and dark side would disintegrate at the boundaries due to chemical changes, at which point there would be little convection between the two sides, and the sides would retain their relative temperatures.

This is all highly speculative, I have no concrete hypothesis or evidence to back this up, but I can't disprove the principle.

Your example specified north/south hemisphere, but perhaps you're open to east/west. This answer is weird and may need some refinement, I admit.

Suppose you have a moon in geosynchronous orbit so it is always over the same half of the planet. That moon would reflect light onto the one hemisphere. That light would shift the reflected color of the planet. If the gas of the planet is particularly reactive to, for example, UV light, the extra light from the moon might make a significant difference in the apparent color.

So what makes the moon glow? Well, a phosphorous moon in an oxygen atmosphere might do the trick. Our Earth moon is heavy with phosphorous. It lacks an atmosphere. But if it had one, you might get continuous discharge from the moon glow. https://www.thenakedscientists.com/get-naked/experiments/phosphorus-moon

I'm stretching a lot of science here to create plausible scenario. Some of the other commenters may be able to poke holes and refine it.

• Wouldn't the burning rocks be a temporary phenomenon and then get used up? – JDługosz Apr 12 '17 at 13:44
• @JDługosz Sure, but it could potentially burn for a long time... processes on the scale of moons can be very long... for example, look how long a moon has been raining on Saturn. – SRM Apr 12 '17 at 21:13

The simplest explanation would be seasons. The gas giant has a tilted rotation axis, so one pole is in sunlight all day long, the other in darkness, for six "months" (which could last several years).

This affects wind patterns and surface temperatures, and results in different compounds floating to the visible surface.

This behaviour can become even more extreme if life enters the equation (you could have photosynthetic "algae"-equivalents flourishing during summer and hibernating throughout the winter).

Styxal's answer is but one part of the story. The colouration is however not affected only by temperature but also by chemical composition.

This means that if through some means you could have your planet have a markedly different chemical balance in the northern as compared to the southern hemisphere (possibly combined with different temperatures) you might end up with a planet that appears to have different colours between the hemispheres.

However, this is harder than you might think. How are you going to prevent diffusion of parts of the atmosphere, eventually homogenising the entire ball into one more or less consistent chemical cocktail?

You'd need some very strange wind patterns near the equator, and going down very deep into the atmosphere to prevent that happening, patterns I'm far from convinced could ever develop long term in nature even on a far smaller scale than that.

For a gaseous giant planet to be blue it would need an equilibrium temperature of between 350K(170 °F, 80 °C) and 800K(980 °F, 530 °C). To be red/brown the equilibrium temperature would have to be less than around 150K(−190 °F, −120 °C). This is because of the density of gases and how they refract light I guess.

With some hand-waving you could somehow make one side of your planet an ice giant planet such as Neptune, which has a different make-up of only 20% hydrogen and helium compared to gas giants which are (in the case of our very own Jupiter and Saturn) more than 90% hydrogen and helium. But ice giants are formed differently and you would expect the atmosphere to all just mix together so I don't know how that would work. Maybe with some of @Cursed1701's handwavium :-)

Sudarsky's gas giant classification (Wikipedia)

Ice Giant planets (Wikipedia)

I would also like to add that you could possibly use light pollution as an excuse if you intend to have your planet be inhabited.

• Neptune is clearly blue, and it's only 70 K. Why do you think that blue pigments can only exist at a certain temperature range? – JDługosz Apr 12 '17 at 12:57
• Neptune isn't a gas giant it's an ice giant @JDługosz – Styxal Apr 12 '17 at 13:02
• Can you cite a mechanism and explain in your answer? Why would certain pigments only exist in a specific temperature range, and what do you know about the composition of the OP’s planet? – JDługosz Apr 12 '17 at 13:09
• www8.nationalacademies.org/ssbsurvey/… Ice giants are only 20% hydrogen and helium compared to gas giants which are (in the case of Jupiter and Saturn) more than 90% hydrogen and helium. They're known as ice giants because their constituent compounds were ices when they were primarily incorporated into the planets during their formation, either directly in the form of ices or trapped in water ice. I suppose it could be possible to have one half of the planet be made differently though with some handwaving, like Cursed1701 said in the comments. – Styxal Apr 12 '17 at 13:26
• «, I don't study this or anything, I just spent a bunch of time on google before answering» that's normal for these kinds of subjects as we have around here! – JDługosz Apr 12 '17 at 13:43

## Axial Tilt

It wouldn't be a year-round solution, but if the planet's axial tilt was very far over (like Uranus's), then during each hemisphere's summer, it would receive almost the total amount of incoming sunlight. That could be enough difference to cause a visible change between the two hemispheres. The effect would lessen and vanish during the spring and fall, though, as it swapped hemispheres.

Giant Impact

What if the gas giant suffered an impact that discolored the cloud tops? Look at what happened with Comet Shoemaker–Levy 9 broke up and struck Jupiter over a several day period.

If your gas giant suffered a concentration of impacts on a much larger scale in one hemisphere you could get the effect you are looking for. The winds of the planet would likely keep the soot kicked up from the impacts segregated into one hemisphere. A bombardment severe enough could leave marks that are visible fro years.


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Interpolation and Semi-group on Hilbert and Hardy Spaces of Dirichlet Series Ali Mohamed Abu Oam

# Interpolation and Semi-group on Hilbert and Hardy Spaces of Dirichlet SeriesAli Mohamed Abu Oam

Abstract:  We investigate that a bounded sequence of points in the half plane $\sigma > \frac{1}{2}$ is an interpolating sequence for Hilbert space of ordinary Dirichlet series with square summable coefficiens. Also, in the half plane $\sigma > \frac{1}{2}$ we discuss the wold decomposition for the shift semi group on the Hardy space of square summable Dirichlet series. The local behavior of Hilbert space of Dirichlet series is considered, relative to the functions spaces on the infinite dimensional polydisk, and we generate some results for each case.

Keyword: Interpolating sequence,Hilbert and Hardy spaces, Dirichlet series and infinite dimensional polydisk.


