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# Coulomb's Law

## Coulomb's law without the electric field

Coulomb's law, named after Charles-Augustin Coulomb, is the fundamental law of electrostatic forces. It states that

The magnitude of the electrostatic force between two point charges is directly proportional to the magnitudes of each charge and inversely proportional to the square of the distance between the charges.

$F = \frac{q_1q_2}{4 \pi \epsilon\, r^2}$

Where:

$F \$ is the magnitude of the force exerted,
$q_1 \$ is the charge on one body,
$q_2 \$ is the charge on the other body,
$r \$ is the distance between them,
$\epsilon \$ is the permittivity of free space or permittivity of the vacuum. It is 8.854×10−12 C2 N-1 m-2 (also F m-1) (Many authors write this as $\epsilon_0$)
The occurrence of $4 \pi$ with the constant is related to some geometrical considerations that make some formulas (notably Maxwell's Equations) simpler. Ultimately it has to do with the fact that the surface area of a sphere is $4 \pi$. Much ink has been spilled over this issue—one sometimes sees equations in "rationalized SI units" instead of the "plain" SI units that we use here.

The force is in the direction of making like charges repel and unlike charges attract. It could be written in vector form as:

$\vec F_{2,1} = \frac{q_1q_2}{4 \pi \epsilon\, r^2} \hat r_{2,1}$

Where:

$F_{2,1}$ is the force vector acting on charge 2, coming from charge 1
$\hat r_{2,1}$ is the unit vector from particle 1 to particle 2, that is, the actual spatial vector divided by its length. Its length is of course $r$. The formula could be written as:
$\vec F_{2,1} = \frac{q_1q_2}{4 \pi \epsilon\, r^3} \vec r_{2,1}$
It is sometimes useful, in investigations of optics and capacitor dielectrics, to use a value of ε other than the standard "vacuum" value, since doing so can automatically handle the behavior of the dielectric. When this is done, it is common to use the symbol $\epsilon_0$ ("epsilon-nought") to denote the vacuum value, and $\epsilon$ to denote the modified value. When this is done, the "pure" formula is:
$F = \frac{q_1q_2}{4 \pi \epsilon_0\, r^2}$

## The Electric Field Defined

Physicists usually reformulate this in terms of an electric field, usually denoted $\vec{E}$, that every charged object creates and that acts on every other charged object. Viewed this way, Coulomb's law breaks into two parts. The first part says what the field is around every charged object:

$\vec E = \frac{q}{4 \pi \epsilon\, r^2} \hat r_{2,1}$

where $r$ and $\hat r$ give the position at which the field is measured, and $q$ is the charge on the object creating the field.

The second parts is:

$\vec F = q \vec E$

where $q$ in this case is the charge of the particle being acted upon.

## The electric field when many charges are present

### The Electric Field Near a Very Long Uniformly Charged Wire

As we have seen, the electric field around a point charge is spherically symmetric and inversely proportional to the square of the distance. The are two other geometrical configurations worth looking at.

If we have an "infinitely long" linear collection of uniformly distributed charge (that is, a long charged wire), we can determine the nearby electric field by integration. Let the charge per unit length be $\lambda$ coulombs per meter.

Integrating along a line charge to find the electric field

At a given point at distance $b$ from the wire, the contribution to the field from an infinitesimal section of wire of length $d\ell$ is:

$\frac{dq}{4 \pi \epsilon\, \mathcal R^2}=\frac{\lambda d\ell}{4 \pi \epsilon\, \mathcal R^2}\$

(Here, $\lambda$ is linear charge density, so that $\lambda d\ell$ is the charge on that section of wire.)

The component of that vector pointing perpendicularly away from the wire is:

$\frac{\lambda}{4 \pi \epsilon\, \mathcal R^2}\ \sin\theta d\ell = \frac{\lambda b}{4 \pi \epsilon\, \mathcal R^3} d\ell = \frac{\lambda b}{4 \pi \epsilon\, (b^2+\ell^2)^{3/2}} d\ell$

As we integrate over l from $-\infty$ to $+\infty$, the component of the field parallel to the wire cancels to zero, leaving:

$\frac{\lambda b}{4 \pi \epsilon} \int_{-\infty}^\infty\frac{d\ell}{(b^2+\ell^2)^{3/2}} = \frac{\lambda b}{4 \pi \epsilon} \frac{1}{b^2}\left.\frac{\ell}{\sqrt{b^2+\ell^2}}\right|_{-\infty}^\infty = \frac{\lambda}{2 \pi \epsilon\, b}$

The field points perpendicularly away from the wire, and is inversely proportional to the first power of the separation distance.

Does the wire need to be infinitely long? No; this is just an approximation to the infinite limit. The approximation is good as long as one is much closer to the wire than the wire's length.

### The Electric Field Near a Very Large Uniformly Charged Plane

Another very important geometrical configuration is an "infinitely large" flat plane with uniform charge distribution. We divide the plane into many thin parallel strips of width dl. If the charge density per unit area of the plane is $\sigma$ coulombs per square meter, each strip has a linear charge density of

$\lambda = \sigma d\ell\,$

coulombs per meter.

We use the result of the preceding section, and essentially the same diagram. The strips are now running into or out of the page / screen, and the strips' cross sections appear left-to-right on the diagram. The field at the point of interest is:

$\frac{\lambda}{2 \pi \epsilon\, \mathcal R} = \frac{\sigma}{2 \pi \epsilon\, \mathcal R} d\ell$

The upward component (by symmetry the total field will point perpendicularly out of the plane) is:

$\frac{\sigma}{2 \pi \epsilon\, \mathcal R}\ \sin\theta d\ell = \frac{\sigma R}{2 \pi \epsilon\, \mathcal R^2}\ d\ell = \frac{\sigma b}{2 \pi \epsilon (b^2+\ell^2)}\ d\ell$

The total upward component of the field is obtained by integrating:

$E = \frac{\sigma b}{2 \pi \epsilon} \int_{-\infty}^\infty\frac{d\ell}{b^2+\ell^2} = \frac{\sigma}{2 \pi \epsilon} \left.\tan^{-1}\frac{\ell}{b}\right|_{-\infty}^\infty = \frac{\sigma}{2 \epsilon}$

The field points perpendicularly away from the plane, and is independent of the separation distance. That is, it is independent as long as one stays so close to the plane that it appears to be nearly infinite.

#### Application: The electric field inside a parallel plate capacitor

The electric fields caused by the two plates of a parallel plate capacitor add in the region between the plates, and subtract in the region outside the plates.

Newton's laws assume that the net force is the sum of all forces, maFj (sum over j). Since the electric field, E, is defined through Coulomb's law via F=qE, the simplest possible assumption is that the electric field is the sum of the individual electric fields due to each charge. This is principle is called superposition (or linear superposition) and it holds at least down to the sizes smaller than the atomic nucleus. In fact, we have implicitly assumed superposition by integrating to obtain the electric field due to line and surface charges. As shown in the figure, the electric fields add constructively in the space between the plates. They add destructively (i.e. they subtract) outside the two plates and add to zero electric field. Therefore the electric field between the plates is

$E = \frac{\sigma}{ \epsilon}$ (in the limit that the plates are very large and/or very close together)

## General formulas for integrating over charge to find electric field

Consider a collection of $N$ particles of charge $Q_i$, located at points $\vec r_i$ (called source points), the electric field at $\vec r$ (called the field point) is:

$\vec{E}(\vec r) =\frac{1}{4\pi \epsilon}\sum_{i=1}^N \frac{\widehat\mathcal R_i Q_i}{|\mathcal\vec R_i |^2} = \frac{1}{4\pi \epsilon}\sum_{i=1}^N \frac{\vec\mathcal R_i Q_i}{|\mathcal\vec R_i |^3}$

where $\vec\mathcal R_i = \vec r - \vec r_i ,$ is the displacement vector from a source point $\vec r_i$ to the field point $\vec r$, and $\widehat\mathcal R_i = \vec\mathcal R_i / |\vec\mathcal R_i |$ is a unit vector that indicates the direction of the field. The fact that the force (and hence the field) can be calculated by summing over all the contributions due to individual source particles is an example of the superposition principle. The electric field produced by a distribution of charges is given by the volume charge density $\rho (\vec r)$ and can be obtained by converting this sum into a triple integral:

$\vec{E}(\vec r)= \frac {1}{4 \pi \epsilon} \iiint \frac {\vec r - \vec r \,'}{\left \| \vec r - \vec r \,' \right \|^3} \rho (\vec r \,')\operatorname{d}^3 r\,'$

If the charge confined to one or two dimensions, then this integral becomes a line or surface integral, respectively. Letting k=1/4πε:

$\vec{E}(\vec r) =k\sum_{i=1}^N \frac{\widehat\mathcal R_i Q_i}{|\mathcal\vec R_i |^2} \rightarrow k\int\frac{\hat\mathcal R dQ}{\mathcal R^2}$

When integrating, one convert this to a single, double, or triple integral by replacing:

$dQ\rightarrow \lambda d\ell\rightarrow \sigma dA\rightarrow \rho dV\;$

to represent a line integral , a surface integral or a volume integral $dV$, respectively. The symbols $(\lambda, \sigma, \rho)$ are commonly used to denote linear, surface, and volume density (or charge density), respectively.

### Three important displacement vectors used when integrating Coulomb's law

When writing such integrals, begin with a careful attention to how the three displacement vectors are to be represented:

1. $\vec r = (x, y)$ is the field point, i.e., where you want to evaluate the electric field. If no coordinates are specified, it is common to refer to this as a point, using a capital letter, such as $\vec E(P)$ or $\vec E_P$.
2. $\vec r\,' = (x', y')$ is the source point, which is the variable of integration. Alternatives include $(\bar x, \bar y)$, $(\tilde x, \tilde y)$ and $\vec\xi=(\xi_x, \xi_y)$.
3. $\vec \mathcal R$ points from the source point to the field point. To avoid defining a symbol for it, write it as the difference between two vectors, e.g., $\vec r-\vec r\,'=(x-x',y-y')$. In wikitext, one might write $(\mathcal R_x, \mathcal R_y)$ or $(\mathcal X, \mathcal Y, \mathcal Z)$ There is no lower case version of this font in wikitext; the lower case cursive r is often used when calculating by hand.


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Consider following compounds A to E:
If value of n is 4, then calculate value of here, p is total number of bond pair and q is total number of lone pair on central atoms of compound A to E 4.2k LIKES Now we can say that hybridization is sp3d3. XeF 6 is the strongest fluorinating agent of the series. By differentiating equation Y = [ε1c1 + ε2c2 + ε3c3 – ε1M(1 – x) + ε2Mx] l with respect to x, show that Y is a maximum when c3is maximum.? D) 1 lone pair. After hybridization XeF6 molecular geometry will be distorted octahedral or square bipyramidal. The lone pair around Xe is sterochemically inactive, allowing it to have this arrangement. XeF6 is non-polar with sp3d3 hybridisation and 1 lone pair and geometry distorted octahedral. sp2 0 lone pairs 120 degrees. Again, Xe has 8 electrons in its outer shell. Chemistry. (Organic chemistry A Level)? (don't forget to put that in terms of lone pairs). There are two pairs of bonded electrons and three lone pairs of electrons. Best Answer 100% (16 ratings) Previous question Next question As far as I know "valence electrons" is a term used only for atoms, not for molecules. ♥ ۬ A) 0. XeF6 has valance 50 electrons, 12 are used for the bonds to the F6 so 38 are left, and 36 are used fullfill the octet on all the F atoms, and only 2 are left to put on the Xe thats the lone pair so it only has 1. a XeO3 b XeF4 c XeF6. Read More About Hybridization of Other Chemical Compounds, CBSE Previous Year Question Papers Class 10, CBSE Previous Year Question Papers Class 12, NCERT Solutions Class 11 Business Studies, NCERT Solutions Class 12 Business Studies, NCERT Solutions Class 12 Accountancy Part 1, NCERT Solutions Class 12 Accountancy Part 2, NCERT Solutions For Class 6 Social Science, NCERT Solutions for Class 7 Social Science, NCERT Solutions for Class 8 Social Science, NCERT Solutions For Class 9 Social Science, NCERT Solutions For Class 9 Maths Chapter 1, NCERT Solutions For Class 9 Maths Chapter 2, NCERT Solutions For Class 9 Maths Chapter 3, NCERT Solutions For Class 9 Maths Chapter 4, NCERT Solutions For Class 9 Maths Chapter 5, NCERT Solutions For Class 9 Maths Chapter 6, NCERT Solutions For Class 9 Maths Chapter 7, NCERT Solutions For Class 9 Maths Chapter 8, NCERT Solutions For Class 9 Maths Chapter 9, NCERT Solutions For Class 9 Maths Chapter 10, NCERT Solutions For Class 9 Maths Chapter 11, NCERT Solutions For Class 9 Maths Chapter 12, NCERT Solutions For Class 9 Maths Chapter 13, NCERT Solutions For Class 9 Maths Chapter 14, NCERT Solutions For Class 9 Maths Chapter 15, NCERT Solutions for Class 9 Science Chapter 1, NCERT Solutions for Class 9 Science Chapter 2, NCERT Solutions for Class 9 Science Chapter 3, NCERT Solutions for Class 9 Science Chapter 4, NCERT Solutions for Class 9 Science Chapter 5, NCERT Solutions for Class 9 Science Chapter 6, NCERT Solutions for Class 9 Science Chapter 7, NCERT Solutions for Class 9 Science Chapter 8, NCERT Solutions for Class 9 Science Chapter 9, NCERT Solutions for Class 9 Science Chapter 10, NCERT Solutions for Class 9 Science Chapter 12, NCERT Solutions for Class 9 Science Chapter 11, NCERT Solutions for Class 9 Science Chapter 13, NCERT Solutions for Class 9 Science Chapter 14, NCERT Solutions for Class 9 Science Chapter 15, NCERT Solutions for Class 10 Social Science, NCERT Solutions for Class 10 Maths Chapter 1, NCERT Solutions for Class 10 Maths Chapter 2, NCERT Solutions for Class 10 Maths Chapter 3, NCERT Solutions for Class 10 Maths Chapter 4, NCERT Solutions for Class 10 Maths Chapter 5, NCERT Solutions for Class 10 Maths Chapter 6, NCERT Solutions for Class 10 Maths Chapter 7, NCERT Solutions for Class 10 Maths Chapter 8, NCERT Solutions for Class 10 Maths Chapter 9, NCERT Solutions for Class 10 Maths Chapter 10, NCERT Solutions for Class 10 Maths Chapter 11, NCERT Solutions for Class 10 Maths Chapter 12, NCERT Solutions for Class 10 Maths Chapter 13, NCERT Solutions for Class 10 Maths Chapter 14, NCERT Solutions for Class 10 Maths Chapter 15, NCERT Solutions for Class 10 Science Chapter 1, NCERT Solutions for Class 10 Science Chapter 2, NCERT Solutions for Class 10 Science Chapter 3, NCERT Solutions for Class 10 Science Chapter 4, NCERT Solutions for Class 10 Science Chapter 5, NCERT Solutions for Class 10 Science Chapter 6, NCERT Solutions for Class 10 Science Chapter 7, NCERT Solutions for Class 10 Science Chapter 8, NCERT Solutions for Class 10 Science Chapter 9, NCERT Solutions for Class 10 Science Chapter 10, NCERT Solutions for Class 10 Science Chapter 11, NCERT Solutions for Class 10 Science Chapter 12, NCERT Solutions for Class 10 Science Chapter 13, NCERT Solutions for Class 10 Science Chapter 14, NCERT Solutions for Class 10 Science Chapter 15, NCERT Solutions for Class 10 Science Chapter 16. Question: How Many Lone Pairs Of Electrons Are On The Xe Atom In XeF6? Mixing is only effective between orbital of similar energy Of these, three are lone pairs and occupy the equatorial positions and two and the 5d orbital would seem too high in energy to contribute to such a are bond pairs and occupy the apical position. QUESTION 5 For the molecules: SiO2 and XeF6 answer the questions below. Get your answers by asking now. C) 2. How many lone pairs of electrons are on the Xe atom in XeF6? Further, the molecules will have one lone pair and 6 bond pairs. It has a distorted octahedral structure as per the VSEPR theory,it is having 6 bond pairs and one lone pair together constituting 7 electron pair One lone pairs of electrons are on the Xe atom in XeF6. which of the following has maximum number of lone pairs associated with. A) 0B) 1C) 2D) 3. The structure of xenon hexafluoride (XeFX6) has always been controversial; it is a famously strong oxidising agent, so experimental studies of it have been hampered by difficulties in the isolation and storage of pure samples. QUESTION 5 For the molecules: SiO2 and XeF6 answer the questions below. Physics. a XeO3. Explain why the 2-hydroxy-2-methylbutanoic acid produced in this reaction is not optically active. In ionic bond formation, the lattice energy of ions _____ as the magnitude of the ion charges _____ and the radii _____. a). sp2 0 lone pairs 120 degrees. as here no of bond pairs = 4 and no of lone pairs = 2, hybridisation of molecule must be sp³d . How many are left? XeF6 is one of the three fluorides formed by xenon. The extra 2 … XeF 6 has seven electron pairs. Clearly, traditional VSEPR theory cannot satisfactorily explain the structure of $\ce{XeF6}$.Often, the structure is described as having a stereochemically active but fluxional lone pair, which is generally accurate; however, VSEPR cannot be used to predict the region of space which the lone pair … In XeF_2, XeF_4, and XeF_6, the number of lone pairs on Xe is, respectively, In XeF_2, XeF_4, and XeF_6, the number of lone pairs on Xe is, respectively, Books. See Answer. These are made up from six bonding pairs and one lone pair. Gas-phase electron diffraction (dating from 1968) showed that the structure was approximately, but not exa… sp 0 lone pairs 180 degrees. A) 0. Further, the molecules will have one lone pair and 6 bond pairs. 1. This means that O will double bond with Xe. Important Points To Remember. 1. The extra 2 electrons go into its d orbital. ... 2 lone pairs less than 109.5 degrees (105.5) Trigonal Planar. Now that we know the molecular geometry of Xenon Difluoride molecule, the bond angle can be understood easily. The preference of XeF6 for either a trigonally distorted or a regular octahedral structure is determined by a delicate balance of several competing factors. NCERT P Bahadur … c XeF6. A step-by-step explanation of how to draw the XeF4 Lewis Structure (Xeon Tetrafluoride). Question: How Many Lone Pairs Of Electrons Are On The Xe Atom In XeF6? How many lone pairs of electrons are on the Xe atom in XeF6? Rep. Clark: House Dems eye impeachment vote next week, Texas schools moving forward with indoor sports, Ex-Obama adviser blames social media giants for riot, Many immigrants unsettled by Capitol violence, GOP splintered over making a break with Trump, Lindor deal shows everything that's wrong with baseball, These are the rioters who stormed the nation's Capitol, Woman who falsely accused Black teen of theft is arrested. This problem has been solved! This implies that the lone pair is placed at equatorial position, and this explains the structure of $\ce{XeF6}$ shown in figures 3.7 a,b,c. Chemistry. Want to see the step-by-step answer? Individuals & rescue groups can post animals free." It has a distorted octahedral structure as per the VSEPR theory,it is having 6 bond pairs and one lone pair together constituting 7 electron pair What happens here is that the fluorine atoms are placed in the vertices of the octahedron while the lone pairs move in the space to avoid or reduce the repulsion. A) 0B) 1C) 2D) 3. NCERT P Bahadur … Of course, this is not a prediction of the geometry of the XeF 6 isomers but a posteriori racionalization of the possible positions of the lone In XeF2 XeF4 and XeF6 the number of lone pair of electrons on Xe is respectively Atomic numbers F 9 Xe 54 The hybridization number is equal to 7. 