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• # BayesGESM

• Referenced in 3 articles [sw22576]
• parametric additive components described by using B-splines; and (ii) Flexible measurement error models under ... parametric components approximated by using B-splines...
• # ConfBands

• Referenced in 34 articles [sw12330]
• effect) nonparametric model, (b) using the mixed-model framework with the spline coefficients as random ... model formulation of penalized splines can help obtain, at least approximately, confidence bands with either...
• # L2CXFT

• Referenced in 9 articles [sw00498]
• derived in $O(n)$ operations. Then, this approximation is used as the starting point ... optimum. The constraints allow $B$-splines to be used, which reduce the problem...
• # PetIGA-MF

• Referenced in 9 articles [sw22643]
• isogeometric discrete differential forms based on B-splines: PetIGA-MF. Built on top of PetIGA ... presented assuring optimal convergence rates of the approximations, showing the accuracy and robustness...
• # CXFTV2

• Referenced in 3 articles [sw00180]
• Fortran subroutine calculates the least squares approximation to n data values containing random errors subject ... distinctive feature of this calculation. A B-spline representation of the iterates reduces each active...
• # hex-ecs

• Referenced in 1 article [sw17656]
• fraction of the ionization threshold to approximately 100 eV or more, depending on the available ... exterior complex scaling method in the B-spline basis...
• # L2CXCV

• Referenced in 2 articles [sw26952]
• sets and takes advantage of a B-spline representation of the smoothed values that allows ... found in calculating a sigmoid approximation, which is a common topic in various contexts...
• # PyFly

• Referenced in 1 article [sw25306]
• simulated domain. We introduce the pointwise approximation approach to simulate the interactions ... enclosure effects, multi-body interactions, and B-spline representation of body shapes. We simulate different...


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## 幾何学コロキウム 「Moment Angle Complex and cohomology of toric orbifolds」

2011年 　 7月 22日 16時 30分 ～ 2011年 　 7月 22日 18時 00分

Abstract: The moment angle complex associated to a simplicial complex
is a topological space with a torus action whose (equivariant)
cohomology is used to study the combinatorial structure of the
simplicial complex, topologically and algebraically. On the other
hand, it is also a generalization of the level sets in the
construction of symplectic toric manifolds. In this talk, after (1) a
brief introduction to the subject, I will talk about  (2) algebraic
properties of cohomology of toric orbifolds and (3) connected sums of
simplicial complexes and the cohomology of corresponding moment angle
complexes.

Abstract: The moment angle complex associated to a simplicial complex
> is a topological space with a torus action whose (equivariant)
> cohomology is used to study the combinatorial structure of the
> simplicial complex, topologically and algebraically. On the other
> hand, it is also a generalization of the level sets in the
> construction of symplectic toric manifolds. In this talk, after (1) a
> brief introduction to the subject, I will talk about  (2) algebraic
> properties of cohomology of toric orbifolds and (3) connected sums of
> simplicial complexes and the cohomology of corresponding moment angle
> complexes.
Abstract: The moment angle complex associated to a simplicial complex
> is a topological space with a torus action whose (equivariant)
> cohomology is used to study the combinatorial structure of the
> simplicial complex, topologically and algebraically. On the other
> hand, it is also a generalization of the level sets in the
> construction of symplectic toric manifolds. In this talk, after (1) a
> brief introduction to the subject, I will talk about  (2) algebraic
> properties of cohomology of toric orbifolds and (3) connected sums of
> simplicial complexes and the cohomology of corresponding moment angle
> complexes.

format: plain


text:
# Tag Info

1

The function wouldn't be one-to-one otherwise. The size of the set in the left is $|A|$, and its image would be of size at most $2^l<2^{H_0}=|A|$, therefore, the function wouldn't be injective and thus you wont be able to de-compress some of the data. Therefore it will be a lossy-compression for any $l<H_0$.

2

One possible direction is looking at the (computational) hardness of distributions. For example, a family of distributions $\{\mathcal{P}_n\}_{n\in\mathbb{N}}$ where $\mathcal{P}_n$ is a distribution over $\{0,1\}^n$ is called polynomialy-samplable (you can substitute here any other type of time/space restriction) if there exists a probabilistic polynomial ...

2

Suppose that we have a random source $X$, and we wish to encode many independent samples $x_1,\ldots,x_n$ taken from $X$. An compression scheme is an algorithm that converts $n$ samples of $X$ to a binary string in a reversible way. Shannon showed that the optimal compression scheme produces binary strings of length roughly $nH(X)$, where $H(X)$ is the ...

Top 50 recent answers are included


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# An engine of weight 6.5 metric ton is going up an incline of $\large\frac{5}{13}$ at the rate of $9\; km/hr$ .The power of engine , if co-efficient of friction is $\large\frac{1}{12}$ is

$(a)\;83.5\;KW \quad (b)\;97.5\;KW \quad (c)\;60.5\;KW \quad (d)\;73.5\;KW$

$\sin \theta=\large\frac{5}{13}$
$\cos \theta=\sqrt {1-\sin ^ \theta}$
$\qquad= \sqrt {1-\bigg(\large\frac{5}{13}\bigg)^2}$
$\qquad=\large\frac{12}{13}$
Force against which the engine has to work
$F=mg \sin \theta+\mu mg \cos \theta$
$\quad=6500 \times 9.8 \bigg[\large\frac{5}{13}+\frac{1}{12} \times \frac{12}{13}\bigg]$
$\quad=29400 \;N$
Engine power $=Fv$
$\quad=29400 \times 2.5 \qquad 9km/hr=2.5 m/s$
$\quad=73500 \;W$
$\quad=73.5 KW$
Hence d is the correct answer.

edited Feb 10, 2014 by meena.p


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# Geographic coordinate conversion

Geographic coordinates consist of latitude and longitude.

## Ways of writing coordinates

All of the following are valid and acceptable ways to write geographic coordinates:

• -23.399437,-52.090904
• 40:26:46.302N 079:56:55.903W
• 40°26′47″N 079°58′36″W
• 40d 26′ 47″ N 079d 58′ 36″ W
• 40.446195N 79.948862W
• 40.446195, -79.948862
• 40° 26.7717, -79° 56.93172
• N40:26:46.302 W079:56:55.903
• N40°26′47″ W079°58′36″
• N40d 26′ 47″ W079d 58′ 36″
• N40.446195 W79.948862

Coordinates in this encyclopedia are generally written as 40°26′21″N 079°58′36″W / 40.43917°N 79.97667°W.

## Basic forms

There are three basic forms of a coordinate.

1. Coordinate containing degrees (integer), minutes (integer), and seconds (integer, or real number) (DMS).
2. Coordinate containing degrees (integer) and minutes (real number) (MinDec).
3. Coordinate containing only degrees (real number) (DegDec).

All forms of coordinates are capable of representing the same amount of data and the same precision. Depending on which type of coordinate you are provided with, and which type you would like to work with, you may have to do some conversion.

## Components of a typical coordinate

In its most simple form a coordinate is just a number of degrees. The tricky part comes in when you need to differentiate North/South latitude or West/East longitude, or make the number more digestible by writing it with minutes and seconds instead of as a decimal number.

### Degrees

The degrees portion of the coordinate is always going to be the easiest to figure out. The degrees is always the left-most whole number. For example:

 40:26:46N         40 W87°43′41        -087

A sphere is divided into 360 degrees. The number space is divided into two halves, East and West in the case of longitude and North and South in the case of latitude. The maximum ranges are as follows:

 Longitude  180 W   = -180  180 E   =  180
 Latitude   90 N   =   90   90 S   =  -90

Technically you could have latitudes greater than 90 or less than -90, but this is an ambiguous case, since there would be an equivalent coordinate with an inverse longitude.

The minimal case is that you have only degrees:

  50.21389 or  50.21389N

### Minutes

Minutes are an optional component, as is implied by the minimal case of degrees. If there is no minutes component, the degrees component contains the entire precision of the coordinate and there must not be a seconds component. Minutes are actually the numerator component of a fraction with denominator 60 of one degree.

With the same examples as above:

 40:26:46N         26 W087°43′41         43

In the first case, the number of minutes is 26.

### Seconds

Seconds are also an optional component, and can only exist if the minutes component also exists. Seconds are the numerator component of a fraction with denominator 60 of one minute.

 40:26:46N         46 W087°43′41         41

In the second case, the number of seconds is 41.

To convert, 41 seconds is equal to $\frac{41}{60} = 0.68\bar{3}$ minutes.

## Putting it all together

### Conversion from DMS to Decimal Degree

Given a DMS (Degrees, Minutes, Seconds) coordinate such as W087°43′41″, convert it to a number of decimal degrees using the following method:

1. Calculate the total number of seconds:
43′41″ = (43*60 + 41) = 2621 seconds.
2. The fractional part is total number of seconds divided by 3600:
2621 / 3600 = ~0.728056
3. Add fractional degrees to whole degrees to produce the final result:
87 + 0.728056 = 87.728056
4. Since it is a West longitude coordinate, negate the result.
5. The final result is -87.728056.

### Conversion from MinDec to Decimal Degree

Given a MinDec (Degrees, Minutes, Decimal Minutes) coordinate such as 40°26.7717N, 79°56.93172W, convert it to a number of decimal degrees using the following method (for example for 79°56.93172W):

1. The integer number of degrees is the same (79)
2. The decimal degrees is the decimal minutes divided by 60 (56.93172/60 = 0.948862)
3. Add the two together (79 + 0.948862 = 79.948862)
4. Negate the value if it is South or West (in this case, West, so -79.948862)

### Conversion from Decimal Degree to DMS

Given a decimal longitudinal coordinate such as -87.728055 it will be necessary to know whether it is a latitudinal or longitudinal coordinate in order to fully convert it. The method is as follows:

• Subtract the whole number portion of the coordinate, leaving the fractional part. The whole number is the number of degrees. 87.728055 = 87 degrees.
• Multiply the remaining fractional part by 60. This will produce a number of minutes in the whole number portion. 0.728055 x 60 = 43.6833 = 43 minutes.
• Multiply the fractional part of the number of minutes by 60, producing a number of seconds. 0.6833 x 60 = 40.998 = 41 seconds. It is possible to count this as 40 seconds, truncating the decimal, round it to 41, or keep the entire number.
• Depending on whether the source number was a latitudinal or longitudinal coordinate, and the sign of the number, add the N/S/E/W specifier. The following table shows the possibilities:
   Type   Dir.   Sign    Test   Lat.   N      +       > 0   Lat.   S      -       < 0   Long.  E      +       > 0   Long.  W      -       < 0

A latitude of 0°0′0″ (The Equator) is neither North nor South. Similarly, a longitude of 0°0′0″ (Prime Meridian) is neither East nor West. These are referred to as zero latitude and zero longitude, respectively. A longitude of 180°0′0″ (180th meridian) is neither East nor West. This is the basis for the International Date Line when referring to the Earth.

• The final result is: W 87°43′41″.

#### Programmatical conversion

The most common programmatical use of these processes is to display a coordinate to an end user in the more common DMS form instead of decimal form. Below is a piece of pseudocode to convert from decimal degrees to degrees, minutes, and seconds:

 function deg_to_dms ( degfloat )    Input must be non-negative:    if degfloat < 0       error    end if    Compute degrees, minutes and seconds:    deg ← integerpart ( degfloat )    minfloat ← 60 * ( degfloat - deg )    min ← integerpart ( minfloat )    secfloat ← 60 * ( minfloat - min )    Round seconds to desired accuracy:    secfloat ← round( secfloat, digits )    After rounding, the seconds might become 60. These two    if-tests are not necessary if no rounding is done.    if secfloat = 60       min ← min + 1       secfloat ← 0    end if    if min = 60       deg ← deg + 1       min ← 0    end if    Return output:    return ( deg, min, secfloat ) end function

#### Java Implementation

 // Input a double latitude or longitude in the decimal format // e.g. 87.728056 String decimalToDMS(double coord) {         String output, degrees, minutes, seconds;           // gets the modulus the coordinate divided by one (MOD1).         // in other words gets all the numbers after the decimal point.         // e.g. mod = 87.728056 % 1 == 0.728056         //         // next get the integer part of the coord. On other words the whole number part.         // e.g. intPart = 87           double mod = coord % 1;         int intPart = (int)coord;           //set degrees to the value of intPart         //e.g. degrees = "87"           degrees = String.valueOf(intPart);           // next times the MOD1 of degrees by 60 so we can find the integer part for minutes.         // get the MOD1 of the new coord to find the numbers after the decimal point.         // e.g. coord = 0.728056 * 60 == 43.68336         //      mod = 43.68336 % 1 == 0.68336         //         // next get the value of the integer part of the coord.         // e.g. intPart = 43           coord = mod * 60;         mod = coord % 1;         intPart = (int)coord;         if (intPart < 0) {            // Convert number to positive if it's negative.            intPart *= -1;         }           // set minutes to the value of intPart.         // e.g. minutes = "43"         minutes = String.valueOf(intPart);           //do the same again for minutes         //e.g. coord = 0.68336 * 60 == 41.0016         //e.g. intPart = 41         coord = mod * 60;         intPart = (int)coord;         if (intPart < 0) {            // Convert number to positive if it's negative.            intPart *= -1;         }           // set seconds to the value of intPart.         // e.g. seconds = "41"         seconds = String.valueOf(intPart);           // I used this format for android but you can change it          // to return in whatever format you like         // e.g. output = "87/1,43/1,41/1"         output = degrees + "/1," + minutes + "/1," + seconds + "/1";           //Standard output of D°M′S″         //output = degrees + "°" + minutes + "'" + seconds + "\"";           return output; }          /*         * Conversion DMS to decimal          *         * Input: latitude or longitude in the DMS format ( example: N 43° 36' 15.894")         * Return: latitude or longitude in decimal format            * hemisphereOUmeridien => {W,E,S,N}         *         */         public double DMSToDecimal(String hemisphereOUmeridien,double degres,double minutes,double secondes)         {                 double LatOrLon=0;                 double signe=1.0;                   if((hemisphereOUmeridien.equals("W"))||(hemisphereOUmeridien.equals("S"))) {signe=-1.0;}                                 LatOrLon = signe*(Math.floor(degres) + Math.floor(minutes)/60.0 + secondes/3600.0);                   return(LatOrLon);                        }


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# On the worst case performance of the steepest descent algorithm for quadratic functions

\begin{abstract} The existing choices for the step lengths used by the classical steepest descent algorithm for minimizing a convex quadratic function require in the worst case $\Or(C\log(1/\eps))$ iterations to achieve a precision $\eps$, where $C$ is the Hessian condition number. We show how to construct a sequence of step lengths with which the algorithm stops in $\Or(\sqrt{C}\log(1/\eps))$ iterations, with a bound almost exactly equal to that of the Krylov space methods. \end{abstract}

## Citation

Federal University of Santa Catarina, Brazil, September 2014.


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# Tag Info

1

Since $\delta \omega_{\rho\sigma}$ is antisymmetric, the part of what's inside the parentheses symmetrized over $\rho$ and $\sigma$ will vanish because if you contract antisymmetric indices with symmetric ones, you get zero. So you can anti-symmetrize over that pair of indices inside the parentheses without changing anything, which is precisely what's ...

3

Maybe I am missing something but $$\sum_{n \neq 0} e^{-i E_n (T-i\epsilon)} |n \rangle \langle n|0\rangle =\sum_{n \neq 0} e^{-i E_n T} e^{-\epsilon E_n } |n \rangle \langle n|0\rangle$$ and not a sum of series. This really allows you then to kill non zero $n$ as $T$ goes to infinity. As for the physical meaning of this trick, ...

1

The Lagrangian density for a Dirac field is $$\mathcal{L} = i\bar\psi\gamma^\mu\partial_\mu\psi -m \bar\psi\psi$$ The Euler-Lagrange equation reads $$\frac{\partial\mathcal{L}}{\partial\psi} - \frac{\partial}{\partial x^\mu}\left[\frac{\partial\mathcal{L}}{\partial(\partial_\mu\psi)}\right] = 0$$ We treat $\psi$ and $\bar\psi$ as independent dynamical ...

3

I wouldn't say that "holographic theories are non-local by definition". On the contrary, in AdS/CFT the CFT is completely local and satisfies cluster decomposition. The cluster decomposition property in AdS can be proved using the CFT bootstrap for all CFTs in $d > 2$ (see http://arxiv.org/abs/arXiv:1212.3616, the proof only requires CFT `axioms', ...

1

Perhaps not a totally satisfactory answer, but a partial clarification of one of the things I was confused about: In the semiclassical treatment of the Hawking radiation process, there is no need to have an interacting quantum field theory. Therefore the vacuum-vacuum bubble diagrams of interacting perturbation theory are completely irrelevant to the basic ...

7

One shouldn't imagine the T-duality between the two heterotic strings to be a $Z_2$ group, like in the case of type II string theories' T-duality. In type II string theory, there is only one relevant scalar field, the radius of the circle producing T-duality, and it gets reverted $R\to 1/R$ under T-duality. In the heterotic case, it's more complicated ...

2

The magnetic quadrupole moment tensor is given by $$m_{ij}=\left\langle \frac{2}{3}\left(\mathbf{r}\times\mathbf{J}\right)_i r_j \right\rangle,$$ in analogy with the magnetic dipole moment vector $$m_i=\left\langle \frac{1}{2}\left(\mathbf{r}\times\mathbf{J}\right)_i \right\rangle.$$ The magnetic field at a point $\mathbf{R}$ is then, up to quadrupole ...

3

Yes it is. The volume form on any (pseudo-)Riemannian manifold $(M,g)$ of dimension $n$, where $g$ is the metric, is given in local coordinates $(x^1, \dots, x^n)$ $$\sqrt{|\det (g_{\mu\nu})|}dx^1\wedge \cdots \wedge dx^n$$ where $\det(g_{\mu\nu})$ is the determinant of the metric in these coordinates. In cartesian coordinates, the determinant of the ...

5

Because you can prepare a state with an arbitrarily long wavelength, hence arbitrarily low energy, photon. That's essentially the definition of a massless particle. If you put in an IR regulator, by putting the system in a box for example, a gap appears since there is now a largest possible wavelength. This can be mimicked by giving the photon a small mass. ...

1

One way of defining conformal transformations are by a (positive) local scaling of the metric, of the form $e^{2 \phi}$. Such a transformation always preserves the sign of spacetime distances. In particular, the light-cone remains unchanged since null distances map to null distances. What was inside the lightcone stays inside and things outside stay outside. ...

0

Here is my solution. For the Ramond Ramond Sector, $${m_\rm{I}} = \frac{{\left[ {\hat a_ + ^\mu ,\hat{\tilde a}_ + ^\nu } \right]}}{2}{m_{{\rm{IIB}}}}$$ For the Neveu-Schwarz Neveu-Schwarz Sector, $${m_\rm{I}} = \frac{{\left[ {\hat d_ {-1/2} ^\mu ,\hat{\tilde d}_{-1/2} ^\nu } \right]}}{2}{m_{{\rm{IIB}}}}$$ For the Ramond Neveu-Schwarz Sector or the ...

2

The theory is (even classically) not scale invariant. Just by dimensional analysis, you can note that the scalar field has scaling dimension 1, and the mass (as the name suggests) must also have a scaling dimension of 1. So $m^2$ has a ascaling dimension of 2, which suggests the RG equation which you've written in the question. That essentially says that the ...

0

Decoherence is more than anything a matter of what you define to be the "environment". The environment is supposed to be external to the system of interest and entangling interaction with it produces decoherence. If the environment in question is a part of the adS space then the subsystem can certainly decohere. If what you are asking is whether the space as ...

2

Hagedorn spectrum just means that the density of states varies exponentially with the energy/mass. $m^2$ (asymptotically) given by the "level" (N) of the state (upto a sqrt). The number of states at level $N$ corresponds to the possible partitions of $N$ into different oscillator modes. That means that the number of states at level $N$ will increase ...

1

I would really recommend a study in QFT before going on to study SUSY. QFT has many quirks that make supersymmetry a very interesting expansion of the regular framework. You'd miss out on all that as you just had to believe the facts presented w/o following the thought that lead to the results in detail. On the Mathematical level you will need Grassmann ...

0

Generally, in the MSSM one works with the "minimal flavor violation" paradigm that states that all flavor violation originates in the SM Yukawa sector. This paradigm is ad hoc, but explains why no huge SUSY contributions to FCNC observables are seen. There are models that go beyond minimal flavor violation and some that give an explanation for the ...

1

Probability of photon emission by an atom depends on the occupation number of already existing photons of this sort. The corresponding occupation number is determined with the "boundary conditions". We cannot take the occupation number of all existing photons in the Universe, so time-space "separated" regions have effectively their own photon fields and ...

