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Definition:Curvature Jump to navigation Jump to search Definition $\kappa = \dfrac {\d \psi} {\d s}$ where: Let $C$ be embedded in a cartesian plane. $\kappa = \dfrac {y''} {\paren {1 + y'^2}^{3/2} }$ where: $y' = \dfrac {\d y} {\d x}$ is the derivative of $y$ with respect to $x$ at $P$ $y'' = \dfrac {\d^2 y} {\d x^2}$ is the second derivative of $y$ with respect to $x$ at $P$. $\begin{cases} x = \map x t \\ y = \map y t \end{cases}$ $\kappa = \dfrac {x' y'' - y' x''} {\tuple {x'^2 + y'^2}^{3/2} }$ where: $x' = \dfrac {\d x} {\d t}$ is the derivative of $x$ with respect to $t$ at $P$ $y' = \dfrac {\d y} {\d t}$ is the derivative of $y$ with respect to $t$ at $P$ $x''$ and $y''$ are the second derivatives of $x$ and $y$ with respect to $t$ at $P$. Also see
Cardinal numbers Cardinality is a measure of the size of a set. Two sets have the same cardinality---they are said to be equinumerous---when there is a one-to-one correspondence between their elements. The cardinality assignment problem is the problem of assigning to each equinumerosity class a cardinal number to represent it. In ZFC, this problem can be solved via the well-ordering principle, which asserts that every set can be well-ordered and therefore admits a bijection with a unique smallest ordinal, an initial ordinal. By this means, in ZFC we are able to assing to every set $X$ a canonical representative of its equinumerosity class, the smallest ordinal bijective with $X$. We therefore adopt the definition that $\kappa$ is a cardinal if it is an initial ordinal, an ordinal that is not equinumerous with any smaller ordinal. Contents Finite and infinite cardinals The set $\omega$ of natural numbers is the smallest inductive set, that is, the smallest set for which $0\in\omega$ and whenever $n\in\omega$ then also $n+1\in\omega$, where $n+1=n\cup\{n\}$ is the successor ordinal of $n$. A set is finite if it is equinumerous with a natural number, and otherwise it is is infinite. In ZFC, the finite sets are the same as the Dedekind finite sets, but in ZF, these concepts may differ. In ZFC, $\aleph$ is a unique order-isomorphism between the ordinals and the cardinal numbers with respect to membership. Countable and uncoutable cardinals A set is countable when it is equinumerous with a subset of $\omega$. This includes all finite sets, including the empty set, and the infinite countable sets are said to be countably infinite. An uncountable set is a set that is not countable. The existence of uncountable sets is a consequence of Cantor's observationt that the set of reals is uncountable. Successor cardinals and limit cardinals Hartog established that for every set $X$, there is a smallest ordinal that does not have an injection into $X$, and this ordinal is now known as the Hartog number of $X$. When $\kappa$ is a cardinal, then the successor cardinal of $\kappa$, denoted $\kappa^+$, is the Hartog number of $\kappa$, the smallest ordinal of strictly larger cardinality than $\kappa$. The existence of successor cardinals can be proved in ZF without the axiom of choice. Iteratively taking the successor cardinal leads to the aleph hierarchy. Although ZF proves the existence of successor cardinals for every cardinal, ZF also proves that there exists some cardinals which are not the successor of any cardinal. These cardinals are known as limit cardinals. Cardinals which are not limit cardinals are known as successor cardinals. The limit cardinals are precisely those which are limit points in the topology of cardinals (hence the name). That is, for any cardinal $\lambda<\kappa$, there is some $\nu>\lambda$ with $\nu<\kappa$. The limit cardinals share an incredible affinity towards the singular cardinals; there does not exist a weakly inaccessible cardinal if and only if the singular cardinals are precisely the limit cardinals. If inaccessibility is inconsistent (which is thought "untrue" by most set theorists, although possible), then ZFC actually proves that any cardinal is singular if and only if it is a limit cardinal. Regular and singular cardinals A cardinal $\kappa$ is regular when $\kappa$ not the union of fewer than $\kappa$ many sets of size each less than $\kappa$. Otherwise, when $\kappa$ is the union of fewer than $\kappa$ many sets of size less than $\kappa$, then $\kappa$ is said to be singular. The axiom of choice implies that every successor cardinal $\kappa^+$ is regular, but it is known to be consistent with ZF that successor cardinals may be singular. The cofinality of an infinite cardinal $\kappa$, denoted $\text{cof}(\kappa)$, is the smallest size family of sets, each smaller than $\kappa$, whose union is all of $\kappa$. Thus, $\kappa$ is regular if and only if $\text{cof}(\kappa)=\kappa$, and singular if and only if $\text{cof}(\kappa)\lt\kappa$. Cardinals in ZF See general cardinal for an account of the cardinality concept arising without the axiom of choice. When the axiom of choice is not available, the concept of cardinality is somewhat more subtle, and there is in general no fully satisfactory solution of the cardinal assignment problem. Rather, in ZF one works directly with the equinumerosity relation. In ZF, the axiom of choice is equivalent to the assertion that the cardinals are linearly ordered. This is because for every set $X$, there is a smallest ordinal $\alpha$ that does not inject into $X$, the Hartog number of $X$, and conversely, if $X$ injects into $\alpha$, then $X$ would be well-orderable. Dedekind finite sets The Dedekind finite sets are those not equinumerous with any proper subset. Although in ZFC this is an equivalent characterization of the finite sets, in ZF the two concepts of finite differ: every finite set is Dedekind finite, but it is consistent with ZF that there are infinite Dedekind finite sets. An amorphous set is an infinite set, all of whose subsets are either finite or co-finite.
This post has been cross-posted on the Quansight LabsBlog. As of November, 2018, I have been working at Quansight. Quansight is a new startup founded by the same people who started Anaconda, which aims to connect companies and open source communities, and offers consulting, training, support and mentoring services. I work under the heading of Quansight Labs. Quansight Labs is a public-benefit division of Quansight. It provides a home for a "PyData Core Team" which consists of developers, community managers, designers, and documentation writers who build open-source technology and grow open-source communities around all aspects of the AI and Data Science workflow. My work at Quansight is split between doing open source consulting for various companies, and working on SymPy. SymPy, for those who do not know, is a symbolic mathematics library written in pure Python. I am the lead maintainer of SymPy. In this post, I will detail some of the open source work that I have done recently, both as part of my open source consulting, and as part of my work on SymPy for Quansight Labs. Bounds Checking in Numba As part of work on a client project, I have been working on contributing codeto the numba project. Numba is a just-in-timecompiler for Python. It lets you write native Python code and with the use ofa simple @jit decorator, the code will be automatically sped up using LLVM.This can result in code that is up to 1000x faster in some cases: In [1]: import numbaIn [2]: import numpyIn [3]: def test(x): ...: A = 0 ...: for i in range(len(x)): ...: A += ix[i] ...: return A ...:In [4]: @numba.njit ...: def test_jit(x): ...: A = 0 ...: for i in range(len(x)): ...: A += ix[i] ...: return A ...:In [5]: x = numpy.arange(1000)In [6]: %timeit test(x)249 µs ± 5.77 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)In [7]: %timeit test_jit(x)336 ns ± 0.638 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)In [8]: 249/.336Out[8]: 741.0714285714286 Numba only works for a subset of Python code, and primarily targets code that uses NumPy arrays. Numba, with the help of LLVM, achieves this level of performance through manyoptimizations. One thing that it does to improve performance is to remove allbounds checking from array indexing. This means that if an array index is outof bounds, instead of receiving an IndexError, you will get garbage, orpossibly a segmentation fault. >>> import numpy as np>>> from numba import njit>>> def outtabounds(x):... A = 0... for i in range(1000):... A += x[i]... return A>>> x = np.arange(100)>>> outtabounds(x) # pure Python/NumPy behaviorTraceback (most recent call last): File "<stdin>", line 1, in <module> File "<stdin>", line 4, in outtaboundsIndexError: index 100 is out of bounds for axis 0 with size 100>>> njit(outtabounds)(x) # the default numba behavior-8557904790533229732 In numba pull request #4432, I amworking on adding a flag to @njit that will enable bounds checks for arrayindexing. This will remain disabled by default for performance purposes. Butyou will be able to enable it by passing boundscheck=True to @njit, or bysetting the NUMBA_BOUNDSCHECK=1 environment variable. This will make iteasier to detect out of bounds issues like the one above. It will work like >>> @njit(boundscheck=True)... def outtabounds(x):... A = 0... for i in range(1000):... A += x[i]... return A>>> x = np.arange(100)>>> outtabounds(x) # numba behavior in my pull request #4432Traceback (most recent call last): File "<stdin>", line 1, in <module>IndexError: index is out of bounds The pull request is still in progress, and many things such as the quality of the error message reporting will need to be improved. This should make debugging issues easier for people who write numba code once it is merged. removestar removestar is a new tool I wrote toautomatically replace import * in Python modules with explicit imports. For those who don't know, Python's import statement supports so-called"wildcard" or "star" imports, like from sympy import * This will import every public name from the sympy module into the currentnamespace. This is often useful because it saves on typing every name that isused in the import line. This is especially useful when working interactively,where you just want to import every name and minimize typing. However, doing from module import * is generally frowned upon in Python. It isconsidered acceptable when working interactively at a python prompt, or in __init__.py files (removestar skips __init__.py files by default). Some reasons why import * is bad: It hides which names are actually imported. It is difficult both for human readers and static analyzers such aspyflakes to tell where a given name comes from when import is used. For example, pyflakes cannot detect unused names (for instance, from typos) in the presence of import . If there are multiple import statements, it may not be clear which names come from which module. In some cases, both modules may have a given name, but only the second import will end up being used. This can break people's intuition that the order of imports in a Python file generally does not matter. import often imports more names than you would expect. Unless the module you import defines __all__or carefully dels unused names at the module level, import will import every public (doesn't start with an underscore) name defined in the module file. This can often include things like standard library imports or loop variables defined at the top-level of the file. For imports from modules (from __init__.py), from module import will include every submodule defined in that module. Using __all__in modules and __init__.pyfiles is also good practice, as these things are also often confusing even for interactive use where import is acceptable. In Python 3, import is syntactically not allowed inside of a function definition. Here are some official Python references stating not to use import * infiles: In general, don’t use from modulename import . Doing so clutters the importer’s namespace, and makes it much harder for linters to detect undefined names. PEP 8 (the official Python style guide): Wildcard imports ( from <module> import ) should be avoided, as they make it unclear which names are present in the namespace, confusing both readers and many automated tools. Unfortunately, if you come across a file in the wild that uses import , itcan be hard to fix it, because you need to find every name in the file that isimported from the and manually add an import for it. Removestar makes thiseasy by finding which names come from * imports and replacing the importlines in the file automatically. As an example, suppose you have a module mymod like mymod/ \| __init__.py \| a.py \| b.py with # mymod/a.pyfrom .b import def func(x): return x + y and # mymod/b.pyx = 1y = 2 Then removestar works like: $ removestar -i mymod/$ cat mymod/a.py# mymod/a.pyfrom .b import ydef func(x): return x + y The -i flag causes it to edit a.py in-place. Without it, it would justprint a diff to the terminal. For implicit star imports and explicit star imports from the same module, removestar works statically, making use ofpyflakes. This means none of the code isactually executed. For external imports, it is not possible to work staticallyas external imports may include C extension modules, so in that case, itimports the names dynamically. removestar can be installed with pip or conda: pip install removestar or if you use conda conda install -c conda-forge removestar sphinx-math-dollar In SymPy, we make heavy use of LaTeX math in our documentation. For example, in our special functions documentation, most special functions are defined using a LaTeX formula, like However, the source for this math in the docstring of the function uses RST syntax: class besselj(BesselBase): """ Bessel function of the first kind. The Bessel `J` function of order `\nu` is defined to be the function satisfying Bessel's differential equation .. math :: z^2 \frac{\mathrm{d}^2 w}{\mathrm{d}z^2} + z \frac{\mathrm{d}w}{\mathrm{d}z} + (z^2 - \nu^2) w = 0, with Laurent expansion .. math :: J_\nu(z) = z^\nu \left(\frac{1}{\Gamma(\nu + 1) 2^\nu} + O(z^2) \right), if :math:`\nu` is not a negative integer. If :math:`\nu=-n \in \mathbb{Z}_{<0}` is* a negative integer, then the definition is .. math :: J_{-n}(z) = (-1)^n J_n(z). Furthermore, in SymPy's documentation we have configured it so that textbetween `single backticks` is rendered as math. This was originally done forconvenience, as the alternative way is to write :math:`\nu` everytime you want to use inline math. But this has lead to many people beingconfused, as they are used to Markdown where `single backticks` produce code. A better way to write this would be if we could delimit math with dollarsigns, like $\nu$. This is how things are done in LaTeX documents, as wellas in things like the Jupyter notebook. With the new sphinx-math-dollarSphinx extension, this is now possible. Writing $\nu$ produces $\nu$, andthe above docstring can now be written as class besselj(BesselBase): """ Bessel function of the first kind. The Bessel $J$ function of order $\nu$ is defined to be the function satisfying Bessel's differential equation .. math :: z^2 \frac{\mathrm{d}^2 w}{\mathrm{d}z^2} + z \frac{\mathrm{d}w}{\mathrm{d}z} + (z^2 - \nu^2) w = 0, with Laurent expansion .. math :: J_\nu(z) = z^\nu \left(\frac{1}{\Gamma(\nu + 1) 2^\nu} + O(z^2) \right), if $\nu$ is not a negative integer. If $\nu=-n \in \mathbb{Z}_{<0}$ is a negative integer, then the definition is .. math :: J_{-n}(z) = (-1)^n J_n(z). We also plan to add support for $$double dollars$$ for display math so that .. math :: is no longer needed either . For end users, the documentation on docs.sympy.org will continue to render exactly the same, but for developers, it is much easier to read and write. This extension can be easily used in any Sphinx project. Simply install it with pip or conda: pip install sphinx-math-dollar or conda install -c conda-forge sphinx-math-dollar Then enable it in your conf.py: extensions = ['sphinx_math_dollar', 'sphinx.ext.mathjax'] Google Season of Docs The above work on sphinx-math-dollar is part of work I have been doing to improve the tooling around SymPy's documentation. This has been to assist our technical writer Lauren Glattly, who is working with SymPy for the next three months as part of the new Google Season of Docs program. Lauren's project is to improve the consistency of our docstrings in SymPy. She has already identified many key ways our docstring documentation can be improved, and is currently working on a style guide for writing docstrings. Some of the issues that Lauren has identified require improved tooling around the way the HTML documentation is built to fix. So some other SymPy developers and I have been working on improving this, so that she can focus on the technical writing aspects of our documentation. Lauren has created a draft style guide for documentation at https://github.com/sympy/sympy/wiki/SymPy-Documentation-Style-Guide. Please take a moment to look at it and if you have any feedback on it, comment below or write to the SymPy mailing list.
Modeling Linear Elastic Materials – How Difficult Can It Be? The most fundamental material model for structural mechanics analysis is the linear elastic model. Trivial as it may sound, there are some important details that may not be obvious at first glance. In this blog post, we will dive deeper into the theory and application of this material model and give an overview of isotropy and anisotropy, allowable values for material data, incompressibility, and interaction with geometric nonlinearity. Isotropic Linear Elasticity In the vast majority of simulations involving linear elastic materials, we are dealing with an isotropic material that does not have any directional sensitivity. To describe such a material, only two independent material parameters are required. There are many possible ways to select these parameters, but some of them are more popular than others. Young’s Modulus, Shear Modulus, and Poisson’s Ratio Young’s modulus, shear modulus, and Poisson’s ratio are the parameters most commonly found in tables of material data. They are not independent, since the shear modulus, G, can be computed from Young’s modulus, E, and Poisson’s ratio, \nu, as Young’s modulus can be directly measured in a uniaxial tensile test, while the shear modulus can be measured in, for example, a pure torsion test. In the uniaxial test, Poisson’s ratio determines how much the material will shrink (or possibly expand) in the transverse direction. The allowable range is -1 <\nu< 0.5, where positive values indicate that the material shrinks in the thickness direction while being pulled. There are a few materials, called Auxetics, which have a negative Poisson’s ratio. A cork in a wine bottle has a Poisson’s ratio close to zero, so that its diameter is insensitive to whether it is pulled or pushed. For many metals and alloys, \nu \approx1/3, and the shear modulus is then about 40% of Young’s modulus. Given the possible values of \nu, the possible ratios between the shear modulus and Young’s modulus are When \nu approaches 0.5, the material becomes incompressible. Such materials pose specific problems in an analysis, as we will discuss. Bulk Modulus The bulk modulus, K, measures the change in volume for a given uniform pressure. Expressed in E and \nu, it can be written as: When \nu= 1/3, the value of the bulk modulus equals the value of Young’s modulus, but for an incompressible material (\nu \to0.5), K tends to infinity. The bulk modulus is usually specified together with the shear modulus. These two quantities are, in a sense, the most physically independent choices of parameters. The volume change is only controlled by the bulk modulus and the distortion is only controlled by the shear modulus. Lamé Constants The Lamé constants \mu and \lambda are mostly seen in more mathematical treatises of elasticity. The full 3D constitutive relation between the stress tensor \boldsymbol \sigma and the strain tensor \boldsymbol \varepsilon can be conveniently written in terms of the Lamé constants: The constant \mu is simply the shear modulus, while \lambda can be written as A full table of conversions between the various elastic parameters can be found here. Incompressibility in Linear Elastic Materials Some materials, like rubber, are almost incompressible. Mathematically, a fully incompressible material differs fundamentally from a compressible material. Since there is no volume change, it is not possible to determine the mean stress from it. The state equation for the mean stress (pressure), p, as function of volume change, \Delta V, as will no longer exist, and must instead be replaced by a constraint stating that Another way of looking at incompressibility is to note that the term (1-2\nu) appears in the denominator of the constitutive equations, so that a division by zero would occur if \nu= 0.5. Is it then a good idea to model an incompressible material approximately by setting \nu= 0.499? It can be done, but in this case, a standard displacement based finite element formulation may give undesirable results. This is caused by a phenomenon called locking. Effects include: Overly stiff models. Checkerboard stress patterns. Errors or warnings from the equation solver because of ill-conditioning. The remedy is to use a mixed formulation where the pressure or volumetric strain is introduced as an extra degree of freedom. In COMSOL Multiphysics, you enable a mixed formulation by selecting one of the methods available in the Use mixed formulation checkbox in the settings for the material model Part of the settings for a linear elastic material with mixed formulation enabled. When Poisson’s ratio is larger than about 0.45, or equivalently, the bulk modulus is more than one order of magnitude larger than the shear modulus, it is advisable to use a mixed formulation. An example of the effect is shown in the figure below. Stress distribution in a simple plane strain model, \nu = 0.499. The top image shows a standard displacement based formulation, while the bottom image shows a mixed formulation. In the solution with only displacement degrees of freedom, the stress pattern shows distortions at the left end where there is a constraint. These distortions are almost completely removed by using a mixed formulation. Orthotropy and Anisotropy In general cases of linear elastic materials, the material properties have a directional sensitivity. The most general case is called anisotropic, which means all six stress components can depend on all six strain components. This requires 21 material parameters. Clearly, it is a demanding task to obtain all of this data. If the stress, \boldsymbol \sigma, and strain, \boldsymbol \varepsilon, are treated as vectors, they are related by the constitutive 6-by-6 symmetric matrix \mathbf D through Fortunately, it is common that nonisotropic materials exhibit certain symmetries. In an orthotropic material, there are three orthogonal directions in which the shear action is decoupled from the axial action. That is, when the material is stretched along one of these principal directions, it will only contract in the two orthogonal directions, but not be sheared. A full description of an orthotropic material requires nine independent material parameters. The constitutive relation of an orthotropic material is easier when written on compliance form, \boldsymbol \varepsilon= \mathbf C \boldsymbol \sigma: \begin{bmatrix} \tfrac{1}{E_{\rm X}} & -\tfrac{\nu_{\rm YX}}{E_{\rm Y}} & -\tfrac{\nu_{\rm ZX}}{E_{\rm Z}} & 0 & 0 & 0 \\ -\tfrac{\nu_{\rm XY}}{E_{\rm X}} & \tfrac{1}{E_{\rm Y}} & -\tfrac{\nu_{\rm ZY}}{E_{\rm Z}} & 0 & 0 & 0 \\ -\tfrac{\nu_{\rm XZ}}{E_{\rm X}} & -\tfrac{\nu_{\rm YZ}}{E_{\rm Y}} & \tfrac{1}{E_{\rm Z}} & 0 & 0 & 0 \\ 0 & 0 & 0 & \tfrac{1}{G_{\rm YZ}} & 0 & 0 \\ 0 & 0 & 0 & 0 & \tfrac{1}{G_{\rm ZX}} & 0 \\ 0 & 0 & 0 & 0 & 0 & \tfrac{1}{G_{\rm XY}} \\ \end{bmatrix} Since the compliance matrix must be symmetric, the twelve constants used are reduced to nine through three symmetry relations of the type Note that \nu_{\rm YX} \neq \nu_{\rm XY}, so when dealing with orthotropic data, it is important to make sure that the intended Poisson’s ratio values are used. The notation may not be the same in all sources. Anisotropy and orthotropy commonly occur in inhomogeneous materials. Often, the properties are not measured, but computed using a homogenization process upscaling from microscopic to macroscopic scale. A discussion about such homogenization — in quite another context – can be found in this blog post. For nonisotropic materials, there are limitations to the possible values of the material parameters similar to those described for isotropic materials. It is difficult to immediately see these limitations, but there are two things to look out for: The constitutive matrix \mathbf D must be positive definite. For a general anisotropic material, the only option is to check if all of its eigenvalues are positive. For an orthotropic material, this is true if all six elastic moduli are positive and \nu_{\rm XY}\nu_{\rm YX}+\nu_{\rm YZ}\nu_{\rm ZY}+\nu_{\rm ZX}\nu_{\rm XZ}+\nu_{\rm YX}\nu_{\rm ZY}\nu_{\rm XZ}<1 If the material has low compressibility, a mixed formulation must be used. It is possible to make an estimate of an effective bulk modulus and the values of the shear moduli. In cases of uncertainty, it is better to take the extra cost of the mixed formulation to avoid possible inaccuracies. Geometric Nonlinearity When working with geometrically nonlinear problems, the meaning of “linear elasticity” is really a matter of convention. The issue here is that there are several possible representations of stresses and strains. For a discussion about different stress and strain measures, see this previous blog post. Since the primary stress and strain quantities in COMSOL Multiphysics are Second Piola-Kirchhoff stress and Green-Lagrange strain, the natural interpretation of linear elasticity is that these quantities are linearly related to each other. Such a material is sometimes called a St. Venant material. Intuitively, one could expect that “linear elasticity” means that there is a linear relation between force and displacement in a simple tensile test. This will not be the case, since both stresses and strains depend on the deformation. To see this, consider a bar with a square cross section. The bar subjected to uniform extension. The original length of the bar is L_0 and the original cross-section area is A_0=a_0^2, where a_0 is the original edge of the cross section. Assume that the bar is extended at a distance \Delta so that the current length is L=L_0+\Delta=L_0(1+\xi). Here, 1+\xi is the axial stretch and \xi can be interpreted as the engineering strain. The new length of the edge of the cross section is a=a_0+d=a_0(1+\eta), where \eta is the engineering strain in the transverse directions. The force can be expressed as the Cauchy stress \sigma_x in the axial direction multiplied by the current cross-section area: To use the linear elastic relation, the Cauchy stress \boldsymbol \sigma must be expressed as the Second Piola-Kirchoff stress \mathbf S. The transformation rule is where \mathbf F is the deformation gradient tensor, and the volume scale is defined as J = det(\mathbf F). Without going into details, for a uniaxial case Since for a St. Venant material in uniaxial extension, the axial stress is related to the axial strain as S_X = E \epsilon_X, we obtain Given that the axial term of the Green-Lagrange strain tensor is defined as the force versus displacement relation is then The linear elastic material furbished with geometric nonlinearity actually implies a cubic relation between force and engineering strain (or force versus displacement, since \Delta =L_0\xi), as shown in the figure below. The uniaxial response of a linear elastic material under geometric nonlinearity. As can be seen in the graph, the stiffness of the material approaches zero at the compression side, \xi = \sqrt{{1}/{3}}-1 \approx -0.42. In practice, this means that the simulation will fail at that strain level. It can be argued that there are no real materials that are linear at large strains, so this should not cause problems in practice. However, linear elastic materials are often used far outside the range of reasonable stresses for several reasons, such as: Often, you may want to do a quick “order of magnitude” check before introducing more sophisticated material models. There are singularities in the model that cause very high strains in a point. Read more about singularities here. In contact problems, the study is always geometrically nonlinear. Often, high compressive strains appear locally in the contact zone at some time during the analysis. In all of these cases, the solver may fail to find a solution if the compressive strains are large. If you suspect this to be the case, it is a good idea to plot the smallest principal strain. If it is smaller than -0.3 or so, we can expect this kind of breakdown. The critical value in terms of the Green-Lagrange strain is found to be -1/3. When this becomes a problem, you should consider changing to a suitable hyperelastic material model. Compression may not be the only problem. In the analysis above, Poisson’s ratio did not enter the equations. So what happens with the cross section? By definition in the uniaxial case, the transverse strain is related to the axial strain by When these strains are Green-Lagrange strains, this is a nonlinear relation stating that Thus, there is a strong nonlinearity in the change of the cross section. Solving this quadratic equation gives the following relation between the engineering strains The result is shown in the figure below. Transverse displacement as a function of the axial displacement for uniaxial tension of a St. Venant material. Five different values of Poisson’s ratio are shown. As you can see, the cross section collapses quickly at large extensions for higher values of Poisson’s ratio. If another choice of stress and strain representation had been made — for example, if the Cauchy stress were proportional to the logarithmic, or “true” strain — it would have resulted in quite a different response. Instead, such a material has a stiffness that decreases with elongation, where the force-displacement response does depend on the value of Poisson’s ratio. Still, both materials can correctly be called “linear elastic”, although the results computed with large strain elasticity can differ widely between two different simulation platforms. Concluding Remarks on Linear Elastic Materials We have illustrated some limits for the use of linear elastic materials. In particular, the possible pitfalls related to incompressibility and to the combination of linear elasticity with large strains have been highlighted. If you are interested in reading more about material modeling in structural mechanics problems, check out these blog posts: Introducing Nonlinear Elastic Materials Obtaining Material Data for Structural Mechanics from Measurements Part 2: Obtaining Material Data for Structural Mechanics from Measurements Fitting Measured Data to Different Hyperelastic Material Models Yield Surfaces and Plastic Flow Rules in Geomechanics Computing Stiffness of Linear Elastic Structures: Part 1 Computing Stiffness of Linear Elastic Structures: Part 2 Editors note: From version 5.4, the settings for mixed formulations look different, since more options were added. Comments (4) CATEGORIES Chemical COMSOL Now Electrical Fluid General Interfacing Mechanical Today in Science
This question already has an answer here: Are there any solutions to $2^n-3^m=1$? 4 answers This question arose as an attempt to answer the following question Relaxed Collatz 3x+1 conjecture. I wanted to show that there is a solution of the equation $2^{k}=3^{z}(2n+1)-1$ for each $n\geq 2$, where $k,z,n\in\mathbb{N}$. But even a special case has put me in a dead end. Are there infinitely many solutions of the equation $2^k=3^z-1$, when $z\rightarrow \infty$? First solution: $2^3 =3^2 -1$.
If we idealize the scenario enough, this is a simple exercise in differential equations, so let's get to work. First, we know that it's initial speed is $150 \text{ m/s}$, but that is by no means its final speed - obviously, the bb slows down as it travels through air! Let's suppose that the moment the bb exits the barrel, it is no longer being pushed (as Steevan pointed out). So, the only force acting on it is air resistance. So the question is, why does the bb slow down significantly with distance traveled - we can determine this exactly, assuming the model is correct. Now, the model you are using (apparently) for air resistance is given as $$F_d = \frac{1}{2} pv^2C_DA.$$ We want to see how the velocity changes as a function of distance! But we know Newton's second law, so we can write that $$F = m \frac{dv}{dt} = m \frac{dv}{dx} \frac{dx}{dt} = m v' v$$ where $v$ is now a function of distance (this uses the chain rule - hope you're comfortable with that!). Now, we can write our differential equation: $$mv'v = -\frac{1}{2} pv^2 C_DA.$$ Note - there is a negative sign there because the force opposes the direction of motion. That is, the force points backwards, and the particle has a positive (forward) velocity. Simplifying, we get $$v' = -\frac{1}{2m} pC_DAv.$$ Now this is a simple differential equation to solve: we separate variables, i.e. $\frac{v'}{v} = -\frac{1}{2m}pC_DA,$ and then doing some more chain rule magic, we end up with $$\frac{dv}{v} = -\frac{1}{2m}pC_DA \, dx.$$ Now we can integrate both sides and find our solution: $$\int_{v(0)}^{v(x)} \frac{dv}{v} = -\frac{1}{2m} pC_DA \int_0^x dx,$$or$$v(x) = v(0)\exp{\left(-\frac{1}{2m} pC_DA x\right)}.$$Finally, we can plug in the initial condition, that at $x=0$, the speed is $150 \text{ m/s}$: $$v(x) = (150 \text{ m/s}) \exp{-\left(\frac{1}{2m} pC_DA x\right)}.$$ Finally, for a numerical answer, you may want to plug in your known constants. Unfortunately, for this you need to know the mass of the bb! For the sake of argument, let's assume a mass of $0.12 \text{ g}$, the most common mass for airsoft bbs, according to Wiki - Airsoft Pellets. So, we can now calculate the speed of the bb as it travels, knowing that $\frac{1}{2} pC_D A = 0.00817 \text{ g/m}$! So now we have a function for velocity: $$v(x) = (150 \text{ m/s}) \exp{(-0.0681x)}.$$ For example, to find the distance at which the speed drops by half, we would solve $$75 \text{ m/s} = (150 \text{ m/s}) \exp{(-0.0681x)},$$ which yields a distance of approximately 10 meters. Now you see why the bb slows down significantly with distance - it's exponential decay, which tends to decrease the quantity a large amount at first, with the amount of decrease decreasing over time (or in this case, distance).
I am following Chapter 8 ("Heteroskedasticity" p. 259) in the 6th edition of Woolridge Introductory Econometrics: A Modern Approach and I don't understand one piece of the transformation of our model. For fGLS, we assume [1] $Var(u\|\boldsymbol{x}) = \sigma^2exp(\delta_0 +\delta_1x_1+\delta_2x_2+...+\delta_kx_k)$, where $x_i, x_2,...,x_k$ are the independent variables appearing in the regression model and the $\delta_j$ are unknown parameters. Then, we use the definition of conditional variance to say $Var(u\|\boldsymbol{x}) = E(u^2\|\boldsymbol{x}$), since our zero conditional mean assumption tells us that $(E(u))^2$ is zero. Now, here is where I'm stuck: Wooldridge says that our assumption [1] above allows us to write $u^2=\sigma^2exp(\delta_0 +\delta_1x_1+\delta_2x_2+...+\delta_kx_k)v$, where $v$ has a mean equal to unity, conditional on $\boldsymbol{x}=x_1,x_2, ...,x_k$ Can someone please help me to develop an intuition for this last step that Wooldridge has taken? Essentially, it seems like we've assumed that $E(u^2\|\boldsymbol{x})=\frac{u^2}{v}$ and I don't understand why or the properties that allow us to do this. I've found this paper https://www.econ.uzh.ch/dam/jcr:e3cddc1b-f89d-4fb4-9474-c2c380355d69/joe_2017.pdf to be useful (specifically, assumption #6 on p. 2), but it doesn't leave me with much intuition for what we're doing and why, especially because I'm fairly new to econometrics. Thanks- Maurus
You want to show that $(n+1)f^\lambda=\sum_{\mu\succ \lambda}f^\mu$, in other words that given an element $i\in\{1,2,\ldots,n+1\}$ and a standard tableau of shape $\lambda$ you can build a standard tableau of some shape obtained by adding a single square to the shape, and do so in a bijective (reversible) way. You may renumber the entries of the tableau monotonically so that it contains all numbers of $\{1,2,\ldots,n+1\}\setminus\{i\}$. Then what you are asking for is exactly achieved by Schensted insertion. This is in fact easy to prove directly, see Lemma 1.3.1. of this paper, which gives the following proof by induction on $n=\|\lambda\|$. Since $f^{(0)}=f^{(1)}=1$ the starting case $n=0$ is OK. Now assuming $n>0$ one has $$ (n+1)f^\lambda =f^\lambda+n\sum_{\mu\prec\lambda}f^\mu =f^\lambda + \sum_{\mu\prec\lambda}\sum_{\nu\succ\mu}f^\nu$$by induction (where the first equality harks back to $f^{\lambda} = \sum_{\mu\prec\lambda}f^\mu$, a consequence of the fact that the standard tableaus of shape $\lambda$ are in bijective correspondence with the standard tableaus of all shapes $\mu$ such that $\mu\prec\lambda$; this correspondence is given by cutting off the square containing $n$). In the double summation we distinguish terms $\nu=\lambda$, of which there are $\#\lambda^-$ where $\lambda^-=\{\,\mu\mid\mu\prec\lambda\,\}$, and terms with $\nu\neq\lambda$. The latter $\nu$ are obtained by removing a square and adding a different square to the shape; this can also be done in the opposite order giving first a shape $\kappa\succ\lambda$ and then $\nu\prec\kappa$. Defining $\lambda^+=\{\,\mu\mid\mu\succ\lambda\,\}$ one has $\#\lambda^+=\#\lambda^-+1$, since for every square that we can remove from the shape there is a square that we can add to the shape in the next row, and we can also always add a square to the first row. Now write our expression as$$ (1+\#\lambda^-)f^\lambda +\sum_{\mu\prec\lambda}\sum_{\textstyle{\nu\succ\mu\atop\nu\neq\lambda}} f^\nu =\#\lambda^+f^\lambda +\sum_{\kappa\succ\lambda}\sum_{\textstyle{\nu\prec\kappa\atop\nu\neq\lambda}} f^\nu = \sum_{\kappa\succ\lambda}\sum_{\nu\prec\kappa}f^\nu = \sum_{\kappa\succ\lambda}f^\kappa,$$where the final equality comes from extending a standard Young tableau of shape $\nu$ to one of shape $\kappa$ by adding the entry $n+1$ in the unique new square. This completes the induction step. The above proof uses exactly the propreties than make Schensted insertion tick, namely that Young's lattice is a $1$-differential poset. In fact if one translates this proof, which is entirely based on bijections, into a recursively defined procedure for extending Young tableaux, the resulting procedure is equivalent, under the natural identifications, to Schensted insertion. You can do the same for any differential poset, giving rise to "Schensted" insertion procedures for each of them. Also the structure of the proof can be made evident by instead of a recursive procedure turning it into a (Fomin) growth diagram construction of Schensted insertion, or more precisely of the full Robinson-Schensted correspondence. Again this extends to arbitrary differential posets.
A-Softmax $L = \frac{1}{N} \sum\limits_{i}{-\log(\frac{e^{ \left\| {x_i} \right\| \psi(\theta_{y_i,i})}}{e^{ \left\| {x_i} \right\| \psi(\theta_{y_i,i})} + \sum\limits_{j\|j \ne y_i}{e^{ \left\| {x_j} \right\| \cos(\theta _{j,i})}}})}$ $\psi(\theta_{y_i,i})=(-1)^k\cos(m\theta_{y_i,i})-2k$, $\theta_{y_i,i}\in[\frac{k\pi}{m}, \frac{(k+1)\pi}{m}]$ and $k \in [0, m-1], \ m \in \mathbb{Z}^+$ The plot of $\psi(\theta_{y_i,i})$ In practice, we set $m$ to $4$. Detailed proof of properties of A-Softmax Loss: https://www.cnblogs.com/heguanyou/p/7503025.html#undefined Normalizing the weight could reduce the prior caused by the training data imbalance Suppose we use a neural network to extract a $1D$ feature $f_i$ for each sample $i$ in the dataset and use Softmax to evaluate our network. To make our analysis easier, we normalize our features. Suppose there are only two classes in the dataset. There are $m$ samples in class 1and $n$ samples in class 2. When our network is strong enough, our features are distributed at both ends of the diameter of the unit circle. Without bias terms, the loss function can be written as: $L = -\sum\limits_{i = 1}^{m + n}\sum\limits_{j = 1}^{2}{a_{i,j} \log(p_{i,j})}$, where $p_{i,j}$ means the probability that sample i belongs to class $j$ (generated by the softmax) and $a_{i,j} = [sample \ i \ belongs \ to \ class \ j]$. Assume that $w_i$ means the weights in the softmax layer for class $i$. Then, $\frac{\partial L}{\partial w_1} = (m – 1) \sum\limits_{i\|a_{i,1}=1}{f_i} = m (m – 1)$. And, $\frac{\partial L}{\partial w_2} = (n – 1) \sum\limits_{i\|a_{i,2}=1}{f_i} = n (n – 1)$. If $m = 100$ while $n = 1$, then $\frac{\partial L}{\partial w_1} / \frac{\partial L}{\partial w_2} \approx 10000$. It’s clear to see that the derivative of $w_1$ is much bigger than the derivative of $w_2$. That’s why the larger sample number a class has, the larger the associated norm of weights tends to be. Biases are useless for softmax In this paper, they use an experiment using MNIST to empirically prove that Biases is not necessary for softmax. In practical, Biases do are useless. Understanding to “the prior that faces also lie on a manifold” As descript in NormFace, the feature distribution of softmax is ‘radial’. So after normalization, features lie on a very thick line on a hypersphere. That’s why Euclidean features failed and $cos$ similarity works well. This is also an interpretation of why the Euclidean margin is incompatible with softmax loss_ Other Points Closed-set FR can be well addressed as a classification problem. Open-set FR is a metric learning problem. The key criterion for metric learning in FR: the maxima intra-class distance is smaller than the minima inter-class distance. Separable $\ne$ discriminative and softmax is only separable.
For an i.i.d sequence of Random Variables $X_1, \dots, X_n$, each with mean $\mu = \mathbb E[X]$, the goal is to estimate some continuous function $f$ evaluated at the mean, $f[\mathbb E[X]]$. If there is some unbiased estimator, $L$, so that $\mathbb E L = f[\mathbb E[X]]$, then can this unbiased estimator always be turned into a consistent series of estimators, where $L' = \frac{1}{n}\sum_1^n L_i$ for $n$ independent samplings of $L$. It looks like $L'$ should converge in probability to $f[\mathbb E[X]]$ by the law of large numbers, and therefore be consistent. Is my reasoning sound here? I tried this out with a few examples and it seems to work but I have little experience so I can't tell, even though it is a basic question. Thanks for the help!
Indiana Jones found ancient Aztec catacombs containing a golden idol. The catacombs consists of $$$n$$$ caves. Each pair of caves is connected with a two-way corridor that can be opened or closed. The entrance to the catacombs is in the cave $$$1$$$, the idol and the exit are in the cave $$$n$$$. When Indiana goes from a cave $$$x$$$ to a cave $$$y$$$ using an open corridor, all corridors connected to the cave $$$x$$$ change their state: all open corridors become closed, all closed corridors become open. Indiana wants to go from cave $$$1$$$ to cave $$$n$$$ going through as small number of corridors as possible. Help him find the optimal path, or determine that it is impossible to get out of catacombs. The first line contains two integers $$$n$$$ and $$$m$$$ ($$$2 \leq n \leq 3\cdot 10^5$$$, $$$0 \leq m \leq 3 \cdot 10^5$$$) — the number of caves and the number of open corridors at the initial moment. The next $$$m$$$ lines describe the open corridors. The $$$i$$$-th of these lines contains two integers $$$u_i$$$ and $$$v_i$$$ ($$$1 \leq u_i, v_i \leq n$$$, $$$u_i \neq v_i$$$) — the caves connected by the $$$i$$$-th open corridor. It is guaranteed that each unordered pair of caves is presented at most once. If there is a path to exit, in the first line print a single integer $$$k$$$ — the minimum number of corridors Indians should pass through ($$$1 \leq k \leq 10^6$$$). In the second line print $$$k+1$$$ integers $$$x_0, \ldots, x_k$$$ — the number of caves in the order Indiana should visit them. The sequence $$$x_0, \ldots, x_k$$$ should satisfy the following: If there is no path, print a single integer $$$-1$$$. We can show that if there is a path, there is a path consisting of no more than $$$10^6$$$ corridors. 4 4 1 2 2 3 1 3 3 4 2 1 3 4 4 2 1 2 2 3 4 1 2 3 1 4 Name
Let $f \in F_p[x]$ be an irreducible polynomial of degree $n$, where $p$ is prime. Prove that its roots are independent over $F_p$. EDIT: It was pointed in the answers that this claim is true only for some irreducible polynomials. My question can be changed to: prove that for any $n$, such irreducible polynomial of degree $n$ exists. Remarks: This is false in characteristic zero: take $x^2+1 \in \mathbb{R}[x]$. I think it's true for any finite field (i.e., $p$ can be a prime power) Let $\theta$ be some root of $f$. $F_p[\theta] = F_{p^n}$, and $\{\theta^i\}_{i=0}^{n-1}$ is a basis. By Frobenius Automorphism and by the identity $x^{p^i}-x = \prod_{a \in F_{p^i}} (x-a)$, one can show that $\{\theta^{p^i}\}_{i=0}^{n-1}$ are all the roots of $f$. It also shows that all of $f$'s roots are in the extension - they are powers of $\theta$. Since $F_p[\theta]$ contains all $f$'s roots, the claim is equivalent to: $Span \{\theta^{p^i}\} = Span \{ \theta^i\}$ The motivation is that I saw this fact used in a cool combinatorics proof, where the bijection $$\vec{a} \to \sum a_i \theta^{p^i}$$ from $(F_p)^n$ to $F_{p^n}$ was used, and I didn't understand why it is a bijection. (For the curious: if we restrict this bijection to Lyndon words of length $n$ with alphabet $F_p$, and send $\sum a_i \theta^{p^i}$ to $\text{minimal polynomial of} \sum a_i \theta^{p^i}$, it becomes a bijection of Lyndon words of length $n$ to irreudcible polynomials of degree $n$)
Definition:Transcendental Element of Ring Extension Definition Let $\struct {D, +, \circ}$ be an integral subdomain of $R$. Let $x \in R$. Then $x$ is transcendental over $D$ if and only if: $\displaystyle \forall n \in \Z_{\ge 0}: \sum_{k \mathop = 0}^n a_k \circ x^k = 0_R \implies \forall k: 0 \le k \le n: a_k = 0_R$ Notation For such an $x$ transcendental over $D$, it is conventional to use the letter $X$. Thus a ring of polynomials over $D$ in such a transcendental is therefore usually denoted $D \sqrt X$. Also see If $x \in R$ is not transcendental over $D$ then it is algebraic over $D$.
A little mixup in notation there. $P_n(x)$ is normally denoting the $n^{th}$ Legendre polynomial, and $L_n(x)$ the $n^{th}$ Laguerre polynomial. Both $\{ P_n\| n \in \mathbb{N} \}$ and $\{ L_n\| n \in \mathbb{N} \}$ form sets of orthogonal functions, which means that when taking an inner product of two of its members which are different, then the result is zero. Here the inner product is some integral of the product of the functions. See for example Legendre Polynomial Orthogonality and Size and Legendre polynomials, Laguerre polynomials: Basic concept. Any function on the relevant interval ($[-1,1]$ for Legendre polynomials, $[0,\infty]$ for Laguerre polynomials) can then be expanded into a series with terms $a_nP_n(x)$ or $a_nL_n(x)$ instead of the 'normal' $a_nx^n$. In particular, any polynomial $P(x)$ can be expanded on $[-1,1]$ as $\sum_{j=0}^{\infty} a_jP_j(x)$ with appropriate coefficients $a_j$. See Using Legendre polynomial to approximate any polynomial.
While reading about refractive index 2 terms popped up, group velocity which alway slows down in a medium and phase velocity which may exceed speed of light. Say in a complete vacuum and using laser with only 1 frequency as an example, which of the two kinds of velocity is defined as standard $c$ in physics? The answer to your question is that in a vacuum for light, the two are equivalent. If you solve Maxwell's equations in a vacuum what you will find is that $$ E(x,t) = E_0 \cos(kx - \omega t) $$ likewise for the magnetic field (in magnitude). You will also find that these solve maxwell's equations only when $$\frac{\omega}{k} \equiv v_p = \frac{1}{\sqrt{\mu_0\epsilon_0}} = c $$ That is, in a vacuum $\omega = c k$. From which is follows that $$ \frac{d\omega}{dk} \equiv v_g = c $$ as well. To answer your question directly, what we usually mean by "speed of light" is the group velocity, as this is the one that Einstein's postulate says should be constant (the max speed at which information can be transmitted). So while we could define the number either way I think it's convention to define the speed of light in vacuum to be referring to the group velocity. Having said that, with respect to your question about what is the speed of light in a medium/vacuum, that's not ontologically the right question to ask. That is, it's not that one exists in any more fundamental sense than the other. There are simply really two differentvelocities and we always need to specify which one we are talking about. Even though the question in the text has a simple answer (phase and group velocity in vacuum are the same, due to the linear dispersion), the general question put in the title ( What really is the speed of light in a medium/vacuum, group or phase velocity? ) has a much more complex answer in the case of a medium. The case of anomalous dispersion makes clear that in some cases even the group velocity may become higher than $c$. In such cases, the relevant velocity (the speed of energy and information transfer) is something else. Quite often, people refer to the signal velocity as the appropriate generalization. However one has to notice that, under such extreme dispersion condition, the deformation with time of a wavepacket may be significant and up to nine different velocities have been identified. In the classical Jackson's textbook on electrodynamics you may find the definition of a couple of these additional velocities (the forerunners velocities).
Periodic solutions in a delayed predator-prey models with nonmonotonic functional response 1. School of Mathematic and Statistics, Lanzhou University, Lanzhou, Gansu 730000, China 2. School of Mathematics and Information Sciences, Ludong University, Yantai, Shandong 264025, China $y^{'}(t)=y(t)[ \frac{\mu (t)x(t-\tau )}{m^2+x^2(t-\tau )} -d(t)]. \]$ is established, where $a(t), b(t), \mu (t)$ and $d(t)$ are all positive periodic continuous functions with period $\omega >0$, $m>0$ and $\tau \geq 0 $ are constants. Keywords:time delay, coincidence degree., positive periodic solution, nonmonotonic functional response, Predator-prey model. Mathematics Subject Classification:Primary: 34K15, 34C25 Secondary: 92D2. Citation:Wan-Tong Li, Yong-Hong Fan. Periodic solutions in a delayed predator-prey models with nonmonotonic functional response. Discrete & Continuous Dynamical Systems - B, 2007, 8 (1) : 175-185. doi: 10.3934/dcdsb.2007.8.175 [1] Hongwei Yin, Xiaoyong Xiao, Xiaoqing Wen. Analysis of a Lévy-diffusion Leslie-Gower predator-prey model with nonmonotonic functional response. [2] Eduardo González-Olivares, Betsabé González-Yañez, Jaime Mena-Lorca, José D. Flores. Uniqueness of limit cycles and multiple attractors in a Gause-type predator-prey model with nonmonotonic functional response and Allee effect on prey. [3] Zengji Du, Xiao Chen, Zhaosheng Feng. Multiple positive periodic solutions to a predator-prey model with Leslie-Gower Holling-type II functional response and harvesting terms. [4] Zhijun Liu, Weidong Wang. Persistence and periodic solutions of a nonautonomous predator-prey diffusion with Holling III functional response and continuous delay. [5] Gianni Gilioli, Sara Pasquali, Fabrizio Ruggeri. Nonlinear functional response parameter estimation in a stochastic predator-prey model. [6] Haiying Jing, Zhaoyu Yang. The impact of state feedback control on a predator-prey model with functional response. [7] [8] Jun Zhou, Chan-Gyun Kim, Junping Shi. Positive steady state solutions of a diffusive Leslie-Gower predator-prey model with Holling type II functional response and cross-diffusion. [9] Shanshan Chen, Junping Shi, Junjie Wei. The effect of delay on a diffusive predator-prey system with Holling Type-II predator functional response. [10] Sze-Bi Hsu, Tzy-Wei Hwang, Yang Kuang. Global dynamics of a Predator-Prey model with Hassell-Varley Type functional response. [11] Mostafa Fazly, Mahmoud Hesaaraki. Periodic solutions for a semi-ratio-dependent predator-prey dynamical system with a class of functional responses on time scales. [12] H. W. Broer, K. Saleh, V. Naudot, R. Roussarie. Dynamics of a predator-prey model with non-monotonic response function. [13] Xiaoling Li, Guangping Hu, Zhaosheng Feng, Dongliang Li. A periodic and diffusive predator-prey model with disease in the prey. [14] Rui Xu, M.A.J. Chaplain, F.A. Davidson. Periodic solutions of a discrete nonautonomous Lotka-Volterra predator-prey model with time delays. [15] Haiyin Li, Yasuhiro Takeuchi. Dynamics of the density dependent and nonautonomous predator-prey system with Beddington-DeAngelis functional response. [16] [17] Wenshu Zhou, Hongxing Zhao, Xiaodan Wei, Guokai Xu. Existence of positive steady states for a predator-prey model with diffusion. [18] [19] Eric Avila-Vales, Gerardo García-Almeida, Erika Rivero-Esquivel. Bifurcation and spatiotemporal patterns in a Bazykin predator-prey model with self and cross diffusion and Beddington-DeAngelis response. [20] 2018 Impact Factor: 1.008 Tools Metrics Other articles by authors [Back to Top]
Hello one and all! Is anyone here familiar with planar prolate spheroidal coordinates? I am reading a book on dynamics and the author states If we introduce planar prolate spheroidal coordinates $(R, \sigma)$ based on the distance parameter $b$, then, in terms of the Cartesian coordinates $(x, z)$ and also of the plane polars $(r , \theta)$, we have the defining relations $$r\sin \theta=x=\pm R^2−b^2 \sin\sigma, r\cos\theta=z=R\cos\sigma$$ I am having a tough time visualising what this is? Consider the function $f(z) = Sin\left(\frac{1}{cos(1/z)}\right)$, the point $z = 0$a removale singularitya polean essesntial singularitya non isolated singularitySince $Cos(\frac{1}{z})$ = $1- \frac{1}{2z^2}+\frac{1}{4!z^4} - ..........$$$ = (1-y), where\ \ y=\frac{1}{2z^2}+\frac{1}{4!... I am having trouble understanding non-isolated singularity points. An isolated singularity point I do kind of understand, it is when: a point $z_0$ is said to be isolated if $z_0$ is a singular point and has a neighborhood throughout which $f$ is analytic except at $z_0$. For example, why would $... No worries. There's currently some kind of technical problem affecting the Stack Exchange chat network. It's been pretty flaky for several hours. Hopefully, it will be back to normal in the next hour or two, when business hours commence on the east coast of the USA... The absolute value of a complex number $z=x+iy$ is defined as $\sqrt{x^2+y^2}$. Hence, when evaluating the absolute value of $x+i$ I get the number $\sqrt{x^2 +1}$; but the answer to the problem says it's actually just $x^2 +1$. Why? mmh, I probably should ask this on the forum. The full problem asks me to show that we can choose $log(x+i)$ to be $$log(x+i)=log(1+x^2)+i(\frac{pi}{2} - arctanx)$$ So I'm trying to find the polar coordinates (absolute value and an argument $\theta$) of $x+i$ to then apply the $log$ function on it Let $X$ be any nonempty set and $\sim$ be any equivalence relation on $X$. Then are the following true: (1) If $x=y$ then $x\sim y$. (2) If $x=y$ then $y\sim x$. (3) If $x=y$ and $y=z$ then $x\sim z$. Basically, I think that all the three properties follows if we can prove (1) because if $x=y$ then since $y=x$, by (1) we would have $y\sim x$ proving (2). (3) will follow similarly. This question arised from an attempt to characterize equality on a set $X$ as the intersection of all equivalence relations on $X$. I don't know whether this question is too much trivial. But I have yet not seen any formal proof of the following statement : "Let $X$ be any nonempty set and $∼$ be any equivalence relation on $X$. If $x=y$ then $x\sim y$." That is definitely a new person, not going to classify as RHV yet as other users have already put the situation under control it seems... (comment on many many posts above) In other news: > C -2.5353672500000002 -1.9143250000000003 -0.5807385400000000 C -3.4331741299999998 -1.3244286800000000 -1.4594762299999999 C -3.6485676800000002 0.0734728100000000 -1.4738058999999999 C -2.9689624299999999 0.9078326800000001 -0.5942069900000000 C -2.0858929200000000 0.3286240400000000 0.3378783500000000 C -1.8445799400000003 -1.0963522200000000 0.3417561400000000 C -0.8438543100000000 -1.3752198200000001 1.3561451400000000 C -0.5670178500000000 -0.1418068400000000 2.0628359299999999 probably the weirdness bunch of data I ever seen with so many 000000 and 999999s But I think that to prove the implication for transitivity the inference rule an use of MP seems to be necessary. But that would mean that for logics for which MP fails we wouldn't be able to prove the result. Also in set theories without Axiom of Extensionality the desired result will not hold. Am I right @AlessandroCodenotti? @AlessandroCodenotti A precise formulation would help in this case because I am trying to understand whether a proof of the statement which I mentioned at the outset depends really on the equality axioms or the FOL axioms (without equality axioms). This would allow in some cases to define an "equality like" relation for set theories for which we don't have the Axiom of Extensionality. Can someone give an intuitive explanation why $\mathcal{O}(x^2)-\mathcal{O}(x^2)=\mathcal{O}(x^2)$. The context is Taylor polynomials, so when $x\to 0$. I've seen a proof of this, but intuitively I don't understand it. @schn: The minus is irrelevant (for example, the thing you are subtracting could be negative). When you add two things that are of the order of $x^2$, of course the sum is the same (or possibly smaller). For example, $3x^2-x^2=2x^2$. You could have $x^2+(x^3-x^2)=x^3$, which is still $\mathscr O(x^2)$. @GFauxPas: You only know $\|f(x)\|\le K_1 x^2$ and $\|g(x)\|\le K_2 x^2$, so that won't be a valid proof, of course. Let $f(z)=z^{n}+a_{n-1}z^{n-1}+\cdot\cdot\cdot+a_{0}$ be a complex polynomial such that $\|f(z)\|\leq 1$ for $\|z\|\leq 1.$ I have to prove that $f(z)=z^{n}.$I tried it asAs $\|f(z)\|\leq 1$ for $\|z\|\leq 1$ we must have coefficient $a_{0},a_{1}\cdot\cdot\cdot a_{n}$ to be zero because by triangul... @GFauxPas @TedShifrin Thanks for the replies. Now, why is it we're only interested when $x\to 0$? When we do a taylor approximation cantered at x=0, aren't we interested in all the values of our approximation, even those not near 0? Indeed, one thing a lot of texts don't emphasize is this: if $P$ is a polynomial of degree $\le n$ and $f(x)-P(x)=\mathscr O(x^{n+1})$, then $P$ is the (unique) Taylor polynomial of degree $n$ of $f$ at $0$.
Hi, in this thread, @rbj mentioned that the MathJax implementation in this forum is erratic and I believe him. I had issued in the past and it is always frustrating. I think most problems are due to conflicts between the wysiwyg editor and Mathjax. If you find a minute or two, would you please use this thread here to reply and try to post an equation or two to possibly help me identify situations where Mathjax is not working as it should. Here is a reminder how to add equations to your post: 1- To post an equation, simply surround TeX code with a couple of dollar signs on each side: $$ E = mc^2 $$ $$ E = mc^2 $$ 2- If you want your equation to be displayed inline, use: $ e^{i\pi}+1=0 $ $ e^{i\pi}+1=0 $ As I said, I suspect that most problems are caused by the editor that might screw up your equations after submitting/editing your post. So if you are able to successfully post equations on your first try, please feel free to 'edit' your post and try to add equations to see if maybe post-editing might be causing problems. Thanks a lot! $$R_{\mu\nu} - \frac{1}{ 2} R g_{\mu\nu} + \Lambda g_{\mu\nu} = \frac{8 \pi G }{c^4} T_{\mu\nu}$$ First time it did not put out as an equation. So I removed the space after the first $$ and before the last $$ and then it worked. I had an error in the first subscript, edited again, and it worked. So for this case, the editing worked. HA! And then I reloaded the page and the double dollar signs went from readable to functional in the reply. So the mood of the editor seems to matter. I think that re-loading the page is always a good idea after submitting a reply, to make sure Mathjax runs on the new code. In your case, you have 'and before the last' surrounded by two sets of dollar signs and Mathjax interprets the in-between as LaTex code. Makes sense? Thanks Mike for your participation to this thread. (I'm doing this from memory, don't sic Rudi Kalman on me): $P_n^- = F^T P_{n-1}^+ F + Q$ This isn't rendering (at least not the first time): $$ \dot x = A x + B (k \hat x) $$ Let me try. $$ \dot x = A x + B (k \hat x)$$ I was able to make the mathjax code work by editing the html (<>). Basically, you had: <p>$$ </p> <p>\dot x = A x + B (k \hat x) </p> <p>$$ </p> After removing the html in between the dollar signs: $$ \dot x = A x + B (k \hat x) $$$$ \dot x = A x + B (k \hat x) $$ I thought the linefeeds might be messing things up. Hmm. Should be $$ \dot x = A x + B\,(k\,\hat x) $$
Scientific posters present technical information and are intended for congress or presentations with colleagues. Since LaTeX is the most natural choice to typeset scientific documents, one should be able to create posters with it. This article explains how to create posters with latex Contents The two main options when it comes to writing scientific posters are tikzposter and beamerposter. Both offer simple commandsto customize the poster and support large paper formats. Below, you can see a side-to-side comparison of the output generated by both packages (tikzposter on the left and beamerposter on the right). Tikzposter is a document class that merges the projects fancytikzposter and tikzposter and it's used to generate scientific posters in PDF format. It accomplishes this by means the TikZ package that allows a very flexible layout. The preamble in a tikzposter class has the standard syntax. \documentclass[24pt, a0paper, portrait]{tikzposter} \usepackage[utf8]{inputenc} \title{Tikz Poster Example} \author{ShareLaTeX Team} \date{\today} \institute{ShareLaTeX Institute} \usetheme{Board} \begin{document} \maketitle \end{document} The first command, \documentclass[...]{tikzposter} declares that this document is a tikzposter. The additional parameters inside the brackets set the font size, the paper size and the orientation; respectively. The available font sizes are: 12pt, 14pt, 17pt, 20pt and 24pt. The possible paper sizes are: a0paper, a1paper and a2paper. There are some additional options, see the further reading section for a link to the documentation. The commands title, author, date and institute are used to set the author information, they are self-descriptive. The command \usetheme{Board} sets the current theme, i.e. changes the colours and the decoration around the text boxes. See the reference guide for screenshots of the available themes. The command \maketitle prints the title on top of the poster. The body of the poster is created by means of text blocks. Multi-column placement can be enabled and the width can be explicitly controlled for each column, this provides a lot of flexibility to customize the look of the final output. \documentclass[25pt, a0paper, portrait]{tikzposter} \usepackage[utf8]{inputenc} \title{Tikz Poster Example} \author{ShareLaTeX Team} \date{\today} \institute{ShareLaTeX Institute} \usepackage{blindtext} \usepackage{comment} \usetheme{Board} \begin{document} \maketitle \block{~} { \blindtext } \begin{columns} \column{0.4} \block{More text}{Text and more text} \column{0.6} \block{Something else}{Here, \blindtext \vspace{4cm}} \note[ targetoffsetx=-9cm, targetoffsety=-6.5cm, width=0.5\linewidth ] {e-mail \texttt{sharelatex@sharelatex.com}} \end{columns} \begin{columns} \column{0.5} \block{A figure} { \begin{tikzfigure} \includegraphics[width=0.4\textwidth]{images/lion-logo.png} \end{tikzfigure} } \column{0.5} \block{Description of the figure}{\blindtext} \end{columns} \end{document} In tikzposter the text is organized in blocks, each block is created by the command \block{}{} which takes two parameters, each one inside a pair of braces. The first one is the title of the block and the second one is the actual text to be printed inside the block. The environment columns enables multi-column text, the command \column{} starts a new column and takes as parameter the relative width of the column, 1 means the whole text area, 0.5 means half the text area and so on. The command \note[]{} is used to add additional notes that are rendered overlapping the text block. Inside the brackets you can set some additional parameters to control the placement of the note, inside the braces the text of the note must be typed. The standard LaTeX commands to insert figures don't work in tikzposter, the environment tikzfigure must be used instead. The package beamerposter enhances the capabilities of the standard beamer class, making it possible to create scientific posters with the same syntax of a beamer presentation. By now there are not may themes for this package, and it is slightly less flexible than tikzpopster, but if you are familiar with beamer, using beamerposter don't require learning new commands. Note: In this article a special beamerposter theme will be used. The theme "Sharelatex" is based on the theme "Dreuw" created by Philippe Dreuw and Thomas Deselaers, but it was modified to make easier to insert the logo and print the e-mail address at the bottom of the poster. Those are hard-coded in the original themes. Even though this article explains how to typeset a poster in LaTeX, the easiest way is to use a template as start point. We provide several in the ShareLaTeX templates page The preamble of a beamerposter is basically that of a beamer presentation, except for an additional command. \documentclass{beamer} \usepackage[english]{babel} \usepackage[utf8]{inputenc} \usepackage{times} \usepackage{amsmath,amsthm, amssymb, latexsym} \boldmath \usetheme{Sharelatex} \usepackage[orientation=portrait,size=a0,scale=1.4,debug]{beamerposter} \title[Beamer Poster]{ShareLaTeX example of the beamerposter class} \author[sharelatexteam@sharelate.com]{ShareLaTeX Team} \institute[Sharelatex University]{The ShareLaTeX institute, Learn faculty} \date{\today} \logo{\includegraphics[height=7.5cm]{SharelatexLogo}} The first command in this file is \documentclass{beamer}, which declares that this is a beamer presentation. The theme "Sharelatex" is set by \usetheme{Sharelatex}. There are some beamer themes on the web, most of them can be found in the web page of the beamerposter authors. The command \usepackage[orientation=portrait,size=a0,scale=1.4,debug]{beamerposter} Imports the beamerposter package with some special parameters: the orientation is set to portrait, the poster size is set to a0 and the fonts are scaled to 1.4. The poster sizes available are a0, a1, a2, a3 and a4, but the dimensions can be arbitrarily set with the options width=x,height=y. The rest of the commands set the standard information for the poster: title, author, institute, date and logo. The command \logo{} won't work in most of the themes, and has to be set by hand in the theme's .sty file. Hopefully this will change in the future. Since the document class is beamer, to create the poster all the contents must be typed inside a frame environment. \documentclass{beamer} \usepackage[english]{babel} \usepackage[utf8]{inputenc} \usepackage{times} \usepackage{amsmath,amsthm, amssymb, latexsym} \boldmath \usetheme{Sharelatex} \usepackage[orientation=portrait,size=a0,scale=1.4]{beamerposter} \title[Beamer Poster]{ShareLaTeX example of the beamerposter class} \author[sharelatexteam@sharelate.com]{ShareLaTeX Team} \institute[Sharelatex University] {The ShareLaTeX institute, Learn faculty} \date{\today} \logo{\includegraphics[height=7.5cm]{SharelatexLogo}} \begin{document} \begin{frame}{} \vfill \begin{block}{\large Fontsizes} \centering {\tiny tiny}\par {\scriptsize scriptsize}\par {\footnotesize footnotesize}\par {\normalsize normalsize}\par ... \end{block} \end{block} \vfill \begin{columns}[t] \begin{column}{.30\linewidth} \begin{block}{Introduction} \begin{itemize} \item some items \item some items ... \end{itemize} \end{block} \end{column} \begin{column}{.48\linewidth} \begin{block}{Introduction} \begin{itemize} \item some items and $\alpha=\gamma, \sum_{i}$ ... \end{itemize} $$\alpha=\gamma, \sum_{i}$$ \end{block} ... \end{column} \end{columns} \end{frame} \end{document} Most of the content in the poster is created inside a block environment, this environment takes as parameter the title of the block. The environment columns enables multi-column text, the environment \column starts a new columns and takes as parameter the width of said column. All LaTeX units can be used here, in the example the column width is set relative to the text width. Tikzposter themes Default Rays Basic Simple Envelope Wave Board Autumn Desert For more information see
Introduction to Modeling Soap Films and Other Variational Problems What do soap films, catenary cables, and light beams have in common? They behave in ways that minimize certain quantities. Such problems are prevalent in science and engineering fields such as biology, economics, elasticity theory, material science, and image processing. You can simulate many such problems using the built-in physics interfaces in the COMSOL Multiphysics® software, but in this blog series, we will show you how to solve variational problems using the equation-based modeling features. Minimizing Over Functions In elementary calculus, we find optimal values for a function of single or multiple variables. We look for a single number or a finite set of numbers \mathbf{x} that minimize (maximize) a function f(\mathbf{x}). In variational calculus, we search for a function u(x) that minimizes (maximizes) a functional E[u(x)]. In some sense, we can think of this as infinite dimensional optimization. Roughly speaking, a functional takes a function and returns a number. For example, a definite integral is a functional. In engineering problems, these functionals commonly represent some kind of energy. For example, in elasticity theory, we can find equilibrium solutions by minimizing the total potential energy. This terminology is often carried over to other variational problems, such as variational image processing. We call the functional the “energy” — even when it doesn’t physically represent energy in the usual sense. A soap film between rings. Consider a soap film between two rings on the yz-plane whose centers are on the x-axis. We want to find the function u(x) that gives us the shape of the soap film when we revolve the graph of the function around the x-axis. This function minimizes the following functional (1) More generally, in variational calculus, we are looking for a function u(x) that minimizes (2) Most engineering problems deal with functionals that contain, at most, a first-order derivative. In the beginning of this series, we will focus on such problems in one spatial dimension. Later on, we will generalize to higher dimensions, higher-order derivatives, and several unknowns. Lastly, maximizing is the same as minimizing the negative, thus we will only talk about minimization in the sequel. The functional we will be dealing with, unless specified otherwise, is (3) Solving Variational Problems Say you find yourself blindfolded in a valley. How do you know when you’ve reached the bottom? (By the way, removing the blindfolds is not an option.) You might feel the ground around you with your hands and feet, and if every area that you test feels higher than where you stand, you are at the bottom of the valley (at least on a local depression). The same idea is used to check minima in both calculus and variational calculus. In calculus, you test neighboring points, whereas in variational calculus, you test neighboring functions. The function u(x) minimizes the functional E[u(x)] if and only if, for every admissible variation \hat{u}(x), it follows that (4) for a small number \epsilon. Not every variation \hat{u} is admissible, since every u+\epsilon \hat{u} has to satisfy constraints on the solution. For example, for a soap film fixed to wires at the ends, every function we compare with the minimal function has to be fixed to the wires as well. Consequently, we only consider those variations with \hat{u}(a) = \hat{u}(b) = 0. We will deal with constraints in detail in later blog posts. A necessary condition for Eq. 4 is (5) assuming sufficient smoothness to allow differentiation. In this series, we are not going to discuss problems with moving or open boundaries. In such cases, we can move the derivative inside the integral and apply the chain rule to obtain (6) Note that we vary only the dependent variable u and its derivatives, not the spatial coordinate x. If you are interested in problems with moving boundaries or interfaces, check out these blog posts on using the level set and phase field methods for free surface problems and modeling free surfaces with moving mesh. As shown in Eq. 1 for soap films, we have F(x,u,u^{\prime}) = u\sqrt{1+u^{\prime}^2} \Rightarrow \frac{\partial F}{\partial u} = \sqrt{1+u^{\prime}^2}, \frac{\partial F}{\partial u^{\prime}} = \frac{uu^{\prime}}{\sqrt{1+u’^2}}. Consequently, the variational problem for soap films is to find u(x) such that (7) Euler-Lagrange Equation In classical variational calculus, we apply integration by parts to move the spatial derivative from the variation \hat{u} to the solution u to obtain the Euler-Lagrange equation (8) and use ordinary differential equation (ODE) methods to find the solution. In higher dimensions, the Euler-Lagrange equation becomes a partial differential equation (PDE). In our case, we do not need to use the Euler-Lagrange equation, so we will not talk about it any further. The reason is that the finite element method works with the variational formulation. In COMSOL Multiphysics, for example, if you use the Coefficient Form PDE or General Form PDE interfaces to specify the Euler-Lagrange equation, the software internally formulates and solves the corresponding variational equation, so why waste the effort? As we will see later, the variational form also provides natural ways of thinking about sophisticated domain and boundary conditions. Implementing a Variational Problem in COMSOL Multiphysics® To specify a variational problem in COMSOL Multiphysics, we use the Weak Form PDE interface. How do we differentiate between the solution u and the corresponding test function \hat{u}? For the latter, we can use the test operator. For example, for the soap film problem, the integrand in the variational formulation has to be entered as sqrt(1+ux^2)test(u) + uux/sqrt(1+ux^2)*test(ux) in the Weak Form PDE node, as shown below. Specifying a variational problem. We consider the simple constraint that the soap film is fixed on wire loops on the left and right sides. The radii of the left and right rings are 1 and 0.9, respectively, thus we know the primary variable u at both ends. The Dirichlet Boundary Condition node is used to specify such boundary conditions. For numerical reasons, in this particular problem, we provide an initial value of 1 instead of the default 0 in Initial Values 1. Using the Dirichlet Boundary Condition node to specify known boundary values. If we compute the solution, we obtain the shape shown below. Profile of a soap film hanging between two vertical circular wire rings. Specifying a Simpler Symbolic Problem In the above example, we carried out the partial differentiation of F with respect to u and u^{\prime} manually. We can avoid unnecessary labor and potential errors by using the symbolic mathematics capabilities of COMSOL Multiphysics. Using symbolic differentiation to reduce manual work. Variational Solution Versus Direct Optimization We can solve a functional minimization problem by direct optimization. In this approach, we do not need to derive the variational problem. On the downside, it requires more computational tools. In COMSOL Multiphysics, for example, direct optimization requires the Optimization Module. If you are interested in direct optimization, check out our blog post on solving the brachistochrone problem using the Optimization Module. Coming Up Next… Today, we showed you how to solve variational problems with simple constraints using the Weak Form PDE interface. This interface comes with every COMSOL Multiphysics installation. You can learn more about the weak form in this blog post. In upcoming posts, we will show how to add more sophisticated constraints, such as point, distributed, and integral constraints. The series will conclude by generalizing to higher spatial dimensions, higher-order derivatives, and multiple fields. Stay tuned! In the meantime, contact us to learn more about the equation-based modeling features of COMSOL Multiphysics via the button below. View More Blog Posts in the Variational Problems and Constraints Series Comments (2) CATEGORIES Chemical COMSOL Now Electrical Fluid General Interfacing Mechanical Today in Science
Approaches There are many methods for Deconvolution (Namely the degradation operator is linear and Time / Space Invariant) out there. All of them try to deal with the fact the problem is Ill Poised in many cases. Better methods are those which add some regularization to the model of the data to be restored. It can be statistical models (Priors) or any knowledge. For images, a good model is piece wise smooth or sparsity of the gradients. But for the sake of the answer a simple parametric approach will be takes - -Minimizing the Least Squares Error between the restored data in the model to the measurements. Model The least squares model is simple. The objective function as a function of the data is given by: $$ f \left( x \right) = \frac{1}{2} \left\\| h \ast x - y \right\\|_{2}^{2} $$ The optimization problem is given by: $$ \arg \min_{x} f \left( x \right) = \arg \min_{x} \frac{1}{2} \left\\| h \ast x - y \right\\|_{2}^{2} $$ Where $ x $ is the data to be restored, $ h $ is the Blurring Kernel (Gaussian in this case) and $ y $ is the set of given measurements. The model assumes the measurements are given only for the valid part of the convolution. Namely if $ x \in \mathbb{R}^{n} $ and $ h \in \mathbb{R}^{k} $ then $ y \in \mathbb{R}^{m} $ where $ m = n - k + 1 $. This is a linear operation in finite space hence can be written using a Matrix Form: $$ \arg \min_{x} f \left( x \right) = \arg \min_{x} \frac{1}{2} \left\\| H x - y \right\\|_{2}^{2} $$ Where $ H \in \mathbb{R}^{m \times n} $ is the convolution matrix. Solution The Least Squares solution is given by: $$ \hat{x} = \left( {H}^{T} H \right)^{-1} {H}^{T} y $$ As can be seen it requires a matrix inversion. The ability to solve this adequately depends on the condition number of the operator $ {H}^{T} H $ which obeys $ \operatorname{cond} \left( H \right) = \sqrt{\operatorname{cond} \left( {H}^{T} H \right)} $. Condition Number Analysis What's behind this condition number? One could answer it using Linear Algebra. But a more intuitive, in my opinion, approach would be thinking of it in the Frequency Domain. Basically the degradation operator attenuates energy of, generally, high frequency. Now, since in frequency this is basically an element wise multiplication, one would say the easy way to invert it is element wise division by the inverse filter. Well, it is what's done above. The problem arises with cases the filter attenuates the energy practically into zero. Then we have real problems... This is basically what's the Condition Number tells us, how hard some frequencies were attenuated relative to others. Above one could see the Condition Number (Using [dB] units) as a function of the Gaussian Filter STD parameter. As expected, the higher the STD the worse the condition number as higher STD means stronger LPF (Values going down at the end are numerical issues). Numerical Solution Ensemble of Gaussian Blur Kernel was created. The parameters are $ n = 300 $, $ k = 31 $ and $ m = 270 $. The data is random and no noise were added. In MATLAB the Linear System was solved using pinv() which uses SVD based Pseudo Inverse and the \ operator. As one can see, using the SVD the solution is much less sensitive as expected. Why is there an error? Looking at a solution (For the highest STD): As one could see the signal is restored very well except for the start and the end. This is due to the use of Valid Convolution which tells us little on those samples. Noise If we added noise, things would look differently! The reason results were good before is due to the fact MATLAB could handle the DR of the data and solve the equations even though they had large condition number. But large condition number means the inverse filter amplify strongly (To reverse the strong attenuation) some frequencies. When those contain noise it means the noise will be amplified and the restoration will be bad. As one could see above, now the reconstruction won't work. Summary If one knows the Degradation Operator exactly and the SNR is very good, simple deconvolution methods will work. The main issue of deconvolution is how hard the Degradation Operator attenuates frequencies. The more it attenuates the more SNR is needed in order to restore (This is basically the idea behind Wiener Filter). Frequencies which were set to zero can not be restored! In practice, in order to have stable results on should add some priors. The code is available at my StackExchange Signal Processing Q2969 GitHub Repository. This answer was given to a different question yet both of them deals with deconvolution of a signal with a Gaussian Kernel. If there are issues not answered, please point me and I will broaden.
This is a question from an old exam, I'm trying to understand the key and comments provided by the examiner. First question is simply determine the Fourier series of the function $f$ defined as $1-x^2$ on $[-\pi,0)$ and as $1+x^2$ on $[0,\pi)$. I got $$f \sim 1+\frac{2}{\pi} \sum \frac{1}{n^3} \left((2-n^2 \pi^2)(-1)^n-2 \right) \sin{nx}$$ while the key gives $$f \sim 1+\frac{2}{\pi} \sum \frac{1}{n^3} \left((n^2 \pi^2-2)(-1)^{n+1}-2 \right) \sin{nx}$$ The only difference is style, but is it of any significance? The key also has the following to say: Note that $f(x) = 1 + f_1(x)$, where $f_1$ is odd. This is the only mention of $f_1$, and I can't figure out what it could possibly refer to. Neither piecewise functions equals $1$ plus the other, so that can't be it. There has to be some significance to this comment but since I don't understand the comment in the first place said significance is lost on me. Finally one is asked to sketch the Fourier series. Specifically: Set $g(x) :=$ (the Fourier series of $f$) for $x$ in $\mathbb{R}$. Sketch the graph of $g$ on the interval $[-2\pi,2\pi]$. Be particularly clear at the jump points! Well, I am dumbfounded. How does one go about finding even, say, $g(1)$, when you have to calculate an infinite number of values and sum them together? Finally, what are the jump joints? Why isn't $g$ continuous everywhere despite $\sin{nx}$ being so? Thank you.
member 545369 1. Homework Statement A space explorer travels in a spaceship with v = 0.9c. She goes from Earth to a distant star that is 4 light years away (again, measured from Earth). What is the distance measured by the explorer and how long will she say it took her to get there? 2. Homework Equations ##L=\frac {L_0} {\gamma}## , ##\Delta t = \Delta t_0 \gamma## 3. The Attempt at a Solution I solved the first part perfectly fine and got the answer of ##L = 1.7## light years. When solving for ##\Delta t## I realized that I didn't have ##\Delta t_0##. No problem! I just solved the equation ##\frac {L_0} v## for ##\Delta t_0## and got a value of ##\frac {40} {9}## years. Since ##\gamma \approx 2.3## I just multiplied ##2.3## by ##\frac {40} {9}## and got a value of ##10.2## years. The actual value is 1.9 years, which happens to also be ##\frac {\Delta t_0} {\gamma}## but I don't see why I'm wrong! NOTE: I know that there is another solution to this problem and perhaps it is a bit simpler. I would like to focus on whats wrong with my thought process and fix THAT rather than try to adapt to another method of solving. EDIT: Figured out my mistake :) Last edited by a moderator:
Partition property A partition property is an infinitary combinatorical principle in set theory. Partition properties are best associated with various large cardinal axioms (all of which are below measurable), but can also be associated with infinite graphs. The pigeonhole principle famously states that if there are $n$ pigeons in $n-m$ holes, then at least one hole contains at least $m$ pigeons. Partition properties are best motivated as generalizations of the pigeonhole principle to infinite cardinals. For example, if there are $\aleph_1$ pigeons and there are $\aleph_0$ many holes, then at least one hole contains $\aleph_1$ pigeons. Contents Definitions There are quite a few definitions involved with partition properties. In fact, partition calculus, the study of partition properties, almost completely either comprisse or describes most of infinitary combinatorics itself, so it would make sense that the terminology involved with it is quite unique. Square Bracket Notation The square bracket notation is somewhat simple and easy to grasp (and used in many other places). Let $X$ be a set of ordinals. $[X]^\beta$ for some ordinal $\beta$ is the set of all subsets $x\subseteq X$ such that $(x,<)$ has order-type $\beta$; that is, there is a bijection $f$ from $x$ to $\beta$ such that $f(a)<f(b)$ iff $a<b$ for each $a$ and $b$ in $x$. Such a bijection is often called an order-isomorphism. $[X]^{<\beta}$ for some ordinal $\beta$ is simply defined as the union of all $[X]^{\alpha}$ for $\alpha<\beta$, the set of all subsets $x\subseteq X$ with order-type less than $\beta$. In the case of $\omega$, $[X]^{<\omega}$ is the set of all finite subsets of $X$. Homogeneous Sets Let $f:[\kappa]^\beta\rightarrow\lambda$ be a function (in this study, such functions are often called partitions). A set $H\subseteq\kappa$ is then called homogeneous for $f$ when for any two subsets $h_0,h_1\subseteq H$ of order type $\beta$, $f(h_0)=f(h_1)$. This is equivalent to $f$ being constant on $[H]^\beta$. In another case, let $f:[\kappa]^{<\omega}\rightarrow\lambda$ be a function. A set $H\subseteq\kappa$ is then called homogeneous for $f$ when for any two finite subsets $h_0,h_1\subseteq H$ of the same size, $f(h_0)=f(h_1)$. The Various Partition Properties Let $\kappa$ and $\lambda$ be cardinals and let $\alpha$ and $\beta$ be ordinals. Then, the following notations are used for the partition properties: $\kappa\rightarrow (\alpha)_\lambda^\beta$ iff for every function $f:[\kappa]^\beta\rightarrow\lambda$, there is set $H$ of order-type $\alpha$ which is homogeneous for $f$. If $\alpha$ is a cardinal (which it most often is), then the requirement on $H$ can be loosened to $H$ having cardinality $\alpha$ and being homogeneous for $f$ without loss of generality. [1] A common abbreviation for $\kappa\rightarrow (\alpha)_2^n$ is $\kappa\rightarrow (\alpha)^n$. $\kappa\rightarrow (\alpha)_\lambda^{<\omega}$ iff for every function $f:[\kappa]^{<\omega}\rightarrow\lambda$, there is set $H$ of order-type $\alpha$ which is homogeneous for $f$. If $\alpha$ is a cardinal (which it most often is), then the requirement on $H$ can be loosened to $H$ having cardinality $\alpha$ and being homogeneous for $f$ without loss of generality. [1] Let $\nu$ be a cardinal. The square bracket partition properties are defined as follows: $\kappa\rightarrow [\alpha]_\lambda^\beta$ iff for every function $f:[\kappa]^\beta\rightarrow\lambda$, there is set $H$ of order-type $\alpha$ and an ordinal $\gamma<\lambda$ such that $f(h)\neq\gamma$ for any $h\in [H]^\beta$. $\kappa\rightarrow [\alpha]_\lambda^{<\omega}$ iff for every function $f:[\kappa]^{<\omega}\rightarrow\lambda$, there is set $H$ of order-type $\alpha$ and an ordinal $\gamma<\lambda$ such that $f(h)\neq\gamma$ for any finite subset $h$ of $H$. $\kappa\rightarrow [\alpha]_{\lambda,<\nu}^\beta$ iff for every function $f:[\kappa]^\beta\rightarrow\lambda$, there is set $H$ of order-type $\alpha$ such that $f$ restricted to $[H]^\beta$ yields less than $\nu$-many distinct outputs. Note that $\kappa\rightarrow[\alpha]_{\lambda,<2}^\beta$ iff $\kappa\rightarrow(\alpha)_\lambda^\beta$. $\kappa\rightarrow [\alpha]_{\lambda,<\nu}^{<\omega}$ iff for every function $f:[\kappa]^{<\omega}\rightarrow\lambda$, there is set $H$ of order-type $\alpha$ such that $f$ restricted to $[H]^{<\omega}$ yields less than $\nu$-many distinct outputs. Theorems and Large Cardinal Axioms There are several theorems in the study of partition calculus. Namely: Ramsey's theorem, which states that $\aleph_0\rightarrow (\omega)_m^n$ for each finite $m$ and $n$. [2] $2^\kappa\not\rightarrow (\kappa^+)^2$ [2] The Erdős-Rado theorem, which states that $\beth_n(\kappa)^+\rightarrow (\kappa^+)_\kappa^{n+1}$. [1] $\kappa\not\rightarrow(\omega)_2^\omega$ [1] For any finite nonzero $n$ and ordinals $\alpha$ and $\beta$, there is a $\kappa$ such that $\kappa\rightarrow(\alpha)_\beta^n$. [1] The Gödel-Erdős-Kakutani theorem, which states that $2^\kappa\not\rightarrow (3)^2_\kappa$. [1] $\kappa\not\rightarrow [\kappa]_\kappa^\omega$. [1] $\lambda^+\not\rightarrow[\lambda+1]^2_{\lambda,<\lambda}$ [1] $\lambda\not\rightarrow[\lambda]^1_{\mathrm{cf}(\lambda),<\mathrm{cf}(\lambda)}$ [1] For any regular $\kappa$, $\kappa^+\not\rightarrow[\kappa^+]^2_{\kappa^+}$. [1] For any inaccessible cardinal $\kappa$, $\kappa\rightarrow(\lambda)_\lambda^2$ for any $\lambda<\kappa$ [3] . However, this may not be an equivalence; if the continuum hypothesis holds at $\kappa$, then $(\kappa^{++})\rightarrow(\lambda)^2_\kappa$ for any $\lambda<\kappa^{++}$. In terms of large cardinal axioms, many can be described using a partition property. Here are those which can be found on this website: Although not a large cardinal itself, Chang's conjecture holds iff $\aleph_2\rightarrow[\omega_1]^{<\omega}_{\aleph_1,<\aleph_1}$, iff $\aleph_2\rightarrow[\omega_1]^{n}_{\aleph_1,<\aleph_1}$ for some $n$, iff $\aleph_2\rightarrow[\omega_1]^{n}_{\aleph_1,<\aleph_1}$ for every finite $n$. [1] A cardinal $\kappa$ is Ramsey iff $\kappa\rightarrow(\kappa)_\lambda^{<\omega}$ for some $\lambda>1$, iff $\kappa\rightarrow(\kappa)_\lambda^{<\omega}$ for every $\lambda<\kappa$. [1][2] A cardinal $\kappa$ is the $\alpha$-Erdős cardinal iff it is the smallest $\kappa$ such that $\kappa\rightarrow(\alpha)^{<\omega}$. A cardinal $\kappa$ is defined to be $\nu$-Rowbottom iff $\kappa\rightarrow[\kappa]_{\lambda,<\nu}^{<\omega}$ for every $\lambda<\kappa$. A cardinal $\kappa$ is Jónsson iff $\kappa\rightarrow[\kappa]_\kappa^{<\omega}$. [1] A cardinal $\kappa$ is weakly compact iff $\kappa\rightarrow(\kappa)^2_\lambda$ for some $\lambda>1$, iff $\kappa\rightarrow(\kappa)^2_\lambda$ for every $\lambda<\kappa$. [2] References Kanamori, Akihiro. Second, Springer-Verlag, Berlin, 2009. (Large cardinals in set theory from their beginnings, Paperback reprint of the 2003 edition) www bibtex The higher infinite. Jech, Thomas J. Third, Springer-Verlag, Berlin, 2003. (The third millennium edition, revised and expanded) www bibtex Set Theory. Drake, Frank. North-Holland Pub. Co., 1974. bibtex Set Theory: An Introduction to Large Cardinals.
Challenge Given an integer, \$s\$, as input where \$s\geq 1\$ output the value of \$\zeta(s)\$ (Where \$\zeta(x)\$ represents the Riemann Zeta Function). Further information \$\zeta(s)\$ is defined as: $$\zeta(s) = \sum\limits^\infty_{n=1}\frac{1}{n^s}$$ You should output your answer to 5 decimal places (no more, no less). If the answer comes out to be infinity, you should output \$\infty\$ or equivalent in your language. Riemann Zeta built-ins are allowed, but it's less fun to do it that way ;) Examples Outputs must be exactly as shown below Input -> Output1 -> ∞ or inf etc.2 -> 1.644933 -> 1.202064 -> 1.082328 -> 1.0040819 -> 1.00000 Bounty As consolation for allowing built-ins, I will offer a 100-rep bounty to the shortest answer which does not use built-in zeta functions. (The green checkmark will still go to the shortest solution overall) Winning The shortest code in bytes wins.
In synthetic differential geometry, one way to define formally étale morphisms is as follows. Say $f:M\to N$ is formally étale if $TM\cong TN\times _N M$, in other words if the unique map from $TM$ to the pullback is an isomorphism. (A generalization of) this approach is taken e.g in definition 8.10 of Kostecki's notes. One could also say $f$ is formally unramified/smooth/étale if the induced arrow $TM\to TN\times _NM$ is injective/surjective/bijective. The classical differential-geometric definition for a smooth map of manifolds says $f$ is an immersion/submersion/étale if each $d_xf:T_xM\to T_{f(x)}N$ is injective/surjective/bijective. Coming back to SDG, I'm pretty sure $T_{f(x)}N\cong T_{f(x)}N\times _N M$ holds in complete generality, so say $f$ is formally unramified/smooth/étale at a point $x:1\to M$ if $d_xf:T_xM\to T_xN$ is monic/epic/invertible. For nonsingular varieties over algebraically closed fields, it's known that $f$ is étale iff it's étale at every point. For general varieties over algebraically closed fields, I see from snooping around the net that we need 'tangent cones', and I don't really know anything about those yet. At any rate, the proofs of these equivalences seem to rely on the Jacobian criterion, and t, so I take it they also work for just formally unramified/smooth maps. In the nonsingular case, this characterization kind of makes sense since the nullstellensatz says points give good information - the singular case I'll have to tackle some time. Anyway, the point of this story is that over algebraically closed fields, we're getting really clean formal equivalences. For example the formally étale case is: $$\require{AMScd} \begin{CD} TM @>{df}>> TN\\ @VVV @VVV\\ M @>>{f}> N \end{CD}\text{ pullback}\iff \require{AMScd} \begin{CD} T_xM @>{d_xf}>> T_xN\\ @VVV @VVV\\ M @>>{f}> N \end{CD}\text{ pullback for all points}$$ and analogously for formally unramified/smooth. At least the equivalence of the formally étale case above looks completely categorical. The only similar property I know of that seems relevant is universal coproducts, which is equivalent to the equivalence below. $$\require{AMScd} \begin{CD} \coprod_iP_i @>>> \coprod_iX_i\\ @VVV @VVV\\ A @>>> B \end{CD}\text{ pullback}\iff \require{AMScd} \begin{CD} P_i @>>> X_i\\ @VVV @VVV\\ A @>>> B \end{CD}\text{ pullback for all }i$$ But tangent bundles are more than just the tangent spaces side by side... My qusetions: What's the categorical property of algebraic closedness that gives the first equivalence above? Is there any relation to universal products, and if so, what is it? Added. Wandering the internet some more, I see strictly Henselian rings pop up in context of inverse function theorems for schemes. In fact, in section 2.3 of the book Néron Models, it's said (strictly) Henselian rings are described geometrically by schemes satisfying "certain aspects of the inverse function theorem". Since the inverse function theorem (I think) is responsible for the equivalence in the case of smooth manifolds, perhaps Henselian properties, and not algebraic closedness, are what makes things tick. I really have no idea about Henselian stuff though, so just throwing it out there...
Forgot password? New user? Sign up Existing user? Log in I came across this question on a site and I didn't find any solution to this.....(except to use calculator) The question was Which digit doesn't occur in the number 2^29 ? Plz help..... Note by Poonayu Sharma 5 years, 4 months ago Easy Math Editor This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science. When posting on Brilliant: italics _italics_ bold __bold__ - bulleted- list 1. numbered2. list paragraph 1paragraph 2 paragraph 1 paragraph 2 [example link](https://brilliant.org) > This is a quote This is a quote # I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world" # I indented these lines# 4 spaces, and now they show# up as a code block.print "hello world" \[ \] 2 \times 3 2^{34} a_{i-1} \frac{2}{3} \sqrt{2} \sum_{i=1}^3 \sin \theta \boxed{123} Sort by: If 2^29=n (mod 9),then 9-n is the answer. Noticing that 2^3=-1(mod 9),2^29=(2^3)^9x2^2=(-1)^9x2^2=-4=5 (mod 9).Therefore the answer is 9-5=4. This can be easily derived using mod. Hope you got it :) Log in to reply Ohh...use of mod didn't strike me :P I was trying to solve it using logs etc.. :P Btw...thanks you sir You're welcome :D....but you're 3 yrs elder, so you must refrain from calling me sir, how could you? And this solution is also inspired by an experiment I did with the cyclicity of digital roots of powers of 2. You could check out my easy-peasy problem on that though :) @Krishna Ar – U deserved "sir" :D.....and I solved the problem u talking about....but not with the (mod 9) methodBut with cyclic sum of digits of powers of 2 @Poonayu Sharma – Oh. Well at first ( when I encountered this problem in class 7) I solved this using cyclicity too :) @Krishna Ar – Ohh...well dont u think it becomes a herculean task for a 7 grade guy to do such sums :P @Poonayu Sharma – @Poonayu Sharma - Woah! I'm mind -blown!!!! How did you level up in Algebra and Number Theory so soon? What resources did you use to gain here? @Krishna Ar – I saved the challenging questions for later. ..then I read all the posts or notes related to them..For some.I used Google to search theories related to sum..and then tried to solve them....though I couldn't solve all(surprisingly I was able to solve 3 lvl 5 question) ;P..I guess you can also do it ...faster than me :D Next target ....JEE ,CALCULUS AND MECHANICS :P @Poonayu Sharma – What did you use to learn all those algebra question- inequalities , Symmetric bounding,,all that. And sorry, I really dont think I can do it faster :(. @Krishna Ar – Frankly speaking. ...The symmetric bounding question was a hit and trial by me..(I think one should not hesitate to use hit and trial if there are no other means at times.).I got it in the second try.. .whenever u see a problem. ..try to find a note about it .(That's the only thing I used to crack sums..also me and my 2 friends did few sums together)..I remember daniel liu once gave a link to a note which could be used to solve a problem...It was useful then ..If u don't find any such notes...Google it...believe me problems do become easier then.. :DHoping ur level soars high :P @Poonayu Sharma – :D @Krishna Ar – Btw...u planning to do iit? :P @Poonayu Sharma – Nope. I do math only as a hobby. I'd wan't to take up medicine as my mother wants me to do it. U wanna go to IIT or somewhere else Mr. Future Cosmologist? @Krishna Ar – I am bad at bio and chemistry right from birth :P..I will do cosmology (wanna go to NASA) whether I get iit or not..currently I'm preparing for iit ...let's see if luck shines upon me and if i get a good rank then ill do iit ...well if not then I'm thinking for PhD in mathematics and then cosmology ( possibly astronaut too) :P But the way medical ppl will have a tough competition because of u :P @Poonayu Sharma – Nice :). I also loved the way you ended it with a joke :). You want to become an astronaut or cosmologist? I must say there is a great difference between the two! But, whatever it be I'm sure you'd do well in it :P. You are preparing for IIT on your own? I feel ICSE curriculum is much better than CBSE. What do you feel? @Krishna Ar – I want to become both actually :P....I have joined classes here called pace for preparation. .I myself have studied from icse and I feel it's standard is better than other boardr...But while you prepare for iit ..It doesn't matter which board ur from ...All become more or less same Thanks for ur compliment though ... And good luck. .. (remember to enjoy life as much as u can now...After 10 ...its hell :( :P) @Poonayu Sharma – YUP. Why do you say life after 10th is hell? In fact, I would consider that heaven. I would have to study only my favorite two subjects. English is passable too. In fact, life now is hell for me!. @Krishna Ar – I used to think the same way bro....You will realise it after u leave ur school and join college or classes ...everything suddenly gets screwed up :P but if u manage ur time and other things (which I failed to)...u wont have troubles ..Hope you enjoy it even after school :D @Poonayu Sharma – Oh! Thanks for your well wishes :) Are there other ways too? Sorry, I can't fathom any. :(... I'm really bad at intuition :-/ You can view the question here with a detailed solution discussion. Thank you sir :D Except there you are given that it's a 9 digit number with distinct digits. The first step in this problem is to show that it is a 9 digit number (using log) and then obviously by the way the problem is stated implies that the digits are distinct Yeah I forgot to mention it because I thought that its understood :P That rises an interesting question. Which powers (or 2) are missing exactly 1 digit (but could have repeats of others)? Unfortunately, I don't think there is an easy way to answer this question. Problem Loading... Note Loading... Set Loading...
Answer The angles are as follows: $A = 61^{\circ}, B = 40.7^{\circ},$ and $C = 78.3^{\circ}$ The lengths of the sides are as follows: $a = 5.4, b = 4,$ and $c = 6$ Work Step by Step Let $b=4$, and let $c = 6$. We can use the law of cosines to find $a$: $a^2 = b^2+c^2-2bc~cos~A$ $a = \sqrt{b^2+c^2-2bc~cos~A}$ $a = \sqrt{4^2+6^2-(2)(4)(6)~cos~61^{\circ}}$ $a = \sqrt{28.729}$ $a = 5.4$ We can use the law of cosines to find $B$: $b^2 = a^2+c^2-2ac~cos~B$ $2ac~cos~B = a^2+c^2-b^2$ $cos~B = \frac{a^2+c^2-b^2}{2ac}$ $B = arccos(\frac{a^2+c^2-b^2}{2ac})$ $B = arccos(\frac{(5.4)^2+6^2-4^2}{(2)(5.4)(6)})$ $B = arccos(0.7586)$ $B = 40.7^{\circ}$ We can find angle $C$: $A+B+C = 180^{\circ}$ $C = 180^{\circ}-A-B$ $C = 180^{\circ}-61^{\circ}-40.7^{\circ}$ $C = 78.3^{\circ}$
Correlation quiz 1 Correlation is a __________ type of statistical analysis. 2 Correlation is: 3 What would you expect the correlation between daily calorie consumption and body weight to be? 4 Estimate the correlation between driving performance and blood alcohol levels: 5 Estimate the correlation between use of contraception and likelihood of pregnancy in heterosexual intercourse: 6 Estimate the correlation between consumer cost and consumer satisfaction: 7 Estimate the correlation between solar flares and suicide: 8 The degrees of freedom for Pearson’s product-moment r = ∑ z x z y n − ? . {\displaystyle r={\frac {\sum z_{x}z_{y}}{n-?}}.} is equal to: 9 Which of the following checks are not relevant to deciding whether the Pearson product-moment correlation is an appropriate indication of the degree of linear association between two variables? 10 The square of the correlation coefficient or r2 is called the
Consider a one dimensional system with fluid in it. Mass and momentum balance equation of the system are (in the absence of external forces and assuming Newtonian behaviour valid for viscocity), \begin{eqnarray} \frac{\partial \rho}{\partial t} + \frac{\partial}{\partial x}(\rho u)= 0\\ \rho\frac{\partial u}{\partial t} = -c_s^2\frac{\partial \rho}{\partial x} + \nu\frac{\partial u}{\partial x^2} \end{eqnarray} What is the total mechanical energy of the system? I would take energy as solely kinetic energy of the system. \begin{eqnarray} E = \sum\frac{1}{2}\rho u^2 \end{eqnarray} It seems that this definition is incorrect as if we start with a system of initial density variation $\rho(x,0) = \rho(x)$ and zero velocity everywhere $u(x,0)= 0$ then we start with zero kinetic energy and zero total energy. But from the mass and momentum balance equations it is clear that there will motion in fluid due to density variations. So we have higher energy than we started with. What is wrong with definition of total energy I am assuming? which part of the energy should be taken into account?
Search Now showing items 1-2 of 2 Search for new resonances in $W\gamma$ and $Z\gamma$ Final States in $pp$ Collisions at $\sqrt{s}=8\,\mathrm{TeV}$ with the ATLAS Detector (Elsevier, 2014-11-10) This letter presents a search for new resonances decaying to final states with a vector boson produced in association with a high transverse momentum photon, $V\gamma$, with $V= W(\rightarrow \ell \nu)$ or $Z(\rightarrow ... Fiducial and differential cross sections of Higgs boson production measured in the four-lepton decay channel in $\boldsymbol{pp}$ collisions at $\boldsymbol{\sqrt{s}}$ = 8 TeV with the ATLAS detector (Elsevier, 2014-11-10) Measurements of fiducial and differential cross sections of Higgs boson production in the ${H \rightarrow ZZ ^{*}\rightarrow 4\ell}$ decay channel are presented. The cross sections are determined within a fiducial phase ...
Search Now showing items 1-2 of 2 Anisotropic flow of inclusive and identified particles in Pb–Pb collisions at $\sqrt{{s}_{NN}}=$ 5.02 TeV with ALICE (Elsevier, 2017-11) Anisotropic flow measurements constrain the shear $(\eta/s)$ and bulk ($\zeta/s$) viscosity of the quark-gluon plasma created in heavy-ion collisions, as well as give insight into the initial state of such collisions and ... Investigations of anisotropic collectivity using multi-particle correlations in pp, p-Pb and Pb-Pb collisions (Elsevier, 2017-11) Two- and multi-particle azimuthal correlations have proven to be an excellent tool to probe the properties of the strongly interacting matter created in heavy-ion collisions. Recently, the results obtained for multi-particle ...
Search Now showing items 1-10 of 24 Production of Σ(1385)± and Ξ(1530)0 in proton–proton collisions at √s = 7 TeV (Springer, 2015-01-10) The production of the strange and double-strange baryon resonances ((1385)±, Ξ(1530)0) has been measured at mid-rapidity (\|y\|< 0.5) in proton–proton collisions at √s = 7 TeV with the ALICE detector at the LHC. Transverse ... Forward-backward multiplicity correlations in pp collisions at √s = 0.9, 2.76 and 7 TeV (Springer, 2015-05-20) The strength of forward-backward (FB) multiplicity correlations is measured by the ALICE detector in proton-proton (pp) collisions at s√ = 0.9, 2.76 and 7 TeV. The measurement is performed in the central pseudorapidity ... Inclusive photon production at forward rapidities in proton-proton collisions at $\sqrt{s}$ = 0.9, 2.76 and 7 TeV (Springer Berlin Heidelberg, 2015-04-09) The multiplicity and pseudorapidity distributions of inclusive photons have been measured at forward rapidities ($2.3 < \eta < 3.9$) in proton-proton collisions at three center-of-mass energies, $\sqrt{s}=0.9$, 2.76 and 7 ... Rapidity and transverse-momentum dependence of the inclusive J/$\mathbf{\psi}$ nuclear modification factor in p-Pb collisions at $\mathbf{\sqrt{\textit{s}_{NN}}}=5.02$ TeV (Springer, 2015-06) We have studied the transverse-momentum ($p_{\rm T}$) dependence of the inclusive J/$\psi$ production in p-Pb collisions at $\sqrt{s_{\rm NN}} = 5.02$ TeV, in three center-of-mass rapidity ($y_{\rm cms}$) regions, down to ... Measurement of pion, kaon and proton production in proton–proton collisions at √s = 7 TeV (Springer, 2015-05-27) The measurement of primary π±, K±, p and p¯¯¯ production at mid-rapidity (\|y\|< 0.5) in proton–proton collisions at s√ = 7 TeV performed with a large ion collider experiment at the large hadron collider (LHC) is reported. ... Two-pion femtoscopy in p-Pb collisions at $\sqrt{s_{\rm NN}}=5.02$ TeV (American Physical Society, 2015-03) We report the results of the femtoscopic analysis of pairs of identical pions measured in p-Pb collisions at $\sqrt{s_{\mathrm{NN}}}=5.02$ TeV. Femtoscopic radii are determined as a function of event multiplicity and pair ... Measurement of charm and beauty production at central rapidity versus charged-particle multiplicity in proton-proton collisions at $\sqrt{s}$ = 7 TeV (Springer, 2015-09) Prompt D meson and non-prompt J/$\psi$ yields are studied as a function of the multiplicity of charged particles produced in inelastic proton-proton collisions at a centre-of-mass energy of $\sqrt{s}=7$ TeV. The results ... Charged jet cross sections and properties in proton-proton collisions at $\sqrt{s}=7$ TeV (American Physical Society, 2015-06) The differential charged jet cross sections, jet fragmentation distributions, and jet shapes are measured in minimum bias proton-proton collisions at centre-of-mass energy $\sqrt{s}=7$ TeV using the ALICE detector at the ... Centrality dependence of high-$p_{\rm T}$ D meson suppression in Pb-Pb collisions at $\sqrt{s_{\rm NN}}$ = 2.76 TeV (Springer, 2015-11) The nuclear modification factor, $R_{\rm AA}$, of the prompt charmed mesons ${\rm D^0}$, ${\rm D^+}$ and ${\rm D^{+}}$, and their antiparticles, was measured with the ALICE detector in Pb-Pb collisions at a centre-of-mass ... K(892)$^0$ and $\Phi$(1020) production in Pb-Pb collisions at $\sqrt{s_{NN}}$ = 2.76 TeV (American Physical Society, 2015-02) The yields of the K*(892)$^0$ and $\Phi$(1020) resonances are measured in Pb-Pb collisions at $\sqrt{s_{NN}}$ = 2.76 TeV through their hadronic decays using the ALICE detector. The measurements are performed in multiple ...
Let $X$ be a nowhere dense set of circle $S^1$. Here $S^1$ is equipped with the standard topology and measure. Q Can we say that $X$ is a finite point set? Is there a counterexample? Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It only takes a minute to sign up.Sign up to join this community Well, you can say it, but it would no be true. Just consider the set$$\left\{\left(\cos\left(\frac1n\right),\sin\left(\frac1n\right)\right)\,\middle\|\,n\in\mathbb N\right\}\subset S^1.$$
If you consider the phase space (the space of initial data) ${\cal{M}}$ of a classical system it can be seen as the cotangent bundle $T^{}Q$ of the configuration space $Q$. As you say this bundle has a natural symplectic structure $\omega:T{\cal{M}}\times T{\cal{M}}\rightarrow \mathbb{R}$. Now given a Hamiltonian $H:{\cal{M}}\rightarrow \mathbb{R}$ as you say using the inverse of the symplectic structure we can obtain the Hamiltonian vector field $X_{H}=\omega^{-1}(dH,.):T^{}{\cal{M}}\rightarrow \mathbb{R}$. Let's consider now coordinates $(q_{1},..,q_{n})$ in $Q$. This set of coordinates give rise to a natural set of coordinates $(q_{1},..,q_{n};p_{1},..,p_{n})$ on ${\cal{M}}$ by taking $(p_{1},..,p_{n})$ to be the components of the cotangent vectors in the coordinate basis associated with $(q_{1},..,q_{n})$. The symplectic form then takes the form $\omega=\sum_{\mu}dp_{\mu}\wedge dq_{\mu}$ and the inverse takes the form $\omega^{-1}=\sum_{\mu}(\frac{\partial}{\partial q}_{\mu})\otimes (\frac{\partial}{\partial p}_{\mu})-(\frac{\partial}{\partial p}_{\mu})\otimes (\frac{\partial}{\partial q}_{\mu})$. Then the hamiltonian vector field is denoted by:$X_{h}=\sum_{\mu}(\frac{\partial H}{\partial q}_{\mu})\otimes (\frac{\partial}{\partial p}_{\mu})-(\frac{\partial H}{\partial p}_{\mu})\otimes (\frac{\partial}{\partial q_{\mu}})$. If you consider now an integral curve of this vector field which means that the curve $\alpha:\mathbb{R}\rightarrow {\cal{M}}$ satisfies $\frac{d \alpha}{dt}=X_{h}$ We obtain $$\frac{dq_{\mu}}{dt}=\frac{\partial H}{\partial p_{\mu}}\\ \frac{dp_{\mu}}{dt}=-\frac{\partial H}{\partial q_{\mu}}$$ which are Hamilton's equation. Moreover, we can defined the poisson bracket of two classical observables as$\{f,g\}=\omega^{-1}(df,dg)$ which satisfies for the coordinates$\{q_{\mu},q_{\nu}\}=0,\{p_{\mu},p_{\nu}\}=0,\{q_{\mu},p_{\nu}\}=\delta_{\nu\mu}$. As you can see these relations are similar to the observables in QM. In fact, there are a lot of quantization procedures from classical theories where this is the starting point. Finally you can define the classical action when the Hamiltonian doesn't depend on time as $S=\int\theta$ with the integral understood to be taken over the manifold defined by holding the energy $E$ constant: $H=E=$const. Here are two pictures that might help from Roger Penrose's The Road of reality: The curves that have as tangent vectors the Hamiltonian flow are the solutions to the equations of motion of the system.
Although I'm not sure that Piketty ever directly discusses the exact definition of $r$, he does make it clear indirectly. On page 52 of the hardcover English-language edition of his book, Piketty declares his "first fundamental law of capitalism": Piketty obtains the share of income $\alpha$ from capital in national income from the income side of the national accounts, dividing "capital income" by national income. The capital share $\alpha$ is just the inverse of the aggregate labor share of income, on which there is an extensive literature. (Of course, there are several methodological choices that must be made when defining these concepts, and some of Piketty's choices differ from choices elsewhere.) When finding $\beta$, Piketty appears to take the aggregate value of capital from respective countries' national accounts as well (though alternative sources are needed for older values). There is lots of information on the relevant choices in the technical appendix of Piketty and Zucman (2014), which is the basis of most discussion of $\alpha$ and $\beta$ in the book. Since Piketty makes clear that $r=\alpha/\beta$, his definition of $r$ for a given year is essentially$$r=\frac{\alpha}{\beta}=\frac{\text{aggregate capital income taken from national accounts}}{\text{aggregate value of capital taken from national accounts}}$$Note that this actually does not (directly) involve tax data. Tax data is the basis of much of Piketty's other work, but not this.
Let $(L,h)\to (X, \omega)$ be a compact toric polarized Kähler manifold of complex dimension $n$. For each $k\in N$, the fibre-wise Hermitian metric $h^k$ on $L^k$ induces a natural inner product on the vector space $C^{\infty}(X, L^k)$ of smooth global sections of $L^k$ by integration with respect to the volume form $\frac{\omega^n}{n!}$. The orthogonal projection $P_k:C^{\infty}(X, L^k)\to H^0(X, L^k)$ onto the space $H^0(X, L^k)$ of global holomorphic sections of $L^k$ is represented by an integral kernel $B_k$ which is called the Bergman kernel (with parameter $k\in N$). The restriction $\rho_k:X\to R$ of the norm of $B_k$ to the diagonal in $X\times X$ is called the density function of $B_k$. On a dense subset of $X$, we describe a method for computing the coefficients of the asymptotic expansion of $\rho_k$ as $k\to \infty$ in this toric setting. We also provide a direct proof of a result which illuminates the off-diagonal decay behaviour of toric Bergman kernels. We fix a parameter $l\in N$ and consider the projection $P_{l,k}$ from $C^{\infty}(X, L^k)$ onto those global holomorphic sections of $L^k$ that vanish to order at least $lk$ along some toric submanifold of $X$. There exists an associated toric partial Bergman kernel $B_{l, k}$ giving rise to a toric partial density function $\rho_{l, k}:X\to R$. For such toric partial density functions, we determine new asymptotic expansions over certain subsets of $X$ as $k\to \infty$. Euler-Maclaurin sums and Laplace's method are utilized as important tools for this. We discuss the case of a polarization of $CP^n$ in detail and also investigate the non-compact Bargmann-Fock model with imposed vanishing at the origin. We then discuss the relationship between the slope inequality and the asymptotics of Bergman kernels with vanishing and study how a version of Song and Zelditch's toric localization of sums result generalizes to arbitrary polarized Kähler manifolds. Finally, we construct families of induced metrics on blow-ups of polarized Kähler manifolds. We relate those metrics to partial density functions and study their properties for a specific blow-up of $C^n$ and $CP^n$ in more detail.
Rhenium osmium dating site Some key papers include: Using Re-Os isotopes to determine the deposition age of petroleum source-rock formation (organic-rich shales) is a major research theme, both in terms of technical development and application.This method has been applied to better understand the origins of global Oceanic Anoxic Events (OAEs), the timing of shale deposition and correlation in Precambrian sedimentary basins and this rise of oxygen on Earth.Re-Os isotopes in natural hydrocarbons is the third major research theme of the laboratory, again approaching this new field from both a systematic evaluation and application perspective. The Re-Os isochron plots the ratio of radiogenic {\displaystyle \left({\frac ((}^{187}\mathrm {Os} }((}^{188}\mathrm {Os} ))\right)_{\mathrm {present} }=\left({\frac ((}^{187}\mathrm {Os} }((}^{188}\mathrm {Os} ))\right)_{\mathrm {initial} }+\left({\frac ((}^{187}\mathrm {Re} }((}^{188}\mathrm {Os} ))\right)\cdot (e^{\lambda t}-1),} This page is based on a Wikipedia article written by contributors (read/edit). Text is available under the CC BY-SA 4.0 license; additional terms may apply. Images, videos and audio are available under their respective licenses. Development of Re-Os geochronology for crustal matrices has been ongoing since 1998 in the laboratory. This work has resulted in major advancements in the direct isotopic dating of crustal sulfide minerals, petroleum source-rocks, and natural hydrocarbons, using Re-Os isotopes.
The way that you ask your question confuses the answer because you say "Will the speed of universe expansion will make the sky bright but the red shift make it invisible to our eyes", because the sky is already "bright" in certain wavelengths, particularly the centimeter cosmic background radiation (CMB). Other than this revision, yes, the observation that the night sky is dark has been a clear argument against an infinitely old universe since long ago. The evidence for the big bang in the form of a consistently increasing red shift pretty much seals the deal for myself, regarding the fact that the universe has an age. Furthermore, over time, you are entirely correct in the assertion that the number of observable objects will increase drastically, and quite possibly infinitely. Consider that we only see $x$ distance away which terminates at the CMB, thus limiting the number of galaxies we can see, with the furthest galaxies being the earliest evolutionary stage of galaxies. The number of "young" galaxies we can see will progressively increase as more of the veil from the CMB is pulled back through the arrival of the new light. The "young" galaxies we can see now will mature and the total number will increase. Whether or not this will increase forever is disputable since dark energy pulls space apart could prevent it but we can't claim to know exactly what the behavior of dark energy far in to the future will be. Additions I started thinking about the problem more and I wanted to formalize things a bit better. Take the most basic case, we'll deal with a flat Newtonian space for now. As before, take $x$ to be the distance to a certain galaxy we are current seeing. Take the present time (after the big bang) to be $t$ and that we're observing that galaxy at $t'$. It follows... $$x=c (t-t')$$ Imagine the universe has a galaxy density of $\rho$ galaxies per unit volume. Then knowing that, we can actually write the rate $r$ at which galaxies older than $t'$ are appearing into our view. This is done knowing the surface of a sphere is $4\pi r^2$. $$r=4 \rho c \pi x^2$$ It's fascinating to consider that in a line connecting every object in the night sky and us, there exists the entire history of the object encoded in the light waves making their way to us. One way to talk about the acceleration of the universe is to say that there is a slowdown in the rate at which we are receiving this information. We are watching the far off objects in slow motion. If we make the obviously incorrect but useful assumption that all objects emit light at the same rate at all times, then the intensity we see will be proportional to $1/x^2$, and given some $S$ which is, say, the number of photons emitted total per unit time, then the intensity of light we receive from a given body would be $S/(4 \pi x^2)$. Multiplying this by the rate, we can get a very nice equation for $s(x)$ which is the contribution to the number of photons we receive from the differential "shell" of stars at $x$. $$s(x) = S \rho c $$ This equation is important because it is cumulative from time at $t'$ to $t$, meaning that the objects that entered our field of vision from the "genesis" of that type of object are still contributing to the population of photons reaching us today. So the number of photons we are receiving could be said to be: $$\int_{t'}^t S \rho c dt = S \rho c (t-t')$$ A more advanced view of the situation simply notes that the "movie" for each of these stars is being played in slow motion. We'll just define a factor for that and put it in the equation. $$l(x) = \frac{\Delta t_{object}}{\Delta t_{Earth}}$$ I should preface this by saying that this isn't actually saying time is going slower for that object, and this isn't even the time dilation as defined by general relativity, this is the time dilation you would measure by watching a clock in a galaxy far away with a space telescope and comparing it to the local time. Yes, these two are different, and yes, I am avoiding advanced relativistic concepts by making it an accounting problem. Now the total # of photons we receive per unit time is the following. $$\int_{t'}^t S \rho c l(x) dt$$ I won't use any calculus chain rules because there's no guarantee that $l(t)$ is any more helpful to you than $l(x)$! But I should also note that the final $x$ you get in this equation at $t$ will be meaningless. It is not the general relativity distance, it's some bastardization of that by using $c t$, which is clearly not how it actually works. Nonetheless, there is some usefulness in the above equation. We can even identify the radiative energy being received by considering the energy of the photon being proportional to it's frequency, with $E_e$ being the energy of the emitted photon and $E_o$ the observed photon. $$\frac{E_e}{E_o} = l(x)$$ And the total energy would then be the following with $h$ the familiar plank constant. $$E = \int_{t'}^t S h \rho c l(x)^2 dt$$ Anyway, my intent is for these to be instructive "kindergarten" equations for the subject. The bottom line is still clear from them - that the # of photons reaching us would increase linearly over time but it's less since $l(x)\le 1$. Similarly, the radiative energy reaching us would be less by even a smaller factor due to the redshift. I hope this is a clear picture.
Division Theorem/Positive Divisor/Uniqueness Theorem $\forall a, b \in \Z, b > 0: \exists! q, r \in \Z: a = q b + r, 0 \le r < b$ In the above equation: $a$ is the dividend $b$ is the divisor $q$ is the quotient $r$ is the principal remainder, or, more usually, just the remainder. It is given by Division Theorem: Positive Divisor: Existence that such $q$ and $r$ exist. Suppose $q_1, r_1$ and $q_2, r_2$ are two pairs of $q, r$ that satisfy $a = q b + r, 0 \le r < b$. That is: $\displaystyle a$ $=$ $\displaystyle q_1 b + r_1, 0 \le r_1 < b$ $\displaystyle a$ $=$ $\displaystyle q_2 b + r_2, 0 \le r_2 < b$ This gives: $0 = b \paren {q_1 - q_2} + \paren {r_1 - r_2}$ Aiming for a contradiction, suppose that $q_1 \ne q_2$. Without loss of generality, suppose that $q_1 > q_2$. Then: $\displaystyle q_1 - q_2$ $\ge$ $\displaystyle 1$ $\displaystyle \leadsto \ \ $ $\displaystyle r_2 - r_1$ $=$ $\displaystyle b \left({q_1 - q_2}\right)$ $\displaystyle $ $\ge$ $\displaystyle b \times 1$ as $b > 0$ $\displaystyle $ $=$ $\displaystyle b$ $\displaystyle \leadsto \ \ $ $\displaystyle r_2$ $\ge$ $\displaystyle r_1 + b$ $\displaystyle $ $\ge$ $\displaystyle b$ This contradicts the assumption that $r_2 < b$. A similar contradiction follows from the assumption that $q_1 < q_2$. Therefore $q_1 = q_2$ and so $r_1 = r_2$. Thus it follows that $q$ and $r$ are unique. $\blacksquare$ It is given by Division Theorem: Positive Divisor: Existence that such $q$ and $r$ exist. Let $a = q b + r$ where $q, r \in \Z$ and $0 \le r < b$. Thus: $\dfrac a b = q + \dfrac r b$ and: $0 \le \dfrac r b \le \dfrac {b - 1} b < 1$ So: $q = \floor {\dfrac a b}$ and so: $r = a - b \floor {\dfrac a b}$ Thus, given $a$ and $b$, the numbers $q$ and $r$ are unique determined. $\blacksquare$ It is given by Division Theorem: Positive Divisor: Existence that such $q$ and $r$ exist. Suppose that: $a = b q_1 + r_1 = b q_2 + r_2$ where both $0 \le r_1 < b$ and $0 \le r_2 < b$. Without loss of generality, suppose $r_1 \ge r_2$. Then: $r_1 - r_2 = b \paren {q_2 - q_1}$ That is: $b \divides \paren {r_2 - r_1}$ where $\divides$ denotes divisibility. But: $r_1 - r_2 < b$ $r_1 - r_2 \ge b$ unless from Integer Divides Zero $r_1 - r_2 = 0$. So $r_1 = r_2$ and it follows directly that $q_1 = q_2$. $\blacksquare$
First steps to infinity and beyond Contents Axioms In math, there are several philosophical liberties that could be taken. The axioms declare the absolute rules of math; what can and can't be done, what exists and what doesn't, and what statements are true and false, are all determined by them. There are many different ways to axiomatize mathematics. These different methods are called Set Theories, and there are quite a few. For example: Zermelo-Fraenkel, Morse-Kelley, Peano Arithmetic, Neumann-Bernays-Gödel, Kripke-Platek, and more. If there is a "universe" in which axioms are true, then that universe is called a model of those axioms. This is an important part of model theory, the study of the formalization of this idea. Finitism Peano Arithmetic is the standard set of axioms which define the natural numbers. Every model of Peano Arithemtic only contains "natural-number like"-things, so the universe of all real numbers or all complex numbers aren't models of Peano Arithmetic. Peano Arithmetic does not declare the existence of an infinite set or number; in fact, the standard model of choice, $\mathbb{N}$, contains only finite numbers and sets. Finitism is the idea that there are no infinite sets or infinite numbers in mathematics. Although this idea may seem somewhat reasonable for most people, if axioms declare there are infinite sets, mathematics becomes incredibly beautiful. More things are provable, things about proving things are provable, model theory is formalized in some way, the strangeness and oddities of infinity become formalized in pure mathematics. The intuition of infinity is often preserved if we let it exist. Letting Infinity Exist When one lets an infinite set exist and powerset exist in the axioms (that is, for any set, there is a set which contains exactly all of its subsets), then this beauty of infinity is shown. It is against intuition to not allow powerset to exist, so this is only assumed natural and absolute. With this, an infinite set and its powerset never have a one-on-one correspondence to each other (this correspondence is also known as a bijection). That is, there is no way to give each element of the infinite set exactly one unique element of its powerset. This intuitively makes the powerset "bigger" than the original set. The proof of this is known as Cantor's theorem. When there is a bijection between two sets, we say the sets have the same cardinality. We then assign numbers to sets based upon their cardinality; a measure of size. Cardinals and Ordinals The ordinals are defined in a way that extends the natural numbers. The smallest ordinal is $0=\{\}$. The next ordinal is $1=\{0\}=\{\{\}\}$. The next ordinal is $2=\{0,1\}=\{\{\},\{\{\}\}\}$. ... Each ordinal is the set of all smaller ordinals. Of course, that begs the question: what is the set of all natural numbers? With this pattern, it should be an ordinal. It is in fact, and it is called $\omega=\{0,1,2,3...\}$. The next ordinal is called $\omega+1$, and then $\omega+2$, and so on. Eventually, one gets to $\omega\cdot 2$, which is simply the the set $\{0,1,2,3...\omega,\omega+1,\omega+2...\}$. Note that $2\cdot\omega$ is not $\omega\cdot 2$, and $1+\omega$ is not $\omega+1$. If you would like a more detailed explanation of ordinal arithmetic, it would be suggested that you should search a reliable source. The cardinality of a set $X$ (denoted $\|X\|$) is the smallest ordinal which has a bijection onto $X$. An ordinal is a cardinal if it is the cardinality of some set. Every finite ordinal is a cardinal. However, no set has cardinality $\omega+1$. In fact, $\omega+1$ has a bijection onto $\omega$. $\omega$ can match to $0$, and then $0$ can match to $1$, $1$ can match to $2$, and so on. Every element of $\omega$ is, with this mapping, given exactly one unique element of $\omega+1$. So, $\|\omega+1\|=\|\omega\|=\omega$. The smallest cardinal is denoted $\aleph_0$. The next smallest cardinal is denoted $\aleph_1$. The next smallest cardinal is denoted $\aleph_2$. This pattern continues. The smallest cardinal larger than all $\aleph_n$ for finite $n$ is $\aleph_\omega$, and etcetera. (The alternative notation $\omega_\alpha$ is also used.) For ordinals, $\alpha,\beta,\gamma,$ and $\delta$ are often used as variables. For cardinals, $\kappa,\lambda,$ and $\mu$ are often used, though $\mu$ is used less often. The Intuition of Large Cardinal Axioms Large cardinal axioms may have some intuition behind them. For example: The correct cardinals $\kappa$ are those who have a specific set of cardinality $\kappa$ which acts almost exactly like the collection of all sets. The worldly cardinals $\kappa$ are those who have a specific set of cardinality $\kappa$ which acts like its own little universe or "world". The inaccessible cardinals $\kappa$ are those who cannot be reached from smaller cardinals by addition of cardinals or taking the cardinality of the powerset of ordinals. The Mahlo cardinals $\kappa$ are those who are inaccessible and have many, many, inaccessibles below them. The weakly compact cardinals $\kappa$ are those for which certain sets of information of cardinality $\kappa$ can be "compacted" to smaller sets of information and still retain the basic properties of the original set. The indescribable cardinals $\kappa$ are those which are so large that one begins to run out of interesting properties for them because for most properties of $\kappa$ there is a smaller ordinal $\alpha$ which has that property. The measurable cardinals $\kappa$ are those who have many very large subsets. The strongly compact cardinals $\kappa$ are those for which certain sets of information of cardinality at least$\kappa$ can be "compacted" to smaller sets of information and still retain the basic properties of the original set. Though it is not a cardinal property, Vopěnka's principle implies that in every collection which has no cardinality (i.e. it is too large for any cardinal), there are two elements which are incredibly similar to each other. It turns out that, although these axioms look intuitive, they cannot be proven to exist with standard mathematics. In fact, standard mathematics proves that they cannot be proven to exist. This is why they are axioms.
Let $C$ denote the positively oriented boundary of the half disk $0 \le r \le 1, 0 \le \theta \le \pi$, and let $f(z)$ be a continuous function defined on that half disk by writing $f(0) = 0$ and using the branch \begin{align} f(z) = \sqrt{r} e^{\frac{i \theta}{2}} \end{align} Show that \begin{align} \int_C f(z) \, dz = 0 \end{align} by evaulating separately the integrals of $f(z)$ over the semicircle and the two radii which make up $C$. Why does the Cauchy-Goursat theorem not apply here? Note: My only question is the last part of the problem statement, where I made the text boldface. (I have figured out everything else in this problem!) I believe the CG Theorem does not apply because the function is not analytic at some point in the half-circle.
I have a question about Galerkin method. I don't understand why the Galerkin method weights the residual by the shape functions and sets it equal to zero. I want to know what is reason of this. Why we must set weights residual functions equal to zero? When I studied the finite element method in graduate school, this notion of multiplying by a weight function was also very alien to me. Eventually, I did find a nice (albeit non-rigorous) analogy that helped me understand it. This analogy is based on 3D vector geometry and an understanding of projections and dot-product. 3D Geometry Imagine a 2D plane lying somewhere in 3D euclidean space. This plane can be though of as the span of two vectors $v_1$ and $v_2$. Thus, any vector $w$ in the plane can be written as a linear combination of these vectors; i.e. $w=c_1v_1+c_2v_2$ Now imagine a point $Q$ in 3D space that is not on the plane. Consider the question: Of all the points on the plane, which point is closest to $Q$? It is the one point $w$ (not shown in the image above) that lies on the line passing through the point $Q$ and perpendicular to the plane. The point $w$ is also known as the orthogonal projection of $Q$ onto the plane. Even though we don't know the coordinates of this point $w$, we do know that the vector between $Q$ and $w$ is perpendicular to all vectors that define the plane, i.e. $v_1$ and $v_2$. Perpendicular also means that the dot product is zero. If we denote the vector between $Q$ and $w$ as $\vec{Q}-\vec{w}$, then forcing $\vec{Q}-\vec{w}$ to be perpendicular to the plane also implies $$(\vec{Q}-\vec{w})\cdot v_1=0$$ and $$(\vec{Q}-\vec{w})\cdot v_2=0$$ This results in a system of equations which we can solve Also notice that to construct $\vec{Q}-\vec{w}$, we must know the coordinate of $Q$ Analogy for the Galerkin Method Let's assume that the solution $u_h$ is a finite linear combination of functions $N_1$,..,$N_k$; thus, $u_h=C_1N_1+...+C_kN_k$. This linear combination acts like the plane in the discussion above. Now, let's assume that there exists some exact solution $u$, which we don't know. This solution $u$ is like the point in 3D space which is not on the plane. In the galerkin method, we're looking for the solution in a space (plane) that is closest to the true solution (point not in the plane). In this sense, the "best solution" is the choice of $u_h$ that the difference $u-u_h$ is perpendicular to the space $u_h$. Note that for function spaces, the "dot-product" is defined by the integral of their product, i.e. $<u,v>=\int_\Omega u v$. So, perpedicular implies that the dot product between the $u-u_h$ and all those basis functions $N_1$, ..., $N_k$ must be zero, i.e. $$\int_\Omega (u-u_h)N_1 = 0$$ $$...$$ $$\int_\Omega (u-u_h)N_k = 0$$ Now, you may be saying to yourself, "This whole setup rests critically on the assumption that we know the exact solution $u$ ahead of time. But the truth is that we generally don't know $u$ a priori. In which case, how can we possibly compute $u-u_h$ in all these equations?" I'm so glad you asked! The truth is that we can't and don't evaluate $u-u_h$ directly. But we do know what $u$ and $u_h$ are supposed to satisfy; i.e. the PDE. Suppose your original PDE problem is $$-\nabla\cdot\left(k(x)\nabla u\right)=f$$ We could also rewrite this as $$Au=f$$ where the operator $A$ is defined by the expression $Au=-\nabla\cdot\left(k(x) \nabla u \right)$ So instead of considering the between the solutions $u-u_h$, we instead consider the absolute difference $Au - Au_h$ in all of the (perpendicular) equations. That is, consider not what $u$ and $u_h$ are, but rather what $u$ and $u_h$ satisfy instead. By replacing the absolute difference with the residual difference in the (perpendicular) equations above, we can write residual difference $$\int_\Omega (Au-Au_h)N_1 = 0$$ $$...$$ $$\int_\Omega (Au-Au_h)N_k = 0$$ Again, we still don't know what $u$ is, so this may not seem very helpful. But in fact, we can replace $Au$ with the known source term $f$ (since $Au=f$). Thus, we obtain the equations $$\int_\Omega (f-Au_h)N_1 = 0$$ $$...$$ $$\int_\Omega (f-Au_h)N_k = 0$$ Thus, enforcing the residual to be orthogonal to the given space results in a system of equations that one can solve for the coefficients $C_1,...,C_k$. Summary The explanation above is a rough "analogy". I haven't really derived anything or given a reasonable proof that $u-u_h$ can be replaced by $Au-Au_h$ and still produce a close approximation. I've also not explained anything about obtaining a weak form of the PDE or how to choose the spaces where $N_1...N_k$ lie in. But the whole idea behind galerkin method as a projection is that for all possible linear combinations of functions in a given (finite dimensional) space (span of $N_1$,...,$N_k$), we are looking for the one that is closest to a solution (point) which generally lies outside of the given space. Closest means that we're looking for the orthogonal projection from the true solution to the given space. If we don't know what the exact solution $u$ is, then we cannot use the absolute difference as a metric in our projection. Thus, we are forced to use the residual difference as the metric of our projection; In other words: The galerkin projection is not about what $u$ is... it's about what $u$ satisfies. Let's say you want to solve the Laplace equation, $-\Delta u = f$. Ideally, of course, you'd like to find a function $u$ so that the residual is zero: $r(u) = -\Delta u - f = 0$. But $u$ is an infinite dimensional object which in general we cannot represent on computers, so we have to find finite dimensional approximations $u_h$. Since $u_h$ is not the exact solution, we cannot expect that $r(u_h)=0$. The question is which set of equations we want $u_h$ to satisfy instead. The Galerkin method chooses $u_h=\sum_i U_i \varphi_i$ to be a linear combination of $N$ shape functions $\varphi_i$ and then determines the coefficients $U_i$ by requiring the residual to satisfy the set of $N$ equations $\left<\varphi_i,r(u_h)\right>=0$. But there are other choices that are possible. For example, the Petrov-Galerkin method requires that $\left<\psi_i,r(u_h)\right>=0$ where the test functions $\psi_i$ are a set of $N$ weighting functions separate from the trial functions $\varphi_i$. I have a bit more material on this issue in lecture 4 at http://www.math.tamu.edu/~bangerth/videos.html . Boris Grigoryevich wants you not to be able to create residuals with the same functions you used to create the solution. While this question is old and has been answered by plenty of smart people, I just want to jot down the intuition I use to explain the Galerkin method to people. The goal in our situation is to find as close of a solution as we can to some continuous residual equation: $$r(u) = 0$$. Let us define the $i^{th}$ basis function as $\phi_{i}(x)$, define the approximate solution as $u_{h} = \sum_{i}^{n} a_{i} \phi_{i}(x)$, and define the residual as $r(u_{h}(x))$, the Galerkin projection ends up being: $$ \int_{\Omega} r(u_{h}(x)) \phi_i(x) dx = 0 \;\;\; \forall i$$ This integral expression can be viewed as an inner product written as: $$ \left< r(u_{h}), \phi_i\right> = 0 \;\;\; \forall i$$ From the perspective of this inner product, the Galerkin Projection forces the residual error to be orthogonal to the choice of basis functions. So while there might be true error associated with using a lower dimensional representation of the solution, the Galerkin Projection aims to minimize the error component associated with the chosen basis. Remember that when you multiply the strong-form equation by the shape function, the shape function is arbitrary. Therefore, by requiring that the residual be orthogonal to any such shape function, such a residual is in fact very close to zero. This is not the same as requiring that the residual be zero exactly, but rather a somewhat weakened requirement that the discrete solution can satisfy.
I have to deal with a situation where I am trying to decompose numbers written as $2K^2, K\ge0$ into sums of squares, i.e. find: $$S_K=\{(x, y)\ \|\ x^2+y^2 = 2K^2\}$$ One of the obvious solutions is $(K, K)$. I have looked into several papers and found a number of algorithms pertaining to this. Basically the solution seems to be to decompose $K$ into its prime factors, decompose said factors as sums of two squares and apply the Brahmagupta-Fibonacci identity to obtain the decomposition, as in this answer. However, the fact that the only primes (aside from 2) that can be decomposed that way are the primes written as $4k+1$, it lead me to the following question: Given the unique decomposition $$K = P*Q\\ P=\prod_{i=1}^{m_1}p_i^{\alpha_i}\\ Q=\prod_{j=1}^{m_2}q_j^{\beta_j}$$ Where $p_i$ and $q_j$ are all primes such as $p_i \equiv 1\ [4]$ and $q_j \equiv -1\ [4]$, and $\alpha_i, \beta_j\ge1$. Can we assert that $S_K = \{(P^2x',P^2y')\ \|\ x'^2+y'^2=2Q^2\}$ (reduced problem)? Or can this miss any decompositions? One inclusion seems right, but I'd like to be sure that this generates all the possible solutions.
Since you already have a method of constructing the needed circle $C$, I assume your question is about uniqueness. Here is an informal analytic approach that could be formalized. Let $M$ be the center of $D$, and $N$ be the center of $C$. Since $C$ passes through $P$ and $Q$, $N$ must lie on the line $\ell = b_{PQ}$. Let $X$ be the center of the center of the circle through $P$ and $Q$ which is tangent to $D$ on opposite side of $\overline{PQ}$ from $M$, and let $Y$ be the center of the circle which is tangent to $D$ on the same side of $\overline{PQ}$ from $M$. Consider the direction from $X$ to $Y$ to be up. Let $Z$ be a point on $\ell$ and consider the circle centered at $Z$ passing through $P$ and $Q$, and the angle formed by the intersections of that circle with $D$ and $M$. Let $\theta$ be the measure on the upper side of that angle. The intersection points will be antipodal if and only if $\theta = \pi$. Let $\theta_0$ be the measure of the upper side of $\angle SMT$ (where, as in the OP, $S$ and $M$ are the intersections of the line through $P$ and $Q$ with $D$.) Because the midpoint of $\overline {PQ}$ is below $M$, $\theta_0 > \pi$. When $Z = X, \theta = 2\pi$. As $Z$ descends, the intersection points move up the sides of $D$, with $\theta$ decreasing. As $Z$ goes to $-\infty$, the intersection points approach $S$ and $T$, and $\theta \to \theta_0$ from above. Since $\theta > \theta_0 >\pi$, No circle in this region has antipodal intersection points. When $Z$ is between $X$ and $Y$, its circle does not intersect $D$. When $Z = Y, \theta = 0$, and as $Z$ rises, the intersection points descend the sides of $D$ and $\theta$ increases continuously. As $Z \to \infty, \theta \to \theta_0-$, so at some point $\theta$ must rise past $\pi$. By the intermediate value theorem, there is a point where it equals $\pi$. But since $\theta$ is strictly increasing in $Z$, it cannot be $\pi$ at more than a single point.
Bug introduced in or after 10.3, persisting through 11.2. I'm trying to solve following PDE (heat equation): $$ \begin{cases} u_t = a \, u_{xx} \\ u(x,0) = 0\\ \lim_{x\to \infty}u(x,t) =0\\ \alpha\, (\theta_0-u(0,t))+\dot{q}_0=-\lambda u_x(0,t) \end{cases}$$ Where basically I have an initial temperature of $0$ everywhere, a constant heat flux at the beginning of the rod, and convection between air and the rod at its beginning ($\theta_0$ is the air temperature which is assumed to be constant). I found following analytical solution for the problem: $$u(x,t) = \frac{\dot{q}_0+\alpha \, \theta_0}{\alpha}\left[ \mathrm{erfc}\, \left(\frac{x}{2\sqrt{a \, t}} \right) -\mathrm{exp}\,\left(\frac{\alpha}{\lambda}x+a \frac{\alpha^2}{\lambda^2}t \right)\mathrm{erfc}\,\left(\frac{x}{2\sqrt{a\, t}} + \frac{\alpha}{\lambda} \sqrt{a\, t} \right) \right] $$ which is physically meaningful. With mathematica, however, I get some meaningless results, probably due to my not that good mathematica skills. This is what I tried to do: λ = 0.8; c = 880; ρ = 1950; a = λ/(c ρ)α = 15; θair = 0;heqn = D[u[x, t], t] == a D[u[x, t], {x, 2}];ic1 = u[x, 0] == 0;bc1 = α (θair - u[0, t]) + 650 == -λ Derivative[1, 0][u][0, t];sol = DSolveValue[{heqn, ic1, bc1}, u[x, t], {x, t}][[1, 1, 1]] Which leads me to a complex solution (plotted below) DensityPlot[sol, {t, 0, 2*3600}, {x, 0, 0.1}, PlotLegends -> Automatic, FrameLabel -> {t[s], x[m]}] Plot[Evaluate[Table[sol, {t, 3600, 7200, 3600}]], {x, 0, .1}, PlotRange -> All, Filling -> Axis, Axes -> True, AxesLabel -> {x[m], θ[°C]}] I'm aware of the fact that I didn't consider the condition at infinity. To do this, I tried to follow this answer without success. Also, mathematica finds the solution without this condition as soon as $\alpha = 0$. This is the plot of the analytical solution I get: Plot[{u[x, 600], u[x, 3600], u[x, 7200]}, {x, 0, .2}, Filling -> Axis, Axes -> True, AxesLabel -> {x[m], θ[°C]}]
ECE662: Statistical Pattern Recognition and Decision Making Processes Spring 2008, Prof. Boutin Collectively created by the students in the class Contents Lecture 10 Lecture notes The perceptron algorithm maps an input to a single binary output value. For a proof of the Perceptron convergence theorem, see this page: First introduced in Lecture 9. The gradient descent algorithm used is discussed in this lecture. Gradient Descent Main article: Gradient Descent Consider the cost function $ J_p(\vec{c}) = \sum -\vec{c}y_i $, where $ y_i $ is the misclassified data. We use the gradient descent procedure to minimize $ J_p(\vec{c}) $. Compute $ \nabla J_p(\vec{c}) = ... = - \sum y_i $. Follow basic gradient descent procedure: - Initial guess $ \vec{c_1} $ - Then, update $ \vec{c_2} = \vec{c_1} - \eta(1) \nabla J_p(\vec{c}) $, where $ \eta(1) $ is the step size - Iterate $ \vec{c_{k+1}} = \vec{c_{k}} - \eta(k) \nabla J_p(\vec{c}) $until it "converges" ( e.g when $ \eta(k) \nabla J_p(\vec{c}) $< threshold ) Gradient Descent in the Perceptron Algorithm Theorem: If samples are linearly separable, then the "batch perceptron" iterative algorithm. The proof of this theorem, Perceptron_Convergence_Theorem , is due to Novikoff (1962). $ \vec{c_{k+1}} = \vec{c_k} + cst \sum y_i $, where $ y_i $ is the misclassified data, terminates after a finite number of steps. But, in practice, we do not have linear separable data. So instead, we use the Least Squares Procedure. We want $ \vec{c} \cdot y_i > 0 $, for all samples $ y_i $. This is a linear inequality problem which is usually hard to solve. Therefore, we need to convert this problem into a linear equality problem. We choose $ b_i $ > 0 and solve $ \vec{c} \cdot y_i = b_i $, for all i The matrix equation has the following form: This can also be written as $ \vec{Y} \cdot \vec{c} = \vec{b} $ If d=n, and $ \vec{y_1} $,..., $ \vec{y_d} $ are "generic" ( i.e. determinant of $ \vec{Y} $ is not 0), then we "can" solve by matrix inversion. If d > n, over-constrained system (there is no solution in the generic case). This is the case where there is more data than you need, and the information is contradictory. In this case, we seek to minimize $ \|\| Y \vec{c} - \vec{b} \|\|_{L_2} $. The solution is given by $ \vec{c} = (Y^{\top}Y)^{-1}Y^{\top}b $, if $ \|Y^{\top}y\| \ne 0 $. If $ \|Y^{\top}y\| = 0 $, $ \vec{c} = lim (Y^{\top}Y + \epsilon1)^{-1}Y^{\top}b $ always exists! Fischer's Linear Discriminant Main article: Fisher Linear Discriminant Fischer's Linear Discriminant solves a dual problem: Traditionally, we have defined a separating hyperplane. Fischer's linear discriminant defines a projection which reduced the data to a single dimension. Fischer's Linear Discriminant optimizes the between class-spread.
Contents If you are old enough you might remember a film called Flubber released in 1997. The film depicts small jelly like characters that were digitally modelled, animated and rendered. The jelly/blobby look of the characters is something you can sort of easily produce in CGI using a technique called metaballs. To say things quickly (we will devote a full lesson to this fun technique some day), this technique was proposed by guru Jim Blinn in 1984. Metaballs go by several other names: soft objects or blobbies for example, though soft objects and metaballs use slightly different equations, so the names are not exactly synonymous but more on that in the future. Why are we interested in soft objects, blobbies or metaballs? Because they are essentially equation-based distance fields (well sort of but we will explain that in a moment), and as you can guess since they are equation-based, they can somehow be rendered using the sphere-tracing method. Furthermore, in the second chapter of this lesson, we spent a great deal of time explaining what is the Lipschitz constant and how this constant could be used to develop distance estimators, however we haven't studied yet a practical example in which this method would be used. Soft objects will give us an opportunity to fill that gap. Finally, metaballs are great fun... Blobbies (rendered as solid objects) just look like simple spheres. It's only when they get close to each other that they start to blend like drops of liquid. The reason for that is quite simple. A metatball is defined by some function that defines a density (like for a gas) that gradually falls off from the center of the blobby. The falloff depends on the equation that is being used, but is generally exponential. When you put two blobbies close to each other, then their densities add up. These ideas are illustrated in the following image. On the left, you have a single blobby. In the middle you have two blobbies. The cyan curves above the blobbies is the profile of the blobbies' density. In the case of the implicit surfaces we studied so far, the isosurface marked the surface to be rendered. But in the case of blobbies there is no isosurface since we deal this time with densities. So how do we go from a gas to a solid so to say? The idea with blobbies is to say that the isosurface this time is not defined by a distance from a point in space to the closest point on the surface of the object, but is defined by all the points in space that have the same density. In the following you can see on the left, you can see what this isoline (in 2D) looks like when we set to 0.4. All points in the image whose density is 0.4 belong or define this isoline (in red). By changing the value, you can modify the profile of the resulting shape. Note that when the threshold of the isoline is low, the two blobbies seem to touch each other. When the threshold is hight (the density values get higher as you get closer to the center of the blobby), they look like two individual drops or spheres. Quite a few different functions to define the profile of metaballs or blobbies have been invented over time. The one that we are going to use in this lesson is the one used by Hart in his paper. The function is:$$F(r) = 2 \dfrac{r^3}{R^3} - 3 \dfrac{r^2}{R^2} + 1.$$ Where $r$ is the distance from a point in space to the center of the blobby, and $R$ is the radius of the blobby itself. Note that the density drops to 0 for any value of $r$ greater than $R$ (figure 2). This is important as we will this property to optimize the rendering process later. The profile of this function can be seen in figure 1 (with $R=1$). The function $F(x)$ returns a density but our sphere-tracing algorithm requires a distance. So how do we from a density to a distance? In a way you can the see the function that returns the distance of a point to the isosurface of an implicit object as a curve that converges to 0 as we get closer to the isosurface (and gets negative once we have passed on the other side of the object's surface). This idea is illustrated in figure 3. As we get closer to the surface, the distance to the actual surface becomes smaller and smaller. This is what happens when our traced-spheres becomes smaller as we approach the surface of an object. We can achieve the same effect with the profile of the blobby density function if subtract some number (that we will call our magic number) from the inverse of that function. You can see what the curve now looks like. As you can see it gets flats for any greater than the blobby maximum radius $R$. This part of the curve is also positive (the blue section), which means that each time we will take a step forward in the direction of the shape, the size of the step will be constant as long as the distance to the blobby center is greater than $r_A$. From $r_A$ to $r_B$ the curve goes down: this when the distance to the blobby surface will start to decrease until it eventually reaches a very small value (close to 0). When we will reach that point, we will have found the point where our ray intersects the surface of the blobby. Unfortunately, it turns out that this simple equation alone can't do the truck, but the final equation will produce a similar shape so hopefully this basic introduction will have helped you to understand the general idea. So why isn't that simple? If chapter 2 we mentioned that Hart had proposed in his paper a solution to develop any "conservative" distance estimation function (what he called a DUF) from any implicit equation. This solution involved to compute the Lipschitz constant of the equation, which we explained requires to compute the function second derivative, then solve the solution to $F''(x) = 0$ and insert this soliton into the function first derivative $F'(x)$. This is how we find the requires Lipschitz constant. The full derivation is given in chapter 2 of this lesson. The DUF for the function can be found by dividing the function itself by the function's Lipschitz constant. We didn't need to use this method to find distance estimators for simple shapes such as a sphere, a curve, or a torus, but we will need to use it for blobbies because there is no solution in this case that can be found on simple geometry deduction. As you can guess, what we have to do now is follow these steps. The function first-order derivative is:$$F'(r) = 6 \dfrac{r^2}{R^3} - 6 \dfrac{r}{R^2}.$$ From which we can derive the function's second-order derivative:$$F''(r) = 12 \dfrac{r}{R^3} - \dfrac{6}{R^2}.$$ Solving for $r$, the solution is:$$ \begin{array}{l} 12 \dfrac{r}{R^3} - \dfrac{6}{R^2} = 0,\\ 12 \dfrac{r}{R^3} = \dfrac{6}{R^2},\\ r = \dfrac{6R^3}{12R^2},\\ r = \dfrac{R}{2}.\\ \end{array} $$ If we now compute $F'(x)$ using the solution we get:$$ \begin{array}{l} F'(\dfrac{R}{2}) = 6 \dfrac{\left( \dfrac{R}{2} \right) ^2}{R^3} - 6 \dfrac{\dfrac{R}{2}}{R^2},\\ F'(\dfrac{R}{2}) = \dfrac{6R^2}{4R^3} - \dfrac{6R}{2R^2},\\ F'(\dfrac{R}{2}) = \dfrac{6}{4R} - \dfrac{6}{2R},\\ F'(\dfrac{R}{2}) = \dfrac{3}{2R} - \dfrac{3}{R},\\ F'(\dfrac{R}{2}) = \dfrac{3}{2R} - \dfrac{6}{2R} = - \dfrac{3}{2R}.\\ \end{array} $$ This is our Lipschitz constant from which we derive our final DUF function:$$d(x, \text{soft object}) = \dfrac {F(r)} {F'(r)} = \dfrac {F(r)} { \dfrac{3}{2R} } = \dfrac {2R} {3} F(x).$$ Where $r$ is equal to $\|\|x - C\|\|$ (the distance between $x$ and the blobby center). The final step is to find an equation that works with more than one blobby (since the point of using metaballs or blobbies is to create interesting liquid-like shapes by creating aggregate of blobbies). The classic solution consists of accumulating their contribution by summing up their density fields. We will do the same for our DUF function. Mathematics tells us that "the Lipschitz constant of a sum is bounded by the sum of the Lipschitz constants", which leads us to our final equation:$$d(x, \text{soft object}) = \dfrac{magic - \sum_{n=0}^{n=i} F(\|\| x - C_i\|\|)} {\sum_{n=0}^{n=i} { \dfrac {3}{2} R_i } }.$$ Where $magic$ is our threshold, $C_i$ are the blobbies centers, and $R_i$ their radii. The follow image shows the process of tracing a ray towards the surface made by two blobbies. You can see that we progress towards the surface of the soft object by regular intervals (the green part of our distance function's curve), and that at some point the circles get smaller and smaller until we eventually reach the surface. Implementation The implementation of soft-object in our simple sphere-tracer is straightforward. We just create a new class derived from the ImplicitShape base class and overwrite the getDistance() method with our equation for blobbies. A few remarks can be made. First notice that a soft object is made of several blobbies in our implementation. In fact you can create as many as you want and store them in the blobbies member variable. In our particular implementation, we created them in the class constructor. Note also that in the getDistance() method we only accumulate the contribution of a blobby if the distance from the point where we evaluate the current blobby's density field and the center of that blobby is lower than the blobby's radius (line 22). Finally if you look at the image above that shows the different spheres that are being traced until we reach the point of intersection with the soft-object's isosurface, you will notice that a great number of spheres are being traced before we can get to that point. This is because, as showed in figure 3, our distance estimation function returns a constant value for all distances of $r$ that are greater than the blobby's radius. And that constant value is rather small. Since we know that the isosurface is contained within the envelope of the individual blobbies, we can accelerate the process by treating the blobbies as a collection of spheres and chose the maximum value between the minimum distance from the point to the spheres and the point to the actual isosurface (line 31). If you simply return the distance to the isosurface, you will notice that the render gets much slower. Here a few results. On the left, just two blobbies and on the right an aggregate of about twenty blobbies. Note that there the transition between the blobbies doesn't seem to be very smooth. This can be caused by the fact that the function that we chose doesn't have second-order derivative continuity (we have already mentioned this problem in the lesson on Perlin's Noise function). Conclusion Hart's paper contains many more interesting techniques and ideas but we can't look at them all in this lesson. Maybe we will come back to it later and complete it if there is an interest from readers. We encourage you to read the paper or other articles on the topic of implicit modelling and try to implement some of the techniques we haven't studied in this lesson yourself. At least now you should have the basic knowledge required to explore this topic further. Some ideas you can try: add colors to the implicit surfaces and try to get them to blend when you use blending or mixing or soft-objects (add a random color to each blobby and see what happens in the blending regions). You can render more shapes, theres is even a method to compute triangles and quads. You can experiment with deforming methods (twisting, bending, etc.). The field of experimentation and exploration is limited but very wild in the case of implicit modeling. We will write another lesson on the topic but that will be devoted this time specifically to Metaballs (the original method proposed by Blinn) and show how to polygonize (create an object) out of the distance field.
Shrewd (All information from [1]) Shrewd cardinals are a generalisation of indescribable cardinals. They are called shrewd because they are bigger in size than many large cardinals which much greater consistency strength (for all notions of large cardinal which do not make reference to the totality of all ordinals). Definitions $κ$ — cardinal, $η>0$ — ordinal, $\mathcal{A}$ — class. $κ$ is $η$-shrewd iff for all $X ⊆ V_κ$ and for every formula $\phi(x_1, x_2)$, if $V_{κ+η} \models \phi(X, κ)$, then $\exists_{0 < κ_0, η_0 < κ} V_{κ_0+η_0} \models \phi(X ∩ V_{κ_0}, κ_0)$. $κ$ is shrewd iff $κ$ is $η$-shrewd for every $η > 0$. $κ$ is $\mathcal{A}$-$η$-shrewd iff for all $X ⊆ V_κ$ and for every formula $\phi(x_1, x_2)$, if $\langle V_{κ+η}, \mathcal{A} ∩ V_{κ+η} \rangle \models \phi(X, κ)$, then $\exists_{0 < κ_0, η_0 < κ} \langle V_{κ_0+η_0}, \mathcal{A} ∩ V_{κ_0+η_0} \rangle \models \phi(X ∩ V_{κ_0}, κ_0)$. $κ$ is $\mathcal{A}$-shrewd iff $κ$ is $\mathcal{A}$-$η$-shrewd for every $η > 0$. One can also use a collection of formulae $\mathcal{F}$ and make $\phi$ an $\mathcal{F}$-formula to define $η$-$\mathcal{F}$-shrewd and $\mathcal{A}$-$η$-$\mathcal{F}$-shrewd cardinals. Properties If $κ$ is $\mathcal{A}$-$δ$-shrewd and $0 < η < δ$, then $κ$ is $\mathcal{A}$-$η$-shrewd. This is a difference between the properties of shrewdness and indescribability. For subtle $\pi$, for every class $\mathcal{A}$, in every club $B ⊆ π$ there is $κ$ such that $\langle V_\pi, \mathcal{A} ∩ V_\pi \rangle \models \text{“$κ$ is $\mathcal{A}$-shrewd .”}$. (The set of cardinals $κ$ below $\pi$ that are $\mathcal{A}$-shrewd in $V_\pi$ is stationary.) there is an $\eta$-shrewd cardinal below $\pi$ for all $\eta < \pi$. ReferencesMain library
Possible Duplicate: Series converges implies $\lim{n a_n} = 0$ Someone can help me? If $(a_n)$ is a decreasing sequence and $\sum a_n$ converges. Then $\lim {(n.a_n)} = 0$. I don't have idea how to solve this. Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It only takes a minute to sign up.Sign up to join this community Possible Duplicate: Series converges implies $\lim{n a_n} = 0$ Someone can help me? If $(a_n)$ is a decreasing sequence and $\sum a_n$ converges. Then $\lim {(n.a_n)} = 0$. I don't have idea how to solve this. By the Cauchy condensation test $$\sum 2^m a_{2^m} < \infty$$ thus $$\lim_n 2^m a_{2^m} =0$$ Now, for each $n$ chose some $m$ so that $2^m \leq n < 2^{m+1}$ and use $$a_{2^m} \geq a_n \geq a_{2^{m+1}}$$
Using method of Lagrange multipliers, I am looking to minimise the function $f(x_1,...,x_n)=\prod_{i=1}^n(1+x_i)$ with the side condition $g(x_1,...,x_n)=\prod_{i=1}^nx_i-q^n=0$. My goal is to show that $f$ is minimal when $x_i=q$ for all $i$. I have $$\nabla g=(\prod_{i=1, i \neq j}^nx_i)_{1\leq j \leq n}$$ $$\nabla f=(\prod_{i=1, i \neq j}^n(1+x_i))_{1\leq j \leq n}$$ Now I know that $\nabla f = \lambda \nabla g$, so for all $j$: $$\prod_{i=1, i \neq j}^n(1+x_i)=\lambda\prod_{i=1, i \neq j}^nx_i$$ and since all $x_i>0$ $$\prod_{i=1, i \neq j}^n(1+\frac{1}{x_i})=\lambda$$ Since the products are equal for all $j$, it follows that the $x_1=x_2=...=x_n$ and thus $x_1...x_n=x_1^n=q^n \iff x_i=q$ for all $i$. So far so good. My problem is to show that it is a minimum: The entries of the Hessian are $$\partial_{x_i} ^2f=0$$ $$\partial_{x_j} \partial_{x_k}f=\prod_{i=1, i \neq j,i \neq k}^n(1+x_i)>0$$ So I have a matrix with zero diagonal and positive entries everywhere else. This matrix is not positive definite, as is easily seen in the $2 \times 2$ case. Did I do any mistakes? How do I argue that it is a minimum?
X Search Filters Format Subjects Library Location Language Publication Date Click on a bar to filter by decade Slide to change publication date range 1. Study of the \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$B_c^+ \rightarrow J/\psi D_s^+$$\end{document}Bc+→J/ψDs+ and \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$B_c^+ \rightarrow J/\psi D_s^{+}$$\end{document}Bc+→J/ψDs∗+ decays with the ATLAS detector The European Physical Journal. C, Particles and Fields, ISSN 1434-6044, 2016, Volume 76 The decays \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy}... Regular - Experimental Physics Regular - Experimental Physics Journal Article 2. (±)‐Hippolide J – A Pair of Unusual Antifungal Enantiomeric Sesterterpenoids from the Marine Sponge Hippospongia lachne European Journal of Organic Chemistry, ISSN 1434-193X, 06/2017, Volume 2017, Issue 24, pp. 3421 - 3426 A pair of enantiomeric sesterterpenoids, (±)‐hippolide J ( 1a and 1b ), were isolated from the marine sponge Hippospongia lachne collected from the South China... Marine sponges \| Natural products \| Terpenoids \| Antifungal agents \| Chiral separation \| SULFATES \| METABOLITES \| METACHROMIA \| MANOALIDE \| NATURAL-PRODUCTS \| CHEMISTRY, ORGANIC \| OKINAWAN SPONGE \| DITERPENOIDS \| INHIBITORS \| DERIVATIVES \| Enantiomers \| Analysis \| Fungi \| Separation \| Nuclear magnetic resonance--NMR \| Fungicides \| Circularity \| Dichroism \| Spectroscopic analysis \| Inhibition \| Functional groups \| Fragmentation Marine sponges \| Natural products \| Terpenoids \| Antifungal agents \| Chiral separation \| SULFATES \| METABOLITES \| METACHROMIA \| MANOALIDE \| NATURAL-PRODUCTS \| CHEMISTRY, ORGANIC \| OKINAWAN SPONGE \| DITERPENOIDS \| INHIBITORS \| DERIVATIVES \| Enantiomers \| Analysis \| Fungi \| Separation \| Nuclear magnetic resonance--NMR \| Fungicides \| Circularity \| Dichroism \| Spectroscopic analysis \| Inhibition \| Functional groups \| Fragmentation Journal Article Physics Letters B, ISSN 0370-2693, 12/2015, Volume 751, Issue C, pp. 63 - 80 An observation of the decay and a comparison of its branching fraction with that of the decay has been made with the ATLAS detector in proton–proton collisions... PARTICLE ACCELERATORS PARTICLE ACCELERATORS Journal Article 4. Long-subchain Janus-dendritic copolymers from locally confined click reaction and generation-dependent micro-phase separationElectronic supplementary information (ESI) available: The data of element analysis, 1H-NMR spectrum of long-chain Janus-dendritic PSt/PtBA with full scan, GPC traces of J-(PSt)2-N3/(PtBA)3-Br and J-(PSt)3-N3/(PtBA)4-Br, GPC-MALLS diagrams of J-(PSt)i/(PtBA)j (i = j), AFM images of J-(PSt)3/(PtBA)3 for different annealing durations. See DOI: 10.1039/c7py00551b ISSN 1759-9954, 7/2017, Volume 8, Issue 26, pp. 3889 - 39 In this study, we put forward a divergent synthetic strategy to prepare long-chain Janus-dendritic copolymers with polystyrene (PSt) and poly( tert -butyl... Journal Article 5. J/ψ production cross section and its dependence on charged-particle multiplicity in p + p collisions at s=200 GeV Physics Letters B, ISSN 0370-2693, 11/2018, Volume 786, pp. 87 - 93 We present a measurement of inclusive production at mid-rapidity ( ) in collisions at a center-of-mass energy of GeV with the STAR experiment at the... [formula omitted] collisions \| Charged-particle multiplicity \| Quarkonium \| Multiple parton interactions \| p+p collisions \| NUCLEAR PHYSICS AND RADIATION PHYSICS \| quarkonium \| charged-particle multiplicity \| multiple parton interactions [formula omitted] collisions \| Charged-particle multiplicity \| Quarkonium \| Multiple parton interactions \| p+p collisions \| NUCLEAR PHYSICS AND RADIATION PHYSICS \| quarkonium \| charged-particle multiplicity \| multiple parton interactions Journal Article PHYSICAL REVIEW LETTERS, ISSN 0031-9007, 03/2015, Volume 114, Issue 12, p. 121801 A search for the decays of the Higgs and Z bosons to J/psi gamma and Upsilon(nS)gamma (n = 1,2,3) is performed with pp collision data samples corresponding to... PHYSICS, MULTIDISCIPLINARY \| PSI \| PHYSICS \| Physics - High Energy Physics - Experiment \| Physics \| High Energy Physics - Experiment \| Fysik \| Physical Sciences \| Naturvetenskap \| Natural Sciences PHYSICS, MULTIDISCIPLINARY \| PSI \| PHYSICS \| Physics - High Energy Physics - Experiment \| Physics \| High Energy Physics - Experiment \| Fysik \| Physical Sciences \| Naturvetenskap \| Natural Sciences Journal Article 7. Measurement of the prompt J/ψ pair production cross-section in pp collisions at √s = 8 TeV with the ATLAS detector The European Physical Journal C: Particles and Fields, ISSN 1434-6052, 2017, Volume 77, Issue 2, pp. 1 - 34 Journal Article 8. Prompt and non-prompt $$J/\psi $$ J/ψ elliptic flow in Pb+Pb collisions at $$\sqrt{s_{_\text {NN}}} = 5.02$$ sNN=5.02 Tev with the ATLAS detector The European Physical Journal C, ISSN 1434-6044, 9/2018, Volume 78, Issue 9, pp. 1 - 23 The elliptic flow of prompt and non-prompt $$J/\psi $$ J/ψ was measured in the dimuon decay channel in Pb+Pb collisions at $$\sqrt{s_{_\text {NN}}}=5.02$$... Nuclear Physics, Heavy Ions, Hadrons \| Measurement Science and Instrumentation \| Nuclear Energy \| Quantum Field Theories, String Theory \| Physics \| Elementary Particles, Quantum Field Theory \| Astronomy, Astrophysics and Cosmology \| PHYSICS OF ELEMENTARY PARTICLES AND FIELDS Nuclear Physics, Heavy Ions, Hadrons \| Measurement Science and Instrumentation \| Nuclear Energy \| Quantum Field Theories, String Theory \| Physics \| Elementary Particles, Quantum Field Theory \| Astronomy, Astrophysics and Cosmology \| PHYSICS OF ELEMENTARY PARTICLES AND FIELDS Journal Article Physical Review Letters, ISSN 0031-9007, 03/2012, Volume 108, Issue 11, p. 112003 A partial wave analysis of the pp¯ mass-threshold enhancement in the reaction J/ψ→γpp¯ is used to determine its J(PC) quantum numbers to be 0(-+), its peak... Journal Article 10. Confirmation of the X(1835) and observation of the resonances X(2120) and X(2370) in J/ψ→γπ + π - η Physical Review Letters, ISSN 0031-9007, 02/2011, Volume 106, Issue 7, p. 072002 With a sample of (225.2±2.8)×10(6) J/ψ events registered in the BESIII detector, J/ψ→γπ(+)π(-)η(') is studied using two η(') decay modes: η(')→π(+)π(-)η and... Journal Article 11. Measurement of the differential cross-sections of inclusive, prompt and non-prompt J / ψ production in proton–proton collisions at s = 7 TeV Nuclear Physics, Section B, ISSN 0550-3213, 2011, Volume 850, Issue 3, pp. 387 - 444 The inclusive production cross-section and fraction of mesons produced in -hadron decays are measured in proton–proton collisions at with the ATLAS detector at... Journal Article 12. Prompt and non-prompt $$J/\psi $$ J/ψ and $$\psi (2\mathrm {S})$$ ψ(2S) suppression at high transverse momentum in $$5.02~\mathrm {TeV}$$ 5.02TeV Pb+Pb collisions with the ATLAS experiment The European Physical Journal C, ISSN 1434-6044, 9/2018, Volume 78, Issue 9, pp. 1 - 28 A measurement of $$J/\psi $$ J/ψ and $$\psi (2\mathrm {S})$$ ψ(2S) production is presented. It is based on a data sample from Pb+Pb collisions at... Nuclear Physics, Heavy Ions, Hadrons \| Measurement Science and Instrumentation \| Nuclear Energy \| Quantum Field Theories, String Theory \| Physics \| Elementary Particles, Quantum Field Theory \| Astronomy, Astrophysics and Cosmology Nuclear Physics, Heavy Ions, Hadrons \| Measurement Science and Instrumentation \| Nuclear Energy \| Quantum Field Theories, String Theory \| Physics \| Elementary Particles, Quantum Field Theory \| Astronomy, Astrophysics and Cosmology Journal Article
My book is An Introduction to Manifolds by Loring W. Tu. The following is an entire subsection (Subsection 22.5) of the section that introduces manifolds with boundary (Section 22, Manifolds with Boundary). Note: I believe that all manifolds with or without boundary referred in this subsection have unique dimensions by some convention (either it's implicit, or it's explicit a I missed it) in the section, contrary to the convention of the book (See here and here). According to an errata by Ehssan Khanmohammadi, the only erratum to be made in this subsection is that $c((0,\varepsilon[) \subset M^\circ$ should be changed to $c(]0,\varepsilon[) \subset M^\circ$. I still have several concerns about this subsection. What exactly is a vector field along$\partial M$, and what is its domain? Asked here. For the local expression of $X$, a vector field along $\partial M$ is the following understanding correct? 2.1 Let $p$ be in the domain of $X$, which is $\partial M$ (see question 1). View $p$ as an element of $M$, which we can do because $\partial M \subseteq M$. We view $p$ as such to obtain a coordinate neighborhood $(U,\varphi) = (U,x^1, ..., x^n)$ of $p$ in $M$ to obtain a local expression for $X$ in $(U, \varphi)$ specifically in $(U \cap \partial M, \varphi\|_{U \cap \partial M})$. 2.2 In the expression $$X_q = \sum_{i=1}^n a_i(q) \frac{\partial}{\partial x^i}\|_q, q \in \partial M,$$ 2.2.1. this is a local expression so "$q \in \partial M$" means "$q \in \partial M \cap U_p$", like in the proof that $\partial M$ is a manifold (without boundary). 2.2.2. the $a_i$'s are $a_i: U \cap \partial M \to \mathbb R$, functions on $U \cap \partial M$ rather than functions on $U$, but I'm not quite sure of this based on the "$a^i$ on $\partial M$" in the smoothness (see question 3) 2.2.3. the "$x_i$"'s are actually the $x_i\|_{U \cap \partial M}$'s from $(U \cap \partial M, \varphi\|_{U \cap \partial M}) = (U \cap \partial M, x^1\|_{U \cap \partial M}, ..., x^n\|_{U \cap \partial M})$, i.e. they are restrictions of the original $x_i$'s rather than the original $x_i$'s themselves, though there is a convention in this book to omit indicating restrictions (see here, here and here). Is this a correct understanding of the smoothness definition? Let $M$ be a manifold with boundary $\partial M$. Let $X$ be a vector field along $\partial M$. For each $p$ in the domain of $X$, which is $\partial M$ (see question 1), $X$ has a local expression at $p$: For any coordinate neighborhood $(U, \varphi) = (U, x^1, ..., x^n)$ of $p$ in $M$, we have a coordinate neighborhood of $p$ in $\partial M$ namely the restriction of $(U, \varphi) = (U, x^1, ..., x^n)$ which is $(U \cap \partial M, \varphi_{U \cap \partial M}) = (U \cap \partial M, x^1\|_{U \cap \partial M}, ..., x^n\|_{U \cap \partial M})$, and hence the local expression is as follows $$X_q = \sum_{i=1}^n a_i(q) \frac{\partial}{\partial x^i}\|_q, q \in \partial M \cap U \tag{1}$$ We define $X$ to be smooth at $p$, if for all coordinate neighborhoods $(U, \varphi) = (U, x^1, ..., x^n)$ of $p$ in $M$, we have that for the (I say "the" because I guess the restriction $(U \cap \partial M, \varphi_{U \cap \partial M})$ is unique for a given $(U, \varphi)$) corresponding coordinate neighborhood $(U \cap \partial M, \varphi_{U \cap \partial M}) = (U \cap \partial M, x^1\|_{U \cap \partial M}, ..., x^n\|_{U \cap \partial M})$ of $p$ in $\partial M$ that gives the local expression for $X$ as in $(1)$ that there exists a neighborhood $W$ of $p$ in $U$ (I'm going to ignore the details of "neighborhood" as in open subset versus "coordinate neighborhood" as in open subset along with homeomorphism and treat "neighborhood" and "coordinate neighborhood" equivalently and thus omit assigning some $\varphi$ or $\psi$ or whatever to $W$) such that the functions $a_i\|_W$ are smooth at p. Now, such neighborhood $W$ of $p$ in $U$ is also a neighborhood of $p$ (in $M$) because $W$ is open in $M$ because $U$ is open in $M$. I guess I take the "$a^i$ on $\partial M$" to mean "$a^i$ on $\partial M \cap W$". I could be missing something that actually suggests that the $a_i$ are originally on $U$ or $M$ or something and then "$a^i$ on $\partial M$" means, respectively, $a_i\|_{U \cap \partial M}$ or $a_i\|_{(M \cap) \partial M}$. $ \ $ Despite the title of the subsection, I don't think there's a definition for outward-pointing vector fields. What is it exactly? Asked here In the proof of Proposition 22.10, is it understood that we cover $\partial M$ by restrictions of the $(U_{\alpha}, x^1_{\alpha}, ..., x^n_{\alpha})$'s like in questions 2 and 3? Asked here Asked here
i need to calculate some expectations which involving the ratio of normal pdf to normal cdf. Specifically, they are $E\{\phi(x)/\Phi(x)\}$ and $E\{x\phi(x)/\Phi(x)\}$ where $x\sim N(0,1)$. Written in integral, they should be $\int_{-\infty }^{+\infty }{\dfrac {\phi^2 \left( x\right) } {\Phi \left( x\right) }dx}$ and $\int_{-\infty }^{+\infty }{\dfrac {x\phi^2 \left( x\right) } {\Phi \left( x\right) }dx}$, respectively. I know these can be numerically evaluated, for example, by the integral function in Matlab. My question is that, can these expectations be evaluated analytically, or be approximated in closed-form?
I am trying different methods for finding the retarded time in an electromagnetism problem. I'm getting different results, but I'm not sure what I'm doing wrong. Consider a charge at the origin that begins moving at time t=0 at uniform velocity in the positive x direction and that it has been moving for quite some time so we can neglect edge effects. In all the below, $\vec{\beta}=\vec{u}/c$ Method 1: In the frame of the charge, The 4-potential is $(V/c, 0)$ where $V=\frac{q}{4\pi\epsilon_0\sqrt{x^2+y^2+z^2}}$. Apply the Lorentz Transform then $V'=\frac{q}{4\pi\epsilon_0\sqrt{(x'-ut')^2+(y'^2)/\gamma^2+(z'^2)/\gamma^2}}$. V' is the potential due to the charge's position in the past, with the distance term in the denominator representing what the distance would be in a corresponding electro static problem. So that distance over c is the retarded time: $$c(t-t_r)=\sqrt{(x'-ut')^2+(y'^2)/\gamma^2+(z'^2)/\gamma^2}$$ Method 2: Let $\vec{x}$ be the observation point and $\vec{x'}=\vec{u}t$ be the source point. Then the retarted time,$t_r$, solves $\|\vec{x}-\vec{u}t_r\|=c(t-t_r)$. Squaring both sides and factoring out $c$ to pair it with the $t_r$ one gets: $$(\beta^2-1)c^2t_r^2+2(ct-\vec{\beta}\cdot\vec{x})ct_r+(x^2-c^2t^2)=0$$ The quadratic equation gives: $$ct_r=\gamma^2(ct-\vec{\beta}\cdot\vec{x}) (-+) \gamma\sqrt{\gamma^2(x-\beta c t)^2+(y^2+z^2)}$$ So : $$c(t-t_r)=\gamma^2(\vec{\beta}\cdot \vec{x}-\beta^2 ct)(+-)\gamma\sqrt{\gamma^2(x-\beta c t)^2+(y^2+z^2)}$$ Method 3: Griffith's coordinate change method: Begin with $V=\int\int\int{\frac{q\delta(\vec{x'}-\vec{u}t')d\tau}{4\pi\epsilon_0\|\vec{x}-\vec{x'}\|}}$ Find the Jacbian associated with the coordinate change, then evaluate the integral: $$V'=\frac{q}{4\pi\epsilon_0(1-\hat{n}\cdot \vec{\beta})R_{eq}}$$ Where $\tau$ is infinitesimal volume, $\vec{x}$ is observation point, $\vec{x'}$ is current location of the charge, i.e. $\vec{u}t'$, and $t'$ is the retarded time , $t_r$. $\vec{u}$ is the uniform velocity. $\vec{\beta}$ is the velocity vector divided by c. $$\vec{R_{eq}}=\vec{x}-\vec{x'}=\vec{x}-\vec{u}t_r$$ $R_{eq}=\|\vec{R_{eq}}\|$, and $\hat{n}=\vec{R_{eq}}/R_{eq}$ Yielding: $$c(t-t_r)=\|\vec{x}-\vec{u}t_r\|-\vec{\beta}\cdot(\vec{x}-\vec{u}t_r)$$ but by definition, doesn't $c(t-t_r)=\|\vec{x}-\vec{u}t_r\|$? If so then the dot product is extraneous.
Better viewed on this Dropbox site: The Hypergeometric Distribution is usually explained via an urn analogy and formulated as the ratio of “favorable outcomes” to all possible outcomes: \[ \displaystyle \boxed{P(x=a;N,A,n,a) = \frac{{A \choose a} \cdot {N-A \choose n-a} }{{N \choose n}}} \] where $N$ is the total number of balls in, $A$ the number of “red” balls in the urn, and $n$ the sample size. The question answered by the above expression is the probability of finding $x=a$ red balls in the sample. However, a different – equally worthy – viewpoint is that of a tree with conditional probabilities. Here is an example for $N=10, A=4, n=3$ So there are 3 leafs with exactly one red ball in the sample. Following the tree we multiply the conditional probabilities of the tree egdes to get to the “and” probability of the leafs. It should be clear that all three probabilites are identical – but for the order of multiplication: \[ P (\mbox{one red}) = 3 \cdot P_{leaf} = 3 \cdot \frac{4}{10} \cdot \frac{6}{9} \cdot \frac{5}{8} = 3 \cdot \frac{4 \cdot 6 \cdot 5}{10 \cdot 9 \cdot 8} \] Alternative formula How many leafs contain exactly one red ball? Exactly ${n \choose a} = {3 \choose 1} = 3$. The probability for the precise event “a red balls followed by n-a blue balls”is: \[ \frac{A}{N} \cdot \frac{A-1}{N-1} \cdots \frac{A-a+1}{N-a+1} \cdot \frac{N-A}{N-a} \cdot \frac{N-A-1}{N-a-1} \cdots \frac{N-A-(n-a-1)}{N-a-(n-a-1)} \] The last denominator is simply $N-n+1$, i.e. the full denominator is $_NV_n=N!/(N-n)!$ The left numerator is simply $_AV_n = A!/(A-a)!$ in analogy the right numerator $_{N-A}V_{n-a} = (N-A)!/(N-A-(n-a))!$ So, all in all \[ {n \choose a} \cdot \frac{_AV_n \cdot _{N-A}V_{n-a}}{_NV_n} = {n \choose a} \cdot \frac{A! \cdot (N-A)! \cdot (N-n)!}{(A-a)! \cdot (N-A-(n-a))! \cdot N!} \] \[ = {n \choose a} \cdot \frac{\frac{A!}{(A-a)!} \cdot \frac{(N-A)!}{(N-A-(n-a))!} }{\frac{N!}{(N-n)!}} = \frac{{A \choose a} \cdot {N-A \choose n-a} }{{N \choose n}} \] which leaves us with \[ \displaystyle \boxed{P(x=a;N,A,n,a) = {n \choose a} \cdot \frac{_AV_n \cdot _{N-A}V_{n-a}}{_NV_n} } \] // add bootstrap table styles to pandoc tables $(document).ready(function () { $('tr.header').parent('thead').parent('table').addClass('table table-condensed'); });
Let $X,Y$ be two disjoint closed sets on $\mathbb{R}$ such that $X\cup Y = [a,b]$. Show that $X= \emptyset$ or $Y = \emptyset$. Here's what I've got by now: Let $k \in X\cup Y$. Therefore $k \in X$ or $k \in Y$. Suppose that $X \neq \emptyset$ and $k \in X$, hence $k \notin Y$ which implies that $k \in \mathbb{R} \setminus Y$. From that it follows that $\exists \epsilon_k > 0$ such that $(k - \epsilon_k, k + \epsilon_k) \subset \mathbb{R} \setminus Y$. Since $k \in X\cup Y = [a,b]$ it follows that $a \leq k \leq b$. Now let's get an $\epsilon > 1$ such that $(k-\frac{\epsilon_k}{\epsilon},k+\frac{\epsilon_k}{\epsilon}) \subset \big((k - \epsilon_k, k + \epsilon_k) \cap [a,b] \big)$. That $\epsilon$ clearly exists, and then it follows that $(k-\frac{\epsilon_k}{\epsilon},k+\frac{\epsilon_k}{\epsilon}) \subset \mathbb{R}\setminus Y \cap [a,b]$. Now see that $\mathbb{R} \setminus Y \cap [a,b] = \mathbb{R} \setminus Y \cap (X \cup Y) = X$. From our previous conclusion, it follows that $X$ is open. Using the same reasoning, by supposing that $Y\neq 0$ it follows that $Y$ is open. So now suppose that $X,Y \neq \emptyset$. Theferore $X,Y$ are open sets. Therefore $X\cup Y$ is also an open set, hence it cannot be a closed interval $[a.b]$ which is closed, therefore $X = \emptyset$ or $Y = \emptyset$. Edit for cases when $k=a$ or $k=b$ As the user 5xum pointed out, to guarantee the existence of that $\epsilon > 0$ we need $k \notin \{a,b\}$. So if I can guarantee that there is such $k \in X$, we're done without loss of generality for the case where $Y \neq \emptyset$. To prove that, suppose that there isn't such $k \in X$. Therefore $X=\{a\}$ or $X=\{b\}$ or $X=\{a,b\}$. In all those three cases $Y$ cannot be closed, because $Y$ would be respectively $(a,b]$, $[a,b)$ and $(a,b)$. Since $Y$ is closed, that $k$ exists. Can someone please check my work? That was a hard lemma for me and even after that proof attempt, I'm not 100% sure if it's fully correct! Thanks and any kind of help is highly appreciated!
The simplest way to do so (for Pearson correlation) is to use Fisher's z-transformation. Let r be the correlation in question. Let n be the sample size used to acquire the correlation. tanh is the hyperbolic tangent atanh or $\tanh^{-1}$ is the inverse hyperbolic tangent. Let z = atanh(r), then z is normally distributed with variance $\frac{1}{n-3}$` Using this, you can construct a confidence interval $ C.I.(\rho) = \tanh\left(\tanh^{-1}(\rho) \pm q \cdot \frac{1}{\sqrt{n-3}}\right) $, where $q$ is the value that describes the level of confidence you want (i.e., the value you would read from a normal distribution table (e.g., 1.96 for 95% confidence)), If zero is in the confidence interval, then you would fail to reject the null hypothesis that the correlation is zero. Also, note that you cannot use this for correlations of $\pm 1$ because if they are one for data that is truly continuous, then you only need 3 data points to determine that. For one-sided values, simply use the z-score you'd use for a 1-sided p-value for it, and then transform it back and see if your correlation is within the range of that interval. Edit: You can use a 1-sided test using the same values. Also, I changed sample values $r$ to theoretical values $\rho$, since that's a more appropriate use of confidence intervals. Source: http://en.wikipedia.org/wiki/Fisher_transformation
A geometric series is a sum of terms in which two successive terms always have the same ratio . For example, 4 + 8 + 16 + 32 + 64 + 128 + 256 ... is a geometric series with common ratio 2. This is the same as 2 * 2 x where x is increacing by one for each number. It is called a geometric series because it occurs when comparing the length, area, volume, etc. of a shape in different dimensions. The sum of a geometric series can be computed quickly with the formula <math>\sum_{k=m}^n x^k=\frac{x^{n+1}-x^m}{x-1}</math> which is valid for all natural numbers m ≤ n and all numbers x ≠ 1 (or more generally, for all elements x in a ring such that x - 1 is invertible). This formula can be verified by multiplying both sides with x - 1 and simplifying. Using the formula, we can determine the above sum: (2 9 - 2 2)/(2 - 1) = 508. The formula is also extremely useful in calculating annuities: suppose you put $2,000 in the bank every year, and the money earns interest at an annual rate of 5%. How much money do you have after 6 years? 2,000 · 1.05 6 + 2,000 · 1.05 5 + 2,000 · 1.05 4 + 2,000 · 1.05 3 + 2,000 · 1.05 2 + 2,000 · 1.05 1 = 2,000 · (1.05 7 - 1.05)/(1.05 - 1) = 14,284.02 An infinite geometric series is an infinite series whose successive terms have a common ratio. Such a series converges if and only if the absolute value of the common ratio is less than one; its value can then be computed with the formula <math>\sum_{k=0}^\infty x^k=\frac{1}{1-x}</math> which is valid whenever \| x \| < 1; it is a consequence of the above formula for finite geometric series by taking the limit for n →∞. This last formula is actually valid in every Banach algebra, as long as the norm of x is less than one, and also in the field of p-adic numbers if \| x\| < 1. p Also useful to mention: <math>\sum_{k=0}^\infty k\cdot x^k=\frac{x}{(1-x)^2}</math> This formula only works for \| x \| < 1, as well. See also: infinite series All Wikipedia text is available under the terms of the GNU Free Documentation License
I don't think your formula is quite right. The two complex samples are the locations (at two successive sampling instants)of the tip of the rotating phasor that represents the analog signal. The information that you need is the angle between phasor positions at thesetwo successive time instants. If $S_{n+1} = r_{n+1}e^{j\theta_{n+1}}$ and$S_n = r_ne^{j\theta_n}$, then you want the value of $\theta_{n+1}-\theta_n$.Now, $S_{n+1}S_n^* = r_{n+1}r_ne^{j(\theta_{n+1}-\theta_n)}$ and if youexpress these complex numbers in rectangular coordinates, then you canget the desired angle as$$\theta_{n+1}-\theta_n = \arctan\left(\frac{\text{Im}(S_{n+1}S_n^)}{\text{Re}(S_{n+1}S_n^)}\right)= \arctan\left(\frac{\text{Re}(S_{n})\text{Im}(S_{n+1})- \text{Im}(S_n)\text{Re}(S_{n+1})}{\text{Re}(S_{n})\text{Re}(S_{n+1})+ \text{Im}(S_{n})\text{Im}(S_{n+1})}\right)$$though usually the four-quadrant arctangent function atan2(y,x) is usedinstead of atan which returns an answer between $-\pi/2$ and $\pi/2$. In digital communications applications such as DQPSK demodulation,the real interest is not in the actual value of the angle, but in whichof the four intervals $(-\pi/4,\pi/4)$, $(\pi/4, 3\pi/4)$, $(3\pi/4, 5\pi/4)$,and $(5\pi/4, 7\pi/4)$ the angle belongs -- in other words,did the phase not change at all, or change by $\pi/2$, or by $\pi$, orby $3\pi/2$ -- and this can be readily determinedby comparing the values of the numerator and denominator inthe argument of the arctangent above, thus saving computationaleffort and speeding up the demodulation process. But themore general formula gives the actual angle, and thus can beused to demodulate a general FM signal as well. The outputwill be a sequence of samples of the original continuous-timesignal that modulated the carrier and thus formed the signalthat was transmitted. D/A conversion will give the reconstructedsignal.
You made a mistake in this last equality $$x[n] = \frac{1}{2\pi jn}\left( e^{-j\pi n/4} -e^{-j\pi n} +2e^{j\pi n/4} -2e^{-j\pi n/4} + e^{j\pi n} -e^{j\pi n/4} \right) \neq \frac{\sin(\pi n/4)}{\pi n}$$ The right way to do this would be: $$\begin{align}x(n)&=\frac{1}{2\pi jn}\left( e^{-j\pi n/4} -e^{-j\pi n} +2e^{j\pi n/4} -2e^{-j\pi n/4} + e^{j\pi n} -e^{j\pi n/4} \right) \\&=\frac{1}{\pi n}\left( \frac{e^{-j\pi n/4}-e^{j\pi n/4}}{2j}+ \frac{e^{j\pi n} -e^{-j\pi n}}{2j} +\frac{2e^{j\pi n/4} -2e^{-j\pi n/4}}{2j} \right) \\&=\frac{1}{\pi n}\left( -\sin\left(\pi n/4\right)+\sin\left(\pi n\right)+2\sin\left(\pi n/4\right) \right)\\&=\frac{1}{\pi n}\left( \sin\left(\pi n\right)+\sin\left(\pi n/4\right) \right)\end{align}$$ Which makes sense as one can see that the DTFT is the sum of two rectangular windows, one of width $2\pi$ and one of width $\pi/2$, each corresponding to each $\mathrm{sinc}()$. EDIT: I've just noticed judging by your comment that the mistake you made was to think that $$\frac{\sin(\pi n)}{\pi n} =0$$ As you already know, the numerator is $0$ for all $n$... except for $n=0$. In that case, the denominator also is zero, so you have an indetermination. The same happens in the continuous case. Remember that we assume $$\mathrm{sinc}(0)=1$$ by taking the limit when $t\to0$. To respect the fact that the discrete $\mathrm{sinc}$ is a sampled version of the continuous one, then it equals $1$ at the origin too. So we can state that: $$\frac{\sin(\pi n)}{\pi n} =\delta(n)$$ Notice that the DTFT you wrote can also be expressed (using the fact that convolution is distributive) as: $$X(\Omega)=\sum_{k=-\infty}^{+\infty}\left(u(\Omega+\pi)-u(\Omega-\pi)\right)\star \delta(\Omega-2k\pi) + \sum_{k=-\infty}^{+\infty}u\left(\left(\Omega+\frac{\pi}{4}\right)-u\left(\Omega-\frac{\pi}{4}\right)\right)\star \delta(\Omega-2k\pi)$$ The first term is a window of width $2\pi$ that is $2\pi$-periodic... So it's basically $1 \ \forall \Omega$. $$X(\Omega)=1 + \sum_{k=-\infty}^{+\infty}u\left(\left(\Omega+\frac{\pi}{4}\right)-u\left(\Omega-\frac{\pi}{4}\right)\right)\star \delta(\Omega-2k\pi)$$ Now it's easier to see that the IDTFT I got at the beginning of the question corresponds indeed to the given DTFT.
Greiner in his book "Field Quantization" page 173, eq.(7.11) did this calculation: ${\mathcal L}^\prime=-\frac{1}{2}\partial_\mu A_\nu\partial^\mu A^\nu+\frac{1}{2}\partial_\mu A_\nu\partial^\nu A^\mu-\frac{1}{2}\partial_\mu A^\mu\partial_\nu A^\nu $ $\space\space\space\space=-\frac{1}{2}\partial_\mu A_\nu\partial^\mu A^\nu+\frac{1}{2}\partial_\mu[A_\nu(\partial^\nu A^\mu)-(\partial_\nu A^\nu) A^\mu]$ The last term is a four-divergence which has no influence on the field equations. Thus the dynamics of the electromagnetic field (in the Lorentz gauge) can be described by the simple Lagrangian ${\mathcal L}^{\prime\prime}=-\frac{1}{2}\partial_\mu A_\nu\partial^\mu A^\nu$ Yes, if it is a four-divergence of a vector whose 0-component doesn't contain time derivatives of the field, indeed according to the variational principle this four-divergence will not influence the field equation. And actually I calculated the time-derivative dependence of 0-component of $[A_\nu(\partial^\nu A^\mu)-(\partial_\nu A^\nu) A^\mu]$, in which only $[A_0(\partial^0 A^0)-(\partial_0 A^0) A^0]$ could possibly contain time-derivative, which vanishes fortunately, so whatever the general case it doesn't matter in this present case. But how can he seem to claim that it holds for a general four-divergence term, The last term is a four-divergence which has no influence on the field equations? EDIT: I only assumed the boundary condition to be $A^\mu=0$ at spatial infinity, not at time infinity. And the variation of the action $S = \int_{t_1}^{t_2}L \, dt$ is due to the variation of fields which vanish at time, $\delta A^\mu(\mathbf x,t_1)=\delta A^\mu(\mathbf x,t_2)=0$, not having the knowledge of $\delta \dot A^\mu(\mathbf x,t_1)$ and $\delta \dot A^\mu(\mathbf x,t_2)$, which don't vanish generally, so the four-divergence term will in general contribute to the action, $$\delta S_j=\delta \int_{t_1}^{t_2}dt\int d^3\mathbf x \partial_\mu j(A(x),\nabla A(x),\dot A(x))^\mu =\delta \int_{t_1}^{t_2}dt\int d^3\mathbf x \dot j^0 =\int d^3\mathbf x [\delta j(\mathbf x, t_2)^0-\delta j(\mathbf x, t_1)^0]$$ which does not vanish in general!
Ion Temperature in Inductively Coupled Plasmas (ICPs) When modeling plasmas, various options exist for choosing an ion temperature. Your choice, however, may strongly influence your model’s results. Let’s discuss the theoretical reason behind this phenomenon and study an example involving an inductively coupled plasma (ICP) to illustrate the influence the different ion temperature options have on your model’s results. Choosing an Ion Temperature Non-equilibrium cold plasmas are characterized by an electron temperature that is much higher than the gas temperature. During plasma modeling, the ion temperature is often set to equal the gas temperature. This is an acceptable approximation, as long as the ions undergo sufficient collisions with neutral gas molecules and then thermalize with the background gas. This is especially true in inductively coupled plasmas (ICP), where the pressure is low and the ions’ mean free path length comes closer to the plasma reactor’s length scale. Moreover, the number of collisions are low, therefore, the ion temperature is somewhere in between the gas and electron temperatures. While COMSOL Multiphysics does not solve for the ion temperature, there are some options available for you to do so. You can choose to set the ion temperature to equal the gas temperature or use a user-defined value or expression. Moreover, you can also elect to define a correlation between the electric field and the ion mobility and employ an Einstein relation to calculate it, using the Local Field Approximation (LFA) — available in the COMSOL software. As mentioned, your choice in ion temperature (especially for low-pressure plasmas) could significantly impact your model’s results. Below, you will find a theoretical reason that helps explains this phenomenon. Theoretical Background For the heavy species transport (and ion transport), a continuity equation with a drift diffusion approximation is solved for each species. The variation of the mass fraction ,w_k, for species k depends on a flux, \mathbf{j}_k, and a reaction term, R_k. In this case, convection and thermal diffusion are neglected for simplicity: To compute the flux, \mathbf j_k, a mixture averaged diffusion coefficient, D_{k,m}, and the ion mobility, \mu_{k,m}, are required: Based on the kinetic theory of gases, binary diffusion coefficients, D_{kj}, are calculated to get the mixture averaged diffusion coefficient, D_{k,m}. You may have already noticed that Lenard-Jones parameters, \sigma and \epsilon / k_B, have to be specified for each plasma species: The ion mobility is then calculated, using an Einstein relation according to: At the reactor walls, the ion flux, \mathbf j_k, to the wall, is computed according to: The ion temperature is needed to compute the ion mobility and flux to the reactor walls, so the choice in ion temperature especially affects the ions’ transport properties within the plasma model. If the migration part of the flux is large, in comparison to the diffusion part, then the choice in ion temperature particularly grows in importance. This is notably true in cases at very low pressures or at high electric field strengths. Using the Local Field Approximation (LFA) To reiterate, you can also compute the ion temperature with the help of the LFA, available in COMSOL Multiphysics. The LFA assumes that the local velocity distribution of the particles is balanced with the local electric field. Therefore, quantities, like ion temperature or ion mobility, can be expressed in terms of (reduced) electric field. The LFA requires that local changes in the electric field are small in comparison to the mean free path length. However, this is not always true in the boundary layer, particularly. The following expression, for the reduced electron mobility as a function of the reduced field, is used in a subsequent ICP example, below. In the equation above, the reduced electric field,E/n, is given in Townsends (Td). Inductively Coupled Plasma Example To demonstrate the impact your ion temperature choice has on an ICP model, let’s take a look at an example. An inductively coupled plasma reactor (similar to the GEC ICP Reactor, Argon Chemistry model) was modeled three times with varying ion temperatures. Because ICPs work at particularly low pressures, the ion temperature choice has to be considered carefully. The ion temperature was: set to 300 K, which corresponds to the gas temperature, in Model 1. set to 0.1 eV (1160 K), which corresponds to a typical literature value, in Model 2. set to D_{k,m} / \mu_{k,m}, while \mu_{k,m} was computed with the LFA , in Model 3. The other model parameters were as follows: Model Parameters Gas Temperature 300 K Coil Power 500 W Pressure 0.02 torr Electron Mobility 4E24 (1/(mVs)) The mean ion temperature from Model 3, which was computed from D_{k,m} / \mu_{k,m}, amounts to 0.22 eV –, or 2515 K. The following figures represent the electron density for all three models after 0.001 seconds. Model 1: Electron density (T_ion = 300 K). Model 2: Electron density (T_ion = 0.1 eV). Model 3: Electron density (T_ion from LFA). As seen above, using a higher ion temperature value significantly increases the electron density. The modeling results are also compared in the table below. The maximum electron density, maximum electron temperature, and the absorbed power are displayed. Max. Electron Density [1/m³] Max. Electron Temperature [eV] Resistive Losses [W] 1. T_i = 300 \text K 4.3E17 4.1 387 2. T_i = 0.1 \text {eV} 2.6E18 2.8 407 3. Local Field Approximation 3.3E18 2.3 41 Based on the table, we can deduce that increasing the ion temperature not only leads to a significant increase in electron density, but also the absorbed power. Additionally, the electron temperature noticeably decreases. The example above illustrates the impact the choice in ion temperature has on the modeling results of an ICP. A comparison of the results with literature values is essential in judging which assumptions give the best outcomes. Comments (0) CATEGORIES Chemical COMSOL Now Electrical Fluid General Interfacing Mechanical Today in Science
Joint work with Øystein Linnebo, University of Oslo. J. D. Hamkins and Ø. Linnebo, “The modal logic of set-theoretic potentialism and the potentialist maximality principles,” to appear in Review of Symbolic Logic, 2018. @ARTICLE{HamkinsLinnebo:Modal-logic-of-set-theoretic-potentialism, author = {Hamkins, Joel David and Linnebo, \O{}ystein}, title = {The modal logic of set-theoretic potentialism and the potentialist maximality principles}, journal = {to appear in Review of Symbolic Logic}, year = {2018}, volume = {}, number = {}, pages = {}, month = {}, note = {}, abstract = {}, keywords = {to-appear}, source = {}, eprint = {1708.01644}, archivePrefix = {arXiv}, primaryClass = {math.LO}, url = {http://wp.me/p5M0LV-1zC}, doi = {}, } Abstract.We analyze the precise modal commitments of several natural varieties of set-theoretic potentialism, using tools we develop for a general model-theoretic account of potentialism, building on those of Hamkins, Leibman and Löwe (Structural connections between a forcing class and its modal logic), including the use of buttons, switches, dials and ratchets. Among the potentialist conceptions we consider are: rank potentialism (true in all larger $V_\beta$); Grothendieck-Zermelo potentialism (true in all larger $V_\kappa$ for inaccessible cardinals $\kappa$); transitive-set potentialism (true in all larger transitive sets); forcing potentialism (true in all forcing extensions); countable-transitive-model potentialism (true in all larger countable transitive models of ZFC); countable-model potentialism (true in all larger countable models of ZFC); and others. In each case, we identify lower bounds for the modal validities, which are generally either S4.2 or S4.3, and an upper bound of S5, proving in each case that these bounds are optimal. The validity of S5 in a world is a potentialist maximality principle, an interesting set-theoretic principle of its own. The results can be viewed as providing an analysis of the modal commitments of the various set-theoretic multiverse conceptions corresponding to each potentialist account. Set-theoretic potentialism is the view in the philosophy of mathematics that the universe of set theory is never fully completed, but rather unfolds gradually as parts of it increasingly come into existence or become accessible to us. On this view, the outer reaches of the set-theoretic universe have merely potential rather than actual existence, in the sense that one can imagine “forming” or discovering always more sets from that realm, as many as desired, but the task is never completed. For example, height potentialism is the view that the universe is never fully completed with respect to height: new ordinals come into existence as the known part of the universe grows ever taller. Width potentialism holds that the universe may grow outwards, as with forcing, so that already existing sets can potentially gain new subsets in a larger universe. One commonly held view amongst set theorists is height potentialism combined with width actualism, whereby the universe grows only upward rather than outward, and so at any moment the part of the universe currently known to us is a rank initial segment $V_\alpha$ of the potential yet-to-be-revealed higher parts of the universe. Such a perspective might even be attractive to a Platonistically inclined large-cardinal set theorist, who wants to hold that there are many large cardinals, but who also is willing at any moment to upgrade to a taller universe with even larger large cardinals than had previously been mentioned. Meanwhile, the width-potentialist height-actualist view may be attractive for those who wish to hold a potentialist account of forcing over the set-theoretic universe $V$. On the height-and-width-potentialist view, one views the universe as growing with respect to both height and width. A set-theoretic monist, in contrast, with an ontology having only a single fully existing universe, will be an actualist with respect to both width and height. The second author has described various potentialist views in previous work. Although we are motivated by the case of set-theoretic potentialism, the potentialist idea itself is far more general, and can be carried out in a general model-theoretic context. For example, the potentialist account of arithmetic is deeply connected with the classical debates surrounding potential as opposed to actual infinity, and indeed, perhaps it is in those classical debates where one finds the origin of potentialism. More generally, one can provide a potentialist account of truth in the context of essentially any kind of structure in any language or theory. Our project here is to analyze and understand more precisely the modal commitments of various set-theoretic potentialist views. After developing a general model-theoretic account of the semantics of potentialism and providing tools for establishing both lower and upper bounds on the modal validities for various kinds of potentialist contexts, we shall use those tools to settle exactly the propositional modal validities for several natural kinds of set-theoretic height and width potentialism. Here is a summary account of the modal logics for various flavors of set-theoretic potentialism. In each case, the indicated lower and upper bounds are realized in particular worlds, usually in the strongest possible way that is consistent with the stated inclusions, although in some cases, this is proved only under additional mild technical hypotheses. Indeed, some of the potentialist accounts are only undertaken with additional set-theoretic assumptions going beyond ZFC. For example, the Grothendieck-Zermelo account of potentialism is interesting mainly only under the assumption that there are a proper class of inaccessible cardinals, and countable-transitive-model potentialism is more robust under the assumption that every real is an element of a countable transitive model of set theory, which can be thought of as a mild large-cardinal assumption. The upper bound of S5, when it is realized, constitutes a potentialist maximality principle, for in such a case, any statement that could possibly become actually true in such a way that it remains actually true as the universe unfolds, is already actually true. We identify necessary and sufficient conditions for each of the concepts of potentialism for a world to fulfill this potentialist maximality principle. For example, in rank-potentialism, a world $V_\kappa$ satisfies S5 with respect to the language of set theory with arbitrary parameters if and only if $\kappa$ is $\Sigma_3$-correct. And it satisfies S5 with respect to the potentialist language of set theory with parameters if and only if it is $\Sigma_n$-correct for every $n$. Similar results hold for each of the potentialist concepts. Finally, let me mention the strong affinities between set-theoretic potentialism and set-theoretic pluralism, particularly with the various set-theoretic multiverse conceptions currently in the literature. Potentialists may regard themselves mainly as providing an account of truth ultimately for a single universe, gradually revealed, the limit of their potentialist system. Nevertheless, the universe fragments of their potentialist account can often naturally be taken as universes in their own right, connected by the potentialist modalities, and in this way, every potentialist system can be viewed as a multiverse. Indeed, the potentialist systems we analyze in this article—including rank potentialism, forcing potentialism, generic-multiverse potentialism, countable-transitive-model potentialism, countable-model potentialism—each align with corresponding natural multiverse conceptions. Because of this, we take the results of this article as providing not only an analysis of the modal commitments of set-theoretic potentialism, but also an analysis of the modal commitments of various particular set-theoretic multiverse conceptions. Indeed, one might say that it is possible ( ahem), in another world, for this article to have been entitled, “ The modal logic of various set-theoretic multiverse conceptions.” For more, please follow the link to the arxiv where you can find the full article. J. D. Hamkins and Ø. Linnebo, “The modal logic of set-theoretic potentialism and the potentialist maximality principles,” to appear in Review of Symbolic Logic, 2018. @ARTICLE{HamkinsLinnebo:Modal-logic-of-set-theoretic-potentialism, author = {Hamkins, Joel David and Linnebo, \O{}ystein}, title = {The modal logic of set-theoretic potentialism and the potentialist maximality principles}, journal = {to appear in Review of Symbolic Logic}, year = {2018}, volume = {}, number = {}, pages = {}, month = {}, note = {}, abstract = {}, keywords = {to-appear}, source = {}, eprint = {1708.01644}, archivePrefix = {arXiv}, primaryClass = {math.LO}, url = {http://wp.me/p5M0LV-1zC}, doi = {}, }
We cannot prove it ... As said in your previous post : in system $\mathcal L(\neg, \to, \bullet)$, $\bullet$ "acts" as a constant truth, i.e. $\bullet \varphi \approx (\varphi \to \varphi)$. If we consider the usual truth functional properties of the conncetives : $\lnot, \rightarrow$, and replace $\bullet \varphi$ with $\top$ ("the true"), we have that : A1) $\neg\neg\bullet\bullet\varphi$ is $\neg \neg (\top \to \top)$, that is always $\top$ A2) $(\neg\bullet\varphi \to \neg \psi)$ is $(\neg \top \to \neg \psi)$, that is $(\bot \to \neg \psi)$, and again it is always $\top$ A3) $((\varphi\to\psi) \to (\neg\psi\to\neg\varphi))$ is a tautology; thus it is also $\top$. Of course, modus ponens preserves validity. But : $\neg\bullet\varphi$ is $\neg \top$ i.e. $\bot$.
I am having trouble interpreting the following question: Let $ \{X, \|\| \cdot \|\|_{X} \} $ and $ \{Y, \|\| \cdot \|\|_{Y} \} $ be Banach spaces. Let $ T(x,y) \colon X \times Y \to \mathbb{R} $ be a functional linear and continuous in each of the two variables. Then $ T $ is linear and bounded with respect to both variables. Does showing that T is linear with respect to both variables mean showing that $T$ is bilinear? If so then it seems that it is by definition. If linearity is actually meant then I don't understand how $ T $ can be linear in both coordinates, for example $ T(ax,ay) = a^2T(x,y) \neq aT(x,y) $? Assuming bilinear is meant I believe the question wants me to use a proposition which states that a family pointwise equi-bounded maps from a Banach space into a normed space are uniformly equi-bounded. It seems the problem would be finished if I could show one of the families $ \{ T(\cdot, y) \}_{y \in Y} $ or $ \{ T(x, \cdot) \}_{x \in X} $ is pointwise equi-bounded. However I don't see any reason why this would be true. My other thought is that maybe there is a way to use the Closed Graph Theorem since the question assumes both X and Y are Banach. However, I think we would need $T$ to be linear instead of bilinear to use its continuity to imply boundedness. Any hints or clarifications on the statement of the problem would be much appreciated.
Definition:Bound Variable Contents Definition Examples In algebra: $x^2 + 2 x y + y^2 = \paren {x + y}^2$ both $x$ and $y$ are bound variables. In the universal statement: $\forall x: P \paren x$ Thus, the meaning of $\forall x: P \paren x$ does not change if $x$ is replaced by another symbol. That is, $\forall x: P \paren x$ means the same thing as $\forall y: P \paren y$ or $\forall \alpha: P \paren \alpha$. And so on. In the existential statement: $\exists x: P \paren x$ Thus, the meaning of $\exists x: P \paren x$ does not change if $x$ is replaced by another symbol. That is, $\exists x: P \paren x$ means the same thing as $\exists y: P \paren y$ or $\exists \alpha: P \paren \alpha$. And so on. Also known as A bound variable is also popularly seen with the name dummy variable. Both terms can be seen on $\mathsf{Pr} \infty \mathsf{fWiki}$. In treatments of pure logic, this is sometimes known as an individual variable. Some sources call it an apparent variable, reflecting the fact that it only "appears" to be a variable. Also see Sources 1946: Alfred Tarski: Introduction to Logic and to the Methodology of Deductive Sciences(2nd ed.) ... (previous) ... (next): $\S 1.4$: Universal and Existential Quantifiers 1955: John L. Kelley: General Topology... (previous) ... (next): Chapter $0$: Sets 1972: A.G. Howson: A Handbook of Terms used in Algebra and Analysis... (previous) ... (next): $\S 1$: Some mathematical language: Variables and quantifiers 1972: Patrick Suppes: Axiomatic Set Theory(2nd ed.) ... (previous) ... (next): $\S 1.2$ Logic and Notation 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra... (previous) ... (next): $\S 3$: Statements and conditions; quantifiers 1980: D.J. O'Connor and Betty Powell: Elementary Logic... (previous) ... (next): $\S \text{III}$: The Logic of Predicates $(1): \ 3$: Quantifiers 1982: P.M. Cohn: Algebra Volume 1(2nd ed.) ... (previous) ... (next): Chapter $1$: Sets and mappings: $\S 1.1$: The need for logic 1996: H. Jerome Keisler and Joel Robbin: Mathematical Logic and Computability: $\S 2.1$ 2008: David Nelson: The Penguin Dictionary of Mathematics(4th ed.) ... (previous) ... (next): Entry: bound: 4. 2014: Christopher Clapham and James Nicholson: The Concise Oxford Dictionary of Mathematics(5th ed.) ... (previous) ... (next): Entry: dummy variable
J. D. Hamkins and M. Kikuchi, “The inclusion relations of the countable models of set theory are all isomorphic,” ArXiv e-prints, 2017. (manuscript under review) @ARTICLE{HamkinsKikuchi:The-inclusion-relations-of-the-countable-models-of-set-theory-are-all-isomorphic, author = {Joel David Hamkins and Makoto Kikuchi}, title = {The inclusion relations of the countable models of set theory are all isomorphic}, journal = {ArXiv e-prints}, editor = {}, year = {2017}, volume = {}, number = {}, pages = {}, month = {}, doi = {}, note = {manuscript under review}, eprint = {1704.04480}, archivePrefix = {arXiv}, primaryClass = {math.LO}, url = {http://jdh.hamkins.org/inclusion-relations-are-all-isomorphic}, abstract = {}, keywords = {under-review}, source = {}, } Abstract.The structures $\langle M,\newcommand\of{\subseteq}\of^M\rangle$ arising as the inclusion relation of a countable model of sufficient set theory $\langle M,\in^M\rangle$, whether well-founded or not, are all isomorphic. These structures $\langle M,\of^M\rangle$ are exactly the countable saturated models of the theory of set-theoretic mereology: an unbounded atomic relatively complemented distributive lattice. A very weak set theory suffices, even finite set theory, provided that one excludes the $\omega$-standard models with no infinite sets and the $\omega$-standard models of set theory with an amorphous set. Analogous results hold also for class theories such as Gödel-Bernays set theory and Kelley-Morse set theory. Set-theoretic mereology is the study of the inclusion relation $\of$ as it arises within set theory. In any set-theoretic context, with the set membership relation $\in$, one may define the corresponding inclusion relation $\of$ and investigate its properties. Thus, every model of set theory $\langle M,\in^M\rangle$ gives rise to a corresponding model of set-theoretic mereology $\langle M,\of^M\rangle$, the reduct to the inclusion relation. In our previous article, J. D. Hamkins and M. Kikuchi, Set-theoretic mereology, Logic and Logical Philosophy, special issue “Mereology and beyond, part II”, vol. 25, iss. 3, pp. 1-24, 2016. we had identified exactly the complete theory of these mereological structures $\langle M,\of^M\rangle$. Namely, if $\langle M,\in^M\rangle$ is a model of set theory, even for extremely weak theories, including set theory without the infinity axiom, then the corresponding mereological reduct $\langle M,\of^M\rangle$ is an unbounded atomic relatively complemented distributive lattice. We call this the theory of set-theoretic mereology. By a quantifier-elimination argument that we give in our earlier paper, partaking of Tarski’s Boolean-algebra invariants and Ersov’s work on lattices, this theory is complete, finitely axiomatizable and decidable. We had proved among other things that $\in$ is never definable from $\of$ in any model of set theory and furthermore, some models of set-theoretic mereology can arise as the inclusion relation of diverse models of set theory, with different theories. Furthermore, we proved that $\langle\text{HF},\subseteq\rangle\prec\langle V,\subseteq\rangle$. After that work, we found it natural to inquire: Question. Which models of set-theoretic mereology arise as the inclusion relation $\of$ of a model of set theory? More precisely, given a model $\langle M,\newcommand\sqof{\sqsubseteq}\sqof\rangle$ of set-theoretic mereology, under what circumstances can we place a binary relation $\in^M$ on $M$ in such a way that $\langle M,\in^M\rangle$ is a model of set theory and the inclusion relation $\of$ defined in $\langle M,\in^M\rangle$ is precisely the given relation $\sqof$? One can view this question as seeking a kind of Stone-style representation of the mereological structure $\langle M,\sqof\rangle$, because such a model $M$ would provide a representation of $\langle M,\sqof\rangle$ as a relative field of sets via the model of set theory $\langle M,\in^M\rangle$. A second natural question was to wonder how much of the theory of the original model of set theory can be recovered from the mereological reduct. Question. If $\langle M,\of^M\rangle$ is the model of set-theoretic mereology arising as the inclusion relation $\of$ of a model of set theory $\langle M,\in^M\rangle$, what part of the theory of $\langle M,\in^M\rangle$ is determined by the structure $\langle M,\of^M\rangle$? In the case of the countable models of ZFC, these questions are completely answered by our main theorems. Main Theorems. All countable models of set theory $\langle M,\in^M\rangle\models\text{ZFC}$ have isomorphic reducts $\langle M,\of^M\rangle$ to the inclusion relation. The same holds for models of considerably weaker theories such as KP and even finite set theory, provided one excludes the $\omega$-standard models without infinite sets and the $\omega$-standard models having an amorphous set. These inclusion reducts $\langle M,\of^M\rangle$ are precisely the countable saturated models of set-theoretic mereology. Similar results hold for class theory: all countable models of Gödel-Bernays set theory have isomorphic reducts to the inclusion relation, and this reduct is precisely the countably infinite saturated atomic Boolean algebra. Specifically, we show that the mereological reducts $\langle M,\of^M\rangle$ of the models of sufficient set theory are always $\omega$-saturated, and from this it follows on general model-theoretic grounds that they are all isomorphic, establishing statements (1) and (2). So a countable model $\langle M,\sqof\rangle$ of set-theoretic mereology arises as the inclusion relation of a model of sufficient set theory if and only if it is $\omega$-saturated, establishing (3) and answering the first question. Consequently, in addition, the mereological reducts $\langle M,\of^M\rangle$ of the countable models of sufficient set theory know essentially nothing of the theory of the structure $\langle M,\in^M\rangle$ from which they arose, since $\langle M,\of^M\rangle$ arises equally as the inclusion relation of other models $\langle M,\in^*\rangle$ with any desired sufficient alternative set theory, a fact which answers the second question. Our analysis works with very weak set theories, even finite set theory, provided one excludes the $\omega$-standard models with no infinite sets and the $\omega$-standard models with an amorphous set, since the inclusion reducts of these models are not $\omega$-saturated. We also prove that most of these results do not generalize to uncountable models, nor even to the $\omega_1$-like models. Our results have some affinity with the classical results in models of arithmetic concerned with the additive reducts of models of PA. Restricting a model of set theory to the inclusion relation $\of$ is, after all, something like restricting a model of arithmetic to its additive part. Lipshitz and Nadel (1978) proved that a countable model of Presburger arithmetic (with $+$ only) can be expanded to a model of PA if and only if it is computably saturated. We had hoped at first to prove a corresponding result for the mereological reducts of the models of set theory. In arithmetic, the additive reducts are not all isomorphic, since the standard system of the PA model is fully captured by the additive reduct. Our main result for the countable models of set theory, however, turned out to be stronger than we had expected, since the inclusion reducts are not merely computably saturated, but fully $\omega$-saturated, and this is why they are all isomorphic. Meanwhile, Lipshitz and Nadel point out that their result does not generalize to uncountable models of arithmetic, and similarly ours also does not generalize to uncountable models of set theory. The work leaves the following question open: Question. Are the mereological reducts $\langle M,\of^M\rangle$ of all the countable models $\langle M,\in^M\rangle$ of ZF with an amorphous set all isomorphic? We expect the answer to come from a deeper understanding of the Tarski-Ersov invariants for the mereological structures combined with knowledge of models of ZF with amorphous sets. This is joint work with Makoto Kikuchi.
I have the following equation: $$D=\frac{1}{64} \pi A ^3 B \sin \left(C\right)-\frac{1}{2} \pi A B \sin \left(C\right)$$ which I want to solve for $A$. The equation is cubic in $A$ so this should give me 3 answers and, potentially, imaginary parts to the answers. I know, from the physical meaning of the parameters, that all parameters ($A$,$B$,$C$ and $D$) are positive and real. My question is: how can I use Solve on this equation and make sure that there will be no imaginary parts popping up in the solution? I have used the suggestion by Chris and tried: Solve[d == 1/64 π a^3 b Sin[c] - 1/2 π a b Sin[c], a, Reals] but I still see that there is a $\imath\sqrt{3}$ term in the answer. Why does this still appear?
Edit: I'm a dumbass. The thing below is supposed to be just the motivation of asking. I want to ask for below and in general, hehe. Assume that we have a general one-period market model consisting of d+1 assets and N states. Using a replicating portfolio $\phi$, determine $\Pi(0;X)$, the price of a European call option, with payoff $X$, on the asset $S_1^2$ with strike price $K = 1$ given that $$S_0 =\begin{bmatrix} 2 \\ 3\\ 1 \end{bmatrix}, S_1 = \begin{bmatrix} S_1^0\\ S_1^1\\ S_1^2 \end{bmatrix}, D = \begin{bmatrix} 1 & 2 & 3\\ 2 & 2 & 4\\ 0.8 & 1.2 & 1.6 \end{bmatrix}$$ where the columns of D represent the states for each asset and the rows of D represent the assets for each state What I tried: We compute that: $$X = \begin{bmatrix} 0\\ 0.2\\ 0.6 \end{bmatrix}$$ If we solve $D'\phi = X$, we get: $$\phi = \begin{bmatrix} 0.6\\ 0.1\\ -1 \end{bmatrix}$$ It would seem that the price of the European call option $\Pi(0;X)$ is given by the value of the replicating portfolio $$S_0'\phi = 0.5$$ On one hand, if we were to try to see if there is arbitrage in this market by seeing if a state price vector $\psi$ exists by solving $S_0 = D \psi$, we get $$\psi = \begin{bmatrix} 0\\ -0.5\\ 1 \end{bmatrix}$$ Hence there is no strictly positive state price vector $\psi$ s.t. $S_0 = D \psi$. By 'the fundamental theorem of asset pricing' (or 'the fundamental theorem of finance' or '1.3.1' here), there exists arbitrage in this market. On the other hand the price of 0.5 seems to be confirmed by: $$\Pi(0;X) = \beta E^{\mathbb Q}[X]$$ where $\beta = \sum_{i=1}^{3} \psi_i = 0.5$ (sum of elements of $\psi$) and $\mathbb Q$ is supposed to be the equivalent martingale measure given by $q_i = \frac{\psi_i}{\beta}$. Thus we have $$E^{\mathbb Q}[X] = q_1X(\omega_1) + q_2X(\omega_2) + q_3X(\omega_3)$$ $$ = 0 + \color{red}{-1} \times 0.2 + 2 \times 0.6 = 1$$ $$\to \Pi(0;X) = 0.5$$ I guess $\therefore$ that we cannot determine the price of the European call using $\Pi(0;X) = \beta E^{Q}[X]$ because there is no equivalent martingale measure $\mathbb Q$ I noticed that one of the probabilities, in what was attempted to be the equivalent martingale measure, is negative. I remember reading about negative probabilities in Wiki and here However the following links mentioned by Wiki seem to assume absence of arbitrage so I think they are not applicable. Or are they? Is it perhaps that this market can be considered to be arbitrage-free under some quasiprobability measure that allows negative probabilities?
Answer 3.2 revolutions Work Step by Step We find: $\Delta \theta = \frac{\omega^2 - \omega_0^2 }{2\alpha}= \frac {0-2.2^2}{2 \times -.12} \approx 20 rads \approx 3.2\ revs $ You can help us out by revising, improving and updating this answer.Update this answer After you claim an answer you’ll have 24 hours to send in a draft. An editorwill review the submission and either publish your submission or provide feedback.
8.8 #54 Does this indefinite integral converge for anyone? Also, if you are having trouble with the integral, take a look at the derivatives of inverse hyperbolic functions. --John Mason It does not converge for me. I used direct comparison to test whether it converges or not. I started by comparing $ \sqrt{x^4-1} $ and $ \sqrt{x^4} $ It was easy from there. --Josh Visigothsandwich Thanks. I'm not really sure that I needed to use some sort of comparison to show it didn't converge, as it did integrate nicely, but its good to have a second opinion, be that mathematical or otherwise. --John Mason It does not converge for me either, but Josh, be careful with your comparison. It's not really valid to use $ 1/\sqrt{x^4} $ as the function for comparison, because $ x/\sqrt{x^4} < x/\sqrt{x^4-1} $, albeit infinitesimally less. I used $ (g(x) = 1/x^{0.99}) > (f(x) = x/\sqrt{x^4-1}) $, in which case P < 1 and diverges. --Randy Eckman 15:44, 26 October 2008 (UTC) That is true. But if you start out with that as your comparison: $ \sqrt{x^4} > \sqrt{x^4 - 1} $ $ \text{For } x > 0 $ $ x\sqrt{x^4} > x\sqrt{x^4 - 1} $ $ \frac{x}{\sqrt{x^4 - 1}} > \frac{x}{\sqrt{x^4}} = \frac{x}{x^2} = \frac{1}{x} $ $ \int^\infty_4 \frac{dx}{x} = \infty $ Certainly much cleaner than working with decimal powers. --John Mason Also, remember, $ \int_1^{\infty}\frac{1}{x^p}dx $ diverges as long as $ p\le 1 $. So if p = 1, it still diverges. That's why it's okay to use the comparison I used. I did what John did, only I multiplied the inequality by x after taking the inverse of both sides and switching the inequality sign.His Awesomeness, Josh Hunsberger Pg. 329, #16 How accurate do we need to make our answers for the roots? After four iterations, I have the first point accurate to four digits. The text doesn't specify a number of correct digits, and out of curiosity I found the precise roots on Mathematica. I don't know how the textbook could expect us to calculate exactly this: {x -> 0.630115} {x -> 2.57327} --Randy Eckman 17:43, 26 October 2008 (UTC) I went to 10 digits, as that was all my calculator could show. And for the record "Reckman" is a very cool name. --John Mason And I went to five digits because that's all my calculator would show me (I think I can change that, but i wasn't sure how). His Awesomeness, Josh Hunsberger Pg.329, #5 I don't really understand how to do #5. It seems like there isn't an actual function. Are we supposed to use maple? Can someone help get me started? --Klosekam 16:19, 27 October 2008 (UTC) Since you need to know where $ e^{-x} = 2x + 1 $ you can just subtract out one side and solve for the roots (aka at what value of x the function takes zero). So you can use $ f(x) = 2x + 1 -e^{-x} $
Frequency domain view of the relationship between a signal and a sampling of that signal Outline Introduction Main Points Conclusion Introduction This Slecture will look into the relationship between a signal and the sampling of that signal in the view of the frequency domain. The signal that will be sampled will be in the time domain $ x(t) $ and after it was Fourier transformed, the sampled signal $ X(f) $ will be in the frequency domain. Here, we will see the relationship between this two signals and what rules need to be obeyed in order to create a good sampling signal within the frequency domain. Main Points The sampling of a signal in the time domain $ x(t) $ is basically the repetition of the sampled signal $ x(t) $ or the $ rep $ function in the frequency domain. The graphical interpretation for this relationship can be seen in the picture below From the picture above, I should point out that this relationship goes both ways, if we try to sample a signal from the frequency domain the sampled signal that will be produced in the time domain will also be in the form a of $ rep $ function. When we sampled a signal in the time domain, we usually sample a continuous time signal. However, we can't just sample the signal without discreeting the points within the continuous time signal. The reason for this is because, each discreet points within the continuous time signal will be the sampled points of the sampled signal in the frequency domain. The graphical interpretation for this relationship can be seen in the picture below We should now look at some of the important facts of a sampled signal. One of the key factor in sampling a signal is to avoid the aliasing effect when sampling. Aliasing effect is when two sampled signals overlap (at least some part of it) with one another. This will produce distortions within a sampled signal. In order for us to avoid aliasing effect, we need to have a sampling frequency which is larger than two times the cut-offs frequencies. This is known as the Nyquist Theorem: $ \begin{align} f_s > 2 \times f_c\\ \end{align} $ The Nyquist theorem can be depicted through the graphical interpretation which can be seen through the picture below Now that we have looked at the graphical interpretation of sampling, we should next see the analytical expression of sampling in the view of the frequency domain. The Fourier transform of sampling is given by: $ \begin{align} X_s(f) &= S_t(f) * X(f)\\ \end{align} $ Since:- $ \begin{align} s_t(f) &= rep_T(\frac{1}{tau}rect(\frac{t}{tau}))\\ \end{align} $ Thus: $ \begin{align} S_t(f) &=\frac{1}{T}comb_\frac{1}{T}(sinc(tau + 1))\\ &= \frac{1}{T}\sum_{k = -\infty}^\infty sinc(tauK/T) \delta(f-\frac{k}{T})\\ \end{align} $ Therefore: $ \begin{align} X_s(f) &= \frac{1}{T}\sum_{k = -\infty}^\infty sinc(tauK/T) \delta(f-\frac{k}{T}) * X(f)\\ &= \frac{1}{T}\sum_{k = -\infty}^\infty sinc(tauK/T)X(f - \frac{k}{T})\\ \end{align} $ From the expression above we can see that the sampled signal is equal to the summation of the sinc function multiplied by the original signal which we wish to sampled and finally divided by the sampling period T. Conclusion To conclude, in order to have a good sampling signal in the frequency domain, the sampling frequency must be at least twice the cut-off frequency. This theorem is known to be as the Nyquist theorem. If this theorem is obeyed, no aliasing would occur on the sampled signals and aliasing would occur if otherwise. Other than that, the sampling signal in the frequency domain is summation of the sinc function multiplied with the original signal which will later be divided by the sampling period T. References [1].Mireille Boutin, "ECE438 Digital Signal Processing with Applications," Purdue University August 26,2009 Questions and comments If you have any questions, comments, etc. please post them on this page.
Those guys only make confusion. I will answer you with a very easy method you can use with piecewise functions. First of all you have two steps functions, which you can easily figure in your mind to be like 2 dimensional boxes of height $2$. Think about them as two boxes, one of which has to move towards the other. Those are two signals, and the convolution is nonzero only when they do intersect. Figure it out as the following picture: Before some obnoxious arsehole points it out: this figure does not represent your problem but it's a great help to figure it out. The first box goes from $0$ to $1$, whilst the second one goes from $\tau -3$ to $\tau$. In your case, you maintain the smallest step function you have, that is the $g$ and you make $f$, the larger step, to move towards the first box. Hence you will follow several steps. Step 1 The signals are like the above figure: separated. In this case, their convolution is simply zero which happens in the range $$\tau - 3 > 1$$ That is $$\tau > 4$$ Step 2 The larger box start to insinuate itself onto the smallest, hence you will have $$\int_{\tau -3}^1 4\ d\tau = 16 - 4\tau$$ And this happens in the range $$\begin{cases} 0 < \tau -3 \\ 1 > \tau -3 \end{cases}$$ Which means $$3 < \tau < 4$$ Step 3 The big box is all passing through the small box, id est the smaller is totally contained into the larger. Hence $$\int_0^1 4\ d\tau = 4$$ In the range $$\begin{cases} 0 > \tau -3 \\ 1 > \tau \end{cases}$$ Which means $$1 < \tau < 1$$ Step 4 Now the larger box fades away, but with a bit to it still inside the smaller box. $$\int_0^{\tau} 4\ d\tau = 4\tau$$ Which happens in the range $$0< \tau < 1$$ Step 5 Finally the big box goes away, and the convolution is zero again for $$\tau < 0$$ At the end you have your result: $$f*g = \begin{cases} 0 & \tau > 4 \\ 16 - 4\tau & 3<\tau < 4 \\ 4 & 1<\tau < 3 \\ 4\tau & 0 < \tau < 1 \\ 0 & \tau < 0 \end{cases}$$ This is a trapezoid, and if you plot it you will get As it has to be.
Let $p \in (1, \infty).$ Then there is a sequence $f_k$ satisfying 1) $f_k \in L^q$ for all $1 \leq q < \infty$ 2) $f_k$ converges to $0$ in $L^q$ for all $q \in [1,p)$ 3) $f_k$ diverges in $L^p$ $\textbf{Attemp}$ Since the integral for $1 \leq q < p$ converges to $0$ and at exactly $p$ it diverges, I think of $\int \frac{1}{x^k}$ which can be either $\log(x)$ and $\frac{1}{x^l}$ where $\log(x)$ diverges at infinity and $\frac{1}{x^l}$ converges to $0$ as $x \rightarrow \infty.$ So I set $f_k(x) = \int_1^k g(x) dx$ such that $f_k(x) = \frac{1}{k^{l_k}}$ for some $l_k > 0$ for all $q < p$ and $f_k(x) = \log(k)$ when $q = p.$ Somehow I feel it is impossible to find such $g$ since $q < p.$ Any suggestion ? Any idea on maybe a better functions ?
ECE662: Statistical Pattern Recognition and Decision Making Processes Spring 2008, Prof. Boutin Collectively created by the students in the class Lecture 21 Lecture notes When the number of categories, c is big, decision tress are particularly good. Example: Consider the query "Identify the fruit" from a set of c=7 categories {watermelon, apple, grape, lemon, grapefruit, banana, cherry} . One possible decision tree based on simple queries is the following: Three crucial questions to answer How do you grow or construct a decision tree using training data? CART Methodology - Classification and Regressive Tree For constructing a decision tree, for a given classification problem, we have to answer these three questions 1) Which question should be asked at a given node -"Query Selection" 2) When should we stop asking questions and declare the node to be a leaf -"When should we stop splitting" 3) Once a node is decided to be a leaf, what category should be assigned to this leaf -"Leaf classification" We shall discuss questions 1 and 2 (3 being very trivial) Need to define 'impurity' of a dataset such that $ impurity = 0 $ when all the training data belongs to one class. Impurity is large when the training data contain equal percentages of each class $ P(\omega _i) = \frac{1}{C} $; for all $ i $ Let $ I $ denote the impurity. Impurity can be defined in the following ways: Entropy Impurity: $ I = \sum_{j}P(\omega _j)\log_2P(\omega _j) $, when priors are known, else approximate $ P(\omega _j) $ by $ P(\omega _j) = \frac{\#\,of\,training\,patterns\,in\,\omega_j}{Total\,\#\,of\,training\,patterns} $ Gini Impurity $ I = \sum_{i\ne j}P(\omega _i)P(\omega _j) = \frac{1}{2}[1- \sum_{j}P^2(\omega _j) $ Ex: when C = 2, $ I = P(\omega _1)P(\omega _2) $ Misclassification Impurity $ I = 1-max P(\omega _j) $ defined as the "minimum probability that a training pattern is misclassified" The following figure shows above-mentioned impurity functions for a two-category case, as a function of the probability of one of the categories.(DHS-399p) Now let us look at each of the three questions in detail. Query Selection Heuristically, want impurity to decrease from one node to its children. We assume that several training patterns are available at node N and they have a good mix of all different classes. I(N) := impurity at node N. Define impurity drop at node N as: $ \triangle I=I(N)-P_{L}I(N_{L})-(1-P_{L})I(N_{R}) $ where $ P_{L} $ and $ (1-P_{L}) $ are estimated with training patterns at node N. A query that miximizes $ \triangle I $ is "probably" a good one. But "finding the query that maximizes" is not a well defined question because we are not doing an exhaustive search over all the possible queries. Rather we narrow down to a set of few queries and find among them, which one maximizes $ \triangle I $. Example: 1. look at separation hyperplane that miximizes $ \triangle I(N) $ 2. look for a single feature threshold (colour or shape or taste) which would maximize $ \triangle I(N) $ Query selection => numerical optimization problem. When to stop splitting ? Key: look for balance. Need to construct a "balanced tree". Many ways to do this. Example: 1. Validation - train with 80% of training data, validate on 20%. Continue splitting until validation error is minimized. 2. Thresholding - stop splitting when threshold $ \beta $ is small (but not too small) $ \beta=0.03 $ Warning: "Horizon Effect" Lack of looking ahead may cause us to stop splitting prematurely. You should keep splitting for a bit, after you meet stopping criteria. How to correct oversplitting? Use "pruning" (or inverse splitting) which implies that take 2 leaves that have a common parent and merge them if "merging helps" i.e. - if I either doesn't change or only increases a little bit. - if validation error either stays the same or decreases. Declare parent to be a leaf. Pruning increases generalization. Idea: Look further than horizon but step back if it is not worth it. Here is an example where increasing the number of nodes typically lowers the impurity. If the stopping condition looks at at a short horizon, then the number of nodes may stop at the first stopping point. If the horizon continues, then the number of nodes can stop at the second stopping point, which appears to be the best location for this example.
Science Advisor Homework Helper 2,559 3 I missed the lectures for this topic, so I don't have the notes, so I was wondering if anyone could give me the idea behind how to solve quartics in radicals. I know its long and messy, so just the basic idea would do. For example: x I recall something about getting rid of the cubic term, so maybe I should substitute x = (u - 1/2), giving: u u u 16u Okay, this one happened to work out nicely, with the degree-1 term going away as well. Now I just have a quadratic in u². So perhaps a different example would be more enlightening. Also, when asked to "solve in radicals" does that mean that the correct answer to the above problem should be given as: [tex]x = \frac{1}{2} \pm \sqrt{\frac{-12 \pm \sqrt{-3504}}{32}}}[/tex] So: [tex]x = \frac{1 \pm \sqrt{\frac{-3 \pm \sqrt{-219}}{2}}}{2}[/tex] x 4+ 2x³ + 3x² + 4x + 5 = 0 I recall something about getting rid of the cubic term, so maybe I should substitute x = (u - 1/2), giving: u 4- 2u³ + 3/4u² - 1/2u + 1/16 + 2(u³ - 3/2u² + 3/4u - 1/8) + 3(u² - u + 1/4) + 4(u - 1/2) + 5 = 0 u 4+ 3/4u² - 1/2u + 1/16 - 3u² + 3/2u - 1/4 + 3u² - 3u + 3/4 + 4u - 2 + 5 = 0 u 4+ 3/4u² + 57/16 = 0 16u 4+ 12u² + 57 = 0 Okay, this one happened to work out nicely, with the degree-1 term going away as well. Now I just have a quadratic in u². So perhaps a different example would be more enlightening. Also, when asked to "solve in radicals" does that mean that the correct answer to the above problem should be given as: [tex]x = \frac{1}{2} \pm \sqrt{\frac{-12 \pm \sqrt{-3504}}{32}}}[/tex] So: [tex]x = \frac{1 \pm \sqrt{\frac{-3 \pm \sqrt{-219}}{2}}}{2}[/tex]
TensorFlow 1 version View source on GitHub Solves one or more linear least-squares problems. Aliases: tf.compat.v1.linalg.lstsq tf.compat.v1.matrix_solve_ls tf.compat.v2.linalg.lstsq tf.linalg.lstsq( matrix, rhs, l2_regularizer=0.0, fast=True, name=None) matrix is a tensor of shape [..., M, N] whose inner-most 2 dimensionsform M-by- N matrices. Rhs is a tensor of shape [..., M, K] whoseinner-most 2 dimensions form M-by- K matrices. The computed output is a Tensor of shape [..., N, K] whose inner-most 2 dimensions form M-by- Kmatrices that solve the equations matrix[..., :, :] * output[..., :, :] = rhs[..., :, :] in the least squaressense. Below we will use the following notation for each pair of matrix and right-hand sides in the batch: matrix=$A \in \Re^{m \times n}$, rhs=$B \in \Re^{m \times k}$, output=$X \in \Re^{n \times k}$, l2_regularizer=$\lambda$. If fast is True, then the solution is computed by solving the normalequations using Cholesky decomposition. Specifically, if $m \ge n$ then$X = (A^T A + \lambda I)^{-1} A^T B$, which solves the least-squaresproblem $X = \mathrm{argmin}_{Z \in \Re^{n \times k}} \|\|A Z - B\|\|_F^2 +\lambda \|\|Z\|\|_F^2$. If $m \lt n$ then output is computed as$X = A^T (A A^T + \lambda I)^{-1} B$, which (for $\lambda = 0$) isthe minimum-norm solution to the under-determined linear system, i.e.$X = \mathrm{argmin}_{Z \in \Re^{n \times k}} \|\|Z\|\|_F^2 $, subject to$A Z = B$. Notice that the fast path is only numerically stable when$A$ is numerically full rank and has a condition number$\mathrm{cond}(A) \lt \frac{1}{\sqrt{\epsilon_{mach}}}$ or$\lambda$is sufficiently large. If fast is False an algorithm based on the numerically robust completeorthogonal decomposition is used. This computes the minimum-normleast-squares solution, even when $A$ is rank deficient. This path istypically 6-7 times slower than the fast path. If fast is False then l2_regularizer is ignored. Args: : matrix Tensorof shape [..., M, N]. : rhs Tensorof shape [..., M, K]. : 0-D l2_regularizer double Tensor. Ignored if fast=False. : bool. Defaults to fast True. : string, optional name of the operation. name Returns: : output Tensorof shape [..., N, K]whose inner-most 2 dimensions form M-by- Kmatrices that solve the equations matrix[..., :, :] * output[..., :, :] = rhs[..., :, :]in the least squares sense. Raises: : linalg.lstsq is currently disabled for complex128 and l2_regularizer != 0 due to poor accuracy. NotImplementedError
When one have a curve $\beta(s)$ which is parametrized by arc length (has natural parametrization) one is able to obtain the tangent, normal and binormal vectors by using Frenet-Serret frame equations: $T = \beta'(s)$, $N=\frac{T'(s)}{\|T'(s)\|}$, $B = T \times N$ But are those formulas valid for non-regular parametrizations when one normalizes the tangent vector? $T=\frac{\beta'(s)}{\|\beta'(s)\|}$, $N$ and $B$ are calculated as above.
Despite the rather recent progress in prime number theory (see the proof of the ternary Goldbach conjecture by H.A. Helfgott, and the striking result of Yitang Zhang improved by Tao, Maynard and others), to my knowledge, we're still unable to prove rigorously that there exists $N>0$ such that for all integer $n>N$, $2n$ is the sum of two primes. What are the main reasons for that? Lack of knowledge about exponential sums? Interesting heuristics that no one is able to turn into a rigorous proof? If so, what are the obstructions to getting a rigorous version of such arguments? Thanks in advance for any insight. Despite the rather recent progress in prime number theory (see the proof of the ternary Goldbach conjecture by H.A. Helfgott, and the striking result of Yitang Zhang improved by Tao, Maynard and others), to my knowledge, we're still unable to prove rigorously that there exists $N>0$ such that for all integer $n>N$, $2n$ is the sum of two primes. What are the main reasons for that? Lack of knowledge about exponential sums? Interesting heuristics that no one is able to turn into a rigorous proof? If so, what are the obstructions to getting a rigorous version of such arguments? As far as I know there are two approaches to Goldbach type problems, the circle method and sieve methods. In the sequel I will restrict myself to the circle method, hoping that someone else writes something about sieves. Define the exponential sum $S(\alpha)=\sum_{n\leq N}\Lambda(n) e(n\alpha)$, where $e(x)=e^{2\pi i x}$. By orthogonality of $e(x)$ we have that the (weighted) number of representations of $N$ as the sum of 3 primes is $$ \int_0^1 S(\alpha)^3e(-N\alpha)\;d\alpha. $$ If $\alpha=\frac{p}{q}$ is rational, then $$ S(\alpha)=\sum_{(a,q)=1} e(\frac{ap}{q})\underset{n\equiv a\pmod{q}}{\sum_{n\leq N}} \Lambda(n), $$ which can be approximated using the prime number theorem for arithmetic progressions. If $\alpha$ is very close to a rational number, evaluation can still be done by partial summation. Hence define the major arcs $\mathfrak{M}\subseteq[0,1]$ as the set of integers very close to rational numbers with small denominator. Then $$ \int_\mathfrak{M}S(\alpha)^3e(-N\alpha)\;d\alpha $$ can approximately be evaluated, it turns out that this integral essentially equals the expected main term. Let $\mathfrak{m}=[0,1]\setminus\mathfrak{M}$ be the so called minor arcs. To prove that every sufficiently large odd integer is the sum of three primes, Vinogradoff showed that for suitably defined arcs the integral over the major arcs is $\sim\mathfrak{S}(N)N^2$, and that for $\alpha\in\mathfrak{m}$ we have $\|S(\alpha)\|\ll\frac{N}{\log^2 N}$. Hence \begin{eqnarray} \int_\mathfrak{m}S(\alpha)^3e(-N\alpha)\;d\alpha & \leq &\int_\mathfrak{m}\|S(\alpha)^2\|\;d\alpha\max_{\alpha\in\mathfrak{m}}\|S(\alpha)\|\\ & \leq & \int_0^1\|S(\alpha)^2\|\;d\alpha\max_{\alpha\in\mathfrak{m}}\|S(\alpha)\|\\ & = & \sum_{n\leq N}\Lambda(n)^2\max_{\alpha\in\mathfrak{m}}\|S(\alpha)\|\\ & \ll & \frac{N^2}{\log N} \end{eqnarray} So for large $N$ the error coming form $\mathfrak{m}$ is of smaller order then the main term, and we obtain an asymptotic for the number of representations. Since proper powers are pretty rare, passing from $\Lambda$ to primes is no problem. All this was known and used by Hardy and Littlewood some 20 years before Vinogradov, however, they could not give an unconditional bound for $S(\alpha)$, and had to assume the generalized Riemann hypothesis. Among the things Helfgott did, was a numerical bound which is non-trivial for rather small values of $N$, which in analytic number theory is usually pretty difficult. However, no matter what progress may come, the argument used for odd $N$ can never prove binary Goldbach. The crucial point in the argument above was that the integral over the major arcs is asymptotically equal to the conjectural main term, and the integral over the minor arcs is of smaller order. The integral over the major arcs can still be evaluated, and is $\mathfrak{S}(N)N$, as it should, however, $$ \int_0^1 \|S(\alpha)\|^2\;d\alpha = \sum_{n\leq N}\Lambda(n)^2\sim N\log N $$ is bigger then the main term. To prove Goldbach for even integers we have to show that $\left\|\int_\mathfrak{m} S(\alpha)^2e(-N\alpha)\;d\alpha\right\|$ is considerably smaller than $\int_\mathfrak{m} \|S(\alpha^2)\|\;d\alpha$, that is, that there is some cancelation within the integral. Still, one can obtain interesting results in this way. For example, Montgomery and Vaughan (The exceptional set in Goldbach's problem, Acta Arith. 27) have shown that all even integers up to $x$ with at most $x^{1-c}$ exceptions can be written as the sum of two primes. They used the fact that $\int_\mathfrak{m} S(\alpha)^2e(-N\alpha)\;d\alpha$ cannot simultaneously be large for many different $N$. Somewhat surprisingly Brüdern, Granville, Perelli, Vaughan, and Wooley, (On the exponential sum over k-free numbers, R. Soc. Lond. Philos. Trans. Ser. A Math. Phys. Eng. Sci. 356) showed that there are interesting binary additive problems which can be solved by the circle method, e.g. problems involving squarefree integers. However, as they explicitly mention, their method fails for sets of density 0, the reason being the same as above.
Tried asking here first, since a similar question had been asked on that site. Seems more relevant for this site however. It is my current understanding that a quantum XOR gate is the CNOT gate. Is the quantum XNOR gate a CCNOT gate? Quantum Computing Stack Exchange is a question and answer site for engineers, scientists, programmers, and computing professionals interested in quantum computing. It only takes a minute to sign up.Sign up to join this community Tried asking here first, since a similar question had been asked on that site. Seems more relevant for this site however. It is my current understanding that a quantum XOR gate is the CNOT gate. Is the quantum XNOR gate a CCNOT gate? Any classical one-bit function $f:x\mapsto y$ where $x\in\{0,1\}^n$ is an $n$-bit input and $y\in\{0,1\}$ is an $n$-bit output can be written as a reversible computation, $$ f_r:(x,y)\mapsto (x,y\oplus f(x)) $$ (Note that any function of $m$ outputs can be written as just $m$ separate 1-bit functions.) A quantum gate implementing this is basically just the quantum gate corresponding to the reversible function evaluation. If you simply write out the truth table of the function, each line corresponds to a row of the unitary matrix, and the output tells you which column entry contains a 1 (all other entries contain 0). In the case of XNOR, we have the standard truth table, and the reversible function truth table $$ \begin{array}{c\|c} x & f(x) \\ \hline 00 & 1 \\ 01 & 0 \\ 10 & 0 \\ 11 & 1 \end{array} \qquad \begin{array}{c\|c} (x,y) & (x,y\oplus f(x)) \\ \hline 000 & 001 \\ 001 & 000 \\ 010 & 010 \\ 011 & 011 \\ 100 & 100 \\ 101 & 101 \\ 110 & 111 \\ 111 & 110 \end{array} $$ Thus, the unitary matrix is $$ U=\left(\begin{array}{cccccccc} 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\ \end{array}\right). $$ This can easily be decomposed in terms of a couple of controlled-not gates and a bit flip or two. The method that I just outlined gives you a very safe way of making the construction that works for any $f(x)$, but it does not perfectly reconstruct the correspondence between XOR and controlled-not. For that, we need to assume a little bit more about the properties of the function $f(x)$. Assume that we can decompose the input $x$ into $a,b$ such that $a\in\{0,1\}^{n-1}$ and $b\in \{0,1\}$ such that for all values of $a$, the values of $f(a,b)$ are distinct for each $b$. In this case, we can define the reversible function evaluation as $$f:(a,b)\mapsto(a,f(a,b)).$$ This means that we're using 1 fewer bits than the previous construction, but from here on the technique can be repeated. So, let's go back to the truth table for XNOR. $$ \begin{array}{c\|c} ab & f(a,b) \\ \hline 00 & 1 \\ 01 & 0 \\ 10 & 0 \\ 11 & 1 \end{array} $$ We can see that, for example, when we fix $a=0$, the two outputs are $1,0$, hence distinct. Similarly for fixing $a=1$. Thus, we can proceed with the reversible function construction $$ \begin{array}{c\|c} ab & af(a,b) \\ \hline 00 & 01 \\ 01 & 00 \\ 10 & 10 \\ 11 & 11 \end{array} $$ and this gives us a unitary $$ U=\left(\begin{array}{cccc} 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right) $$ You can easily check that this is the same as $\text{cNOT}\cdot(\mathbb{1}\otimes X)$. The quantum XNOR is not a CCNOT. CCNOT would take 3 bits as input, whereas XOR, XNOR, and CNOT take in only 2 bits or qubits as input. The reason why we say the XOR can be thought of as a CNOT is explained here, and the same reasoning can be used to construct the (2 qubit) XNOR.
Currently, I am constructed a bayesian multilevel model to analyze a panel data set which now basically looks like the following: $y_{ijt} = \beta_{0ij} + X\beta + \epsilon_{ijt}$. So, now only a individual specific intercept but I want to extend this to other parameters. I estimate the model using bayesian econometrics. Now, to increase predictive power I want to add a lagged dependent variable in my model, so it looks like this: $y_{ijt} = \beta_{0ij} + X\beta + \rho y_{ij(t-1)} + \epsilon_{ijt}$. I was wondering whether I should take care of endogeneity by incorporating the lagged dependent variable in my model using the bayesian approach? In the frequentist approach including the lagged dependent variable will lead to severe inconsistency of the parameter $\rho$, so I think that I also have to take this problem into account using bayesian analysis. Could someone give me some explanation about this, since I cannot find any explanation on this subject usinng bayesian analysis. In this case could someone also help me on how to model the initial value $y_{ij0}$ in this case?
If $f,g$ are uniformly continuous, then is $\alpha f+\beta g$ uniformly continuous? So far, I looked at here If $f,g$ are uniformly continuous prove $f+g$ is uniformly continuous, but I didn't understand something. I know that from what I've been told,: 1. $\forall\epsilon >0$ $\exists\delta_1 >0$, if $\|x-y\|<\delta_1$ then $\|f(x)-f(y)\|<\epsilon$ 2. $\forall\epsilon >0$ $\exists\delta_2 >0$, if $\|x-y\|<\delta_2$ then $\|g(x)-g(y)\|<\epsilon$ I need to show that for every $\epsilon$ there exists some $\delta$ such that if $\|x-y\|<\delta$ then $\|(\alpha f+ \beta g)(x) - (\alpha f+ \beta g)(y)\|<\epsilon$. So I know the method, I need to show that from $\|(\alpha f+ \beta g)(x) - (\alpha f+ \beta g)(y)\|<\epsilon$ - I need to do some manipulations and get $\|x-y\|<\delta$ (I need to find that $\delta$). So, why in the answers there it just shows that $\|(\alpha f+ \beta g)(x) - (\alpha f+ \beta g)(y)\|<\epsilon$? Thanks!
Unconventional analogs of single-parametric method of iterational aggregation Abstract When we solve practical problems that arise, for example, in mathematical economics, in the theory of Markov processes, it is often necessary to use the decomposition of operator equations using methods of iterative aggregation. In the studies of these methods for the linear equation $x = Ax + b$ the most frequent are the conditions of positiveness of the operator $A$, constant $b$ and the aggregation functions, and also the implementation of the inequality $\rho(A) <1$ for the spectral radius $ \rho (A) $ of the operator $ A $. In this article, for an approximate solution of a system composed of the equation $x=Ax+b$ represented in the form $ x = A_1x + A_2x + b, $ where $ b \in E, $ $ E $ is a Banach space, $ A_1, A_2 $ are linear continuous operators that act from $ E $ to $ E $ and the auxiliary equation $ y = \lambda y - (\varphi, A_2x) - (\varphi, b) $ with a real variable $y$, where $ (\varphi, x) $ is the value of the linear functional $ \varphi \in E ^ * $ on the elements $ x \in E $, $ E^* $ is conjugation with space $ E $, an iterative process is constructed and investigated $$ \begin{split} x^{(n+1)}&=Ax^{(n)}+b+\frac{\sum\limits_{i=1}^{m}A^i_1x^{(n)}}{(\varphi, x^{(n)})\sum\limits_{i=0}^{m}\lambda^i}(y^{(n)}-y^{(n+1)}) \quad (m<\infty),\\ y^{(n+1)}&=\lambda y^{(n+1)}-(\varphi,A_2x^{(n)})-(\varphi,b). \end{split} $$ The conditions are established under which the sequences $ {x ^ {(n)}}, {y ^ {(n)}}$, constructed with the help of these formulas, converge to $ x ^ , y ^ $ as a component of solving the system constructed from equations $ x = A_1 x + A_2 x + b $ and the equation $ y = \lambda y - (\varphi, A_2 x) - (\varphi, b) $ not slower than the rate of convergence of the geometric progression with the denominator less than $1$. In this case, it is required that the operator $ A $ be a compressive and constant by sign, and that the space $ E $ is semi-ordered. The application of the proposed algorithm to systems of linear algebraic equations is also shown. In this article, for an approximate solution of a system composed of the equation $x=Ax+b$ represented in the form $ x = A_1x + A_2x + b, $ where $ b \in E, $ $ E $ is a Banach space, $ A_1, A_2 $ are linear continuous operators that act from $ E $ to $ E $ and the auxiliary equation $ y = \lambda y - (\varphi, A_2x) - (\varphi, b) $ with a real variable $y$, where $ (\varphi, x) $ is the value of the linear functional $ \varphi \in E ^ * $ on the elements $ x \in E $, $ E^* $ is conjugation with space $ E $, an iterative process is constructed and investigated $$ \begin{split} x^{(n+1)}&=Ax^{(n)}+b+\frac{\sum\limits_{i=1}^{m}A^i_1x^{(n)}}{(\varphi, x^{(n)})\sum\limits_{i=0}^{m}\lambda^i}(y^{(n)}-y^{(n+1)}) \quad (m<\infty),\\ y^{(n+1)}&=\lambda y^{(n+1)}-(\varphi,A_2x^{(n)})-(\varphi,b). \end{split} $$ The conditions are established under which the sequences $ {x ^ {(n)}}, {y ^ {(n)}}$, constructed with the help of these formulas, converge to $ x ^ , y ^ $ as a component of solving the system constructed from equations $ x = A_1 x + A_2 x + b $ and the equation $ y = \lambda y - (\varphi, A_2 x) - (\varphi, b) $ not slower than the rate of convergence of the geometric progression with the denominator less than $1$. In this case, it is required that the operator $ A $ be a compressive and constant by sign, and that the space $ E $ is semi-ordered. The application of the proposed algorithm to systems of linear algebraic equations is also shown. Keywords aggregating functional, decomposition, iterative aggregation 3 :: 10 Refbacks There are currently no refbacks. The journal is licensed under a Creative Commons Attribution-NonCommercial-NoDerivs 3.0 Unported.
Seminar Parent Program: Location: MSRI: Simons Auditorium The $L^p$-Brunn--Minkowski theory for $p\geq 1$, proposed by Firey and developed by Lutwak in the 90's, replaces the Minkowski addition of convex sets by its $L^p$ counterpart, in which the support functions are added in $L^p$-norm. Recently, B\"{o}r\"{o}czky, Lutwak, Yang and Zhang have proposed to extend this theory further to encompass the range $p \in [0,1)$. In particular, they conjectured an $L^p$-Brunn--Minkowski inequality for origin-symmetric convex bodies in that range, which constitutes a strengthening of the classical Brunn-Minkowski inequality. Our main result confirms this conjecture locally for all (smooth) origin-symmetric convex bodies in $\Real^n$ and $p \in [1 - \frac{c}{n^{3/2}},1)$. In addition, we confirm the local log-Brunn--Minkowski conjecture (the case $p=0$) for small-enough $C^2$-perturbations of the unit-ball of $\ell_q^n$ for $q \geq 2$, when the dimension $n$ is sufficiently large, as well as for the cube, which we show is the conjectural extremal case. For unit-balls of $\ell_q^n$ with $q \in [1,2)$, we confirm an analogous result for $p=c \in (0,1)$, a universal constant. It turns out that the local version of these conjectures is equivalent to a minimization problem for a spectral-gap parameter associated with a certain differential operator, introduced by Hilbert (under different normalization) in his proof of the Brunn--Minkowski inequality. Joint work with Alexander Kolesnikov (Moscow).No Notes/Supplements Uploaded No Video Files Uploaded
I've been reviewing proofs for a couple of calculus theorems and as I was trying to recall the proof of the Riemann Sum Theorem which uses Lower Sums and Upper Sums I came up with an idea to prove it using The Mean Value Theorem for Integrals. I just wanted to know if my proof is good since I'm not sure if I've done everything properly. If you've got any hints for me, I'd be delighted. My manipulations in the 3. step might be based a " little bit" too much ( just a little bit I hope!) on my " intuitions". Excuse me, if this violation of Mathematical principals send shiver down your spine today. Proof Just to make sure there is no misunderstanding - this is what I call the " Riemann Sum Theorem": Assuming that: $f: [a;b]\to\mathbb{R}$. ${P_n}$ is a sequence of partitions of $[a;b]$ such that $\|P_n\|\to0$ where $\|P_n\|$ is the length of the longest partition. $f$ continuous. $S_n$ is a Riemann Sum of $f$ and $P_n$. We know that $S_n(f, P_n)=\sum_{j=1}^nf(y_j)(x_{j}-x_{j-1})$ where $y_j\in[x_{j-1};x_j]$, $j=\{1,\dots,n\}$ and $P_n=\{x_0,\dots,x_n\}$. $S_n(f, P_n) = f(y_1)(x_1-x_0)+\dots+f(y_n)(x_n-x_{n-1})$ and it starts to resemble something we can apply the Mean Values Theorem to. Especially, that $$ \forall_{k \in [1;n]}\forall_{\epsilon>0}\exists_{\delta>0}\forall_{u,v \in [x_{k-1};x_k]}\left(\|u-v\|<\delta \implies \|f(u)-f(v)\|<\epsilon\right) $$ So, as $n \to \infty$:$$\begin{cases}f(y_1)(x_1-x_0) \to \int_{[x_0;x_1]}f\\f(y_2)(x_2-x_1) \to \int_{[x_1;x_2]}f\\\vdots\\f(y_n)(x_n-x_{n-1}) \to \int_{[x_{n-1};x_n]}f\\\end{cases}$$which " sums up" to $$ \int_{[x_0;x_n]}f = \int_{[a;b]}f $$ $\square\dots$ is it? Note: Sorry if I use unpopular and weird names for theorems. I don't study Mathematics in English and it is difficult to find the exact translation from time to time. Feel free to edit my posts if necessary.
It´s known that for first order theories, it holds $\mathbf{ZFC} \vdash T \vdash \varphi \leftrightarrow T \models \varphi$. Why does this not hold in the higher order case (any simple example?)? Furthermore if I change models by interpretations, does it hold that $T \vdash \varphi$ iff for all interpretations (in the sense defined by Tarski) $M$ of the language $L (T)$ (the underlying language of the theory $T$) in a theory $T'$, $T' \vdash\varphi^M$ for the higher order case (I´m interested mainly in the second order case)? Thanks in advance. EDIT For first order languages: An interpretation $I$ of a language $L$ in a language $L'$ is a correspondence that associates for each predicate symbol $P$ of $L$ a predicate symbol $P_I$ of $L'$. And for each function symbol $f$ in $L$ a function symbol $f_I$ of $L'$. Furthermore, given a (first order) theory $T'$ in $L'$, then I is said to be a interpretation of $L$ in $T'$ if $1)$ There is a fixed predicate symbol in $\mathfrak{U}_I$, called the domain. $2)$For each function symbol $f$ in $L$, $T'\vdash\mathfrak{U}_I x_1 \rightarrow … \rightarrow \mathfrak{U}_Ix_n \rightarrow \mathfrak{U}_If_I x_1…x_n$ Moreover, given a theory $T$ in $L$, $I$ is said to be an interpretation of $T$ in $T'$ if for each formula $\varphi$ it holds that $T'\vdash\mathfrak{U}_I x_1 \rightarrow … \rightarrow \mathfrak{U}_Ix_n \rightarrow \varphi^I $ where $\varphi^I$ is defined inductively as: $\exists x (\mathfrak{U}_I x \wedge \psi^I)$ if $\varphi$ is $\exists x\psi$ and the other cases (negation, conjunction, predicate symbol, function symbol etc) in the obvious way. The main difference between models and interpretations are that in a model the predicate $\mathfrak{U}_I$ defines a set and that $T' \vdash \forall y (y \ \text{axiom of T}) \rightarrow I \models y$. While, in an interpretation, it just holds that for every axiom $\varphi$ of $T$, $T' \vdash \varphi^I$.
Unpolarized light is more properly called light with random polarization. That makes it more clear what it means: the polarization state (circular, linear, elliptic) varies randomly over space, wavelength, and time. Consider the scenario below, where a diffuse light source is converted to a collimated beam with a narrow range of wavelengths $\lambda\pm\Delta\lambda$. The beam is then split by a polarizer. If you took an intensity profile of the two polarization components with a sufficiently short exposure time, you would get two light/dark patterns that would clearly show that some parts of the beam are clearly s-polarized (polarization vector perpendicular to the drawing plane), others clearly p-polarized, and yet others something in between or even dark. However, if you took another picture, the pattern would look completely different. If the exposure time is too long, then you would get two images that are both a uniform 50% gray image. The maximum shutter time $\tau$ for which you can see a snapshot of the polarization would depend on the bandwidth $\Delta\lambda$ as $$\tau\approx \frac{\lambda^2}{2\pi c\Delta\lambda},$$where $c=3\times10^8~\mathrm{m/s}$ is the speed of light. For example, if you do this with narrowband red light ($\lambda = 650\pm1$ nm), you would need an exposure time of 200 femtoseconds (2E-13 s). So, even though the light has a definite polarization at a particular point in time, you will not notice it in practice. There is another effect that makes it difficult to notice the instantaneous polarization of randomly polarized light: the length scale of the spatial intensity fluctuations in the two sensor images will depend on the size of the pinhole. For a beam with 1 cm diameter and a lens with 10 cm focal length, the pinhole would need to be 20 $\mu$m or so to see the pattern clearly. If you increase the size of the pinhole, then the patterns will become finer and finer until you cannot resolve them anymore. And if the pinhole is so small and the exposure time is so short, you would need to start with a pretty bright light source in order to see anything at all.
My favourite one: $[0, 1]$ is compact, i.e. every open cover of $[0, 1]$ has a finite subcover. Proof: Suppose for a contradiction that there is an open cover $\mathcal{U}$ which does not admit any finite subcover. Thus, either $\left[ 0, \frac{1}{2} \right]$ or $\left[ \frac{1}{2}, 1 \right]$ cannot be covered with a finite number of sets from $\mathcal{U}$ - name it $I_1$. Again, one of the two $I_1$'s subintervals of length $\frac{1}{4}$ can't be covered with a finite number of sets from $\mathcal{U}$. Continuing, we get a descending sequence of intervals $I_n$ of length $\frac{1}{2^n}$ each, every of which cannot be finitely covered. By the Cantor Intersection Theorem, $$\bigcap_{n=1}^{\infty} I_n = \{ x \}$$ for some $x \in [0, 1].$ But there is such $U \in \mathcal{U}$ that $x \in U$ and so $I_n \subseteq U$ for some sufficiently large $n$. That's a contradiction. But given an arbitrary cover $\mathcal{U}$, I think finding a finite subcover may be a somewhat tedious task. :p P.S. There actually comes a procedure from the proof above: See if $[0, 1]$ itself is covered by one set from $\mathcal{U}$. If so, we're done. If not, execute step 1. for $\left[ 0, \frac{1}{2} \right]$ and $\left[ \frac{1}{2}, 1 \right]$ to get their finite subcovers, then unite them. The proof guarantees you will eventually find a finite subcover (i.e. you'll never end up going downwards infinitely), but you cannot tell how long it will take. So it is not as constructive as one would expect.
Let $n$ be a natural number. I want to calculate $$\int_0^{\infty} \frac{x^n}{1+x^{2n}}$$ using contour integration. I declare $f(z) = \frac{z^n}{1+z^{2n}}$. This function has $2n$ simple poles of the form $exp(i\frac{-\pi}{2n} + i\frac{\pi k}{n})$ for $k = 0,1,2, ... , 2n-1$. My problem is to find which contour to use. I thought about using the following: $\gamma_1(t) = t$, $t \in [0,R]$ $\gamma_2(t) = Re^{it}$, $t \in [0,\frac{\pi}{n}]$ $\gamma_3(t) = te^{i\frac{\pi}{n}}$, $t \in [0,R]$ This way I have only one pole in the contour and I can use the residue theorem. Is this a good approach? I somehow have problem with the calculations here. Help would be appreciated
Contents DFT ( Discrete Fourier Transform ) The Discrete Time Fourier Tranform is a good way to analyze discrete time signals in the frequency domain in theory, but in application the infinite range of time and frequency in the conversion formulas can make analysis difficult, especially on the computer. The Discrete Fourier Transform is very similar to the DFT, but uses a finite time signal, allowing its formulas to be finite sums which a computer can easily calculate. Signals must be discrete and time-limited (or truncated) for use with the DFT. A discrete periodic signal can be used when only one period of the signal is analyzed. The DFT of a signal will be discrete and have a finite duration. Definition DFT $ X[k] = \sum_{n=0}^{N-1}{x[n]e^{-j \frac{2{\pi}}{N}kn}}, \mbox{ }k = 0, 1, 2, ..., N-1 $ Inverse DFT (IDFT) $ x[n] = \frac{1}{N}\sum_{k=0}^{N-1}{X(k)e^{j \frac{2{\pi}}{N}kn}}, \mbox{ }n = 0, 1, 2, ..., N-1 $ Properties Linearity For all $ a,b $ in the complex plane, and all $ x_1[n],x_2[n] $ with the same period N $ ax_1[n] + bx_2[n] \longleftrightarrow aX_1[k] + bX_2[k] $ Time-Shifting For all $ n_0 $ , and all x[n] with period N $ x[n - n_0] \longleftrightarrow X[k]e^{-j \frac{2{\pi}}{N} n_0 k} $ Modulation $ x[n]e^{j \frac{2 \pi}{N}k_0n} \longleftrightarrow X[k-k_0] $ Duality $ X[n] \longleftrightarrow Nx[-k] $, where X[n] is the DFT of a DFT Parseval's Relation $ \sum_{n=0}^{N-1} \|x[n]\|^2 = \frac{1}{N} \sum_{k=0}^{N-1} \|X[k]\|^2 $ Initial Value $ \sum_{n=0}^{N-1} x[n] = X[0] $ Periodicity $ X[k + N] = X[k] $ for all k. X[k] is periodic with the same period N as x[n]. Relation to DTFT $ X[k] = Y(k \frac{ 2 \pi}{N}) $ where Y(w) is the DTFT of signal $ y[n] = \left\{ \begin{array}{c l} x[n] & n=0,...,N-1 \\ 0 & \mbox{ else} \end{array} \right. $ Important DFT Pairs $ x[n] = \delta [n],\mbox{ }0 \le n < N \longleftrightarrow X[k] = 1,\mbox{ } 0 \le k < N $ both repeat with period N $ x[n] = 1,\mbox{ } 0 \le n < N \longleftrightarrow X[k] = N \delta [n],\mbox{ } 0 \le k < N $ both repeat with period N $ x[n] = e^{j2 \pi k_0n}, \mbox{ }0 \le n < N \longleftrightarrow X[k] = N \delta [k-k_0],\mbox{ } 0 \le k < N $ both repeat with period N $ x[n] = cos( \frac{2 \pi}{N} k_0n) \longleftrightarrow \frac{N}{2}(\delta [k-k_0] + \delta[k-(N-k_0)],\mbox{ } 0 \le k < N $ both repeat with period N Spectral Analysis via DFT There are two sources of inaccuracies in the spectral analysis of a DFT signal. These are leakage, which is caused by signal truncation, and the picket fence effect, which is caused by frequency sampling. Leakage is caused by the convolution by the DTFT of a window function, which is a sinc. This causes copies of the original spectrum with decreasing amplitude to be placed next to the original spectrum. A graphic analysis of this can be seen in Some material for the lecture on DFT (re:leakage effect). The picket fence effect is caused by samples of a signal not necessarily occurring at important points, such as local maximums and minimums. Only the points that are sampled are available for data, which can cause some problems and inaccuracies. There is a reconstruction formula to obtain the DTFT of the truncated x[n] from the DFT. It assumes $ x_p[n] $ is the periodic repetition (with period N, the duration of x[n]) of x[n]. The formula is $ X_{tr}(w) = \sum_{k=0}^{N-1} X[k]P(w - \frac{2 \pi}{N}k) $, where $ P(w) = \left\{ \begin{array}{c l} 1 & \mbox{if }w \mbox{ a multiple of 2}\pi \\ \frac{sin(w \frac{N}{2})}{Nsin( \frac{N}{2})} & \mbox{ else} \end{array} \right. $
Contents What Will We Study in this Chapter? In the first chapter of this lesson, we said that projection matrices were used in the GPU rendering pipeline. We mentioned that there were two GPU rendering pipelines: the old one, called the fixed-function pipeline and the new one which is generally referred to as the programmable rendering pipeline. We also talked about how clipping, the process that consists of discarding or trimming primitives that are either outside or straddling across the boundaries of the frustum, happens somehow while the points are being transformed by the projection matrix. Finally, we also explained that in fact, projection matrices don't convert points from camera space to NDC space but to homogeneous clip space. It is time to provide more information about these different topics. Let's explain what it means when we say that clipping happens while the points are being transformed. Let's explain what clip space is. And finally let's review how the projection matrices are used in the old and the new GPU rendering pipeline. Clipping and Clip Space Let's recall quickly that the main purpose of clipping is to essentially "reject" geometric primitives which are behind the eye or located exactly at the eye position (this would mean a division by 0 which we don't want) and more generally trim off part of the geometric primitives which are outside the viewing area (more information on this topic can be found in chapter 2). This viewing area is defined by the truncated pyramid of the perspective or viewing frustum. Any professional rendering system actually somehow needs to implement this step. Note that the process can result into creating more triangles as shown in figure 1 than the scenes initially contained. The most common clipping algorithms are the Cohen-Sutherland algorithm for lines and the Sutherland-Hodgman algorithm for polygons. It happens that clipping is more easily done in clip space than in camera space (before vertices are transformed by the projection matrix) or screen space (after the perspective divide). Remember that when the points are transformed by the projection matrix, they are first transformed as you would with any other 4x4 matrix, and the transformed coordinates are then normalized: that is, the x- y- and z-coordinates of the transformed points are divided by the transformed point z-coordinate. Clip space is the space points are in just before they get normalized. In summary, what happens on a GPU is this. Points are transformed from camera space to clip space in the vertex shader. The input vertex is converted from Cartesian coordinates to homogeneous coordinates and its w-coordinate is set to 1. The predefined gl_Position variable, in which the transformed point is stored, is also a point with homogeneous coordinates. Though when the input vertex is multiplied by the projection matrix, the normalized step is not yet performed. gl_Position is in homogeneous clip space. When all the vertices have been processed by the vertex shader, triangles whose vertices are now in clip space are clipped. Once clipping is done, all vertices are normalized. Their x- y- and z-coordinates of each vertex are divided by their respective w-coordinate. This is where and when perspective divide occurs. Let's recall, that after the normalization step, points which are visible to the camera are all contained in the range [-1,1] both in x and y. This happens in the last part of the point-matrix multiplication process, when the coordinates are normalized as we just said: $$\begin{array}{l} -1 \leq \dfrac{x'}{w'} \leq 1 \\ -1 \leq \dfrac{y'}{w'} \leq 1 \\ -1 \leq \dfrac{z'}{w'} \leq 1 \\ \end{array}$$ Or: $0 \leq \dfrac{z'}{w'} \leq 1$ depending on the convention you are using. Therefore we can also write:$$\begin{array}{l} -w' \leq x' \leq w' \\ -w' \leq y' \leq w' \\ -w' \leq z' \leq w' \\ \end{array}$$ Which is the state x', y' and z' are before they get normalized by w' or to say it different, when coordinates are in clip space. We can add a fourth equation: $0 \lt w'$. The purpose of this equation is to guarantee that we will never divide any of the coordinates by 0 (which would be a degenerate case). These equations mathematically works. You don't really need though to try to represent what vertices look like or what it means to work with a four-dimensional space. All it says is that the clip space of a given vertex whose coordinates are {x, y, z} is defined by the extents [-w,w] (the w value indicates what the dimensions of the clip space are). Note that this clip space is the same for each coordinate of the point and the clip space of any given vertex is a cube. Though note also that each point is likely to have its own clip space (each set of x, y and z-coordinate is likely to have a different w value). In other words, every vertex has its own clip space in which it exists (and basically needs to "fit" in). This lesson is only about projection matrices. All we need to know in the context of this lesson, is to know where clipping occurs in the vertex transformation pipeline and what clip space means, which we just explained. Everything else will be explained in the lessons on the Sutherland-Hodgman and the Cohen-Sutherland algorithms which you can find in the Advanced Rasterization Techniques section. The "Old" Point (or Vertex) Transformation Pipeline Vertex is a better term when it comes to describe how points (vertices) are transformed in OpenGL (or Direct3D, Metal or any other graphics API you can think of). OpenGL (and other graphics APIs) had (in the old fixed-function pipeline) two possible modes for modifying the state of the camera: GL_PROJECTION and GL_MODELVIEW. GL_PROJECTION allowed to set the projection matrix itself. As we know by now (see previous chapter) this matrix is build from the left, right, bottom and top screen coordinates (which are computed from the camera's field of view and near clipping plane), as well as the near and far clipping planes (which are parameters of the camera). These parameters define the shape of the camera's frustum and all the vertices or points from the scene contained within this frustum are visible. In OpenGL, these parameters were passed to the API through a call to glFrustum (which we show an implementation of in the previous chapter): GL_MODELVIEW mode allowed to set the world-to-camera matrix. A typical OpenGL program set the perspective projection matrix and the model-view matrix using the following sequence of calls: First we would make the GL_PROJECTION mode active (line 1). Next, to set up the projection matrix, we would make a call to glFrustum passing as arguments to the function, the left, right, bottom and top screen coordinates as well as the near and far clipping planes. Once the projection matrix was set, we would switch to the GL_MODELVIEW mode (line 4). Actually, the GL_MODELVIEW matrix could be seen as the combination of the "VIEW" transformation matrix (the world-to-camera matrix) with the "MODEL" matrix which is the transformation applied to the object (the object-to-world matrix). There was not concept of world-to-camera transform separate from the object-to-world transform. The two transforms were combined in the GL_MODELVIEW matrix.$${GL\_MODELVIEW} = M_{object-to-world} * M_{world-to-camera}$$ First a point $P_w$ expressed in world space was transformed to camera space (or eye space) using the GL_MODELVIEW matrix. The resulting point $P_c$ was then projected onto the image plane using the GL_PROJECTION matrix. We ended up with a point expressed in homogeneous coordinates in which the coordinate w contained the point $P_c$'s z coordinate. The Vertex Transformation Pipeline in the New Programmable GPU Rendering Pipeline The pipeline in the new programmable GPU rendering pipeline is more or less the same than the old pipeline, but what is really different in this new pipeline, is the way you set things up. In the new pipeline, there is no more concept of GL_MODELVIEW or GL_PROJECTION mode. This step can now be freely programmed in a vertex shader. As mentioned in the first chapter of this lesson, the vertex shader, is like a small program. You can program this vertex shader to tell the GPU how vertices making up the geometry of the scene should be processed. In other words, this is where you should be doing all your vertex transformations: the world-to-camera transformation if necessary but more importantly the projection transformation. A program using an OpenGL API doesn't produced an image if the vertex and its correlated fragment shader are not defined. The simplest form of vertex shader looks like this: This program doesn't even transform the input vertex with a perspective projection matrix, which in some cases can produce a visible result depending on the size and the position of the geometry as well as how the viewport is set. But this is not relevant in this lesson. What we can see by looking at this code is that the input vertex is set to be a vec4 which is nothing else than a point with homogeneous coordinates. Note that gl_Position too is a point with homogeneous coordinates. As expected, the vertex shader output the position of the vertex in clip space (see diagram of the vertex transformation pipeline above). In reality you are more likely to use a vertex shader like this one: It uses both a world-to-camera and projection matrix to transform the vertex to camera space and then clip space. Both matrices are set externally in program using some calls (glGetUniformLocation to find the location of the variable in the shader and glUniformMatrix4fv to set the matrix variable using the previously found location) that are provided to you by the OpenGL API: In fact things get even more confusing if you look at the order in which the matrices are used in the OpenGL vertex shader. You will notice we write $Proj * View * vtx$ instead of $vtx * View * Proj$. The former form is used when you deal with column-major matrices (because it implies that you multiply the matrix by the point rather than the point by the matrix as explained in our lesson on Geometry. Conclusion? OpenGL assume matrices are column-major (so this is how you need to use them in shaders) yet coefficients are mapped in memory using a row-major order form. Confusing? Remember that matrices in OpenGL (and vectors) use column-major order. Thus if you use a row vectors like we do on Scratchapixel, you will need to transpose the matrix before setting up the matrix of the vertex shader (line 2). They are other ways of doing this in modern OpenGL but we will skip them in this lesson which is not devoted to that topic. This information can easily be found on the Web anyway.
Get the noise calibration. That is, the ratio \[ \frac{\sigma_{i}}{g_{i}k} \] where $ k$ is a constant determined by the electronics of units DAC/MIP, and $ \sigma_i, g_i$ are the noise and gain of the $ i $ strip respectively. This correction is needed because some of the reconstructed data (what which have an AliESDFMD class version less than or equal to 3) used the wrong zero-suppression factor. The zero suppression factor used by the on-line electronics was 4, but due to a coding error in the AliFMDRawReader a zero suppression factor of 1 was assumed during the reconstruction. This shifts the zero of the energy loss distribution artificially towards the left (lover valued signals). So let's assume the real zero-suppression factor is $ f$ while the zero suppression factor $ f'$ assumed in the reconstruction was (wrongly) lower. The number of ADC counts $ c_i'$ used in the reconstruction can be calculated from the reconstructed signal $ m_i'$ by \[ c_i' = m_i \times g_i \times k / \cos\theta_i \] where $\theta_i$ is the incident angle of the $ i$ strip. This number of counts used the wrong noise factor $ f'$ so to correct to the on-line value, we need to do \[ c_i = c_i' - \lfloor f'\times n_i\rfloor + \lfloor f\times n_i\rfloor \] which gives the correct number of ADC counts over the pedestal. To convert back to the scaled energy loss signal we then need to calculate (noting that $ f,f'$ are integers) \begin{eqnarray} m_i &=& \frac{c_i \times \cos\theta_i}{g_i \times k}\\ &=& \left(c_i' - \lfloor f'\times n_i\rfloor + \lfloor f\times n_i\rfloor\right)\frac{\cos\theta}{g_i \times k}\\ &=& \left(\frac{m_i'\times g_i\times k}{\cos\theta} - \lfloor f'\times n_i\rfloor + \lfloor f\times n_i\rfloor\right) \frac{\cos\theta}{g_i \times k}\\ &=& m_i' + \frac{1}{g_i \times k} \left(\lfloor f\times n_i\rfloor- \lfloor f'\times n_i\rfloor\right)\cos\theta\\ &=& m_i' + \frac{\lfloor n_i\rfloor}{g_i \times k} \left(f-f'\right)\cos\theta \end{eqnarray} Definition at line 63 of file AliFMDCorrNoiseGain.h.
I was confused on how to find the derivative of this function: y = log 2(x/(x-1) If anyone could help me, please and thank you!!! $y=\log_2\left(\dfrac{x}{x-1}\right) = \dfrac{1}{\ln(2)}\ln\left(\dfrac{x}{x-1}\right)\\ \dfrac{dy}{dx} = \dfrac{1}{\ln(2)}\cdot \dfrac{1}{\dfrac{x}{x-1}}\cdot \left(\dfrac{(x-1)-x}{(x-1)^2}\right)= \\ -\dfrac{1}{\ln(2)}\cdot \dfrac{x-1}{x}\cdot \dfrac{1}{(x-1)^2} = \\ -\dfrac{1}{\ln(2)}\dfrac{1}{x(x-1)} =\\ \dfrac{1}{\ln(2)}\dfrac{1}{x(1-x)}$.
J. Bagaria, J. D. Hamkins, K. Tsaprounis, and T. Usuba, “Superstrong and other large cardinals are never Laver indestructible,” Arch. Math. Logic, vol. 55, iss. 1-2, pp. 19-35, 2016. (special volume in memory of R.~Laver) @ARTICLE{BagariaHamkinsTsaprounisUsuba2016:SuperstrongAndOtherLargeCardinalsAreNeverLaverIndestructible, AUTHOR = {Bagaria, Joan and Hamkins, Joel David and Tsaprounis, Konstantinos and Usuba, Toshimichi}, TITLE = {Superstrong and other large cardinals are never {L}aver indestructible}, JOURNAL = {Arch. Math. Logic}, FJOURNAL = {Archive for Mathematical Logic}, note = {special volume in memory of R.~Laver}, VOLUME = {55}, YEAR = {2016}, NUMBER = {1-2}, PAGES = {19--35}, ISSN = {0933-5846}, MRCLASS = {03E55 (03E40)}, MRNUMBER = {3453577}, MRREVIEWER = {Peter Holy}, DOI = {10.1007/s00153-015-0458-3}, eprint = {1307.3486}, archivePrefix = {arXiv}, primaryClass = {math.LO}, url = {http://jdh.hamkins.org/superstrong-never-indestructible/}, } Abstract. Superstrong cardinals are never Laver indestructible. Similarly, almost huge cardinals, huge cardinals, superhuge cardinals, rank-into-rank cardinals, extendible cardinals, $1$-extendible cardinals, $0$-extendible cardinals, weakly superstrong cardinals, uplifting cardinals, pseudo-uplifting cardinals, superstrongly unfoldable cardinals, $\Sigma_n$-reflecting cardinals, $\Sigma_n$-correct cardinals and $\Sigma_n$-extendible cardinals (all for $n\geq 3$) are never Laver indestructible. In fact, all these large cardinal properties are superdestructible: if $\kappa$ exhibits any of them, with corresponding target $\theta$, then in any forcing extension arising from nontrivial strategically ${\lt}\kappa$-closed forcing $\mathbb{Q}\in V_\theta$, the cardinal $\kappa$ will exhibit none of the large cardinal properties with target $\theta$ or larger. The large cardinal indestructibility phenomenon, occurring when certain preparatory forcing makes a given large cardinal become necessarily preserved by any subsequent forcing from a large class of forcing notions, is pervasive in the large cardinal hierarchy. The phenomenon arose in Laver’s seminal result that any supercompact cardinal $\kappa$ can be made indestructible by ${\lt}\kappa$-directed closed forcing. It continued with the Gitik-Shelah treatment of strong cardinals; the universal indestructibility of Apter and myself, which produced simultaneous indestructibility for all weakly compact, measurable, strongly compact, supercompact cardinals and others; the lottery preparation, which applies generally to diverse large cardinals; work of Apter, Gitik and Sargsyan on indestructibility and the large-cardinal identity crises; the indestructibility of strongly unfoldable cardinals; the indestructibility of Vopenka’s principle; and diverse other treatments of large cardinal indestructibility. Based on these results, one might be tempted to the general conclusion that all the usual large cardinals can be made indestructible. In this article, my co-authors and I temper that temptation by proving that certain kinds of large cardinals cannot be made nontrivially indestructible. Superstrong cardinals, we prove, are never Laver indestructible. Consequently, neither are almost huge cardinals, huge cardinals, superhuge cardinals, rank-into-rank cardinals, extendible cardinals and $1$-extendible cardinals, to name a few. Even the $0$-extendible cardinals are never indestructible, and neither are weakly superstrong cardinals, uplifting cardinals, pseudo-uplifting cardinals, strongly uplifting cardinals, superstrongly unfoldable cardinals, $\Sigma_n$-reflecting cardinals, $\Sigma_n$-correct cardinals and $\Sigma_n$-extendible cardinals, when $n\geq 3$. In fact, all these large cardinal properties are superdestructible, in the sense that if $\kappa$ exhibits any of them, with corresponding target $\theta$, then in any forcing extension arising from nontrivial strategically ${\lt}\kappa$-closed forcing $\mathbb{Q}\in V_\theta$, the cardinal $\kappa$ will exhibit none of the large cardinal properties with target $\theta$ or larger. Many quite ordinary forcing notions, which one might otherwise have expected to fall under the scope of an indestructibility result, will definitely ruin all these large cardinal properties. For example, adding a Cohen subset to any cardinal $\kappa$ will definitely prevent it from being superstrong—as well as preventing it from being uplifting, $\Sigma_3$-correct, $\Sigma_3$-extendible and so on with all the large cardinal properties mentioned above—in the forcing extension. Main Theorem. Superstrong cardinals are never Laver indestructible. Consequently, almost huge, huge, superhuge and rank-into-rank cardinals are never Laver indestructible. Similarly, extendible cardinals, $1$-extendible and even $0$-extendible cardinals are never Laver indestructible. Uplifting cardinals, pseudo-uplifting cardinals, weakly superstrong cardinals, superstrongly unfoldable cardinals and strongly uplifting cardinals are never Laver indestructible. $\Sigma_n$-reflecting and indeed $\Sigma_n$-correct cardinals, for each finite $n\geq 3$, are never Laver indestructible. Indeed—the strongest result here, because it is the weakest notion—$\Sigma_3$-extendible cardinals are never Laver indestructible. In fact, each of these large cardinal properties is superdestructible. Namely, if $\kappa$ exhibits any of them, with corresponding target $\theta$, then in any forcing extension arising from nontrivial strategically ${\lt}\kappa$-closed forcing $\mathbb{Q}\in V_\theta$, the cardinal $\kappa$ will exhibit none of the mentioned large cardinal properties with target $\theta$ or larger. The proof makes use of a detailed analysis of the complexity of the definition of the ground model in the forcing extension. These results are, to my knowledge, the first applications of the ideas of set-theoretic geology not making direct references to set-theoretically geological concerns. Theorem 10 in the article answers (the main case of) a question I had posed on MathOverflow, namely, Can a model of set theory be realized as a Cohen-subset forcing extension in two different ways, with different grounds and different cardinals? I had been specifically interested there to know whether a cardinal $\kappa$ necessarily becomes definable after adding a Cohen subset to it, and theorem 10 shows indeed that it does: after adding a Cohen subset to a cardinal, it becomes $\Sigma_3$-definable in the extension, and this fact can be seen as explaining the main theorem above. Related MO question \| CUNY talk
also, if you are in the US, the next time anything important publishing-related comes up, you can let your representatives know that you care about this and that you think the existing situation is appalling @heather well, there's a spectrum so, there's things like New Journal of Physics and Physical Review X which are the open-access branch of existing academic-society publishers As far as the intensity of a single-photon goes, the relevant quantity is calculated as usual from the energy density as $I=uc$, where $c$ is the speed of light, and the energy density$$u=\frac{\hbar\omega}{V}$$is given by the photon energy $\hbar \omega$ (normally no bigger than a few eV) di... Minor terminology question. A physical state corresponds to an element of a projective Hilbert space: an equivalence class of vectors in a Hilbert space that differ by a constant multiple - in other words, a one-dimensional subspace of the Hilbert space. Wouldn't it be more natural to refer to these as "lines" in Hilbert space rather than "rays"? After all, gauging the global $U(1)$ symmetry results in the complex line bundle (not "ray bundle") of QED, and a projective space is often loosely referred to as "the set of lines [not rays] through the origin." — tparker3 mins ago > A representative of RELX Group, the official name of Elsevier since 2015, told me that it and other publishers “serve the research community by doing things that they need that they either cannot, or do not do on their own, and charge a fair price for that service” for example, I could (theoretically) argue economic duress because my job depends on getting published in certain journals, and those journals force the people that hold my job to basically force me to get published in certain journals (in other words, what you just told me is true in terms of publishing) @EmilioPisanty > for example, I could (theoretically) argue economic duress because my job depends on getting published in certain journals, and those journals force the people that hold my job to basically force me to get published in certain journals (in other words, what you just told me is true in terms of publishing) @0celo7 but the bosses are forced because they must continue purchasing journals to keep up the copyright, and they want their employees to publish in journals they own, and journals that are considered high-impact factor, which is a term basically created by the journals. @BalarkaSen I think one can cheat a little. I'm trying to solve $\Delta u=f$. In coordinates that's $$\frac{1}{\sqrt g}\partial_i(g^{ij}\partial_j u)=f.$$ Buuuuut if I write that as $$\partial_i(g^{ij}\partial_j u)=\sqrt g f,$$ I think it can work... @BalarkaSen Plan: 1. Use functional analytic techniques on global Sobolev spaces to get a weak solution. 2. Make sure the weak solution satisfies weak boundary conditions. 3. Cut up the function into local pieces that lie in local Sobolev spaces. 4. Make sure this cutting gives nice boundary conditions. 5. Show that the local Sobolev spaces can be taken to be Euclidean ones. 6. Apply Euclidean regularity theory. 7. Patch together solutions while maintaining the boundary conditions. Alternative Plan: 1. Read Vol 1 of Hormander. 2. Read Vol 2 of Hormander. 3. Read Vol 3 of Hormander. 4. Read the classic papers by Atiyah, Grubb, and Seeley. I am mostly joking. I don't actually believe in revolution as a plan of making the power dynamic between the various classes and economies better; I think of it as a want of a historical change. Personally I'm mostly opposed to the idea. @EmilioPisanty I have absolutely no idea where the name comes from, and "Killing" doesn't mean anything in modern German, so really, no idea. Googling its etymology is impossible, all I get are "killing in the name", "Kill Bill" and similar English results... Wilhelm Karl Joseph Killing (10 May 1847 – 11 February 1923) was a German mathematician who made important contributions to the theories of Lie algebras, Lie groups, and non-Euclidean geometry.Killing studied at the University of Münster and later wrote his dissertation under Karl Weierstrass and Ernst Kummer at Berlin in 1872. He taught in gymnasia (secondary schools) from 1868 to 1872. He became a professor at the seminary college Collegium Hosianum in Braunsberg (now Braniewo). He took holy orders in order to take his teaching position. He became rector of the college and chair of the town... @EmilioPisanty Apparently, it's an evolution of ~ "Focko-ing(en)", where Focko was the name of the guy who founded the city, and -ing(en) is a common suffix for places. Which...explains nothing, I admit.
Answer $65$ Work Step by Step We know that $ n!=1 \cdot 2 \cdot 3 .....(n-1)n $ Thus, we have $\dfrac{40!}{38!}=\dfrac{40 \cdot 39 \cdot 38!}{4 \cdot 3 \cdot 2 \cdot 1 \times 38! }$ $=\dfrac{40 \cdot 39}{24}$ $=65$ You can help us out by revising, improving and updating this answer.Update this answer After you claim an answer you’ll have 24 hours to send in a draft. An editorwill review the submission and either publish your submission or provide feedback.
Mössbauer spectroscopy is a versatile technique used to study nuclear structure with the absorption and re-emission of gamma rays, part of the electromagnetic spectrum. The technique uses a combination of the Mössbauer effect and Doppler shifts to probe the hyperfine transitions between the excited and ground states of the nucleus. Mössbauer spectroscopy requires the use of solids or crystals which have a probability to absorb the photon in a recoilless manner, many isotopes exhibit Mössbauer characteristics but the most commonly studied isotope is 57Fe. Introduction Rudolf L. Mössbauer became a physics student at Technical University in Munich at the age of 20. After passing his intermediate exams Mössbauer began working on his thesis and doctorate work in 1955, while working as an assistant lecturer at Institute for Mathematics.In 1958 at the age of 28 Mössbauer graduated, and also showed experimental evidence for recoilless resonant absorption in the nucleus, later to be called the Mössbauer Effect.In 1961 Mössbauer was awarded the Nobel Prize in physics and, under the urging of Richard Feynman, accepted the position of Professor of Physics at the California Institute of Technology. Mössbauer Effect The recoil energy associated with absorption or emission of a photon can be described by the conservation of momentum.In it we find that the recoil energy depends inversely on the mass of the system. For a gas the mass of the single nucleus is small compared to a solid.The solid or crystal absorbs the energy as phonons, quantized vibration states of the solid, but there is a probability that no phonons are created and the whole lattice acts as the mass, resulting in a recoilles emission of the gamma ray. The new radiation is at the proper energy to excite the next ground state nucleus. The probability of recoilles events increases with decreasing transition energy. \[P_R = P_{\gamma} \tag{1}\] \[P^2_\gamma = P^2_{\gamma}\tag{2}\] \[2 M E_R = \dfrac{E^2_{\gamma}}{c^2}\tag{3}\] \[E_R = \dfrac{E^2_\gamma}{2M{c^2}}\tag{4}\] Doppler Effect The Doppler shift describes the change in frequency due to a moving source and a moving observer.$f$ is the frequency measured at the observer, $v$ is the velocity of the wave so for our case this is the speed of light $c$, $v_r$ is the velocity of the observer, $v_s$ is the velocity of the source which is positive when heading away from the observer, and $f_0$ is the initial frequency. \[f = {\left (\dfrac{v+v_r}{v+v_s}\right)} f_0 \tag{5}\] \[f = {\left (\dfrac{c}{c+v_s}\right)} f_0\tag{6}\] In the case where the source is moving toward a stationary observer the perceived frequency is higher.For the opposite situation where the source travels away from the observer frequencies recorded at the observer will be of lower compared to the initial wave. The energy of a photon is related to the product of Planck's constant and the frequency of the electromagnetic radiation. Thus for increasing frequencies the corresponding energy also increase, and the same is true in the reverse case where frequencies decrease and therefore energy decreases. \[E = \dfrac{hc}{\lambda} = hv \tag{7}\] The energy differences between hyperfine states are minimal (fractions of an eV) and the energy variation is achieved by the moving the source toward and away from the sample in an oscillating manner, commonly at a velocity of a few mm/s.The transmittance is then plotted against the velocity of the source and a peak is seen at the energy corresponding to the resonance energy. In the above spectrum the emission and absorption are both estimated by the Lorentzian distribution. Mössbauer Isotopes By far the most common isotopes studied using Mössbauer spectroscopy is 57Fe, but many other isotopes have also displayed a Mössbauer spectrum. Two criteria for functionality are The excited state is of very low energy, resulting in a small change in energy between ground and excited state. This is because gamma rays at higher energy are not absorbed in a recoil free manner, meaning resonance only occurs for gamma rays of low energy. The resolution of Mössbauer spectroscopy depends upon the lifetime of the excited state. The longer the excited state lasts the better the image. Both conditions are met by 57Fe and it is thus used extensively in Mössbauer spectroscopy. In the figure to the right the red colored boxes of the periodic table of elements indicate all elements that have isotopes visible using the Mössbauer technique. Hyperfine Interactions Mössbauer spectroscopy allows the researcher to probe structural elements of the nucleus in several ways, termed isomer shift, quadrupole interactions, and magnetic splitting. These are each explained by the following sections as individual graphs, but in practice Mössbauer spectrum are likely to contain a combination of all effects. Isomer Shift An isomeric shift occurs when non identical atoms play the role of source and absorber, thus the radius of the source, Rs, is different that of the absorber, Ra, and the same holds that the electron density of each species is different. The Coulombic interactions affects the ground and excited state differently leading to a energy difference that is not the same for the two species.This is best illustrated with the equation: \[R_A \neq R_S \tag{8}\] \[\rho_S \neq \rho_S \tag{9}\] \[E_A \neq E_S \tag{10}\] \[\delta = E_A-E_S = \dfrac{2}{3}nZ{e^2}{(\rho_A - \rho_S)}(R^2_{es} - R^2_{gs}) \tag{11}\] Where delta represents the change in energy necessary to excite the absorber, which is seen as a shift from the Doppler speed 0 to V 1. The isomer shift depends directly on the s-electrons and can be influenced by the shielding p,d,f electrons.From the measured delta shift there is information about the valance state of the absorbing atom The energy level diagram for $\delta$ shift shows the change in source velocity due to different sources used. The shift may be either positive or negative. Quadrupole Interaction The Hamiltonian for quadrupole interaction using ${}^{57}Fe$ nuclear excited state is given by \[H_Q = \dfrac{eQV_{ZZ}}{12}[3I^2_Z-I(I+1) + \eta(I^2_X-I^2_y)] \tag{12}\] where the nuclear excited states are split into two degenerate doublets in the absence of magnetic interactions. For the asymmetry parameter $\eta = 0$ doublets are labeled with magnetic quantum numbers $m_{es} = {}^+_-3/2$ and $m_{es} = {}^+_-1/2$, where the $m_{es} = {}^+_-3/2$ doublet has the higher energy. The energy difference between the doublets is thus \[\Delta{EQ} = \dfrac{eQV_{zz}}{2}\sqrt{1+\dfrac{\eta^2}{3}} \tag{13}\] The energy diagram and corresponding spectrum can be seen as Magnetic Splittin g Magnetic splitting of seen in Mössbauer spectroscopy can be seen because the nuclear spin moment undergoes dipolar interactions with the magnetic field \[E(m_I) = -g_n{\beta_n}{B_{eff}}m_I \tag{14}\] where $g_n$ is the nuclear g-factor and $\beta_n$ is the nuclear magneton. In the absence of quadrupole interactions the Hamiltonian splits into equally spaced energy levels of The allowed gamma stimulated transitions of nuclear excitation follows the magnetic dipole transition selection rule: \[ \Delta I = 1 and \Delta m_I = 0, {}^+_-1 \tag{15}\] \[m_I\] is the magnetic quantum number and the direction of $\beta$ defines the nuclear quantization axis. If we assume $g$ and $A$ are isotropic (direction independent) where $g_x = g_y = g_z$ and B is actually a combination of the applied and internal magnetic fields: \[H = g\beta{S}\centerdot{B}+AS\centerdot{I} - g_n\beta_nB\centerdot{I} \tag{16}\] The electronic Zeeman term is far larger then the nuclear Zeeman term, meaning the electronic term dominates the equation so $S$ is approximated by $ \langle S \rangle$ and \[ \langle S_z\rangle = m_s = {}^+_-\dfrac{1}{2} \tag{17a}\] and \[\langle S_x \rangle = \langle S_y \rangle \approx 0 \tag{17b}\] $H_n = A \langle S \rangle \centerdot{I} - g_n\beta_nB\centerdot{I} \tag{18}$ Pulling out a $-g_n\beta_n$ followed by $I$ leaves \[H_n = -g_n\beta_n \left( -\dfrac{A \langle S \rangle}{g_n\beta_n} + B\right){I} \tag{19}\] Substituting the internal magnetic field with \[B_{int} = -\dfrac{A \langle S \rangle }{g_n\beta_n}\tag{20}\] results in a combined magnetic field term involving both the applied magnetic field and the internal magnetic field \[H_n = -g_n\beta_n(B_{int} + B)\centerdot{I} \tag{21}\] which is simplified by using the effective magnetic field $B_{eff}$ \[H_n = -g_n\beta_nB_{eff}\centerdot{I} \tag{22}\] Outside links References Gütlich, P., Link, R., & Trautwein, A. (1978). Mössbauer spectroscopy and transition metal chemistry. Inorganic chemistry concepts, v. 3. Berlin: Springer-Verlag. G J Long and F Grandjean (1993) Mössbauer Spectroscopy Applied to Magnetism and Materials. New York., Science Vol. 1,, eds. Lawrence Que, J., Ed. (2000). Physical Methods in Bioinorganic Chemistry. Sausalito, CA, University Science Books. Filatov, M. (2007). "On the calculation of Mössbauer isomer shift." The Journal of Chemical Physics 127(8): 8. Introduction to mossbauer spectroscopy. 2013]. Available from http://www.rsc.org/Membership/Networking/InterestGroups/MossbauerSpect/part2.asp. Bubek, Moritz, and Dennis Rettinger. 2004. Mossbauer-effekt. Haskins, J. R. 1965. Advanced mossbauer-effect experiments. American Journal of Physics: 646. Kistner, O. C., and A. W. Sunyar. 1960. Evidence for quadrupole interaction of Fe57m, and influence of chemical binding on nuclear gamma-ray energy. Physical Review Letters 4 (8): 412. Preston, R. S., S. S. Hanna, and J. Heberle. 1962. Mössbauer effect in metallic iron. Physical Review 128 (5): 2207. Problems The magnetic splitting of $m_I = 0$ intensity of transition is related to $sin^2(\theta)$ of the angle between the incoming gamma ray and the effective magnetic field. When is the intensity of transition at max? Why is it important for the sample to be in solid or crystalline state? What case will result in a delta shift of 0.00 mm/s? Why is the Doppler effect important to Mössbauer spectroscopy? Why are both the emission and absorption distributions the same? (both estimated with Lorentzian functions) Contributors Kevin MacDow (UCD)
It is well known to us that the Solubility of solute in a solution increases with the increase in the temperature because, when the temperature increases the molecules of the solvent gain more kinetic energy. Thus the molecules move randomly and having greater distance from each other which is responsible for the large voids between them and it gives more space for the solute molecules between them to come in. However, there a few salts like cerium sulphate , lithium carbonate sodium carbonate monohydrate, etc. whose solubility decreases with the increase in temperature. How it is so ? A saturated solution is at equilibrium (rate of dissolution is equal to rate of crystallization) with some equilibrium constant $K_1$. If you change the temperature of the system at equilibrium, you will observe a different equilibrium constant $K_2$. Whether the equilibrium constant increase or decreases is described by the Van't Hoff equation: $$\ln \left(\frac{K_2}{K_1}\right) = \frac{-\Delta H}{R}\left(\frac{1}{T_2}-\frac{1}{T_1}\right)$$ So it depends on the sign of the enthalpy of reaction, in this case the enthalpy of dissolution. It is well known to us that the Solubility of solute in a solution increases with the increase in the temperature because, when the temperature increases the molecules of the solvent gain more kinetic energy. The statement about solubility is not always true, and the explanation leaves out something. A lot of things change when you increase the temperature. The solvent gains kinetic energy, the solute gains kinetic energy, and the solid gains kinetic energy. How this influences the solubility depends on the specific system, and is hard to predict. However, there a few salts like cerium sulphate , lithium carbonate sodium carbonate monohydrate, etc. whose solubility decreases with the increase in temperature. How it is so ? Now that we introduced the Van't Hoff equation, we know it must have to do with the enthalpy of dissolution. Usually, you expect the interactions in the solid that need to be broken to be stronger than the gains from solvating the solute. In these cases, it is apparently the opposite. What you refer to, the decreased in a solid's solubility in a liquid with increased temperature, is frequently called retrograde or inverse solubility, and occurs when the dissolution of the solute is exothermic. The explanation for this can be viewed as a manifestation of Le Chatelier's principle. Essentially, since evolved heat can be viewed as a product of an exothermic reaction, the addition of more heat (e.g. a higher temperature) is equivalent to adding a product to the product side of the chemical equation for dissolution, driving the equilibrium back toward the reactants, in this case toward the undissolved compound. Take the exothermic dissolution of calcium sulfate in water for example: $$\ce{CaSO4_{(s)} <--> Ca^{2+}_{(aq)} + SO4^{2-}_{(aq)} + heat}$$ In this case, the solubility decreases with increasing temperature because by increasing the temperature you are adding heat to the product side. Inversely, if you pull heat from this system, e.g. cool it, you drive the equilibrium toward the products side and solubility is increased. As others noted, the effect of temperature is governed completely by $\Delta H$ of dissolution. The reason that dissolution can have either positive or negative $\Delta H$ is because the favorability is determined by the Gibbs free energy $\Delta G = \Delta H - T \Delta S$. If the $\Delta S$ term is negative (before multiplication by -T) and $\Delta H$ is also negative, dissolution is favorable only if $\|\Delta H\|>\|T\Delta S\|$. It is important to note also that the magnitude of $\Delta G^\circ$ does NOT correlate with the solubility, since the change in concentration per change in $\Delta G^\circ$ is not constant. That is why $\Delta S^\circ$ does not determine the effect of temperature. This can be seen by noting that the important value is $K$. Since $\Delta G^\circ=-RT\ln K=\Delta H^\circ - T\Delta S^\circ$, we can rearrange and find that $\ln K = -\frac{\Delta H^\circ}{R}\left(\frac{1}{T}\right)+\frac{\Delta S^\circ}{R}$, so a plot of $\ln K$ vs $\frac{1}{T}$ (which is the temperature dependence of solubility) has a slope whose sign is determined only by $\Delta H^\circ$.
Let $$y''-q(x)y=0$$be a differential equation with initial conditions on $0\leq x<\infty,$ as $y(0)=1,y'(0)=1$ where $q(x)$ is a positive monotonically increasing continuous function. Then which of the following are true? $y(x)\rightarrow\infty$ as $x\rightarrow\infty$. $y'(x)\rightarrow\infty$ as $x\rightarrow\infty$. $y(x)$ has finitely many zeros in $[0,\infty)$. $y(x)$ has infinitely many zeros in $[0,\infty)$. Please don't mind, actually I am new in differential equation. I only know that by Picard's theorem the above differential equation has the unique solution, but I don't know what is the solution as I tried by direct hit and trial method. According to me the solution of above differential equation will be some thing in exponential form, so according to me its answer will be $a$, $b$, $c$. But I don't know the exact method. Please help me to solve the above problem. Thanks in advance.

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