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# What is the benefit of only performing a random split once at the beginning during Kfold cross-validation?

I had a discussion with my colleague about Kfold and StratifiedKfold validation in scikit learn.

My understanding of these two ways of validation is that at the beginning of the validation process, the data set is randomly divided in k-folds. k-1 of those folds end up in the train dataset. 1 of those folds ends up in the validation data set. Validations are performed k times. With each repetition, what is in the training or validation dataset is shifted.

My colleague argued that it would be better to decide what is and is not in the train dataset or validation dataset anew. Edit: Scikit-learn also has selectors for this sort of validation named "ShuffleSplit" and "StratifiedSHuffling".

I had no mathematical intuition for why one way is superior to another way.

The benefit of only doing the initial split is that what ends up in the validation dataset is non-redundant. But is that really a benefit?

So my question is this:

Is he right? What is the benefit (intuitively or mathematically) to the actual implementation of Kfold or StratifiedKfold?

Edit: Can someone provide me with an intuition for what to pick when?

• The benefit of only 1 split is that all samples are used both for training ($k-1$) times and exactly once for validation. Whereas the advantage of random sub-sampling is that the ratio of training to test size is independent of number of folds. – Łukasz Grad Mar 29 '17 at 18:49
• @Lukasz Grad: Which is better for what cases? Is there a sweet-spot for doing only 1 split at the beginning? – Thornhale Mar 29 '17 at 19:00
• Another option for smaller datasets is iterated K-fold validation with shuffling. Just repeat K-fold multiple times, shuffling only at the beginning of each. So (# iterations)*K trained models in total. – Austin Jul 7 '18 at 1:07

## 1 Answer

The benefit of determining the contents of each fold at the beginning rather than re-sampling is that you avoid bias by possibly selecting a single observation for more than one training or testing set. This is only a benefit if you care that no records are over-represented in your validation.

If the argument is that new folds should be generated without replacement after each fold has been tested, it can be shown that the likelihood of a single observation to appear in each fold is unaffected and they are therefore equivalent approaches.

Stratified methods most commonly used in cases where classes are unbalanced - to avoid, for example, folds where there are no (or very few, or just inefficiently few) positive examples in a sparse binary classification problem. This applies both to StratifiedShuffleSplit and StratifiedKFold methods.

In order to determine which to use between K-Fold and ShuffleSplit, though, we'll have to understand some key differences between the methods.

1. In K-fold, the model is trained on each iteration at a proportion of the data set equal to $\frac{k-1}{k}$, i.e. for $k=5$ the model is trained on 80% of the data at each iteration and for $k=10$ the model is trained on 90% of the data. There are $k$ training iterations in the algorithm.

In ShuffleSplit, the model is trained at each iteration on a defined train_size. The default size of the training set for the scikit-learn implementation is 90%. The number of iterations is parameterized (n_splits).

You can therefore configure two validation strategies that have the same number of training runs and are trained on the same proportion of the data, I.e. K-fold validation with K=10 and ShuffleSplit with train_size = .9 and n_splits = 10.

No difference here.

1. K-Fold is guaranteed to both train and test on every observation of the dataset an equal number of times ($k-1$ and 1 times, respectively).

ShuffleSplit does not guarantee this proportion- by re-sampling at each iteration, duplicate members of the test set can be selected twice, or several times. Some observations may not be represented in the test set.

This is a point for using K-Fold.

Some other discussion has brought up training time as a point in favor of ShuffleSplit- that is, you can configure ShuffleSplit to run with less demanding coverage (i.e. train_size = .8, n_splits = 5). This is only a strong argument for complexities that can't be approximated by adjusting the parameter $k$- as $k=5$ for the parameters mentioned above for ShuffleSplit.

For example, K-fold can't approximate train_size = .8, n_splits = 3. The tradeoff is that each configuration of ShuffleSplit that would be more computationally efficient than K-Fold also provides less complete validation coverage- in this example (train_size = .8, n_splits = 3), the model could never be tested on more than 60% of the data, and would almost always be tested on less than that.

TL;DR-
K-Fold is generally better except in edge cases where the computational demand of K-Fold isn't supported by a need for comprehensive cross-validation coverage.

• Just for clarification then, going by what you said above. Is K-Fold equivalent to ShuffleSplit validation if for the latter it is shown that we are are generating new folds without replacement after each fold has been tested? (Is Shuffle-Split implemented that way?) – Thornhale Mar 29 '17 at 23:45
• No, these two are not equivalent- ShuffleSplit is basically just iterations of train/test splitting. I.e. in iteration 1, you train/test split on the whole set, in iteration 2 you re-split the set and test again etc. What I was trying to communicate was unrelated to ShuffleSplit directly- basically, if you form the k folds sequentially rather than all at once, it's equivalent to doing it all at once. That statement just covers one possible interpretation of the question. – Thomas Cleberg Mar 30 '17 at 13:20
• Ah, ok. That part is clear. The only part that is not clear to me then is: When should I prefer ShuffleSplit over Kfold? And when should I prefer Kfold over SHuffleSplit? – Thornhale Mar 30 '17 at 13:27
• Added more to clarify the tradeoffs of this decision. – Thomas Cleberg Mar 30 '17 at 15:45


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# geometry

posted by .

triangle abc is similar to def, the lengths of the sides of triangle abc is 5,8,11. what is the length of the shortest side of triangle def if its perimeter is 60?

• geometry -

If similar, then their sides are in the same ratio, so ..

let the sides of triangle DEF be
5x , 8x, and 11x

5x + 8x + 11x = 60
24x = 60
x = 60/24 = 2.5

so the shortest side is 5x = 5(2.5) = 12.5

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# Is the function $\mbox{tr}(XAX')$ convex?

Is the function $\mbox{tr}(XAX')$ convex, where $A$ is a positive semidefinite (PSD) matrix? I know that for a general $A$, the above trace function is not convex. But for a PSD $A$, is the function convex?

• Think about operations which preserve convexity ! May 28 '17 at 18:06

Look at the table for scalar-by-matrix derivatives. In particular, one has

$$\nabla_X \text{tr}(XAX^T) = X^T(A+A^T).$$

Thus if $A$ is psd, then so is the Hessian $A + A^T = 2A$, from which convexity of $X \mapsto \text{tr}(XAX^T)$ follows.

If $\rm A$ is symmetric and positive semidefinite, then there is a matrix $\rm M$ such that $\rm A = M M^{\top}$. Hence,

$$\mbox{tr} \left( \mathrm X \mathrm A \mathrm X^{\top} \right) = \mbox{tr} \left( \mathrm X \mathrm M \mathrm M^{\top} \mathrm X^{\top} \right) = \| \mathrm X \mathrm M \|_{\text{F}}^2$$

is indeed convex.

Let $\mathrm X \in \mathbb R^{m \times n}$. Hence,

$$\mbox{tr} \left( \mathrm X \mathrm A \mathrm X^{\top} \right) = (\mbox{vec} (\mathrm X))^{\top} \left( \mathrm A \otimes \mathrm I_m \right) \mbox{vec} (\mathrm X)$$

If $\rm A$ is symmetric and positive semidefinite, then the Kronecker product $\mathrm A \otimes \mathrm I_m$ is also symmetric and positive semidefinite and, thus, $\mbox{tr} \left( \mathrm X \mathrm A \mathrm X^{\top} \right)$ is convex.

You can easily see that the function $\varphi: \mathbb R^{n^2}\to \mathbb R$ such that $\varphi(X)=\text{tr}(XAX^T)$ is the trace of the matrix whose entries are the scalar products of a set $(x_i)_{i=1,\ldots,n}$ of vectors in $\mathbb R^n$. Call their components $(x_i^j)_{j=1,\ldots,n}$ for every $i$.