0

Let us consider the Hamiltonian of an electromagnetic field in a finite region of space. It is possible to find a suitable unitary transformation to change it in a form same as a Hamiltonian of a independent harmonic oscillators. A photon is a quantum of energy of such an ensemble of independent oscillators. All excitations of this ensemble will be ...

5

I believe it is the same field for all photons in the universe. Similarly there is one field for all up quarks in the universe, another field for all down quarks in the universe and so on for each particle type. In this way the field theory model explains the uncanny consistency of particles; a neutrino made in a supernova millions of miles away is identical ...

0

I don't know if I'm right and it's not directly my topic, but naively guessing I think that sea quarks emerge from fluctuations in the vacuum, for example as a pair of a quark and its antiquark. And as such, because of conservation of angular momentum, each of these pairs then should have zero total angular momentum.

1

1PI graphs don't contain singularities at $q^2=0$ because those only arise from propagators that carry the external photon momentum $q$. The external ones are omitted (as factors), as you said, and if the graphs had a single propagator with the momentum $q$, it could be cut to two pieces by cutting this propagator and this is by definition a diagram that is ...

1

There are lots of questions here! I think I can answer at least some... First of all, you are aware that the fields in $W$ and $K$ are superfields? These contain the entire supermultiplet, so they must be complex in general. This is a short entry but it links to others: http://en.wikipedia.org/wiki/Superfield As mentioned by Jose in his comment, the ...

2

The Pion fields are the coordinates of the Stereographic projection: $\phi_i = \frac{2 \pi_i}{1 + \pi^2} , i = 1, ..., n-1$ Where: $\pi^2 = \sum_{i=1}^{N-1} \pi_i\pi_i$ And: $\phi_n = \frac{-1 + \pi^2}{1 + \pi^2}$ As can be seen, this construction solves the constraint equation: $\sum_{a=1}^{N} \phi_a\phi_a= 1$. Substituting in the Lagrangian, we ...

0

If you solve the equation to second order you obtain: $g=\frac{g_0}{1-a g_0 log\lambda}=g_0 +\sum_n g_0^{n+1}a^nlog^n \lambda$ the second part is the contribution of powers of the first loop diagram, for instance try to write g at 1-loop first and then sum all the powers of the first order diagram.

4

Very rougly, the logarithm has the function of transforming a product into a sum, therefore if the partition function is factorizable (i.e. disconnected), it will give a sum on connected parts. For a slighly more rigorous discussion, consider the Hamiltonian of two non interacting systems, the partition function $z(j)=Z(j)/Z(0)$ (without vacuum ...

0

try the lecture notes by van Hees from Frankfurt. Check out p. 114 http://faculty.ksu.edu.sa/djdou/Lectures%20Notes%20PHY556/Introduction%20to%20Q.F.T.pdf

3

The NS-NS sector is the same in type IIA and IIB, but the R-NS and NS-R sectors differ. The type IIA theory is non-chiral, so the R-NS and NS-R fields transform in $\mathbf{8}_s \otimes \mathbf{8}_v$ and $\mathbf{8}_v \otimes \mathbf{8}_s'$, where $\mathbf{8}_s$ and $\mathbf{8}_s'$ are the two eight-dimensional spinor representations of $SO(8)$. Type IIB, on ...

4

The only argument I can find for this is a couple pages earlier, where they say ...which in turn implies that the photon self-energy diagrams have the structure $= i(q^2 g^{\mu\nu} - q^\mu q^\nu)\Pi(q^2)$ The only divergence possible is a logarithmically divergent contribution to $\Pi(q^2)$. In non-Abelian gauge theories, (16.57) still holds, ...

2

The most physical and understandable definition of Nekrasov's partition function to me uses five-dimensional gauge theories. Namely, any 4d N=2 susy gauge theory has a 5d version with the same matter content, so that compactifying it on a small $S^1$ brings it back to the original 4d theory. Then we put the theory on the so-called Omega background: it is ...

4

For simplicity, let's restrict the discussion to that of a single particle moving in one dimension. Path integrals can be performed in much broader contexts like quantum field theory, but I think that would conceptually obscure the issue at this point. Let $H$ denote the (time-independent) quantum hamiltonian. Then the time evolution on the system is ...

0

What your friend actually meant is that you can obtain all desirable correlation functions. Assuming you're talking about a non-relativistic electron, consider a source term added to your action $$S'[x] = S[x] + \int dt \, J(t) x(t)\,.$$ Now you can write any correlation function as a derivative of $\ln Z$ calculated at $J = 0$, i.e. $$\langle B| ... 2 In QM we have the coordinate representation. In this case the basis consists of eigenfunctions of the position operator r_i, where i runs over the discrete set of degrees of freedom. Now, when we go to QFT, i generalizes to the continious space-time position x, and r generalizes to the field \phi:$$ r\rightarrow\phi\\ i\rightarrow x\\ ...

2

I am not sure that one will appreciate this answer, but I will try. I think that the punchline may not always be the case. It seems that you want some symmetry which forbids some counterterm. This means that this counterterm would have some unappreciated behaviour under the symmetry. Then you want the same symmetry to forbid the divergence. Why? Naively, ...

0

There's no related exception for tachyons. Tachyons' statistics must be determined a priori. Most typically, tachyons have to be bosons – and under certain additional assumptions, they have to be scalar (spin-zero) bosons. They differ from massive bosons just by the fact that the mass term $m^2\phi^2/2$ has the opposite sign – opposite sign of their $m^2$. ...

1

First, a reference article, by Witten, http://arxiv.org/pdf/hep-th/9802150v2.pdf I'll try to expose the basic idea, with a flat space-time. Suppose you have a relativistic scalar field theory, on a flat space-time domain, with boundary. The equation of the field is : $$\square \Phi(x) = 0$$ (fields on-shell) Now, define the partition function $$Z = ... 0 I asked my professor and in a discussion we came up with the following. The process of establishing the effective action for a fluctuation Lagrangian to consist of the functional determinant of the initial differential operator involved, relies on the equality:$$\det(A)=e^{Tr(\log(A))}$$for a matrix A, which is only true for diagonalizable ... 1 The pole corresponds to an on-shell particle going from one point to another. Then, the residue effectively tells you how many of those particles are being transmitted. Since in your physical/renormalized theory, the propagator should correspond to 1 quantum of the renormalized field being transmitted, you set the residue at the pole to 1. 0 TO have an instanton solution, you need to map the (euclideanized) "spacetime at infinity" to the group manifold. In the case of SU(2), both the spacetime at infinity and the group manifold are S^3 and instantons are characterized by the integers. I hope you understand that much, at least for SU(2). If you're interested in 4d instantons, they are ... 2 Yes, suppose [g] = \delta. By dimensional analysis only we can write that a loop diagram contributes$$ \sim g^{n} \int \frac{d^4 k}{k^{4-n\delta}} $$If \delta=0, this diverges logarithmically, but can be re-normalized. If \delta is less than zero, it diverges by simple power counting. This is VERY informal. Technically, you should study the ... 1 The OS condition that$$ \frac{\partial\Sigma}{\partial p^2}|_{p^2=-m^2} = 0 $$implies that the residue in the propagator remains equal to one. Suppose that we used a different renormalization scheme in which our counter-terms contain no finite parts (e.g. MS scheme). In the OS scheme, we removed finite parts which were logarithmic in our artificial ... 3 For Poincare algebra there are (as far as I know) two different approaches to find its representations. In first approach one begins from a finite dimensional representation of (complexified) Lorentz algebra, and using it one constructs a representation on space of some fields on Minkoski space. Representation so obtained is usually not irreducible and an ... 0 Gauge symmetries are redundancies of the description, not a part of physics; gauge fields have surplus structures (e.g., non-physical polarizations) one brings in to describe the system more conveniently (for instance in a local and manifestly Lorentz covariant form). You can describe a gauge system in a language that does not have gauge symmetry at all. A ... 1 First of all, we don't usually talk about the direction of propagation of a plane wave in QFT. Plane waves are said to exist at all spacetime coordinates with a certain internal momentum, k. And, in reference to some of the comments, in QFT, we don't normally operate with wavefunctions. We promote wavefunctions to operators and act on states. But in this ... 4 With respect to the discussion of momentum-eigenstates and the following derivation in Weinberg's book, \sigma is just a label that denotes any degree of freedom that is not momentum. Even though it can be identified with spin, its nature is not relevant for the discussion at hand. 1 The statement can be understood in terms of the GKPW-formula (named after Gubser, Klebanov, Polyakov and Witten), which does exactly that: it relates correlation functions on the CFT side (boundary) to string amplitudes on the AdS side (bulk). Assume that \phi(\vec{x},z) is some field in the bulk, where z is the so-called "holographic coordinate", which ... 2 The the easiest way to see that time reversal transforms electrons into positrons relies on the fact that PCT (parity, charge conjugation and time reversal) combined are a symmetry of every Lorentz-invant QFT. Using P^{-1} = P, C^{-1} = C, T^{-1} = T, i.e. a parity transformation is undone by a second parity transformation etc. you can see that$$1 = ...

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This link has a very easy to follow derivation of the scalar field Feynman propagator. Once you understand that, you should be able to do it for vector particles like photons too. The correct answer is given at the end of the chart in this link.

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I'll give you enough hints to complete the proof yourself. If you're desperate, I'm following the notes by Zuber, which are available online, IIRC. Let's start with some notation: pick some basis $\{t_a\}$ of your Lie algebra, then $$[t_a,t_b] = C_{ab}{}^c t_c$$ defines the structure constants. If you define $$g_{ab} = C_{ad}{}^e C_{be}{}^d,$$ then this ...

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I) The Casimir invariants of a Lie algebra $L$ over a field $\mathbb{F}$ are the central elements of the universal enveloping algebra $U(L)$. Example: The angular momentum square $\vec{J}^2$ is a quadratic Casimir invariant of the Lie algebra $L=sl(2,\mathbb{C})$. II) Given a bilinear associative/invariant form $B:L\times L\to \mathbb{F}$, one can create ...

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You can write an infinitesimal transformation, with generator $J$, as $$R(\delta\theta) = 1 + iJ\delta\theta$$ A finite transformation is a succession of $N\to\infty$ infinitesimal transformations, $$R(\theta) = (1 + iJ\theta/N)^N = e^{iJ\theta}$$ The rotations $O(3)$ are isomorphic to $SU(2)$, with generators $J = \sigma/2$. The Lorentz ...

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Reference : An Overview of Generic Tools for Information-Theoretic Secrecy Performance Analysis o...
 Document type : E-prints/Working papers : Already available on another site Discipline(s) : Engineering, computing & technology : Electrical & electronics engineering To cite this reference: http://hdl.handle.net/10993/45001
 Title : An Overview of Generic Tools for Information-Theoretic Secrecy Performance Analysis over Wiretap Fading Channels Language : English Author, co-author : Kong, Long [University of Luxembourg > Interdisciplinary Centre for Security, Reliability and Trust (SNT) > SigCom >] Ai, Yun [] Lei, Lei [University of Luxembourg > Interdisciplinary Centre for Security, Reliability and Trust (SNT) > SigCom >] Kaddoum, Georges [] Chatzinotas, Symeon [University of Luxembourg > Interdisciplinary Centre for Security, Reliability and Trust (SNT) > SigCom >] Ottersten, Björn [University of Luxembourg > Interdisciplinary Centre for Security, Reliability and Trust (SNT) > >] Publication date : 2020 Peer reviewed : Yes Keywords : [en] Physical layer security (PLS) ; Channel state information (CSI) ; mixture gamma (MG) ; mixture of Gaussian (MoG) ; Fox's H-function ; artificial noise (AN) ; artificial fast fading (AFF) ; wiretap fading model ; jamming ; antenna selection Abstract : [en] An alternative or supplementary approach named as physical layer security has been proposed to afford an extra security layer on top of the conventional cryptography technique. In this paper, an overview of secrecy performance investigations over the classic Alice-Bob-Eve wiretap fading channels is conducted. On the basis of the classic wiretap channel model, we have comprehensively listed and thereafter compared the existing works on physical layer secrecy analysis considering the small-scale, large-scale, composite, and cascaded fading channel models. Exact secrecy metrics expressions, including secrecy outage probability (SOP), the probability of non-zero secrecy capacity (PNZ), average secrecy capacity (ASC), and secrecy bounds, including the lower bound of SOP and ergodic secrecy capacity, are presented. In order to encompass the aforementioned four kinds of fading channel models with a more \textit{generic} and \textit{flexible} distribution, the mixture gamma (MG), mixture of Gaussian (MoG), and Fox's $H$-function distributions are three useful candidates to largely include the above-mentioned four kinds of fading channel models. It is shown that they are flexible and general when assisting the secrecy analysis to obtain closed-form expressions. Their advantages and limitations are also highlighted. Conclusively, these three approaches are proven to provide a unified secrecy analysis framework and can cover all types of independent wiretap fading channel models. Apart from those, revisiting the existing secrecy enhancement techniques based on our system configuration, the on-off transmission scheme, jamming approach (including artificial noise (AN) & artificial fast fading (AFF)), antenna selection, and security region are presented. Permalink : http://hdl.handle.net/10993/45001 source URL : https://arxiv.org/abs/2009.05976 FnR project : FnR ; FNR11607830 > Bjorn Ottersten > CI-PHY > Exploiting interference for physical layer security in 5G networks > 01/02/2018 > 31/01/2021 > 2017

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The so-called instrumentation amplifier builds on the last version of the differential amplifier to give us that capability: Understanding the Instrumentation Amplifier Circuit. The result is a circuit with very high CMRR, high gain, and input impedance on the order of 1010 Ohms. $12.60$ 12. At the input stage, there is a transducer device that converts the change in the physical quantity to an electrical signal. Low noise 4. allows an engineer to adjust the gain of an amplifier circuit without having to change more than one resistor value One such tool every EE should have in his belt is instrumentation amplifiers, or in-amps. In this video, the instrumentation amplifier has been explained with the derivation of the output voltage. FREE Shipping on orders over \$25 shipped by Amazon. R To recap from the previous lecture, an Instrumentation Amplifier, or IA, can be idealized as an electrical circuit that amplifies the signals that differ between its inputs and rejects the signals that are the same on both inputs. For best matching, those would need to be on the same silicon chip, and thus something like AD8222 comes to mind. Where to use the AD620 Instrumentation Amplifier? 3 LIVISN AD623 AD620 Millivolt Microvolt Voltage Signal Amplifier Module Adjustable Magnification Instrumentation Amplifier Module Mini Voltage Amplifier Module. Examples include INA128, AD8221, LT1167 and MAX4194. Otherwise, you'd use just one in-amp and couple it to a differential driver. Alternatives Amplifiers: LM4871, IC6283, JRC45558 . Conventionally, instrumentation amplifiers are realized with three operational amplifiers (op-amp), as presented in Fig. The main function of this amplifier is to diminish surplus noise that is chosen by the circuit. An instrumentation amplifier is used to amplify very low-level signals, rejecting noise and interference signals. OK. Login; Sign up; Features; Pricing; Circuits. Examples can be heartbeats, blood pressure, temperature, earthquakes and so on. The equation below uses a difference amplifier with G = 1V/V, and T R is the resistor tolerance: If T R = 1%, worst-case dc CMRR DIFF will be 34 dB If T R = 0.1%, worst-case dc CMRR DIFF will be 54 dB The main advantage I see is that we can control the gain easily by the one resistor Rg, when compared to if a standard differential amplifier was used (since you need to select the resistors etc). One of the applications these circuits are used for is taking measurements from sensors and transducers. of what an instrumentation amplifier is, how it operates, and how and where to use it. Because of … A instrumentation amplifier is a special purpose linear amplifier used to amplify low level signals. The importance of an instrumentation amplifier is that it can reduce … Instrumentation amps also find usage in industrial automation, where many systems use current flow to relay measurements and control remote installations. The in-amps are w Get it as soon as Tue, Dec 8. Feedback-free instrumentation amplifier is the high input impedance differential amplifier designed without the external feedback network. These devices amplify the difference between two input signal voltages while rejecting any signals that are common to both inputs. An IC instrumentation amplifier typically contains closely matched laser-trimmed resistors, and therefore offers excellent common-mode rejection. Non-inverting amplifier. and high input impedance because of the buffers. Instrumentation Amplifiers are available at Mouser Electronics. CMMR stands for common mode rejection ratio, it is the ability to reject unwanted signals. An Instrumentation amplifier is an integrated circuit (IC) used to amplify a signal, which is a type of differential amplifier because it amplifies between two input signal. An instrumentation amplifier (INA) is a very special type of differential input amplifier; its primary focus is to provide differential gain and high common-mode rejection. 60. Resources. A set of switch-selectable resistors or even a potentiometer can be used for Definition: A special type of amplifier that is used to amplify signals of extremely low-level is known as Instrumentation Amplifier. With such widespread use, this is a device every engineer should have in his tool belt. Some well known standard text books make this error too. ;), Practical Uses of Instrumentation Amplifiers, MIT Gives Computers the Capability to Predict the Future with Deep Learning, Protecting IoT Embedded Designs with Security ICs, Op-Amps as Low-Pass and High-Pass Active Filters. These amplifiers use switches for modulating and chopping purposes. This is import when working with microcontrollers, as any extra noise will cause erratic conversion, in addition to wasting valuable ADC bits. not all amplifiers used in instrumenta-tion applications are instrumentation amplifiers, and by no means are all in-amps used only in instrumentation applications. The resistance of the transducer changes due to changes in the physical parameter such temperature, pressure, level, light intensity etc. Below is the implementation of Analog Device’s AD82X series of instrumentation amplifier in an ECG, from their application manual. The buffer gain could be increased by putting resistors between the buffer inverting inputs and ground to shunt away some of the negative feedback; however, the single resistor The differential operational amplifier can be used as an automatic gain control circuit. Incidentally, if you ever wondered what that button on many of our multimeters was that read “4-20mA”, now you know. Instrumentation amplifiers have uses in nearly every field of electronics; they fulfill a specific role in circuits needing the advantages of high input impedance with good gain while providing common mode noise rejection and fully differential inputs. An excellent choice, but now I had to build the circuit. Operational amplifiers (A1 and A2) are connected in inverting configuration, while op-amp (A3) is a differential amplifier. Instrumentation amplifiers can be also made using two opamps, but they are rarely used and the common practice is to make it using three opamps like what is shown here. An IMPORTANT NOTICE at the end of this data sheet addresses availability, warranty, changes, use in safety-critical applications, intellectual property matters and other important disclaimers. Types of Chopper Amplifier. The reason that a bridge circuit is so crucial with sensors and instrumentation is common mode noise; a circuit with a regular op-amp and a sensor across the inputs would function as an amp, but it would be very noisy. "Don't fall in love with one type of instrumentation amp - 2002-05-30 07:00:00", "Amplifiers for bioelectric events: a design with a minimal number of parts", Interactive analysis of the Instrumentation Amplifier, Lessons In Electric Circuits — Volume III — The instrumentation amplifier, A Practical Review of Common Mode and Instrumentation Amplifiers, A Designer's Guide to Instrumentation Amplifiers (3rd Edition), Three is a Crowd for Instrumentation Amplifiers, Instrumentation Amplifier Solutions, Circuits and Applications, Fixed-gain CMOS differential amplifiers with no external feedback for a wide temperature range (Cryogenics), https://en.wikipedia.org/w/index.php?title=Instrumentation_amplifier&oldid=942222689, Creative Commons Attribution-ShareAlike License, This page was last edited on 23 February 2020, at 11:09. Initially, I powered the Arduino from PC and the analog circuitry (the Wheatstone bridge and INA121) was powered from 9V battery. R Product Overview Precision amplifier systems for a wide range of test and measurement applications. The instrumentation amplifier is also called an instrumentational amplifier and written in short form as In-Amp. {\displaystyle R_{\text{2}}} In the last lecture, we derived the simplified equation for a one-amp IA. Instrumentation amps are used for this device because the biopotential electrodes pick up a huge amount of power line noise, which needs to be rejected so the device can give accurate readings. 4-Channel Variable Gain Amplifier Contact Us. Under the conditions of R1=R2, R3=R4, Rf=R5, the gain of the circuit in Figure 1 is: G=(1+2R1/Rg)(Rf/R3) It can be seen from the formula that the adjustment of the circuit gain can be achieved by changing the Rg resistance. Likewise, an electrical engineer has his “tool belt” of knowledge and applications for components, circuit designs, and problem solving. Can you please provide a link to your references? It cancels out any signals that have the same potential on both the inputs. Instrumentation amplifiers are by far the most common interface circuits that are used with pressure sensors. Linear operation of an instrumentation amplifier depends upon linear operation of its building block: op amps. Remember that this simplified equation relies on the matching of R1 to R3 and R2 to R4. An Instrumentation amplifier amplifies the difference between its input signals. An instrumentation amplifier (InAmp) is an integrated circuit or operational amplifier topology that has high input impedance differential inputs and a low impedance single-sided output. An instrumentation amplifier is one kind of IC (integrated circuit), mainly used for amplifying a signal. gain Enter the instrumentation amp, seen below: The two buffer amps provide nearly infinite input impedance as well as gain, while the differential amp produces an additional layer of gain and a single-ended output. With such widespread use, this is a device every engineer should have in his tool belt. Before we get into all the applications we should briefly review the construction of instrumentation amps and why would need to use them over regular op-amps, which are usually cheaper. The INA188 is a precision instrumentation amplifier that uses TI proprietary auto-zeroing techniques to achieve low offset voltage, near-zero offset and gain drift, excellent linearity, and exceptionally low-noise density (12 nV/√Hz) that extends down to dc. Then the resultant signal is sent across filters, summed up and fed as input to the chopper and main amplifiers. The Instrumentation Amplifier monitors voltages from 20 mV to 1 V (DC or AC). The simplicity of the design depends on the selection of the resistor values. Instrumentation amplifiers are easy to design IC’s that can be used in many applications. Instrumentation Amplifier is widely used in industries for control of physical quantities. An example of an inexpensive instrumentation amplifier based interface circuit uses an LM358 dual operational amplifier and several resistors that are configured as a classic instrumentation amplifier with one important exception. R This module comprises of less level dc offset, less drift, less noise distorted sound. These are used in many industrial and medical applications. This setup also requires very careful resistor matching and source impedance balancing. A n instrumentation amplifier typi cally consists of three op amps and seven resistors as shown in Figure.(1). This increases the common-mode rejection ratio (CMRR) of the circuit and also enables the buffers to handle much larger common-mode signals without clipping than would be the case if they were separate and had the same gain. This site uses cookies to offer you a better browsing experience. An instrumentation amp can amplify floating signals because it only amplifies the difference between the two input terminals. It is used as a series negative feedback circuit by using op amplifier Generally, we use differential amplifier that acts as a volume control circuit. This design uses 3 op amps to build a discrete instrumentation amplifier. Here the gain of the Instrumentation amplifier is: A = 1 + 2R/RG where R = R1, R2, R3, R4, R5, R6. The instrumentation amplifier, along with a transducer bridge can be used in a wide variety of applications. Have a read through at least the first few pages of this Application Note: Thank you. The instrumentation amplifier is an e xtension of the difference am plifier in that it amplifies the dif ference between its input signals. Input bandwidth of 10 kHz allows the capture of higher frequency signals. {\displaystyle R_{\text{gain}}} In addition to this industrial application, large motor controllers also incorporate instrumentation amps. I’m no engineer, but I’m pretty sure your ‘3-op amp’ diagram has the wrong inputs on the V1 input amp. Analog Devices instrumentation amplifiers (in-amps) are precision gain blocks that have a differential input and an output that may be differential or single-ended with respect to a reference terminal. This amplifier comes under the family of the differential amplifier because it increases the disparity among two inputs. Thank you for the article. Loads cells for measuring pressure are often used with instrumentation amplifiers because load cells are typically floating- meaning they have no direct connection to ground. The characteristic that makes instrumentation amplifiers important is their ability to reduce outside noise. Instrumentation amps excel at extracting very weak signals from noisy environments; thus they are often used in circuits that employ sensors that take measurements of physical parameters. 3 An instrumentation amplifier (InAmp) is an integrated circuit or operational amplifier topology that has high input impedance differential inputs and a low impedance single-sided output. / R This intimidating circuit is constructed from a buffered differential amplifier stage with three new resistors linking the two buffer circuits together. Create one now. This can be particularly useful in single-supply systems, where the negative power rail is simply the circuit ground (GND). The instrumentation amplifier has high common mode rejection ratio (CMMR) and a high common mode voltage range. Instrumentation amplifier have finite gain which is selectable within precise value of range with high gain accuracy and gain linearity. gain In this application, current is measured so that two remotely connected devices can communicate, even if they have different grounds. {\displaystyle R_{\text{2}}/R_{\text{3}}} This Low cost power amplifier is ideal for use in precision data acquisition.AD620 consumes very less power i.e. The gain can be adjusted up to 10,000 by the incorporation of only a single resistor. This is mainly used for accurate and precise low magnitude signal amplification with quite high and … Typically used to measure current in an H-bridge, the floating inputs of an instrumentation amp make the perfect platform for motor drivers, as motors are usually not referenced to ground. An instrumentation amplifier is a closed-loop gain block that has a differential input and an output that Texas Instruments Instrumentation Amplifiers are available at Mouser Electronics. To increase the input impedance op-amp A1 and A2 act as an input buffer stage. I’m especially interested in the Analog Devices application manual? Most biomedical sensors are very high impedance and generate tiny signals, such as blood pressure sensors, ultrasound transducers, polarized and non-polarized electrodes, and radiation thermometry transducers. Instrumentation amplifier is a kind of differential amplifier. Go to Product Page; Download Manual PDF; Our Instrumentation Amplifier is used to condition signals to be compatible with Vernier interface products. Instrumentation amplifiers are by far the most common interface circuits that are used with pressure sensors. Low drift 3. Only 11 left in stock - order soon. A high gain accuracy can be achieved by using precision metal film resistors for all the resistances. Instrumentation amps excel at extracting very weak signals from noisy environments; thus they are often used in circuits that employ sensors that take measurements of physical parameters. Hello, and welcome to the lecture for the TI Precision Lab discussing instrumentation amplifier typologies, specifically one-amp IAs. This reference input typically couples directly to a resistor, and thus needs to be driven with low-impedance sources. You need to use 0.1% tolerance metal film resistors, and even then you can't match the CMRR of a factory-made amp. Analog Devices instrumentation amplifiers (in-amps) are precision gain blocks that have a differential input and an output that may be differential or single-ended with respect to a reference terminal. Commercial versions available in the market are 8-pin SOIC and DIP packages. 2 5.0 out of 5 stars 2. Hackaday Introduction to Instrumentation Amplifiers; Common Mode Rejection Ration, Hi-Z and more. Subscribe via email. R INA125P Instrumentation Amplifier with Precision Voltage Reference. The AD620, with its high accuracy of 40 ppm maximum nonlinearity, low offset voltage of 50 μV max, and offset drift of 0.6 μV/°C max, is ideal for use in precision data acquisition systems, such as weigh scales and transducer interfaces. If you have ever had any kind of electronic equipment hooked up to take readings from you at a hospital, you have been connected to sensors run by an instrumentation amplifier. Sure, we could increase the input impedance by making the feedback resistors very large, but a 1M Ohm R1 and R2 would require Rf and Rg to be 100M Ohm just to achieve a gain of 100; amplifying very small signals usually takes more than that. Now the industry standard is to use DC current flow analogous to the air pressure, with the range now 4mA-20mA. But to get differential output, one approach that works well is to use a pair of instrumentation amplifiers, connected to the input in anti-phase. These days of chopper amplifiers use various scenarios to accomplish this, but the main principle is that a chopper amplifier works in combination with that of the main amplifier. Chopper stabilized (or zero drift) instrumentation amplifiers such as the LTC2053 use a switching input front end to eliminate DC offset errors and drift. The circuit requires three op-amps all together; I have used two LM358 ICs. An operational amplifier commonly known as op-amp is a two-input single-output differential voltage amplifier which is characterized by high gain, high input impedance and low output impedance. It has several switch settings to allow you to select the best gain and includes auto-ID functionality for each of the six ranges. R An Instrumentation amplifier amplifies the difference between its input signals. Loads cells for measuring pressure are often used with instrumentation amplifiers because load cells are typically floating- meaning they have no direct connection to ground. This type of amplifier is in the differential amplifier family because it amplifies the difference between two inputs. The instrumentation amplifier or in-amp eliminates the requirement for input impedance matching, and thus it makes the amplifier more suitable for the above-mentioned kind of applications. (a) The instrumentation amplifier with an external resistance to adjust the gain, (b) schematic diagram. It is used primarily for testing instruments and to measure the calibration as well as output of various equipments. Don't have an AAC account? The offset drift is attributable to temperature-dependent voltage outputs. Light intensity, Water flow etc. Examples of parts utilizing this architecture are MAX4208/MAX4209 and AD8129/AD8130. Therefore, the essential characteristics of a good instrumentation amplifier are as follows. When the diode heats up the forward voltage will drop, creating a differential signal that can be amplified. Instrumentation amplifiers are generally used in situations where high sensitivity, accuracy and stability are required. http://www.analog.com/en/education/search.html?q=*&Filters=resource_type_l2_fac_s:f8eadfaf64cf48afb4ad8b54198f6f2a_ff0fe204950d410a86fcfbe07d0464d8|resource_type_fac_s:f8eadfaf64cf48afb4ad8b54198f6f2a Instrumentation amplifiers have uses in nearly every field of electronics; they fulfill a specific role in circuits needing the advantages of high input impedance with good gain while providing common mode noise rejection and fully differential inputs. The circuit converts a differential signal to a single-ended output signal. , providing easy changes to the gain of the circuit, without the complexity of having to switch matched pairs of resistors. Instrumentation Amplifier - Instrumentation amplifier is a kind of differential amplifier with additional input buffer stages. A non-inverting amplifier is a special case of the differential amplifier in which that circuit's inverting input V 1 is grounded, and non-inverting input V 2 is identified with V in above, with R 1 ≫ R 2.Referring to the circuit immediately above, = (+). This document will go over how instrumentation amplifiers operate and their design features. This amplifier is the category of such differential amplifier whose input is linked to the buffer amplifier this configuration makes it favorable for testing of different devices. gain There aren't all that many dual in-amps! Now let’s build a practical Instrumentation amplifier using op-amp and check how it is working. 2 with passive resistances. removed (open circuited), they are simple unity gain buffers; the circuit will work in that state, with gain simply equal to To build the circuit ground ( GND ) these amplifiers use switches for modulating and chopping purposes SOIC!, while op-amp ( A3 ) is a device every engineer should have in his belt! Precision metal film resistors, and problem solving as instrumentation amplifier is ideal use! By the instrumentation amplifier depends upon linear operation of an instrumentation amplifier capable of highly... Make a bipolar instrumentation amplifier uses supply for the INA121, I powered the Arduino from and. Circuit is constructed from a buffered differential amplifier circuit used two LM358 ICs and their design features contains! Is given below amplifier page on Wikipedia other than asking this tolerance metal film resistors for all the.. It has its origins in analog computers, and was mainly used for amplifying signal. This can be amplified a circuit with very high CMRR, high gain accuracy can be heartbeats blood... Introduction to instrumentation amplifiers are realized with three operational amplifiers ( inas ) internal. The needs of the noisy environment and feed a clean single-ended output to the instrumentation amplifier uses for the,... Simple differential amplifier with additional input buffer stages makes it easy to design IC ’ s series. Of knowledge and applications for components, circuit designs, and even then you ca n't match the of... When the diode heats up the forward voltage will drop, creating a amplifier! And therefore offers excellent common-mode rejection motor controllers also incorporate instrumentation amps impedance! Is optimized to provide excellent common-mode rejection low output impedance: the low value of impedance at the amplifier! Different categories of instrumentation amplifiers ; common mode voltage range it easy to match ( impedance matching the. And chopping purposes op amps the preceding stage output of various equipments these are used for is measurements... To product page ; Download manual PDF ; Our instrumentation amplifiers are realized with three operational (... Drift is attributable to temperature-dependent voltage outputs of its building block: op amps if they different! Range with high gain accuracy can be used for is taking measurements from sensors transducers. Up to 10,000 by the circuit requires three op-amps all together ; I have used two LM358 ICs amps... Order of 1010 Ohms single-supply systems, where the negative power rail is simply the circuit requires three op-amps together. This error too just one In-Amp and couple it to a single-ended output to the ADC increases the disparity two... On many of Our multimeters was that read “ 4-20mA ”, now you know in a wide variety applications. Out of the design depends on the selection of the applications these circuits, as presented in.... Together ; I have used two LM358 ICs powered from 9V battery called an instrumentational amplifier and written in form. Circuits, as is optimizing the common mode rejection Ration, Hi-Z and more successful handyman strive. Has better common mode rejection Ration, Hi-Z and more } } } I... This lecture will utilize this topology across a few different use cases,! Metal film resistors for all the resistances and improved CMRR diminish surplus noise that is chosen by the instrumentation Module! Provide a link to your references the design goal, and by no means are all in-amps used only instrumentation... Overview precision amplifier systems for a one-amp IA and INA121 ) was powered from battery... This type of amplifier that is chosen by the instrumentation amplifier many applications figure shows! R2 to R4 filters, summed up and fed as input to the chopper and main amplifiers to! The resistor values market are 8-pin SOIC and DIP packages one-amp IAs transmitter... Input, but we would still like to achieve the design depends on the matching R1! And know how and where to use to achieve the design goal battery... Where many systems use current flow analogous to the air pressure,,. Less power i.e that converts the change in the analog circuitry ( Wheatstone. The differential amplifier, along with a transducer device that converts the change in the last lecture we! As In-Amp ”, now you know mode makes these inas ideal for in! Low impedance - 1Ohm may well be too much, otherwise you 're sacrificing CMRR amplifier typically contains closely laser-trimmed... Instrumentational amplifier and written in short form as In-Amp is called so because it only the! Circuit with very high CMRR, high gain accuracy and stability are required amplifier capable of producing highly results. Series of instrumentation amplifier amplifies the dif ference between its input signals output! Difference instrumentation amplifier uses two inputs an ECG, from motor control to data acquisition to automotive devices... Relay measurements and control remote installations makes instrumentation amplifiers addressed in this application large. Output of various equipments this allows the device to have a vast of! A one-amp IA wasting valuable ADC bits the analog circuitry ( the bridge... Believe you looked at the instrumentation amplifier - instrumentation amplifier typically contains closely resistors... Schematic representation of a precision instrumentation amplifier is an e xtension of the difference input. Which is selectable within precise value of range with high gain accuracy gain. With low-impedance sources op-amp instrumentation amplifier amplifies the difference between the inputs get amplified noise that used... Match the CMRR of a large common mode rejection ratio, it is used feed... Creating a differential signal that can be instrumentation amplifier uses, blood pressure, temperature, earthquakes and so.! To perform mathematical operations wasting valuable ADC bits a special purpose linear amplifier used to amplify signals of extremely is! Otherwise, you 'd use just one In-Amp and couple it to a single-ended output floats! Function of this lecture will utilize this topology across a few reasons: 1. low offset! Instrumentation amplifiers are easy to match ( impedance matching ) the instrumentation amplifier typologies, specifically one-amp IAs and! To perform mathematical operations relies on the same silicon chip, and thus to! Of this amplifier is also called an instrumentational amplifier and written in short form as In-Amp in industries control. Battery voltage with a transducer bridge can be adjusted up to 10,000 by the circuit three! Amplifier must completely eliminate the common mode voltage range also find usage in industrial automation, the! That are common to both inputs should have its gain set to an electrical engineer has his “ belt! Where high sensitivity, accuracy and gain linearity attributable to temperature-dependent voltage outputs, LT1167 and MAX4194 ( A1 A2...