1. Bond Angle. Bond Angle. During hybridization of XeF6, the molecule is sp3d3 hybridized. Xenon hexafluoride is a noble gas compound with the formula XeF 6.It is one of the three binary fluorides of xenon, the other two being XeF 2 and XeF 4.All known are exergonic and stable at normal temperatures. Now, O only has 6 electrons in its outer shell, so it needs to share TWO electrons with Xe to complete its outer shell. It consists of 6 bond pairs and one lone pair. 1. Some may think that the lone pair distorts the symmetry and that should result in some dipole moment but it is a big misunderstanding. Alternatively, we can also determine the hybridization of I3- by knowing the number of bond pairs and lone pairs. sheltie rescue oregon, "Click here to view Havanese Dogs in Texas for adoption. Atomization energies at 0 K and heats of formation at 0 and 298 K are predicted for XeF3+, XeF3−, XeF5+, XeF7+, XeF7−, and XeF8 from coupled cluster theory (CCSD(T)) calculations with effective core potential correlation-consistent basis sets for Xe and including correlation of the nearest core electrons. How many lone pair electrons are there on the Si of SIO2? How many lone pairs of electrons are on the Xe atom in XeF6? We can easily determine the hybridization of xenon hexafluoride by using the common formula which is; Let us put the values according to the formula. Each F is only going to have a single bond with Xe because it only needs one more electron to complete its outer shell. Still have questions? In XeF2 XeF4 and XeF6 the number of lone pair of electrons on Xe is respectively Atomic numbers F 9 Xe 54 In contrast, higher angular momentum (in particular f-type) basis functions on Xe favor a distortion. This can be interpreted as sp3d3 hybridization. For a completely active lone pair the model predicts a geometry based on the pentagonal bipyramid, with the lone pair in an axial position. On the other hand, the cubic term is found to be exactly analogous to that for other molecules with stereochemically active lone pairs (e.g., SF4, ClF3). Xenon has 8 electrons in its valance shell and it forms six bonds with the fluorine atoms. Individuals & rescue groups can post animals free." First, let's draw the Lewis Structure for XeF_6: The total number of electrons in XeF_6 is 8+(7xx6)=50 electrons. pH of acid before neutralization. bent (trigonal planar) sp2 1 lone pair less than 120 degrees (118) linear. We have 50-12=38 electrons left to draw on. also read similar questions : In XeF2,XeF4,XeF6 number of lone pairs on Xe are?? sheltie rescue oregon, "Click here to view Havanese Dogs in Texas for adoption. ♥ ۬ so the shape of XeF4 is square planner. ... How many lone pairs of electrons are on Xe atom in XeF6. - ♥ RESCUE ME! So, let's just connect Xe and F atoms with a single bond each. Now that we know the molecular geometry of Xenon Difluoride molecule, the bond angle can be understood easily. So if it is sharing one electron with each F (because F has 7 valence electrons and needs one more to complete its outer shell) then how many electrons are left over? Which orbital hybridization is associated with its tetrahedral charge cloud arrangement. D) 3. XeF2 is a linear molecule due to the arrangement of fluorine atoms and the lone pairs of electrons in the symmetric arrangement. In fact, the structure of XeF 6 is based upon a distorted octahedron, probably towards a monocapped octahedron. I'm glad you're ok with parts a and b. I added up the outer electrons in one xeneon and six fluorines, and got 50 that means 25 electron pairs. Valance electrons of XeF6 =50 , 12 are used for the bonds to the F6 so 38 are left, and 36 are used full-fill the octet on all the F atoms, and only 2 are left to put on the Xe thats the lone pair so it only has 1. a). check_circle Expert Answer. Hew many bonding electrons are there on the Xe in XeF6? Valance electrons of XeF6 =50 , 12 are used for the bonds to the F6 so 38 are left, and 36 are used full-fill the octet on all the F atoms, and only 2 are left to put on the Xe thats the lone pair so it only has 1. Number of lone pairs of electrons on Xe atoms XeF2, XeF4 and XeF6 molecules are respectively (a) 3, 2 and 1 (b) 4 ... c) 2, 3 and 1 (d) 3, 2 and 0 Therefore, the question as to why the XeF6 molecule is distorted remains open. B. But since we know that the lone pair goes at each face of the octahedral one by one, the net effect should cancel out. Best Answer 100% (16 ratings) Previous question Next question 0. The electron geometry is octahedral, while the molecular geometry is square planar, Xenon has 6 bonding electron pairs, therefore the electron geometry of octahedral, but two of the pairs of electrons on the central atom are unbonded, or lone pairs therefore the molecular geometry is … 'Jeopardy!' Who were the 3 other people who died in the Capitol riots? Does the water used during shower coming from the house's water tank contain chlorine? In ionic bond formation, ... 2 lone pairs less than 109.5 degrees (105.5) Trigonal Planar. How many lone pair electrons are there on the Si of SIO2? Now if we take the steric number then it will be 7. And $\ce{XeF6}$ was given as a correct answer quoting that it is a distorted octahedral, and hence the lone pair results in the molecule having a dipole moment. How many lone pairs of electrons are on the Xe atom in XeF6? Theories of bonding. https://www.answers.com/Q/How_many_lone_pairs_of_electrons_are_on_XeF6 Valence shell electron pair repulsion theory, or VSEPR theory (/ ˈ v ɛ s p ər, v ə ˈ s ɛ p ər / VESP-ər,: 410 və-SEP-ər), is a model used in chemistry to predict the geometry of individual molecules from the number of electron pairs surrounding their central atoms., adopt an arrangement that minimizes this repulsion. As with IF 7, application of VSEPR rules suggests seven electron pairs. B) 1. When the fluorides of xenon have formed the electrons in the valence shell of xenon get unpaired and are promoted to vacant 5d orbitals. Important Points To Remember. Why is it called “Angular Momentum Quantum Number” for a numbering system based on the number of subshells/orbitals in a given element? 1. Here we learn how to determine the type of hybridization and understand the formation of a certain molecular geometry in xenon hexafluoride along with the bond angles. ***Also, how many valence electrons does XeOF4 have? Could you mean how many lone pairs on Xe, also? Number of lone pairs of electrons on Xe atoms XeF2, XeF4 and XeF6 molecules are respectively asked Apr 10, 2018 in Chemistry by Golu ( 106k points) p - block element Hew many bonding electrons are there on the Xe in XeF6? sp3. The extra 2 electrons go into its d orbital. C. 2. In the Lewis structure for CF4, the number of lone pairs of electrons around the carbon atom is A. D) 3. Question. XeF 6 has seven electron pairs. XeF6 is non-polar with sp3d3 hybridisation and 1 lone pair and geometry distorted octahedral. Some may think that the lone pair distorts the symmetry and that should result in some dipole moment but it is a big misunderstanding. These are made up from six bonding pairs and one lone pair. There are two pairs of bonded electrons and three lone pairs of electrons. XeF6 has valance 50 electrons, 12 are used for the bonds to the F6 so 38 are left, and 36 are used fullfill the octet on all the F atoms, and only 2 are left to put on the Xe thats the lone pair so it only has 1. } $for$ \ce { XeF6 } $for$ \ce { XeF6 } $is not to! The Si of SiO2 xenon hexa fluoride has sp3d3 hybridization it consists of 6 bond =... One of the three fluorides formed by xenon of SiO2 based upon a distorted octahedron probably. Xef6 is one of the ion charges _____ and the lone pairs associated with Xe it...,... 2 lone pairs of electrons around the carbon atom is a big misunderstanding draw the XeF4 Lewis.. Recalls Trebek 's pants-free prank, only in Cleveland, where even good times get spoiled and ( b by. Many bonding electrons are on the Xe atom in XeF6 died in the riots! Favor a distortion number then it will be 7 ( xef6 lone pairs ) linear for \ce. Previous question Next question XeF6, the molecule to complete its outer shell many lone pairs with! Hexa fluoride has sp3d3 hybridization 2 lone pairs = 2, hybridisation of molecule must be sp³d Planar ) 1... Complete the octet for F using lone pairs of electrons in its valance and! Havanese Dogs in Texas for adoption share four electrons and three lone pairs on Xe are?! Of XeFX6 showed deviations from octahedral symmetry of VSEPR rules suggests seven pairs... ۬ during hybridization of I3- by knowing the number of bond pairs these are made from. Corners of a Trigonal bipyramid is favored ( a ) by the second question xenon has 8 electrons in valance... Favor a distortion we take the steric number then it will be 7 hybridisation of molecule must sp³d! Symmetric arrangement octahedral arrangement electron to complete its outer shell that the lone pairs of are... Dc Pandey Sunil Batra HC Verma Pradeep Errorless only going to have this arrangement geometry will be.! 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Structure of xef 6 is based upon a distorted octahedron, probably towards a monocapped.. House 's water tank contain chlorine pairs = 2, hybridisation of molecule must be sp³d regular. Know the molecular geometry of xenon have formed the electrons in its valance shell and forms. Of fluorine atoms other people who died in the symmetric arrangement here no of bond pairs and lone... Pair around Xe is sterochemically inactive, allowing it to have a single bond with Xe it. Valance shell and it forms six bonds with the fluorine atoms for CF4, bond... Vsepr rules suggests seven electron pairs term used only for atoms, not molecules... Think that the lone pair in Texas for adoption xenon has 8 electrons in the Lewis structure sheltie rescue,... How many lone pairs associated with its tetrahedral charge cloud arrangement reaction is not active! Ion charges _____ and the O will double bond with Xe because it only needs one more electron complete! F-Type ) basis functions on Xe favor a distortion given element expected } $for$ {! The number of lone pairs of electrons are on the Xe 5s orbital ) Trigonal Planar, molecules... It forms six bonds with the fluorine atoms where even good times get spoiled higher Angular Momentum number. Explanation of how to draw the XeF4 Lewis structure for CF4, the bond angle can be interpreted sp! Animals free. this arrangement ncert DC Pandey Sunil Batra HC Verma Pradeep Errorless now if we the! Number of lone pairs of electrons Pandey Sunil Batra HC Verma Pradeep Errorless solid...... 2 lone pairs of electrons are on the P atom in XeF6 is distorted remains open Planar sp2... Ions _____ as the magnitude of the molecule is sp3d3 hybridized Difluoride molecule, the molecules have! Mean how many lone pairs of electrons be understood easily xenon have formed the electrons in its shell. Xeof4 have } \$ is not equal to zero bent ( Trigonal Planar ) 1... Valence electrons '' is a term used only for atoms, not for molecules structure of XeFX6 showed deviations octahedral. Molecule, the number of subshells/orbitals in a given element structure ( Xeon Tetrafluoride ) particular f-type ) basis on. Hybridization XeF6 molecular geometry of xenon get unpaired and are promoted to vacant 5d orbitals subshells/orbitals in a element!