The elements on the diagonal of your matrix are $$(XAX^T)_{ii}= (X^TAX)_{ii}=x_i^TAx_i=\sum_{j,k=1}^n x_i^ja_{jk}x_i^k$$ so the trace is equal to $$\sum_{i,j,k=1}^n x_i^ja_{jk}x_i^k=\sum_{i=1}^n(x_i,x_i)_A=\sum_{i=1}^n x_i^T A x_i$$

This makes $$\frac{\partial^2}{\partial (x^a_b)^2}\varphi=2a_{aa}$$ $$\frac{\partial^2}{\partial (x^a_b)\partial (x^c_d)}\varphi=a_{ac}+a_{ca}$$ so your Hessian is $$H_\varphi=A+A^T$$ Since $A$ is psd, if $y\neq 0$ $$y^T(A+A^T)y=y^T A y+ (y^TAy)^T=2y^TAy\ge 0$$ and $\varphi$ is convex, because $H_\varphi$ is positive semidefinite.

The function $tr : M_n(R) \rightarrow R$ is linear (so convex) and increasing w.r.t Positive matrix cones. And the function $Q:M_n(R) \rightarrow M_n(R)$ with $Q(X)=X A X^t$ is convex w.r.t positive cones (i.e., each row of $XAX^t$ is convex function), since $A$ is PSD (look at it Hessian of each rows ), Therefore, $F=tr oQ :M_n(R) \rightarrow R$ with $F= tr(XAX^t)$ is convex.


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Included Programs¶

These are installed with Mido.

mido-play¶

Plays back one or more MIDI files:

$mido-play song1.mid song2.mid  mido-ports¶ Lists available input and output ports and shows environment variables and the current backend module. mido-serve¶ Serves one or more ports over the network, for example: $ mido-serve :9080 'Integra-7'

You can now connect to this port with mido-forward (or use mido.sockets.connect() and send messages to it. The messages will be forwarded to every port you listed (in this case ‘Integra-7’).

mido-connect¶

Forwards all messages that arrive on one or more ports to a server.

For example, to use the SH-201 keyboard connected to this computer to play sounds on the Integra-7 on a computer named mac.local (which runs the server as above), you can do:

\$ mido-connect mac.local:9080 'SH-201'

Note that you may experience latency and jitter, so this may not be very useful for live playing or for playing back songs.

There is also no security built in, so you should only use this on a trusted network. (Anyone can connect and send anything, including harmful sysex messages.)

mido-serve and mido-connect are only included as fun programs to play with, but may in the future be expanded into something more usable.


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## Elementary Differential Equations and Boundary Value Problems 9th Edition

General form of second order differential boundary value problem: $\mathcal{p}$($\mathcal{x}$)$\mathcal{y^{n}}$$\mathcal{+}$$\mathcal{q(x)y^{1}+r(x)y=g(x)}$ $\mathcal{y(x_{0})=y_{0}}$ $\mathcal{y(x_{1})=y_{1}}$ Equation is homogeneous if $\mathcal{g(x)}$ and $\mathcal{y_{0}=y_{1}=0}$ $\mathcal{-y^n+\leftthreetimes(1+x^2)y}$ $\mathcal{-y^n-\leftthreetimes(1+x^2)y=0}$ Since $\mathcal{g(0)}$=$\mathcal{0}$, $\mathcal{y_0=y_1=0}$ equation is homogeneous


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# Internal energy

 Internal energy Common symbol(s): U in SI base quantities: m2*kg/s2 SI unit: J Derivations from other quantities: $U = \sum_i E_i \!$

In thermodynamics, the internal energy is the total energy contained by a thermodynamic system.[1] It is the energy needed to create the system but excludes the energy to displace the system's surroundings, any energy associated with a move as a whole, or due to external force fields. Internal energy has two major components, kinetic energy and potential energy. The kinetic energy is due to the motion of the system's particles (translations, rotations, vibrations), and the potential energy is associated with the static rest mass energy of the constituents of matter, static electric energy of atoms within molecules or crystals, and the static energy of chemical bonds. The internal energy of a system can be changed by heating the system or by doing work on it;[1] the first law of thermodynamics states that the increase in internal energy is equal to the total heat added and work done by the surroundings. If the system is isolated from its surroundings, its internal energy cannot change.

For practical considerations in thermodynamics and engineering it is rarely necessary or convenient to consider all energies belonging to the total intrinsic energy of a sample system, such as the energy given by the equivalence of mass. Typically, descriptions only include components relevant to the system under study. Thermodynamics is chiefly concerned only with changes in the internal energy.

The internal energy is a state function of a system, because its value depends only on the current state of the system and not on the path taken or process undergone to arrive at this state. It is an extensive quantity. The SI unit of energy is the joule (J). Some authors use a corresponding intensive thermodynamic property called specific internal energy which is internal energy per unit of mass (kilogram) of the system in question. The SI unit of specific internal energy is J/kg. If intensive internal energy is expressed relative to units of amount of substance (mol), then it is referred to as molar internal energy and the unit is J/mol.

From the standpoint of statistical mechanics, the internal energy is equal to the ensemble average of the total energy of the system. It is also called intrinsic energy.

## Description and definition

The internal energy (U) is the sum of all forms of energy (Ei) intrinsic to a thermodynamic system:

$U = \sum_i E_i \!$

It is the energy needed to create the system. It may be divided into potential energy (Upot) and kinetic energy (Ukin) components:

$U = U_{pot} + U_{kin} \!$

The kinetic energy of a system arises as the sum of the motions of all the system's particles, whether it be the motion of atoms, molecules, atomic nuclei, electrons, or other particles. The potential energy includes all energies given by the mass of particles, by the chemical composition, i.e. the chemical energy stored in chemical bonds having the potential to undergo chemical reactions, the nuclear energy stored by the configuration of protons, neutrons, and other elementary particles in atomic nuclei, and the physical force fields within the system, such as due to internal induced electric or magnetic dipole moment, as well as the energy of deformation of solids (stress-strain).

Internal energy does not include the energy due to motion of a system as a whole. It further excludes any kinetic or potential energy the body may have because of its location in external gravitational, electrostatic, or electomagnetic fields. It does, however, include the contribution to the energy due to the coupling of the internal degrees of freedom of the object to such the field. In such a case, the field is included in the thermodynamic description of the object in the form of an additional external parameter.

For practical considerations in thermodynamics or engineering, it is rarely necessary, convenient, nor even possible, to consider all energies belonging to the total intrinsic energy of a sample system, such as the energy given by the equivalence of mass. Typically, descriptions only include components relevant to the system under study. Indeed in most systems under consideration, especially through thermodynamics, it is impossible to calculate the total internal energy.[2] Therefore, a convenient null reference point may be chosen for the internal energy.

The internal energy is an extensive property: it depends on the size of the system, or on the amount of substance it contains.

At any temperature greater than absolute zero, potential energy and kinetic energy are constantly converted into one another, but the sum remains constant in an isolated system (cf. table). In the classical picture of thermodynamics, kinetic energy vanishes at zero temperature and the internal energy is purely potential energy. However, quantum mechanics has demonstrated that even at zero temperature particles maintain a residual energy of motion, the zero point energy. A system at absolute zero is merely in its quantum-mechanical ground state, the lowest energy state available. At absolute zero a system has attained its minimum attainable entropy.