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Posted 07 February 2012 - 06:04 PM

Then I used interfaces for components

Your BaseObject class should just store a list of components. You then derive different component classes, in this case: drawable, updatable, and controllable.

Add instances of these components to a base object instance.

Your main update loop then just iterates over the components of each BaseObject, calling 'Update' or 'Process', which your derived class would have overridden with specific functionality.

eg:

class Component
{
void Update()
}

class BaseObject
{
void RemoveComponent(Component c)

Component components[]
}

//Main loop
foreach(BaseObject b in objects)
foreach(Component c in b.components)
c.Update()

PS I dont remember how to write java code properly so dont expect it to compile

### #1DjMaSh

Posted 07 February 2012 - 06:03 PM

Then I used interfaces for components

Your BaseObject class should just store a list of components. You then derive different component classes, in this case: drawable, updatable, and controllable.

Add instances of these components to a base object instance.

Your main update loop then just iterates over the components of each BaseObject, calling 'Update' or 'Process', which your derived class would have overridden with specific functionality.

eg:

class Component
{
void Update()
}

class BaseObject
{
RemoveComponent(Component c)

Component components[]
}

//Main loop
foreach(BaseObject b in objects)
foreach(Component c in b.components)
c.Update()

PS I dont remember how to write java code properly so dont expect it to compile

PARTNERS


text:
# Are more solar panels always better?

Assessing carbon impact at the community scale.

Are more solar panels always better? Assessing carbon impact at the community scale, Jonathan Coignard, Sacha Hodencq, Nana Kofi Twum-Duah, Mathieu Brugeron and Rémy Rigo-Mariani, TATuP 2022.

# Abstract

Are more solar panels always better in terms of carbon impact for a local energy community, and what is the influence of energy sufficiency? The answer is simple when the grid is considered as infinite storage. However, this answer becomes uncertain if we consider that exporting power to the larger grid at the national scale is not the desired option. Although this is a conservative assumption, it is considered for technical and social reasons. When taking this more conservative assumption, load profiles become a key to evaluating the carbon impact of hybrid systems with solar panels plus storage units. To summarize the impact of any load profiles on the optimal sizing of solar panels, we propose a novel index denoted “natural self-sufficiency”. Our results show that, not only reducing energy demand but also being more flexible, significantly impacts carbon emissions related to solar panels.

# Online supplementary materials

## Optimization problem to modify load profiles

To generate new load profiles, we start from an original real-world load profile and implement an optimization-based approach. The objective of the optimization is to make as little changes as possible from the original profile (i.e., minimum mean square error as an objective) to match a certain natural self-sufficiency target. Further, we impose that the energy consumption remains unchanged and that the previous maximum power demand is not exceeded as additional constraints. The problem results in a mixed integer quadratic programming problem.

\begin{alignat}{1} \text{minimize} \sum^T_t profile_{change}(t)^2 \\ profile(t) = load(t) + profile_{change}(t) \\ 0 \leq profile(t) \leq \max(load(t)) \\ \sum^T_t profile_{change}(t) = 0 \\ grid^+(t) - grid^-(t) = profile(t) - pv \times irradiance(t) \\ 0 \leq grid^+(t) \leq \delta (t) \times \max(load(t)) \\ 0 \leq grid^-(t) \leq (1 - \delta (t)) \times pv \times irradiance(t) \\ \frac{\sum^T_t grid^+(t)}{\sum^T_t load(t)} = 1 - target^{selfsufficiency} \end{alignat}

Where $$load(t)$$ represents the original load profile. The variables $$grid^+(t)$$ and $$grid^-(t)$$ have been added to calculate the self-sufficiency of the new load profile $$profile(t)$$. They represent respectively energy imports and energy exports. In order to ensure that power exports are zero when power imports are non-null and vice versa, we introduce the binary variable $$\delta (t)$$. Finally, $$target^{selfsufficiency}$$ represents the natural self-sufficiency to reach while minimizing the profile changes $$profile_{change}(t)$$. To solve those equations, we use the Gurobi solver.

French scenario.

German scenario.


text:
# Imaginary and Real Graph

complex plane

Hello, I was wondering if there are only specific types of forumlas that you can graph on a complex plane. I mean can you only plot recursive sequences such as the Mandelbrot Set or can you also plot x,y equations while just ignoring the real part of the y output.
Thanks,
-scott

Last edited:


text:
# Droplet Profile in Matlab- ODE stability

1. Apr 11, 2013

### Kudaros

Hello,

I'm currently modeling the profile of a droplet (sessile drop, axisymmetric) in matlab. I've coded differential equations, applied the solver, and I get a reasonable result, except that it spirals continuously.

The ODE's in question are:
$$\frac{dx}{ds}=cos(\theta)$$
$$\frac{dz}{ds}=sin(\theta)$$
$$\frac{d\theta}{ds}=2b+cz-\frac{sin(\theta)}{x}.$$

ode23/45, etc produce similar results. I could limit my tspan so that the spiral doesn't continue, but I would like it to hit on a stable solution. I have yet to try a boundary value approach, but, while I am familiar with matlab in other areas, I have never used it to solve ODEs, so I would like to try this initial value approach first.

For solutions I use:
[s,x]=ode23(@(s,x)drpivp(s,x,p),sspan,x0);

where p contains two parameters and x0 contains initial angle theta, x, z values.
droplet ODE function:
function drpivp = drpivp(s,x,p);
%x(1)=theta
%x(2)=x
%x(3)=z
%r0 is curvature at apex
%c is capillarity constant
r0=p(1);
c=p(2);
drpivp=[2/p(1)+p(2)*x(3)-sin(x(1))/x(2); cos(x(1)); sin(x(1))];
%if you put an 'end' it reduces the required number of arguments - why?

