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https://www.physicsforums.com/threads/increase-in-electric-potential-energy.754028/ | # Increase in electric potential energy
1. May 16, 2014
### Miraj Kayastha
If a positive charge A moves towards another stationary positive charge B then the A's electric potential energy increases. But shouldn't the electric potential energy of B also increase as it is also in a way moving towards the A inside the A's electric field?
So shouldn't the total increase in electric potential of the system be the double of the increase in any one of the charges?
2. May 16, 2014
### Matterwave
1) The potential energy might decrease if the charges A and B are opposite.
2) Assuming A and B are of like charge, the increase in potential energy is held in the configuration of the system A and B together. The potential energy is not attributable to either A or B independently. This is also true for the gravitational potential energy. However, in the case of two objects of disproportionately different sizes, it is often convenient to neglect the motion of one of the objects (the larger one, since it accelerates so little). And in this case, one often talks about "the potential energy of the smaller object" since that's the only motion we care about, when in fact the potential energy is contained in the configuration of the system.
3. May 16, 2014
### Staff: Mentor
The potential energy of the system (of charges A and B) in its final configuration equals the total work that some outside agent has to do, in order to bring both charges from infinity to their final locations.
Bring charge A to its final location. This requires no work, because charge B is still infinitely far away.
Bring charge B to its final location, a distance rAB from A. This requires work kqAqB/rAB.
The total work and the potential energy are therefore 0 + kqAqB/rAB.
4. May 16, 2014
### Miraj Kayastha
But doesnt the potential of A and B increase as A or B gets closer because both of the particles a simultaneously changing their positions in the elctric field of each other?
5. May 16, 2014
### Matterwave
There's just one potential. The potential due to the configuration of A and B. There's no two separate potentials that you are thinking of. There isn't a potential of A and a potential of B, there is just potential of A and B. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9468075037002563, "perplexity": 422.2966506456101}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-13/segments/1521257647885.78/warc/CC-MAIN-20180322131741-20180322151741-00091.warc.gz"} |
https://www.physicsforums.com/threads/shm-w-coulumb-forces.154780/ | # SHM w/ Coulumb Forces
• Start date
• #1
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This is a question I already solved but was curious about something. There are two positive point charges of the same maginitude on the same axis. They are some distance appart with third point positive charge placed slightly off the midpoint between the two original charges. The question is to find the period of the SHM. In solving this I basically applied Coulumb's law as the restoring force. After looking at the answer the book gave I noticed it still had a variable for mass included. My question is why would mass even come into play for a system composed of point particles with electric force? If gravity is neglected, couldn't I just apply the the particle's charges in place of mass?
• #2
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Are you refering to how the electric force is generally much stronger than the gravitational force, so why we even care about the gravitational force? In general, if I am understanding what you are asking about, you are right and the gravitational force probably won't have much of an effect. Nonetheless, if you want to examine the particle's motion of long periods of time the gravitational force might need to be included; for example, gravity might damp the SHM.
• #3
168
1
The question ignores gravity and does not even define a mass for any of the charges (they are all treated as positive point particles). The solution in the book still included a "m" for mass as though it would matter for a system such as this one.
• #4
1,860
0
Hmm, I would have to see the problem, but I don't see anything irregular with having mass constants in your solutions - particularly if they are in the angular frequency.
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1K | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8137197494506836, "perplexity": 662.3548920172419}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-49/segments/1637964363437.15/warc/CC-MAIN-20211208022710-20211208052710-00299.warc.gz"} |
https://physics.stackexchange.com/questions/167448/properties-of-controlled-z-rotations/184288 | # Properties of controlled z-rotations
Given n qubit gate of the form c-z-z-z... (shorthand for c-z between qubit 1 and 2 followed by c-z between 1 and 3 and so on up to n qubits) it seems to be possible to find local unitaries which will change this gate into one that preserves phase when flipping all the qubits. That is |1111..1> will have the same phase as |000..0> ans so on.
An example would be c-z. All input states remain unchanged except |11>, which picks up a - sign. However, if you put an S gate on each qubit (a pi/2 phase gate) you will find that |00> and |11> are left unchanged, and |01> and |10> pick up an i. That is, each basis state has the same quantum amplitude after the gate as its bitwise flipped counterpart.
I haven't proven this, but I have tested it for 2, 3 and 4 qubits. However I haven't been able to do the same thing for gates with more than one control qubit, such as c-c-z. Is anyone aware of any reason why this might be, or any even slightly relevant research which might shed light on this?
Btw, if it helps, the reason I ask is that I am looking at using a coherent optical state to mediate entangling interactions between qubits. A coherent state is defined by a complex number, and in this particular system, when it interacts with a qubit, this complex number can be displaced in an arbitrary direction conditional on the state of the qubit, |0> and |1> displace the coherent state in opposite directions. When you use a series of these controlled displacements to take the coherent state in a closed path in the complex plane, this leaves a a phase on the qubits of the form e^(±iA) where A is the area enclosed by the path, and the sign is dependant on whether the path is travelling clockwise or anti-clockwise. You can use this effect to build entangling gates, specifically c-not as shown in http://arxiv.org/abs/quant-ph/0509202.
I have been trying to describe a c-c-z gate using fewer qubit-bus interactions than building it out of many 2-qubit gates.
The types of paths I am looking at (if not all paths in general) all seem to have this property where flipping all the qubits does not affect the phase, but it seems not all multiqubit gates have this property (including c-c-z), but I haven't worked out a way of determining which do and which don't apart from trial and error.
Any help would be appreciated.
• Might TheoreticalComputerScience.SE be a better place for this? – Jim Feb 27 '15 at 15:45
• Possibly, I guess its the kind of thing that could be physics, maths or CS. I guess I'll leave it here for the moment and if no one answers I'll have a look at the CS stackexchange. – Leo Rogers Feb 27 '15 at 18:05
• Can you describe more precisely what exactly is the property you want after the local unitaries are applied? Do you mean c-c-c-z in the first line rather than c-z-z-z? – Māris Ozols Feb 28 '15 at 8:26
• I mean c-z-z-z, i guess as a short hand for meaning multiple c-z's, so the first qubit is the only control qubit, and all the others are targets. – Leo Rogers Feb 28 '15 at 18:58
• I'll edit the main post to ad an example. – Leo Rogers Feb 28 '15 at 18:59
## 1 Answer
I've done a lot more work on this since asking, and now have the answers to my questions. First, I mentioned that it seemed that with single qubit gates you can always change the state given by c-z-z-z... (from the equal superposition) to one with this 'bit-flip' property, but that I hadn't proven it. Turns out this is true for any circuit made up of c-z gates, and this can be proven to be true. It works for one c-z gate, if you start with a circuit producing a state with the bit-flip property, and concatenating another circuit (such as one containing only one c-z) that does this will produce another circuit with this property.
Second, it turns out that there are no single qubit gates that only alter the phases on the qubits to produce the bit-flip property for the c-c-z gate. I only have a proof regarding c-c-z, not any general criteria about which circuits you can do this with and which you can't, except that it is always possible to do it with any circuit using only c-z gates and single qubit gates that commute with c-z.
I'm still working on this. I'll update when I have more general results. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8012597560882568, "perplexity": 304.0816931242221}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-26/segments/1560627999130.50/warc/CC-MAIN-20190620004625-20190620030625-00438.warc.gz"} |
http://math.stackexchange.com/questions/188289/characterising-continuous-maps-between-metric-spaces | # Characterising continuous maps between metric spaces
Let $f:(X,d)\to (Y,\rho)$.
Prove that $f$ is continuous if and only if $f$ is continuous restricted to all compact subsets of $(X,d)$. I could do the left to right implication but couldn't do the reverse implication. Please help.
-
If $x_n \to x$ in $X$, then $\{x_n\}\cup \{x\}$ is a compact subset of $X$, I guess... – Siminore Aug 29 '12 at 9:41
assume that $f$ is continuous on every compact subset of $X$.
Suppose that a sequence $x_n$ of elements of $X$ is such that $x_n\rightarrow x$ for some $x\in X$. Observe that a set $\{x_n:n\in\mathbb{N}\}\cup\{x\}$ is a compact subset of the space $X$. Thus, along our assumption, $F$ is continuous on that set and it does simply means, that $f(x_n)\rightarrow f(x)$. Since the choice of a sequence was arbitrary we have proved that $f:X\rightarrow Y$ is a continuous function. Q.E.D. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9647459983825684, "perplexity": 89.93745842730131}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-10/segments/1394678664178/warc/CC-MAIN-20140313024424-00040-ip-10-183-142-35.ec2.internal.warc.gz"} |
https://arxiv-export-lb.library.cornell.edu/abs/2102.06939?context=cs | cs
(what is this?)
# Title: Optimal Streaming Algorithms for Graph Matching
Abstract: We present parameterized streaming algorithms for the graph matching problem in both the dynamic and the insert-only models. For the dynamic streaming model, we present a one-pass algorithm that, with high probability, computes a maximum-weight $k$-matching of a weighted graph in $\tilde{O}(Wk^2)$ space and that has $\tilde{O}(1)$ update time, where $W$ is the number of distinct edge weights and the notation $\tilde{O}()$ hides a poly-logarithmic factor in the input size. For the insert-only streaming model, we present a one-pass algorithm that runs in $O(k^2)$ space and has $O(1)$ update time, and that, with high probability, computes a maximum-weight $k$-matching of a weighted graph. The space complexity and the update-time complexity achieved by our algorithms for unweighted $k$-matching in the dynamic model and for weighted $k$-matching in the insert-only model are optimal.
A notable contribution of this paper is that the presented algorithms {\it do not} rely on the apriori knowledge/promise that the cardinality of \emph{every} maximum-weight matching of the input graph is upper bounded by the parameter $k$. This promise has been a critical condition in previous works, and lifting it required the development of new tools and techniques.
Subjects: Data Structures and Algorithms (cs.DS); Computational Complexity (cs.CC) Cite as: arXiv:2102.06939 [cs.DS] (or arXiv:2102.06939v2 [cs.DS] for this version)
## Submission history
From: Ge Xia [view email]
[v1] Sat, 13 Feb 2021 14:55:50 GMT (38kb)
[v2] Thu, 25 Feb 2021 18:31:14 GMT (39kb)
Link back to: arXiv, form interface, contact. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.859566330909729, "perplexity": 1279.8472595107125}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-06/segments/1674764500357.3/warc/CC-MAIN-20230206181343-20230206211343-00383.warc.gz"} |
https://gitlab.unige.ch/Conor.Mccoid/UNIGE/-/commit/43f184313783c4d055c1b0e49423610446ee9882 | Commit 43f18431 by conmccoid
### updates to notes for PSIM and stochastic networks
\documentclass{article} \usepackage{/home/mccoid/LaTeX/preamble} \usepackage{preamble} \usepackage{float} \floatstyle{boxed} \newfloat{algorithm}{t}{} \floatname{algorithm}{Algorithm} \title{Collocation matrices representing inverse operators - notes} \author{Conor McCoid} ... ... @@ -57,14 +62,14 @@ These matrices will be denoted by: For the Wronskians of $E \setminus E_i$ we present the following lemma: \begin{lemma}[Wronskians of exponential functions] \begin{lem}[Wronskians of exponential functions] Let $\set{\lambda_k}_{k=1}^m \in \mathbb{R}$ and $\Lambda$ the matrix defined earlier for such a set, then \begin{equation*} \W{\set{e^{\lambda_k x}}_{k=1}^m}{x} = \abs{\Lambda} e^{x \sum_{k=1}^m \lambda_k} . \end{equation*} \label{lem:exp} \end{lemma} \end{lem} \begin{proof} The case of any two $\lambda_k$ being equal is trivially true as both sides are necessarily zero. ... ... @@ -113,14 +118,14 @@ The special case is equivalent to $M=1$ and $\lambda_1 = 0$. As such, the set $E = \set{\Poly{j}}_{j=0}^{m-1}$. The following lemma presents the Wronskians for such polynomials: \begin{lemma}[Wronskians of polynomials] \begin{lem}[Wronskians of polynomials] \begin{align*} (i) && \Wpoly{k=0}{m} & = 1 \\ (ii) && \Wpoly{k=0,k \neq j}{m} & = \Wpoly{k=1}{m-j} \\ (iii) && \Wpoly{k=1}{m} & = \Poly{m} \end{align*} \label{lem:poly} \end{lemma} \end{lem} \begin{proof} \begin{description} ... ... @@ -218,12 +223,12 @@ The case for one root with multiplicity $m$, represented by $M=1$, is a generali The set $E = E_1 = \set{\Poly{k} e^{\lambda_1 x}}_{k=1}^m$ which is the set $E$ from the previous case multiplied by $e^{\lambda_1 x}$. The following lemma then makes the generalization simple. \begin{lemma} \begin{lem} \begin{equation*} W(\{ f_k g \}_{k=1}^m ; x) = g^m W(\{ f_k \}_{k=1}^m ; x ) \end{equation*} \label{lem:group} \end{lemma} \end{lem} \begin{proof} It is trivially true for $m=1$. ... ... @@ -436,4 +441,78 @@ One may write out equation (\ref{eq:Wronskian system}) with this in mind: where $\tilde{I}_k$ indicates the selection of those rows corresponding to $\lambda_k$. Note that all $F_k^{-1}(x)$ are principal submatrices of $F_{k^*}^{-1}$, where $m_{k^*} \geq m_k$ for all $k=1,...,M$. \section{Other attempts} \newcommand{\Lcal}{\mathcal{L}} \newcommand{\ddx}{\frac{d}{dx}} Let $\hat{\Lcal}$ be the operator for the set $\set{\Poly{k} e^{\lambda_j x}}_{j \neq q}$ and $\tilde{\Lcal}$ the one for $\set{\Poly{k} e^{\lambda_q x}}_{k \neq n}$, then \begin{align*} \Lcal & = \hat{\Lcal} \tilde{\Lcal} \\ & = \left [\left ( \ddx \right )^{m-l} +\sum_{j \neq q} \lambda_j \left ( \ddx \right )^{m-l-1} + \dots + \prod_{j \neq q} \lambda_j \right ] \left [ \left ( \ddx \right )^l + r(x) \left ( \ddx \right )^{l-1} + \dots + s(x) \right ] \\ & = \left ( \ddx \right )^m + \left ( \sum_{j \neq q} \lambda_j + r(x) \right ) \left ( \ddx \right )^{m-1} + \dots \end{align*} but then $\Lcal$ is the operator for \begin{itemize} \item $\set{\Poly{k} e^{\lambda_j x} } \cup \set{\hat{\Lcal} \left ( \Poly{k} e^{\lambda_q x} \right )}_{k \neq n}$ or \item $\set{\tilde{\Lcal} \left ( \Poly{k} e^{\lambda_j x} \right )}_{k,j \neq q} \cup \set{\Poly{k} e^{\lambda_q x}}_{k \neq n}$. \end{itemize} \begin{align*} E & = \set{P_k(x)}_{k=1}^m \\ & = \set{\Poly{j} e^{\lambda_i x}}_{j=0, i=1}^{m_i-1,M} \\ & = \cup_{i=1}^M \set{\Poly{j} e^{\lambda_i x}}_{j=0}^{m_i-1} \\ & = \cup_{i=1}^M E_i \end{align*} \begin{itemize} \item $E \to \Lcal$ such that $\forall f \in E$ $\Lcal f = 0$. \item $\cup_{i \neq k} E_i \to \tilde{\Lcal}$ in the same way: $\tilde{\Lcal} f = f^{(m-m_k)}(x) + \sum_{i \neq k} \lambda_i^{m_i} f^{(m - m_k - 1)}(x) + \dots$. \item $E_k \setminus \set{\Poly{k} e^{\lambda_i x} } \to \hat{\Lcal}$, $\hat{\Lcal}f = f^{(m_k-1)}(x) + r(x) f^{(m_k-2)}(x) + \dots$. \item $E \setminus \set{\Poly{k} e^{\lambda_i x} } \to \bar{\Lcal}$, $\bar{\Lcal} f = f^{(m-1)}(x) + \left ( r(x) + \sum_{i \neq k} \lambda_i^{m_i} \right ) f^{(m-2)}(x) + \dots$. \end{itemize} Need $W(\hat{\Lcal} \left ( E_k \setminus \set{\Poly{k} e^{\lambda_i x}} ; x\right)$. \begin{align*} \hat{\Lcal} \left ( E_k \setminus \set{\Poly{n} e^{\lambda_k x}} \right ) & = \prod_{i \neq k} \left ( \ddx - \lambda_i \right )^{m_i} \Poly{j} e^{\lambda_k x}, \quad j \neq n \\ & \neq \prod_{i \neq k} \left ( \ddx - \lambda_i \right )^{m_i - 1} \left ( \poly{j-1}+ \lambda_k \Poly{j} - \lambda_i \poly{j} \right ) e^{\lambda_k x} \end{align*} \begin{align*} \tilde{\lambda}_l = \begin{cases} \lambda_1 & 1 \leq l \leq m_1 \\ \lambda_i & \sum_{j < i, j \neq k} m_j < l \leq \sum_{j \leq i,j \neq k} m_j \end{cases} \end{align*} \begin{align*} \prod_{l=1}^{m-m_k} & \left ( \ddx - \tilde{\lambda}_l \right ) \Poly{j} e^{\lambda_k x} = \\ \prod_{l=2}^{m-m_k} & \left ( \ddx - \tilde{\lambda}_l \right ) \left ( \poly{j-1} + (\lambda_k - \tilde{\lambda}_l ) \Poly{j} \right ) e^{\lambda_k x} = \\ \prod_{l=3}^{m-m_k} & \left ( \ddx - \tilde{\lambda}_l \right ) \left ( \poly{j-2} + (2 \lambda_k - \tilde{\lambda}_1 - \tilde{\lambda}_2 ) \poly{j-1} + (\lambda_k^2 - \lambda_k \tilde{\lambda}_1 - \tilde{\lambda}_2 \lambda_k + \tilde{\lambda}_1 \tilde{\lambda}_2 ) \Poly{j} \right ) e^{\lambda_k x} \end{align*} \begin{align*} \Lcal P_k(x) & = P_k^{(m)}(x) - \sum_{i=1}^{M} m_i \lambda_i P_k^{(m-1)}(x) + \dots = 0 \\ \hat{\Lcal} P_k(x) & = P_k^{(m-1)}(x) + r(x) P_k^{(m-2)}(x) + \dots = 0 & \forall k \neq j \\ \tilde{\Lcal} P_j(x) & = P_j'(x) + q(x) P_j(x) = 0 \\ \bar{\Lcal} P_k(x) & = P_k^{(m-1)}(x) + p(x) P_k^{(m-2)}(x) + \dots & \text{such that } \Lcal P_k(x) = \bar{\Lcal} \tilde{\Lcal} P_k(x) \\ \breve{\Lcal} P_k(x) & = P_k'(x) + s(x) P_k(x) & \text{such that } \Lcal P_k(x) = \breve{\Lcal} \hat{\Lcal} P_k(x) \\ \implies -\sum_{i=1}^M m_i \lambda_i & = q(x) + p(x) = r(x) + s(x) \end{align*} $\Omega$ (in this document, pretty sure $\Lambda$ in the current version) has $\sum_{k=1}^M \left [ \frac{m_k (m_k+1)}{2} + (m-m_k)^2 \right ]$ elements. To make $\Omega$ make each piece separately. Separate into coefficients and roots. \begin{algorithm} $\lambda_k$, $m_k$ $\Omega_{\lambda_k} = \text{spdiags} \left ( 0:-1:1-m, \lambda_k^{(0:m-1)}, m, m_k \right )$; $P_T$ (Pascal's triangle in lower triangular form, size $m \times m$) $\Omega_k = P_T(:,1:m_k) \otimes \Omega_{\lambda_k}$; ($\otimes$ is the Hadamard matrix product) $\Omega = \begin{bmatrix} \Omega_1 & \Omega_2 & \dots & \Omega_M \end{bmatrix}$; \caption{Algorithm to construct $\Omega$} \end{algorithm} \end{document}
... ... @@ -2,7 +2,8 @@ \usepackage{preamble} \usepackage{tikz} \usetikzlibrary{positioning} \usetikzlibrary{positioning,calc} \usepackage{pgfplots} \begin{document} ... ... @@ -53,42 +54,24 @@ Define $\dagger \bbp = \bbp \cup \bbr_- \cup \set{-\infty}$. $(\dagger \bbp, \lor)$ is a commutative group. \end{lemma} \subsection{The Probabilistic Semiring} \subsection{The Probabilistic Concentric Ring??} In addition to the OR operation, one can equip $\bbp$ with the AND operation. This operation is equivalent to standard multiplication and shall be referred to as such. \begin{lemma} $(\bbp, \lor, \cdot)$ is a commutative semiring. \end{lemma} \subsection{The Probabilistic Field} Unfortunately, the OR and AND operations do not distribute, meaning they do not interact in any meaningful way. Essentially, one can set up two groups that overlap in the region $\bbp$ and each operation considers consequences only on their respective group. We can call $(\bbp,\lor,\cdot)$ a concentric ring, though such an object has little use. The probadd inverses have already been included in $\dagger \bbp$. The multiplicative inverses that remain lie in $(1,\infty)$. Taken together, this forms the extended real numbers, $\bbr^*$. \begin{lemma} $(\bbr^*, \lor, \cdot)$ is a field. \end{lemma} \begin{proof} Not true, since the operations do not allow for distributivity. \end{proof} \subsection{The Probabilistic Inner Product Space} Since $(\bbr^*, \lor, \cdot)$ is a field, the space $({\bbr^*}^n, \bbr^*, \lor, \cdot)$ is a vector space. -not true, see proof of prev. lemma Define the inner product over this vector space as Define the inner product over this space as \langle \vec{p}, \vec{q} \rangle = p_1 q_1 \lor p_2 q_2 \lor \dots \lor p_n q_n . \begin{lemma} $({\bbr^*}^n, \bbr^*, \lor, \cdot, \langle \cdot, \cdot \rangle)$ is an inner product space. \end{lemma} \subsection{Probabilistic Matrices} To represent the linear transformations of probabilistic vectors, one can construct matrices containing elements of $\bbr^*$. ... ... @@ -98,6 +81,20 @@ This is made trivial by using the probabilistic inner product: A \vec{p} = \begin{bmatrix} \vec{a}_1^\top \\ \vdots \\ \vec{a}_n^\top \end{bmatrix} \vec{p} = \begin{bmatrix} \langle \vec{a}_1, \vec{p} \rangle \\ \vdots \\ \langle \vec{a}_n, \vec{p} \rangle \end{bmatrix} . \end{equation*} \begin{figure} \centering \begin{tikzpicture} \begin{axis}[samples=500, domain=-4:4, restrict y to domain=-4:4, xlabel={$p$}, ylabel={$\dagger p = \frac{-p}{1-p}$}] \addplot[thick] plot (\x, {-\x/(1-\x)}); \addplot[dotted] plot (\x, 1); \addplot[dotted] plot (1, \x); \end{axis} \end{tikzpicture} \caption{Visualization of the inverses $\dagger p$.} \end{figure} \section{Stochastic Networks} We define a stochastic network as a graph with vertices taking on a discrete set of $k$ values. ... ... @@ -383,6 +380,23 @@ roundnode/.style={circle, fill=white, draw=black, very thick}] The paths follow the relationship $\mathcal{D} \supset \mathcal{C} \subset \mathcal{A} \subset \mathcal{B}$.} \end{table} \subsection{Notes} If two paths cross and one arrives at the intersection before the other than the probability of infection is: \begin{equation*} p_1(t) + p_2(t) (1 - p_1(t)) = p_1(t) \lor p_2(t). \end{equation*} If two paths collide at the same time then the probability of infection is either $$1 - (1 - p_1(t))(1-p_2(t)) = p_1(t) \lor p_2(t)$$ or $$p_1(t) (1-p_2(t)) + p_2(t) (1-p_1(t)) = \left ( p_1(t) \lor p_2(t) \right ) - \left (p_1(t) \land p_2(t) \right ).$$ I think the first option makes more sense. The total probability for a node is then the probsum of all paths that run through it. Keywords: Bayesian networks, MCMC \appendix \section{Notes} ... ... | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 1.0000044107437134, "perplexity": 2482.443584914562}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-06/segments/1674764499946.80/warc/CC-MAIN-20230201144459-20230201174459-00212.warc.gz"} |
http://www.maplesoft.com/support/help/Maple/view.aspx?path=DifferentialGeometry/JetCalculus/ZigZag | Calling Sequences - Maple Programming Help
Home : Support : Online Help : Mathematics : DifferentialGeometry : JetCalculus : DifferentialGeometry/JetCalculus/ZigZag
JetCalculus[ZigZag] - lift a ${d}_{H}$-closed form on a jet space to a $d$-closed form
Calling Sequences
ZigZag(${\mathbf{ω}}$)
Parameters
$\mathrm{ω}$ - a differential bi-form on the jet space of a fiber bundle
Description
• Let be a fiber bundle, with base dimension $n$ and fiber dimension $m$ and let be the infinite jet bundle of $E$. The space of $p$-forms on decomposes as a direct sum of bi-forms
Let be a bi-form of degree $\left(r,s\right)$. I f suppose that or, if , that . See HorizontalExteriorDerivative, VerticalExteriorDerivative, and IntegrationByParts for the definitions of the space ${\mathrm{\Omega }}^{\left(r,s\right)}\left({J}^{\infty }\left(E\right)\right)$, the horizontal exterior derivative ${d}_{H}$, the vertical exterior derivative and the integration by parts operator
Given that or define a degree form by
where and ${d}_{V}\left({\mathrm{θ}}_{i}\right)={d}_{H}\left({\mathrm{θ}}_{i+1}\right).$
The forms ${\mathrm{θ}}_{i}$ are of bi-degree The forms can be calculated inductively using the horizontal homotopy operators . The fundamental property of this construction is that the form is always closed with respect to the standard exterior derivative, that is,
• If $\mathrm{ω}$ is a bi-form of degree $\left(r,s\right),$then ZigZag(${\mathbf{\omega }}$) returns the differential form of degree $r+s$.
• The command ZigZag is part of the DifferentialGeometry:-JetCalculus package. It can be used in the form ZigZag(...) only after executing the commands with(DifferentialGeometry) and with(JetCalculus), but can always be used by executing DifferentialGeometry:-JetCalculus:-ZigZag(...).
Examples
> with(DifferentialGeometry): with(JetCalculus):
Example 1.
Create the jet space ${J}^{3}\left(E\right)$for the bundle with coordinates $\left(x,y,u\right)\to \left(x,y\right)$.
> DGsetup([x, y], [u], E, 3):
Define a type (1, 0) form and show that it is ${d}_{H}$ -closed.
E > omega1 := evalDG((u[1, 2]*u[1, 1, 1] + u[1 ,1]*u[1, 1, 2])*Dx + (u[1, 2]*u[1 ,1, 2] + u[1, 1]*u[1, 2, 2])*Dy);
${\mathrm{ω1}}{≔}\left({{u}}_{{1}{,}{1}}{}{{u}}_{{1}{,}{1}{,}{2}}{+}{{u}}_{{1}{,}{2}}{}{{u}}_{{1}{,}{1}{,}{1}}\right){}{\mathrm{Dx}}{+}\left({{u}}_{{1}{,}{1}}{}{{u}}_{{1}{,}{2}{,}{2}}{+}{{u}}_{{1}{,}{2}}{}{{u}}_{{1}{,}{1}{,}{2}}\right){}{\mathrm{Dy}}$ (2.1)
E > HorizontalExteriorDerivative(omega1);
${0}{}{\mathrm{Dx}}{}{\bigwedge }{}{\mathrm{Dy}}$ (2.2)
Apply the ZigZag command to ${\mathrm{ω}}_{1}$ to obtain a form ${\mathrm{θ}}_{1}$.
E > theta1 := ZigZag(omega1);
${\mathrm{θ1}}{≔}{{u}}_{{1}{,}{2}}{}{{\mathrm{du}}}_{{1}{,}{1}}{+}{{u}}_{{1}{,}{1}}{}{{\mathrm{du}}}_{{1}{,}{2}}$ (2.3)
Check that ${\mathrm{θ}}_{1}$ is $d$-closed and that its [1, 0] component matches ${\mathrm{ω}}_{1}$.
E > ExteriorDerivative(theta1);
${0}{}{\mathrm{dx}}{}{\bigwedge }{}{\mathrm{dy}}$ (2.4)
E > convert(theta1, DGbiform, [1, 0]);
$\left({{u}}_{{1}{,}{1}}{}{{u}}_{{1}{,}{1}{,}{2}}{+}{{u}}_{{1}{,}{2}}{}{{u}}_{{1}{,}{1}{,}{1}}\right){}{\mathrm{Dx}}{+}\left({{u}}_{{1}{,}{1}}{}{{u}}_{{1}{,}{2}{,}{2}}{+}{{u}}_{{1}{,}{2}}{}{{u}}_{{1}{,}{1}{,}{2}}\right){}{\mathrm{Dy}}$ (2.5)
Example 2.
Define a type (2, 0) form ${\mathrm{ω}}_{2}$ and show that its Euler-Lagrange form is 0.
E > omega2 := evalDG((- u[2, 2]*u[1, 2] - u[2]*u[1, 2, 2] - u[1]*u[1, 2, 2] - u[1, 2]^2)*Dx &w Dy);
${\mathrm{ω2}}{≔}{-}\left({{u}}_{{1}}{}{{u}}_{{1}{,}{2}{,}{2}}{+}{{u}}_{{2}}{}{{u}}_{{1}{,}{2}{,}{2}}{+}{{u}}_{{1}{,}{2}}^{{2}}{+}{{u}}_{{2}{,}{2}}{}{{u}}_{{1}{,}{2}}\right){}{\mathrm{Dx}}{}{\bigwedge }{}{\mathrm{Dy}}$ (2.6)
E > EulerLagrange(omega2);
${0}{}{\mathrm{Dx}}{}{\bigwedge }{}{\mathrm{Dy}}{}{\bigwedge }{}{{\mathrm{Cu}}}_{\left[\right]}$ (2.7)
Apply the ZigZag command to ${\mathrm{ω}}_{2}$ to obtain a 2-form ${\mathrm{θ}}_{2}$.
E > theta2 := ZigZag(omega2);
${\mathrm{θ2}}{≔}{-}\left(\frac{{{u}}_{{1}{,}{2}}}{{3}}{+}\frac{{{u}}_{{2}{,}{2}}}{{6}}\right){}{\mathrm{dx}}{}{\bigwedge }{}{{\mathrm{du}}}_{{1}}{-}\frac{{2}{}{{u}}_{{1}{,}{2}}}{{3}}{}{\mathrm{dx}}{}{\bigwedge }{}{{\mathrm{du}}}_{{2}}{-}\left(\frac{{{u}}_{{1}}}{{3}}{+}\frac{{2}{}{{u}}_{{2}}}{{3}}\right){}{\mathrm{dx}}{}{\bigwedge }{}{{\mathrm{du}}}_{{1}{,}{2}}{-}\frac{{{u}}_{{1}}}{{6}}{}{\mathrm{dx}}{}{\bigwedge }{}{{\mathrm{du}}}_{{2}{,}{2}}{+}\frac{{{u}}_{{2}{,}{2}}}{{3}}{}{\mathrm{dy}}{}{\bigwedge }{}{{\mathrm{du}}}_{{1}}{+}\left(\frac{{{u}}_{{1}{,}{2}}}{{3}}{+}\frac{{{u}}_{{2}{,}{2}}}{{6}}\right){}{\mathrm{dy}}{}{\bigwedge }{}{{\mathrm{du}}}_{{2}}{+}\frac{{{u}}_{{2}}}{{3}}{}{\mathrm{dy}}{}{\bigwedge }{}{{\mathrm{du}}}_{{1}{,}{2}}{+}\left(\frac{{{u}}_{{2}}}{{6}}{+}\frac{{{u}}_{{1}}}{{3}}\right){}{\mathrm{dy}}{}{\bigwedge }{}{{\mathrm{du}}}_{{2}{,}{2}}{-}\frac{{1}}{{3}}{}{{\mathrm{du}}}_{\left[\right]}{}{\bigwedge }{}{{\mathrm{du}}}_{{1}{,}{2}}{+}\frac{{1}}{{6}}{}{{\mathrm{du}}}_{\left[\right]}{}{\bigwedge }{}{{\mathrm{du}}}_{{2}{,}{2}}{+}\frac{{1}}{{3}}{}{{\mathrm{du}}}_{{1}}{}{\bigwedge }{}{{\mathrm{du}}}_{{2}}$ (2.8)
Check that ${\mathrm{θ}}_{2}$ is $d-$closed and that its [2, 0] component matches ${\mathrm{ω}}_{2}$.
E > ExteriorDerivative(theta2);
${0}{}{\mathrm{dx}}{}{\bigwedge }{}{\mathrm{dy}}{}{\bigwedge }{}{{\mathrm{du}}}_{\left[\right]}$ (2.9)
E > convert(theta2, DGbiform, [2, 0]);
${-}\left({{u}}_{{1}}{}{{u}}_{{1}{,}{2}{,}{2}}{+}{{u}}_{{2}}{}{{u}}_{{1}{,}{2}{,}{2}}{+}{{u}}_{{1}{,}{2}}^{{2}}{+}{{u}}_{{2}{,}{2}}{}{{u}}_{{1}{,}{2}}\right){}{\mathrm{Dx}}{}{\bigwedge }{}{\mathrm{Dy}}$ (2.10)
Example 3.
Define a type (2, 1) form ${\mathrm{ω}}_{3}$ and show that .
E > omega3 := EulerLagrange(u[1]*u[2]^2*Dx &w Dy);
${\mathrm{ω3}}{≔}{-}\left({2}{}{{u}}_{{1}}{}{{u}}_{{2}{,}{2}}{+}{4}{}{{u}}_{{2}}{}{{u}}_{{1}{,}{2}}\right){}{\mathrm{Dx}}{}{\bigwedge }{}{\mathrm{Dy}}{}{\bigwedge }{}{{\mathrm{Cu}}}_{\left[\right]}$ (2.11)
E > IntegrationByParts(VerticalExteriorDerivative(omega3));
${0}{}{\mathrm{Dx}}{}{\bigwedge }{}{\mathrm{Dy}}{}{\bigwedge }{}{{\mathrm{Cu}}}_{\left[\right]}{}{\bigwedge }{}{{\mathrm{Cu}}}_{{1}}$ (2.12)
Apply the ZigZag command to ${\mathrm{ω}}_{3}$ to obtain a form ${\mathrm{θ}}_{3}$.
E > theta3 := ZigZag(omega3);
${\mathrm{θ3}}{≔}{-}{2}{}{{u}}_{{2}}^{{2}}{}{\mathrm{dx}}{}{\bigwedge }{}{\mathrm{dy}}{}{\bigwedge }{}{{\mathrm{du}}}_{{1}}{-}{4}{}{{u}}_{{1}}{}{{u}}_{{2}}{}{\mathrm{dx}}{}{\bigwedge }{}{\mathrm{dy}}{}{\bigwedge }{}{{\mathrm{du}}}_{{2}}{+}{2}{}{{u}}_{{2}}{}{\mathrm{dx}}{}{\bigwedge }{}{{\mathrm{du}}}_{\left[\right]}{}{\bigwedge }{}{{\mathrm{du}}}_{{1}}{+}{2}{}{{u}}_{{1}}{}{\mathrm{dx}}{}{\bigwedge }{}{{\mathrm{du}}}_{\left[\right]}{}{\bigwedge }{}{{\mathrm{du}}}_{{2}}{-}{2}{}{{u}}_{{2}}{}{\mathrm{dy}}{}{\bigwedge }{}{{\mathrm{du}}}_{\left[\right]}{}{\bigwedge }{}{{\mathrm{du}}}_{{2}}$ (2.13)
Check that ${\mathrm{θ}}_{3}$ is $d-$closed and that its [2, 1] component matches ${\mathrm{ω}}_{3}.$
E > ExteriorDerivative(theta3);
${0}{}{\mathrm{dx}}{}{\bigwedge }{}{\mathrm{dy}}{}{\bigwedge }{}{{\mathrm{du}}}_{\left[\right]}{}{\bigwedge }{}{{\mathrm{du}}}_{{1}}$ (2.14)
E > convert(theta3, DGbiform, [2, 1]);
${-}\left({2}{}{{u}}_{{1}}{}{{u}}_{{2}{,}{2}}{+}{4}{}{{u}}_{{2}}{}{{u}}_{{1}{,}{2}}\right){}{\mathrm{Dx}}{}{\bigwedge }{}{\mathrm{Dy}}{}{\bigwedge }{}{{\mathrm{Cu}}}_{\left[\right]}$ (2.15) | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 52, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9945780038833618, "perplexity": 4455.345780242189}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-30/segments/1500549448146.46/warc/CC-MAIN-20170728083322-20170728103322-00605.warc.gz"} |
https://cs.stackexchange.com/questions/13887/complexity-of-deciding-if-a-formula-has-exactly-1-satisfying-assignment/13888 | # Complexity of deciding if a formula has exactly 1 satisfying assignment
The decision problem
Given a Boolean formula $\phi$, does $\phi$ have exactly one satisfying assignment?
can be seen to be in $\Delta_2$, $\mathsf{UP}$-hard and $\mathsf{coNP}$-hard. Is anything more known about its complexity?
Your problem is known as the $\text{UNIQUE-SAT}$ problem which is $\mathsf{US}$-complete. The problem is in $\mathsf{D^p}$ but not known to be $\mathsf{D^p}$-hard under deterministic polynomial time reductions, where the class $\mathsf{D^p} = \{ L_1 \cap \overline{L_2} \mid L_1,L_2 \in \mathsf{NP} \}$.
It was shown by Papadimitriou and Yannakis [1] that the set of uniquely satisfiable formulas is contained in $\mathsf{D^p}$. This follows by the definition of $\mathsf{D^p}$: let $L_1$ be SAT, and let $L_2$ be the set of formulas with $2$ or more satisfying assignments. Regarding $\mathsf{D^p}$-hardness of $\text{UNIQUE-SAT}$, Blass and Gurevich [2] gave a partial answer. For one, they showed a non-relativizing proof technique would be needed to settle the question. However, Valiant and Vazirani [3] gave a randomized polynomial time reduction from $\text{SAT}$ showing $\mathsf{D^p}$-hardness of $\text{UNIQUE-SAT}$ under randomized polynomial time reductions .
When it is known that the problem has at most one assignment or no assignments, the promise problem is called $\text{UNAMBIGUOUS-SAT}$. The Valiant–Vazirani theorem states that if there is a polynomial time algorithm for $\text{UNAMBIGUOUS-SAT}$, then $\mathsf{NP}=\mathsf{RP}$. To prove their theorem they showed that the promise problem $\text{UNAMBIGUOUS-SAT}$ is $\mathsf{NP}$-hard under randomized polynomial time reductions. A corollary that follows from the Valiant–Vazirani theorem is that $\text{UNIQUE-SAT}$ is complete for $\mathsf{D^p}$ under randomized polynomial time reductions. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9913694262504578, "perplexity": 247.40136964552232}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-43/segments/1570986673538.21/warc/CC-MAIN-20191017095726-20191017123226-00229.warc.gz"} |
https://ysharifi.wordpress.com/tag/commutative-ring/ | ## Rings satisfying x^3 = x are commutative
Posted: December 13, 2010 in Elementary Algebra; Problems & Solutions, Rings and Modules
Tags: , , , , ,
Throughout $R$ is a ring.
Theorem (Jacobson). If for every $x \in R$ there exists some $n > 1$ such that $x^n=x,$ then $R$ is commutative.
The proof of Jacobson’s theorem can be found in standard ring theory textbooks. Here we only discuss a very special case of the theorem, i.e. when $x^3=x$ for all $x \in R.$
Definitions. An element $x \in R$ is called idempotent if $x^2=x.$ The center of $R$ is
$Z(R)=\{x \in R: \ xy=yx \ \text{for all} \ y \in R \}.$
It is easy to see that $Z(R)$ is a subring of $R.$ An element $x \in R$ is called central if $x \in Z(R).$ Obviously $R$ is commutative iff $Z(R)=R,$ i.e. every element of $R$ is central.
Problem. Prove that if $x^3=x$ for all $x \in R,$ then $R$ is commutative.
Solution. Clearly $R$ is reduced, i.e. $R$ has no nonzero nilpotent element. For every $x \in R$ we have $(x^2)^2=x^4 = x^2$ and so $x^2$ is idempotent for all $x \in R.$ Hence, by Remark 3 in this post, $x^2$ is central for all $x \in R.$ Now, since
$(x^2+x)^2=x^4+2x^3+x^2=2x^2+2x$
we have $2x=(x^2+x)^2-2x^2$ and thus $2x$ is central. Also, since
$x^2+x=(x^2+x)^3=x^6+3x^5+3x^4+x^3=4x^2+4x,$
we have $3x=-3x^2$ and hence $3x$ is central. Therefore $x = 3x-2x$ is central. $\Box$
A similar argument shows that if $x^4=x$ for all $x \in R,$ then $R$ is commutative (see here!).
## Units in polynomial rings
Posted: September 2, 2010 in Elementary Algebra; Problems & Solutions, Rings and Modules
Tags: , , ,
Lemma. Let $R$ be a commutative ring with 1. If $a \in R$ is nilpotent and $b \in R$ is a unit, then $a+b$ is a unit.
Proof. So $a^n = 0$ for some integer $n \geq 1$ and $bc = 1$ for some $c \in R.$ Let
$u = (b^{n-1}- ab^{n-2} + \ldots + (-1)^{n-2}a^{n-2}b + (-1)^{n-1}a^{n-1})c^n$
and see that $(a+b)u=1. \ \Box$
Problem. Let $R$ be a commutative ring with 1. Let $p(x) = \sum_{j=0}^n a_j x^j, \ a_j \in R,$ be an element of the polynomial ring $R[x].$ Prove that $p(x)$ is a unit if and only if $a_0$ is a unit and all $a_j, \ j \geq 1,$ are nilpotent.
First Solution. ($\Longrightarrow$) Suppose that $a_1, \cdots , a_n$ are nilpotent and $a_0$ is a unit. Then clearly $p(x)-a_0$ is nilpotent and thus $p(x)=p(x)-a_0 + a_0$ is a unit, by the lemma.
($\Longleftarrow$) We’ll use induction on $n,$ the degree of $p(x).$ It’s clear for $n = 0.$ So suppose that the claim is true for any polynomial which is a unit and has degree less than $n.$ Let $p(x) = \sum_{j=0}^n a_jx^j, \ n \geq 1,$ be a unit. So there exists some $q(x)=\sum_{j=0}^m b_jx^j \in R[x]$ such that $p(x)q(x)=1.$ Then $a_0b_0=1$ and so $b_0$ is a unit. We also have
$a_nb_m = 0, \ a_nb_{m-1}+ a_{n-1}b_m = 0, \ \cdots , a_nb_0 +a_{n-1}b_1 + \cdots = 0.$
So $AX=0,$ where
$A=\begin{pmatrix}a_n & 0 & 0 & . & . & . & 0 \\ a_{n-1} & a_n & 0 & . & . & . & 0 \\ . & . & . & & . & . & . \\ . & . & . & & . & . & . \\ . & . & . & & . & . & . \\ * & * & * & . & . & . & a_n \end{pmatrix}, \ \ X=\begin{pmatrix}b_m \\ b_{m-1} \\ . \\ . \\ . \\ b_0 \end{pmatrix}.$
Thus $a_n^{m+1}X =(\det A)X = \text{adj}(A)A X = 0.$ Therefore $a_n^{m+1}b_0=0$ and hence $a_n^{m+1} = 0$ because $b_0$ is a unit. Thus $a_n,$ and so $-a_nx^n,$ is nilpotent. So $p_1(x)=p(x) -a_nx^n$ is a unit, by the lemma. Finally, since $\deg p_1(x) < n,$ we can apply the induction hypothesis to finish the proof. $\Box$
Second Solution. ($\Longrightarrow$) This part is the same as the first solution.
($\Longleftarrow$) Let $p(x) = \sum_{j=0}^n a_jx^j, \ a_n \neq 0,$ be a unit of $R[x]$ and let $q(x)=\sum_{i=0}^m b_i x^i \in R[x], \ b_m \neq 0,$ be such that $p(x)q(x)=1.$ Then $a_0b_0=1$ and so $a_0$ is a unit. To prove that $a_j$ is nilpotent for all $j \geq 1,$ we consider two cases:
Case 1 . $R$ is an integral domain. Suppose that $n > 0.$ Then from $p(x) q(x)=1$ we get $a_n b_m = 0,$ which is impossible because both $a_n$ and $b_m$ are non-zero and $R$ is an integral domain. So $n=0$ and we are done.
Case 2 . $R$ is arbitrary. Let $P$ be any prime ideal of $R$ and let $\overline{R}=R/P.$ For every $r \in R$ let $\overline{r}=r+P.$ Let
$\overline{p(x)}=\sum_{j=0}^n \overline{a_j}x^j, \ \ \overline{q(x)}=\sum_{i=0}^m \overline{b_i}x^i.$
Then clearly $\overline{p(x)} \cdot \overline{q(x)}=\overline{1}$ in $\overline{R}[x]$ and thus, since $\overline{R}$ is an integral domain, $\overline{a_j}=\overline{0}$ for all $j \geq 1,$ by case 1. Hence $a_j \in P$ for all $j \geq 1.$ So $a_j, \ j \geq 1,$ is in every prime ideal of $R$ and thus $a_j$ is nilpotent. $\Box$
## Representation of polynomials
Posted: May 23, 2010 in Elementary Algebra; Problems & Solutions, Rings and Modules
Tags: ,
This is a generalization of the ordinary representation of polynomials:
Problem. Let $R$ be a commutative ring with $1$ and $A \in R[x]$ have degree $n \geq 0$ and let $B \in R[x]$ have degree at least $1$. Prove that if the leading coefficient of $B$ is a unit of $R$, then there exist unique polynomials $Q_0,Q_1,...,Q_n \in R[x]$ such that $\deg Q_i < \deg B,$ for all $i$, and $A = Q_0+Q_1B+...+Q_nB^n$
SolutionUniqueness of the representation : Since the leading coefficient of $B$ is a unit, for any $C \in R[x]$ we have $\deg (BC)=\deg B + \deg C.$ Now suppose that $Q_0 + Q_1B + \cdots + Q_nB^n = 0,$ with $Q_n \neq 0.$ Let $\alpha, \ \beta$ be the leading coefficients of $Q_n$ and $B$ repectively. Then the leading coefficient of $Q_0 + Q_1B + \cdots +Q_nB^n$ is $\alpha \beta^n.$ Thus $\alpha \beta^n = 0.$ Since $\beta$ is a unit, we’ll get $\alpha = 0,$ which contradicts $Q_n \neq 0.$ Therefore $Q_0 = Q_1= \cdots = Q_n=0.$
Existence of the representation : We only need to prove the claim for $A=x^n.$ The proof is by induction over $n.$ It is clear for $n = 0,$ Suppose that the claim is true for any $k < n.$ If $n < \deg B,$ then choose $A=Q_0$ and $Q_1 = \cdots = Q_n=0.$ So we may assume that $n \geq \deg B.$ Let $B=b_mx^m + b_{m-1}x^{m-1}+ \cdots + b_0.$ Therefore, since $b_m$ is a unit, we will have $x^m=b_m^{-1}B-b_m^{-1}b_{m-1}x^{m-1} - \cdots - b_m^{-1}b_0,$ which will give us $x^n = b_m^{-1}x^{n-m}B - b_m^{-1} b_{m-1}x^{n-1} - \cdots - b_m^{-1}b_0 x^{n-m}.$ Now apply the induction hypothesis to each term $x^{n-k}, \ 1 \leq k \leq m,$ to finish the proof.
## “Almost Boolean” rings are commutative
Posted: March 12, 2010 in Elementary Algebra; Problems & Solutions, Rings and Modules
Tags: , , ,
It is easy to prove that if every element of a ring is idempotent, then the ring is commutative. This fact can be generalized as follows.
Problem. 1) Let $R$ be a ring with identity and suppose that every element of $R$ is a product of idempotent elements. Prove that $R$ is commutative.
2) Give an example of a noncommutative ring with identity $R$ such that every element of $R$ is a product of some elements of the set $\{r \in R: \ r^n=r, \ \text{for some} \ n \geq 2 \}.$
Solution. 1) Obviously we only need to prove that every idempotent is central. Suppose first that $ab = 1,$ for some $a,b \in R.$ We claim that $a = b = 1.$ So suppose the claim is false. Then $a = e_1e_2 \cdots e_k,$ where $e_j$ are idempotents and $e_1 \neq 1.$ Let $e = e_2 \cdots e_kb.$ Then $e_1e = 1$ and hence $1 - e_1 = (1 - e_1)e_1e = 0.$ Thus $e_1 = 1.$ Contradiction! Now suppose that $x^2 = 0,$ for some $x \in R.$ Then $(1 - x)(1 + x) = 1$ and therefore $x = 0$, by what we just proved. Finally, since $(ey-eye)^2=(ye-eye)^2=0$ for any idempotent $e \in R$ and any $y \in R,$ we have $ey = ye$ and so $e$ is central.
2) One example is the ring of $2 \times 2$ upper triangular matrices with entries from $\mathbb{Z}/2\mathbb{Z}.$
## Module-finiteness of Laurent polynomial rings
Posted: March 11, 2010 in Elementary Algebra; Problems & Solutions, Rings and Modules
Tags: , , ,
Let $R$ be a commutative ring with identity and $S=R[x,x^{-1}],$ the ring of Laurent polynomials with coefficients in $R.$ Obviously $S$ is not a finitely generated $R$-module but we can prove this:
Problem. There exists $f \in S$ such that $S$ is a finitely generated $R[f]$-module.
Solution. Let $f=x+x^{-1}.$ Then $x=f - x^{-1}$ and $x^{-1}=f-x.$ Now an easy induction shows that $x^n \in xR[f]+R[f]$ for all $n \in \mathbb{Z}.$ Hence $S=xR[f] + R[f]. \ \Box$ | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 200, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9914846420288086, "perplexity": 128.23211393887226}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-26/segments/1498128320070.48/warc/CC-MAIN-20170623151757-20170623171757-00675.warc.gz"} |
http://mathhelpforum.com/calculus/97053-find-area-region-enclosed.html | Math Help - Find the area of the region enclosed
1. Find the area of the region enclosed
Find the area of the region enclosed between and from to .
can anyone help with this one.
2. $
dA = f(x) - g(x) <=> A = \int {f(x) - g(x)}\$
where f(x) is the top-function
So, figure out what function is above the other and get the anti derivative, then put in your limits
3. so the anti derivative is -3cos(x)-2sin(x)
what do i do after i plug in the limits
4. Originally Posted by dat1611
Find the area of the region enclosed between and from to .
can anyone help with this one.
first find the intersection of the two functions in the given interval ...
$3\sin{x} = 2\cos{x}$
$\tan{x} = \frac{2}{3}$
$x = \arctan\left(\frac{2}{3}\right)$
finding the area will consist of two integrals because the "top" function changes ...
let $c = \arctan\left(\frac{2}{3}\right)$
$A = \int_0^c 2\cos{x} - 3\sin{x} \, dx + \int_c^{\frac{7\pi}{10}} 3\sin{x} - 2\cos{x} \, dx
$
integrate and use the Fundamental Theorem of Calculus to find the area value.
note that this area problem is probably better suited for completion with a calculator.
5. Originally Posted by dat1611
Find the area of the region enclosed between and from to .
can anyone help with this one.
So you have ${y_1} = 3\sin x,{\text{ }}{y_2} = 2\cos x,{\text{ }}0 \leqslant x \leqslant \frac{{7\pi }}{{10}},{\text{ }}S = ?$
${y_1} = {y_2} \Leftrightarrow 3\sin x = 2\cos x \Leftrightarrow \tan x = \frac{2}{3} \Leftrightarrow x = \arctan \frac{2}{3} + \pi k{\text{ }}\left( {k \in \mathbb{Z}} \right).$
$S = {S_1} + {S_2}.$
${S_1} = \int\limits_0^{\arctan \frac{2}{3}} {\left( {2\cos x - 3\sin x} \right)dx} = \left. {\left( {2\sin x + 3\cos x} \right)} \right|_0^{\arctan \frac{2}{3}} =$
$= 2\sin \left( {\arctan \frac{2}{3}} \right) + 3\cos \left( {\arctan \frac{2}{3}} \right) - 3 =$
$= 2 \cdot \frac{{{2 \mathord{\left/{\vphantom {2 3}} \right.\kern-\nulldelimiterspace} 3}}}{{\sqrt {1 + {{\left( {{2 \mathord{\left/
{\vphantom {2 3}} \right.\kern-\nulldelimiterspace} 3}} \right)}^2}} }} + 3 \cdot \frac{1}{{\sqrt {1 + {{\left( {{2 \mathord{\left/{\vphantom {2 3}} \right.\kern-\nulldelimiterspace} 3}} \right)}^2}} }} - 3 =$
$= \frac{4}{{\sqrt {13} }} + \frac{9}{{\sqrt {13} }} - 3 = \frac{{13}}{{\sqrt {13} }} - 3 = \sqrt {13} - 3.$
${S_1} = \int\limits_{\arctan \frac{2}{3}}^{{{7\pi } \mathord{\left/{\vphantom {{7\pi } {10}}} \right.\kern-\nulldelimiterspace} {10}}} {\left( {3\sin x - 2\cos x} \right)dx} = - \left. {\left( {3\cos x + 2\sin x} \right)} \right|_{\arctan \frac{2}{3}}^{{{7\pi } \mathord{\left/{\vphantom {{7\pi } {10}}} \right.\kern-\nulldelimiterspace} {10}}} =$
$= - \left( {3\cos \frac{{7\pi }}{{10}} + 2\sin \frac{{7\pi }}{{10}} - \sqrt {13} } \right) = \sqrt {13} - 3\cos \left( {\pi - \frac{{3\pi }}{{10}}} \right) - 2\sin \left( {\pi - \frac{{3\pi }}{{10}}} \right) =$
$= \sqrt {13} + 3\cos \frac{{3\pi }}{{10}} - 2\sin \frac{{3\pi }}{{10}} = \sqrt {13} + 3 \cdot \frac{{\sqrt {5 - \sqrt 5 } }}{{2\sqrt 2 }} - 2 \cdot \frac{{\sqrt 5 + 1}}{4} =$
$= \sqrt {13} + \frac{{3\sqrt 2 \sqrt {5 - \sqrt 5 } }}{4} - \frac{{\sqrt 5 + 1}}{2} = \sqrt {13} + \frac{3}{4}\sqrt {10 - 2\sqrt 5 } - \frac{1}{2}\left( {\sqrt 5 + 1} \right).$
Finally
$S = \sqrt {13} - 3 + \sqrt {13} + \frac{3}{4}\sqrt {10 - 2\sqrt 5 } - \frac{1}{2}\left( {\sqrt 5 + 1} \right) =$
$= \frac{1}{2}\left( {4\sqrt {13} - \sqrt 5 } \right) + \frac{3}{4}\sqrt {10 - 2\sqrt 5 } - \frac{7}{2} \approx {\text{4}}{\text{.35642}}$
See this picture | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 19, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9128332138061523, "perplexity": 415.64478240358227}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-22/segments/1433195036613.12/warc/CC-MAIN-20150601214356-00084-ip-10-180-206-219.ec2.internal.warc.gz"} |
https://standards.globalspec.com/std/166046/ieee-c57-113 | # IEEE C57.113
## Guide for Partial Discharge Measurement in Liquid-Filled Power Transformers and Shunt Reactors
inactive
Organization: IEEE Publication Date: 5 December 1991 Status: inactive Page Count: 27
##### scope:
Foreword
(This foreword is not a part of IEEE C57.113-1991, IEEE Guide for Partial Discharge Measurement in Liquid-Filled Power Transformers and Shunt Reactors.)
As a long-term trend beginning in the 1950's, the combination of lower insulation levels and higher system voltages brought about increased interest in the detection and measurement of partial discharge activity within the insulation structure of a transformer. As the name implies, a partial discharge is the breakdown of a small section of the insulation path and is undesirable because of deterioration of the insulation and formation of gas that may accumulate at a critical stress area. In general, electrical measurements of partial discharge activity should be made on the basis of the resultant momentary change in the voltage at the terminals of the transformer. Such change may be expressed as a voltage change or, by suitable calibration, as an apparent charge. An apparent charge is that charge in coulombs that, if injected between terminals, would cause the same voltage change as that resulting from the partial discharge.
The initial efforts at measuring partial discharge levels, particularly in regard to acceptance criteria between the user and the manufacturer, utilized NEMA 107-1940, which provided radio influence voltage (RIV) readings in microvolts (mV) on a quasi-peak basis at or near 1.0 MHz. Later, this approach was modified by using the bushing tap instead of a separate coupling capacitor and eventually was included as the standard method of measuring partial discharges in transformers in IEEE C57.12.90-1980.
Meanwhile, however, the industry has recognized that measuring partial discharges in terms of apparent charge has many advantages over the RIV approach. Two advantages are (1) the differences in internal capacitance between transformers are compensated by the calibration procedure, thus the measured level is more closely related to the true level of the partial discharge, and (2) a generally lower specified measuring frequency provides for less attenuation of partial discharge located deep within the transformer insulation structure.
The problems that have delayed a change to apparent charge measurements as an industry standard have been that (1) the RIV approach has had the advantage of being based on a recognized and established circuit, and (2) the industry has gained a great deal of experience with the circuit including appropriate acceptance levels. To take advantage of the apparent charge approach required first that a standard circuit be developed. This was undertaken by the Task Force for the Measurement of Apparent Charge within the IEEE Transformers Committee. This document is the result of that effort. Apparent charge measurements may be made on a wide-band or narrow-band basis, as both systems are recognized and widely used. Without giving preference to one or the other, it is the object of this document to describe the wide-band method. General principles of partial discharge measurements including the narrow-band method are covered in IEC 270 (1981), IEC 76-3 (1980), and IEEE Std 454-1973. The Task Force that prepared this document hopes and expects that users and manufacturers of transformers will make apparent charge measurements on new transformers so that an adequate base of experience can be developed and so that the apparent charge method will eventually become the standard method of measuring partial discharge within transformers. This document was developed by the Task Force for the Measurement of Apparent Charge and the Working Group on Partial Discharge Tests in Transformers of the IEEE Transformers Committee.
Scope
This test procedure applies to the detection and measurement by the wide-band apparent charge method of partial discharges occurring in liquid-filled power transformers and shunt reactors during dielectric tests, where applicable.
### Document History
June 17, 2010
Recommended Practice for Partial Discharge Measurement in Liquid-Filled Power Transformers and Shunt Reactors
This recommended practice describes the test procedure for the detection and measurement by the wideband apparent charge method of partial discharges (PDs) occurring in liquid-filled power...
IEEE C57.113
December 5, 1991
Guide for Partial Discharge Measurement in Liquid-Filled Power Transformers and Shunt Reactors
Foreword (This foreword is not a part of IEEE C57.113-1991, IEEE Guide for Partial Discharge Measurement in Liquid-Filled Power Transformers and Shunt Reactors.) As a long-term trend beginning in the...
January 1, 1991
Guide for Partial Discharge Measurement in Liquid-Filled Power Transformers and Shunt Reactors
A description is not available for this item.
January 1, 1988
Guide for partial discharge measurement in liquid-filled power transformers and shunt reactors
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https://www.physicsforums.com/threads/number-theory-largest-impossible-score.157983/ | # Homework Help: Number Theory - Largest Impossible Score
1. Feb 25, 2007
### Frillth
Edit: I finally figured this problem out, although I didn't use residue systems, so a solution is no longer necessary.
1. The problem statement, all variables and given/known data
For this problem, I need to find a fomula for the largest unattainable score in a game where you can either score m or n points at a time (m and n are relatively prime) and prove that this formula will always work. Complete residue systems are suggested to be used in the proof.
2. Relevant equations
{0, n, 2n, 3n, ... , (m-1)n} is a complete residue system mod m when m and n are relatively prime.
3. The attempt at a solution
Through trial and error, I have found that the largest impossible score is mn - (m+n). I'm stuck, however, on proving this. I don't even know how to start, so any help would be greatly appreciated.
Last edited: Feb 25, 2007 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8827778100967407, "perplexity": 394.68804950401955}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-43/segments/1539583509336.11/warc/CC-MAIN-20181015163653-20181015185153-00127.warc.gz"} |
https://www.themathdoctors.org/a-closer-look-at-a-limit-proof/ | # A Closer Look at a Limit Proof
#### (A new question of the week)
A recent question asked about one of our explanations of the limit of x2 (which we have discussed at least five times). This led to a deeper examination of what was said; and as I have looked through this and other pages, I have realized that it would be worthwhile to look at the details more carefully.
The question referred to this page, from 1998:
A Limit Proof Using EstimationHow do I show that the limit of x^2 as x->(-2) is 4, using the delta-epsilon definition of a limit?
Here is the first part of Doctor Rob’s reply:
The definition says that for any epsilon > 0, no matter how small, you can find a delta > 0 such that: |x-(-2)| = |x+2| < delta implies that |x^2-4| < epsilonThe idea is to start with what you want to show, that |x^2-4| < epsilon, and to manipulate this until you can get it into the form |x+2| < some expression in epsilon. Then picking delta to be this expression in epsilon will do, and the proof is to work backwards through the steps of the manipulation. In this case: |x^2-4| < epsilon <==> |(x+2)^2 - 4*(x+2)| < epsilon <== |x+2|^2 + 4*|x+2| < epsilon (by the triangle inequality) <==> |x+2|^2 + 4*|x+2| - epsilon < 0Now you can use the Quadratic Formula to solve for |x+2|, and thus find an upper bound on |x+2| in terms of epsilon. That will be what you choose for delta. One tricky part is that each step needs an implication arrow in one direction (<==) but not necessarily in the other.
Before I look at the new question, we should read between the lines to see if we can follow this, which is not the method I am most familiar with.
First, note the structure of what he is doing: The goal is to show that, given a fact about delta, we can make a conclusion about epsilon; but first we have to find the appropriate delta. So the proof actually has two parts, which shouldn’t be confused. The search (exploration) part looks as if we were solving an inequality; but because of the ultimate goal, we need each step not to imply the next, but to be implied by it. Note the arrows, <==> and <==, the latter meaning that the new line will imply the one before.
Now, how does he get $$|x^2-4| < \epsilon \Leftrightarrow |(x+2)^2 – 4(x+2)| < \epsilon$$? This is not obvious from what was said; but we can check it by expanding $$(x+2)^2 – 4(x+2)$$ as $$x^2 + 4x + 4 – 4x – 8 = x^2 – 4$$ as claimed. One way to obtain that result (without just guessing) is to make a substitution: We want an expression in terms of $$x+2$$, so we let $$u = x+2$$, and replace $$x$$ with $$u-2$$:
$$x^2 – 4 = (u – 2)^2 – 4 = u^2 – 4u + 4 – 4 = u^2 – 4u = (x+2)^2 – 4(x + 2)$$.
The benefit of this is that we can use our knowledge of $$|x+2$$ in subsequent steps.
How about the next line, which refers to the triangle inequality? This is the fact that, for any a and b, $$|a – b| \le |a| + |b|$$, just as any side of a triangle is less than the sum of the other two sides. Specifically, it tells us that
$$|(x+2)^2 – 4(x+2)| \le |(x+2)^2| + |4(x+2)|$$. As a result, if the right side is less than epsilon, we can conclude that the left side is also less than epsilon (since it is even smaller), which is what we will want to conclude when this chain of statements is reversed to make the final proof. This is why we were looking for an expression greater than the expression we had.
Now we have a quadratic inequality, $$|x+2|^2 + 4|x+2| – \epsilon < 0$$, that we can solve. Doctor Rob suggests using the quadratic formula to do this, and this is where the recent question came in.
Juares wrote,
In this link, http://mathforum.org/library/drmath/view/53357.html , have example:
$$\displaystyle\lim_{x\rightarrow -2}x^2 = 4$$:
$$|x+2|^2 + 4|x+2| < \epsilon$$
$$|x+2|^2 + 4|x+2| – \epsilon < 0$$
and solve by quadratic formula for any ε
$$\displaystyle\frac{-b \pm\sqrt{b^2-4ac}}{2a}$$
$$\displaystyle\frac{-4|x+2| \pm\sqrt{(4|x+2|)^2-4(|x+2|)(-\epsilon)}}{2|x+2|}$$
He showed more work, but this is enough to see the error. Doctor Rick responded:
Hi, Juares. It looks like you haven’t understood what Doctor Rob meant when he said:
” |x^2-4| < epsilon
<==> |(x+2)^2 – 4(x+2)| < epsilon
<== |x+2|^2 + 4|x+2| < epsilon (by the triangle inequality)
<==> |x+2|^2 + 4*|x+2| – epsilon < 0
Now you can use the Quadratic Formula to solve for |x+2|, and thus find an upper bound on |x+2| in terms of epsilon.”
In your work, you are taking a = |x + 2|, b = 4|x + 2|, and c = – ε, which implies that you are trying to solve the “quadratic equation” |x + 2|x2 + 4|x + 2|x – ε = 0. That’s not a quadratic equation!
When Doctor Rob said to solve for |x + 2|, he meant that you treat |x + 2| as the variable. If we let u = |x + 2|, then the inequality becomes
u2 + 4u – ε < 0
Now solve this inequality for u. That is, solve the related equation u2 + 4u – ε = 0 for u, which gives the endpoints of the solution interval(s), and decide whether the solution set for the inequality lies between these points or outside them.
Now Juares correctly applied the quadratic formula to solve the equation:
$$\displaystyle\frac{-4\pm\sqrt{16+4\epsilon}}{2} = -2+\sqrt{4+\epsilon}$$
This is δ. Let ε = 0.01. Then $$\displaystyle\delta =-2+\sqrt{4,1} = 0,025$$
This was followed by some long calculations presented as an image, so I will not attempt to reproduce it all. The important thing is that he has found a value for delta in general, and then is observing what happens for a specific value of epsilon, which is an excellent way to get a better feel for what is happening.
Hi again, Juares. Let me see whether I understand what you have done here.
We are working on solving the inequality u2 + 4u – ε < 0, where u = |x + 2|. You apply the Quadratic Formula to solve the related equation u2 + 4u – ε = 0, and this gives you the result
-2 – √(4 + ε) < |x + 2| < -2 + √(4 + ε)
Now, the quantity on the left is clearly negative (for any positive value of ε), whereas an absolute value (or modulus, as you probably call it), |x + 2|, must be non-negative, so we can further restrict the possible values for |x + 2|:
0 ≤ |x + 2| < -2 + √(4 + ε)
As Doctor Rob said, “The definition says that for any ε > 0, no matter how small, you can find a delta > 0 such that: |x-(-2)| = |x+2| < δ implies that |x2 – 4| < ε.” We have in fact found that:
For any ε > 0, we can choose δ = -2 + √(4 + ε)
and then |x – (-2)| < δ implies that |x2 – 4| < ε.
So we have found what is needed for the proof.
The rest of what you have written appears to be a check of this result by testing one possible value for ε, namely, ε = 0.1. In this case, we choose
δ = -2 + √(4 + 0.1) = 0.024845…
0 < |x – 2| < 0.024845…You have rounded that last number to 0.025; I have just kept more digits.
What I find, next, is
2 – 0.025 < x < 2 + 0.025
1.975 < x < 2.025
(I used the idea that |x – a| < b means that the distance between x and a is less than b, that is, x is between a-b and a+b.)
We want to demonstrate that, for any x in this interval, |x2 – 4| < 0.1. I suppose that you may be trying to use the fact that |x2 – 4| = |x + 2| |x – 2| to prove this. I would prefer to note that f(x) = |x2 – 4| is an increasing function on [1.975, 2.025], so that it obtains its greatest value at x = 2.025; and in that case, we find
If x = 2.025 then |x2 – 4| = 0.10065
If we hadn’t rounded our δ up, we would find that when x = 2 + δ, |x2 – 4| = 0.1; and for any x between 2 – δ and 2 + δ, |x2 – 4| < 0.1. This is what we expected to find, if our formula for δ in terms of ε is good.
So you have found a valid proof, and I hope I have helped you see that it is valid.
Note that since delta is an upper bound, we don’t want to increase it! So rather than rounding up, it would have been appropriate to round down, to ensure the required conclusion.
Now I want to continue and look at the last part of Doctor Rob’s answer, where he gives an alternative approach (which is what I am familiar with). It is important to observe that there is not just one valid delta in these proofs; not only can Doctor Rob’s first delta be replaced by anything smaller (since that will still imply the required conclusion), but he hasn’t determined that his delta is the largest or best possible number to use. In the next part, he obtains an entirely different formula for delta:
Another tricky part is that there isn't necessarily a unique answer. In this case, you could have proceeded like this instead: |x^2-4| < epsilon <==> |(x+2)*(x-2)| < epsilon <==> |x+2|*|x-2| < epsilon <== |x+2|*(4 + |x+2|) < epsilon (by the triangle inequality) <== 5*|x+2| < epsilon (provided |x+2| <= 1)and so on. At the end, you pick delta to be the minimum of 1 and the expression involving epsilon, and this ensures the "provided ..." part.Other expressions for delta in terms of epsilon may also work.
Finishing this work, we see that if $$|x+2| < \epsilon/5$$, then $$|x^2-4| < \epsilon$$, so we can take delta to be $$\epsilon/5$$ or anything smaller.
So this version of the proof takes $$\delta=min(\epsilon/5, 1)$$. In Juares’ example with ε = 0.01, we got δ = 0.024845… . This new version finds $$\delta = min(0.01/5, 1) = 0.002$$. This is a much smaller value (and therefore also works). The first method obtained a less restrictive value for delta, but both obtain values that provably lead to the required limit.
This same method (though for x approaching 2 rather than -2) is discussed in detail by two Math Doctors here:
Epsilon/Delta Definition of Limits
We also saw Doctor Fenton’s version of the same approach on Monday, as part of this answer:
Formal Definition of a Limit
In the following page, I looked at the same problem with a numerical value of delta (similar to Juares’ check):
Definition of the Limit
Finally, I examined a specific feature of this proof here:
Delta-Epsilon Proofs and Arbitrary Epsilon Choice
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https://www.bradfordlynch.com/blog/2015/12/04/InvestmentPortfolioOptimization.html | Investment Portfolio Optimization | Bradford (Lynch) Levy’s Blog
# Investment Portfolio Optimization
###### Originally Posted: December 04, 2015
The need to make trade-offs between the effort exerted on specific activities is felt universally by individuals, organizations, and nations. In many cases, activities are mutally-exclusive so partaking in one option excludes participation in another. Deciding how to make these trade-offs can be immensely difficult, especially when we lack quantitative data about risk and reward to support our decision.
In the case of financial assets, there is a wealth of numerical data available. So in this post, I explore how historical data can be leveraged to choose specific mixes of assets based on investment goals. The tools used are quite simple and rely on the mean-variance of assets’ returns to find the efficient frontier of a portfolio.
Note: The code and data used to generate the plots in this post are available here.
## Historical Stock Price Data
Most people have probably interacted with Google Finance or Yahoo! Finance to access financial information about stocks. The Python library Pandas provides an exceedingly simple interface for pulling stock quotes from either of these sources:
This integration makes pulling historical pice data very easy, which is great since we want to estimate portfolio performance using past pricing data.
## Basic Portfolio Theory
Investopedia defines “Portfolio theory” as:
Modern Portfolio Theory - MPT A theory on how risk-averse investors can construct portfolios to optimize or maximize expected return based on a given level of market risk, emphasizing that risk is an inherent part of higher reward. “Modern Portfolio Theory - MPT.” Investopedia.com. Investopedia, 2015. Web. 5 December 2015.
The idea is the fundamental concept of diversification and essentially boils down to the idea that pooling risky assets together, such that the combined expected return is aligned with an investor’s target, will provide a lower risk-level than carrying just one or two assets by themselves. The technique is even more effective when the assets are not well correlated with one another.
Portfolio theory’s efficacy can be seen by examining the underlying statistical thoughts behind the concept. For a single asset, the expected mean-variance of returns ‘r’ over ‘n’ observations is given by the standard formulas below.
\begin{align*} & Mean(Returns) = \frac{1}{n} \sum_{i=1}^{n}r_i \\ & Var(Returns) = \sum_{i=1}^{n}(r_i-\mu)^2 \\ \end{align*}
Expanding this to the mean-variance of a portfolio of assets that are potentially correlated to one another, we have the weighted average of the mean returns, and the sum of the terms in the covariance matrix for the assets:
\begin{align*} & Mean(Portfolio\ Returns) = \sum_{i=1}^{n}w_i r_i \\ & Var(Portfolio\ Returns) = \sum_{i=1}^{n}\sum_{j=1}^{n}w_i w_j \rho_{i,j} \sigma_i \sigma_j = \mathbf{w}^T \bullet \big(\mathbf{cov} \bullet \mathbf{w}\big) \\ \end{align*}
The Python package NumPy provides us with all of the functions necessary to calculate the mean-variance of returns for individual assets and portfolios, including vector math operations to keep the code clean.
This is great, if we have a set of assets and their weights within a portfolio, abut 20 lines of Python will retrieve historical quotes, calculate the mean-variance of the assets, and return the expected portfolio performance. But what if we didn’t know which assets to purchase and in what proportion to one another? Surely there are multiple ways to achieve a target return and perhaps those different asset blends will yield varying amounts of risk while providing the same return.
## Monte Carlo Method
One approach to optimizing a portfolio is application of the Monte Carlo Method. For unfamiliar readers, this is the idea of carrying out repeated trials using randomly generated inputs and observing the outcomes. A physical example of this would be flipping a coin 100 times and counting the number of heads and tails. Based on the results, the observer could estimate whether the coin is fair or not. In the digital world, computers can rapid generate random numbers extremely quickly, enabling observation of outcomes from complex scenarios that are based on the probabilities of certain events occurring. The code below illustrates how simple it is to implement a Monte Carlo simulation using Python:
In the plot below, 25,000 portfolios with randomly varying weights of the following assets were generated and evaluated. Their expected annual return is then plotted versus the historical volatility of the portfolio. Further, each point representing a portfolio has been shaded according to the “Sharpe Ratio.”
From Figure 1 it is obvious that changing the weight of each asset in the portfolio will have a dramatic effect on the expected return and the level of risk that the investor is exposed to. Most notably, if the investor is targeting a 12% return it is evident that the volatility could be reduced to 11.5% but some portfolios share that same expected return with as high as 17.5% volatility. Figure 1 makes it very clear that we must be thoughtful when choosing how much weight each asset in our portfolio should carry.
## Efficient Frontier and Portfolio Optimization
Based on the insights from Figure 1, it is evident that a target return can be achieved with a wide range of risk levels. This introduces the concept of the “Efficient Frontier.” The Efficient Frontier is the set of portfolios that achieve a given return with the minimum amount of risk for that return. In Figure 1, these were the portfolios furthest to the left for each expected return.
Keeping this concept in mind, a more structured approach can be applied to the selection of asset weights such that we consider only these efficient portfolios which meet a criteria important to the investor. First, she could optimize the weights to target a metric such as the Sharpe Ratio. Alternatively, the investor could opt to find the minimum volatility portfolio and accept the return that that provides. The code below consists of several helper functions that make use of SciPy’s optimization library to solve these two problems.
Figure 2 shows results from these optimizations, the portfolios with the highest Sharpe Ratio and lowest volatility are denoted by the red and yellow stars respectively.
This structured approach can be taken a step further to find the Efficient Frontier across a range of desired target returns. The set of portfolios that provide a target return with minimum volatility is found by iteratively applying the SciPy optimizer as shown in the code below. Figure 3 shows the results for this optimization.
## Optimum Portfolio of Many Assets
The framework developed up to the point can readily be applied to find the optimum portfolio when an investor is faced with choosing from many assets. Figure 4 below shows the results for a portfolio consisting of Apple, Amazon, BMW, Coke, Ford, Google, Microsoft, Pepsi, and Toyota. Figure 5 shows the weights of each asset for the Sharpe Ratio maximizing and minimum volatility portfolios. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8254583477973938, "perplexity": 1083.0371530021005}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-21/segments/1652662595559.80/warc/CC-MAIN-20220526004200-20220526034200-00735.warc.gz"} |
https://math.stackexchange.com/questions/2234190/proving-the-inequality-0-leq-frac-sqrtxy1-p-fracx-frac1p-1-y-f | # Proving the inequality $0\leq \frac{\sqrt{xy}}{1-p}\frac{x^{\frac{1}{p}-1}-y^{\frac{1}{p}-1}}{x^{\frac{1}{p}}-y^{\frac{1}{p}}} \leq 1$
Suppose $p\in(0,1)$. How might one show that $$\tag{1} 0\leq \frac{\sqrt{xy}}{1-p}\frac{x^{\frac{1}{p}-1}-y^{\frac{1}{p}-1}}{x^{\frac{1}{p}}-y^{\frac{1}{p}}} \leq 1$$ for all $x,y\in[0,1]$? It is clearly non-negative, so the hard part is to show that it is never greater than 1.
I was hoping to use a technique similar to the one to prove that $$0\leq \sqrt{xy}\frac{\log x - \log y}{x-y}\leq 1$$ for all $x,y\in[0,1]$. We can use an integral representation and see that \begin{align*} \sqrt{xy}\frac{\log x - \log y}{x-y} &= \int_{0}^{\infty} \frac{\sqrt{xy}}{(x+t)(y+t)}dt\\ &\leq \int_{0}^{\infty} \frac{\sqrt{xy}}{(\sqrt{xy}+t)^2}dt\\ & = 1. \end{align*} Is there a suitable integral representation that can prove (1)?
• You must have a typo in (1), because the numerator is zero. Did you mean $x^{1/p-1}-y^{1/p-1}$? Also the title doesn't format correctly. – Χpẘ Apr 14 '17 at 16:53
I've figured out the correct integral representation to use here. For $a\in(-1,1)$, consider the following integral representations: \begin{align*} \frac{x^a-y^a}{x-y} &= \frac{\sin(a\pi)}{\pi}\int_{0}^{\infty}\frac{t^a}{(x+t)(y+t)}dt\\ \text{and}\qquad ax^{a-1} &= \frac{\sin(a\pi)}{\pi}\int_{0}^{\infty}\frac{t^a}{(x+t)^2}dt. \end{align*} Similar to the example in the original post, we have \begin{align*} \frac{1}{a}\frac{x^a-y^a}{x-y} &\leq \frac{\sin(a\pi)}{a\pi}\int_{0}^{\infty}\frac{t^a}{(\sqrt{xy}+t)^2}dt\\ & = (\sqrt{xy})^{a-1}. \end{align*} Thus, if we let $a=1-p$, we have \begin{align*} \frac{1}{1-p}\frac{x^{\frac{1}{p}-1}-y^{\frac{1}{p}-1}}{x^{\frac{1}{p}}-y^{\frac{1}{p}}} = \frac{1}{1-p}\frac{x^{\frac{1-p}{p}}-y^{\frac{1-p}{p}}}{x^{\frac{1}{p}}-y^{\frac{1}{p}}} &= \frac{1}{a}\frac{x^{\frac{a}{p}}-y^{\frac{a}{p}}}{x^{\frac{1}{p}}-y^{\frac{1}{p}}}\\ &\leq \left(\sqrt{x^{\frac{1}{p}}y^{\frac{1}{p}}}\right)^{a-1} \\ & = \left(\sqrt{x^{\frac{1}{p}}y^{\frac{1}{p}}}\right)^{-p}\\ &=\frac{1}{\sqrt{xy}} \end{align*} which proves the desired result.
Hence, even though I only originally conjectured it for $p\in(0,1)$, the claim holds for $p\in(1,2)$ as well!
• Also, the example with the logarithm is recovered in the limit $p\rightarrow 1$. – luftbahnfahrer Apr 21 '17 at 18:00
Here is what I have done so far. This is incomplete, but might help someone else.
The inequality is the same as $\sqrt{xy}(x^{\frac{1}{p}-1}-y^{\frac{1}{p}-1}) \leq (1-p)(x^{\frac{1}{p}}-y^{\frac{1}{p}})$ or, letting $\frac1{p} = q$, $\sqrt{xy}(x^{q-1}-y^{q-1}) \leq (1-1/q)(x^{q}-y^{q})$ where $q > 1$.
Since $\int t^{q-1}dt =\frac{t^q}{q}$, or $t^q =q\int t^{q-1}dt$, this becomes
$\sqrt{xy}(x^{q-1}-y^{q-1}) \leq (1-1/q)q\int_y^x \int t^{q-1}dt = (q-1)\int_y^x t^{q-1}dt$.
Letting $r = q-1$, this is
$\sqrt{xy}(x^r-y^r) \le r\int_y^x t^rdt$ or $\sqrt{xy}\frac{x^r-y^r}{x-y} \le \frac{r}{x-y}\int_y^x t^rdt$ where $r > 0$.
As a check, letting $y \to x$, this becomes $x r x^{r-1} \le r x^r$ which is true.
For another check, if $r=1$ this is $\sqrt{xy}(x-y) \le \frac12(x^2-y^2)$ or $\sqrt{xy} \le \frac12(x+y)$ which is true.
I don't know where to go from here. $t^r$ is not always concave or convex for different values of $r$, so perhaps considering $r < 1$ and $r > 1$ separately might work. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 1, "x-ck12": 0, "texerror": 0, "math_score": 0.9994897246360779, "perplexity": 203.4521743045477}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-43/segments/1570986660231.30/warc/CC-MAIN-20191015182235-20191015205735-00483.warc.gz"} |
http://mathhelpforum.com/algebra/57796-solve-positive-real-x.html | Math Help - Solve for positive real x
1. Solve for positive real x
Solve for positive real x
$\lfloor \sqrt{x}\lfloor \sqrt{x} \rfloor \rfloor +\lfloor \sqrt{x}\rfloor +1=x$
2. Originally Posted by perash
Solve for positive real x
$\lfloor \sqrt{x}\lfloor \sqrt{x} \rfloor \rfloor +\lfloor \sqrt{x}\rfloor +1=x$
It should be obvious from the definition of "floor" that $\lfloor x\rfloor+ 1> x$ for all positive x so there is no positive x satisfying that equation.
3. try x=3, and x=10000
I think they work
floor(x) +1 is less than x but that is not the case in this problem because we are taking the sq root of x.
I couldnt solve this, but it doesnt work for any perfect squares and x has to be a whole number.
4. Originally Posted by perash
Solve for positive real x
$\lfloor \sqrt{x}\lfloor \sqrt{x} \rfloor \rfloor +\lfloor \sqrt{x}\rfloor +1=x$
There may be a quick proof using fixed-point theorems, but let’s consider the following algebraic method.
Since the LHS is an integer, all solutions are (positive) integers. I claim that all solutions are of the form $x=n^2+2n,\ n\in\mathbb{Z}^+.$
Note that $\lfloor\sqrt{n^2+2n}\rfloor=n.$ This is because $n<\sqrt{n^2+2n} for all $n\in\mathbb{Z}^+.$
And $\lfloor n\sqrt{n^2+2n}\rfloor=n^2+n-1$ because $\forall n\in\mathbb{Z}^+,$
$n^2+n-1\ <\ n\sqrt{n^2+2n}\ <\ n^2+n$
i.e. $n^4+2n^3-n^2-2n+1\ <\ n^4+2n^3\ <\ n^4+2n^3+n^2$
This proves that all $x=n^2+2n,\ n\in\mathbb{Z}^+,$ are solutions. Now we must show that there are no other solutions.
Any other solution would have to be of the form $x=n^2+2n-k,$ where $1\leqslant k\leqslant2n.$ Suppose there is such a solution.
Again we have $\lfloor\sqrt{n^2+2n-k}\rfloor=n$ since $n\leqslant\sqrt{n^2+2n-k} for all $1\leqslant k\leqslant2n.$
Then, in order for the solution to hold, we would need to have $\lfloor n\sqrt{n^2+2n-k}\rfloor=n^2+n-k-1.$ This would mean that we would need
$n\sqrt{n^2+2n-k}\ <\ n^2+n-k$
i.e. $n^4+2n^3-kn^2\ <\ n^4+2n^3+n^2-2kn^2-2kn+k^2$
i.e. $k^2-(n^2+2n)k\ >\ -n^2$
i.e. $\left[k-\frac{n^2+2n}2\right]^2\ >\ \frac{\left(n^2+2n\right)^2}4-n^2=\frac{n^4+4n^3}4\quad\ldots\fbox{1}$
Now $1\leqslant k\leqslant2n$ $\Rightarrow$ $k-\frac{n^2+2n}2\leqslant\frac{2n-n^2}2$ and $\frac{n^2+2n}2-k\leqslant\frac{n^2+2n-2}2.$
If $k-\frac{n^2+2n}2\geqslant0,$ then $\left[k-\frac{n^2+2n}2\right]^2\leqslant\frac{\left(2n-n^2\right)^2}4=\frac{n^4-4n^3+4n^2}4<\frac{n^4+4n^3}4.$
If $\frac{n^2+2n}2-k\geqslant0,$ then $\left[\frac{n^2+2n}2-k\right]^2\leqslant\frac{\left(n^2+2n-2\right)^2}4=\frac{n^4+4n^3-8n+4}4<\frac{n^4+4n^3}4.$
In either case, we have a clear contradiction of $\fbox{1}.$
This completes the proof. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 32, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9535092711448669, "perplexity": 506.9266752071432}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-10/segments/1394678705235/warc/CC-MAIN-20140313024505-00005-ip-10-183-142-35.ec2.internal.warc.gz"} |
http://mathhelpforum.com/number-theory/205951-primitive-root-modulo-p-print.html | # primitive root modulo p
• Oct 23rd 2012, 01:51 PM
maximus101
primitive root modulo p
Let p,q be odd primes with $p = 2q + 1$
Show that 2 is a primitive root modulo p if and only if $q \equiv 1 (mod 4)$
• Oct 30th 2012, 09:04 PM
Salahuddin559
Re: primitive root modulo p
I guess this proves a part of it, http://mathhelpforum.com/number-theo...tive-root.html.
Salahuddin
Maths online | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 2, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9864258766174316, "perplexity": 2539.196729852644}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-34/segments/1502886105187.53/warc/CC-MAIN-20170818213959-20170818233959-00159.warc.gz"} |
http://www.ktustudents.in/2015/10/calculus-lecture-note-about-infinite.html | MA101 CALCULUS :Infinite Series
Here is some advanced study material for MA101 Calculus. This will be useful for students of KTU. Although it is not strictly the same syllabus of ktu it will give a thorough knowledge for you students
The purpose of this section is to discuss sums that contain infinitely many terms. The most familiar examples of such sums occur in the decimal representations of real numbers. For example, when we write 1 3 in the decimal form 1 3 = 0.3333 . . . , we mean 1 3 = 0.3 + 0.03 + 0.003 + 0.0003 + · · · which suggests that the decimal representation of 1 3 can be viewed as a sum of infinitely many real numbers. SUMS OF INFINITE SERIES Our first objective is to define what is meant by the “sum” of infinitely many real numbers. We begin with some terminology. 9.3.1 definition An infinite series is an expression that can be written in the form X` k=1 uk = u1 + u2 + u3 + · · · + uk + · · · The numbers u1, u2, u3, . . . are called the terms of the series. Since it is impossible to add infinitely many numbers together directly, sums of infinite series are defined and computed by an indirect limiting process. To motivate the basic idea, consider the decimal 0.3333 . . .
sum of the series in an interval of length 0.001 (or less). Find the smallest value of n such that the interval containing π 4/90 in part (a) has a length of 0.001 or less. (c) Approximate π 4/90 to three decimal places using the midpoint of an interval of width at most 0.001 that contains the sum of the series. Use a calculating utility to confirm that your answer is within 0.0005 of π 4/90. 40. We showed in Section 9.3 that the harmonic seriesP` k=1 1/k diverges. Our objective in this problem is to demonstrate that although the partial sums of this series approach +`, they increase extremely slowly. (a) Use inequality (2) to show that for n ≥ 2 ln(n + 1) < sn < 1 + ln n (b) Use the inequalities in part (a) to find upper and lower bounds on the sum of the first million terms in the series. (c) Show that the sum of the first billion terms in the series is less than 22. (d) Find a value of n so that the sum of the first n terms is greater than 100. 41. Use a graphing utility to confirm that the integral test applies to the series P` k=1 k 2 e −k , and then determine whether the series converges. C 42. (a) Show that the hypotheses of the integral test are satisfied by the series P` k=1 1/(k3 + 1). (b) Use a CAS and the integral test to confirm that the series converges. (c) Construct a table of partial sums for n = 10, 20, 30, . . . , 100, showing at least six decimal places. (d) Based on your table, make a conjecture about the sum of the series to three decimal-place accuracy. (e) Use part (b) of Exercise 36 to check your conjecture
Part One
Part Two
Part Three | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9593312740325928, "perplexity": 231.11318023585463}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-13/segments/1490218190181.34/warc/CC-MAIN-20170322212950-00323-ip-10-233-31-227.ec2.internal.warc.gz"} |
http://carterkaplan.blogspot.com/2020/04/the-naturalistic-fallacy.html | ## Wednesday, April 15, 2020
### The Naturalistic Fallacy
Wherever you look, G. E. Moore's Naturalistic Fallacy is not described very clearly. Moore's own explanation is too-involved, and his illustrations of the idea are tiresome. When she explains it in her paper "Modern Moral Philosophy," Elizabeth Anscombe, for some reason, embarks upon a long-winded description of an idea from Hume that resembles the Naturalistic Fallacy. The Ethics textbooks are similarly confusing, or boring.
It is, however, a pretty easy idea. First, recall what we already know about propositions. Next, consider how the Naturalistic Fallacy helps us to understand why (I believe) that Logic cannot prove--like a mathematical or geometric proof--a moral proposition.
Consider the following propositions:
1) Two and two is four.
2) Stealing is wrong.
In each of these two examples, is has a different kind of meaning.
In the first example, "Two and two is four", is means equivalence. It is true in an analytic sense. That is, it is logically true: 2 + 2 = 4. We don't usually call mathematical equations "propositions," but note that the example here is a grammatical sentence, and it is proved by the relationship of (2 + 2) and (4). That is, two and two is four.
In the second example, "Stealing is wrong," is means "I think" or "some people think" stealing is wrong. Although many people agree, and even though I (the consummate highbrow) and you (card-carrying members of the Highbrow Commonwealth) think (or believe) stealing is wrong, in this case, is is not the same as equivalence or conclusive proof.
Stealing is wrong because we say so.
George Edward Moore | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9301053285598755, "perplexity": 3543.950365883035}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-06/segments/1674764499899.9/warc/CC-MAIN-20230201013650-20230201043650-00324.warc.gz"} |
https://mathoverflow.net/questions/346393/finding-littlewood-richardson-coefficients-without-using-identities | # Finding Littlewood-Richardson coefficients without using identities
The Littlewood-Richardson coefficients $$C^{R}_{QP}$$ for some partitions $$R, Q, P$$ can usually be dealt with using identities like for example $$C^{R}_{QP} = 0 \quad \text{ if } \quad|R| \neq |Q| + |P|$$ or $$C^{R}_{Q\bullet} = \delta^{R}_{Q}.$$ However, how does one actually calculate something like $$C^{(2,1)}_{(1), (1,1)}?$$
Since $$|R| = |Q| + |P|$$ and $$P \neq \bullet$$ I cannot use the above identities and I struggle to find any method of actually calculating the coefficients.
• Your coefficient has both $P$ and $Q$ with either one row or one column. If either holds then the Littlewood–Richardson Rule can be replaced with the easier Pieri's Rule. (Sometimes called Young's rule for the single part case, i.e. $C^R_{P (q)}$.) I often see the LR rule applied in the literature when this easier result suffices. Nov 19 '19 at 18:00
## 1 Answer
There are several approaches.
Use linear algebra. Compute Schur polynomials (using the Jacobi-Trudi identity, say), and then use the fact that the coefficient of $$s_\nu$$ in the product $$s_\lambda s_\mu$$ is a Littlewood-Richardson coefficient.
Use a combinatorial interpretation. One can count so called Littlewood-Richardson tableaux, or lattice points in so-called Berenstein-Zelevinsky polytopes. The latter can be done rather efficient using some lattice-point counting program, such as lattE.
However, note that it has been proved that computing Littewood-Richardson coefficients is #$$P$$-complete, meaning that there is no super-efficient algorithm for computing these (that is, no nice closed formula).
If you want to compute several coefficients, a recursive approach might be suitable. One can show that the multiplicative constants for so called shifted Schur functions are generalizations of Littlewood-Richardson coefficients, and these do satisfy a very nice recursive formula. See the paper and proposition 3.4 in A Littlewood-Richardson Rule for Factorial Schur Functions by Alexander I. Molev and Bruce E. Sagan.
See also this answer. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 10, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9929695725440979, "perplexity": 443.2729880888812}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-49/segments/1637964363312.79/warc/CC-MAIN-20211206194128-20211206224128-00638.warc.gz"} |
https://www.arxiv-vanity.com/papers/1707.03021/ | # The observational evidence for horizons: from echoes to precision gravitational-wave physics
Vitor Cardoso Paolo Pani
###### Abstract
The existence of black holes and of spacetime singularities is a fundamental issue in science. Despite this, observations supporting their existence are scarce, and their interpretation unclear. We overview how strong a case for black holes has been made in the last few decades, and how well observations adjust to this paradigm. Unsurprisingly, we conclude that observational evidence for black holes is impossible to come by. However, just like Popper’s black swan, alternatives can be ruled out or confirmed to exist with a single observation. These observations are within reach. In the next few years and decades, we will enter the era of precision gravitational-wave physics with more sensitive detectors. Just as accelerators require larger and larger energies to probe smaller and smaller scales, more sensitive gravitational-wave detectors will be probing regions closer and closer to the horizon, potentially reaching Planck scales and beyond. What may be there, lurking?
“The crushing of matter to infinite density by infinite tidal gravitation forces is a phenomenon with which one cannot live comfortably. From a purely philosophical standpoint it is difficult to believe that physical singularities are a fundamental and unavoidable feature of our universe […] one is inclined to discard or modify that theory rather than accept the suggestion that the singularity actually occurs in nature.” Kip Thorne, Relativistic Stellar Structure and Dynamics (1966)
“No testimony is sufficient to establish a miracle, unless the testimony be of such a kind, that its falsehood would be more miraculous than the fact which it endeavors to establish.” David Hume, An Enquiry concerning Human Understanding (1748)
## 1 Introduction
The discovery of the electron and the known neutrality of matter led in 1904 to J. J. Thomson’s “plum-pudding” atomic model. Data from new scattering experiments was soon found to be in tension with this model, which was eventually superseeded by Rutherford’s, featuring an atomic nucleus. The point-like character of elementary particles opened up new questions. How to explain the apparent stability of the atom? How to handle the singular behavior of the electric field close to the source? What is the structure of elementary particles? Some of these questions were elucidated with quantum mechanics and quantum field theory. Invariably, the path to the answer led to the understanding of hitherto unknown phenomena.
The history of elementary particles is a timeline of the understanding of the electromagnetic (EM) interaction, and is pegged to its characteristic behavior (which necessarily implies that other structure has to exist on small scales within any sound theory).
Arguably, the elementary particle of the gravitational interaction are black holes (BHs). Within General Relativity (GR), BHs are indivisible and are in fact the simplest macroscopic objects that one can conceive. The uniqueness theorems – establishing that the two-parameter Kerr family of BHs describes any vacuum, stationary and asymptotically flat, regular BH solution – have turned BHs into somewhat of a miracle elementary particle [1].
Can BHs, as envisioned in vacuum GR, hold the same surprises that the electron and the hydrogen atom did when they started to be experimentally probed? Are there any parallels that can be useful guides? The BH interior is causally disconnected from the exterior by an event horizon. Unlike the classical description of atoms, the GR description of the BH exterior is self-consistent and free of pathologies. The “inverse-square law problem” – the GR counterpart of which is the appearance of pathological curvature singularities – is swept to inside the horizon and therefore harmless for the external world. There are good indications that classical BHs are perturbatively stable against small fluctuations, and attempts to produce naked singularities, starting from BH spacetimes, have failed. BHs are not only curious mathematical solutions to Einstein’s equations, but also their formation process is well understood. In fact, there is nothing spectacular with the presence or formation of an event horizon. The equivalence principle dictates that an infalling observer crossing this region (which, by definition, is a global concept) feels nothing extraordinary: in the case of macroscopic BHs all of the local physics at the horizon is rather unremarkable.
Why, then, should one question the existence of BHs? 111It is ironic that one asks this question more than a century after the first BH solution was derived. Even though Schwarzschild and Droste wrote down the first nontrivial regular, asymptotic flat, vacuum solution to the field equations already in 1916 [2, 3], several decades would elapse until such solutions became accepted. The dissension between Eddington and Chandrasekhar over gravitational collapse to BHs is famous – Eddington firmly believed that nature would find its way to prevent full collapse – and it took decades for the community to overcome individual prejudices. Progress in our understanding of gravitational collapse, along with breakthroughs in the understanding of singularities and horizons, contributed to change the status of BHs. Together with observations of phenomena so powerful that could only be explained via massive compact objects, the theoretical understanding of BHs turned them into undisputed kings of the cosmos. The irony lies in the fact that they quickly became the only acceptable solution. So much so, that currently an informal definition of a BH might well be “any dark, compact object with mass above three solar masses.” There are a number of important reasons to do so.
• Even if the BH exterior is pathology-free, the interior is not. The Kerr family of BHs harbors, singularities and closed timelike curves in its interior, and more generically it features a Cauchy horizon signaling the breakdown of predictability of the theory. In fact, the geometry describing the interior of an astrophysical spinning BH is currently unknown. A possible resolution of this problem may require accounting for quantum effects. It is conceivable that these quantum effects are of no consequence whatsoever on physics outside the horizon. Nevertheless, it is conceivable as well that the resolution of such inconsistency leads to new physics that resolves singularities and does away with horizons, at least in the way we understand them currently. Such possibility is not too dissimilar from what happened with the atomic model after the advent of quantum electrodynamics.
• In a related vein, quantum effects around BHs are far from being under control, with the evaporation process itself leading to information loss. The resolution of such problems could include changing the endstate of collapse [4, 5, 6, 7] (perhaps as-yet-unknown physics can prevent the formation of horizons), or altering drastically the near-horizon region [8]. The fact of the matter is that there is no tested nor fully satisfactory theory of quantum gravity, in much the same way that one did not have a quantum theory of point particles at the beginning of the 20th century.
• It is tacitly assumed that quantum gravity effects become important only at the Planck scale. At such lengthscales , the Schwarzschild radius is of the order of the Compton wavelength of the BH. In the orders of magnitude standing between the Planck scale and those accessible by current experiments many surprises can hide (to give but one example, extra dimensions would change the numerical value of the Planck scale).
• Horizons are not only a rather generic prediction of GR, but their existence is in fact necessary for the consistency of the theory at the classical level. This is the root of Penrose’s (weak) Cosmic Censorship Conjecture, which remains one of the most urgent open problems in fundamental physics. In this sense, the statement that there is a horizon in any spacetime harboring a singularity in its interior is such a remarkable claim, that (in an informal description of Hume’s statement above) it requires similar remarkable evidence. This is the question we will entertain here, is there any observational evidence for the existence of BHs?
• It is in the nature of science that paradigms have to be constantly questioned and subjected to experimental and observational scrutiny. Most specially because if the answer turns out to be that BHs do not exist, the consequences are so extreme and profound, that it is worth all the possible burden of actually testing it. Within the coming years we will finally be in the position of performing unprecedented tests on the nature of compact dark objects, the potential pay-off of which is enormous. As we will argue, the question is not just whether the strong-field gravity region near compact objects is consistent with the Kerr geometry, but rather to quantify the limits of observations in testing the event horizon.
• If we don’t entertain alternatives, we won’t find them. We need to know how to model alternatives to understand how consistent the data is with accepted paradigms.
• Finally, there are sociological issues, sometimes invoked to claim that one should invest taxpayers money in sure-fire science. The reality is that taxpayers, all of us, want to know. More frequently than not, the thrill lies in the intellectual possibilities rather than on merely checking known facts. This is, after all, why pure science is relevant. Thus, it is a waste of resources not to use (built and already paid for) detectors to investigate issues that lie at the heart of fundamental questions.
Known physics all but exclude BH alternatives. Nonetheless, the Standard Model of fundamental interactions is not sufficient to describe the cosmos – at least on the largest scales – and also leaves all of the fundamental questions regarding BHs open. It may be wise to keep an open mind. As we will argue, from the point of view of observers merely testing the spacetime structure without a priori theoretical bias, the evidence for horizons is nonexistent. In fact, the question to be asked is not if there is an event horizon in the spacetime, but how close to it do experiments or observations go.
## 2 Setting the stage: escape cone, photospheres, quasinormal modes, and tidal effects
“Alas, I abhor informality.”
That Mitchell and Webb Look, Episode 2
Our framework will be that of GR. Let us start by focusing on spherical symmetry for simplicity. Birkhoff’s theorem guarantees that any vacuum spacetime is described by the Schwarzschild geometry, which in standard coordinates reads
ds2=−fdt2+f−1dr2+r2dΩ2, (1)
where is the total mass of the spacetime (we use geometrical units, except if otherwise stated) and
f=1−2Mr. (2)
The existence of a null hypersurface at is the defining feature of a BH. This hypersurface is called the event horizon of a Schwarzschild BH. In the following, we will compare the properties of the latter with those of an ultracompact object with a effective surface at
r0=2M(1+ϵ), (3)
having often in mind the case where (for instance, for Planckian corrections of an astrophysical BH, ). Although the above definition is coordinate-dependent, the proper distance between the surface and scales like . Most of the results discussed below show a dependence on , making the distinction irrelevant. There are objects for which the effective surface is ill-defined, because the matter fields are smooth everywhere. In such cases, can conservatively be taken to be the geometric center of the object.
### 2.1 Geodesics
The geodesic motion of timelike or null particles in the geometry (1) can be described with the help of two conserved quantities, the specific energy and angular momentum , where a dot stands for a derivative with respect to proper time [9, 10]. The radial motion can be computed via a normalization condition,
˙r2=E2−f(L2r2+δ1)≡E2−Vgeo, (4)
where for timelike or null geodesics, respectively. The null limit can be approached letting and re-scaling all quantities appropriately. At large distances, , the motion of matter resembles that in Newton’s theory. For instance, circular orbits, defined by , exist and are stable at large distance: small fluctuations in the motion restore the body to its original position. However, circular trajectories are stable only when , and unstable for smaller radii. The surface defines the innermost stable circular orbit (ISCO), and has an important role in controlling the inner part of the accretion flow onto compact objects.
### 2.2 Escape trajectories
There are other GR effects on the motion of particles around compact objects, beyond the appearance of ISCOs in spacetimes. One of them concerns the behavior of light rays and how they are bent by the spacetime curvature. There exists a critical value of the angular momentum for a light ray to be able to escape to infinity. By requiring that a light ray emitted at a given point will not find turning points in its motion, Eq. (4) yields [9]. Suppose that the light ray is emitted by a locally static observer at . In the local rest frame, the velocity components of the photon are [11]
vlocalφ = MKr0√f0,vlocalr=√1−K2M2f0r20, (5)
where . With this, one can easily compute the escape angle, . In other words, the solid angle for escape is
ΔΩesc=2π⎛⎝1− ⎷1−27M2(r0−2M)r30⎞⎠∼27π(r0−2M8M), (6)
where the last step is valid for . For angles larger than these, the light ray falls back and either hits the surface of the object, if there is one, or will be absorbed by the horizon. The escape angle is depicted in Fig. 1 for different emission points . The rays that are not able to escape reach a maximum coordinate distance,
rmax∼2M(1+4f0M2r20sin2ψ). (7)
This result is accurate away from , whereas for the photon approaches the photosphere () discussed in the next section. The coordinate time that it takes for photons that travel initially outward, but eventually turn back and hit the surface of the object, is shown in Fig. 1 as a function of the locally measured angle , and is of order for most of the angles , for . We find a closed form expression away from , which describes well the full range (see Fig. 1),
troundtrip∼8Mlog(cot(ψ/2)). (8)
When averaging over , the coordinate roundtrip time is then , for any , where “Cat” is Catalan’s constant.
### 2.3 Photospheres, ECOs and ClePhOs
Another truly relativistic feature is the existence of circular null geodesics, i.e., of circular motion for high-frequency electromagnetic or gravitational waves (GWs). In the Schwarzschild geometry, this can happen only at . This location defines a surface called the photosphere, or, in the context of equatorial slices, a light ring. The photosphere has a number of interesting properties, and is useful to understand certain features of spacetimes. For example, it controls how BHs look like when illuminated by accretion disks or stars, thus defining their so-called “shadow”. Imaging these shadows for the supermassive dark source SgrA is the main goal of the Event Horizon Telescope [12, 13]. The photosphere also has a bearing on the spacetime response to any type of high-frequency waves, and therefore describes how high-frequency gravitational waves linger close to the horizon.
At the photosphere, . Thus, circular null geodesics are unstable: a displacement of null particles grows exponentially [14, 10]
δ(t)∼δ0eλt,λ=√−f2V′′r2E2=13√3M. (9)
A geodesic description anticipates that light or GWs may persist at or close to the photosphere on timescales . Because the geodesic calculation is local, these conclusions hold irrespectively of the spacetime being vacuum all the way to the horizon or not. An ultracompact object with surface at , with , would feature exactly the same geodesics and properties close to its photosphere, provided that on timescales the null geodesics did not have time to bounce off the surface. We are requiring three -fold times for the instability to dissipate more than of the energy of the initial pulse. This amounts to requiring that
ϵ≲ϵcrit∼0.0165. (10)
Thus, the horizon plays no special role in the response of high frequency waves, nor could it: it takes an infinite (coordinate) time for a light ray to reach the horizon. The above threshold on is a natural sifter between two classes of compact, dark objects. For objects characterized by , light or GWs can make the roundtrip from the photosphere to the object’s surface and back, before dissipation of the photosphere modes occurs. For objects satisfying (10), the waves trapped at the photosphere relax away by the time that the waves from the surface hit it back.
We can thus use the properties of the ISCO and photosphere to distinguish between different classes of models: an object is defined as compact if it features an ISCO, or in other words if its surface satisfies . Accretion disks around compact objects of the same mass should have similar characteristics. It is customary to define Exotic Compact Objects (ECOs) as those horizonless compact objects more massive than a neutron star. Among the compact objects, some feature photospheres. These could be called ultracompact objects (UCOs). Finally, those objects which satisfy condition (10) have a “clean” photosphere, and will be designated by ClePhOs. The early-time dynamics of ClePhOs is expected to be the same as that of BHs. At late times, ClePhOs should display unique signatures of their surface. The zoo of compact objects is summarized in Fig. 2 222This diagram can be misleading in special cases. Some objects with a surface at finite can nevertheless behave as ClePhOs. For example, constant density stars near the Buchdahl limit behave as ClePhOs for axial modes: these do not couple to the fluid and travel unimpeded to the center of the star. Their travel time can be considerably long, hence these stars act, for all purposes, as ClePhOs for these fluctuations. A specific example is discussed in footnote 3. Nevertheless, it should be clear that these are contrived examples which are not meant to describe quantum corrections..
### 2.4 Quasinormal modes
Low-frequency gravitational or EM waves are not well described by null particles. In the linearized regime, massless waves are all described by a master partial differential equation of the form [15]
f2∂2Ψ(t,r)∂r2+ff′∂Ψ(t,r)∂r−∂2Ψ(t,r)∂t2−V(r)Ψ(t,r)=S(t,r). (11)
The source term contains information about the cause of the disturbance . The information about the angular dependence of the wave is encoded in the way the separation was achieved, and involves an expansion in scalar, vector or tensor harmonics for different spins of the perturbation. These angular functions are labeled by an integer , and the effective potential is
V=f(l(l+1)r2+(1−s2)2Mr3), (12)
with for scalar, vector or axial tensor modes. The equation does not describe completely gravitational perturbations, the potential for polar gravitational perturbations is more complicated [16, 17, 15].
It is hard to extract general information from the PDE (11), in particular because its solutions depend on the source term and initial conditions. We can gain some insight by studying the source-free equation in Fourier space. By defining the Fourier transform through , one gets the following ODE
d2ψdz2+(ω2−V)ψ=0, (13)
where is implicitly written in terms of the new “tortoise” coordinate
z=r+2Mlog(r2M−1), (14)
such that diverges logarithmically near the horizon. In terms of , Eq. (13) is equivalent to the time-independent Schrödinger equation in one dimension and it reduces to the wave equation governing a string when . To understand a string of length with fixed ends, one imposes Dirichlet boundary conditions and gets an eigenvalue problem for . The boundary conditions can only be satisfied for a discrete set of normal frequencies, (). The corresponding wavefunctions are called normal modes and form a basis onto which one can expand any configuration of the system. The frequency is purely real because the associated problem is conservative.
If one is dealing with a BH spacetime, the appropriate conditions (required by causality) correspond to having waves traveling outward to spatial infinity ( as ) and inwards to the horizon ( as ) [see Fig. 3]. Due to backscattering off the effective potential (12), the eigenvalues are not known in closed form, but they can be computed numerically [16, 17, 15]. The fundamental mode (the lowest dynamical multipole in GR) of gravitational perturbations reads [18]
Mω≡M(ωR+iωI)≈0.373672−i0.0889623. (15)
Remarkably, the entire spectrum is the same for both the axial or the polar gravitational sector [16]. The frequencies are complex and are therefore called quasinormal frequencies. Their imaginary component describes the decay in time of fluctuations on a timescale , and hints at the stability of the geometry. Unlike the case of a string with fixed end, we are now dealing with an open system: waves can travel to infinity or down the horizon and therefore it is physically sensible that any fluctuation damps down. The corresponding modes are the quasinormal modes (QNMs), which in general do not form a complete set.
Boundary conditions play a crucial role in the structure of the QNM spectrum. If a reflective surface is placed at , where (say) Dirichlet boundary conditions have to be imposed, the spectrum changes considerably. For the fundamental mode of the eigenspectrum reads
Mωpolar ≈ 0.13377−i2.8385×10−7 (16) Mωaxial ≈ 0.13109−i2.3758×10−7. (17)
Not only are the two gravitational sectors no longer isospectral but, more importantly, the perturbations have smaller frequency and are much longer lived, since a decay channel (the horizon) has disappeared.
More generically, for Dirichlet or Neumann boundary conditions at , the QNMs in the limit read [19]
MωR ≃ M2|z0|(pπ−δ)∼|logϵ|−1, (18) MωI ≃ −βlsM|z0|(2MωR)2l+2∼−|logϵ|−(2l+3), (19)
where , is an odd (even) integer for Dirichlet (Neumann) boundary conditions, is the phase of the wave reflected at , and [20, 21]. The above scaling can be understood in terms of modes trapped between the peak of the potential (12) at and the “hard surface” at [22, 23, 24, 25] [see Fig. 3]. Low-frequency waves are almost trapped by the potential, so their frequency scales as the size of the cavity (in tortoise coordinates), , just like the normal modes of a string. The (small) imaginary part is given by waves which tunnel through the potential and reach infinity. The tunneling probability can be computed analytically in the small-frequency regime and scales as [20]. After a time , a wave trapped inside a box of size is reflected times, and its amplitude reduces to . Since, in this limit, we immediately obtain
ωR∼1/z0,ωI∼|A|2/z0∼ω2l+3R. (20)
This scaling agrees with exact numerical results and is valid for any and any type of perturbation.
Clearly, a perfectly reflecting surface is an idealization. In certain models, only low-frequency waves are reflected, whereas higher-frequency waves probe the internal structure of the specific object [26, 27]. In general, the location of the effective surface and its properties (e.g., its reflectivity) can depend on the energy scale of the process under consideration.
### 2.5 Quasinormal modes, photospheres, and echoes
The effective potential for geodesic motion reduces to that of wave propagation in the high-frequency, high-angular momentum (i.e., eikonal) regime. Thus, some properties of geodesic motion have a wave counterpart [10]. The instability of light rays along the null circular geodesic translates into some properties of waves around objects compact enough to feature a photosphere. A wave description needs to satisfy “quantization conditions.” Since GWs are quadrupolar in nature, the lowest mode of vibration should satisfy
MωgeoR=2˙φ˙t=23√3∼0.3849. (21)
In addition the mode is damped, as we showed, on timescales . Overall then, the geodesic analysis predicts
Mωgeo∼0.3849−i0.19245. (22)
This crude estimate, valid in principle only for high-frequency waves, matches rather well even the fundamental mode of a Schwarzschild BH, Eq. (15).
Nevertheless, QNMs frequencies can be defined for any dissipative system, not only for compact objects or BHs. Thus, the association with photospheres has limits. Such an analogy is nonetheless enlightening in the context of objects so compact that they have photospheres and resemble Schwarzschild deep into the geometry, in a way that condition (10) is satisfied [22, 23, 28].
For a BH, it becomes clear that the excitation of the spacetime modes happens at the photosphere [14]. The vibrations excited there travel outwards to possible observers or down the event horizon. The structure of GW signals at late times is therefore expected to be relatively simple. This is shown in Fig. 4, which refers to the scattering of a Gaussian pulse of axial quadrupolar modes off a BH. The pulse crosses the photosphere, and excites its modes. The ringdown signal, a fraction of which travels to outside observers, is to a very good level described by the lowest QNMs. The other fraction of the signal generated at the photosphere travels downwards and into the horizon. It dies off and has no effect on observables at large distances.
Contrast the previous description with the dynamical response of ultracompact objects for which condition (10) is satisfied (i.e., a ClePhO) [cf. Fig. 4]. The initial description of the photosphere modes still holds, by causality. Thus, up to timescales of the order (the roundtrip time of radiation between the photosphere and ) the signal is identical to that of BHs [22, 23]. At later times, however, the pulse traveling inwards is bound to see the object and be reflected either at its surface or at its center. In fact, this pulse is semi-trapped between the object and the light ring. Upon each interaction with the light ring, a fraction exits to outside observers, giving rise to a series of echoes of ever-decreasing amplitude. From Eqs. (18)–(19), repeated reflections occur in a characteristic echo delay time [22, 23]
τecho∼4M|logϵ|. (23)
However, the main burst is typically generated at the photosphere and has therefore a frequency content of the same order as the BH QNMs (15). The initial signal is of high frequency and a substantial component is able to cross the potential barrier. Thus, asymptotic observers see a series of echoes whose amplitude is getting smaller and whose frequency content is also going down 333 An interesting, nontrivial example was worked out years ago in a different context [29]. The example concerns axial GWs emitted during the scatter of point particles around ultracompact, constant-density stars. The GW signal shows a series of visible, well-space echoes after the main burst of radiation, which are associated to the quasi-trapped -modes of ultracompact stars [30]. It turns out that all the results of that work fit very well our description: axial modes do not couple to the fluid and travel free to the geometrical center of the star, which is therefore the effective surface in this particular case. The time delay of the echoes in Fig. 1 of Ref. [29] is very well described by the GW’s roundtrip time to the center, , where is the radius of the star and is the Buchdahl’s limit [31]. . At very late times, the QNM analysis is valid, and the signal is described by Eqs. (18)–(19).
A formal description of the the frequency content of the sequence of echoes, and how they can be obtained with the help of the BH response, can be found in Ref. [25]. A method to reconstruct the potential of the echo source, after the modes have been extracted was proposed in Ref. [32].
### 2.6 The role of the spin
When spherical symmetry is broken by angular momentum , Birkhoff’s theorem does not apply, so the exterior metric of a spinning object is not unique and technically much more difficult to compute.444Nonetheless, there are indications that all multipole moments of the external spacetime approach those of a Kerr BH as [33, 34]. In this limit, it is natural to expect that the exterior geometry is extremely close to Kerr, unless some discontinuity occurs in the BH limit. It would be extremely important to confirm or confute this conjecture. Furthermore, BH uniqueness theorems do not apply in the presence of matter. In particular, a stationary object is not necessarily axisymmetric, unlike the Kerr solution which uniquely describes a BH in stationary configurations. Even restricting to axisymmetric configurations, the linear response of a spinning object depends also on the azimuthal number (such that ), which characterizes the angular dependence of the perturbations and yields a Zeeman-like splitting of the QNM frequencies. The relation between null geodesics and BH QNMs in the eikonal limit is more involved but conceptually similar to the static case [35].
Assuming that the exterior spacetime approaches the Kerr geometry when , the linear response of a ClePhO can be studied with a toy model [36, 37, 38, 39] in which the Kerr horizon is replaced by a reflective surface at , where are the locations of the horizons and is the dimensionless spin. In this case, Eq. (18) is modified by the replacement on the left-hand side [19] (where is the angular velocity of the object when ), whereas Eq. (19) is replaced by [20]
MωI≃−βls|z0|(2M2r+r+−r−)[ωR(r+−r−)]2l+1(ωR−mΩ), (24)
where now . The result (24) shows how angular momentum can bring about substantial qualitative changes. The spacetime is unstable for (i.e., in the superradiant regime [21]), on a timescale . This phenomenon is called ergoregion instability [40, 21, 41]. In the limit and for sufficiently large spin, and . Notice that such description is only valid at small frequencies, and therefore becomes increasingly less accurate at large spins and away from the instability threshold (where ). For large spin the result is more complex and can be found in Ref. [42]. Furthermore, in the superradiant regime the “damping” factor, , so that, at very late times (when the pulse frequency content is indeed described by these formulas), the amplitude of the QNMs increases due to the instability. Such increase is anyway small, for example when , and .
In the spinning case, the echo delay time (23) reads
τecho∼2M[1+(1−χ2)−1/2]logϵ, (25)
which corresponds to the period of the corotating mode, .
Note that, as we explained earlier, the signal can only be considered as a series of well-defined pulses at early stages. In this stage, the pulse still contains a substantial amount of high-frequency components which are not superradiantly amplified. Thus, amplification will occur at late times; the early-time evolution of the pulse generated at the photosphere is more complex.
### 2.7 BHs in binaries: tidal heating and tidal deformability
The properties of an event horizon have also important consequences for the dynamics of binary systems containing a BH. As previously discussed, a spinning BH absorbs radiation of frequency , but amplifies radiation of smaller frequency. In this respect, BHs are dissipative systems which behave just like a Newtonian viscous fluid [43, 44]. Dissipation gives rise to various interesting effects in a binary system – such as tidal heating [45, 46], tidal acceleration, and tidal locking – the Earth-Moon system being no exception due to the friction of the oceans with the crust.
For low-frequency circular binaries, the energy flux associated to tidal heating at the horizon corresponds to the rate of change of the BH mass [47, 48],
˙M=˙EH∝Ω5KM2(ΩK−Ω), (26)
where is the orbital angular velocity and the (positive) prefactor depends on the masses and spins of the two bodies. Thus, tidal heating is stronger for highly spinning bodies relative to the nonspinning by a factor .
The energy flux (26) leads to a potentially observable phase shift of GWs emitted during the inspiral. Thus, it might be argued that an ECO binary can be distinguished from a BH binary, because for the former [49]. However, the trapping of radiation in ClePhOs can efficiently mimic the effect of a horizon [49]. In order for absorption to affect the orbital motion, it is necessary that the time radiation takes to reach the companion, , be much longer than the radiation-reaction time scale due to heating, , where is the binding energy of the binary (assuming equal masses). For BHs, because of time dilation, so that the condition is always satisfied. For ClePhOs, is of the order of the GW echo delay time, Eq. (23), and therefore increases logarithmically as . Thus, an effective tidal heating might occur even in the absence of an event horizon if the object is sufficient compact. The critical value of increases strongly as a function of the spin. For orbital radii larger than the ISCO, the condition requires for , and therefore even Planck corrections at the horizon scale are not sufficient to mimic tidal heating. This is not necessarily true for highly spinning objects, for example at the ISCO requires for .
Finally, the nature of the inspiralling objects is also encoded in the way they respond when acted upon by the external gravitational field of their companion – through their tidal Love numbers [50]. An intriguing result in classical GR is the fact that the tidal Love numbers of a BH are precisely zero [51, 52, 53]. On the other hand, those of ECOs are small but finite [54]. In particular, the tidal Love numbers of ClePhOs vanish logarithmically in the BH limit. Thus, any measurement of the tidal Love number translates into an estimate of the distance of the ECO surface from its Schwarzschild radius,
ϵ∼e−1/k. (27)
Owing to the above exponential dependence, a measurement of the tidal Love number of an ECO at the level of already probes Planck distances away from the gravitational radius [54]. This level of accuracy is in principle within reach future GW detectors [49].
## 3 Beyond vacuum black holes
“Mumbo Jumbo is a noun and is the name of a grotesque idol said to have been worshipped by some tribes. In its figurative sense, Mumbo Jumbo is an object of senseless veneration or a meaningless ritual.” Concise Oxford English Dictionary
To entertain the possibility that dark, massive, compact objects are not BHs, requires one to discuss some outstanding issues. One can take two different stands on this topic:
• a pragmatic approach of testing the spacetime close to compact, dark objects, irrespective of their nature, by devising model-independent observations that yield unambiguous answers.
• a less ambitious and more theoretically-driven approach, which starts by constructing objects that are very compact, yet horizonless, within some framework. It proceeds to study their formation mechanisms and stability properties; then discard solutions which either do not form or are unstable on relatively short timescales; finally, understand the observational imprints of the remaining objects, and how they differ from BHs’.
In practice, when dealing with outstanding problems where our ignorance is extreme, both approaches should be used simultaneously. Indeed, using concrete models can sometimes be a useful guide to learn about broad, model-independent signatures. As it will hopefully become clear, one could design exotic horizonless models which mimic all observational properties of a BH with arbitrary accuracy. While the statement “BHs exist in our Universe” is fundamentally unfalsifiable555Arguably, this is true for any statement about the existence of a physical entity in our Universe, but this does not prevent us to accumulate more and more evidence to support a given model or theory, nor to rule out competitors., alternatives can be ruled out or confirmed to exist with a single observation, just like Popper’s black swans.
### 3.1 Are there alternatives?
A nonexhaustive summary of possible objects which could mimic BHs is shown in Table 1. Stars made of constant density fluids are perhaps the first known example of compact configurations, the literature on the subject is too long to list. In GR, isotropic static spheres made of ordinary fluid satisfy the Buchdahl limit on their compactness, [31], and can thus never be a ClePhO. Interestingly, polytrope UCO stars always have superluminal sound speed [56].
Compact solutions can be built with fundamental, massive bosonic fields. The quest for self-gravitating bosonic configurations started in an attempt to understand if gravity could produce “solitons” through the nonlinearities of the field equations. Such configurations are broadly referred to as boson stars. Boson stars have attracted considerable interest as light scalars are predicted to occur in different scenarios, and ultralight scalars can explain the dark matter puzzle [135]. When fastly spinning, such stars can be extremely compact. There seem to be no studies on the classification of such configurations (there are solutions known to display photospheres, but it is unknown whether they can be as compact as ClePhOs).
The other objects in the list require either unknown matter or large quantum effects. For example, a specific model of a wormhole was discussed carefully in the context of “BH foils” [101]. In the nonspinning case, the geometry of this model is described by
ds2=−(f+λ)dt2+dr2f+r2dΩ2. (28)
The constant is assumed to be extremely small, where is the Planck length. In the context of GR, this solution requires matter violating the null energy condition. Other objects, such as fuzzballs, were introduced as a microstate description of BHs in string theory. In this setup the individual microstates are horizonless and the horizon arises as a coarse-grained description of the microstate geometries. Phenomena such as Hawking radiation can be recovered, in some instances, from classical instabilities [110, 136]. “Gravitational-vacuum stars” or gravastars [88, 4] are ultracompact configurations supported by a negative pressure, which might arise as an hydrodynamical description of one-loop QFT effects in curved spacetime, so they do not necessarily require exotic new physics [137]. In these models, the Buchdahl limit is evaded because the internal effective fluid is anisotropic and also violates the energy conditions [138]. Gravastars have been recently generalized to include anti-de Sitter cores, in what was termed AdS bubbles, and which may allow for holographic descriptions [97].
For most objects listed in Table 1 which are inspired by quantum-gravity corrections, the changes in the geometry occur deep down in the strong field regime. Some of these models – including also black stars [123], superspinars [114] or collapsed polymers [118, 119] – predict that large quantum backreaction should affect the horizon geometry even for macroscopic objects. In these models, our parameter is naturally of the order for masses in the range . The -hole model predicts even more compact objects, with [117]. In all these cases, both quantum-gravity or microscopic corrections at the horizon scale select ClePhOs as well-motivated alternatives to BHs.
Despite a number of supporting arguments – some of which urgent and well founded – it is important to highlight that there is no horizonless ClePhO for which we know sufficiently well the physics at the moment.
### 3.2 Formation and evolution
Although supported by sound arguments, the vast majority of the alternatives to BHs are, at best, incompletely described. Precise calculations (and often even a rigorous framework) incorporating the necessary physics are missing.
One notable exception to our ignorance are boson stars. These configurations are known to arise, generically, out of the gravitational collapse of massive scalars. Their interaction and mergers can be studied by evolving the Einstein-Klein-Gordon system, and there is evidence that accretion of less massive boson stars makes them grow and cluster around the configuration of maximum mass. In fact, boson stars have efficient gravitational cooling mechanisms that allow them to avoid collapse to BHs and remain very compact after interactions [67, 72, 74].
The remaining objects listed in Table 1 were built in a phenomenological way or they arise as solutions of Einstein equations coupled to exotic matter fields. For example, models of quantum-corrected objects do not include all the (supposedly large) local or nonlocal quantum effects that might prevent collapse from occurring. In the absence of a complete knowledge of the missing physics, it is unlikely that a ClePhO forms out of the merger of two ClePhOs. These objects are so compact that at merger they will be probably engulfed by a common apparent horizon. The end product is, most likely, a BH. On the other hand, if large quantum effects do occur, they would probably act on short timescales to prevent apparent horizon formation. In some models, Planck-scale dynamics naturally leads to abrupt changes close to the would-be horizon, without fine tuning [117]. Likewise, in the presence of (exotic) matter or if GR is classically modified at the horizon scale, Birkhoff’s theorem no longer holds, and a star-like object might be a more natural outcome than a BH. In summary, with the exception of boson stars, we do not know how or if UCOs and ClePhOs form in realistic collapse or merger scenarios.
### 3.3 On the stability problem
“There is nothing stable in the world; uproar’s your only music.” John Keats, Letter to George and Thomas Keats, Jan 13 (1818)
Appealing solutions are only realistic if they form and remain as long-term stable solutions of the theory. In other words, solutions have to be stable when slightly perturbed or they would not be observed. There are strong indications that the exterior Kerr spacetime is stable, although a rigorous proof is still missing [139]. There are good reasons to believe that some – if not all – horizonless compact solutions are linearly or nonlinearly unstable.
Some studies of linearized fluctuations of ultracompact objects are given in Table 1. We will not discuss specific models, but we would like to highlight two general results. Linearized gravitational fluctuations of any nonspinning UCO are extremely long-lived and decay no faster than logarithmically [140, 55, 111, 112]. Indeed, such perturbations can be again understood in terms of modes quasi-trapped within the potential barrier shown in Fig. 3: they require a photosphere but are absent in the BH case. For a ClePhO, these modes are very well approximated by Eqs. (18)–(19). The long damping time of these modes has led to the conjecture that any UCO is nonlinearly unstable and may evolve through a Dyson-Chandrasekhar-Fermi type of mechanism [140, 55]. The endstate is unknown, and most likely depends on the equation of the state of the particular UCO: some objects may fragment and evolve past the UCO region into less compact configurations, via mass ejection, whereas other UCOs may be forced into gravitational collapse to BHs.
The above mechanism is supposed to be active for any spherically symmetric UCO, and also on spinning solutions. However, it is nonlinear in nature. On the other hand, UCOs (and especially ClePhOs) can develop negative-energy regions once spinning. In such a case, there is a well-known linear instability, dubbed as ergoregion instability. The latter affects any horizonless geometry with an ergoregion [40, 41, 78, 141, 38] and is deeply connected to superradiance [21].
In summary, there is good evidence that UCOs are linearly or nonlinearly unstable. Unfortunately, the effect of viscosity is practically unknown [55], and so are the timescales involved in putative dissipation mechanisms that might quench this instability. On the other hand, even unstable solutions are relevant if the timescale is larger than any other dynamical scale in the problem 666After all, “we are all unstable”.. The nonlinear mechanism at work for spherically symmetric spacetimes presumably acts on very long timescales only; a model problem predicts an exponential dependence on the size of the initial perturbation [142]. At least in some portion of the parameter space the ergoregion-instability timescale is very large [40, 78, 38]. From Eq. (24), we can estimate the timescale of the instability of a spinning ClePhO,
τ≡1ωI∼−|logϵ|1+(1−χ2)−1/22βls(r+−r−r+)[ωR(r+−r−)]−(2l+1)ωR−mΩ. (29)
As previously discussed, a spinning ClePhO is (superradiantly) unstable only when . For example, for and , the above formula yields
τ∈(5,1)(M106M⊙)yrwhen ϵ∈(10−45,10−22). (30)
Generically, the ergoregion instability acts on timescales which are parametrically longer than the dynamical timescale, , of the object. Given such long timescales, it is likely that the instability can be efficiently quenched by some dissipation mechanism of nongravitational nature, although this effect would be model-dependent [38]. Furthermore, it is possible that the endstate of the instability is simply an ECO spinning at the superradiant threshold, . In such case, the instability would only rule out highly-spinning ECOs in a certain compactness range. Likewise, EM or GW observations indicating statistical prevalence of slowly-spinning compact objects, across the entire mass range, could be an indication for the absence of an horizon.
Finally, there are indications that instabilities of UCOs are merely the equivalent of Hawking radiation for these geometries, and that therefore there might be a smooth transition in the emission properties when approaching the BH limit [110, 101].
## 4 Have exotic compact objects been already ruled out by electromagnetic observations?
“And here - ah, now, this really is something a little recherché.” Sherlock Holmes, The Musgrave Ritual, Sir Arthur Conan Doyle
Before the GW revolution, questions about the observational evidence for BHs were initially posed in the context of EM observations. A concise overview of the arguments is presented in Refs. [143, 144], concluding that “it is fundamentally impossible to give an observational proof for the existence of a BH horizon.” Surprisingly however, there have been recent claims [145, 146, 147] that observations of some objects (most notably the compact radio source Sgr A) all but rule out other candidates, including those whose surface is a Planck-distance away from the horizon (i.e., for Sgr A, thus qualifying as ClePhOs according to our classification). The argument can, roughly, be described as follows:
• There are systems with a very low accretion rate and very low luminosity, for which the central object is very dark.
• The coordinate time that a particle in the accretion disk takes to reach the surface scales as . Even for Planck-sized this time is very small.
• Assumption: because the above timescale is so small, a thermodynamic and dynamic equilibrium must be established between the accretion disk and the central object, on relatively short timescales.
• Assumption: Because there is equilibrium, the central object must be returning in EM radiation most of the energy that it is taking in from the disk.
• Because we see no such radiation, the premise that there is a surface must be wrong.
The arguments above fail to take into account important physics:
#### Timescale to reach equilibrium.
Strong lensing can prevent equilibrium from being achieved around ultracompact objects in sufficiently short time. Consider matter in the accretion disk falling in, releasing scattered radiation on the surface of the object which is then observed by our detectors. Suppose, for the sake of the argument, that once hitting the surface, it is scattered isotropically. Then, as discussed in Section 2.2, only a fraction is able to escape during the first interaction with the star, cf. Eq. (6). The majority of the radiation, will fall back onto the surface after a time given by the average of Eq. (8777One might wonder if the trapped radiation bouncing back and forth the surface of the object might not interact with the accretion disk. As we showed in Section 2.2, this does not happen, as the motion of trapped photons is confined to within the photosphere.. Suppose one injects, instantaneously, an energy onto the object. Then, after a time , the energy emitted to infinity during interactions reads
ΔE∼[1−(1−ϵ)N]δM. (31)
Clearly, as since all energy will eventually escape to infinity. However, if , i.e. when
ϵ≪10−16(M106M⊙)(tHubbleT), (32)
where we have been extremely conservative in normalizing by the age of the Universe, . The luminosity of ClePhOs satisfying the above property is roughly
˙E∼10−17(ϵ10−16)(δMM). (33)
When , this quantity is minute and it is impossible to achieve equilibrium on any meaningful timescale. Indeed, any injected energy is eventually radiated in a timescale longer than whenever Eq. (32) is satisfied. Thus, gravastars, fuzzballs, and other objects inspired by quantum effects at the horizon scale () are not ruled out as exotic supermassive objects.
Cascade into other channels. Even if the object were returning all of the incoming radiation on a sufficiently short timescale, a sizeable fraction of this energy could be in channels other than EM. Let us estimate the center-of-mass energy in collisions between matter being accreted and the compact object. The infalling matter is assumed to be on radial free fall with four-velocity . For particles at the surface of the object . When the particles collide, their CM energy reads [148],
ECM=m0√2√1−gμνvμ(1)vμ(2)∼m0√2Eϵ1/4, (34)
with the mass of the colliding particles (assumed to be equal). Therefore, the Lorentz factor of the colliding particles, in the CM frame, is for , and for Planck-scale modifications of supermassive objects. Therefore, is orders of magnitude above the maximum energy achievable at the LHC. At these CM energies, it is possible that new degrees of freedom are excited (some of which might even be the unknown physics that forms these compact objects in the first place). It is well possible that if new fields exist they carry a substantial fraction of the luminosity (while going undetected in our telescopes)888Incidentally, gravitational radiation makes only negligible contribution to the total flux, even for such large CM energies. The ratio of EM to gravitational radiation generated during these processes was estimated to be [149]
with the charge-to-mass ratio of the colliding particles. Thus, even at such large CM energies, EM radiation is vastly larger than gravitational radiation, because for protons. . Even without advocating new physics beyond the scale, extrapolation of known hadronic interactions to large energies suggests that about of the collision energy goes into neutrinos, whose total energy is a sizeable fraction of that of the photons emitted in the process [150].
In summary, we believe that some observations of supermassive objects most likely are compatible with ClePhOs (in particular they do not rule out gravastars nor quantum corrections at the horizon scale); these same observations are probably not compatible with the remaining of the UCO parameter space.
No argument whatsoever can be made to rule out exotic compact objects of arbitrary compactness, simply because the transition to the BH limit should be continuous. Therefore, any argument suggesting that exotic compact objects are excluded for any has to be taken with great care.
## 5 Testing the nature of BHs with GWs
“It is well known that the Kerr solution provides the unique solution for stationary BHs in the universe. But a confirmation of the metric of the Kerr spacetime (or some aspect of it) cannot even be contemplated in the foreseeable future.” S. Chandrasekhar, The Karl Schwarzschild Lecture, Astronomische Gesellschaft, Hamburg (September 18, 1986)
Testing the nature of dark, compact objects with EM observations is plagued with several difficulties. Some of these are tied to the incoherent nature of the EM radiation in astrophysics, and the amount of modeling and uncertainties associated to such emission. Other problems are connected to the absorption by the interstellar medium. As discussed in the previous section, testing quantum or microscopic corrections at the horizon scale with EM probes is very challenging.
The historical detection of GWs by aLIGO [151] opens up the exciting possibility of testing gravity in extreme regimes with unprecedented accuracy [152, 153, 131, 154, 155, 49]. GWs are generated by coherent motion of massive sources, and are therefore subjected to less modeling uncertainties (they depend on far fewer parameters) relative to EM probes. The most luminous GWs come from very dense sources, but they also interact very feebly with matter, thus providing the cleanest picture of the cosmos, complementary to that given by telescopes and particle detectors.
Compact binaries are the preferred sources for GW detectors (and in fact were the source of the first detected events [151, 156]). The GW signal from compact binaries is naturally divided in three stages, corresponding to the different cycles in the evolution driven by GW emission [157, 158, 159]: the inspiral stage, corresponding to large separations and well approximated by post-Newtonian theory; the merger phase when the two objects coalesce and which can only be described accurately through numerical simulations; and finally, the ringdown phase when the merger end-product relaxes to a stationary, equilibrium solution of the field equations [159, 15, 160].
All three stages provide independent, unique tests of gravity and of compact GW sources. The GW signal in the two events reported so far is consistent with them being generated by BH binaries, and with the endstate being a BH. To which level, and how, are alternatives consistent with current and future observations? This question was recently addressed by several authors and may be divided into two different schemes (see Table 2).
### 5.1 Smoking guns: echoes and resonances
For binaries composed of ClePhOs, the GWs generated during inspiral and merger is expected to be very similar to those by a corresponding BH binary with the same mass and spin. The arguments were detailed in Section 2. However, clear distinctive features appear due to the absence of a horizon. These features are:
• The appearance of late-time echoes in the waveforms [22, 23, 28, 39]. After the merger, ClePhOs give rise to two different ringdown signals: the first stage is dominated by the photosphere modes, and is indistinguishable from a BH ringdown. However, after a time (see Section 2.4), the waves trapped in the “photosphere+ClePhO” cavity start leaking out as echoes of the main burst. This is a smoking-gun for new physics, potentially reaching microscopic or even Planckian corrections at the horizon scale [22, 23].
Strategies to dig GW signals containing “echoes” out of noise are not fully under control, first efforts are underway [162, 163]. The ability to detect such signals depends on how much energy is converted from the main burst into echoes (i.e., on the relative amplitude between the first echo and the prompt ringdown signal in Fig. 4). Define the ratio of energies to be . Then, the signal-to-noise ratio necessary for echoes to be detectable separately from the main burst, for a detection threshold of , is . In such case the first echo is detectable with a simple ringdown template (an exponentially damped sinusoid [164]). Space-based detectors will see prompt ringdown events with very large [165]. For , we estimate that the planned space mission LISA [166] will see at least one event per year, even for the most pessimistic population synthesis models used to estimate the rates [165]. The proposed Einstein telescope [167] or Voyager-like [168] third-generation Earth-based detectors will also be able to distinguish ClePhOs from BHs with such simple-minded searches. The event rates for LIGO are smaller, and more sophisticated searches need to be implemented. Preliminary analysis of GW data using the entire echoing sequence was reported recently [37, 169, 170]. More detailed characterization of the echo waveform (e.g., using Green’s function techniques [25] and accounting for spin effects [39]) is necessary to reduce the systematics in data analysis.
Model-dependent fluid modes are also excited. Due to redshift effects, these will presumably play a subdominant role in the GW signal.
• Certain models of ECOs arise naturally in effective theories with extra gravitational degrees of freedom [171]. In such case, the detection of extra polarizations (as achievable in the future with a global network of interferometers) might provide evidence for new physics at the horizon scale.
• Resonant mode excitation during inspiral. The echoes at late times are just vibrations of the ClePhO. These vibrations have relatively low frequency and can, in principle, also be excited during the inspiral process itself, leading to resonances in the motion as a further clear-cut signal of new physics [95, 86, 79].
• Spin-mass distribution of compact objects skewed towards low spin, across the mass range. The development of the ergoregion instability depletes angular momentum from spinning ClePhOs, independently of their mass. Although the effectiveness of such process is not fully understood, it would lead to slowly-spinning objects as a final state.
• It is also possible that, at variance with the BH case, the merger of two ClePhOs can be followed by a burst of EM radiation associated with the presence of high-density matter during the collision. Although such emission might be strongly redshifted, searches for EM counterparts of candidate BH mergers can provide another distinctive signature of new physics at the horizon scale.
### 5.2 Precision physics: QNMs, tidal deformability, heating, and multipolar structure
The GW astronomy era will also gradually open the door to precision physics, for which smoking signs may not be necessary to test new physics.
• If the product of the merger is an ECO but not a ClePhO, it will simply vibrate differently from a BH. Thus, precise measurements of the ringdown frequencies and damping times allow one to test whether or not the object is a BH [172, 173]. Such tests are in principle feasible for wide classes of objects, including boson stars [173, 86, 79], gravastars [94, 138, 96], wormholes [174, 175], or other quantum-corrected objects [124, 120].
• The merger phase can also provide information about the nature of the coalescing objects. ECOs which are not sufficiently compact will likely display a merger phase resembling that of a neutron-star merger rather than a BH merger [23, 176]. The situation for ClePhOs is unclear since no simulations of a full coalescence are available.
• The absence of a horizon affects the way in which the inspiral stage proceeds. In particular, three different features may play a role, all of which can be used to test the BH-nature of the objects.
• No tidal heating for ECOs. Horizons absorb incoming high frequency radiation, and serve as sinks or amplifiers for low-frequency radiation able to tunnel in (see Section 2.7). UCOs and ClePhOs, on the other hand, are not expected to absorb any significant amount of GWs. Thus, a “null-hypothesis” test consists on using the phase of GWs to measure absorption or amplification at the surface of the objects [49]. LISA-type GW detectors [166] will place stringent tests on this property, potentially reaching Planck scales near the horizon and beyond [49].
• Nonzero tidal Love numbers for ECOs. In a binary, the gravitational pull of one object deforms its companion, inducing a quadrupole moment proportional to the tidal field. The tidal deformability is encoded in the Love numbers, and the consequent modification of the dynamics can be directly translated into GW phase evolution at higher-order in the post-Newtonian expansion [177]. It turns out that the tidal Love numbers of a BH are zero [51, 52, 178, 179, 180, 181], allowing again for null tests. By devising these tests, existing and upcoming detectors can rule out or strongly constraint boson stars [182, 54, 49, 87] or even generic ClePhOs [54, 49].
• Different multipolar structure. Spinning objects in a binary are also expected to possess multipole moments that differ from those of the Kerr geometry. This property is true at least for UCOs and for less compact objects. The impact of the multipolar structure on the GW phase will allow one to estimate and constrain possible deviations from the Kerr geometry [183].
## 6 Summary
“The important thing is not to stop questioning. Curiosity has its own reason for existing.” Albert Einstein, From the memoirs of William Miller, an editor, quoted in Life magazine, May 2, 1955; Expanded, p. 281
Thomson’s atomic model was carefully constructed, and tested theoretically for inconsistencies999It is amusing that some alternative models – which we now know to be closer to Rutherford’s model – were studied, shown to be unstable and therefore ruled out [184].. Rutherford’s incursion into scattering of particles were not meant to disprove the model, they were aimed at testing its accuracy. According to Marsden, “…it was one of those ’hunches’ that perhaps some effect might be observed, and that in any case that neighbouring territory of this Tom Tiddler’s ground might be explored by reconnaissance. Rutherford was ever ready to meet the unexpected and exploit it, where favourable, but he also knew when to stop on such excursions” [185].
All the current observational evidence gathered around massive, compact and dark objects is compatible with the BH hypothesis. There is no equally-well-motivated alternative, satisfying known laws of physics and composed of ordinary matter, which is compatible with those same observations. Despite this, there are long-standing problems associated with horizons and singularities, which hint at some inconsistency between classical gravity and quantum mechanics at the scale of the horizon. Furthermore, since the majority of the matter in our Universe is unknown, we cannot exclude that dark compact objects made of exotic matter are lurking in the cosmos. Even, and if, these issues are resolved within classical physics, some outstanding questions remain. What evidence do we have that compact objects are BHs? How deep into the potential can observations probe, and up to where are they still compatible with the current paradigm? At one hand’s reach, we have the possibility to dig deeper and deeper into the strong-field region of compact objects, for free. As the sensitivity of EM and GW detectors increases, so will our ability to probe regions of ever increasing redshift. Perhaps the strong-field region of gravity holds the same surprises that the strong-field EM region did?
Acknowledgments. We are indebted to Carlos Barceló, Silke Britzen, Ram Brustein, Raúl Carballo-Rubio, Bob Holdom, Luis Garay, Marios Karouzos, Gaurav Khanna, Joe Keir, Kostas Kokkotas, Claus Laemmerzahl, José Lemos, Caio Macedo, Samir Mathur, Emil Mottola, Ken-ichi Nakao, Richard Price, Ana Sousa, Bert Vercnocke, Frederic Vincent, Sebastian Voelkel and Aaron Zimmerman for providing detailed feedback, useful references and for suggesting corrections to an earlier version of the manuscript. We thank also the many other colleagues who provided constructive criticism and comments, including all the participants of the Models of Gravity: Black Holes, Neutron Stars and the structure of space-time meeting in Oldenburg, of the Foundations of the theory of Gravitational Waves meeting in Stockholm, of the Gravitational Waves and Cosmology workshop at DESY, of the Workshop on Modern aspects of Gravity and Cosmology in Orsay, of the WE-Heraeus-Seminar on Do Black Holes Exist? - The Physics and Philosophy of Black Holes in Bad Honnef, of the GW161212: The Universe through gravitational waves workshop at the Simons Center, and of the Quantum Vacuum and Gravitation: Testing General Relativity in Cosmology workshop at the MITP. V.C. acknowledges financial support provided under the European Union’s H2020 ERC Consolidator Grant “Matter and strong-field gravity: New frontiers in Einstein’s theory” grant agreement no. MaGRaTh–646597. Research at Perimeter Institute is supported by the Government of Canada through Industry Canada and by the Province of Ontario through the Ministry of Economic Development Innovation. This article is based upon work from COST Action CA16104 “GWverse”, and MP1304 “NewCompstar” supported by COST (European Cooperation in Science and Technology). This work was partially supported by FCT-Portugal through the project IF/00293/2013, by the H2020-MSCA-RISE-2015 Grant No. StronGrHEP-690904. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8079454898834229, "perplexity": 948.7860921195331}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-21/segments/1652662658761.95/warc/CC-MAIN-20220527142854-20220527172854-00589.warc.gz"} |
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## Post-edited Pathwise regularisation by noise in PDEs Gubinelli, Massimiliano (Auteur de la Conférence) | CIRM (Editeur )
We discuss some examples of the "good" effects of "very bad", "irregular" functions. In particular we will look at non-linear differential (partial or ordinary) equations perturbed by noise. By defining a suitable notion of "irregular" noise we are able to show, in a quantitative way, that the more the noise is irregular the more the properties of the equation are better. Some examples includes: ODE perturbed by additive noise, linear stochastic transport equations and non-linear modulated dispersive PDEs. It is possible to show that the sample paths of Brownian motion or fractional Brownian motion and related processes have almost surely this kind of irregularity. (joint work with R. Catellier and K. Chouk) We discuss some examples of the "good" effects of "very bad", "irregular" functions. In particular we will look at non-linear differential (partial or ordinary) equations perturbed by noise. By defining a suitable notion of "irregular" noise we are able to show, in a quantitative way, that the more the noise is irregular the more the properties of the equation are better. Some examples includes: ODE perturbed by additive noise, linear ...
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## Post-edited The Onsager Theorem De Lellis, Camillo (Auteur de la Conférence) | CIRM (Editeur )
In the fifties John Nash astonished the geometers with his celebrated isometric embedding theorems. A folkloristic explanation of his first theorem is that you should be able to put any piece of paper in your pocket without crumpling or folding it, no matter how large it is.
Ten years ago László Székelyhidi and I discovered unexpected similarities with the behavior of some classical equations in fluid dynamics. Our remark sparked a series of discoveries and works which have gone in several directions. Among them the most notable is the recent proof of Phil Isett of a long-standing conjecture of Lars Onsager in the theory of turbulent flows. In a joint work with László, Tristan Buckmaster and Vlad Vicol we improve Isett's theorem to show the existence of dissipative solutions of the incompressible Euler equations below the Onsager's threshold.
In the fifties John Nash astonished the geometers with his celebrated isometric embedding theorems. A folkloristic explanation of his first theorem is that you should be able to put any piece of paper in your pocket without crumpling or folding it, no matter how large it is.
Ten years ago László Székelyhidi and I discovered unexpected similarities with the behavior of some classical equations in fluid dynamics. Our remark sparked a series of ...
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## Multi angle Ill-posedness for Leray solutions of the ipodissipative Navier-Stokes equations De Lellis, Camillo (Auteur de la Conférence) | CIRM (Editeur )
In a joint work with Maria Colombo and Luigi De Rosa we consider the Cauchy problem for the ipodissipative Navier-Stokes equations, where the classical Laplacian $-\Delta$ is substited by a fractional Laplacian $(-\Delta)^\alpha$. Although a classical Hopf approach via a Galerkin approximation shows that there is enough compactness to construct global weak solutions satisfying the energy inequality à la Leray, we show that such solutions are not unique when $\alpha$ is small enough and the initial data are not regular. Our proof is a simple adapation of the methods introduced by Laszlo Székelyhidi and myself for the Euler equations. The methods apply for $\alpha < \frac{1}{2}$, but in order to show that they produce Leray solutions some more care is needed and in particular we must take smaller exponents. In a joint work with Maria Colombo and Luigi De Rosa we consider the Cauchy problem for the ipodissipative Navier-Stokes equations, where the classical Laplacian $-\Delta$ is substited by a fractional Laplacian $(-\Delta)^\alpha$. Although a classical Hopf approach via a Galerkin approximation shows that there is enough compactness to construct global weak solutions satisfying the energy inequality à la Leray, we show that such solutions are not ...
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## Multi angle Mean field type control with congestion Laurière, Mathieu (Auteur de la Conférence) | CIRM (Editeur )
The theory of mean field type control (or control of MacKean-Vlasov) aims at describing the behaviour of a large number of agents using a common feedback control and interacting through some mean field term. The solution to this type of control problem can be seen as a collaborative optimum. We will present the system of partial differential equations (PDE) arising in this setting: a forward Fokker-Planck equation and a backward Hamilton-Jacobi-Bellman equation. They describe respectively the evolution of the distribution of the agents' states and the evolution of the value function. Since it comes from a control problem, this PDE system differs in general from the one arising in mean field games.
Recently, this kind of model has been applied to crowd dynamics. More precisely, in this talk we will be interested in modeling congestion effects: the agents move but try to avoid very crowded regions. One way to take into account such effects is to let the cost of displacement increase in the regions where the density of agents is large. The cost may depend on the density in a non-local or in a local way. We will present one class of models for each case and study the associated PDE systems. The first one has classical solutions whereas the second one has weak solutions. Numerical results based on the Newton algorithm and the Augmented Lagrangian method will be presented.
This is joint work with Yves Achdou.
The theory of mean field type control (or control of MacKean-Vlasov) aims at describing the behaviour of a large number of agents using a common feedback control and interacting through some mean field term. The solution to this type of control problem can be seen as a collaborative optimum. We will present the system of partial differential equations (PDE) arising in this setting: a forward Fokker-Planck equation and a backward Hamilto...
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## Multi angle On the analysis of a class of thermodynamically compatible viscoelastic fluids with stress diffusion Málek, Josef (Auteur de la Conférence) | CIRM (Editeur )
We first summarize the derivation of viscoelastic (rate-type) fluids with stress diffusion that generates the models that are compatible with the second law of thermodynamics and where no approximation/reduction takes place. The approach is based on the concept of natural configuration that splits the total response between the current and initial configuration into the purely elastic and dissipative part. Then we restrict ourselves to the class of fluids where elastic response is purely spherical. For such class of fluids we then provide a mathematical theory that, in particular, includes the long-time and large-data existence of weak solution for suitable initial and boundary value problems. This is a joint work with Miroslav Bulicek, Vit Prusa and Endre Suli. We first summarize the derivation of viscoelastic (rate-type) fluids with stress diffusion that generates the models that are compatible with the second law of thermodynamics and where no approximation/reduction takes place. The approach is based on the concept of natural configuration that splits the total response between the current and initial configuration into the purely elastic and dissipative part. Then we restrict ourselves to the class ...
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http://clay6.com/qa/31949/copper-is-an-unreactive-metal-and-doesn-t-react-in-normal-circumstances-wit | Browse Questions
# Copper is an unreactive metal and doesn’t react in normal circumstances with dilute acids. However it does react with nitric acid. Which of the following equations represent how Copper reacts with DILUTE nitric acid?
$\begin{array}{1 1}(A) 3Cu + 8HNO_3 \rightarrow 3Cu(NO_3)_2 + 2NO + 4H_2O \\ (B) Cu + 4HNO_3 \rightarrow Cu(NO_3)_2 + 2NO + 2H_2O \\ (C) 3Cu + 8HNO_3 \rightarrow Cu(NO_3)_4 + NO + 2H_2O \\ (D) Cu + 4HNO_3 \rightarrow 3Cu(NO_3)_2 + 2NO + 4H_2O \end{array}$
There are two equations for the reaction of copper with nitric acid. It depends on whether the nitric acid is concentrated or not.
If it is concentrated and in excess then the ratio is 1:4 copper to nitric acid. If it is dilute then the ratio is 3:8.
For example, copper reacts with DILUTE nitric acid at ambient temperatures with a 3:8 stoichiometry.
$(A) 3Cu + 8HNO_3 –> 3Cu(NO_3)_2 + 2NO + 4H_2O$ which is the answer
$(B) Cu + 4HNO_3 –> Cu(NO_3)_2 + 2NO + 2H_2O$ is the reaction with CONCENTRATED Nitric acid. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8410776257514954, "perplexity": 3709.341885734794}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-13/segments/1490218187225.79/warc/CC-MAIN-20170322212947-00110-ip-10-233-31-227.ec2.internal.warc.gz"} |
https://www.physicsforums.com/threads/complex-eigenvalue-proof.867382/ | # Complex eigenvalue proof
• Start date
• #1
144
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## Homework Statement
Suppose the matrix A with real entries has the complex eigenvalue λ=α+iβ, β does not equal 0. Let Y0 be an eigenvector for λ and write Y0=Y1 +iY2 , where Y1 =(x1, y1) and Y2 =(x2, y2) have real entries. Show that Y1 and Y2 are linearly independent.
[Hint: Suppose they are not linearly independent. Then (x2, y2)=k(x1, y1[/SUB) for some constant k. Then Y0=(1+ik)Y1. Then use the fact that Y0 is an eigenvector of A and that AY1 contains no imaginary part.
AY=λY
## The Attempt at a Solution
Honestly, not too sure where to start for this one. I know I should begin by considering the scenario where Y1 and Y2 are not linearly independent, but I do not know where I should begin with this information. Thanks for your help :)
• #2
Samy_A
Homework Helper
1,241
510
## Homework Statement
Suppose the matrix A with real entries has the complex eigenvalue λ=α+iβ, β does not equal 0. Let Y0 be an eigenvector for λ and write Y0=Y1 +iY2 , where Y1 =(x1, y1) and Y2 =(x2, y2) have real entries. Show that Y1 and Y2 are linearly independent.
[Hint: Suppose they are not linearly independent. Then (x2, y2)=k(x1, y1[/SUB) for some constant k. Then Y0=(1+ik)Y1. Then use the fact that Y0 is an eigenvector of A and that AY1 contains no imaginary part.
AY=λY
## The Attempt at a Solution
Honestly, not too sure where to start for this one. I know I should begin by considering the scenario where Y1 and Y2 are not linearly independent, but I do not know where I should begin with this information. Thanks for your help :)
You have ##A Y_0 = \lambda Y_0##, ##\lambda = \alpha +i \beta## and ##Y_0= Y_1+iY_2##.
Start by expressing ##A Y_0 = \lambda Y_0## in terms of the real numbers ##\alpha, \beta## and the real vectors ##Y_1, Y_2##.
• #3
144
1
okay so I would have A(Y1 + iY2)=(α+iβ)(Y1 + iY2)
and then I suppose I would multiply this out
• #4
Samy_A
Homework Helper
1,241
510
okay so I would have A(Y1 + iY2)=(α+iβ)(Y1 + iY2)
and then I suppose I would multiply this out
Yes, do that. Remember that ##A, Y_1, Y_2, \alpha, \beta## are all real. This will give you two equations (by setting the real and imaginary parts of the resulting equation equal to each other).
Then assume that ##Y_1, Y_2## are linearly dependent, and see what that gives. At this stage, remember that ##\beta \neq 0##.
(Or you could reverse the order, first assume that ##Y_1, Y_2## are linearly dependent, and then do the multiplication.)
Last edited:
• #5
144
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okay so the two equations I get (by setting the real and imaginary parts of the resulting equation equal to each other) is:
AiY2= iβY1 +αiY2 and AY1Y1Y2
but now what from here?
• #6
Samy_A
Homework Helper
1,241
510
okay so the two equations I get (by setting the real and imaginary parts of the resulting equation equal to each other) is:
AiY2= iβY1 +αiY2 and AY1Y1Y2
but now what from here?
So you have ##AY_2=\beta Y_1 +\alpha Y_2##, ##AY_1=\alpha Y_1 - \beta Y_2## (*).
You want to prove that ##Y_1, Y_2## are linearly independent.
Assume that they are linearly dependent: that means that ##\exists k \in \mathbb R, k\neq 0## such that ##Y_1=k Y_2##.
Plug this in into the equations (*).
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http://www.chegg.com/homework-help/questions-and-answers/unstable-high-energy-particle-created-laboratory-moves-speed-0988c-relative-stationary-ref-q1250427 | An unstable high-energy particle is created in the laboratory, and it moves at a speed of 0.988c. Relative to a stationary reference frame fixed to the laboratory, the particle travels a distance of 1.16*10^-3 m before disintegrating.
(a) What is the proper distance traveled? ____ m
(b) What is the distance measured by a hypothetical person traveling with the particle? ____ m
(c) Determine the particle's the proper lifetime. ____s
(d) Determine the particle's dilated lifetime. _____s | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9493733644485474, "perplexity": 1268.097827315552}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-22/segments/1464049278887.13/warc/CC-MAIN-20160524002118-00051-ip-10-185-217-139.ec2.internal.warc.gz"} |
https://researchoutput.ncku.edu.tw/zh/publications/terahertz-negative-refraction-in-a-high-temperature-superconducti | # Terahertz negative refraction in a high-temperature superconducting material
Meng Ru Wu, Heng Tung Hsu, Chien Jang Wu, Shoou Jinn Chang
2 引文 斯高帕斯（Scopus）
## 摘要
The phenomenon of negative refraction at terahertz (THz) frequency for a high-temperature superconducting material is theoretically investigated based on the inhomogeneous wave theory. Negative refraction, which arises from the negative refractive angle, can occur when a p-polarized wave is incident from a dielectric to a superconductor. The dependences of negative refractive angle on the angle of incidence, the wave frequency, and the temperature have been analytically studied. The conditions for obtaining a pronounced result in negative refraction are elucidated. The study of negative refraction is of potential use in designing a tunable polarizer. In addition, it is an extended study of a transmission problem that is not treatable from the Fresnel equations in electromagnetics and optics.
原文 English 7008526 230-235 6 IEEE Transactions on Terahertz Science and Technology 5 2 https://doi.org/10.1109/TTHZ.2014.2384275 Published - 2015 三月 1
• 輻射
• 電氣與電子工程 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9011787176132202, "perplexity": 2940.9044052364093}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-43/segments/1634323585522.78/warc/CC-MAIN-20211022212051-20211023002051-00144.warc.gz"} |
http://mathhelpforum.com/algebra/109460-making-n-subject-equation.html | # Math Help - Making n the subject in this equation.
1. ## Making n the subject in this equation.
Hi this is my first thread so i hope i have the right forum, ive been trying to make n the subject in the equaion
Z = X-2n/Sqrt nt^2 + nt^2
the squareroot goes over the entire bottom part of the fraction, and t stands for symbol theta (not sure how to get greek characters) and the reason why there are 2 of the nt^2 is because theta is subscripted a and b.
if anyone couls shed any light on this it would be a big help!!
thanks
2. ques is not clear,,,,,where are "a" and "b"
3. Originally Posted by jeremyparker10
ques is not clear,,,,,where are "a" and "b"
nt(subscript a)^2 and nt(subscript b)^2
the subscripts are bottom right of t | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8127710819244385, "perplexity": 2002.649708064849}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-52/segments/1419447546506.80/warc/CC-MAIN-20141224185906-00000-ip-10-231-17-201.ec2.internal.warc.gz"} |
http://mathhelpforum.com/advanced-algebra/86131-solved-order-elements-multiplicative-group.html | # Math Help - [SOLVED] order of elements in multiplicative group
1. ## [SOLVED] order of elements in multiplicative group
13) Find the order of each of the following elements in the multiplicative group of units U(sub)p.
a. [2] for p=13
b. [5] for p=13
c. [3] for p=17
d. [8] for p=17
If you can tell me what units mean, and how to solve at least one of the problems, I think I should be able to do the rest. At this point, I don't know where to begin.
Thanks!
2. Originally Posted by yvonnehr
13) Find the order of each of the following elements in the multiplicative group of units U(sub)p.
a. [2] for p=13
b. [5] for p=13
c. [3] for p=17
d. [8] for p=17
If you can tell me what units mean, and how to solve at least one of the problems, I think I should be able to do the rest. At this point, I don't know where to begin.
Thanks!
$\mathbb{U}_n = \{ [k] : 1 \le k < n \text{ and } (k,n) = 1 \}$
that is, the set of all equivalence classes in $\mathbb{Z}_n$, where the representatives are less than $n$ and relatively prime to $n$ (including 1)
example, $\mathbb{U}_{12} = \{ [1], [5], [7], [11] \}$
since 5,7, and 11 are the integers between 1 and 12 that are relatively prime to 12.
$\mathbb{U}_n$ is an Abelian group with respect to $\odot$ (you know what $\odot$ means, right?).
so what would $\mathbb{U}_p$ look like for some prime $p$?
can you find the order of the elements? do you remember how order is defined?
3. Originally Posted by Jhevon
$\mathbb{U}_n = \{ [k] : 1 \le k < n \text{ and } (k,n) = 1 \}$
that is, the set of all equivalence classes in $\mathbb{Z}_n$, where the representatives are less than $n$ and relatively prime to $n$ (including 1)
example, $\mathbb{U}_{12} = \{ [1], [5], [7], [11] \}$
since 5,7, and 11 are the integers between 1 and 12 that are relatively prime to 12.
$\mathbb{U}_n$ is an Abelian group with respect to $\odot$ (you know what $\odot$ means, right?).
so what would $\mathbb{U}_p$ look like for some prime $p$?
can you find the order of the elements? do you remember how order is defined?
$\mathbb{U}_p$ would equal $\mathbb{Z}_p$ since each element would be relatively prime to p. Right?
If I am not mistaken, the order of a group is the number of elements in the group. I don't know how to find the order of an element. And I don't know how being an Abelian group relates to finding the solution.
Thanks. Can someone please clarify? I really want to solve this homework problem.
4. Originally Posted by yvonnehr
$\mathbb{U}_p$ would equal $\mathbb{Z}_p$ since each element would be relatively prime to p. Right?
yes. well, not including [0]. you want $\mathbb{Z}_p^*$
If I am not mistaken, the order of a group is the number of elements in the group.
yes, for finite groups. but you want the order of the given element.
I don't know how to find the order of an element.
Let $(G,*)$ be a group and $a \in G$. The order of $a$, denoted $\circ (a)$, is defined as the smallest positive integer $n$ such that $a^n = e$ ( $e$ is the identity element. $a^n = \underbrace {a*a* \cdots *a}_{\text{product of }n~a\text{'s}}$)
so going back to $\mathbb{U}_{12}$. Lets find the order of $[7]$. Since, as i said, the operation is $\odot$, lets do this operation with $[7]$ and itself until we get the identity (which is $[1]$).
$[7] \odot [7] = [7 \cdot 7] = [49] = [1]_{12}$.
thus, $[7]^2 = [1]$, and so $\circ ([7]) = 2$
what about $[5]$?
$[5] \odot [5] = [25] = [1]$, so again, $\circ ([5]) = 2$
now do your problems. the fact that you are working with $\mathbb{Z}_p^*$ helps
And I don't know how being an Abelian group relates to finding the solution.
you asked what "units" mean, so i gave you the definition. whether all parts of the definition come in handy in solving the problem is another story
Thanks. Can someone please clarify? I really want to solve this homework problem.
you should be able to get somewhere now
5. ## Thanks!
Thank you so much. After seeing your example, I realized that I was generating for an additive and not a multiplicative group. <:-)
Thank you so much!!! | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 45, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8688656091690063, "perplexity": 190.1506587247277}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-41/segments/1410657135080.9/warc/CC-MAIN-20140914011215-00285-ip-10-234-18-248.ec2.internal.warc.gz"} |
https://www.zora.uzh.ch/id/eprint/70009/ | # Octopus-inspired sensorimotor control of a ulti-arm soft robot
Li, Tao; Nakajima, Kohei; Calisti, Marcello; Laschi, Cecilia; Pfeifer, Rolf (2012). Octopus-inspired sensorimotor control of a ulti-arm soft robot. In: 2012 International Conference on Mechatronics and Automation (ICMA), Chengdu, Sichuan, China, 5 August 2012 - 8 August 2012, 948-955.
## Abstract
Soft robots have significant advantages over traditional rigid robots because of their morphological flexibility. However, the use of conventional engineering approaches to control soft robots is difficult, especially to achieve autonomous behaviors. With its completely soft body, the octopus has a rich behavioral repertoire, so it is frequently used as a model in building and controlling soft robots. However, the sensorimotor control strategies in some interesting behaviors of the octopus, such as octopus crawling, remain largely unknown. In this study, we review related biological studies on octopus crawling behavior and propose its sensorimotor control strategy. The proposed strategy is implemented with an echo state network on an octopus-inspired, multi-arm crawling robot. We also demonstrate the control strategy in the robot for autonomous direction and speed control. Finally, the implications of this study are discussed.
## Abstract
Soft robots have significant advantages over traditional rigid robots because of their morphological flexibility. However, the use of conventional engineering approaches to control soft robots is difficult, especially to achieve autonomous behaviors. With its completely soft body, the octopus has a rich behavioral repertoire, so it is frequently used as a model in building and controlling soft robots. However, the sensorimotor control strategies in some interesting behaviors of the octopus, such as octopus crawling, remain largely unknown. In this study, we review related biological studies on octopus crawling behavior and propose its sensorimotor control strategy. The proposed strategy is implemented with an echo state network on an octopus-inspired, multi-arm crawling robot. We also demonstrate the control strategy in the robot for autonomous direction and speed control. Finally, the implications of this study are discussed. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8000032305717468, "perplexity": 3030.8542882005063}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-50/segments/1606141692985.63/warc/CC-MAIN-20201202052413-20201202082413-00478.warc.gz"} |
https://www.futurestarr.com/blog/other/4x15-8-calculator | FutureStarr
4X(1+5-8) Calculator
## 5 8 Calculator
This is a calculator that uses the eight corner algorithm to display the eight digits in a calculator.
### Calculator
Unlike adding and subtracting integers such as 2 and 8, fractions require a common denominator to undergo these operations. One method for finding a common denominator involves multiplying the numerators and denominators of all of the fractions involved by the product of the denominators of each fraction. Multiplying all of the denominators ensures that the new denominator is certain to be a multiple of each individual denominator. The numerators also need to be multiplied by the appropriate factors to preserve the value of the fraction as a whole. This is arguably the simplest way to ensure that the fractions have a common denominator. However, in most cases, the solutions to these equations will not appear in simplified form (the provided calculator computes the simplification automatically). Below is an example using this method.
frobenius method calculator For. Newton's method. But P and Q cannot be arbitrary: (x−x 0)P(x) and (x−x 0)2Q(x) must be analytic at x 0. Note that values different from ‘frobenius’ (or 2) and ‘kullback-leibler’ (or 1) lead to significantly slower fits. For R s = (r ij(s)) where (r ij(s)) denotes the matrix elements for some i and jthe second formula is equivalent to r ik(st) = X j r ij(s)r jk(t): With the above notation, we can go from a linear representation to a matrix Solution for Use the method of Frobenius to find the solution of the following differential equation near the point x = 0. When the number of coins is two, there is explicit formula if GCD is not 1. norm. The Method of Frobenius M1. Method of Frobenius ODE Calculator. I’ve found it SO exhausting trying to keep up with Stewart’s algebraic operations. Largest singular value of A. 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Answered: Rolando Molina on 13 Apr 2018 If r=0, then the Frobenius method is just the ordinary power series method (expanding around an ordinary point) and I every worked example of the same problem I'm attempting to solve do use the ordinary power series method without mentioning the indicial equation or the name of Frobenius. We use cookies to improve your experience on our site and to show you relevant advertising. Method of Frobenius. the rate of profit associated with A 1 is greater than the 6. but going with r =1 I get. While it is out of the scope of the present study, it is noteworthy that the multiple time scales method can also be applied directly to the beam’s partial differential equations of motion frobenius-method-calculator. jjyjj 1: You can think of this as the operator norm of xT. A calculator for solving differential equations. m. Ordinary differential equations can be a little tricky. Learning math takes practice, lots of practice. 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Three kinds geometry of time-momentum space seems to be used by Babylonians to accurately calculate Jupiter’s orbit [101]. 2 Frobenius Series Expansion Example 4: We revisit Example 1 by using a Frobenius series to solve the equations directly. 38) at very large values of x, we make the change of variable x : 1 . ” First, you assume a To do this, first, we calculate the squares of all elements in a given matrix and then add them to get a total. The orbit of Jupiter is a graph in coordinate method that uses the characteristic polynomial to recursively calculate the inverse, though it requires calculating the coe cients of the characteristic polynomial when initiated, which is costly, and thus the method has fallen into disuse. We. STEPS. 4 and begin with Deï¬nition 7. In this case, the analog of Theorem 1 is more complicated. Ps. Andrews, Scotland This approach to solving differential equations consists of four general steps. 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Find more Mathematics widgets in Wolfram|Alpha. The Frobenius method is a method to identify an infinite series solution for a second-order ordinary differential equation. A recurrence relation – a formula determining a n using a i, i and b = <2, 7, 1> a ⋅ b = (a 1 * b 1) + (a 2 * b 2) + (a 3 * b 3) a ⋅ b = (3 * 2) + (5 * 7) + (8 * 1) a ⋅ b = 6 + 35 + 8 The first method is used to reduce the equation of order three from a three-variable Frobenius-related problem to be a system of equations with two fixed variables. â‹® . 0. L0(u) = 0, where the differential operator L0 is made from (a1) by retaining only the leading terms. For negative b … Solving systems of linear equations using Gauss Seidel method calculator - Solve simultaneous equations 2x+y+z=5,3x+5y+2z=15,2x+y+4z=8 using Gauss Seidel method, step-by-step online. Questionnaire. Rex Frobenius. ACM95b/100b Lecture Notes Caltech 2004 The Method of Frobenius Consider the equation x2y00 +xp(x)y0 +q(x)y = 0; (1) where x = 0 is a regular singular point. more. As usual, the generic form of a power series is theoretically calculate the ensemble average severalof statistics in the newly introduced statistical testsby using the Perron-Frobenius integral operator. 4 4. . For speeding up the computation there are two modes that calculate the over-time errors. From Power series(or can say Taylor Series expansion)or Polynom Power series Calculator. np. Abstract. Answer: There is an explicit formula for the Frobenius number when there are only two different coin denominations, x and y : xy − x − y(This formula was Bookmark this question. Just enter the matrix, choose what you want to calculate, push the button and let the matrix calculator do the job for you! unit vector along this line. \displaystyle \infty ∞. 1 Frobenius Method In this section, we consider a method to find a general solution to a second order ODE about a singular point, written in either of the two equivalent forms below: x2y′′+xb(x)y′+c(x)y= 0 (1. Finally, we can formulate the method of Frobenius series as follows. 3. Write y(x) = X n=0 ∞ a n xn. \) matrix A {a ij} Matrix norm. The Frobenius number of , …, is the largest integer for which the Frobenius equation has no frobenius-method-calculator. The basic idea is to approximate the solution with a power series of the form: (1) X1 m=0 a See Joseph L. The best t line is the one maximizing jAvj2 and hence minimizing the sum of the squared distances of the points to the line. A character indicating the type of norm desired. The Set-Up The Calculations and Examples The Main Theorems The Di erential Equation and Assumptions The DE Our DE is L [y ]=x 2 d 2 y L. Step 2 Let v 2 = u 2 – u 2, v 1 ‖ v 1 ‖ 2 v 1 . Calculate by the Power Series Method the value of y as general solution (term-by-term from n = O to n = 4) of the following ordinary differential equation: (10 – x2)y"' + 5xy + y = 0 = 2. Frobenius-Perron. Questions? Ask them below!Prerequisites: Regular series solutions of O Equal Roots Example - Creighton University In this section we define ordinary and singular points for a differential equation. This ODE could not be solved by power series method, and it requires the Frobenius method. Hence, we can now easily answer the above question by following the below steps: Let us only consider the method at the point x = 0 for simplicity. Given an arbitrary basis { u 1, u 2, …, u n } for an n -dimensional inner product space V, the. Gram-Schmidt algorithm. To assess the performance of Frobenius Norm Filtering method, it is compared with the Adaptive Median Filtering (AMF) method and the conventional Median filtering method. Related Symbolab blog posts. Finally, 3 The Method We start with stating the problem we want to solve again: Let A be an arbitrary 3 3 matrix. Mcdougal littell integrated mathematics 3 online book, FORMULA MATH PROGRAMING, basic algebra step by step, functions worksheet, 6th grade, biology book prentice hall-teachers version, orleans From this example we see that the method have the following steps: 1. The basic idea is to look for solutions of the form (x − x0)r X n=0 ∞ a n (x − x0)n. (8) It is easy to show that x = 0 is a regular singular point of Eq. This work is related to a conjecture due to Dubrovin … We show that the image of the Chebyshev-Frobenius homomorphism is either transparent or skew-transparent. some. in the generalized power Welcome to the Gram-Schmidt calculator, where you'll have the opportunity to learn all about the Gram-Schmidt orthogonalization. By browsing this website, you agree to our use of cookies. Let us get started. While it is out of the scope of the present study, it is noteworthy that the multiple time scales method can also be applied directly to the beam’s partial differential equations of motion Differential Transform Method To Theare changing over time, and there is a relationship between the variables . to compare the distance from pA to the set of points sP: sP = set (points) pA = point distances = np. This gives the second (linearly independent) solution to the ODE, and we have the general solution y = c 1y 1 +c 2y 2 = c 1cos √ 2x +c′ 2sin √ 2x (x >0). Let A = LU, where L is the lower triangular matrix and U is the upper triangular matrix Using the method of paired comparisons to evaluate the BCS rankings and the NCAA football championship game 2007–08. Frobenius method with imaginary powers. You write down problems, solutions and notes to go back Caveat: There are some instances when only one Frobenius solution can be constructed. 1 The BFGS Method In this Section, I will discuss the most popular quasi-Newton method,the BFGS method, together with its precursor & close relative, the DFP algorithm. Example. 1 Substitute y(x) = P A number of methods, from several areas of mathematics, have been used in the hope of finding a formula giving the Frobenius number and algorithms to calculate it. If you want 1. Table 5 shows four nonlinear solutions of the rotating beam problem using the mode shapes obtained with the Frobenius approximation method (φ F r η). Just like running, it takes practice and dedication. 1-norm, the largest column sum of the absolute values of A. This may not have significant meaning to us at face value, but Laplace transforms are extremely useful in mathematics, engineering, and science. HUD's PD&R (Office of Policy Development and Research) is responsible for maintaining current information on housing needs, market conditions, and existing programs, as well as conducting research on priority housing and community development issues. the solution of the given ordinary differential equation by the method of Frobenius is the sum of the two Following is an example of min method in non historical velocity. This small but powerful calculator was designed to be software compatible with the HP-41C. By default np linalg norm method calculates nuclear norms. See Method section for more information. 3), then the Frobenius root associated with the second matrix (call it l 2 *) is greater than the Frobenius root associated with the first (l 1 *), i. Other resources: Basic differential equations and solutions. Asymmetry measure based on Frobenius norm In realistic physical scenarios, there are some states which are invariant under a set of symmetric operations or The vector field $$E= \sum _{i=1}^n d_i t^i \partial _{t_i}$$ is known as Euler vector field, and it defines the degrees $$d_i$$ and the charge d of M. The rst two properties are straightforward to prove. norm(array_2d,"fro") Full Code We can calculate the dot product for any number of vectors, however all vectors must contain an equal number of terms. Subject Marks = 85 Marks. com provides practical strategies on online solver frobenius, syllabus for intermediate algebra and multiplying and dividing rational and other math topics. The Frobenius method 134 Chapter 5. Math notebooks have been around for hundreds of years. 287], but it is not shown how to arrive at this solution. Proof Frobenius Series Method Frobenius Series Method . The Frobenius number is the largest value for which the Frobenius equation (1) has no solution, where the are positive integers, is an integer, and the solutions are nonnegative integer. But the Hessenberg method makes it possible to obtain good The best known method for computing pairings is based on Miller’s algo- rithm. Suppose that. Although this is called a "norm" and works on matrices, it's not really a "matrix norm" in our deï¬nition. beta_loss float or {‘frobenius’, ‘kullback-leibler’, ‘itakura-saito’}, default=’frobenius’. Frobenius Method If is an ordinary point of the ordinary differential equation, expand in a Taylor series about . But the transformation matrix to compute exact resolutions via Rødseth’s method for ï¬nding the Frobenius number when n = 3, to investigate Algebraic Geometric codes via the properties of special semigroups and their corresponding conductors and to study tiling problems. basic properties. The primary reference is Lucas Pseudoprimes by Baillie and Wagstaff (1980). D. π. count method from itertools is kind of a smart counter which will automatically update itself inside the same kernel. Thus, x 0 = 0 is a singular point. Ly = 2x2y00 ¡xy0 +(1+x)y = 0 … In this tutorial, I will show how to calculate the Frobenius Norm of a given matrix in Java. colwise(). Power series method 1 2. \square! \square! . n = norm (X) returns the 2-norm or maximum singular value of matrix X , which is approximately max (svd (X)). , it is conserved or invariant under a unitary transformation (such as a rotation) : 0 is a regular singular point of the ODE ⇒ use Frobenius theory. def calculate_frobenius_norm (channel, u, s_diagonalized, vt): """ method to calculate the frobenius norm of the reconstructed matrix: Done by taking 1 to k (k=rank of the channel) components in the reconstruction process. l 1 * < l 2 * As r 1 = (1-l 1 *)/l 1 * and r 2 = (1-l 2 *)/l 2 *, then this implies that: r 1 > r 2. es. Try to sum back and ï¬nd out a closed form formula for y. 5. But You can easily calculate Frobenius norms using passing the abbreviation of it that fro. Further, xp(x) = −1/2andx2q(x) = (1 +x)/2. Then we can calculate qP efficiently using this identity so that we can reduce the number of elliptic additions to compute mP for an integer m and P ∈ En . Electrical Network Theorems - Engineering Articles Here is a simple calculator to calculate Current from Resistance and Voltage using Ohm's law. But the Hessenberg method makes it possible to obtain good Differential Transform Method To Theare changing over time, and there is a relationship between the variables . a = = " The Method of Frobenius If the conditions described in the previous section are met, then we can find at least one solution to a second order differential equation by assuming a solution of the form: ∑∑ ∞ = ∞ = = = + nn00 n r n n n r y x a x a x (4) where r and an are constants to be determined, and n = 0, 1, 2, 3,…While n is always an The Method of Frobenius I In this section we begin to study series solutions of a homogeneous linear second order differential equation with a regular singular point at x 0 = 0, so it can be written as x 2 A ( x) y ″ + x B ( x) y ′ + C ( x) y = 0, where A, B, C are polynomials and A ( 0) ≠ 0 . Then formulas (7) would still hold, now for r = r2 = ¡4: In particular, from (7) with m = 2, we would have a2(¡2)(0)+ a0 = 0: But this means a0 = 0, which contradicts the initial assumption a0 6= 0. Then p(x) and q(x) are analytic at the origin and have convergent Thus, the method of Frobenius applies in the case that the origin is a regular singular point. What is obtained through the formula above is the volume of blood by the proportion of cellular parts divided by the number of erythrocytes. In Chapter 11, the method of separation of Page 12/32 The best known method for computing pairings is based on Miller’s algo- rithm. Then there is a constant r such that the ODE has a solution of the form y =(x− a)r X∞ n=0 b n bessel_frobenius. Included are most of the standard topics in 1st and 2nd order differential equations, Laplace transforms, systems of differential eqauations, series solutions as well as a brief introduction to boundary value problems, Fourier series and partial differntial … Firstly - this function is designed to work over a list and return all of the values, e. For the next trigonometric identities we start with Pythagoras' Theorem: The Pythagorean Theorem says that, in a right triangle, the square of a plus the square of b is equal to the square of c: a 2 . The second order Euler equationis discussed in Section 7. Whenever you need assistance on syllabus for college or maybe long division, Algebra-equation. 4. p(x)y ″ + q(x)y ′ + r(x)y = 0. ). 169. Thus, it still has two problems: 1) it is not so efficient when the base a A couple weeks ago I sought out some resources to learn calculus ahead of time, and I was recommended Stewart’s Calculus. Here b(t) = 1 and c(t) = t2 − v2 are analytic at t = 0. one of its eigenvalues is positive and greater than or equal to (in absolute value) all other eigenvalues. This simple algorithm is a way to read out the orthonormal basis of the space spanned by a bunch of random vectors. The simplest systems to study would be nonlinear and parametric oscillators. To illustrate the method of Frobenius we ï¬rst consider an example. Suppose that X is a scheme of characteristic p > 0. The return value depends on the value of the given parameter. We … Cauchy-Euler Equations and Method of Frobenius June 28, 2016 Certain singular equations have a solution that is a series expansion. Get the free "General Differential Equation Solver" widget for your website, blog, Wordpress, Blogger, or iGoogle. Solving Frobenius Equations and Computing Frobenius Numbers. Total Time (in Minutes) = 180 Minutes. Somewhat related, a matrix with just one column (or one row, but not both) different from the identity matrix is also sometimes called a Frobenius matrix; see, e. pdf, determining the solution of the system of equations with two variables/math help. Elastic solution of axisymmetric thick truncated conical shells Solve your math problems using our free math solver with step-by-step solutions. Step 1. In the ï¬rst section of this paper, we study the properties of Ext1 Gr (L(µ),H0(λ)), which turns out to be semisimple and to have a good ï¬ltration for large pand pr- this problem on elliptic curves is the binary method which repeats ‘doublings’ and ‘additions’ of points. 288): We have learned to use Laplace transform method to solve ordinary differ ential equations in Section 6. Explanation. Fixed typos, two new figures, cosmetic (This Frobenius norm is implemented in Matlab by the function norm(A,'fro'). Practice your math skills and learn step by step with our math solver. Regular Sturm–Liouville problems 155 An example is the Frobenius norm. The Method of Frobenius We now approach the task of actually finding solutions of a second-order linear dif ferential equation near the regular singular point x = 0. Here is a set of notes used by Paul Dawkins to teach his Differential Equations course at Lamar University. The Euler–Cauchy equation can be solved by taking the guess z … Given an M * N matrix, the task is to find the Frobenius Norm of the matrix. Commonly, the expansion point can be taken as , resulting in the Maclaurin series (1) Plug back into the ODE and group the coefficients by power. fro_norms = np. The discovery of interactions and parallels with the Frobenius morphism has been an impetus for many new results, including new invariants attached to singularities but also D- and F-module GATE Maths Syllabus Exam Pattern 2022. Substitute into the equation and determine a n. This method more decreased the number of multiplica-tions; however, the number of Frobenius mappings inversely increased. We also show who to construct a series solution for a differential equation about an ordinary point. Use the code given below. norm() method takes arr, ord, axis, and keepdims … Euler Method: In mathematics and computational science, the Euler method is a first-order numerical procedure for solving ordinary differential equations with a given initial value. ∞. The procedure to use the power series calculator is as follows: Step 1: Enter the function, variable, point, order in the respective input field. Input variable is node's non-historical container's VELOCITY and output variable VECTOR_3D_NORM will store minimum and TIME will store the time minimum occured for each node. (3) Compute Different Types of Norms of Matrix. :return: list of frobenius norms calculated by the first k signular values of the matrix """ The Solutions of the -Hypergeometric Differential Equation. Practice Makes Perfect. Answer: Equations of the form a_2(z)f''(z)+a_1(z)f'(z)+a_0(z)f(z)=0, are very common in dynamical systems (nonlinear diff equations) theory. Is there any other way to get the indicial equation? Derivative Calculator This simple and convenient derivative calculator will help you solve any problem, just enter the value of the function and you will immediately get a solution with a detailed step-by-step description. This method requires at aver-age 3l=2 elliptic operations to compute mP for l bit integer m and a point P in an elliptic curve. 1: Frobenius’ Method. Frobenius method 7 1. The ring A is an F p-algebra, so it admits a Frobenius endomorphism. Step 3 … Calculus questions and answers. In this section, we solve the following ordinary linear second-order -hypergeometric differential equation defined by Mubeen [ 13] using Frobenius method. For example, A [2] [2] = { {1, 3}, {2, 4} } Frobenius Norm of A = sqrt (1 2 … The absolute Frobenius morphism. As shown in Figure 5, at least one root exist in the interval [ ð‘™, ] … By Perron-Frobenius theorem (G. We know from the previous section that this equation will have series solutions which both converge and solve the differential equation everywhere. frobenius method xy. In the following we solve the second-order differential equation called the hypergeometric differential equation using Frobenius method, named after Ferdinand Georg Frobenius. Perron-Frobenius theorem. (3 redLices when p() P0 and q(x) qo are constants. Step 2: add equations and solve a new equation for . This online calculator implements Newton's method (also known as the Newton–Raphson method) for finding the roots (or zeroes) of a real-valued function. yields the solution y(x) = ce1=x, which could not be captured by a Frobenius expansion about x0 = 0. The Lagrange–Charpit equations (see (2)) for the above equation can be written as dx 2pu = dy 2q = du 2p2u+2q2 = dp −p3 Free derivatives calculator (solver) that gets the detailed solution of the first derivative of a function. Our math solver supports basic math, pre-algebra, algebra, trigonometry, calculus and more. operator. • The call Frobenius (A) mod p computes the Frobenius form of A modulo p which is a prime integer. Infinity norm, the largest row sum of the absolute values of A. ) # 'distances' is a list. See Table 1. B) Use Gauss’ theorem to calculate the same integral as a volume integral the Frobenius method would not work. zs. Goldfarb [11] uses Broyden’s method [4] for iteratively inverting matrices. The method used to solve this boundary value problem converges as . of. This equation is given in [2, Problem 7, p. The method of Frobenius is a modiï¬cation to the power series method guided by the above observation. This method makes use of successive approximations to find a fixed point of a nonlinear map. While it is out of the scope of the present study, it is noteworthy that the multiple time scales method can also be applied directly to the beam’s partial differential equations of motion C 2 ( P ) is the second Frobenius companion matrix associated to the polynomial P ( X ) : In order to change respectively Γ 1 to Ω 1 and then Γ 2 to Ω 2 , some steps are required. Theorem 1 (Frobenius k-ary Method). If they can be found, then every solution of the ODE can be expressed in the form y(x) = c 1y 1(x) + c 2y 2(x); where c 1 and c MATH-342 Frobenius Method MATH 342 Notes 1. The method is called the Frobenius method, named after the mathematicianFerdinand Georg Frobenius. With dx= dz z2 (19) 3. 244]; the solution is given in [2, p. Assuming that the singular point is , we can calculate as follows: and Derivation. Home / Linear Algebra / Matrix Transform; Calculates the L1 norm, the Euclidean (L2) norm and the Maximum(L infinity) norm of a matrix. proof. 1) (1. Broyden’s Good Method. In this video we apply the method of Frobenius to solve a differential equationxy'' + y' + 2xy = 0with a power series expanded about the regular singular poi The Frobenius method is a generalization of the treatment of the simpler Euler–Cauchy equation. Matrix Calculator. norm() method is used to get one of eight different matrix norms or vector norms. For all s;t2Gˆ st= ˆ sˆ t if R st= R sR t. for the results, with d=0. The table below shows results of using the improved Euler method with step sizes and to find approximate values of the solution of the initial value problem at . Method of Frobenius To ï¬nd a series solution about a regular singular point x 0, use the following procedure. Choose an open affine subset U = Spec A of X. If p = 2, then n is approximately max (svd The Solutions of the -Hypergeometric Differential Equation. It is very basic and just involves a beginner level knowledge of how matrices work in programming. The main intention of this book is to highlight such methods, ideas, viewpoints and applications to a broader audience. The resulting series can be used to study the solution to problems for which direct calculation is di cult. It lacked the HP-41C's expandability but it offered, a two-line dot matrix screen with customizable menus, greater speed, smaller size, and a lower price as compensation. This general problem for n coins is known as classic Forbenius coin problem. If the second argument is not given, p = 2 is used. Evaluate the following Laplace transforms online matrix LU decomposition calculator, find the upper and lower triangular matrix by factorization example of how the method can be used. Beta divergence to be minimized, measuring the distance between X and the dot product WH. Matrix norm [1-5] /5: Disp-Num [1] 2019/12/09 04:44 20 years old level / High-school/ University/ Grad student Solving systems of linear equations using Gauss Seidel method calculator - Solve simultaneous equations 2x+y+z=5,3x+5y+2z=15,2x+y+4z=8 using Gauss Seidel method, step-by-step online. The emphasis mainly has been to optimize the Miller’s loop and the final exponentiation in the algorithm [ [3], [1], [14], [31], [38],[5],[22]]. For comparison, it also shows the corresponding approximate values obtained with Euler’s method in example:3. frobenius method 4xy. g. , [a1] , p. Calculate least common multiple, download master aptitude test questions and answer + GMAT, free online algebra pdf, substitution method of algebra. Knowing the GATE Mathematics Exam Pattern 2022 will help the candidates to prepare better for the exams. Bessel Function Calculator: x: n: J 0 (x): J 1 (x): