text:
# \lowercase inside \url

Why is this snippet:

\newcommand{\Subject}{Skeleton}
\url{http://www.cheatsheet.jim.pm?subject=\lowercase{\Subject{}}}

rendered to this?

http://www.cheatsheet.jim.pm?subject=\lowercase{Skeleton}

and not the intended:

http://www.cheatsheet.jim.pm?subject=skeleton

-
I believe that url makes use of verbatim. – Sigur Apr 8 '13 at 14:32
You are probably right @Sigur, but why does \Subject{} get expanded but not \lowercase? – SlimJim Apr 8 '13 at 14:33
\lowercase is not an expandable command. – David Carlisle Apr 8 '13 at 14:34
@sigur: url proves a \verb-like command but does not use verbatim to do so. – Matthew Leingang Apr 8 '13 at 15:52

\lowercase (and {}) are not expandable so you need to change the order of execution:

\documentclass{article}
\usepackage{url}

\newcommand{\Subject}{Skeleton}

\newcommand\lcURL[1]{%
\edef\tmp{\lowercase{\noexpand\url{http://www.cheatsheet.jim.pm?subject=#1}}}%
\tmp}

\begin{document}

\lcURL{\Subject}

\end{document}

If you need to remove space you need to use expansion only if you are not going to interfere with \url. Fortunately LaTeX has a built-in command to remove spaces via expansion:

\newcommand{\Subject}{Skeleton bones}
\makeatletter
\newcommand\lcURL[1]{%
\edef\tmp{\lowercase{\noexpand\url{http://www.cheatsheet.jim.pm?subject=\zap@space#1 \@empty}}}%
\tmp}
\makeatother

produces

http://www.cheatsheet.jim.pm?subject=skeletonbones

-
it works but I have no clue how, know I want to remove spaces from the \Subject with \StrSubstitute{#1}{ }{} but I cannot get it to compile and show properly, I tried putting inside the lcURL at different places as well as a function RemoveSpace that mimics the magic from lcURL but all in vain. The most common error is ! Argument of \reserved@a has an extra }. – SlimJim Apr 8 '13 at 21:41
I don't think \StrSubstitute works via expansion so it will interfere with \url, see update. – David Carlisle Apr 8 '13 at 22:04

Just to see how one can use \StrSubstitute in this context:

\documentclass{article}
\usepackage{url,xstring}

\newcommand{\Subject}{Ske Le Ton}

\newcommand\lcURL[1]{%
\begingroup
\StrSubstitute{#1}{ }{}[\tmp]%
\edef\tmp{%
\lowercase{%
\endgroup
\noexpand\url{http://www.cheatsheet.jim.pm?subject=\tmp}%
}%
}%
\tmp
}

\begin{document}

X\lcURL{\Subject}X

X\url{http://www.cheatsheet.jim.pm?subject=skeleton}X

\end{document}

With \begingroup and \endgroup we make sure that any definition of \tmp vanishes after usage.

-


text:
Show that if a is any m x n matrix, then ImA = A and AIn = A

Homework Statement

Show that if A is any m x n matrix, then the m x m identity multiplied by A = A, and A multiplied by the n x n identity = A.

The Attempt at a Solution

I know how to prove this by writing out a general m x n matrix, and multiplying it by the identity, but is there a better way of showing this? It just seems kind of silly when I write it out...

$$(AI)_{i,j } = \sum_{r= 1}^n A_{i, r} I_{r, j} = a_{i, j}$$

format: plain


text:
# Should 4 fundamental forces really be 3 because of electroweak unification?

Well, the so called "electroweak unification" is really more of an "electroweak mixing". I want to show you how the mixing is done, so that you yourself can decide whether you prefer to call it unification or mixing. You do not need to completely understand the equations, I will try to highlight the important points.

The standard model is written in the language of quantum field theory (QFT). In QFT one usually starts from a Lagrangian, which is a function of the fields that we suppose to be the elementary constituents of the world, or, to be even more precise, the elementary objects of our description of the world, and then one quantizes it by using the appropriate techniques. Given a Lagrangian, the whole theory can be derived from it. The Lagrangian of the electromagnetic field and a fermionic (charged) field, such as the electron field, can be written as $$\mathcal{L}=\bar{\psi}\gamma^{\mu}(i\partial_{\mu}-m-eA_{\mu})\psi-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}$$ Here $e$ is the electric charge of the fermion, $m$ its mass, $\psi$ its field, $A_{\mu}$ the photonic field and $F_{\mu\nu}F^{\mu\nu}$ is something like the product of the electric and magnetic fields. The two terms $$\mathcal{L}_{\psi}=-\bar{\psi}\gamma^{\mu}(i\partial_{\mu}-m)\psi$$ and $$\mathcal{L}_{A}=-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}$$ tell you how the fields behave independently of each other. The term $$\mathcal{L}_{int}=e\bar{\psi}\gamma^{\mu}\psi A_{\mu}$$ contain both $\psi$ and $A_{\mu}$, and it tells you how the fields interact.  Now, the full Lagrangian of the electroweak theory is much more complex. It contains the Higgs field, a symmetry breaking potential and six generations of fermions, so I won't write it in its full form. Nevertheless, the interaction term between a single fermion and the "electroweak" field can be schematically written as

$$\mathcal{L}_{int}=-\bar{\psi}\gamma^{\mu}(g'B_{\mu}+g\tau^{a}W_{\mu}^{a})\psi$$ Here $a=1,2,3$, $B_{\mu}$ and $W_{\mu}^{a}$ are the four "electroweak" fields and $\tau^{a}$ are three appropriate matrices ($\psi$ is a column vector and $\bar{\psi}$ is a row vector, so they can have matrices in between). $g$ and $g'$ are the coupling constants of the theory, and they can be chosen different (and in fact they are). This is very important to the distinction between unification and mixing. Now, first of all, the $W$'s and $B$ cannot be directly identified with the photonic field and the $W^{\pm}$, $Z$ fields. As I told you, the full Lagrangian contains other terms. These terms, through the mechanism of spontaneous symmetry breaking, give mass to linear combinations of the $W$'s and the $B$. The combinations are as follows: $$W^{\pm}_{\mu}=\frac{1}{\sqrt{2}}(W^{1}_{\mu}\mp i W^{2}_{\mu})\\ Z_{\mu}=\frac{1}{\sqrt{g^{2}+g'^{2}}}\ (gW^{3}_{\mu}-g'B_{\mu})\\ A_{\mu}=\frac{1}{\sqrt{g^{2}+g'^{2}}}\ (g'W^{3}_{\mu}+gB_{\mu})$$ $W^{\pm}_{\mu}$ are the fields of the $W^{\pm}$ bosons. $Z_{\mu}$ is the field of the $Z$ boson. $A_{\mu}$ is the field of the photon. The former are given mass. $A_{\mu}$ turns out to be given no mass. When you rearrange the terms in the Lagrangian in terms of these combinations of fields, you get the Lagrangian of electrodynamics plus the Lagrangian of the weak theory. As you can see, the photonic field is a mixing between the $B$ field and the $W^{3}$ field. The $Z$ field is too. So one gets electrodynamics and the weak theory by mixing fields.  Now, as I told you, the coupling constants $g$ and $g'$ can be chosen to be different. This is because in the theory which contains $B$ and the $W$'s, symmetry principles (together with something called the renormalizability principle) don't constrain the $B$ field to interact directly with the $W$'s. So you can interpret the starting point of the electroweak theory to be a theory in which two different kind of interactions exist. But because the theory contains the symmetry breaking terms, at low energies the two mix to give the theories of electromagnetic and weak interactions. On the other hand, let's suppose that to begin with we had a theory of a truly single interaction. Then would $g$ and $g'$ be constrained to be chosen equal? The answer is still: no. The theory would have the very same form as the one with two different interactions. This has to do with the symmetry group of the theory, which in this case is called $U(1)\times SU(2)$ ($U(1)\equiv B, SU(2)\equiv W$'s). When you have a $U(1)$ multiplying some other group, it is difficult to give an interpretation to the multiplication itself. $U(1)$ doesn't change the structure of the second group in any case, i.e. it behaves as a somewhat distinct portion of the product group (and this is why the coupling constants can be chosen to be different).

In conclusion, if one wants to call the electroweak theory a "unification" of the electromagnetic and weak interactions, one is allowed to do so. On the other hand, given the behavior of the fields contained in the initial Lagrangian of the theory, one could also say that the electroweak theory is a "mixing" between two different kinds of interactions, such that this mix gives back the electromagnetic and weak interactions. You can see why we still say that there are four fundamental interactions.