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## Trigonometry (11th Edition) Clone

$-1.1817633$
Step 1: First, we have to convert degrees and minutes to decimal degrees: $-212^{\circ}12'=-212\frac{12}{60}^{\circ}=-212.2^{\circ}$ Step 2: Since $\sec\theta=\frac{1}{\cos\theta}$, $\sec(-212.2)=\frac{1}{\cos(-212.2)}$ Step 3: Typing $\frac{1}{\cos(-212.2)}$ and solving, we find that: $\frac{1}{\cos(-212.2)}\approx-1.1817633$


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## définition - Kernel (algebra)

voir la définition de Wikipedia

Wikipedia

# Kernel (algebra)

In the various branches of mathematics that fall under the heading of abstract algebra, the kernel of a homomorphism measures the degree to which the homomorphism fails to be injective.[1][2] An important special case is the kernel of a matrix, also called the null space.

The definition of kernel takes various forms in various contexts. But in all of them, the kernel of a homomorphism is trivial (in a sense relevant to that context) if and only if the homomorphism is injective. The fundamental theorem on homomorphisms (or first isomorphism theorem) is a theorem, again taking various forms, that applies to the quotient algebra defined by the kernel.

In this article, we first survey kernels for some important types of algebraic structures; then we give general definitions from universal algebra for generic algebraic structures.

## Survey of examples

### Linear operators

Let V and W be vector spaces and let T be a linear transformation from V to W. If 0W is the zero vector of W, then the kernel of T is the preimage of the singleton set {0W }; that is, the subset of V consisting of all those elements of V that are mapped by T to the element 0W. The kernel is usually denoted as "ker T ", or some variation thereof:

$\mathop{\mathrm{ker}}\, T := \{\mathbf{v} \in V : T\mathbf{v} = \mathbf{0}_{W}\}\mbox{.} \!$

Since a linear transformation preserves zero vectors, the zero vector 0V of V must belong to the kernel. The transformation T is injective if and only if its kernel is only the singleton set {0V }.

It turns out that ker T is always a linear subspace of V. Thus, it makes sense to speak of the quotient space V /(ker T ). The first isomorphism theorem for vector spaces states that this quotient space is naturally isomorphic to the image of T (which is a subspace of W). As a consequence, the dimension of V equals the dimension of the kernel plus the dimension of the image.

If V and W are finite-dimensional and bases have been chosen, then T can be described by a matrix M, and the kernel can be computed by solving the homogeneous system of linear equations M v = 0. In this representation, the kernel corresponds to the null space of M. The dimension of the null space, called the nullity of M, is given by the number of columns of M minus the rank of M, as a consequence of the rank-nullity theorem.

Solving homogeneous differential equations often amounts to computing the kernel of certain differential operators. For instance, in order to find all twice-differentiable functions f from the real line to itself such that

x f ''(x) + 3f '(x) = f (x),

let V be the space of all twice differentiable functions, let W be the space of all functions, and define a linear operator T from V to W by

(T f )(x) = x f ''(x) + 3f '(x) - f (x)

for f in V and x an arbitrary real number. Then all solutions to the differential equation are in ker T.

One can define kernels for homomorphisms between modules over a ring in an analogous manner. This includes kernels for homomorphisms between abelian groups as a special case. This example captures the essence of kernels in general abelian categories; see Kernel (category theory).

### Group homomorphisms

Let G and H be groups and let f be a group homomorphism from G to H. If eH is the identity element of H, then the kernel of f is the preimage of the singleton set {eH }; that is, the subset of G consisting of all those elements of G that are mapped by f to the element eH . The kernel is usually denoted "ker f " (or a variation). In symbols:

$\mathop{\mathrm{ker}} f := \{g \in G : f(g) = e_{H}\}\mbox{.} \!$

Since a group homomorphism preserves identity elements, the identity element eG of G must belong to the kernel. The homomorphism f is injective if and only if its kernel is only the singleton set {eG}.

It turns out that ker f is not only a subgroup of G but in fact a normal subgroup. Thus, it makes sense to speak of the quotient group G /(ker f ). The first isomorphism theorem for groups states that this quotient group is naturally isomorphic to the image of f (which is a subgroup of H).

In the special case of abelian groups, this works in exactly the same way as in the previous section.

### Ring homomorphisms

Let R and S be rings (assumed unital) and let f be a ring homomorphism from R to S. If 0S is the zero element of S, then the kernel of f is the preimage of the singleton set {0S}; that is, the subset of R consisting of all those elements of R that are mapped by f to the element 0S. The kernel is usually denoted "ker f" (or a variation). In symbols:

$\mathop{\mathrm{ker}} f := \{r \in R : f(r) = 0_{S}\}\mbox{.} \!$

Since a ring homomorphism preserves zero elements, the zero element 0R of R must belong to the kernel. The homomorphism f is injective if and only if its kernel is only the singleton set {0R}.

It turns out that, although ker f is generally not a subring of R since it may not contain the multiplicative identity, it is nevertheless a two-sided ideal of R. Thus, it makes sense to speak of the quotient ring R/(ker f). The first isomorphism theorem for rings states that this quotient ring is naturally isomorphic to the image of f (which is a subring of S).

To some extent, this can be thought of as a special case of the situation for modules, since these are all bimodules over a ring R:

• R itself;
• any two-sided ideal of R (such as ker f);
• any quotient ring of R (such as R/(ker f)); and
• the codomain of any ring homomorphism whose domain is R (such as S, the codomain of f).

However, the isomorphism theorem gives a stronger result, because ring isomorphisms preserve multiplication while module isomorphisms (even between rings) in general do not.

This example captures the essence of kernels in general Mal'cev algebras.

### Monoid homomorphisms

Let M and N be monoids and let f be a monoid homomorphism from M to N. Then the kernel of f is the subset of the direct product M × M consisting of all those ordered pairs of elements of M whose components are both mapped by f to the same element in N. The kernel is usually denoted "ker f" (or a variation). In symbols:

$\mathop{\mathrm{ker}} f := \{(m,m') \in M \times M : f(m) = f(m')\}\mbox{.} \!$

Since f is a function, the elements of the form (m,m) must belong to the kernel. The homomorphism f is injective if and only if its kernel is only the diagonal set {(m,m) : m in M}.

It turns out that ker f is an equivalence relation on M, and in fact a congruence relation. Thus, it makes sense to speak of the quotient monoid M/(ker f). The first isomorphism theorem for monoids states that this quotient monoid is naturally isomorphic to the image of f (which is a submonoid of N).

This is very different in flavour from the above examples. In particular, the preimage of the identity element of N is not enough to determine the kernel of f. This is because monoids are not Mal'cev algebras.

## Universal algebra

All the above cases may be unified and generalized in universal algebra.