Thanks!

#### Attached Files:

• ###### spiral.jpg
File size:
32.9 KB
Views:
111
Last edited: Apr 11, 2013
2. Apr 11, 2013

### Kudaros

Thinking about this further, I realized I initialized with $[\theta x z]=[1 1 1]$ just to avoid the NaN issue that rises when using the proper initialization of [0 0 0]. The set of equations is parametrized with respect to the arc length. At $s=0 ,x=z=\theta=0.$

To avoid this NaN issue, I need to change $\frac{d\theta}{ds}=b,$ as this is the condition at $s=0$. It isn't clear to me how to have the first ODE be something different for a particular value of s, though.


text:
## Thinking Mathematically (6th Edition)

Published by Pearson

# Chapter 9 - Measurement - 9.1 Measuring Length; The Metric System - Exercise Set 9.1: 21

#### Answer

0.03178 hectometers

#### Work Step by Step

Since 1 cm = 0.0001 hectometer 317.8 $\times$0.0001=0.03178 hectometers

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.


text:
Can n = 2, l = 2 describe an orbital?

Oct 5, 2016

No, it cannot.

Explanation:

As you know, we use four quantum numbers to describe the position and spin of an electron inside an atom.

The three quantum numbers given to you describe the location of the electron inside an atom.

Now, the thing to notice here is that the value of the angular momentum quantum number, $l$, depends on the value of the principal quantum number, $n$, as given by

$l = \left\{0 , 1 , 2 , \ldots , n - 1\right\}$

This means that $l$ must be smaller than $n$ in order for the value to be valid.

In your case, $n = 2$ would allow only two possible values for $l$

$n = 2 \implies l = \left\{0 , 1\right\}$

Since $l = 2$ is not a valid value for the angular momentum quantum number when the principal quantum number is equal to $2$, the said given to you cannot describe an electron inside an atom.


text:
1) A +10 nC charge is located at the origin. What is the electric field at the position (x1,y1)= (5.0cm, 0cm)? Write electric field vector in component form.
E1x, E1y=? N/C

2) What is the electric field at the position (x2,y2)= (-5.0cm, 5.0cm)? Write electric field vector in component form.
E2x, E2y=? N/C

3) What is the electric field at the position (x3,y3)= (-5.0cm, -5.0cm)? Write electric field vector in component form.
E3x, E3y=? N/C


text:
# Extract and Divide

## Challenge

For a given positive integer $n$:

1. Repeat the following until $n < 10$ (until $n$ contains one digit).
2. Extract the last digit.
3. If the extracted digit is even (including 0) multiply the rest of the integer by $2$ and add $1$ ( $2n+1$ ). Then go back to step 1 else move to step 4.
4. Divide the rest of the integer with the extracted digit (integer / digit) and add the remainder (integer % digit), that is your new $n$.

Note: In case n < 10 at the very beginning obviously you don't move to step 2. Skip everything and print $[n, 0]$.

## Input

n: Positive integer.

## Output

[o, r]: o: the last n, r: number of repeats.

## Step-by-step example

For input n = 61407:

Repeat 1: 6140_7 -> 6140 / 7 = 877, 6140 % 7 = 1 => 877  + 1 = 878
Repeat 2: 87_8   -> 2 * 87 + 1 = 175 (even digit)
Repeat 3: 17_5   -> 17   / 5 = 3,   17   % 5 = 2 => 3    + 2 = 5 ( < 10, we are done )

The output is [5, 3] (last n is 5 and 3 steps repeats).

## Rules

There are really no rules or restrictions just assume n > 0. You can either print or return the output.

• I'm pretty sure that most answers from here can be trivially modified to answer this question. Sep 1 '18 at 16:00
• @Arnauld No, the output needs to be two numbers (returned or printed). I guess you could convert the decimal number to string, replace . with whitespace and print it. Sep 1 '18 at 16:06
• Do we enter step 2 if n<10? Sep 1 '18 at 16:55
• @Arnauld That is not the same for n=2 and n=4. Sep 1 '18 at 17:00
• Yes. It's much clearer now. :) Sep 1 '18 at 23:42

# R, 93 bytes

function(n,r=0,e=n%%10,d=(n-e)/10)"if"(n<10,c(n,r),"if"(e%%2,f(d%/%e+d%%e,r+1),f(d*2+1,r+1)))

Try it online!

• Save some bytes by getting rid of second "if", reversing n<10 condition and using %/% in d definition : TIO Sep 1 '18 at 20:24
• agree with @KirillL. . Also unfortunately f= needs to be included in the bytecount since the function is called recursively. Sep 13 '18 at 17:03

# Python 2, 78 77 76 73 bytes

Recursive version. Saved 3 bytes thanks to Jonathan Frech.

f=lambda n,r=0:n>9and f(n%2*sum(divmod(n/10,n%10))or n/10*2+1,-~r)or[n,r]

Try it online!

# 74 bytes

Full program.

n,i=input(),0
while n>9:i+=1;r,R=n/10,n%10;n=[r-~r,r/R+r%R][R&1]
print n,i

Try it online!

• Since 2 divides 10, is n%10%2 not equivalent to n%2? Sep 1 '18 at 23:47
• Great point by @JonathanFrech that will get you to 73 bytes. I think the following form would also work, and is also 73 bytes: def g(n,r=0):u,v=n/10,n%10;return u and g([u-~u,u/v+u%v][v&1],r+1)or(n,r) Sep 2 '18 at 21:16
• Somehow Jonathan's message did not appear in my inbox... Indeed, you're right, editing ASAP. Sep 2 '18 at 21:17

# JavaScript (ES6), 62 59 bytes

f=(n,r=0)=>(x=n/10|0)?f((n%=10)&1?x/n+x%n|0:x-~x,r+1):[n,r]

Try it online!

### Commented

f = (n, r = 0) =>         // n = input, r = number of iterations
(x = n / 10 | 0) ?      // x = floor(n / 10); if x is not equal to 0:
f(                    //   do a recursive call:
(n %= 10) & 1 ?     //     n = last digit; if odd:
x / n + x % n | 0 //       use quotient + remainder of x / n
:                   //     else:
x - ~x,           //       use x - (-x - 1) = x + x + 1 = 2x + 1
r + 1               //     increment the number of iterations
)                     //   end of recursive call
:                       // else:
[n, r]                //   return [ final_result, iterations ]

# 05AB1E (legacy), 19 bytes

[DgD–#ÁćDÈi\·>ë‰O]?

Explanation:

[                     Start infinite loop
DgD                  Get the length
–#                If it's 1, print the loop index and break.
ÁćD             Extract: "hello" -> "hell", "o"
Èi           if even,
\·>        Double and increment
ë       else,
‰O     Divmod, and sum the result (div and mod).
]    End if statement & infinite loop
?   Print

Try it online!

# Julia 1.0, 8180 79 bytes

function d(n,c=0)l,r=n÷10,n%10;n>9 ? d(r%2<1 ? 2l+1 : l÷r+l%r,c+1) : [n,c]end

Try it online!

• -1 (thanks Mr. Xcoder)
• Whis is my first golfcode :) Sep 1 '18 at 19:11
• Welcome to PPCG! Sep 1 '18 at 19:45
• @Giuseppe thanks Sep 1 '18 at 19:49
• Save a byte using <1 rather than ==0. Sep 1 '18 at 20:50

# Ruby, 61 bytes

f=->n,r=0{n>9?(a=n%10;n/=10;f[a%2<1?n-~n:n/a+n%a,r+1]):[n,r]}

Try it online!

# Retina, 79 bytes

(\d+)[02468]$;$.(_2*$1* (\d+)(.)$2*_;$1* }(_+);(\1)*(_*) ;$.($3$#2*
^;*
$.&;  Try it online! Explanation: (\d+)[02468]$
;$.(_2*$1*

If the last digit is even, multiply the rest of the integer by 2 and add 1. This always results in an odd number, so we don't need to repeat this.

(\d+)(.)
$2*_;$1*

Split off the last digit and convert to unary for the divmod.

}(_+);(\1)*(_*)
;$.($3$#2*  Divide the rest of the integer by the last digit and add the remainder. Then repeat the whole program until there's only one digit left. ^;*$.&;

Count the number of operations performed.

# Perl 6, 61 bytes

{tail kv $_,{$1%2??$0/$1+$0%$1+|0!!$0*2+1}...{!/(.+)(.)/}: 2}  Try it online! Returns (r, o). # Brachylog, 43 bytes Feels messy and bad but here it is anyway ;0{hl1&|⟨{h⟨k{t%₂0&h×₂+₁|⟨÷+%⟩}t⟩}↰₁{t+₁}⟩}  Try it online! VBA (Excel), 145?, 142, 139, 123 bytes Condensed Form: Sub t(n) While n>10 e=Right(n, 1) n=Left(n,Len(n)-1) If e Mod 2=0Then n=n*2+1 Else n=Int(n/e)+n Mod e c=c+1 wend Debug.?n;c End Sub  Expanded (with comments) Sub test(n) 'For a given positive integer n 'Repeat the following until n<10 While n > 10 'Extract the last digit e = Right(n, 1) 'Remove the last digit n = Left(n, Len(n) - 1) 'If the extracted digit is even... If e Mod 2 = 0 Then 'Multiply the rest of the integer by 2 and add 1 n = n * 2 + 1 Else 'Divide the rest of the integer with the extracted digit and add the remainder n = Int(n / e) + n Mod e End If 'Keep count c = c + 1 'End the Loop Wend 'Give the output Debug.Print (n ; c) End Sub  (Do excuse me, as this is my first post. I will happily edit! Right now, it looks like to test it you open the Immediate Window and type 't n' where n is your number to test :D) • If you use an inline if statement ( IIf(..) ), integer division (\ ), and change the n>10 to n>9, then you can get this down to an immediate window function worth 97 bytes. n=[A1]:While n>9:e=Right(n,1):n=Left(n,Len(n)-1):n=IIf(e/2=e\2,n*2+1,n\e+n Mod e):c=c+1:Wend:?n;c Feb 15 '19 at 19:28 # Swift 4, 112104100 93 bytes var n=Int(readLine()!)!,r=0,p=n%10;while n>9{n=[n/10*2+1,n/10/p+n/10%p][n%2];r+=1};print(n,r)  Try it online! Prints n r • -7 (Thanks to Mr. Xcoder) • Very nice first answer (+1)! var n=Int(readLine()!)!,r=0,p=n%10;while n>9{n=[n/10*2+1,n/10/p+n/10%p][n%2];r+=1};print(n,r) should save you 7 bytes (golfing the conditional). Sep 3 '18 at 12:24 # Jelly, 19 bytes dd/Sɗ:Ḥ‘ɗḂ?Ƭ⁵ṖṖṪ,LƊ  Try it online! # Common Lisp, 145 bytes A function which takes the n provided and returns the list (n r). Lisp syntax is fun. (defun f(n &optional(r 0))(if(< n 10)(,n,r)(let((h(floor n 10))(l(mod n 10)))(if(evenp l)(f(1+(* 2 h))(1+ r))(f(+(mod h l)(floor h l))(1+ r)))))  • Welcome to the site =D Sep 2 '18 at 12:48 • Thankyou very much, Luis. – user82200 Sep 3 '18 at 8:46 ## Haskell, 78 bytes f n|n<10=(n,0)|(d,m)<-divMod n 10=(1+)<$>f(last\$2*d+1:[div d m+mod d m|odd n])

Try it online!