A more practical answer is that in many cases it is more useful to consider them separately. You could compare with electromagnetism. If I want to design a motor, it is much easier to work with the magnetic field generated by the coils than to invoke the whole glory of Maxwell's equations. Similarly, if I want to explain the propagation of light waves, there is no need to worry about the weak interaction. If I want to study low energy beta decay, the electromagnetic force is not important. There are substantial regions of the world where electromagnetism is isolated from the weak force. Electroweak theory is beautiful and important but for most applications they are distinct.


text:
# Accessing neural network variables to configure Transfer Functions¶

In some cases, it may be beneficial to access parameters of the neural network to configure Transfer Functions. These variables may be specified as global variables of the neural network module that is to be loaded. These variables may contain size information of a dynamically sizable (typically engineered) neural network.

There are two conceptually different ways to access neural network variables.

## Accessing neural network variables in the body of a Transfer Function¶

Accessing neural network parameters in the body of a Transfer Function is quite simple through the nrp.config.brain_root variable. This variable allows direct access to the Python module used as neural network (in case of a Python network) or the generated network object in case a H5 network is used.

## Accessing neural network variables to configure Transfer Functions¶

The problem when accessing variables of the neural network in the specification of a Transfer Function (i.e., in the decorators) is that the value of these variables may change when the neural network is changed. However, the specification of a Transfer Function is cached, i.e. rather than re-executing the code of a Transfer Function for example on a reset, we execute required operations on parts of Transfer Functions. This allows us to do partial resets, for example to keep recorded csv data when the brain is changed during a simulation.

Therefore, the nrp.config.brain_root variable must not be used to configure Transfer Function devices. The CLE simply would not recognize any changes to the network variable.

Instead, such a variable access can be implemented in a very similar fashion to the access of neural network populations, as these also change whenever the brain is reloaded. That is, one simply has to access the variable through the provided API nrp.brain. This variable tracks its access and thus allows the NRP to track that the configuration depends on the neural network variable and to recompute the current value in case the neural network changed during the simulation.

For an example, consider that you have two variables in your network called ImageNumPixelRows and ImageNumPixelColumns that correspond to the resolution of an image. Based on these variables, a device group has to be used in a Transfer Function that contains as many devices as there are pixels in the converted image. In this situation, one would typically want to easily change this image resolution without having to adjust several magic numbers in all affected Transfer Functions.

To solve this, the supported way is to specify the access to neural network variables through the nrp.brain variable. You can simply do arithmetic calculations, navigate through the network model using attribute or index accesses or call methods. However, some parts of the Python language such as the in operator or list comprehension are not supported. The usage is exemplified in the following listing.

@nrp.MapSpikeSource("LGNBrightInput", nrp.map_neurons(nrp.nrange(0, nrp.brain.ImageNumPixelRows * nrp.brain.ImageNumPixelColumns), lambda i: nrp.brain.LGNBright[i]), nrp.dc_source)
@nrp.Robot2Neuron()
def my_transfer_function(t, LGNBrightInput):
...

The way how this works is that any access to the nrp.brain variable implicitly defines a function that is executed lazily when the neural network is loaded or changed.

A consequence of this is that only very few functions are supported by this type of API. This is also the reason that the regular range function can not be used if the arguments depend on the neural network and a custom function nrp.nrange has to be used that extends the functionality of the builtin range by a support of this lazy evaluation API.

If a use case requires a complicated access beyond the currently supported subset of the Python language, we offer a way to insert a blackbox in this procedure where you can define a lazy evaluation function yourself. This is done through the nrp.resolve method. You have to supply a function to this method that either takes no arguments, one argument that is the readily loaded neural network or two arguments consisting of the neural network module and the brain communication adapter.

The usage is depicted in the following listing:

@nrp.MapSpikeSource("LGNBrightInput", nrp.map_neurons(nrp.nrange(0, nrp.resolve(lambda brain: brain.ImageNumPixelRows * brain.ImageNumPixelColumns)), lambda i: nrp.brain.LGNBright[i]), nrp.dc_source)
@nrp.Robot2Neuron()
def my_transfer_function(t, LGNBrightInput):
...

However, we believe that where possible, the first type of this specification is more understandable and therefore encourage its usage.


text:
# Machine Learning on dask

The data science team at Comtravo uses dask to coordinate fairly complex machine learning workloads, both for training and running them in production. Our NLP pipeline has a lot of cross-dependencies between the different predictive models and I find it really useful to have an easy, lightweight, and purely ‘pythonic’ way of encoding and executing model dependencies. The direct use of the dask computational graph was not something I was familiar with before joining Comtravo. I hope you’ll enjoy this overview of what the graph is and how to use it to coordinate custom workloads.

# What is dask?

Dask is a Python library usually marketed as “out-of-core pandas”. Meaning that dask is able to run computations on pandas data frames that do not fit into memory. I won’t talk about that here, as there are lots of tutorials that demonstrate that use case (see References at the bottom of the article).

I want to highlight that dask is much more than just out-of-core pandas, it also offers dask.bag, which is a Map-Reduce-like abstraction over any data on disk (remote or local). I’ve found dask.bag especially useful for log parsing; I can read a large number of log files directly from S3 into a dask.bag. There’s also dask.array, which is an out-of-core abstraction over large numpy arrays.

So, in general, dask provides out-of-core abstractions over existing functionality in pandas and numpy. The part I find really interesting is the way in which these out-of-core abstractions are done. Rather than reimplementing a lot of the pandas API, dask creates wrappers that split certain operations into aggregates that work on small chunks of the original data. For instance, the arithmetic mean ($$\frac{1}{N}\sum_{i=1}^N a_i$$) can easily be parallelised, since the sum(s) as well as the overall count of values can be computed on small samples and then aggregated together. Below is a visualisation of what that looks like for a dask.dataframe with 3 partitions:

While the arithmetic mean is rather trivial to parallelise, other computations are not. The exact way in which some specific computation should be parallelised is encoded into a computational graph, an example of one is above.

# Computational Graphs 101

A computational graph is a data structure that represents operations (functions) and dependencies between operations. Specifically, the graph needs to be directed and acyclic, in other words a DAG. Each node in the graph is a unit of computation that can be executed individually. Formally, computational graphs are a part of $$\lambda$$-calculus (lambda-calculus), which is a form of function abstraction. $$\lambda$$-calculus is a universal model of computation (in the sense of Turing complete), so although many of the examples below look simple, the paradigm they follow is very powerful.

The foundational layer of dask is the computational graph. This encodes the work to be done (executable functions) and the dependencies those pieces of work depend on. Let’s see how a simple dependency graph is encoded in dask. Since dask is pure python, the graph is just encoded as a python dict. The example below is taken from the dask documentation:

def inc(i):
return i+1

return a + b

dsk = {'x': 1,
'y': (inc, 'x'),

Let’s unpack what’s happening in the code above. The graph has three nodes named x, y and z. The node x just contains data, in this case the integer 1. Node y is an executable function inc which takes one parameter. Notice how the syntax dask has adopted for executable functions in the graph resembles a regular python function call with the opening bracket moved left by one token, i.e, inc('x') $$\rightarrow$$ (inc, 'x').

The function inc should probably take an integer, though, not a string. The parameter, or parameters, for an executable function are interpreted by dask. If the parameter is a name that refers back to the graph, like x in this case, dask will de-reference that argument and pass in the output of the named node. In this case, since x just contains the integer 1, it’ll be passed into y when y is called.

The node named z is where interesting things start to happen. It’s also an executable function and takes two parameters: x and y. The x we’re already familiar with; the value of the node named y, on the other hand, is an executable, so dask will call the function and pass the output of that function call as the second parameter to z. This chaining of nodes allows us to encode very complex dependencies.

Let’s get back to machine learning.

# Machine learning pipelines on dask.

Machine learning tends to break down into easily parallelisable tasks. For instance, cross-validation and hyperparameter search are both cases where each individual task can be executed independent of any other task: Computing the cross-validation results of fold n is independent of computing the cross-validation results of any other fold. There’s also a number of machine learning algorithms that are inherently parallel. An obvious example are model ensembles such as random forests where each individual tree is trained on a separate data sample. As a side note, there’s dask-ml (docs) which has parallel implementations of a number of machine learning algorithms.