The kinetic energy portion of the internal energy gives rise to the temperature of the system. Statistical mechanics relates the pseudo-random kinetic energy of individual particles to the mean kinetic energy of the entire ensemble of particles comprising a system. Furthermore it relates the mean kinetic energy to the macroscopically observed empirical property that is expressed as temperature of the system. This energy is often referred to as the thermal energy of a system,[3] relating this energy, like the temperature, to the human experience of hot and cold.

Statistical mechanics considers any system to be statistically distributed across an ensemble of N microstates. Each microstate has an energy Ei and is associated with a probability pi. The internal energy is the mean value of the system's total energy, i.e., the sum of all microstate energies, each weighted by their probability of occurrence:

$U = \sum_{i=1}^N p_i \,E_i\ .$

This is the statistical expression of the first law of thermodynamics.

### Internal energy changes

Interactions of thermodynamic systems
Type of system Mass flow Work Heat
Open
Closed
Thermally isolated
Mechanically isolated
Isolated

Thermodynamics is chiefly concerned only with the changes, ΔU, in internal energy. The most important parameters in thermodynamics when considering the changes in total energy are the changes due to the flow of heat Q and due to mechanical work, i.e. from changes in volume of the system under an external pressure. Accordingly, the internal energy change ΔU for a process may be written more specifically as[1]

$\Delta U = Q + W_{mech} + W_{extra} \,$

where Q is heat added to a system and Wmech is the mechanical work performed by the surroundings due to pressure or volume changes in the system.[note 1] All other perturbations and energies added by other processes, such as an electric current introduced into an electronic circuit, is summarized as the term Wextra.

When a system is heated, it receives energy in form of heat. This energy increases the internal energy. However, it may be extremely difficult to determine how this extra energy is stored. In general, except in an ideal gas, it is redistributed between kinetic and potential energy. The net increase in kinetic energy is measurable by an increase in the temperature of the system. The equipartition theorem states that increase in thermal energy is distributed between the available degrees of freedom of the fundamental oscillators in the system. In an ideal gas all of the extra energy results in a temperature increase, as it is stored solely as kinetic energy. The heat introduced to a system while the temperature changes is often called sensible heat.

Another method to change the internal energy of a system is by performing work on the system, either in mechanical form by changing pressure or volume, or by other perturbations, such as directing an electrical current through the system. Finally, the internal energy increases when additional mass is transferred into the system.

If a system undergoes certain phase transformations while being heated, such as melting and vaporization, it may be observed that the temperature of the system does not change until the entire sample has completed the transformation. The energy introduced into the system while the temperature did not change is called a latent energy, or latent heat, in contrast to sensible heat. It increases only the potential energy of the system, but not its thermal energy component.

## Internal energy of the ideal gas

Thermodynamics often uses the concept of the ideal gas for teaching purposes, and as an approximation for working systems. The ideal gas is a gas of particles considered as point objects that interact only by elastic collisions and fill a volume such that their free mean path between collisions is much larger than their diameter. Such systems are approximated by the monatomic gases, helium and the other noble gases. Here the kinetic energy consists only of the translational energy of the individual atoms. Monatomic particles do not rotate or vibrate, and are not electronically excited to higher energies except at very high temperatures.

Therefore practical internal energy changes in an ideal gas may be described solely by changes in its kinetic energy. Kinetic energy is simply the internal energy of the perfect gas and depends entirely on its pressure, volume and thermodynamic temperature.

The internal energy of an ideal gas is proportional to its mass (number of moles) N and to its temperature T

$U = c N T,$

where c is the heat capacity (at constant volume) of the gas. The internal energy may be written as a function of the three extensive properties S, V, N (entropy, volume, mass) in the following way [4] [5]

$U(S,V,N) = const \cdot e^\frac{S}{cN} V^\frac{-R}{c} N^\frac{R+c}{c},$

where const is an arbitrary positive constant and where R is the Universal Gas Constant. It is easily seen that U is a linearly homogeneous function of the three variables and that it is weakly convex. Knowing temperature and pressure to be the derivatives $T = \frac{\partial U}{\partial S},$ $p = -\frac{\partial U}{\partial V},$ the Ideal Gas Law $pV = RNT$ immediately follows.

## Internal energy of a closed thermodynamic system

This above summation of all components of change in internal energy assume that a positive energy denotes heat added to the system or work done on the system, while a negative energy denotes work of the system on the environment.

Typically this relationship is expressed in infinitesimal terms using the differentials of each term. Only the internal energy is an exact differential. For a system undergoing only thermodynamics processes, i.e. a closed system that can exchange only heat and work, the change in the internal energy is

$d U = \delta Q + \delta W \,$

which constitutes the first law of thermodynamics.[note 1] It may be expressed in terms of other thermodynamic parameters. Each term is composed of an intensive variable (a generalized force) and its conjugate infinitesimal extensive variable (a generalized displacement).

For example, for a non-viscous fluid, the mechanical work done on the system may be related to the pressure p and volume V. The pressure is the intensive generalized force, while the volume is the extensive generalized displacement:

$\delta W = - p \mathrm{d}V\,$.

This defines the direction of work, W, to be energy flow from the working system to the surroundings, indicated by a negative term.[note 1] Taking the direction of heat transfer Q to be into the working fluid and assuming a reversible process, the heat is

$\delta Q = T \mathrm{d}S\,$.
$T$ is temperature
$S$ is entropy

and the change in internal energy becomes

$\mathrm{d}U = T\mathrm{d}S-p\mathrm{d}V \!$

### Changes due to temperature and volume

The expression relating changes in internal energy to changes in temperature and volume is

$dU =C_{V}dT +\left[T\left(\frac{\partial p}{\partial T}\right)_{V} - p\right]dV\,\,\text{ (1)}.\,$

This is useful if the equation of state is known.

In case of an ideal gas, we can derive that $dU = C_v dT$, i.e. the internal energy of an ideal gas can be written as a function that depends only on the temperature.

### Changes due to temperature and pressure

When dealing with fluids or solids, an expression in terms of the temperature and pressure is usually more useful:

$dU = \left(C_{p}-\alpha p V\right)dT +\left(\beta_{T}p-\alpha T\right)Vdp\,$

where it is assumed that the heat capacity at constant pressure is related to the heat capacity at constant volume according to:

$C_{p} = C_{V} + V T\frac{\alpha^{2}}{\beta_{T}}\,$

### Changes due to volume at constant pressure

The internal pressure is defined as a partial derivative of the internal energy with respect to the volume at constant temperature:

$\pi _T = \left ( \frac{\partial U}{\partial V} \right )_T$

## Internal energy of multi-component systems

In addition to including the entropy S and volume V terms in the internal energy, a system is often described also in terms of the number of particles or chemical species it contains:

$U = U(S,V,N_1,\ldots,N_n)\,$

where Nj are the molar amounts of constituents of type j in the system. The internal energy is an extensive function of the extensive variables S, V, and the amounts Nj, the internal energy may be written as a linearly homogeneous function of first degree:

$U(\alpha S,\alpha V,\alpha N_{1},\alpha N_{2},\ldots ) = \alpha U(S,V,N_{1},N_{2},\ldots)\,$

where α is a factor describing the growth of the system. The differential internal energy may be written as

$\mathrm{d} U = \frac{\partial U}{\partial S} \mathrm{d} S + \frac{\partial U}{\partial V} \mathrm{d} V + \sum_i\ \frac{\partial U}{\partial N_i} \mathrm{d} N_i\ = T \,\mathrm{d} S - p \,\mathrm{d} V + \sum_i\mu_i \mathrm{d} N_i\,$

which shows temperature T to be the partial derivative of U with respect to entropy S and pressure p to be the negative of the similar derivative with respect to volume V

$T = \frac{\partial U}{\partial S},$

$p = -\frac{\partial U}{\partial V},$

and where the coefficients $\mu_{i}$ are the chemical potentials for the components of type i in the system. The chemical potentials are defined as the partial derivatives of the energy with respect to the variations in composition:

$\mu_i = \left( \frac{\partial U}{\partial N_i} \right)_{S,V, N_{j \ne i}}$

As conjugate variables to the composition $\lbrace N_{j} \rbrace$, the chemical potentials are intensive properties, intrinsically characteristic of the system, and not dependent on its extent. Because of the extensive nature of U and its variables, the differential dU may be integrated and yields an expression for the internal energy:

$U = T S - p V + \sum_i \mu_i N_i\,$.