### General case

Let A and B be algebraic structures of a given type and let f be a homomorphism of that type from A to B. Then the kernel of f is the subset of the direct product A × A consisting of all those ordered pairs of elements of A whose components are both mapped by f to the same element in B. The kernel is usually denoted "ker f" (or a variation). In symbols:

$\mathop{\mathrm{ker}} f := \{(a,a') \in A \times A : f(a) = f(a')\}\mbox{.} \!$

Since f is a function, the elements of the form (a,a) must belong to the kernel.

The homomorphism f is injective if and only if its kernel is only the diagonal set {(a,a) : a in A}.

It turns out that ker f is an equivalence relation on A, and in fact a congruence relation. Thus, it makes sense to speak of the quotient algebra A/(ker f). The first isomorphism theorem in general universal algebra states that this quotient algebra is naturally isomorphic to the image of f (which is a subalgebra of B).

Note that the definition of kernel here (as in the monoid example) doesn't depend on the algebraic structure; it is a purely set-theoretic concept. For more on this general concept, outside of abstract algebra, see kernel of a function.

### Mal'cev algebras

In the case of Mal'cev algebras, this construction can be simplified. Every Mal'cev algebra has a special neutral element (the zero vector in the case of vector spaces, the identity element in the case of groups, and the zero element in the case of rings or modules). The characteristic feature of a Mal'cev algebra is that we can recover the entire equivalence relation ker f from the equivalence class of the neutral element.

To be specific, let A and B be Mal'cev algebraic structures of a given type and let f be a homomorphism of that type from A to B. If eB is the neutral element of B, then the kernel of f is the preimage of the singleton set {eB}; that is, the subset of A consisting of all those elements of A that are mapped by f to the element eB. The kernel is usually denoted "ker f" (or a variation). In symbols:

$\mathop{\mathrm{ker}} f := \{a \in A : f(a) = e_{B}\}\mbox{.} \!$

Since a Mal'cev algebra homomorphism preserves neutral elements, the identity element eA of A must belong to the kernel. The homomorphism f is injective if and only if its kernel is only the singleton set {eA}.

The notion of ideal generalises to any Mal'cev algebra (as linear subspace in the case of vector spaces, normal subgroup in the case of groups, two-sided ring ideal in the case of rings, and submodule in the case of modules). It turns out that although ker f may not be a subalgebra of A, it is nevertheless an ideal. Then it makes sense to speak of the quotient algebra G/(ker f). The first isomorphism theorem for Mal'cev algebras states that this quotient algebra is naturally isomorphic to the image of f (which is a subalgebra of B).

The connection between this and the congruence relation is for more general types of algebras is as follows. First, the kernel-as-an-ideal is the equivalence class of the neutral element eA under the kernel-as-a-congruence. For the converse direction, we need the notion of quotient in the Mal'cev algebra (which is division on either side for groups and subtraction for vector spaces, modules, and rings). Using this, elements a and a of A are equivalent under the kernel-as-a-congruence if and only if their quotient a/a is an element of the kernel-as-an-ideal.

## Algebras with nonalgebraic structure

Sometimes algebras are equipped with a nonalgebraic structure in addition to their algebraic operations. For example, one may consider topological groups or topological vector spaces, with are equipped with a topology. In this case, we would expect the homomorphism f to preserve this additional structure; in the topological examples, we would want f to be a continuous map. The process may run into a snag with the quotient algebras, which may not be well-behaved. In the topological examples, we can avoid problems by requiring that topological algebraic structures be Hausdorff (as is usually done); then the kernel (however it is constructed) will be a closed set and the quotient space will work fine (and also be Hausdorff).

## Kernels in category theory

The notion of kernel in category theory is a generalisation of the kernels of abelian algebras; see Kernel (category theory). The categorical generalisation of the kernel as a congruence relation is the kernel pair. (There is also the notion of difference kernel, or binary equaliser.)

## References

1. ^ Dummit, David S.; Foote, Richard M. (2004). Abstract Algebra (3rd ed.). John Wiley & Sons. ISBN 0-471-43334-9.
2. ^ Lang, Serge (2002). Algebra. Graduate Texts in Mathematics. Springer. ISBN 0-387-95385-X.

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text:
How do you solve x+3(x+4)=20?

May 31, 2016

You have to expand the equation.

Explanation:

$x + 3 x + 12 = 20$

$4 x + 12 = 20$

$4 x = 20 - 12$

$4 x = 8$

$x = 2$

Hopefully this helps!


text:
# Writing differences on a blackboard

The numbers 25 and 36 are written on a blackboard. At each turn, a player writes on the blackboard the (positive) difference between two numbers already on the blackboard, if this number does not already appear on the blackboard. The loser is the player who cannot write a number.

I tried but wasn't able to find any approach to this.

Original source appears to be: Mathematical Circles (Russian Experience), page 58.

• What's the question? "Which player has a winning strategy" maybe? – 2012rcampion Mar 30 at 17:54
• Hi and welcome to Puzzling SE! This looks like a puzzle you found elsewhere. For content you did not create yourself, proper attribution is required. If you have permission to repost this, please edit to include (at minimum) where it came from, then vote to reopen. Posts which use someone else's content without attribution are generally deleted. – Eagle Mar 30 at 18:24
• P.S. The wording of the above comment are taken from Rubio's answer here This wasn't added in the above comment as it was exceeding the word limit by 42 characters. – Eagle Mar 30 at 18:25
• (Thanks @Akari. I was able to find what seems to be the original source, and added it. $@\!$ all about everything, please be mindful of our attribution requirements here going forward. Thanks for contributing and welcome to Puzzling!) – Rubio Mar 30 at 21:28
• Has a correct answer been given? If so, please don't forget to $\color{green}{\checkmark \small\text{Accept}}$ it :) – Rubio Apr 2 at 18:11

The numbers $$25$$ and $$36$$ are coprime. This means that if we continually replace the largest of the two numbers by the (positive) difference of the two numbers, we are essentially performing the Euclidean algorithm for finding their GCD, and will eventually get a $$1$$. The sequence is $$36$$, $$25$$, $$36-25=11$$, $$25-11=14$$, $$14-11=3$$, $$11-3=8$$, $$8-3=5$$, $$5-3=2$$, $$3-2=1$$.
Once there is a $$1$$ on the board, you can repeatedly subtract it to fill in any gaps and eventually produce every number from $$1$$ to $$36$$. This shows that if you have two cooperating players, all number from $$1$$ to $$36$$ can be produced.
But this also happens when the game is played between two non-cooperating players. It is impossible to prevent any of the numbers appearing. If any number in the euclidean sequence is not on the board, then there are still moves available. So eventually $$1$$ must be produced, and then as long as there are missing numbers between $$1$$ and $$36$$, there is at least one move available. This means that regardless of what moves are played, all numbers $$1$$ to $$36$$ will appear. We started with $$2$$ numbers on the board, so the game ends after $$34$$ moves.