Let’s look at how cross-validation can easily be parallelised using low-level dask operations. First, we do some imports and set up the data structures

import sklearn

dsk = {}        # create an empty graph
dsk['X'] = X    # add training data
dsk['y'] = y    # add training data
dsk['SVM'] = sklearn.linear_model.SGDClassifier(loss='hinge', penalty='l2', alpha=1e-5, max_iter=25)

def fit_model(mdl: sklearn.base.Estimator,
X: np.ndarray,
y: np.ndarray) -> sklearn.base.Estimator:
"""Take an unfitted model, and fit its weights on data X and y.
"""
return sklearn.base.clone(mdl).fit(X, y)

def predict(mdl: sklearn.base.Estimator,
X: np.ndarray) -> Iterable[bool]:
"""Take a fitted model and produce predictions for the X.
"""
return mdl.predict(X)

def evaluate(y_true: Iterable[bool]],
y_pred: Iterable[bool]) -> str:
"""Evaluate the quality of predictions y_pred against the ground truth y_true.
"""
return sklearn.metrics.classification_report(y_true, y_pred)

Then we create some number of cross-validation folds and add .fit, .predict and .evaluate nodes for every fold into the graph.

kfold = sklearn.model_selection.KFold(random_state=348347)
for i_fold, (trn, tst) in enumerate(kfold.split(X)):
# add the train / test splits of each fold into the graph
dsk[f'X/cv/transform-tst-{i_fold:02d}'] = X[tst]
dsk[f'y/cv/transform-trn-{i_fold:02d}'] = y[trn]
dsk[f'X/cv/transform-trn-{i_fold:02d}'] = X[trn]
dsk[f'y/cv/transform-tst-{i_fold:02d}'] = y[tst]

cv_model_key = f'svm/cv/fit-{i_fold:02d}'               # output is a binary blob (the fitted model)
cv_pred_trn_key = f'svm/cv/transform-trn-{i_fold:02d}'  # output is a numpy array of predictions on the training data
cv_pred_tst_key = f'svm/cv/transform-tst-{i_fold:02d}'  # output is a numpy array of predictions on the test data
cv_report_trn_key = f'svm/cv/report-trn-{i_fold:02d}'   # a textual report of the trained classifier
cv_report_tst_key = f'svm/cv/report-tst-{i_fold:02d}'   # a textual report of the trained classifier

dsk[cv_model_key] = (fit_model, 'SVM', f'X/cv/transform-trn-{i_fold:02d}', f'y/cv/transform-trn-{i_fold:02d}')
dsk[cv_pred_trn_key] = (predict, cv_model_key, f'X/cv/transform-trn-{i_fold:02d}')
dsk[cv_pred_tst_key] = (predict, cv_model_key, f'X/cv/transform-tst-{i_fold:02d}')
dsk[cv_report_trn_key] = (evaluate, cv_pred_trn_key, f'y/cv/transform-trn-{i_fold:02d}')
dsk[cv_report_tst_key] = (evaluate, cv_pred_tst_key, f'y/cv/transform-tst-{i_fold:02d}')

dsk[f'svm/prod'] = (fit_model, 'SVM', 'X', 'y')

This somewhat cumbersome looking code creates a graph that parallelises 3-fold cross-validation of an SVM classifier. The key to how the graph encodes the dependencies is in the last five lines on code inside the for loop. The first node cv_model_key just calls the .fit_model function with an unfitted model and some data. However, notice that the data we pass in comes from the *-trn-* key for that fold, i.e. the training data.

The next two nodes cv_pred_trn_key and cv_pred_tst_key take the trained model from the previous step and produce some predictions from it on the test data. Finally, the cv_report_trn_key and cv_report_tst_key nodes take the predictions from the previous two steps and the ground truth data and evaluate the quality of the predictions.

This might seem like a lot of work for something that sklearn can already parallelise. The key, however, is that this method is much more general than sklearn’s parallel cross-validation operators. The above graph could contain hundreds of nodes and given the compute infrastructure, dask could execute the workload on a distributed cluster.

There’s a few things to note here. Normally, to fit a sklearn model you would call svm.fit(X, y), here we’re calling fit_model(svm, X, y). We’ve stored the base model under the key 'SVM', but we can’t encode in the graph a node 'SVM'.fit because the string 'SVM' does not have a method .fit. This unfortunately requires us to introduce proxies for each callable we wish to execute.

# Execute where ever

With the graph now encoded, we can execute those jobs on a local machine using all the cores the computer has.

from dask.distributed import Client, LocalCluster

cluster = LocalCluster()
client = Client(cluster)

# only retrieve the fitted model from the first fold
svm_00 = client.get(dsk, keys=['svm/cv/fit-00'])

# only retrieve the fitted models from the all the folds
svms = client.get(dsk, keys=['svm/cv/fit-00', 'svm/cv/fit-01', 'svm/cv/fit-02'])

We’re not constrained to the local machine, though. We could just as easily use an existing Kubernetes or yarn cluster.

from dask_kubernetes import KubeCluster

cluster = KubeCluster.from_yaml('worker-spec.yml')
client = Client(cluster)

These additional cluster implementations are external projects usually not developed by the core dask team and the level of support and maturity will vary. I haven’t used dask_kubernetes (yet), but it was developed by the UK MET office.

# A few tricks to ease the pain

By default, dask will hold results that are still required by other tasks in memory. This can become problematic, but luckily there’s an easy workaround. Since all the nodes in the graph are essentially Python callables, we can wrap them into a caching callable that serialises and deserialises and inputs/outputs of tasks and only hand over the correct path of those results to dask.

Wrapping the results in a thin caching layer has another additional benefit: dask will try to move computation to where data resides as the callable functions tend to be faster to move over the network than large data sets. In fact, dask tracks the size of the results and tries to make trade-offs between moving data or moving computation. However, there are cases where moving even large, several hundred megabyte data sets around is not an issue, e.g. in machine learning cases where the model fitting can easily take several hours or even days. Caching the results to disk allows dask to distribute the computational load more evenly as the “result” of each computation is just a path reference. Furthermore, the cached results can be used to great effect when restarting a partially completed computation.

# Some References

### Matti Lyra

Natural Language Processing | Machine Learning | Data Science | Research

Matti has been an active member of the data science community in Berlin, he has led the PyData Berlin committee since 2016. Matti holds a PhD in Natural Language Processing from the University of Sussex. His thesis explored using topical bias in large text corpora to build better ensemble models for text classification. He's interested in language and linguistics, the philosophy of AI, cycling, photography and coffee.

Updated


text:
# Separability of a subring and of a pre-image

Let $A \subseteq B \subseteq C$ be commutative rings.

A known result, which can be found in De Meyer and Ingraham's book, says that separability of $A \subseteq C$ implies separability of $B \subseteq C$.

When separability of $A \subseteq C$ implies separability of $A \subseteq B$? By 'when' I mean what additional condition guarantees that implication.

Another, somewhat similar, question: Let $A \subseteq B$, $A \subseteq C$, $B \twoheadrightarrow C$ be commutative rings.

When separability of $A \subseteq C$ implies separability of $A \subseteq B$?

Thank you very much for any help!

Edit: (1) By separable I mean the definition as in the above mentioned book on page 40 (this page is available in the preview), namely: $A \subseteq B$ is separable if $B$ is a projective $B \otimes_A B$-module. The definition of separability also appears here. (2) There is an answer to a special case of my first question, due to Adjamagbo's transfer theorem. In that special case, 'when'= $B \subseteq C$ is flat, or 'when'= $B$ is normal.

I do not know what definition of separability you are using. $A\subset C$ is separable means every element of $C$ is separable over $A$ in the usual definition. Thus, for any subring $A\subset B\subset C$, clearly $B$ is separable over $A$, since every element of $B$ is an element of $C$ and thus separable.
• If we deal with three fields, then the second question also has a positive answer: By assumption $B \twoheadrightarrow C$ (surjective) and trivially it is injective ($B$ is a field), so $B$ and $C$ are isomorphic, hence separability of $A \subseteq C$ clearly implies separability of $A \subseteq B$. Jul 19, 2017 at 12:49


text:
Physics

# If in the formula $X= 3YZ^2$, the dimensions of $X$ and $Z$ are those of the capacitance and magnetic induction, then the dimensions of Y are :

$M^{-3}L^{-2}T^8A^4$

##### SOLUTION
$\left[ Y \right] =\dfrac { \left[ X \right] }{ { \left[ Z \right] }^{ 2 } } =\dfrac { \left[ M^{ -1 }{ L }^{ -2 }{ T }^{ 4 }{ A }^{ 2 } \right] }{ { \left[ M^{ 1 }{ { T }^{ -2 }A }^{ -1 } \right] }^{ 2 } }$
$\left[ Y \right] =\left[ M^{ -3 }{ L }^{ -2 }{ T }^{ 8 }{ A }^{ 4 } \right]$

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Single Correct Medium Published on 18th 08, 2020
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• D. $L\, , LT$ and $\displaystyle T^{2}$

Asked in: Physics - Units and Measurement

1 Verified Answer | Published on 18th 08, 2020


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The ionization energy of hydrogen like atom is 9 times of hydrogen atom the radiator of 2nd bohr orbit of this atom

Sujit Kumar
111 Points
3 years ago
The ionization energy of an atom is given by
$E=13.6\frac{Z^{2}}{n^{2}}$
Therefore for Hydrogen atom Z=1 (Atomic Number) ; n=1 (Outermost shell is the first shell)
$\rightarrow E=13.6\ e.v$
Therefore for the hydrogen like atom
$E=13.6*9\ e.v$               (* is the multiplication symbol)
Therefore Z (Atomic Number) of the Hydrogen like atom is 3
Radius of Bohr Orbit is given by
$r=0.529\frac{n^{2}}{Z}$

$\rightarrow r=0.529\frac{2^{2}}{3}$
(n=2 [second bohr orbit] Z=3)
$\rightarrow r=0.80533 \AA$
$(\AA \ is\ 10^{-10} meters)$


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# The number of BC and IC

1. Nov 12, 2007

### Glass

So I'm confused on how we know we need x many BC and/or IC. For example, regarding the heat equation, we need 2 BC, and 1 IC (supposedly because there are two spatial derivatives and one time derivative). And similarly, for the Wave equation we need 2 BC and 2 IC. And another thing, why is it that when we look at D'Alembert's solution of the infinite string wave equation, why we only need 2 IC and no IC? Thanks, I'm very confused.