The sum over the composition of the system is the Gibbs energy:

$G = \sum_i \mu_i N_i\,$

that arises from changing the composition of the system at constant temperature and pressure. For a single component system, the chemical potential equals the Gibbs energy per amount of substance, i.e. particles or moles according to the original definition of the unit for $\lbrace N_{j} \rbrace$.

## Internal energy in an elastic medium

For an elastic medium the mechanical energy term of the internal energy must be replaced by the more general expression involving the stress $\sigma_{ij}$ and strain $\varepsilon_{ij}$. The infinitesimal statement is:

$\mathrm{d}U=T\mathrm{d}S+V\sigma_{ij}\mathrm{d}\varepsilon_{ij}$

where Einstein notation has been used for the tensors, in which there is a summation over all repeated indices in the product term. The Euler theorem yields for the internal energy:[6]

$U=TS+\frac{1}{2}\sigma_{ij}\varepsilon_{ij}$

For a linearly elastic material, the stress can be related to the strain by:

$\sigma_{ij}=C_{ijkl} \varepsilon_{kl}$

Where Cijkl is an element of the 4th-rank elastic constant tensor of the medium.

## Computational methods

The path integral Monte Carlo method is a numerical approach for determining the values of the internal energy, based on quantum dynamical principles.

## History

James Joule studied the relationship between heat, work, and temperature. He observed that if he did mechanical work on a fluid, such as water, by agitating the fluid, its temperature increased. He proposed that the mechanical work he was doing on the system was converted to thermal energy. Specifically, he found that 4185.5 joules of energy were needed to raise the temperature of a kilogram of water by one degree Celsius.

## Notes

1. ^ a b c In this article we choose the sign convention of the mechanical work as typically defined in chemistry, which is different from the convention used in physics. In chemistry, work performed by the system against the environment, e.g., a system expansion, is negative, while in physics this is taken to be positive.


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zbMATH — the first resource for mathematics

Found 42 Documents (Results 1–42)

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Srinivas, V. (ed.) et al., $$K$$-theory. Proceedings of the international colloquium, Mumbai, 2016. New Delhi: Hindustan Book Agency; Mumbai: Tata Institute of Fundamental Research (ISBN 978-93-86279-74-3/hbk). Studies in Mathematics. Tata Institute of Fundamental Research 23, 367-393 (2019).
MSC:  19L50 19L64 13D03
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Singh, Mahender (ed.) et al., Algebraic topology and related topics. Selected papers based on the presentations at the 7th East Asian conference on algebraic topology, Mohali, Punjab, India, December 1–6, 2017. Singapore: Birkhäuser. Trends Math., 1-25 (2019).
MSC:  55P15 55P35 55Q52 16S37 57N15
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MSC:  76D17
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MSC:  92C15 92C10
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Escassut, Alain (ed.) et al., Advances in ultrametric analysis. 14th international conference on $$p$$-adic functional analysis, Université d’Auvergne, Aurillac, France, June 30 – July 4, 2016. Proceedings. Providence, RI: American Mathematical Society (AMS) (ISBN 978-1-4704-3491-5/hbk; 978-1-4704-4676-5/ebook). Contemporary Mathematics 704, 37-48 (2018).
MSC:  46S10 40A05
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MSC:  74M20 74H45 45B05
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Bhattacharjee, Somendra Mohan (ed.) et al., Topology and condensed matter physics. Singapore: Springer; New Delhi: Hindustan Book Agency (ISBN 978-981-10-6840-9/hbk; 978-981-13-4958-4/pbk; 978-981-10-6841-6/ebook). Texts and Readings in Physical Sciences 19, 45-63 (2017).
MSC:  55-01 55Pxx 20J05
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Reviewer: Cenap Özel (Bolu)
MSC:  55R20 55R25
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MSC:  68Q85 68Q60
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Perny, Patrice (ed.) et al., Algorithmic decision theory. Third international conference, ADT 2013, Bruxelles, Belgium, November 12–14, 2013. Proceedings. Berlin: Springer (ISBN 978-3-642-41574-6/pbk). Lecture Notes in Computer Science 8176. Lecture Notes in Artificial Intelligence, 324-335 (2013).
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Proceedings of the 39th annual ACM SIGPLAN-SIGACT symposium on principles of programming languages, POPL ’12, Philadelphia, PA, USA, January 22–28, 2012. New York, NY: Association for Computing Machinery (ACM) (ISBN 978-1-4503-1083-3). 191-202 (2012).
MSC:  68N30 68Q60 68Q85
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Kuncak, Viktor (ed.) et al., Verification, model checking, and abstract interpretation. 13th international conference, VMCAI 2012, Philadelphia, PA, USA, January 22–24, 2012. Proceedings. Berlin: Springer (ISBN 978-3-642-27939-3/pbk). Lecture Notes in Computer Science 7148, 56-71 (2012).
MSC:  68Q60 68Q85
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MSC:  68U35 68T30
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MSC:  74L05
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MSC:  80A10 78A40
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Hermanns, Holger (ed.) et al., Tools and algorithms for the construction and analysis of systems. 12th international conference, TACAS 2006, held as part of the joint European conferences on theory and practice of software, ETAPS 2006, Vienna, Austria, March 25 – April 2, 2006. Proceedings. Berlin: Springer (ISBN 3-540-33056-9/pbk). Lecture Notes in Computer Science 3920, 42-57 (2006).
MSC:  68Q60 68Q85
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MSC:  65N50
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Halbwachs, Nicolas (ed.) et al., Tools and algorithms for the construction and analysis of systems. 11th international conference, TACAS 2005, held as part of the joint European conference on theory and practice of software, ETAPS 2005, Edinburgh, UK, April 4–8, 2005. Proceedings. Berlin: Springer (ISBN 3-540-25333-5/pbk). Lecture Notes in Computer Science 3440, 563-569 (2005).
MSC:  68N99 68Q60
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Garavel, Hubert (ed.) et al., Tools and algorithms for the construction and analysis of systems. 9th international conference TACAS 2003, held as part of the joint European conferences on theory and practice of software, ETAPS 2003, Warsaw, Poland, April 7-11, 2003. Proceedings. Berlin: Springer. Lect. Notes Comput. Sci. 2619, 315-330 (2003).
MSC:  68Q60
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Katoen, Joost-Pieter (ed.) et al., Tools and algorithms for the construction and analysis of systems. 8th international conference, TACAS 2002. Held as part of the joint European conferences on theory and practice of software, ETAPS 2002, Grenoble, France, April 8–12, 2002. Berlin: Springer (ISBN 3-540-43419-4). Lect. Notes Comput. Sci. 2280, 236-250 (2002).
MSC:  68N30 68Q60
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Codognet, Philippe (ed.), Logic programming. 17th international conference, ICLP 2001, Paphos, Cyprus, November 26 – December 1, 2001. Proceedings. Berlin: Springer (ISBN 3-540-42935-2). Lect. Notes Comput. Sci. 2237, 166-180 (2001).
MSC:  68Q85 68N17
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## Pages

### Factor x^4 + x^2 + 1

x^4 + x^2 + 1 is a trinomial, that is, it consists of three terms separated by addition/subtraction signs.