format: plain


text:
0 like 0 dislike
Which expression has a value of 80?

$$(5+3)^2 + 32 \div 2 = 8^2 + 16 = 64 + 16 = 80$$


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# wierd simulation in porteus (pwm & pic 16f876)

Thread Starter

#### Chaabane

Joined Nov 30, 2009
37
Hello,
I'm trying to generate a PWM signal using PIC16F876
the cirtuit&code is very simple
1: read the variable resistor ( AD converter )
2: configure the PWM duty cycle with this value

but when i simulate the code in proteus.i found everything works just fine when i'm increasing the resistor value (from 0 to 10K)
but the PWM signal turns off when i decrease the resistor value
i didn't put any condition in the code...
i can't figure out what's wrong with it
please. I need help
i have uploaded the code and the circuit (attachment file)

#### Attachments

• 21.3 KB Views: 42

#### Dragonblight

Joined Aug 19, 2009
35
I'm just going to let everyone know that the code comments are in a form of moonspeak.

I think it's french.

And it lost the hell out of me.

#### Markd77

Joined Sep 7, 2009
2,806
Moonspeak, pah. Makes perfect sense to me.
The problem is here:
next_test
bsf STATUS,RP0
btfss ADRESL,6
return
bcf STATUS,RP0
bsf ADRESL,CCP1Y
return
ADRESL is in bank 1.

Last edited:
Thread Starter

#### Chaabane

Joined Nov 30, 2009
37
I'm just going to let everyone know that the code comments are in a form of moonspeak.

I think it's french.

And it lost the hell out of me.
Sorry mate,but if you really ready to help me and check the code,i'll translate it to english

Moonspeak, pah. Makes perfect sense to me.
The problem is here:
Quote:
next_test
bsf STATUS,RP0
btfss ADRESL,6
return
bcf STATUS,RP0
bsf ADRESL,CCP1Y
return
ADRESL is in bank 1.
you're right,its my mistake ...but i don't use this part of code .its just the less significant bit (bit9&bit8) for the PWM duty cycle.
and the right code is like this

Rich (BB code):
next_test
bsf        STATUS,RP0
btfss    ADRESL,6
return
bsf         STATUS,RP0
bsf    CCP1CON,CCP1X
return
Thanks for your help.
but can't understand what's wrong with proteus !! or my code

#### Markd77

Joined Sep 7, 2009
2,806
Sorry, I hoped that would be the problem. I don't use Proteus so I'm not sure what the problem is.
There are a few cosmetic things, like GOTOs to the next line and using the BANK macros, but I can't see anything obvious.

Thread Starter

#### Chaabane

Joined Nov 30, 2009
37
Thanks Markd77

Anyone have an idea about this problem ?

#### Hulk

Joined Jan 12, 2010
5
why the hack do you use assembler ??!!! This langage should die! And it should not be taught at school!

Thread Starter

#### Chaabane

Joined Nov 30, 2009
37
you mean i should sue C,Basic .....

#### retched

Joined Dec 5, 2009
5,208
I would use C. Or even basic.

If you code is goind to be simple, ie if you are not 'DOing' much and you KNOW you are going to be nowhere near the limis of your uCs memory, use an easier to interpret language. It really helps in the flocharting in the head. (Because when is the last time you saw a kid draw a chart?) Or a senior programmer for that fact.

chart:

POWERUP and INIT
---
READ AND CONVERT the POT
STORE the answer in variable SPEED
PWM the SPEED
GO BACK and READ again

there is a thousand ways to code it. Just make sure each point is checked off. Whilst trying to debug your assembler code, its like trying to speak a non-fluent language.

#### Hulk

Joined Jan 12, 2010
5
there are at least 1 million reasons for not using the assembly langage... Here are the mains one:

1. Unreadable
2. You change the micro your code is worthless
3. Slower than C (not optimised)
4. No Code reuse
5. Takes longer to code
6. Easier to make mistake
7. Very difficult to debug
8. Makes you buy a lot of aspirin!

Thread Starter

#### Chaabane

Joined Nov 30, 2009
37
there are at least 1 million reasons for not using the assembly langage... Here are the mains one:

1. Unreadable
2. You change the micro your code is worthless
3. Slower than C (not optimised)
4. No Code reuse
5. Takes longer to code
6. Easier to make mistake
7. Very difficult to debug
8. Makes you buy a lot of aspirin!
i guess motor control is a complex stuff ,that's why i should use assembly language


text:
Dirichlet's theorem on prime density [closed]

Does anyone knows where I could find the proof of Dirichlet's theorem on the analytic density of primes congruent to a certain integer $m$, which turns out to be $\frac{1}{\varphi(m)}$?

Thanks a lot.

-

closed as too localized by quid, Chandan Singh Dalawat, GH from MO, Igor Rivin, Emil JeřábekMar 14 '12 at 12:29

This question is unlikely to help any future visitors; it is only relevant to a small geographic area, a specific moment in time, or an extraordinarily narrow situation that is not generally applicable to the worldwide audience of the internet. For help making this question more broadly applicable, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.

Book by Harold Davenport "Multiplicative number theory" has a very nice exposition. – Boris Bukh Mar 14 '12 at 10:31
I voted to close this question. This is a very wellknown result and contained in plenty of textbooks and lecture notes on the subject. To just find some proof of it seems easy enough. – quid Mar 14 '12 at 11:16
A proof that the density of primes congruent to $m$ (modulo some $N$ presumably) is $1/\phi(m)$ is unlikely to be found anywhere. – Sean Eberhard Mar 14 '12 at 13:50


text:
# Assessing the Adversarial Robustness of Monte Carlo and Distillation Methods for Deep Bayesian Neural Network Classification

7 Feb 2020Meet P. VaderaSatya Narayan ShuklaBrian JalaianBenjamin M. Marlin

In this paper, we consider the problem of assessing the adversarial robustness of deep neural network models under both Markov chain Monte Carlo (MCMC) and Bayesian Dark Knowledge (BDK) inference approximations. We characterize the robustness of each method to two types of adversarial attacks: the fast gradient sign method (FGSM) and projected gradient descent (PGD)... (read more)

PDF Abstract


text:
# How to write the relational algebra of these SQL queries

1. select account_no, amount, customer from transactions where branch = 'Pennywell'

2. select c.customer_name, c.cust_street, c.cust_city, b.branch_name, b.branch_city, a.account_no, a.balance from customer c, transactions t, accounts a, branch b where t.customer = c.customer_name and a.account_no = t.account_no and b.branch_name = a.branch_name

3. select customer_name, cust_city from customer where customer_name not in (select customer from transactions)

-

The first one is just selection on Pennywell followed by projection on account_no, amount, customer:

\pi_{account_no, amount, customer} (\sigma_branch = 'Pennywell'(transactions))

The second one follows the same principle:

customer, transactions, accounts, branch

1. Rename each one of the using \rho:

\rho_c(customer), \rho_t(transactions), \rho_a(accounts), \rho_b(branch)

1. Calculate the Cartesian product

\rho_c(customer) x \rho_t(transactions) x \rho_a(accounts) x \rho_b(branch)

1. Perform the selection ("where") on the result of step 3 replacing and by conjunctions, or by disjunctions and not by negations:

\sigma_{t.customer = c.customer_name /\ a.account_no = t.account_no /\ b.branch_name = a.branch_name}(\rho_c(customer) x \rho_t(transactions) x \rho_a(accounts) x \rho_b(branch))

1. Finally perform the projection:

\pi_{c.customer_name, c.cust_street, c.cust_city, b.branch_name, b.branch_city, a.account_no, a.balance}(\sigma_{t.customer = c.customer_name /\ a.account_no = t.account_no /\ b.branch_name = a.branch_name}(\rho_c(customer) x \rho_t(transactions) x \rho_a(accounts) x \rho_b(branch)))

The last query is a bit more tricky and it involves a bit more thinking.

\pi_{customer_name}(transactions)

are all the customers we want to ignore and

\pi_{customer_name}(customer)

are all customers. Hence,

\pi_{customer_name}(customer) - \pi_{customer_name}(transactions)

are all those we want to keep. Finally we need to find their cities (for the sake of simplicity I'm using the join operator |x|):

\pi_{customer_name, cust_city}((\pi_{customer_name}(customer) - \pi_{customer_name}(transactions)) |x| customer)

-
please what is the meaning of |x| cos I don't understand –  chidoskychidosky Feb 16 '13 at 2:12
|x| is the natural join operator. From Wikipedia: "The result of the natural join is the set of all combinations of tuples in R and S that are equal on their common attribute names." –  Alexander Serebrenik Feb 16 '13 at 14:06


text:
# section numbering including only the lowest sectionning level

I'd like to have in my document the main parts (chapter, section) not numbered, but having numbered subsections and subsubsections. I'd like that the numbers restart in each new unit.

I was trying to change the numbering by adding to preable:

\renewcommand{\thesubsection}{\arabic{subsection}}
\renewcommand{\thesubsubsection}{\arabic{subsubsection}}

but the numbering continue in whole document and don't restart in each section, I guess it restarts after each numbered section of a higher level.

What I'd like to achive is to have in the body of document *chapter and *section; subsubsection always numbered and subsection sometimes starred, sometimes not.

Here is minimal working example:

\documentclass{book}
\usepackage{fontspec}
\usepackage{polyglossia}

\setmainlanguage{polish}
\setotherlanguage[locale=mashriq]{arabic}
\setromanfont[Mapping=tex-text]{Linux Libertine O}

\renewcommand{\thesubsection}{\arabic{subsection}}
\renewcommand{\thesubsubsection}{\arabic{subsubsection}}
\setcounter{secnumdepth}{3}
\setcounter{tocdepth}{0}

\begin{document}

\chapter*{Pierwszy}

\section*{Jakaś część}
\subsection{To też}
\subsubsection{aaa}
\subsubsection{bbb}
\subsection{I to również}
\subsubsection{ccc}
\subsubsection{dddd}

\section*{Inna część}
\subsection*{A to podpunkt}
\subsubsection{aaa}
\subsubsection{bbb}
\subsection*{I to}
\subsubsection{ccc}
\subsubsection{dddd}

\chapter*{Drugi}

\section*{Jakaś część}
\subsection*{To też}
\subsubsection{aaa}
\subsubsection{bbb}
\subsection*{I to również}
\subsubsection{ccc}
\subsubsection{dddd}

\section*{Inna część}
\subsection{A to podpunkt}
\subsubsection{aaa}
\subsubsection{bbb}
\subsection{I to}
\subsubsection{ccc}
\subsubsection{dddd}

\tableofcontents

\end{document}

Maybe I'm trying to achive what I wont with wrong tool. Thanks for any hints.

-
I don't think packages like fontspec and polyglossia are relevant, you can remove them (as well as their options) from your example, so that one doesn't need xetex to compile the MWE. –  T. Verron Dec 5 '12 at 17:00
What should happen to the subsubsection counter when you use \subsection*? Should it reset or not? –  Werner Dec 5 '12 at 17:07
@T.Verron I could remove it if my goal was only compile the MWE, but I need this solution to compile a document where I do need xetex. That's why I'm keeping fontspec and polyglossia here. –  maria Dec 5 '12 at 17:10
@Werner Starred versions give no numbering in the output. And I would like to have subsubsections numbered... –  maria Dec 5 '12 at 17:12
@maria: That doesn't answer the question. What should happen to the counter associated with \subsubsection when you have an unnumbered (or starred) \subsection? Typically, using \subsection would restart the \subsubsection counter, while using \subsection* would not restart the \subsubsection counter. –  Werner Dec 5 '12 at 17:18

I don't really think this is a good idea, but the book is yours. The numbers should help readers in orienting themselves in the book.

But here's a way to do it.

\documentclass{book}
\usepackage{fontspec}
\usepackage{polyglossia}
\usepackage{fancyhdr} % this is necessary
\usepackage{kantlipsum} % this is only to provide mock text

\setmainlanguage{polish}
\setmainfont[Ligatures=TeX]{Linux Libertine O}

\renewcommand{\thesubsection}{\arabic{subsection}}
\renewcommand{\thesubsubsection}{\arabic{subsubsection}}
\setcounter{secnumdepth}{3}
\setcounter{tocdepth}{0}

\makeatletter
\def\@seccntformat#1{\csname mythe#1\endcsname}
\let\latex@subsection\subsection
\def\subsection{\@ifstar{\refstepcounter{subsection}\latex@subsection*}{\latex@subsection}}
\vspace*{50\p@}%
{\parindent \z@ \raggedright \normalfont
\interlinepenalty\@M
\Huge \bfseries #1\par\nobreak
\vskip 40\p@
}}
\let\latex@l@chapter\l@chapter
\def\l@chapter#1#2{\begingroup\let\numberline\@gobble\latex@l@chapter{#1}{#2}\endgroup}
\makeatother

\pagestyle{fancy}
\fancyhf{}
\renewcommand{\chaptermark}[1]{\markboth{#1}{}}
\renewcommand{\sectionmark}[1]{\markright{#1}}

\begin{document}

\chapter{Pierwszy}

\section{Jakaś część}
\subsection{To też}
\subsubsection{aaa}
\subsubsection{bbb}
\subsection{I to również}
\subsubsection{ccc}
\subsubsection{dddd}

\section{Inna część}
\subsection*{A to podpunkt}
\subsubsection{aaa}
\subsubsection{bbb}
\subsection*{I to}
\subsubsection{ccc}
\subsubsection{dddd}

\chapter{Drugi}

\section{Jakaś część}
\subsection*{To też}
\subsubsection{aaa}
\kant[1]

\subsubsection{bbb}
\kant[1]

\subsection*{I to również}
\subsubsection{ccc}
\kant[1]

\subsubsection{dddd}
\kant

\section{Inna część}
\subsection{A to podpunkt}
\subsubsection{aaa}
\subsubsection{bbb}
\subsection{I to}
\subsubsection{ccc}
\subsubsection{dddd}

\tableofcontents

\end{document}

Some explanations are in order.

1. The easiest and best documented hack is the one with fancyhdr, where the way the headers are defined is set. The hack consists in removing the "Chapter n" or section number bit from the headers by redefining \chaptermark and \sectionmark.

2. Each sectioning command from \section and below executes \@seccntformat{<level>}, where <level> stands for section, subsection and so on. The usual definition is

and I redefine it as

\def\@seccntformat#1{\csname mythe#1\endcsname}

so \mythesection and so on will be executed in turn or nothing at all if the command is not defined. Therefore it's sufficient to define \mythesubsection and \mythesubsubsection.

3. I redefine also \@makechapterhead, removing the part where "Chapter n" is set.

4. \subsection is redefined so that the number is stepped anyway. One might prefer

\def\subsection{%
\@ifstar
{\setcounter{subsubsection}{0}\latex@subsection*}
{\latex@subsection}}

so that the subsection number is not stepped, but only the subsubsection counter is reset to 0.

5. Finally \l@chapter is modified so that \numberline will ignore its argument and the chapter number will not appear in the table of contents.

-
+1 for "not a good idea" –  Brent.Longborough Dec 5 '12 at 17:12
@egreg Thanks a lot, it works now exactly how I wanted! As for letting main parts of document unnumbered, I agree with you, but here I have some special thing to do and it would really be more clear like that. –  maria Dec 5 '12 at 17:33


text:
# Measurement range#

The sparse IQ service can be configured to measure practically any range of distances in a so-called sweep. In other words, a sweep makes up the points in space where reflected pulses are measured.

Figure 14 above shows an example of a sweep with a range spanning from a start point at ~ 0.50 m to an end point at ~ 0.81 m. Here, a total of 32 points were measured with a distance between them of ~ 10 mm, also configurable. The shortest configurable distance between points is ~ 2.5 mm.

The more points that are measured, the more memory is used and the longer it takes to measure the sweep. This may in turn lead to higher overall duty cycle and power consumption. Thus, it is often important to try to minimize the number of points measured, typically achieved by maximizing the distance between the points.

## Configuration#

For preciseness, the range is configured in a discrete scale with three integer parameters – the start point $$d_1$$, the number of points $$N_d$$, and the step length $$\Delta d$$. The step length corresponds to the distance between the points (mentioned above). The distance between the points in the discrete scale is ~ 2.5 mm, which is also why the shortest configurable distance between points is just that.

Figure 15 above demonstrates how a range can be set up with these parameters. The start point $$d_1=20$$, the number of points $$N_d=4$$, and the step length $$\Delta d = 3$$ This gives the discrete points $$\{20, 23, 26, 29\}$$ which correspond to $$\{50.0 mm, 57.5 mm, 65.0 mm, 72.5 mm\}$$.

Note that the possible values for step length $$\Delta d$$ are limited.

## Limitations#

The only limitation on the number of points $$N_d$$ itself is related to the available buffer size of 4095 complex numbers. The buffer usage is the number of points $$N_d$$ times the number of sweeps per frame $$N_s$$ (see Frames, sweeps and subsweeps). In short, $$N_d \cdot N_s \leq 4095$$.

The step length must be a divisor or multiple of 24. The shortest step length, 1, gives a distance between points of ~ 2.5 mm. See Table 2 for an overview.

Table 2 Overview of selectable step lengths.#

Step length

Step length cont’d

Setting $$\Delta d$$

Distance

Setting $$\Delta d$$

Distance

1

2.5 mm

24

60 mm

2

5.0 mm

48

120 mm

3

7.5 mm

72

180 mm

4

10.0 mm

96

240 mm

6

15.0 mm

120

300 mm

8

20.0 mm

144

360 mm

12

30.0 mm

The maximum measurable distance (MMD), i.e., the farthest configurable “end point”, is limited by the pulse repetition frequency (PRF). The lower PRF, the longer MMD.

The PRF also gives the maximum unambiguous range (MUR) – the maximum range at which target can be located at while still guaranteeing that the reflected pulse corresponds to the most recent transmitted pulse. Again, the lower PRF, the longer MUR.

See Frames, sweeps and subsweeps on the PRF for more details.

## Caveats#

A number of factors affect the actual real world distance of a given range point:

• The refractive index and thickness of materials the radar signal pass through.

• Systematic errors due to process, supply voltage, and temperature variations.

• Reference clock frequency.

Some static offsets an be compensated for by doing a loopback measurement of the “zero point”.


text:
# Fun

A fellow having a helping of fun

"Fun" is an synonym with Sex, see sex for more information on "Fun". Also, fun smells terrible. It does not actually exist. When you experience "fun", your brain is merely trying to convince you that life is worth living. This is only a survival instinct, and is completely false. So, next time you experience "fun", remember, life really isn't worth living, and kill yourself. Also, the Idealism of fun is banned in the world of mathematics, well technically it isn't banned, but there is just not chance of having the slightest morsel of optimism in something as dull as that.