2. Nov 12, 2007

### Ben Niehoff

Each integration supplies an arbitrary constant (or in partial differential equations, an arbitrary function of the other independent variables). Thus, for the most general solution, there are N extra degrees of freedom for each Nth-order derivative (in the heat equation, 2 for space + 1 for time; in the wave equation, 2 + 2, etc.). Therefore, in order to get to an equation that describes a specific physical situation, we need to supply additional conditions to narrow down these extra degrees of freedom.

As for exactly which additional conditions make the most sense, that depends on the physical situation. Often, we are interested in the time evolution of a system, and so it makes sense to give an initial condition for every order in the time derivatives. But it is not strictly necessary that it be so; we could, for example, give final conditions instead, if we were interested in calculating things that way.


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# Help with Game Theory question

• May 17th 2013, 06:53 AM
entrepreneurforum.co.uk
Help with Game Theory question
Okay kind of a basic one here, just going through the Gibbons book question 1.5,

I'm trying to calculate the both payoffs in the monopoly game

I have my profit function
$\pi = q_i [a - (q_i + q_j) - c]$

And my quantities in monopoly are qm - after maximising profits

$q_i * = (a + q_j* - c )/2 \textnormal{and} q_j * = (a + q_i* - c )/2$

since in monopoly

$(q_m / 2) = q_i = q_j = [((a-c)/2) / 2] = [(a-c)/4]$

my profits are (a-c)^2 / 4

which is wrong and should be divided by 8 - not sure where I have gone wrong...
• May 17th 2013, 12:29 PM
Shakarri
Re: Help with Game Theory question
I cant tell where you went wrong because you jumped from the expression for qi and qj to the final profit. Here's the solution.

$q_i=q_j=\frac{a-c}{4}$

$\pi=q_i(a-q_i-q_j-c)$

$\pi=q_i(a-\frac{a-c}{4}-\frac{a-c}{4}-c)$

$\pi=q_i(a-\frac{a-c}{2}-c)$

$\pi=q_i(\frac{a-c}{2})$

$\pi=\frac{a-c}{4}(\frac{a-c}{2})$


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# Is it appropriate to email an eminent researcher in your field?

Sometimes, I feel to contact some of the eminent researchers in my field for any of the following reasons:

1. Appreciating their research publication (recent times). They publish in top conferences, which are usually not hosted in my country or nearby.

2. Requesting comments on some of my research hypothesis

3. Sometimes just because I am a die hard fan of them. For example, probably the only reason I continued with research in Computer Science was due to Don Knuth.

4. Sometimes, to know what they think on some specific research area that has propagated due to there work. (Probably looks like some journalistic work)

5. For knowing how did they tackle the pressures or certain situation during their PhD or research. (Yes, it's vague but these questions come to mind and probably should be answered by oneself or personal interaction, but adding it for the sake of completion)

Since, most of them are located outside my country, I can't visit or phone them. So, how are such emails perceived. Is it appropriate to send such emails, given that they are expected to have very busy schedule and it would probably waste their time?

Though I have mentioned my field as Computer Science, the question should be applicable to all the fields.

• A polite e-mail can't hurt. Whether you get an answer probably depends on many factors. – gerrit Oct 29 '13 at 16:27
• Of course, Knuth is a bad example, since he does not read email. Of course that page does note two resticted purpose email address that will be read, but there is zero chance he would reply to any of the above sent to them. – Kevin Cathcart Oct 29 '13 at 20:13
• You may be interested in this blog post: fourhourworkweek.com/blog/2007/12/10/… – JoelFan Oct 30 '13 at 1:45
• @KevinCathcart yeah, I knew that. Hence saving money and waiting for a good opportunity to meet him in person :) – krammer Oct 30 '13 at 5:53
• Many of the criteria you list don't require a famous correspondent. You can learn also learn a great deal from the people in their area who are experts but non preeminent. – Zach H Oct 30 '13 at 15:08

Here are some of my strayed thoughts.

Think from return of investments, of yours and your idol's

If you ever perceived that your e-mail would be a "waste of their time," then why send it? I feel that most eminent researchers have a trait of "ignore everyone and head for their goal;" getting acknowledgement and acceptance is probably not their primary concern. A specific e-mail describing how their work has inspired your study/project is probably fine, but I wouldn't go so far to expect they would reply and give specific comments on your hypotheses.

From your point of view, instead of using the energy and bandwidth to send the e-mails, there are a lot more you can do:

There are many ways to show your appreciation

First, they would probably like to see their work being formally cited and, more importantly, applied to the field or crossed into other fields. Each idea geminated from their work is an appreciation by itself, and in the mean time you can also enhance your publication and research paradigm. The plus is: if you have done enough of it, the big shot may actually contact you and give comments.

Second, you can help preaching the researcher's ideas and agenda. You can write blogs, answer other people questions, use their works in your journal clubs or lectures, etc. to subtly introduce the researcher's teaching to the public. Better yet, refine the researcher's ideas, and incorporate into yours. Become a spiritual successor with your own unique approach. And let your career be inspired by the researcher.

You can learn from someone without establishing communication

For some more senior researchers, look for their auto-biography, biography, interviews, and documentaries that feature them. I will probably never be able to talk to Itzhak Perlman, but I learned a lot about him through books, websites, documentaries, and musics that he plays. (And actually, he has a Youtube channel as well, but I am suffering from too much fanboy shyness to write any comment.)

For younger researchers, try look for their blogs, Youtube channels, open courses, or even biographies of their mentors. All these may help you become more familiar with them.

Another way is to indirectly know them. Most of these researchers would have a lot of students or proteges, who may be closer to your rank and more likely to communicate with you. You may build a relationship with them, and learn a thing or two about their interaction with their mentor.

Try technology

For their new publications... nowadays most online journals allow leaving comments online. You may try to say a nice thing or two there. If they write a blog, that's even easier. Some researchers maintain a LinkedIn page or a Twitter account, try connect with them and follow them. Hope for the best.

Use other famous people as leverage

If you really want to communicate with them, also try using other organizations. For instance, you can write to some online radio station and suggest an interview topic and some guests, which of course will include your idolized researcher. You can also write to some prominent podcast hosts and give them a couple reason to invite so and so for an interview. Make good use of crowdsourcing, invite your peers and friends to support your petition.

Some heroes/heroines are better left a bit mysterious

This is sad but occasionally painfully true. Some famous people are better left not known at personal level. They could be immensely arrogant, they may not have a nanogram of social skill, they may be a jerk... Unless I have reconstructed a pretty concrete and reliable image about the researcher from different sources, I would probably want to keep them as what they are in my mind, and as an inspiration for my work.

While many of these topics are excellent points to start a 1-1 conversation, I seriously doubt whether you'd get much of a response over email. This is not to say that they wouldn't appreciate your praise, but that there may not be anything to reply to in particular.

I suspect an email will get a better response if you have specific questions about research content: ideally, a question related to something in a recent paper that isn't entirely obvious. For example, in my field it's sometimes the case that someone sketches a proof idea in a conference paper without a full version, and fleshing this out requires some clarification from the authors.

Of course, you can always add in a question from your list as extra cargo. Then it's a little more likely that someone will reply.

More generally, think about it this way: if you never met someone before, and they came up to you and asked you questions that might be construed as personal (especially 5), or that require you to come up with opinions on the fly (2,4), you might hesitate to respond. The same thing, but without the pressure to say something, will happen over email.

In a nutshell: Sending a polite email on an appropriate topic is pretty much always acceptable. However, don’t automatically expect an answer.

The second part is important. The eminent researcher is probably a busy person, and you have no particular moral claim on their time and attention. Don’t make it sound like you feel entitled to their assistance — “Hey Mr. Tao! I need to understand the Green–Tao theorem for my masters thesis. Please can you explain it to me in simple terms? P.S. my thesis is due next Tuesday.” And don’t be disappointed if they don’t reply.