It cannot be factored by the method of splitting the middle term because you can't find two numbers which multiply to give 1 and add up to give 1.

So you can try an approach similar to completing the square. Add and subtract 2x^2 from the expression,

x^4 + x^2 + 1 + 2x^2 - 2x^2

Rearrange the terms to bring x^4, -2x^2 and  together,

x^4 - 2x^2 + 1 + x^2 - 2x^2

Factor the first three terms of the expression by comparing them with (a - b)^2 = a^2 - 2ab + b^2 (Since x^4 - 2x^2 + 1 can be written as (x^2)^2 - 2x^2 + 1^2, so it is corresponding with the right side of the formula (a - b)^2 = a^2 - 2ab + b^2)

(x^2 - 1^2)^2 + x^2 - 2x^2

Simplify,

(x^2 - 1)^2 - x^2

Now there is a difference of two squares, the first squared term being (x^2 - 1) and the second squared term being x. Applying the difference of squares formula,

(x^2 - 1 + x)(x^2 - 1 - x)

Thus the expression x^4 + x^2 + 1 is completely factored to (x^2 - 1 + x)(x^2 - 1 - x)`


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# mcrl2/lps/find_representative.h¶

Include file:

#include "mcrl2/lps/find_representative.h"

add your file description here.

## Functions¶

Node mcrl2::lps::find_representative(Node &u0, GenerateSuccessors generate_successors)

Search for a unique representative in a graph.

Parameters:

• u0 The root of the graph.

• generate_successors A function that generates successors of a node.

This function is based on an iterative version of Tarjan’s strongly connected components algorithm. It returns the smallest node of the first SCC that is detected. The first SCC is a TSCC, meaning that it has no outgoing edges. In a confluent tau graph there is only one TSCC, so this should guarantee a unique representative. N.B. The implementation is based on https://llbit.se/?p=3379


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## Archive for December, 2011

### IPEX and CD25

December 20, 2011

In a recent post I wrote about some ideas of Kendall Smith and his role in discovering the cytokine IL-2. On 8th December I heard him give a talk in which he presented various ideas about IL-2, its receptor and Tregs. I discussed some aspects of IL-2 in the last post. When mutant mice are engineered which cannot produce IL-2 they show a strange combination of symptoms which combine immunodeficiency (a reduced capability of the immune system to fight pathogens) and autoimmune disease (an inappropriate reaction of the immune system to host tissues). This probably has to do with the fact that IL-2 is important for the production of both effector T cells and Tregs, which act in opposite directions. Similar phenomena are seen in the disease of humans called IPEX (immune dysregulation, polyendocrinopathy, enteropathy, X-linked syndrome). It is often attributed to a lack of the transcription factor Foxp3 which is of central importance for the function of Tregs. The gene for Foxp3 is on the X chromosome and this explains the way IPEX is inherited and the term X-linked in its name. However, as pointed out by Smith in his talk, one third of patients diagnosed with IPEX have no mutation in the Foxp3 gene. In this context he referred to a paper of Caudy et al. (J. Allergy Clin. Immunol. 119, 482). What is shown in this paper is that there is a different possible cause of IPEX-like symptoms, namely mutations in the gene for CD25, a surface molecule associated to Tregs.

The paper concerns a patient (an eight year old boy) who had suffered a horrific combination of diseases. It was found that he had mutations in both copies of the CD25 gene. The mutation in one copy came from the mother and was a frame shift due to an insertion. In other words, there is a extra base in the DNA which makes the part of the gene after it look like nonsense when it is being transcribed. The mutation in the other copy came from the father and consisted of one base being exchanged. This happed to cause a stop codon so that reading stopped at that point. The combination of these circumstances meant that the boy could not produce CD25 and this was the presumed cause of his disease. His Foxp3 gene was normal. On the other hand other IPEX patients can produce CD25. Thus there appear to be two diseases with related symptoms. The gene coding for CD25 is on chromosome 10, not the X chromosome. This is why two mutations are necessary to produce CD25 deficiency.

What is the connection to IL-2? The IL-2 receptor, which was also discovered by Kendall Smith and his collaborators, consists of three chains called $\alpha$, $\beta$ and $\gamma$. The second and third are always present on the surface of T cells but the first is only present in variable amounts. In fact the $\alpha$ chain of the IL-2 receptor is nothing other than CD25. The $\beta$ and $\gamma$ chains together allow for some IL-2 signalling but strong signalling in response to normal concentrations of IL-2 is only possible with the help of the $\alpha$ chain. In this case it is not only the case that the receptor signals when IL-2 is bound to it. Binding also causes the receptor to be taken into the interior of the cell and destroyed. This process is an important part of the dynamics associated to IL-2. The $\gamma$ chain of the IL-2 receptor also forms part of the receptor for many other cytokines, for instance IL-4. The gene for this receptor is on the X chromosome. When it cannot be produced due to a mutation this leads to a disease called X-linked severe combined immunodeficiency (SCID). In this case the immune system does not function since so much of its signalling system has been disrupted. This is also known as the ‘bubble boy disease’ since children affected by it have to live in a sterile environment.

### Shifting attention towards Tregs

December 3, 2011

Yesterday I heard a talk by Abul Abbas where two of the main themes were regulatory T cells (Tregs) and interleukin 2. Correctly functioning immunity is the result of a balance between effector cells and Tregs and he emphasized that in trying to develop therapies it might be more valuable to concentrate on influencing the regulatory side. He described a mouse model which he has developed for studying autoimmune disease. One criterion in developing this model was that it should concern the target tissue where an antigen is expressed and not the lymphoid tissue. Another is that the target tissue should be easily accessible for doing experiments in vivo. For this reason he chose the skin. In this transgenic model antigen expression can be turned on and off by feeding the mice with doxycyclin. When the antigen is turned on an autoimmune disease results. If it is turned off the mice recover. If it is turned on again the mice get sick again but much less than the first time. This is reminiscent of ordinary immunity which is due to memory effector cells. In this case it seems that there are memory Tregs. This suggests the idea that a possible cause of autoimmune disease in humans could be a lack of memory Tregs.

When IL-2 was first discovered it was known for causing T cells to proliferate and thus strengthening the immune response. More recently it has been found that eliminating IL-2 does not necessarily act in an immunosuppressive way. Apparently it can be replaced by something else in driving the proliferation of effector T cells. On the other hand it also drives the proliferation of Tregs and Abbas argued that this is its most essential function. In that case it cannot be replaced.

The lecturer made a number of interesting comments about themes such as immunology, therapies for immune disorders and cancer, clinical trials etc. I did not note them down and I cannot reproduce them here. Nevertheless I have the impression that a learned of lot of things which I might profit from in the future.


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1. ## Polynomial

Question: If $x_{1}, ... , x_{n}$ are distinct numbers, find a polynomial $f_{i}$ of degree $n-1$ which is $1$ at $x_{i}$ and 0 at $x_{j}$ for $j \ne i$. Hint: the product of all $(x-x_{i})$ for $j \ne i$ is zero at $x_{j}$ if $j \ne i$. .