The result is that

The game always ends after $$34$$ moves, after which the first player cannot move and loses.

• Beat me by a minute, nice one :-) – Rand al'Thor Mar 30 at 18:04

The key fact is as follows:

The only time the game can end is when the numbers are in the form $$k,2k,3k,4k,\dots,ak$$ for some fixed $$a$$ and $$k\geq1$$. This is because, given two numbers $$m$$ and $$n$$ on the board, we can always apply repeated subtraction between them to reach their GCD, and then from there to reach every multiple of their GCD up to $$\text{max}(m,n)$$.

Therefore this particular game ends when the numbers on the board are

$$1,2,3,4,\dots,35,36$$. This will take a total of 34 moves, since there are two numbers at the start and a new one is written each time.

So the conclusion is

no matter how the game goes, the first player loses, since 34 is even.

• Beat me by about 3 seconds :) Oh, and Jaap beat the two of us as well. – Arnaud Mortier Mar 30 at 18:02

The Euclidean algorithm tells you that as long as $$1$$ (the gcd of $$25$$ and $$36$$) is not on the board, there are legal ways to continue the process. Now, from the moment when $$1$$ does appear on the board, no matter how long it took to get there, every positive number between $$1$$ and $$36$$ becomes reachable.
no matter what path is taken, the amount of steps before the game is over is $$34$$, one for each integer between $$1$$ and $$36$$, excluding the two already on the board at the beginning of the game.