That said, don’t despair either — don’t feel “oh, it’s not worth writing because they’ll never reply to a nobody like me”! Some may well be too busy or dismissive, but many are also genuinely kind people, and very generous with their help. Others (I have a specific rather famous person in mind here) are simply incorrigible curiosity-hounds, and will happily get caught up for hours by an intriguing problem from a random stranger when they’re supposed to be grading final exams. So they may well reply — just don’t presume it.

On the other hand: what’s an appropriate topic?

• Questions about their research can be fine, provided you’ve done your homework. For instance, I had a question on mathoverflow recently which looked like it might have been answered by a particular researcher, but nobody could find the specific paper, and it was possibly unpublished. Writing to ask e.g. “Do you recall which of your papers on $\tau$-categories might contain or imply a result something like […]?” is certainly OK.

• Questions about practical matters that may be their responsibility are good — at worst they can always hand it off to someone else. “I’m applying to your department’s PhD programme, and it wasn’t clear to me from the website if students are expected to have a potential advisor in advance. If so, are you potentially accepting new students at the moment? I am very interested in your work because…”

• Plain fan mail is OK, as long as it’s sincere and you’re not trying to get something for it! Nobody’s going to mind hearing “Please excuse my writing out of the blue; I just wanted to let you know how much I enjoyed your talk at such-and-such conference. As a non-specialist, I really appreciate the effort you put into making it accessible to a wide audience.” However, if this is followed up by “Incidentally, I am applying for a job at such-and-such department. Do you have any insider advice you can share?” then the initial compliments will look a lot less sincere. Don’t do this!

What’s not appropriate:

• Asking for significant favours (unless you have a personal connection of some sort).

• Asking them to explain their research to you because you don’t want to read their papers. (That’s what helpful strangers in internet forums are for!)

• Personal questions (unless you’re a serious journalist writing e.g. a book on drug use among academics; but in that case you should know about how to get in touch with sources appropriately already).

We can be exahustive here.

If they feel like writing that kind of things to strangers then they most likely have a blog that you can follow, that saves a lot of time for them, by broadcasting. If they do, you can write comments in the blog, that could be appropriate and extend posts and discussion, which is good for them (a chance to clarify something), for you (getting answers, yay!) and for anyone that reads the post and then the comments (and doesn't need to ask the same again...).

If they don't feel like writing that kind of things to strangers then most probably they won't reply. In that case you will be wasting their time (that's bad) and your time (that's worse; selfsteem, please :P ). Thus that's probably not very wise.

If you have questions not covered in the aforementioned blogs, you can ask here, to a crowd of anonymous people. There are many people giving great answers (really, I know my profile is the closest one in distance, but you should check somewhere else). I know we are not so famous and important and wise and everything, but we try to do our best and even if we can understand that you may prefer the advise of other people, we still have a little heart and that hurts. :(

:P Personally, I think it is normal to feel admiration for people that basically are, in some aspect or another, what you want to be and struggle to be in the future. It is probably useful and interesting to get some insight from them, and understand their perspective (personal and professional) on many questions, if you are going to idealize someone and have an idol, probably a rock star is a worse option than a researcher, but I'm not sure whether idealizing and idolizing researchers is any good anyway...

This answer may be biased because I tend to underestimate the relevance of mentoring and I consider more important talent and hard work. I'm also to some extent opposite to you, I admire and appreciate some people, but I'm more introverted, so my natural reaction is thinking to myself: "Stay quiet and don't bother them, you despicable maggot!" (This is clearly worse than idolizing them).

As a final advise, keep your extroversion, it's good for you. That means you will have plenty of things to speak about when you go to a conference and chances to do so. You will love conferences, so focus on publishing to go there (you should focus on publishing anyway). If you like visiting places then that's another big plus.

So work hard and remember, we (a bunch of anonymous people) will always be here whenever you need us, remember us when you get to be famous. Also, write a blog.


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## Sunday, May 15, 2011

### What Android tablets could do better than the Ipad

How can Android beat the iPad 2?  Tough.  The iPad install base is large, and the iPad is simply a great product.  But the iPad 2 is not perfect.  In any case, if Android does not a better product, it will never catch on.  So, from the perspective of an adult user (i.e., not for primary use as a gaming machine), what could an Android tablet do better than the iPad 2, which would lead me to trade mine in?

1. Better cameras.  The ipad 2 cameras are awful, even for skype.  I don't mean (just) the resolution.  I mean the lens angle.  It is not wide enough.  You have to hold the Ipad about 4 feet away from your face in order not to look like a moon face.  It also means that inevitable shaking makes the system worse.
2. Better software developers.  For example, tap into better system software through outside ideas.  Allow awesome third-party system software to take control of the iPad--but only if it has been carefully checked and vetted.  (For example, I would buy a tablet that allowed me to set up a "point system" for my kids: if you do educational games for x hours, you get to play any kinds of games for y hours.  No one other than Apple can implement this on the iPad.)  Offer a venue for Android developers.  Oh, and build faith among your developers.  I don't mean be static in terms of always sticking religiously to legacy interfaces.  See, most external Apple developers do not trust Apple, and rightly so.  They understand that Apple may pull the rug out from under them at any moment if Apple finds it in its own interest.  Apple's predatory behavior vis-a-vis its developers is Android's single biggest asset, and Apple's single-biggest weakness.  If Android were just competitive, every developer would prefer putting their stakes with Android, and not with iOS.
3. Better web browsing.  Safari sucks.  I want real tabs.  Background loading, even if I exit it.  (Flash is not half as important as the basic experience.)  Since I am at it, make sure the other base software for adults is better, too.  Offer a better skype and email client than what exists on the iPad.  (Skype, where is the iPad client?  Why am I running an iPhone skype??)  Lure all the magazine and book publishers to go to Android.  Work with Amazon.
4. Offer some better technology.  Take some risks to lock up something unique.  Offer a truly foldable tablet, where the fold is seamless.  Or a flexible tablet.   Or a sun viewable tablet. Wireless charging. (Retina display?  Who cares.  The ipad is plenty readable, even today.  The resolution battle and battery life battles, like the CPU battles of old, are mostly over.  1000x1000 on a 10" tablet is decent.  8 hour battery life is decent.  Yes, you can be better, but this is not what will make or break the next system.)   There are plenty of better tech solutions that have failed.  But there are few solutions without a compelling advantage that were able to overtake a market leader.  Apple is a smart leader.  They learned how important market share is fighting Intel and Microsoft.  Apple now has the iOS software base.  They have the user base.  Business as usual just won't work for Android.  Android just has to become better than the iPad, or it will never catch on.
5. Offer an optional "pen" mode for more accurate drawing with a stylus.  Offer voice recognition software deeply embedded in the system.

Of course, the whole tablet experience has to be right, too.  Be as good when you can be.  The device should be as thin and nice (and crapware-free) as the iPad.

PS: This is also why google TV failed---it has to be simple and integrated.  A TV that has everything seamlessly integrated, without cables, complex menus, etc.  The DVD, Bluray, DVR, etc., all seamless.  One remote control.  My grandmother would have to be able to operate it.

### The Secret To Making Great Movies

Sadly, there is no secret to making great movies.

The "secret" is having a good story.  A story that is interesting.  A story in which it is not obvious what will happen next, yet you can hardly wait to see what will happen next.  A story in which everything makes sense (in the end).  A story which is believable.  A world.

This is why good theater works, even though it is on a small stage in an obviously unbelievable setting.  The story must be engaging.  Think the "Usual Suspects."  Or "The Lives of the Others."  Or "No Country for Old Men."  Or "Snatch."  Or "The Wire."  Or "Downton Abbey."  Or many other movies and series that did not cost an arm and a leg to make.

You don't need a great director, great actors, high production budgets, or special effects.  Yes, these can help.  But a great story will make mediocre "everything else" appear great.  A boring story will make great "everything else" appear mediocre.

Of course, some stories may intrinsically require a great director, great actors, huge production budgets, or great special effects.  For Bladerunner, the feeling of future LA was vital.  For the series Rome, a believable Rome 2000 years ago was vital.  For the liquid metal robot in Terminator II, the special effect was vital.  (But note that Terminator I, which is just as good, was made on a shoestring.)  For Lawrence of Arabia, how could you film this, if not in the desert with hundreds of actors and extras?   For the Godfather, it had to be Martin Scorcese.  And Al Pacino.  And...

But, in the end, nothing other than a good story really matters.

So, why does Hollywood--and, worse, network television--produce so much shit?  It's because Hollywood is not out to make great movies.  It's out to sell movies.  If movies like "Independence Day" and reality TV sells, then this is what will be produced.