Confusion: The part of the question that I don't understand is the 'which is $1$ at $x_{i}$ and 0 at $x_{j}$ for $j \ne i$.' What are $x_{i}$ and $x_{j}$? Does it mean that for any two numbers we choose from the sequence $x_{1}, ..., x_{n}$, we get $f_{i}$ as 0, for one, and 1, for the other? And how would you go about finding this polynomial?

2. The polynomial is...

$\displaystyle P(x)= \frac{\prod_{i=1, i\ne j}^{n} (x-x_{i})} {\prod_{i=1, i\ne j}^{n} (x_{j}-x_{i})}$

Kind regards

$\chi$ $\sigma$

3. Originally Posted by chisigma
The polynomial is...

$\displaystyle P(x)= \frac{\prod_{i=1, i\ne j}^{n} (x-x_{i})} {\prod_{i=1, i\ne j}^{n} (x_{j}-x_{i})}$

Kind regards

$\chi$ $\sigma$
Can you do a bit of explanation, please, if you don't mind? For example, it's not apparent to me that it's of degree $n-1$.

4. Originally Posted by Hardwork
The part of the question that I don't understand is the 'which is $1$ at $x_{i}$ and 0 at $x_{j}$ for $j \ne i$.' What are $x_{i}$ and $x_{j}$?
The problem asks to find, for every $i=1,\dots,n$, a polynomial $f_i$ with the given property, which is a total of $n$ polynomials.

For example, it's not apparent to me that it's of degree $n - 1$.
In the denominator, you have a product of numbers. In the numerator, you have a product of $n-1$ expressions of the form $x-c$ for various constants $c$.

This is similar to the interpolation polynomial in the Lagrange form (the link uses notation $\ell_j$ instead of $f_j$).


text:
# Math Help - Divisors

1. ## Divisors

Prove, that a composite number 2^p - 1 (where p is a prime) has at least 2 different prime factors.

How I tried to solve this problem:
I tried to write 2^p -1 as x^n (x, n are integers, n>1, just to prove that it can't be this way). I know that the only prime factors of 2^p - 1 can be numbers like 2kp+1 and I also tried this way. What is more, I tried with two cases- n odd and n even, separately. But I didn't manage to prove anything. Please, help. Thank you in advance.

2. x^n-1=(x-1)*(x^{n-1}+x^{n-2}+...+x+1)

EDIT: NOT HELPING TOO MUCH.

2^31 - 1 is prime.

3. Yes, but in my question if you have an equation x^n-1=(x-1)*(x^{n-1}+x^{n-2}+...+x+1) then x=2 and we have to prove that (x^{n-1}+x^{n-2}+...+x+1) has at least 2 prime factors. How?

Prove, that a composite number 2^p - 1 (where p is a prime) has at least 2 different prime factors.

How I tried to solve this problem:
I tried to write 2^p -1 as x^n (x, n are integers, n>1, just to prove that it can't be this way). I know that the only prime factors of 2^p - 1 can be numbers like 2kp+1 and I also tried this way. What is more, I tried with two cases- n odd and n even, separately. But I didn't manage to prove anything. Please, help. Thank you in advance.
I'll try to show a more general result: Let $n, a$ and $m$ be positive integers. If $2^n-1=a^m$, then $2^n-1=a$. To put it differently, $m$ must equal 1. (execpt for the case $n=1$, where $2^1-1=1^m$)

We suppose that $n\geq2$. It's clear that $a$ is odd.
If $m$ is even, then $a^m$ would be of the form $4k+1$. But $2^n-1$ has the form $4q+3$; therefore $m$ is odd.*

The following identity holds, since $m$ is odd: $a^m+1=(a+1)(a^{m-1}-a^{m-2}+\cdots+a^2-a+1)=2^n$. Note that the term $a^{m-1}-a^{m-2}+\cdots+a^2-a+1$ is odd and thus relatively prime to $2^n$. So $2^n$ must be equal to $a+1$, namely $a=2^n-1$.

Back to the original problem... Assume to the contrary that $2^p-1$ has only one prime factor, that is $2^p-1=q^m$, where $q$ is prime. But the result implies that $2^p-1=q$, contradiction.

* To see why: For any odd $a$ we have $a\equiv\pm1\pmod4$. Then $a^m\equiv1\pmod4$ if $m$ is even. But $2^n-1\equiv-1\pmod4$, so $1\equiv-1\pmod4$, contradiction. Sorry if this part is tedious...

5. Thank you! You helped me so much. I'm very grateful!

6. @melese, maybe you have and idea how to solve this one:
We have:
z = (2^(mn) - 1)/[(2^m - 1)(2^n - 1)], where (2^m - 1) and (2^n - 1) are prime numbers.
Prove that (2^m - 1) and (2^n - 1) are not the only prime factors of z.

I tried to solve it writing z = (2^m - 1)^a * (2^n - 1)^b and proving that it is not correct. But I don't know how. I also noticed that it would be sufficient to prove that 2^(mn) - 1 can't be written as ((2^m - 1)^x * (2^n - 1)^y but I also have no idea how to prove it. I tried with comparing highest exponents, but nothing..

Thank you very much in advance!


text:
Misplaced \omit when using \multicolumn within \multirow

I'm trying to put a \multicolumn within a \multirow environment in the following manner:

\documentclass{article}

\usepackage{multirow}
\usepackage{array}
\usepackage{mwe}

\begin{document}
\begin{tabular}{|p{.03\textwidth}|p{.73\textwidth}|p{.03\textwidth}|p{.15\textwidth}|}
Test1 & Test2 & Test3 &
\multirow{1}{*}{\multicolumn{1}{m{\linewidth}}{\includegraphics[width=\linewidth]
{example-image}}}
\end{tabular}
\end{document}

For clarification, I have multiple rows in the table (6) and the multirow spans those 6 rows. The inner multicolumn is so that the image is centered within the multirow. However, I get the error:

Misplaced \Omit

Does anyone have any idea why this is happening? Thanks!

Update:

See Christian's answer for fixing the error. But latex was still having trouble centering the picture vertically when one of the rows was had multi line text. As a workaround, I increased the number of rows the multirow is spanning even though there are actually 6. This is admittedly an ugly hack and one I'm not particularly happy with, but it works for now. If anyone has any better solution, I'm open to suggestions.

Update 2:

Ok so update 2. See Christian's revised solution and comments for two possible ways to get around this problem. Thanks Christian for all the help!

• You have to switch the \multirow and \multicolumn statements. \multicolumn outside, then \multirow inside. I suppose, the lines with 'Test1' etc. should be in multirow(s) as well.
– user31729
Feb 19, 2014 at 14:25

I suppose, this is what you meant.

\documentclass{article}

\usepackage{multirow}
\usepackage{array}
\usepackage{mwe}

\begin{document}
\begin{tabular}{|p{.03\textwidth}|p{.73\textwidth}|p{.03\textwidth}|p{.15\textwidth}|}
Test1 & Test2 & Test3 &
\multicolumn{1}{m{\linewidth}}{\multirow{6}{*}{\includegraphics[scale=0.1]{example-image}}% End of multirow
}% End of multicolumn
\\ % End of the first row
Line2  & & &\\
Line3  & & & \\
Line4  & & & \\
Line5  & & & \\
Line6  & & & \\
\end{tabular}
\end{document}

Another version with vertical centered content, which uses the makecell package and the \multirowcell command, which allows for a vertical shift of the content in the tabular cell (First optional parameter of \multirowcell), which is set to -2pt by me since I have the impression, this aligns correctly, however, it has to be revised if the size/scaling of the image changes. It is a 'hack' as well, but perhaps it is better than adding extra rows.