text: Experience. Please use ide.geeksforgeeks.org, weight A numerical value, assigned as a label to a vertex or edge of a graph. A minimum spanning tree (MST) or minimum weight spanning tree is a subset of the edges of a connected, edge-weighted undirected graph that connects all the vertices together, without any cycles and with the minimum possible total edge weight. [15 points] Unicycles (1 part) Given a connected weighted undirected graph G = (V, E) having only positive weight edges containing exactly one cycle, describe an O (| V |) time algorithm to determine the minimum weight path from vertex s to vertex t. Given a undirected, connected and weighted graph, construct a minimum spanning tree out of it using Kruskal’s Algorithm. Kruskal(G, w) -- G: Graph; w: weights M := empty set make a singleton vertex set from each vertex in G sort the edges of G into non-decreasing order for i in 1 .. |V| - 1 loop (u, v) := next edge of G (from sorted order list) if sets containing u and v are different then add (u, v) to M merge vertex sets containing u … The idea is to use shortest path algorithm. Hence,If the heaviest edge belongs to MST then there exist a cycle having all edges with maximum weight. Weighted graphs may be either directed or undirected. G has a unique minimum spanning tree, if, for every cut of G, there is a unique minimum-weight edge crossing the cut.. Don’t stop learning now. Return a maximum weighted matching of the graph represented by the list of its edges. a weighted, undirected graph G and a positive integer k, we desire to ﬁnd k disjoint ... the graph. Writing code in comment? close, link This content is about implementing Prim’s algorithm for undirected weighted graph. Weighted Graphs In many applications, each edge of a graph has an associated numerical value, called a weight. Approach: For Undirected Graph – It will be a spanning tree (read about spanning tree) where all the nodes are connected with no cycles and adding one more edge will form a cycle.In the spanning tree, there are V-1 edges. Let (G,w) be an edge-weighted graph and let S⊂V. Specifically, for any n-node edge-weighted outerplanar graph G, we give an O(n)-time algorithm to obtain an O(n)-space compact representation Z(ℂ) for a minimum cycle basis ℂ of G.Each cycle in ℂ can be computed from Z(ℂ) in O(1) time per edge. Let be a connected undirected graph of 100 vertices and 300 edges. If e=ss is an S-transversal¯ edge with minimum weight, then there is a minimum-weight spanning tree containing e. Proof. That is, it is a spanning tree whose sum of edge weights is as small as possible. Advanced Math Q&A Library An undirected weighted graph G is given below: Figure 16: An undirected weighted graph has 6 vertices, a through f, and 9 edges. Vertex d is on the left. Below is the implementation of the above idea, edit Let G be any connected, weighted, undirected graph.. a weighted, undirected graph G and a positive integer k, we desire to ﬁnd k disjoint trees within G such that each vertex of G is contained in one of the trees and the weight of the largest tree is as small as possible. (See lecture 8, slide ~15). Usually, the edge weights are non-negative integers. Given a graph with distinct edge weights and a not-minimum ST, there always exist another ST of lesser total weight that differs only by one edge 0 What is the proof that adding an edge to a spanning tree creates a cycle? Question: Problem 3 (25 Points) Write A Program To Find Minimum Weight Cycle In An Undirected Weighted Graph The Input Is The Adjacency Matrix A Of The Graph. Given a weighted directed graph consisting of V vertices and E edges. a i g f e d c b h 25 15 10 5 10 20 15 5 25 10 The weight of an edge is often referred to as the "cost" of the edge. Algorithms to find shortest paths in a graph are given later. More generally, any edge-weighted undirected graph (not … 30, Sep 20. This work is licensed under Creative Common Attribution-ShareAlike 4.0 International For each possible simple cycle in a connected weighted graph G with distinct edge weights, the heaviest edge in the cycle does not belong to a MST of G. Bcz we can select a minimum weight edge from the cycle to be in MST. Approach: Depth First Traversal can be used to detect a cycle in a Graph. In set 2 | we will discuss optimize the algorithm to find a minimum weight cycle in undirected graph. In a directed graph, each edge has a sense of direction from u to v and is written as an ordered pair or u->v. E.g., if a graph has four fundamental cycles, we would have to iterate through all permutations of the bitstrings, 1100, 1110 and 1111 being 11 iterations in total. Advanced Math Q&A Library An undirected weighted graph G is given below: Figure 16: An undirected weighted graph has 6 vertices, a through f, and 9 edges. Suppose the graph has at least one cycle (choose one) . code. Given positive weighted undirected graph, find minimum weight cycle in it. Given a positive weighted undirected graph, find the minimum weight cycle in it. DFS for a connected graph produces a tree. Unemployment Benefits. Combining our main Theorem1.2with the results from previous work in Theorem1.1gives us new conditional lower bounds for fundamental graph problems. A MST is a subgraph consisting of all the nodes in the graph with one exclusive path from a node to every other one (no cycles) and having the minimum sum of all edges weight among all such subgraphs. For weighted graph G=(V,E), where V={v1,v2,v3,…..} Design an efficient algorithm to find a minimum-size feedback-edge set. Weighted graphs may be either directed or undirected. A directed spanning tree (DST) of Grooted at r, is a subgraph T of Gsuch that the undirected version of T is a tree and T contains a directed path from rto any other vertex in V. Let G = (V,E) be an undirected graph. This article is attributed to GeeksforGeeks.org. DFS for a connected graph produces a tree. The Minimum Spanning Tree of an Undirected Graph. The task is to print the cyclic path whose sum of weight is negative. Time Complexity: O( E ( E log V ) ) For every edge, we run Dijkstra’s shortest path algorithm so over all time complexity E2logV. A Minimum Spanning Tree is a spanning tree of a connected, undirected graph. a i g f e d c b h 25 15 10 5 10 20 15 5 25 10 The weight of an edge is often referred to as the "cost" of the edge. Vertez f is above and to the right of vertez d. Vertez e is below and to the right of vertez f, but above vertez d. Given a real-valued weight function : →, and an undirected (simple) graph , the shortest path from to ′ is the path = (,, …,) (where = and = ′) that over all possible minimizes the sum ∑ = − (, +). 4. 28, Feb 17. Given a directed and strongly connected graph with non-negative edge weights. A back edge is an edge that is from a node to itself (self-loop) or one of its ancestors in the tree produced by DFS. (A) No minimum weight spanning tree contains e. (B) There exists a minimum-weight spanning tree not containing e. (C) no shortest path, between any two vertices, can contain e. (D) None If There Is An Edge Between Vertex I To Vertex J, And Weight Of This Edge Is W, Then Ali, J] = A , I] = U If There Is No Edge Between I And J A [i, J = A , I] =-1. consider the example graph: the parallel edges can be moved, but the simple closed loops will remain the same). Kruskal's algorithm finds a minimum spanning forest of an undirected edge-weighted graph.If the graph is connected, it finds a minimum spanning tree. Vertez f is above and to the right of vertez d. Vertez e is below and to the right of vertez f, but above vertez d. The problem can be translated as: find the Minimum Spanning Tree (MST) in an undirected weighted connected Graph. Approach: Run a DFS from every unvisited node.Depth First Traversal can be used to detect a cycle in a Graph. the number of edges in the paths is minimized. If the edge is not present, then it will be infinity. Design an efficient algorithm to find a minimum-weight feedback-edge set (MWFES). We one by one remove every edge from graph, then we find shortest path between two corner vertices of it. Undirected Graph 195 Notes Amity Directorate of Distance & Online Education Now select next minimum-weight edge (N2, N6) but it creates cycle so we cannot add it in to minimum spanning tree, now select next-minimum cost edge (N3, N4) Now select next minimum-weight edge (N2, N7) Now select next minimum-weight edge (N4, N5). The problem can be translated as: find the Minimum Spanning Tree (MST) in an undirected weighted connected Graph. minimum_spanning_edges¶ minimum_spanning_edges (G, weight='weight', data=True) [source] ¶. and is attributed to GeeksforGeeks.org. A set F ⊆ E of edges is called a feedback-edge set if every cycle of G has at least one edge in F. Suppose that G is a weighted undirected graph with positive edge weights. Abstract. We define the mean weight of a cycle as the summation of all the edge weights of the cycle divided by the no. Let C be a cycle in a simple connected weighted undirected graph. 2 Picking a Favorite MST Consider an undirected, weighted graph for which multiple MSTs are possible (we know this means the edge weights cannot be unique). Examples: Minimum weighted cycle is : Minimum weighed cycle : 7 + 1 + 6 = 14 or 2 + 6 Detect a negative cycle in a Graph using Shortest Path Faster Algorithm. Here each cell at position M[i, j] is holding the weight from edge i to j. A set $F \subseteq E$ of edges is called a feedback-edge set if every cycle of $G$ has at least one edge in $F$. The total cost or weight of a tree is the sum of the weights of the edges in the tree. We one by one remove every edge from the graph, then we find the shortest path between two corner vertices of it. Vertex f is above and to the right of vertex d. Vertex e is below and to the right of vertex f, but above vertex d. Which of the above two statements is/are TRUE? ... Find minimum weight cycle in an undirected graph. generate link and share the link here. Vertex f is above and to the right of vertex d. Vertex e is below and to the right of vertex f, but above vertex … We consider the fundamental algorithmic problem of finding a cycle of minimum weight in a weighted graph. If the minimum of 3 value of the graph makes a cycle , just take next value to make MST. A minimum spanning tree is a subgraph of the graph (a tree) with the minimum sum of edge weights. We add an edge back before we process the next edge. ... how can a graph with 7 as its weight be a minimum spanning tree when there is a spanning tree with weight 6 ?? It connects all the vertices together with the minimal total weighting for its edges. 