Of course, I think that Hollywood is also too short-sighted.  Making a good-story movie is a larger risk than making "The Matrix 5" or "Spiderman 8."  But, a new world with a new story can itself create more spinoffs.  Of course, even a good film (like Rocky 1) will then warp into a bad one (like Rocky 14), but I can live with this.

And, of course, convincing the folks providing the cash is easier said than done...

## Wednesday, January 19, 2011

• In wide use.  Universal.  Free.  Easy to install everywhere.
• Good Infrastructure (emacs support, typesetting stage, pdf output; many users)
• Beautiful output
• The LaTeX companion book
• The only structured text-based word processing system (in wide use?) with good math support
• "Easy typing" oriented---XML is painful
• auctex support in emacs with nice highlighting
• Mature --- no bugs in base.  few if any bugs in packages
• Many, many packages on ctan
• Many helpful souls on comp.text.tex
• Defining simple user macros is easy
•
There simply is no alternative with its feature set.  (There are alternatives for smaller tasks, e.g., WYSIWYG for letters, etc.)

• Based on ancient macro language, which only very few wizards still understand.  When they disappear, TeX will die.  It is already happening slowly.
• Insane syntax (or shall we say non-syntax).  User documents could instead be sanely defined with a grammar, while keeping TeX bowels hidden.  User text difficult to parse: not clear how many arguments each command has (see below).
• Weird catcodes
• Incomprehensible error messages
• No easy to understand end user programming language for complex tasks
• No clear hooks for external programs (e.g., a preprocessor)
• No stdin support
• Poor namespace in macros (no '.', digits, '_' in macro names)
• Strange meta characters.  '%' for comment, '#' for macro expansion.  $for math---why not \m{math} to free up the$ sign for what it is?
• No clear separation of content and markup
• Syntax changes require knowledge of both emacs auctex and tex macro language, neither of which are very easy to learn.  error messages would have to be sane.  usage would have to be wide.
• Painful font installation  (No "drop font here and it will work.")
• No multithreading and multiprocessor support (even for multiple \include{} files)
• No definitive authority.  Knuth has pretty much abandoned it, and TUG does not have the resources (nor believes that it has the right) to abandon old and obsolete features.  It's as if Larry had disappeared and perl remained stuck forever at version 1 or version 2.
Unfortunately, conTeXt is not ready (we spent a year trying to get a complex book to typeset, but ultimately gave up.)  conTeXt is also based on the TeX messy macro language.  thanks to Hans Hagen and his team for pushing the envelope, though.

Small Syntax Gripes

• $should be the dollar symbol. It is common enough. • math should be typeset with  or \m{ math }, instead of '$$' and '$'
• # is an uncommon character.  good.  one or two of these should be the comment character, not %.  % should be percent.  It is a fairly common character.
• I should know what an argument to a macro is by looking at it.  where is the argument to \sqrt3 ?  you can know this only if you know the definition of \sqrt.  what is the argument to \mymacro{1} ?  Is '{1}' an argument or a block following a macro?  Again, you need to find the definition of \mymacro.

What I Need

A sane-syntax markup language designed for easy typing (i.e., not XML) with an auctex emacs plugin, with a syntax check BEFORE the typeset, with an extended namespace for commands (or macros),  with an ability to use most ctan packages..  Separation of markup from formatting.  HTML or pdf generation.  A user community for this.  Books written for it.

Your mileage may vary, of course.  And to be clear, it is easy to point out weaknesses.  It is hard to put something together.  The folks who have made TeX and LaTeX what it is have to be commended.


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# Libre-Office Writer Pasting with source formatting

I try to paste my code in my libre office writer document with source formatting, that means I want the blue lines in my code to stay blue, italic to stay italic etc..

Sometimes this works automatically. Other times, the Document arbitrarily decides that it will convert source formatting to document's format, so everything is pasted as plain text.

I am trying to find a solution but cannot, also my paste special option seems to be grayed out. This shouldn't be this difficult and time-consuming.

[edit: removed redundant tag]

edit retag close merge delete

Sort by » oldest newest most voted

Hi Gravity

There is a bug hidden within all this somewhere, although I'm not certain precisely where it is.

Mine is LO-4.4.3.2 under Debian 6.0.

A quick refresher:-

Copy (Ctrl-C or Ctrl-Insert) copies text (plus more) to the Clipboard.

Paste (Ctrl-V or Shift-Insert) pastes clipboard text at the cursor

The above is true within a text/plain editor but can vary in a GUI editor such as LO-Writer. That is because the copy process often puts much more than just text/plain text into the Clipboard. It can also add a whole menagerie of other MIME artists (such as text/rtf, and so on). Those can be accessed individually by using Paste Special..., though the options then available depend both on how they were first placed within the Clipboard plus the compatibility of LO-Writer with the OS.

From experience, the Writer default is to copy to the clipboard:-

1. LO-Writer special-format (default on paste)
2. text/rtf
3. text/html
4. text/plain

In general this has worked faultlessly for me. With one exception (and it does vary): ...

Paste Special... is a particular bugbear. I've been able to establish that when Copy puts only plain-text in the clipboard then Paste Special... will be greyed-out. That is obviously dependant on the source, but even some source-html will result in text/plain on copy back in writer. I do not know Debian well enough to know the defaults for clipboard copy from html.code element.

more


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# The special subgroups of a finite abelian group of rank two

Let $G=\langle a_{1}\rangle\times\langle a_{2}\rangle$ such that $|a_{i}|=2^{k_{i}}$ and $k_{1}>k_{2}$ and $H$ be a subgroup of $G$ that there exists an automorphism of $G$ such that fix only the elements of $H$. How do we can find all subgroups of $G$ with this property?

Note: Any subgroup $H$ with the above property must contain the element $a_{1}^{2^{k_{1}-1}}$.

• I suppose this is related to your MSE question math.stackexchange.com/questions/851243/… . Is this follow-up purely from interest, or from a similar exercise? Jul 10, 2014 at 7:56
• I think that when $H \le \langle a_1 \rangle$ it is possible if and only if $|H| \ge 2^{k_1-k_2}$. Jul 10, 2014 at 8:31
• @DerekHolt: Yes, very nice. All examples that i investigated, confirm your comment. Jul 10, 2014 at 9:33

As your example below shows, we cannot assume in all cases that $H \le \langle a_1 \rangle$ as I asserted previously, but I'll explain how to do that case anyway.
To avoid all the notation, I will consider two specific cases, one where it is not possible and one where it is possible. That should be enough to enable you to solve it in general. I am going to call the generators $a,b$ rather than $a_1,a_2$.
First suppose that $|a|=32$, $|b|=4$, and $|H|=4$, so $H = \langle a^8 \rangle$. Suppose $\phi \in {\rm Aut}(G)$ with ${\rm Fix}(\phi) = H$. Since $\phi(b^2) \ne b^2$ or $a^{16}$, we must have $\phi(b^2) = a^{16}b^2$. Also, since $\phi(a^8) = a^8$, we must have $\phi(a^4)^2 = a^8$, and so $\phi(a^4) = a^4, a^{20}, a^4b^2$ or $a^{20}b^2$. The first is not possible, since $a^4 \not\in {\rm Fix}(\phi)$. If $\phi(a^4) = a^{20}$, then $\phi(a^4b^2) = a^4b^2$, which is not possible because $a^4b^2 \not\in {\rm Fix}(\phi)$. The other two possibilities are ruled out by the fact that $a^4$ is a $4$-th power, but neither $a^4b^2$ nor $a^{20}b^2$ is a $4$-th power. So there i no such $\phi$ in this case.
Now suppose that $|a|=32$, $|b|=4$, and $|H|=8$, so $H = \langle a^4 \rangle$. Then we can define $\phi$ with ${\rm Fix}(\phi) = H$ by $\phi(a) = a^{25}b$, $\phi(b)=a^8b^3$.
• Thank you. Only one question. for any subgroup $H$ we can we can choose the generators $a_{1}, a_{2}$ of $G$ such that $H=H_{1}\times H_{2}$ with $H_{i}\leq ⟨a_{i}⟩$. For my better understand of this, please explain this special example: Let $G=\langle a\rangle\times\langle b\rangle$ $|a|=16$ , $|b|=2$ and $H=\langle a^4b\rangle$.Then how do can choose the generators $a_{1}$ and $a_{2}$ for $G$ such that $G=\langle a_{1}\rangle\times\langle a_{2}\rangle$ and $H=H_{1}\times H_{2}$ with $H_{i}\leq ⟨a_{i}⟩$. please guide me. Jul 10, 2014 at 15:28


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text:
# Unique Features of Aqueous Solutions

An aqueous solution is one that is occurring in water. What makes water significant is that it can allow for substances to dissolve and/or be dissociated into ions within it.

### Electrolytes

Water is generally the solvent found in aqueous solution, where a solvent is the substance that dissolves the solute. The solute is the substance or compound being dissolved in the solvent. A solute has fewer number of particles than a solvent, where it's particles are in random motion. Interestingly, aqueous solutions with ions conduct electricity to some degree. Pure water, having a very low concentration of ions, cannot conduct electricity. When a solute dissociates in water to form ions, it is called an electrolyte, due to the solution being a good electrical conductor. When no ions are produced, or the ion content is low, the solute is a non-electrolyte. Non-electrolytes do not conduct electricity or conduct it to a very small degree. In an aqueous solution a strong electrolyte is considered to be completely ionized, or dissociated, in water, meaning it is soluble. Strong acids and bases are usually strong electrolytes. A weak electrolyte then is considered to be one that is not completely dissociated, therefore still containing whole compounds and ions in the solution. Weak acids and bases are generally weak electrolytes. In other words, strong electrolytes have a better tendency to supply ions to the aqueous solution than weak electrolytes, and therefore strong electrolytes create an aqueous solution that is a better conductor of electricity.

Things to note:

• Most soluble ionic compounds and few molecular compounds are strong electrolytes.
• Most molecular compounds are weak or non electrolytes.

 Example

Here's an example of MgCl2 in water:

MgCl2 --H2O Mg2+(aq) + 2Cl- (aq)

The ionic compound dissociates completely to form ions in water, therefore, it is a strong electrolyte.

Now let's look at a weak electrolyte:

HC2H3O2(aq) ↔ H+(aq) + C2H3O2-(aq)

The ionic compound, HC2H3O2 in this situation, only partially dissociates, as expressed by the double arrows in the reaction. This means that the reaction is reversible and never goes to completion.

The H+ cation is a proton that interacts with the H2O molecules that it is submerged in. The interaction is called hydration. The actual H+ ion does not exist in the aqueous solution. It is the hydronium ion, H3O+ that interacts with water to create additional species like H5O2+, H9O4+, and H7O3+. H3O+ can simply be described as the hydration of one H+ and one water molecule. For nonelectrolytes, all that needs to be done is write the molecular formula because no reaction or dissociation occurs. One example of a nonelectrolyte is sugar: written as C₆H₁₂O₆ (aq).

### Ion Concentrations

In an aqueous solution the amount of ions of a species is related to the number of moles of that species per concentration of the substance in the aqueous solution. Molarity is the number of moles of a solute (n) divided by the total volume (V) of the solution:

M= n/V

Molarity, or concentration can also be represented by placing the solute within brackets.  []

 Example

Determine the concentration of K+ in 0.238M KNO3in water.

Since there is one mole of potassium in KNO3, multiply the concentration of the species by the number of moles of the atom to obtain [K+] = 0.238 M KNO3 x 1 mol K+= 0.238 M. Since there is also only one mole of nitrite in KNO3, its concentration is also 0.238M.

For more complex situations, the stoichiometric factors from the equation determine the concentration. For instance, if we have the equation:

Al²(SO₄)₃ → 2 Al3+(aq) + 3 SO₄²⁻(aq)

If the concentration of Al²(SO₄)₃ is 0.019M, what is the concentration of Al⁺? Simply multiply [Al²(SO₄)₃] 0.016M x the stoichiometric factor of Al over Al²(SO₄)₃, which is 2/1. The concentration of Al then becomes 0.038 M.

### Precipitation Reactions

Precipitation reactions occur when the resulting product of an aqueous solution is insoluble. This means that there is a solid produced, called the precipitate. The precipitate is a combination of cation and anions forming an ionic bond. Precipitates are used in manufacturing chemicals, so that certain ions can be isolated by forming precipitates with them. We can predict if a precipitate is produced using the solubility rules for common ionic solids:

1. Salts of group 1 cations (with some exceptions for Li+) and the NH4+ cation are soluble.
2. Nitrates, acetates, and perchlorates are soluble.
3. Salts of silver, lead, and mercury(I) are insoluble.
4. Chlorides, bromides, and iodides are soluble.
5. Carbonates, phosphates, sulfides, oxides, and hydroxides are insoluble (sulfides of group 2 cations and hydroxides of Ca2+, Sr2+, and Ba2+ are slightly soluble).
6. Sulfates are soluble except for those of calcium, strontium, and barium.

When the reaction does not produce a precipitate or the product is soluble, there is no reaction.

 Example

Here's an example of a precipitation reaction:

AgNO3(aq) + NaI(aq) → AgI(s) + NaNO3(aq)

This is considered the "whole" reaction, because all of the species involved are recognized. However, in an aqueous solution, the particles look more like this:

Ag+(aq) + NO3-(aq) + Na+(aq) + I-(aq) → AgI(s) + Na+(aq) + NO3-(aq)

This is called the "ionic" form, because all the ions in the solution are shown. If we now remove the ions that are not involved in creating the solid, or the spectator ions, we can generate a net ionic equation:

Ag+(aq) + I-(aq) → AgI(s)

What are the spectator ions in this reaction?

NaOH(aq) + MgCl2(aq) → NaCl(aq) + Mg(OH)2(s)

As you can see, it is critical to include the symbols for each ion to identify what state they're in: either gaseous (g), solid (s), liquid (l) or aqueous (aq). It is also important to remember that these types of ionic equations also must have a balanced charge. You can use stoichiometric coefficients to ensure that both sides of the equation have equal net charges.

### Acid Base Reactions

An acid is a substance that gives off H+ ions in an aqueous solution, while a base gives of OH- ions. Strong acids almost completely dissociate to become H+ ions, and strong bases dissociate to become OH-. There are only a few strong acids and bases. Most are weak, meaning they produce few H+ or OH- ions in aqueous solution. Here is a list of common strong acids and bases:

Strong Acids Strong Bases
HCl LiOH
HBr NaOH
HI KOH
HClO4 RbOH
HNO3 CsOH
H2SO4 Ca(OH)2
Sr(OH)2
Ba(OH)2

In a neutralization reaction an acid and a base are combined to produce water and an aqueous salt. The acid and base neutralize/balance each other to get a result that is the salt.

For example:

HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)

Can you determine which is the acid, base, and salt?

### References

1. General Chemistry: Principles & Modern Applications, Ninth Edition. Petrucci, Harwood, Herring, Madura. Pearson Education, Inc 2007.  Prentice Hall.

### Problems

Predict if a reaction is likely to occur, and the resulting product:

a. HI(aq) + Zn(NO3)2(aq)

b. CuSO4(aq) + Na2CO3(aq)

c. AgNO3(aq) + CuCl2(aq)

Determine the concentration:

[Al3+] in 0.078 M Al2(SO4)3

### Contributors

• Keerith Singh (UCD), Tenaya Natov (UCD), Zachary Mar (UCD), Mandeep Sohal (UCD)

18:32, 1 Oct 2013

## Tags

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### Wikitexts

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text:
Welcome to symthic forums! We would love if you'd register!
You don't have to be expert in bit baking, everyone is more than welcome to join our community.

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## Transports are definitely bugged on a few vanilla maps.

Hey! If this is your first visit on symthic.com, also check out our weapon damage charts.
Currently we have charts for Battlefield 3, Call of Duty: Black Ops 2, Medal of Honor: Warfighter and Call of Duty: Modern Warfare 3

Holy War? No Thanks.

Posts: 2,425

Date of registration
: Jul 25th 2013

Platform: PC

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Tuesday, April 18th 2017, 5:26pm

Ha that is interesting. Does this apply to any other vehicle weapons, too?

Posts: 3

Date of registration
: Feb 26th 2017

Platform: PS4

Location: Gensokyo (AL)

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Wednesday, April 19th 2017, 1:14am

I think some maps the Armored Car is useful (Sinai and Suez on occasions) but other times its wasted for a YOLO run mostly, but now that you mentioned it I think the Armored Car on Amiens or Empires edge (Operations) for the attackers only spawn at the first set of flags, after that set I never see it.

Posts: 3,103

Date of registration
: Apr 26th 2013

Platform: PS4

Location: Arizona, USA

Reputation modifier: 14

Wednesday, April 19th 2017, 6:49am

In a testing session with SomeRandomGuy, I may have stumbled onto a glitch that prevents new transports from spawning in. We were on an empty server on Fao Fortress and I was using the jeep to continually return back to the C flag where we were conducting the experiment (it involved me getting shot repeatedly). After each ride over we'd destroy the jeep to ensure one would spawn in again, however on one occasion we forgot and two jeeps ended up existing at the same time. To save on AT ammunition I put one right on top of the other and blew them both up. The next time I respawned a jeep refused to materialize. At this point I remembered this thread and put two and two together: I think if both/ all types of the same transport (provided there's more than 1) are destroyed before a player spawns at the HQ that this glitch will appear and prevent further spawns. Of course this is just a hunch, but it can probably be replicated if need be.
To Aim Assist or not to Aim Assist, that is the question.
For 'skill cannons,' that is.

Nope, Aim Assist or bust.

the Sebstalder is quiet good since it can 3hit kill at any distanc ,but In my opinion i actually thikn the sweeper is better, its got a really really fast firerate that can beat alll those Noobmaticos, Helregall adn shitguns in close quarters , and its also really accurate out to like l;ong range,. overall great allround gun, jsut my 2$tho Posts: 1,197 Date of registration : Dec 7th 2011 Platform: PC Battlelog: Reputation modifier: 11 Wednesday, April 19th 2017, 8:39pm Speaking of transport bugs, has anyone experienced the bug where the car would sometimes suddenly de-spawn while being driven lately? I haven't had it happen in a fair while and wonder if DICE could have "fixed" it in a way that prevents the cars from respawning.. bob Salt Miner Posts: 3,136 Date of registration : Mar 19th 2014 Platform: Xbox One Location: Canada Battlelog: Reputation modifier: 13 Wednesday, April 19th 2017, 8:52pm ### Quoted from "Spectacle" Speaking of transport bugs, has anyone experienced the bug where the car would sometimes suddenly de-spawn while being driven lately? I haven't had it happen in a fair while and wonder if DICE could have "fixed" it in a way that prevents the cars from respawning.. I haven't personally experienced that in retail, but I remember this was a common issue in the Beta. The cause was that any flag-spawned assets would despawn if that flag was captured by the other team. Posts: 738 Date of registration : Dec 3rd 2014 Platform: PS4 Battlelog: Reputation modifier: 8 Wednesday, April 19th 2017, 9:25pm ### Quoted from "Spectacle" Speaking of transport bugs, has anyone experienced the bug where the car would sometimes suddenly de-spawn while being driven lately? I haven't had it happen in a fair while and wonder if DICE could have "fixed" it in a way that prevents the cars from respawning.. Are you talking about the AC glitch where the car just falls into the ground and you're left standing there looking silly? I've had it happen several times, once recently. It only ever happens to me on Sinai. I've even seen it happen to another player right in front of me. Pretty hilarious. As for the possible reasons mentioned for the glitch, it would somewhat match my experiences if vehicles don't respawn when all of a certain type are destroyed. That would explain why Amiens works well because both teams get 2. The odd thing is that sometimes I can actually get a good run going on Empire's Edge where I get blown up, respawn and it just appears right away. Other times it won't. I have had times where the car on Ballroom or Suez does actually respawn. At first I thought it might have to do with where I was standing, or maybe how many flags our team has (maybe it spawns if a team is losing?). There really doesn't seem to be much consistency though. I'd love to test it but I don't have anyone to test with Posts: 3,103 Date of registration : Apr 26th 2013 Platform: PS4 Location: Arizona, USA Reputation modifier: 14 Thursday, April 20th 2017, 11:05am There's been a new development in determining what glitches out transport spawns. This time SomeRandomGuy and I attempted our experiment in a live scenario because the result of the empty server test was too limited to be conclusive. However, there is a certain radius that a player must be in for it to spawn, that much is confirmed. It began on Ballroom Blitz, and as you may know only one Armored Car spawns on this map. At the beginning of the round a few Blueberries hopped into it and b-lined straight to E where they died within 30 seconds. I was monitoring their journey through the large map at spawn and as soon as it was destroyed it reappeared at the Deployment; nothing wrong here. I jumped into it, drove it into the trenches and proceeded to shower it with explosives. Upon my return to the Deployment it did not spawn even when inside the spawn radius. It wasn't until a different player spawned in that it came back. Interesting. The next map was Empire's Edge which has 2 Jeeps and 1 Armored Car available for transport, my target being the Armored Car. So SomeRandomGuy entered it and drove it off into the water while I was waiting within the spawn range. Lo and behold as soon as it blew up there it was at the Deployment again. Just as I had anticipated. Next we both got in a drove it to the Out of Bounds. I jumped out before dying and sprinted back to the radius looking to see if it would respawn. It didn't. Then SomeRandomGuy respawned only to come across the same result. Then a Blueberry tank spawned in, entered the radius and boom - the Armored Car materialized then and there. We replicated this same result several times afterwards. Transports can, in fact, get their spawns glitched out. Possible Conclusion 1: When a player enters* a transport from the Deployment and dies in it they will be unable to trigger a second spawn of that same exact transport (not the same model or type, but the vehicle that spawns in the same position); another player is required to do so. Possible Conclusion 2: When a player enters* a transport from the Deployment and they destroy it themselves, they will be unable to trigger a second spawn of the same exact transport; another player is required to do so. *This is key. Only those who have previously entered the transport will encounter the glitch. There was a Scout hanging out near the Deployment within the spawn range and the Armored Car continued to respawn while he remained within that radius. As soon as he moved away, the Armored Car remained absent. To Aim Assist or not to Aim Assist, that is the question. For 'skill cannons,' that is. Nope, Aim Assist or bust. ### Prepare your laughbox ### Quoted from "Zer0Cod3x" the Sebstalder is quiet good since it can 3hit kill at any distanc ,but In my opinion i actually thikn the sweeper is better, its got a really really fast firerate that can beat alll those Noobmaticos, Helregall adn shitguns in close quarters , and its also really accurate out to like l;ong range,. overall great allround gun, jsut my 2$ tho

Posts: 1,601

Date of registration
: Jan 12th 2014

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Thursday, April 20th 2017, 12:08pm

While not a transport bug, I've found another vehicle spawn bug.

On Amiens you get a second tank when you take the gimme. If one of those tanks dies, the tank count will NOT go past one unless the gimme flag is flipped. At which point you get another tank.

Haven't seen this on other maps (ie where planes spawn on captured flags).

Holy War? No Thanks.

Posts: 2,425

Date of registration
: Jul 25th 2013

Platform: PC

Battlelog:

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Thursday, April 20th 2017, 12:53pm

These spawn issues can be quite game-breaking for amour, and really annoying in case of transports, which see too little use anyway.

Kudos to everyone for actively looking for this.

Must be quite some fun to watch you guys explode transports, driving them into water, out of bounds etc. All for the sake of science!

Posts: 3,103

Date of registration
: Apr 26th 2013

Platform: PS4

Location: Arizona, USA

Reputation modifier: 14

Thursday, April 20th 2017, 12:59pm

It's funny, that's what SomeRandomGuy and I would shout when we hijacked the transports and sent them to watery graves or blew them to smithereens when Blueberries were involved. "It's in the name of science!" XD
To Aim Assist or not to Aim Assist, that is the question.
For 'skill cannons,' that is.

Nope, Aim Assist or bust.


text:
# Finishing part of DE

Discussion in 'Physics & Math' started by neelakash, May 30, 2007.

1. ### neelakashRegistered Senior Member

Messages:
491
Given DE is D^2(y)+5Dy+4y=cos 2x ........where D=>(dy/dx)

We are to find the general solution.(I have done it already)
Is the particular integral unique?If not,find another PI and show that the general solution is the same for all PI.

I did the first part and got the general solution as

y=C1 exp(-x)+C2 exp(-4x)+(1/10)cos 2x

I cannot understand why the PI would be non-unique?Can anyone please check if I am correct in having that answer?

3. ### temurman of no wordsRegistered Senior Member

Messages:
1,330
Think of the equation as Ly=f where L is a linear operator. To find a general solution you probably solved the homogeneous problem Ly=0 first then found a particular solution by trying it with a probable form. Let us write it formally: let h be a solution of Lh=0, and let p be some particular solutioin of Lp=f. Then you can see that p+h is also a particular solution: L(p+h)=Lp+Lh=f+0=f. A nice way of saying it is that the particular solution is not unique because L has a nontrivial kernel. You are writing the general solution for Ly=0 means you finding the kernel of L, and adding a particular solution to it means you are parallel shifting this kernel. The kernel of L is a linear space so the soluton set of the nonhomogeneous equation is an affine space, and this can be found by shifting the kernel of L by any vector (part.sol.) of this affine space. Perhaps you can see the situation clearly if you think L as a matrix and y,h,p,f etc as vectors.

Last edited: May 31, 2007

5. ### neelakashRegistered Senior Member

Messages:
491
Oh!!!I was wrong to write that.Using method of undetermined co-efficients,I assumed the form of solution: a sin 2x+b cos 2x
and got a=(1/10) and b=0.

Thank you very much for the conceptual help.


text:
## anonymous 5 years ago Let f(x) be a function such that f(1)=1 and for x≥1 f'(1)=i/(x2+f2(x)). Prove that the limit f(x) exists and the limit is less than 1+(π/4)

1. anonymous

Just want to clear up a few things. First, is that supposed to say (since you can't have an i in the derivative) $f'(1)=\frac{1}{x^2+f(x)^2}$ Second, does what limit exist? The limit as x approaches infinity?

2. anonymous