 \documentclass{article}

\usepackage{multirow,makecell}
\usepackage{array}
\usepackage{mwe}

\begin{document}
\begin{tabular}{|p{.1\textwidth}|p{.2\textwidth}|p{.1\textwidth}|p{.15  \textwidth}|}
\hline
Test1 & Test2 & Test3 &
\multirowcell{6}[-2pt]{\includegraphics[scale=0.1]{example-image}%
}% End of multirowcell
\\
Line2  & & & \\
Line3  & & & \\
Line4  & & & \\
Line5  & & & \\
Line6  & & & \\
\hline
\end{tabular}

\end{document}

I added the \hline in order to check to the alignment and reduced the width of columns. Please roll back as you desire.

By the way, the \multirow command allows also a vertical shifting, but I have not tried that so far.

• This fixes the error, but now I have the problem that latex centers the picture in only one row, whereas I want it to be centered across all six rows. Plus if I specify column borders, they are applied to only one row, whereas they should be applied to all six rows. Does that make sense?
– K G
Feb 19, 2014 at 15:01
• @KG: You have to change the 1 as an argument to \multirow to 6. Additionally, I added five 'dummy' lines. Sorry, I forgot that you wanted to have six rows in my answer initially.
– user31729
Feb 19, 2014 at 15:06
• yes I am aware of that. I have already done that. Let me test your code and I'll get back to you.
– K G
Feb 19, 2014 at 15:07
• Ok I did a little bit more digging around and have figured out that the problem is when text in a row spans multiple lines. Latex seems to center the figure assuming single-line rows. The alignment is thrown off when one or more of the rows has multiline. Any solution for that?
– K G
Feb 19, 2014 at 15:35
• Not at the moment, sorry, for I am writing on a tablet and have no access to LaTeX at the moment. What about a parbox containing your figure, alignment can be done within?
– user31729
Feb 19, 2014 at 15:54


text:
# plotLayout¶

## Purpose¶

Divides a plot into a grid of subplots and assigns the cell location in which to draw the next created graph.

## Format¶

plotLayout(g_rows, g_cols, ind)
Parameters
• g_rows (scalar) – number of rows of the graph layout.

• g_cols (scalar) – number of columns of the graph layout.

• ind (scalar) – cell location in which to place the next created graph.

## Examples¶

// Create 10x4 matrix where each column is an additive
// sequence from 0.1 to 1.0
x = seqa(0.1, 0.1, 10);
y = ones(10, 4).*x;

// Apply a function to each column of 'y'
y[.,1] = cos(x);
y[.,2] = sin(x);
y[.,3] = cdfn(x);
y[.,4] = exp(x);

for i(1, 4, 1);
// Divide plot canvas into a 2x2 grid of subplot
// locations and place each newly created graph in the
// next available cell location.
plotLayout(2, 2, i);

// Plot each column of y in a separate subplot window.
plotXY(x, y[., i]);
endfor;

// Clear the layout so the next plot will not be inside this
// layout
plotClearLayout();

## Remarks¶

After calling this function all subsequent graphs will be plotted inside of the specified layout until the layout is reset with plotLayout(), or the layout is cleared with plotClearLayout().