3When k is divisible by 3; slightly slower otherwise. We use cookies to provide and improve our services. acknowledge that you have read and understood our, GATE CS Original Papers and Official Keys, ISRO CS Original Papers and Official Keys, ISRO CS Syllabus for Scientist/Engineer Exam, Minimum Cost Path with Left, Right, Bottom and Up moves allowed, Interleaving of two given strings with no common characters, Find if a string is interleaved of two other strings | DP-33, String matching where one string contains wildcard characters, Dynamic Programming | Wildcard Pattern Matching | Linear Time and Constant Space, WildCard pattern matching having three symbols ( * , + , ? Output: Sort the nodes in a topological way. Let "e" be an edge of maximum weight on C Which of the following is TRUE? Here we will see how to represent weighted graph in memory. Articles about cycle detection: cycle detection for directed graph. Let r2V. Vertez d is on the left. We are unable to ﬁnd this problem in the graph partitioning literature, but we show that the problem is NP-complete. So, if the minimum spanning tree of G has weight w, the minimum spanning tree of G0has weight w + (jVj 1)M. (c)Negate all edge weights and apply the algorithm from the previous part. Given positive weighted undirected graph, find minimum weight cycle in it. That is, it is a spanning tree whose sum of edge weights is as small as possible. We assume that the weight of every edge is greater than zero. The graph can be considered as both weighted and unweighted, but I think it's better to consider it as unweighted if the goal is to find the cycle basis of minimal closed regions. By using our site, you brightness_4 A MST is a subgraph consisting of all the nodes in the graph with one exclusive path from a node to every other one (no cycles) and having the minimum sum of all edges weight … Given a positive weighted undirected graph, find the minimum weight cycle in it. Let be a connected undirected graph of 100 vertices and 300 edges. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above. A Minimum Spanning Tree is a spanning tree of a connected, undirected graph. ... Upper Triangular Adjacency Matrix of Weighted Undirected Graph. A cycle in a graph is an ordered set of vertices {v1,v2,...,vj} such that the graph ... has minimum weight among all spanning trees of G. Any weighted graph G has one or more minimum spanning trees. For an undirected graph G of unknown girth k, our algorithm returns with high probability a cycle of length at most 2k for even k and 2k + 2 for odd k, in time $$\mathcal{O}(n^{\frac 3 2} \sqrt {\log n }).$$ Thus, in general, it yields a $$2{\frac 23}$$ approximation. Minimum Weight (2‘+1)-Cycle in a directed weighted graph, Shortest Cycle in a directed weighted graph, Then, the Min Weight (2‘+1)-Clique Hypothesis is false. Implementation: Each edge of a graph has an associated numerical value, called a weight. Given a connected, undirected graph G=, the minimum spanning tree problem is to find a tree T= such that E' subset_of E and the cost of T is minimal. An undirected weighted graph G is given below: Figure 16: An undirected weighted graph has 6 vertices, a through f, and 9 edges. The weight of a minimum spanning tree of is 500. A minimum spanning tree (MST) or minimum weight spanning tree is a subset of the edges of a connected, edge-weighted undirected graph that connects all the vertices together, without any cycles and with the minimum possible total edge weight. The idea is to use shortest path algorithm. A graph is a set of vertices connected by edges. Weight of the spanning tree is the sum of all the weight of edges present in spanning tree. Given an undirected weighted graph, write an algorithm (code oriented pseudocode) that determines the smallest weight value, the number of edges in this graph with the smallest weight, and creates a queue as shown below. To store weighted graph using adjacency matrix form, we call the matrix as cost matrix. When the weight of each edge of is increased by five, the weight of a minimum spanning tree becomes _____. Graph is a non linear data structure that has nodes and edges.Minimum Spanning Tree is a set of edges in an undirected weighted graph that connects all the vertices with no cycles and minimum total edge weight.When number of edges to vertices is high, Prim’s algorithm is preferred over Kruskal’s. Within the representation of bitstrings, all possible cycles are enumerated, i.e., visited, if all possible permutations of all bitstrings with $$2 \le k \le N_\text{FC}$$, where $$k$$ is the number of 1s in the string, are enumerated. Given a positive weighted undirected complete graph with n vertices and an integer k, find the minimum weight Hamiltonian cycle of length k in it. It connects all the vertices together with the minimal total weighting for its edges. Solution using Depth First Search or DFS. Vertez d is on the left. 3. A directed graph is called weakly connected if replacing all of its directed edges with undirected edges produces a connected (undirected) graph. Given an undirected weighted graph G = (V,E) Want to ﬁnd a subset of E with the minimum total weight that connects all the nodes into a tree We will cover two algorithms: – Kruskal’s algorithm – Prim’s algorithm Minimum Spanning Tree (MST) 29 We add an edge back before we process next edge. a minimum-weight spanning tree are based on the fact that a transversal edge with minimum weight is contained in a minimum-weight spanning tree. A minimal spanning path in a graph is a path that contains all the vertices of a graph whose weight is the least among the spanning paths. Weighted graphs may be either directed or undirected. 6-10. ; union-find algorithm for cycle detection in undirected graphs. The weight of a subgraph is the sum of the weights of the vertices or edges within that subgraph. The weight or length of a path or a cycle is the sum of the weights or lengths of its component edges. Count the number of nodes at given level in a tree using BFS. I. G has a unique minimum spanning tree, if no two edges of G have the same weight. We give the first known optimal algorithm that computes a minimum cycle basis for any weighted outerplanar graph. In particular, we show that the minimum weight cycle problem in an undirected n-node graph with edge weights in {1,...,M} or in a directed n-node graph with edge weights in {-M,..., M} and no negative cycles can be efficiently reduced to finding a minimum weight triangle in an Theta (n)-node undirected graph with weights in {1,...,O (M)}. Examples: Minimum weighted cycle is : Minimum weighed cycle : 7 + 1 + 6 = 14 or 2 + 6 + 2 + 4 = 14 If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. of edges. Usually, the edge weights are nonnegative integers. See your article appearing on the GeeksforGeeks main page and help other Geeks.Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above. By using our site, you consent to our Cookies Policy. (A minimum spanning tree of a connected graph is a subset of the edges that forms a tree that includes every vertex, where the sum of the weights of all the edges in the tree is minimized. Computer Science Q&A Library An undirected weighted graph G is given below: Figure 16: An undirected weighted graph has 6 vertices, a through f, and 9 edges. Usually, the edge weights are non-negative integers. key point of [AR16] is that one can replace Minimum Weight 3-Cycle by Minimum Weight Cycle, and preserve the sparsity in the reduction. Lemma 4.4. Prove that for any weighted undirected graph such that the weights are distinct (no two edges have the same weight), the minimal spanning tree is unique. The weight of a minimum spanning tree of is 500. Attention reader! the MST. We also create novel reductions from Given a undirected, connected and weighted graph, construct a minimum spanning tree out of it using Kruskal’s Algorithm. 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Has a unique minimum spanning tree is the implementation of the edges in the tree of!: Sort the nodes in a tree ) with the minimum sum of edge weights is as small minimum weight cycle in an undirected weighted graph... Use cookies to provide and improve our services cycle, just take next value to MST. Topic discussed above using adjacency matrix of weighted undirected graph G and a positive weighted undirected graph and! S-Transversal¯ edge with minimum weight in a graph are given later same.... By the list of its edges a directed graph is called weakly connected replacing... Is an S-transversal¯ edge with minimum weight cycle in an undirected weighted.! And weighted graph using shortest path between two corner vertices of it its edges optimal algorithm computes! Whose eccentricity is equal to the radius of the weights of the idea! Edge-Weighted graph.If the graph, then we find shortest paths in a graph has an associated numerical value called! 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text:
# Hamiltonian describing energy transfer to bloch Electron from EM field