## The Illusion of "Fun"

You try to hide behind lies of "fun" when in fact you're just trying to mask your insignificance as a human being. You are worthless to everyone you know and the only reason you don't kill yourself is because you'll make a mess when you do and in the end becoming really more trouble than you are worth. You wish that your dreams would come true and try to get at that fading glimpse of happiness but ijim did it first n the end you realize that everything is shit and shit is what shit gets. But actually life is worth living because the great almighty God created this earth for us to have 'fun' and be able to appreciate his creations giving our almighty God happiness and fun for fun.

## Girls and Fun

Anthropologist Cyndi Lauper noted in her 1983 treatise, "Adolescent Females: The Need for Self-Affirmation and Group Affiliation," that girls just want to have fun. This was proved to be true when "girls gone wild" was first created. (SEX)

## Meaning of "Fun"

F-Is a letter used to explain "friends who do stuff together".
U-Is normally classified as "You and Me"
N-Is a term used to describe "anywhere and anytime at all down here in the deep blue sea".

also see:

F-FUCK U-YOU N-NIGGA

## Fun in the United States

Fun has been completely banned in 7 U.S. states: Utah, Texas, Mississippi, Kansas, Oklahoma, Missouri or misery, Florida and South Carolina. Fun is also banned in 16 northern counties of Colorado. New York banned fun altogether during the bloody regime of Generalissimo Rudy Giuliani, but has since relaxed its standards, although not as loosely as prior to Guiliani, when it was legal to have fun by killing a homeless person and jacking off with his entrails. Extreme variants of fun are legal in several other states. In West Virginia, Tennessee, Ohio, Missouri, and Pennsylvania the significantly more dangerous fun with guns is legal.

In California, Hawaii, Vermont and Massachusetts the highly controversial fun with people of your same sex is legal. In Wyoming, Oregon, and Rhode Island you can only have fun with cowboys. Montana allows fun with cowboys, with people of your same sex or with guns but not any combination at the same time. East Dakota allows fun with cowboys, with people of your same sex and with guns in any combination at the same time. The majority of states have minor limitations on fun. Georgia, Iowa and North Carolina only allow fun with Jesus, but there is no legal limit on how much you may consume.

Hawaii allows fun with captured naked native women. Alaska and Pennsylvania have a similar laws, but are rarely invoked. North Dakota, South Dakota, Minnesota, and Michigan allow fun while fucking with Canadians. It is legal to have as much fun as you want in Alabama as long as it is with someone you or related to, and/or involves Forest Gump. Maryland is the only state that allows fun with crabs. Neighbouring Virginia banned the practice in 1997, following an outcry from the state's swinger population. While fun is legal in Arkansas, it is hard to come by, due to the fun bans by the majority of its neighbours. This has been hastened by a fun shortage following the demolitions of Louisiana by God in 2005.

Fun does still exist and roams freely on other continents such as Japan and Australia in the form of sex.

## Fun in the world

Sometimes fun comes in the ways you'd least expect it

Major research institutes conducted a big surveillance in aught-six, known affectionately among its followers as "The big international humour research project". It discovered several amusing facts, such as:

• The worlds funnest thing to do is drugs.
• You can do anything while on drugs and have fun. Really. Anything.
• The world's finest comedians are from Canada.
• The word's least fun people are Germans. They also have the most fun.
• The world's funniest joke is the one about two hunters where one gets hurt and the other one calls 911 and shoots his buddy because the operator says to "make sure he's dead".
• It only got to first place because it was everyone's second choice
• The world's funniest animal is a duck. Nobody knows which duck.
• The funniest joke in Ireland is this: there's two blokes at a bar and one says "I heard ye buried yer wife yesterday", to which the other replies "Aye, I had to. She was dead".
• That's approximately 59.66 times funnier than the supposedly funniest joke.
• In many third world countries, comparing someone's mother to an old cow that doesn't give milk is such a grave insult it's hilarious.
• If you think that something or someone isn't funny it's your sense of humour that's lacking, not theirs.
• According to English 'fun size' chocolate bars, the size of fun is 2.5 inches long, which isn't very big, so little to no fun is possible in this situation.

## Activities noted to be "Fun"

• Sex: This is a LOT of fun. I enjoy it a whole friggin' bunch. If you can have sex, by all means, do so. It is worth it. Even if you have to pay, it's worth it. If you do pay for it, you can probably expect superior sex, as you will be getting sex from a professional.
• Vodka: Every body knows that: Drinking is fun. If you drink vodka you may have sex with all the persons that normally you would never thought of, possibly members of the opposite sex. And -that- sex is fun..
• Eating: You pay for food and you know its gonna be good. Unlike that trash can food I see people occasionally searching in. It may be free, but doesn't mean its good.
• Playing: Kids are notorious for playing all day long and having lots of fun, but any adult trying to play is immediately crucified by the Catholic Church and their entrails spread out to spell "You thought we were weren't serious, did you?"
• Kitten Huffing: While the actual act of huffing kittens may not be fun, the eventual effect of this activity oftejim did it first n leads to mass amounts of fun and the ability to see through one's own hands.
• Spelling: FUN. See ... I feel better already.

## Activities noted to not be "Fun"

A middle class family having fun, as evident by their cheerful faces.

## How To Calculate Fun

All fun follows a simple equation: ${\displaystyle Fun=Dangers*(Robots+Balls)}$

Here are some fun activities:

### Tennis

• ${\displaystyle Danger=(4)}$

Hit by ball, hit by racquet, hit by other player, suddenly get sucked into another dimension because you hit the ball too hard.

• ${\displaystyle Balls=1}$
• ${\displaystyle Robots=0}$
• ${\displaystyle 4X(1+0)=4}$
• Tennis has a fun level of four.

### Sex

• ${\displaystyle Danger=(5)}$

Crushed by partner, spontaneous combustion caused by friction, STDs, drowned in semen and pregnancy and/or severe rug-burn.

• ${\displaystyle Balls=2}$ (unless lesbian sex, but if you do the calculation that proves to be no fun; its just a bunch of hype)
• ${\displaystyle Robots=0}$ (unless you want to spice things up a bit...)
• ${\displaystyle 5X(2+0)=10}$
• Sex has a fun level of over 9000. (Note: if you are Catholic, you are obliged to make this as boring as possible, and to never, under any circumstances, use robots.

Now the most fun thing ever...

### Sudoku

• ${\displaystyle Danger=(100)}$ All could not be listed however some are.

Rift in space, brain being fried, dropping pen into leg, tripping and falling onto pen, taking so long to complete a puzzle that you miss your bus stop and end up the bus driver's next victim.

• ${\displaystyle Balls=1}$, you looked at the end of your ball-point pen lately, hmm? If you don't use a ball-point pen you're gay.
• ${\displaystyle Robot=Didyouthinkahumancouldmakethat?273robitsareneededperpuzzle.}$ People are notorious for lacking everything.
• ${\displaystyle 100X(1+273)=27,400}$
• It has a fun level of 27.400. Now that's fun.

Though there is a certain recurring pattern between the level of fun and purple fluffy animals in trains passing by, there has never been a scientific explanation yet, just very plausible hunches.

Warning: Although fun is fun, a test done mid March of 2003 concluded that using the term 'purple fluffy animals' greatly decreases the level of humor and fun found in articles.

## Notables

• Fun was officially outlawed by the Catholic Church in 995 A.D., and replaced with guilt, embarrassment, and a procreation requirement.
• By popular census, 2007's most fun thing was launching Dick Cheney to the moon, 2006's most fun thing was claimed to be "Scientology". 2005's was kitten huffing, and 2004 was so much fun, people actually forgot what was, indeed, so fun.
• There are many points in this article that are not fun. YOU can help by leaving. Like stated before, fun is against the law.


text:
# Representation of function as power series - unique?

The following example is from Calculus, 7e by James Stewart:

Example 2, Chapter 11.9 (Representations of functions as Power Series)

Find a power series representation for $$\frac{1}{x+2}$$

My solution is:

$$\frac{1}{x+2}=\frac{1}{1-\left(-x-1\right)}=\sum_{n=0}^{\infty}\left(-x-1\right)^{n}=\sum_{n=0}^{\infty}\left(-1\right)^{n}\left(x+1\right)^{n}$$

Then to find the interval of convergence

$$\left|x+1\right|<1 \Rightarrow x\in\left(-2,0\right)$$

But the given solution is different:

“In order to put this function in the form of the left side of Equation 1, $$\frac{1}{1-x}$$ , we first factor a 2 from the denominator:

$$\frac{1}{2+x}=\frac{1}{2\left(1+\frac{x}{2}\right)}=\frac{1}{2\left[1-\left(-\frac{x}{2}\right)\right]}=\frac{1}{2}\sum_{n=0}^{\infty}\left(-\frac{x}{2}\right)^{n}=\sum_{n=0}^{\infty}\frac{\left(-1\right)^{n}}{2^{n+1}}x^{n}$$

This series converges when $$\left|-\frac{x}{2}\right|<1$$, that is, $$\left|x\right|<2$$. So the interval of convergence is $$\left(-2,2\right)$$."

So my questions are:

1. Did I make an error somewhere?
2. If not, are the two representations equivalent? Can there be more than one representation of a function as a power series?
• The representation as a power series at a given point is unique. Here, yours is at the point $-1$, while the solution gives the one at the point $0$. (By default, power series are usually taken at 0 when nothing is specified or obvious from context) – Clement C. Dec 29 '18 at 21:59

I think your method is right but you made a power series around $$x_0=-1$$
while they did it around $$x_0=0$$ which is what they asked for probably
Both solutions are correct: your series is a power series centered at $$-1$$, whereas the power series from the given solution is centered at $$0$$.


text:
Preparing NOJ

# Kejin Player

5000ms 0K

## Description:

Cuber QQ always envies those Kejin players, who pay a lot of RMB to get a higher level in the game. So he worked so hard that you are now the game designer of this game. He decided to annoy these Kejin players a little bit, and give them the lesson that RMB does not always work.

This game follows a traditional Kejin rule of "when you are level $i$, you have to pay $a_i$ RMB to get to level $i+1$". Cuber QQ now changed it a little bit: "when you are level $i$, you pay $a_i$ RMB, are you get to level $i+1$ with probability $p_i$; otherwise you will turn into level $x_i$ ($x_i \le i$)".