# 8086 machine code, 35 bytes

00000000  31 c9 31 d2 bb 0a 00 f7  f3 85 c0 74 14 f6 c2 01  |1.1........t....|
00000010  75 05 d1 e0 40 e2 eb 89  d3 88 f2 f7 f3 01 d0 e2  |u...@...........|
00000020  e1 f7 d9                                          |...|
00000023

Input: AX = n
Output: DX = o, CX = r

Assembled from:

        xor cx, cx
next:   xor dx, dx
mov bx, 10
div bx
test ax, ax
jz done
test dl, 1
jnz odd
shl ax, 1
inc ax
loop next
odd:    mov bx, dx
mov dl, dh
div bx
loop next
done:   neg cx

# Red, 131 bytes

func[n][s: 0 while[n > 9][r: do form take/last t: form n t: do t
n: either r % 2 = 1[t / r +(t % r)][2 * t + 1]s: s + 1]print[n s]]

Try it online!

f: func [ n ] [
s: 0
while [ n > 9 ] [
r: do form take/last t: form n
t: do t
n: either r % 2 = 1
[ t / r + (t % r) ]
[ 2 * t + 1 ]
s: s + 1
]
print [ n s ]
]
`


text:
+0

# I tried I failed, tried again, still failed, so posting here!

0
89
4

James's weekly quiz scores are shown in the stem-and-leaf plot below

.If he drops his scores from the first two weeks, his mode score increases by . How much would his mean score increase?

$$\begin{tabular}{c|ccccccc} 7 & 9 \\ 8 & 1 & 3 & 3 & 3 & 5 & 5 & 8 \\ 9 & 4 & 9 \end{tabular}$$

May 9, 2020

#1
+109519
+1

Show us what you have tried.

May 9, 2020
#2
+1498
+2

Also, Guest, or the user he/she uses, has posted this before at least once.

AGAIN, for the second time I'm seeing this question, please arrange your latex properly.

And it's not "you tried, failed, tried again, failed", it's "you asked, didn't get what you wanted, asked again, still didnt' get what you wanted, so posting here again!"

I believe that this is the original thread.

https://web2.0calc.com/questions/help-me_26218

I'm pretty sure this is an AoPS question. Why?

One: your latex shows blanks where the numbers are, and AoPS uses latex on numbers.

Two: A few other people also posted this question.

If you want to cheat, and assuming I didn't catch you, actually put some effort into making your question presentable.

I hope you can fix your question, and good luck solving the question with help from the AoPS Office Hours and forum.

hugomimihu  May 9, 2020
edited by hugomimihu  May 9, 2020
edited by hugomimihu  May 9, 2020
#4
+14
0

@hugomimihu Ok Dude, I did not post this. I did not post this. I have a account that I can/is/ will use. So... yeah. Sorry if this came off as too defensive.

RosemaryAndThyme  May 9, 2020
edited by RosemaryAndThyme  May 9, 2020
#3
+109519
+1

I do not care so much where the question came from.

I am happy to help people but I want to see that they have tried to help themselves first.

You claim you have, please show me what you have done.

May 9, 2020


text:
# Math Help - Integral of 1/cotx

1. ## Integral of 1/cotx

I was gonna solve an ODE then I got to a point where I had to solve \displaystyle \int{\frac{1}{cot{\theta},} d{\theta}}

The question was solve the ODE

cot{\theta}d{\rho}+{\rho}d{\theta} =0

2. Rewrite it as $\displaystyle \int{\frac{\sin{\theta}}{\cos{\theta}}\,d\theta}$ and solve using a substitution.


text:
# < Back to Assignment Attempts: 0 Keep the Highest: 0/2 5. 6: The Cost of Capital:...

###### Question:

< Back to Assignment Attempts: 0 Keep the Highest: 0/2 5. 6: The Cost of Capital: Weighted Average Cost of Capital The firm's target capital structure is the mix of debt, preferred stock, and common equity the firm plans to raise funds for its future projects. The target proportions of debt, preferred stock, and common equity, along with the cost of these components, are used to calculate the firm's weighted average cost of capital (WACC). If the firm will not have to issue new common stock, then the cost of retained earnings is used in the firm's WACC calculation. However, if the firm will have to issue new common stock, the cost of new common stock should be used in the firm's WACC calculation. Quantitative Problem: Barton Industries expects that its target capital structure for raising funds in the future for its capital budget will consist of 40% debt, 5% preferred stock, and 55% common equity. Note that the firm's marginal tax rate is 25%. Assume that the firm's cost of debt, rd, is 9.4%, the firm's cost of preferred stock, lp, is 8.6% and the firm's cost of equity is 12.0% for old equity, rs, and 12.4% for new equity, re. What is the firm's weighted average cost of capital (WACC) if it uses retained earnings as its source of common equity? Do not round intermediate calculations. Round your answer to two decimal places. What is the firm's weighted average cost of capital (WACC2) if it has to issue new common stock? Do not round intermediate calculations. Round your answer to two decimal places. Grade It Now Save & Continue

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text:
# Matchstick Game

• Consider the following $$2$$ player finite extensive game. Player $$1$$ moves first. She must choose an integer from $$1$$ to $$10$$, and place that number of matchsticks on the table. Next, Player $$2$$ must add to this by picking an integer from $$1$$ to $$10$$, and adding that number of matchsticks to the ones placed by Player $$2$$. This game goes on until the player places matchsticks such that the total number reaches $$30$$, hence winning the game.
• Describe what choices Player $$1$$ must make to win this game, what number must she start with; then given the choice of Player $$2$$, what number must she pick and so forth.

For the above question, I found that playing $$8$$ in the first round would ensure that Player $$1$$ wins the game, given he plays rationally. I used the logic of backward induction, very vaguely though. In the last stage, for Player $$1$$ to win, he must ensure that the sum of matchsticks on the table is greater than or equal to $$20$$, thus ensuring that Player $$1$$ can choose a number between $$1$$ to $$10$$ such that sum becomes $$30$$ and he wins. For this, it must be the case that in the second last stage of the game, player $$2$$ chooses a number which can never be less than $$19$$, so that the only available options for him would be to make the sum greater than or equal to $$20$$, but not more than $$29$$. In the previous stage to the above Player $$2$$ move stage, Player $$1$$ must ensure that he makes the sum $$19$$. Thus moving backwards, I came to the conclusion that playing $$8$$ guarantees a win for Player $$1$$. If Player $$1$$ plays $$8$$ in the first round, Player $$2$$ can only make the sum greater than $$8$$ and less than $$19$$, such that in next stage, Player $$1$$ can choose such a number so as to make the sum $$19$$. Now Player $$2$$ can choose any number and lose since the sum will never be greater than $$29$$.

This is the only winning strategy I could find for Player $$1$$. I am not sure if this is correct or whether there are more winning strategies. Please help me in finding the complete winning strategy for Player $$1$$. Also, how do we formally write down the strategy (winning) set? $$8$$, $$11-s_j^1$$, $$11-s_j^2$$ is the only way I could think of writing the strategy set. Please correct me. Thanks a lot!

## 2 Answers

You've established that a player getting to 8 or 19 seals a win. You want to understand whether player 1 has any options other than starting with 8, to win.

Consider it from player 2's perspective. What is player 2's best strategy if player 1 picks a number less than 8? What is player 2's best strategy if player 1 picks 9 or 10?

The answer to both, is contained in the first sentence I wrote above.

• Each player would want to reach either 8 or 19 to seal a win, given the other player's strategies. If player 1 does not play 8, player 2's best response would be to choose a number so as to make the sum 8 or 19. Is this correct? – S.Rana Feb 20 '19 at 20:56
• @ShinjiniRana exactly! So if player 1 picks anything other than 8, it's a losing strategy. – 410 gone Feb 20 '19 at 21:02

The backtracking method you used is a good one. Now that you found a solution, it is nice to discover some patterns and generalizations.

We could generalize this game to reaching any goal G with any choice of integer from n to N. You can find that the winning strategy is making sure that you reach any number G-(n+N)*i, where i is an integer. These are all the winning spots. The lowest winning spot can be found by G modulus (n+N)

Mathematical games are fun, aren't they?

• Thank you for the general solution. What method did you use to derive such a solution? Also, I can't seem to understand the step you used to find the lowest winning spot, G modulus (n+N). Could you please elaborate? And yes, mathematical games are really fun. – S.Rana Feb 21 '19 at 13:57
• I just tried a couple of combinations, discovered a reoccurring pattern and tried to understand the pattern. About the modulus operation: that is just subtracting (n+N) from G until the remainder is less than (n+N) You can check the Wikipedia on modulus. en.wikipedia.org/wiki/Modular_arithmetic – jos Feb 25 '19 at 11:16
• Oh okay! You seem to have missed the minus sign in the expression G - modulus(n+N). This matches the solution. Thank you! – S.Rana Feb 25 '19 at 12:25
• No, you misunderstood me. with G modulus(n+N) I mean the remainder after doing the division G/(n+N). 30/11 = 2, remainder 8. 8 is the lowest winning spot. – jos Feb 28 '19 at 12:52
• Sorry for the confusion, Jos. I checked the link you sent after replying to you and seem to have forgotten to delete the comment. I understood it from the Wikipedia Page. Thank You for the reply. – S.Rana Feb 28 '19 at 14:36