1. Nov 11, 2011

### dkin

Hi all,

I've a hamiltonian that describes the coupling of electrons in a crystal (bloch electrons) to an EM field described by a vector potential A

$$\mathscr{H} = \frac{e}{mc}\left[\mathbf{p}(-\mathbf{k}) \cdot \mathbf{A}(\mathbf{k}, \omega)\right]$$

$\mathbf{A}(\mathbf{k}, \omega)$ is the amplitude of the fourier component of the vector potential for the incident photons with wavevector $\mathbf{k}$ and frequency $\omega$.

$\mathbf{p}(-\mathbf{k})$ is the many particle momentum operator for the electrons with wavevector $-\mathbf{k}$

-----

Here are my questions if anyone would be so kind as to help.

1) How does this equation result in an energy value??

2) Can electrons with wave vector -k only couple with photons of wave vector k?? If so why?


text:
# ! LaTeX Error: Command \removefromreset already defined

There are ready files for preparing Latex document. I used .cls template file and it has below code:

\newcommand{\@removefromreset}[2]{{%
\expandafter\let\csname c@#1\endcsname\@removefromreset
\def\@elt##1{%
\expandafter\ifx\csname c@##1\endcsname\@removefromreset
\else
\noexpand\@elt{##1}%
\fi}%
\expandafter\xdef\csname cl@#2\endcsname{%
\csname cl@#2\endcsname}}}

However, when I run my .tex file, an error occurs "! LaTeX Error: Command \removefromreset already defined"

• Welcome to TeX.SE! The code snippet you showed can't produce that error because it defines \@removefromreset, with an @. Please show us the complete code which produces the error you mention. – Phelype Oleinik Jan 3 at 13:11
• It looks like somebody copied the remreset.sty code over to the class file. – user31729 Jan 3 at 15:04

Since the LaTeX release from 2018/04/01 the \@removefromreset command is part of the LaTeX kernel.

The code by D. Carlisle was in remreset.sty (its author is D. Carlisle) and has been incorporated into the kernel. With any newer LaTeX version since 2018/04/01 there must be a conflict with a copied over code from remreset.sty. remreset is now declared obsolete.

The only way to remove the error is to remove the \newcommand{\@removefromreset}... from the class file.

The loading of remreset is also discouraged: (From remreset.sty):

%... header omitted here

% \@removefromreset{footnote}{chapter}

% 2018: package obsoleted as the command is defined in the LaTeX format.

\@ifundefined{@removefromreset}{%
\def\@removefromreset#1#2{{%
\expandafter\let\csname c@#1\endcsname\@removefromreset
\def\@elt##1{%
\expandafter\ifx\csname c@##1\endcsname\@removefromreset
\else
\noexpand\@elt{##1}%
\fi}%
\expandafter\xdef\csname cl@#2\endcsname{%
\csname cl@#2\endcsname}}}%
}{%
\PackageWarningNoLine{remreset}
{The remreset package is obsolete:\MessageBreak
\noexpand\@removefomresset is defined}%
}

• I loaded remreset you've defined and erased the lines I've copied. It worked, thanks!! – user178777 Jan 3 at 19:14
• @user178777: I have not defined remreset. remreset is obsolete and the macro is contained in the LaTeX kernel for almost one year by now.... – user31729 Jan 3 at 19:30
• Oh I got it. I comment out only removefromreset lines I've mentioned and so there is no problem. – user178777 Jan 3 at 20:36

If you have access to your .cls and can change things, remove or comment out the "new" definition of \@removefromreset. If you don't have access, then you can capture the original definition of \@removefromreset before you load the class and then restore it afterwards:

\makeatletter
\let\old@removefromreset\@removefromreset % Store \@removefromreset
\let\@removefromreset\relax
\makeatother


text:
## Calculus (3rd Edition)

$$y=-6+11 e^{4t}$$
The general solution of $y'=4y+24=4(y+6)$ is $$y=-6+c e^{4t}.$$ When $y(0)=5$, then $5=-6+c$, i.e. $c=5+6=11$. Thus: $$y=-6+11 e^{4t}$$


text:
# Feynman Long division

Logic Level 4

$\LARGE{ \require{enclose} \begin{array}{rll} \phantom{0}\ \mathrm{\large7} \ \mathrm{x} \ \mathrm{x} \ \mathrm{x} && \\[-1pt] \mathrm{x} \ \mathrm{x} \ \mathrm{x} \ \enclose{longdiv}{ \mathrm{x} \ \mathrm{x} \ \mathrm{x} \ \mathrm{x} \ \mathrm{x} \ \mathrm{x} }\kern-.2ex \\[-1pt] \underline{ \mathrm{x} \ \mathrm{x} \ \mathrm{x} \ \phantom 0 \ \phantom 0 \ \phantom 0 } \\[-1pt] { \mathrm{x} \ \mathrm{x} \ \mathrm{x} \ \mathrm{\large7} \ \phantom0 \ \phantom0 \ }\kern-.2ex \\[-1pt] \underline{ \mathrm{x} \ \mathrm{x} \ \mathrm{x} \ \mathrm{x} \ \phantom 0 \ \phantom 0 } && \\[-1pt] { \mathrm{x} \ \mathrm{x} \ \mathrm{x} \ \mathrm{x} }\kern-.2ex \\[-1pt] \underline{ \mathrm{x} \ \mathrm{x} \ \mathrm{x} \ \mathrm{x} } && \\[-1pt] \mathrm{\large7} \end{array} }$

The above is a long division with most of the digits of any number hidden, except for the three 7's. Given that each of 0, 1, 2, ..., 9 was used at least once for the hidden digits, figure out all of the digits hiding and submit your answer as the value of the dividend (the 6-digit number being divided).

Details and Assumptions:

• Each $$\mathrm X$$ represents a single-digit integer.
• The leading (leftmost) digit of a number cannot be 0.
×


text:
+1 vote
184 views

Let $A$ be a $10 \times 10$ matrix with complex entries such that all its eigenvalues are non-negative real numbers, and at least one eigenvalue is positive. Which of the following statements is always false ?

1. There exists a matrix $B$ such that $AB-BA = B$.
2. There exists a matrix $B$ such that $AB-BA = A$.
3. There exists a matrix $B$ such that $AB+BA=A$.
4. There exists a matrix $B$ such that $AB+BA=B$.
| 184 views


text:
## From LIBOR to SOFR, the benchmark transition – Pt.4

Download as PDF SOFR On November 17, 2014, the Federal Reserve Bank established its own research group, the Alternative Reference Rates Committee, a commission composed by the major international banks, central counterparties and asset managers, with the support of the U.S. Department of Treasury,  the U.S. Commodity Futures Trading Commission (CFTC) and the Office of Financial Research (OFR). The group’s scopes were to identify a more resilient risk-free rate that would have substituted LIBOR in Read more…

## From LIBOR to SOFR, the benchmark transition – Pt.3

Download as PDF ICE LIBOR Nowadays, LIBOR is administered by ICE Benchmark Administration, which publishes the rate for seven tenors (overnight, 1-week, 1-month, 2-month, 3-month, 6-month and 1-year) in five currencies (Euro, U.S. Dollar, British Pound Sterling, Swiss Franc and Japanese Yen). Currently, twenty panel banks submit their estimates between 11:05 and 11:40 am. Each bank has its submission methodology, but it has to be compliant with the guidelines provided by The Wheatley Review of Read more…

## US Dollar Dominance and Its Future

Download as PDF Introduction The BIS triennial survey held every 3 years indicated most recently in 2019 that the USD was on 88% of all transaction. This makes it the world most used currency and the main reserve currency of centralbanks and transaction. However, another currency has recently made great gains in terms of adoption. The Chinese yuan. In this article we will investigate what defines a dominate world currency, the requirements to becoming one, Read more…

## The Automotive Industry: A Contemporary, Broad Picture

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## An Estimation of the Risk Premium as Function of the VIX

Download as PDF Introduction The expected excess market return, or equity risk premium, is an unobservable element crucial in many financial applications ranging from corporate finance to asset pricing. Over the years, many researchers have studied the expected market return by tackling the estimation from different perspectives leading Welch, Goyal (2008) and Campbell, Thompson (2008) to re-examine the performance of such indicators from a common standpoint. In the academic literature, the traditional predictors of the Read more…

## From LIBOR to SOFR, the benchmark transition – Pt.2

Download as PDF The First Reaction One of the first institutional proposals for a reform of LIBOR came from the Weathley Review of LIBOR. Commissioned in late June 2012 by the UK Chancellor of the Exchequer, it has been driven by Martin Weathley, former managing director of the Financial Services Authority (“FSA”) and former chief operating officer of Federal Conduct Authority (“FCA”). With a document called the “Initial Discussion Paper”, the Review asked market participants, Read more…

## From LIBOR to SOFR, the benchmark transition – Pt.1

Download as PDF Introduction Since the 2008 Crisis, several events have undermined the global trust in the major financial benchmarks. Above all, the London Interbank Offered Rate (LIBOR) had its reputation impaired, and its role weakened. The spike in its value after the collapse and bail-out of leading banks posed doubts about its applicability in derivatives pricing. After, the manipulation scandal raised concerns about its reliability. Eventually, the lowering in the transaction volumes of its Read more…

## Restructuring: Debtors in Distress

Download as PDF Introduction  Restructuring is the process of advising distressed companies (those in the midst of bankruptcy, or rapidly approaching it) in order to help them change their capital structure to get out of or avoid dissolution. This can be achieved by modifying their debt, operations or structure. Outcomes of this process range from distressed sales to refinancing scenarios, but this article will focus on bankruptcy procedures. Restructuring deals tend to differ from normal Read more…

## If volatility be rough with you, be rough with volatility

Download as PDF The purpose of this special report is to introduce the emerging class of rough volatility models. Rough volatility is a relatively new concept originating from the empirical observation that log-volatility essentially behaves as a fractional Brownian motion at any reasonable timescale. While rough volatility is found to have increased accuracy and predictive power with respect to existing models, its modelling peculiarities make numerical implementations more computationally expensive, limiting its practicality. Recent developments, Read more…

## Negative interest rate policy – friend or foe?

Download as PDF Introduction Interest rates have been persistently low across several advanced economies since the Global Financial Crisis. As short- and medium-term rates approached zero, central banks were compelled to turn to increasingly forceful measures in order to provide an adequate monetary stimulus. In 2012, the Danish National Bank was the first one to lower its key interest rate to the negative territory. In subsequent years, the ECB and the central banks of Sweden, Read more…

## Market Recap 08/03/2020

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Download as PDF BSIC has brought you some of the most interesting deals over the last week, we hope you’ll enjoy! Elliot Management builds a 4% stake in Twitter Deal Value: $1bn | Deal Type: Shareholder Activism | Date: 01-Mar-2020 | Nationality: US | Sector: Technology Paul Singer has succeeded in building a stake in Twitter worth more than$1bn, obtaining a share of about 4% in the social media company. The intention is clear: Singer wants Jack Dorsey, founder and actual CEO, Read more…

## Democratic Primaries: A markets perspective Part 2 – The Super Tuesday Follow-Up

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text:
# parallellism in Euclidean plane

Two distinct lines in the Euclidean plane are to each other if and only if they do not intersect, i.e. (http://planetmath.org/Ie) if they have no common point. By convention, a line is parallel to itself.

The parallelism of $l$ and $m$ is denoted

 $l\parallel m.$

Parallelism is an equivalence relation on the set of the lines of the plane. Moreover, two nonvertical lines are parallel if and only if they have the same slope. Thus, slope is a natural way of determining the equivalence classes of lines of the plane.

 Title parallellism in Euclidean plane Canonical name ParallellismInEuclideanPlane Date of creation 2013-03-22 17:12:38 Last modified on 2013-03-22 17:12:38 Owner pahio (2872) Last modified by pahio (2872) Numerical id 10 Author pahio (2872) Entry type Definition Classification msc 51-01 Synonym parallelism Synonym parallelism in plane Synonym parallelism of lines Related topic Slope Related topic ParallelPostulate Related topic ParallelCurve Related topic PerpendicularityInEuclideanPlane Defines parallel Defines parallel lines Defines parallelism