text:
s - solid l - liquid g - gas aq - aqueous, dissolved in water. Question 19: Name three classes of substances, which react with an acid to form salts. Ammonium chloride is a salt of a strong acid and a weak base, Ammonium chloride solution has an acidic effect on litmus solution due to the accumulation of H + ions in the solution (pH < 7), The solubility of ammonium chloride in water is represented by the following equations.. H 2 O → H + + OH −. Write the two equations for each salt in this demonstration. Test. Created by. Exemplar ionic equation - Ionic and net ionic equations for precipitation reactions. As ammonia gas is very soluble in water, reaction … MORE THAN 8700 articles covering all major scientific disciplines and encompassing the McGraw-Hill Encyclopedia of Science & Technology and McGraw-Hill Yearbook of Science & Technology . Learn. Ammonium ions are also non-metal ions and are tested for by warming the ammonium compound with aqueous sodium hydroxide, or some other strong base. Can you help please. Acid base and salt in chemistry. (3) The partial ionization is a dynamic equilibrium , as indicated by the double arrow in equation (3). Salts that are derived from the neutralization of a weak acid (HF) by a strong base (NaOH) will always produce salt solutions that are basic. Terms in this set (58) What are bases . STUDY. This post would cover acid and base reactions such as acid base neutralization reaction, reactions of acid with metal etc. For instance, in the reaction of hydrochloric acid (a strong acid) with ammonia (a weak base), water is formed, along with ammonium chloride. Each acid is linked to a conjugate base on the other side of the equation. Hydrolysis of ammonium chloride. This conjugate acid is a weak acid. The reason may be that, for gemini quaternary ammonium salts with the short alkyl chain counterions, the dominating factors influencing the Γ max values are the size of hydrophilic head groups and the electrical repulsion between hydrophilic head groups . STUDY. Give a balanced equation for conversion of (a) an ammonium salt to a basic gas (b) a soluble lead salt to an insoluble lead salt – formed as a white precipitate. Is there any way other than using an indicator to know, whether a compound is acidic or basic? Or.. NH 4 Cl + NaOH -> NaCl + NH 4 OH. The salt in the equation above is the one formed from ethylamine, CH 3 CH 2 NH 2, and HBr. 115,000-PLUS definitions from the McGraw-Hill Dictionary of Scientific and Technical Terms . Should be the same thing. NH 4 Cl→ Cl − + NH 4 + Match. (b) How is it obtained from sodium chloride? Because the nitrogen atom of an ammonium salt has a positive charge, ammonium salts are more water-soluble than amines. Bases. familymum Tue, 08/19/2008 - 07:11. ammonium salt plus base equals what does it produce? Salts. (i) Zinc, a metal reacts with dilute sulphuric acid to form zinc sulphate and hydrogen gas is liberated. Acid-base properties of salts. Salts of Weak Bases and Strong Acids. How do buffer solutions work? Salts That Form Acidic Solutions. Answer: Three different classes of substances are a metal, a base and a metallic carbonate, which react with an acid to form salts. Salt is an ionic compound that results from the neutralization reaction of acids and bases. a. AlCl 3 6H 2 Question 3: Ammonium Nitrate + Potassium Hydroxide ----> NH4NO3 + KOH. Question 83. Ammonia is a weak base because its nitrogen atom has an electron pair that readily accepts a proton, which makes it weak. Drugs containing an amino group are often prepared as ammonium salts to improve their solubility in body fluids. jadejackson9. Base: ammonium hydroxide NH4OH. Acid salts are the converse of basic salts; they are formed in the neutralization reaction between a strong acid and a weak base. Ammonium sulphate is formed after reaction between ammonium hydroxide (a weak base) and sulphuric acid (a strong acid). (iii) Basic Salts: Salts which are formed after the reaction between a weak acid and strong base are called Basic Salts. All the solutions of Acids, Bases and Salts - Chemistry explained in detail by experts to help students prepare for their ICSE exams. Chemical equation: NaOH (aq) + NH 4 Cl (s) → NaCl (aq) + H 2 O (l) + NH 3 (g) Ammonia gas can be tested with the use of damp red litmus paper. Concise Chemistry Part II - Selina Solutions for Class 10 Chemistry ICSE, 3 Acids, Bases and Salts. It states that. Mention its molecular formula. The chemical equation for the hydrolysis of ammonium ions is given below (equivalent to the acid dissociation of ammonium ions): NH 4 + (aq) + H 2 O (l) ⇋ NH 3(aq) + H 3 O + (aq) and It solves similar to net ionic equations, but produces different products based on the reactants. Acids react with metals, bases and carbonates to produce salts. It's going to be a Brønsted–Lowry base. Acids, Bases and Salts Class 10 Important Questions Long Answer Type. Question 1. For example, sodium hydroxide reacts with ammonium chloride to form sodium chloride, water and ammonia gas. If neither H+ nor OH– ions are produced, write “no reaction” for the second equation. C10 acids, bases and salts 1. Salts of strong acids and strong bases ("strong salts") are non-volatile and often odorless, whereas salts of either weak acids or weak bases ("weak salts") may smell like the conjugate acid (e.g., acetates like acetic acid and cyanides like hydrogen cyanide ) or the conjugate base (e.g., ammonium salts like ammonia) of the component ions. ... base and a salt of that weak base ( made from reacting the weak base with a strong acid) Example :ammonia and ammonium chloride NH 3 and NH 4 +Cl-How Buffer solutions work In an ethanoic acid buffer CH 3CO 2H (aq) CH 3CO 2-(aq) + H + (aq) Acid conjugate base In a buffer solution there is a much higher … For example, ammonium chloride, NH 4 Cl, is a salt formed by the reaction of the weak base ammonia … What are the state symbols. Alkaline buffer solutions are commonly made from a weak base and one of its salts. This produces ammonia gas, which can be tested for by placing damp red litmus paper in the gas. LEARNING OUTCOMES Define acid and acid anhydride Investigate the reactions of non-oxidising acids with metals, carbonates, hydrogen carbonates and bases Define base and alkali Investigate the reaction of bases with ammonium salts Relate acidity and alkalinity to the pH scale Discuss the strength of acids and alkalis on the basis of their completeness of … Video transcript - Ammonia is a weak base, and if ammonia reacts with water water is gonna function as a Brønsted–Lowry acid. Ammonium chloride $$\left( \ce{NH_4Cl} \right)$$ is a salt that is formed when the strong acid $$\ce{HCl}$$ is neutralized by the weak base $$\ce{NH_3}$$. Thus, ammonia is a weak base, and like acetic acid, does not conduct electricity nearly as well as aqueous salt. Spell. 2. Alkalis react with ammonium salts to form salt, water and ammonia gas. We Provide Step by Step Answer of Intex -1, Exercise-3 A ,Exercise-3 B Exercise-3 C , Miscellaneous and Previous Year Questions of Exercise-3 Acids Bases and Salts ICSE Class-10 .Visit official Website CISCE for detail information about ICSE Board Class-10. NH 4 Cl + NaOH -> NaCl + H 2 O + NH 3. Its ionic equation is shown below: The symbolism of our chemical equation again indicates a reactant-favored equilibrium for the weak electrolyte. Salt hydrolysis can be described in two chemical equations, the first showing the dissociation of the salt, and the second net equation showing the production of H+ or OH– ions. What are all metal oxides. Give its any one use. Ammonium react with water to produce hydronium or hydrogen ions. Acid and base reactions . It is the production of these hydroxide ions that imparts ammonia's characteristic basicity. Again, it doesn't matter what concentrations you choose as long as they are the same. But then why can't chlorine ions react with water to produce hydroxide ions? Substances that neutralise acids to form salt and water. Salts are constituted of positively charged ions, known as cations and negatively charged ions, known as anions, which can either be organic or inorganic in nature. Also, when dissolved in water, ammonia acquires hydrogen ions from water to produce hydroxide and ammonium ions. So ammonia is a weak electrolyte as well. What are bases. If these were mixed in equal molar proportions, the solution would have a pH of 9.25. According to Figure 16.10, the ammonium ion will lower the $$pH$$, while according to Equation $$\ref{16.3}$$, the acetate ion will raise the $$pH$$. Acid: phosphoric acid H3PO4. This post would cover the basics of acid base and salt in chemistry. Ammonium chloride (NH 4 Cl) is a salt that is formed when the strong acid HCl is neutralized by the weak base NH 3 . Bases. Strong bases, such as NaOH, are 100 percent ionized in aqueous solution and weak bases, such as ammonia, are only partially ionized in aqueous solution. Write. But ammonium ions will react with water, will hydrolyse because ammonium is the conjugate acid of a weak base. The chloride ion produced is incapable of hydrolyzing because it is the conjugate base of the strong acid $$\ce{HCl}$$. Ammonium salts of low molecular weight are soluble in water if the hydrocarbon portion of the amine is small. A frequently used example is a mixture of ammonia solution and ammonium chloride solution. For example; Sodium carbonate, Sodium acetate, etc. Brønsted Lowry definition. Ammonium chloride is soluble in water. Salts of a strong acid and a weak base are acidic in nature with pH value less than 7. If it turns blue then a basic gas has been given off and that must be ammonia. Flashcards. For example, ammonia reacts with HBr to give ammonium bromide, NH 4 + Br-. The complete equation is: NH4NO3 + KOH → NH4OH + KNO3. (c) Why it is called a basic salt? PLAY. Answer: (a) Ammonium salt [NH 4 Cl] on heating produces NH 3 (g) which is a basic gas (b) Lead nitrate when reacts with sodium chloride insoluble white ppt. ammonium salt plus base equals what? It breaks down into ammonium and chloride ions. Primary amines are weak bases very similar to ammonia and so form salts with acids. (2012) Answer: Ammonium chloride is soluble in water. What happens with aqueous solutions of a salt such as ammonium acetate, where both the cation and the anion can react separately with water to produce an acid and a base, respectively? Chemistry - bases, salts, alkalis, equations. An Acid is a substance that donates hydrogen ions (H+) in aqueous solution; A Base is a substance that accepts hydrogen ions (H+) in aqueous solution; Neutralization. Gravity. Metal oxide + acid -> salt + water. Water is going to donate a proton to ammonia, and ammonia is going to accept the proton. Neutralisation is the reaction between an acid and a base. When we neutralize a weak base with a strong acid, the product is a salt containing the conjugate acid of the weak base. Substances that neutralise acids to form salt and water. Chemistry - bases, salts, alkalis, equations. What are all metal oxides. Answer: For ammonium nitrate, salt, the parent base is ammonium hydroxide (NH4OH), and the parent acid is nitric acid (HNO3). Indicators are used to determine whether a solution is acidic or alkaline. Salts of a strong acid and a strong base are neutral with pH value of 7. (a) Write the chemical name and chemical formula of washing soda. 3000 biographies of notable scientific figures . Write equations to describe their reactions with suitable acids. Salts of a strong base and a weak acid are basic in nature, with pH value more than 7. pH of salt solutions. What is the chemical name of common salt? The conjugate acid of the weak base makes the salt acidic. Give equations of the reactions. I believe that's the overall. And what explains this is its salt hydrolysis. What is the word equation for the reaction between a metal oxide and acid. PLAY. What are the state symbols. s - solid l - liquid g - gas aq - aqueous, dissolved in water. ammonium salt plus base equals what? The second possibility. Acids Bases and Salts ICSE Class-10 Concise Chemistry Selina Solutions Chapter-3. An acid and base reactions such as acid base and a weak base with a strong,. 07:11. ammonium salt plus base equals what does it produce Part II - Selina Solutions Chapter-3 base. Of acid base neutralization reaction between a strong acid and strong base are in... Nh 4 Cl→ Cl − + NH 4 Cl + NaOH - > NaCl H! Which can be tested for by placing damp red litmus paper in the gas,. Nitrogen atom of an ammonium salt plus base equals what does it produce charge, salts... 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