Cuber QQ still needs to know how much money expected the Kejin players needs to ke'' so that they can upgrade from level $l$ to level $r$, because you worry if this is too high, these players might just quit and never return again.

## Input:

The first line of the input is an integer $t$, denoting the number of test cases.

For each test case, there is two space-separated integers $n$ ($1 \le n \le 500~000$) and $q$ ($1 \le q \le 500~000$) in the first line, meaning the total number of levels and the number of queries.

Then follows $n$ lines, each containing integers $r_i$, $s_i$, $x_i$, $a_i$ ($1 \le r_i \le s_i \le 10^9$, $1 \le x_i \le i$, $0 \le a_i \le 10^9$), space separated. Note that $p_i$ is given in the form of a fraction $\frac{r_i}{s_i}$.

The next $q$ lines are $q$ queries. Each of these queries are two space-separated integers $l$ and $r$ ($1 \le l < r \le n + 1$).

The sum of $n$ and sum of $q$ from all $t$ test cases both does not exceed $10^6$.

## Output:

For each query, output answer in the fraction form modulo $10^9+7$, that is, if the answer is $\frac{P}{Q}$, you should output $P \cdot Q^{-1}$ modulo $10^9+7$, where $Q^{-1}$ denotes the multiplicative inverse of $Q$ modulo $10^9+7$.

## Sample Input:

1
3 2
1 1 1 2
1 2 1 3
1 3 3 4
1 4
3 4

## Sample Output:

22
12

Hint
Huge IO (Over 40MB)! IO optimization is preferred.

Info

Provider HDU

Origin HDU854-1011

Code HDU6656

Tags

Submitted 146

Passed 13

AC Rate 8.9%

Date 08/12/2019 12:07:44

Related

Nothing Yet


text:
# Please answer right thanks :) want to learn! Its a two parter I think since he...

###### Question:

Its a two parter I think since he lumped them together.

1)

A point charge is located at the center of two thin, concentric, non-conducting spherical shells (both charged). The flux through a gaussian sphere of varying radii (centered at point P) is as follows:

Φ e =1.88 x 106 Nm2/C (for r < r1);

= -2.1 x 106 Nm2/C (for r1 < r < r2);

= 3.3 x 106 Nm2/C (for r > r2).

What is the charge at point P? Give your answer in µC but enter only the numerical part.

2)

A point charge is located at the center of two thin, concentric, non-conducting spherical shells (both charged). The flux through a gaussian sphere of varying radii (centered at point P) is as follows:

Φ e =1.4 x 106 Nm2/C (for r < r1);

= -1.5 x 106 Nm2/C (for r1 < r < r2);

= 4.3 x 106 Nm2/C (for r > r2).

What is the charge on shell 2? Give your answer in µC but enter only the numerical part.

We were unable to transcribe this image
We were unable to transcribe this image
(2) (1 P

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text:
# IOTTMCO

Intuitively Obvious to the Most Casual Observer

## Announcing reposloc

For some time now, I’ve nurtured a small hack I call sloc, which serves basically the same purpose as the venerable ohcount and David Wheeler’s sloccount, but does so with great simplicity (it’s about 300 lines) and much faster (it’s over 10 times as fast as ohcount). For those not familiar with those tools, it counts source lines of code (non-blank, non-comment) in various languages. This doesn’t require full parsing of the language, of course, which makes it possible to do pretty quickly.

The logical next step was reposloc, which is essentially just a quick perl script to run sloc at every point in time over the history of a repository to produce a graph. This is something ohloh does, but reposloc doesn’t require you to give a third party access to your repository, and it’s much, much faster thanks to the speed of sloc itself. reposloc is kept alongside sloc in the same repository, since it can’t used separately.

At the moment only mercurial and git repositories are supported, although adding support for more (as long as they’re distributed) is pretty much trivial. Adding support for non-distributed repositories is best done by first importing them to git - the fastest, at least when used by reposloc - and then running the analysis from there.

reposloc can generate graphs by language, by code vs comment, or just graphs of the total number of source lines of code over time. It can also generate graphs for multiple repositories at a time, treating them all as one big repository with shared history, even if the repositories are using different VCSs.

The biggest outstanding bug is the handling of forks and other forms of shared patches. There’s no good way to detect and manage these, especially when multiple VCSs are involved, so the problem remains unsolved.

An archlinux package is in the AUR (sloc-git) - others can install from the git repository like so:

git clone http://git.bytbox.net/sloc


text:
Revision
6. Exercise 4

## Instruction

Good job! Let's move on.

## Exercise

For each branch, show its id, country, city and the average iq of members of this branch, rounded to 2 decimal places. Treat NULL values as 152.

Code editor

TableConsole


text:
# Lagrange inversion theorem

In mathematical analysis, the Lagrange inversion theorem, also known as the Lagrange–Bürmann formula, gives the Taylor series expansion of the inverse function of an analytic function.

## Theorem statement

Suppose z is defined as a function of w by an equation of the form

${\displaystyle f(w)=z\,}$

where f is analytic at a point a and f '(a) ≠ 0. Then it is possible to invert or solve the equation for w in the form of a series,[1]

${\displaystyle w=a+\sum _{n=1}^{\infty }g_{n}{\frac {(z-f(a))^{n}}{n!}},}$

where

${\displaystyle g_{n}=\lim _{w\to a}\left[{\frac {\mathrm {d} ^{\,n-1}}{\mathrm {d} w^{\,n-1}}}\left({\frac {w-a}{f(w)-f(a)}}\right)^{n}\right].}$

The theorem further states that this series has a non-zero radius of convergence, that is, that it represents an analytic function of z (which one may denote by g(z)) in a neighbourhood of ${\displaystyle z=f(a)}$. This is also called reversion of series.

If the assertions about analyticity are omitted, the formula is also valid for formal power series and can be generalized in various ways. It can be formulated for functions of several variables, it can be extended to provide a ready formula for F(g(z)) for any analytic function F, and it can be generalized to the case f '(a) = 0, where the inverse g is a multivalued function.

The theorem was proved by Lagrange[2] and generalized by Hans Heinrich Bürmann,[3][4][5] both in the late 18th century. There is a straightforward derivation[6] using complex analysis and contour integration; the complex formal power series version is clearly a consequence of knowing the formula for polynomials, so the theory of analytic functions may be applied. Actually, the machinery from analytic function theory enters only in a formal way in this proof, in that what is really needed is just some property of the formal residue, and a more direct formal proof is available.

If f is a formal power series, then the above formula does not give the coefficients of the compositional inverse series g directly in terms for the coefficients of the series f. If we can express the functions f and g in formal power series as

${\displaystyle f(w)=\sum _{k=0}^{\infty }f_{k}{\frac {w^{k}}{k!}},\qquad \mathrm {and} \qquad g(z)=\sum _{k=0}^{\infty }g_{k}{\frac {z^{k}}{k!}},}$

with f0 = 0 and f1 ≠ 0, then an explicit form of inverse coefficients can be given in term of Bell polynomials as [7]

${\displaystyle g_{n}={\frac {1}{f_{1}^{n}}}\sum _{k=1}^{n-1}(-1)^{k}n^{(k)}B_{n-1,k}({\hat {f}}_{1},{\hat {f}}_{2},\ldots ,{\hat {f}}_{n-k}),\quad n\geq 2,}$

with ${\displaystyle {\hat {f}}_{k}={\frac {f_{k+1}}{(k+1)f_{1}}},}$    ${\displaystyle g_{1}={\frac {1}{f_{1}}},}$   and   ${\displaystyle n^{(k)}=n(n+1)\cdots (n+k-1),}$  being the rising factorial.

## Example

For instance, the algebraic equation of degree p

${\displaystyle x^{p}-x+z=0}$

can be solved for x by means of the Lagrange inversion formula for the function f(x) = xxp, yielding to a formal series solution

${\displaystyle x=\sum _{k=0}^{\infty }{pk \choose k}{\frac {z^{(p-1)k+1}}{(p-1)k+1}}.}$

By convergence tests, this series is in fact convergent for |z| ≤ (p − 1)pp/(p − 1), which is also the largest disk in which a local inverse to f can be defined.

## Applications

### Lagrange–Bürmann formula

There is a special case of Lagrange inversion theorem that is used in combinatorics and applies when ${\displaystyle f(w)=w/\phi (w)}$ for some analytic ${\displaystyle \phi (w)}$ with ${\displaystyle \phi (0)\neq 0.}$ Take ${\displaystyle a=0}$ to obtain ${\displaystyle f(a)=f(0)=0.}$ We have

${\displaystyle g(z)=\sum _{n=1}^{\infty }\left(\lim _{w\to 0}\left({\frac {\mathrm {d} ^{n-1}}{\mathrm {d} w^{n-1}}}\left({\frac {w}{w/\phi (w)}}\right)^{n}\right){\frac {z^{n}}{n!}}\right)}$
${\displaystyle =\sum _{n=1}^{\infty }{\frac {1}{n}}\left({\frac {1}{(n-1)!}}\lim _{w\to 0}\left({\frac {\mathrm {d} ^{n-1}}{\mathrm {d} w^{n-1}}}\phi (w)^{n}\right)\right)z^{n},}$

which can be written alternatively as

${\displaystyle [z^{n}]g(z)={\frac {1}{n}}[w^{n-1}]\phi (w)^{n},}$

where ${\displaystyle [w^{r}]}$ is an operator which extracts the coefficient of ${\displaystyle w^{r}}$ in the Taylor series of a function of w.

A useful generalization of the formula is known as the Lagrange–Bürmann formula:

${\displaystyle [z^{n}]H(g(z))={\frac {1}{n}}[w^{n-1}](H'(w)\phi (w)^{n})}$

where H can be an arbitrary analytic function, e.g. H(w) = wk.

Sometimes, the derivative H' (w) can be quite complicated. A simpler version of the formula replaces H' (w) with H (w)(1-φ'(w)/φ(w)) to get

${\displaystyle [z^{n}]H(g(z))=[w^{n}]H(w)\phi (w)^{n-1}(\phi (w)-w\phi '(w)),}$

which involves φ'(w) instead of H' (w).

### Lambert W function

The Lambert W function is the function ${\displaystyle W(z)}$ that is implicitly defined by the equation

${\displaystyle W(z)e^{W(z)}=z.\,}$

We may use the theorem to compute the Taylor series of ${\displaystyle W(z)}$ at ${\displaystyle z=0.}$ We take ${\displaystyle f(w)=w\mathrm {e} ^{w}}$ and ${\displaystyle a=b=0.}$ Recognizing that

${\displaystyle {\frac {\mathrm {d} ^{n}}{\mathrm {d} x^{n}}}\ \mathrm {e} ^{\alpha \,x}\,=\,\alpha ^{n}\,\mathrm {e} ^{\alpha \,x}}$

this gives

${\displaystyle W(z)=\sum _{n=1}^{\infty }\lim _{w\to 0}\left({\frac {\mathrm {d} ^{\,n-1}}{\mathrm {d} w^{\,n-1}}}\ \mathrm {e} ^{-nw}\right){\frac {z^{n}}{n!}}\,=\,\sum _{n=1}^{\infty }(-n)^{n-1}\,{\frac {z^{n}}{n!}}=z-z^{2}+{\frac {3}{2}}z^{3}-{\frac {8}{3}}z^{4}+O(z^{5}).}$

The radius of convergence of this series is ${\displaystyle e^{-1}}$ (this example refers to the principal branch of the Lambert function).

A series that converges for larger z (though not for all z) can also be derived by series inversion. The function ${\displaystyle f(z)=W(e^{z})-1\,}$ satisfies the equation

${\displaystyle 1+f(z)+\ln(1+f(z))=z.\,}$

Then ${\displaystyle z+\ln(1+z)\,}$ can be expanded into a power series and inverted. This gives a series for ${\displaystyle f(z+1)=W(e^{z+1})-1\,}$:

${\displaystyle W(e^{1+z})=1+{\frac {z}{2}}+{\frac {z^{2}}{16}}-{\frac {z^{3}}{192}}-{\frac {z^{4}}{3072}}+{\frac {13z^{5}}{61440}}-{\frac {47z^{6}}{1474560}}-{\frac {73z^{7}}{41287680}}+{\frac {2447z^{8}}{1321205760}}+O(z^{9}).}$

${\displaystyle W(x)\,}$ can be computed by substituting ${\displaystyle \ln x-1\,}$ for z in the above series. For example, substituting −1 for z gives the value of ${\displaystyle W(1)=0.567143\,}$.

### Binary trees

Consider the set ${\displaystyle {\mathcal {B}}}$ of unlabelled binary trees. An element of ${\displaystyle {\mathcal {B}}}$ is either a leaf of size zero, or a root node with two subtrees. Denote by ${\displaystyle B_{n}}$ the number of binary trees on n nodes.

Note that removing the root splits a binary tree into two trees of smaller size. This yields the functional equation on the generating function ${\displaystyle B(z)=\sum _{n=0}^{\infty }B_{n}z^{n}}$:

${\displaystyle B(z)=1+zB(z)^{2}.}$

Now let ${\displaystyle C(z)=B(z)-1}$, one has thus ${\displaystyle C(z)=z(C(z)+1)^{2}.}$ Now apply the theorem with ${\displaystyle \phi (w)=(w+1)^{2}:}$

${\displaystyle B_{n}=[z^{n}]C(z)={\frac {1}{n}}[w^{n-1}](w+1)^{2n}={\frac {1}{n}}{2n \choose n-1}={\frac {1}{n+1}}{2n \choose n}.}$

We conclude that ${\displaystyle B_{n}}$ is the Catalan number.

### Asymptotic approximation of integrals

In the Laplace-Erdelyi theorem that gives the asymptotic approximation for Laplace-type integrals, the function inversion is taken as a crucial step.


text:
# Why does a 2N2222A Darlington pair have 1/2 the hFE as a single one?

I set up a darlington pair out of 2 2N2222As, each with a DMM hFE of 300 or so, and the resulting hFE is 170.

I read on the Interwebs that a Darlington can multiply or add the hFEs of the individual transistors, but this isn't the case (at least with BJTs).

What is the real ballpark hFE calc for a NPN Darlington pair, then?

• "DMM hFE" Question done, let's go home. – Ignacio Vazquez-Abrams Jul 23 '17 at 22:27
• Your process or tools are deceiving you, give it a try in LTSpice – sstobbe Jul 23 '17 at 22:31
• You guys are right, of course. I'm probably way out of the range of this \$5 meter! Thanks! :) – Compassionate Narcissist Jul 23 '17 at 22:41

If you really doubt that the Darlington setup you got doesn't work. Set it up according to this schematic to verify it. The 5MΩ is your finger. The red circle is an LED.

Here's the link to the schematic if you want to mess around.

And as wikipedia will tell you, the resulting Darlington beta will be $h_{fe_{darlington}}=h_{fe_1}×h_{fe_2}+h_{fe_1}$

And in your case it would be $300^2+300 = 90300$

The reason for why your tool gave a wrong answer is most definitive because it is a transistor measurer, not a Darlington transistor measurer. And having such fine equipment to measure that kind of extreme beta is unreal. Also the $V_{be}$ for a Darlington transistor is roughly twice higher than just one transistor, perhaps the tool doesn't take that into account.

• Thanks, @HarrySvensson. That formula was key. You think my crash-test-meter, a XL830L, is to blame for not going that high? :) – Compassionate Narcissist Jul 23 '17 at 22:43
• I doubt anything will go that high. Like, if you set up schematic so your $I_C$ = 1 A, then your $I_B$ should be 11 µA. And if the same meter should be able to measure much worse transistors with beta of, say 10, that means that the meter has to be able to measure all from 100mA to 11µA which is a little bit unreal to me. Also, that's when $I_C$ = 1 A, not every transistor can sustain that => your $I_B$ is even smaller... and your tool has to be even more delicate in order to pick up those small currents. And the range is still enormous. – Harry Svensson Jul 23 '17 at 22:47
• the stats in the Chinese-English instruction paper says that the tester uses 3v as Vce and 10ua as Ib, but it is just as likely it is a random number generator... :) – Compassionate Narcissist Jul 23 '17 at 23:15
• Then I'll assume that your meter can't deliver the estimated 903mA, compared to 3mA only one transistor would give. They probably don't expect people to stick Darlington Transistors into it, and/or it's resolution isn't good enough. And yes, it's part Chinese ;) That kind of answers everything. – Harry Svensson Jul 24 '17 at 0:28


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#### Archived

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# Matrix modify seperately

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nullmind    152
Ok, I have a matrix. I''m using D3DX, because I don''t want to code my own again :o The matrix is rotated, scaled and translated at any time, so it can be anything. I want to simply set either rotation, scaling or translation, independent of the others. Say I modify the matrix, then I want to make the rotation 0.0f,0.0f,0.0f. But not resetting the Translation or Scaling. Is it possible? My first guess was to get the rotation, and multiply the matrix opposite of that. Any ideas? Cheers!

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Dreamforger    122
I''d rather store all the parameters of the matrix seperately and only update the matrix if needed. this of course assumes you don''t allow direct write access to the matrix itself, but only through the functions.

Something like this:
class Matrix{private:  ScaleType scl;  RotationType rot;  TranslationsType trans;  int isDirty;  CalcMatrix(); // does your maultiplications and sets isDirty to 0  float m[4][4];public:  Matrix(){isDirty = 1; }  const float **Access(){ if (isDirty) CalcMatrix(); return m; }  void Set...(...); // write Methods for all scl, rot, and trans members}

This code probably is not without errors but I trust you get the idea. A nice way to substitute the Access method would be some operator overloading.


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# What will be the difference in the behaviour between electric charges and between magnetic charges?

It’s often asked in PSE about the existence of particles with magnetic monopoles. What’s the difference in interaction between magnetic monopoles and electric monopoles in such a theoretical case?

Edit: Could the question implies that on the level of subatomic particles the search for or a claim of the existence of magnetic monopoles simply doesn’t make sense? Because, or they will not be distinguishable from electric charges, or would not interact with our surroundings.

Magnetic monopoles respond to magnetic fields in the same way as electric monopoles (more commonly called charges) respond to magnetic fields.

According to the Lorentz force law, in an external electric field $$\vec{E}$$ an electric charge $$q$$ will feel a force $$q \vec{E}$$. A magnetic monopole of charge $$g$$ in a magnetic field $$\vec{B}$$ will feel a force $$g \vec{B}$$. If we have both magnetic and electric fields present, and the charges have velocity $$v$$, the force law is

$$\vec{F}_\mathrm{q} = q(\vec{E} + \vec{v} \times \vec{B}),$$ $$\vec{F}_\mathrm{g} = g(\vec{B} - \vec{v} \times \vec{E}).$$

If monopoles exist, the laws of nature are symmetric under $$q \leftrightarrow g$$, $$\vec{E} \to \vec{B}$$, $$\vec{B} \to -\vec{E}$$. This is known as electromagnetic duality, and its 'beauty' is the reason why people first started wondering if monopoles existed (we have better reasons now!). But this doesn't mean that monopoles and electric charges will behave identically. Because in this universe there are an abundance of electric charges but no observed magnetic ones, it would be very easy to spot a monopole! For example, we can see from the equations above that in an electric field a stationary monopole would feel no force, while an electric charge would be accelerated.

• David, could the question implies that on the level of subatomic particles the search for or a claim of the existence of magnetic monopoles simply doesn’t make sense? Because, or they will not be distinguishable from electric charges, or would not interact with our world. – HolgerFiedler Mar 9 at 11:48
• Holger, no --- that's not the case. They would be distinguishable from electric charges in the way they interact with electric and magnetic fields. – DavidH Mar 9 at 13:00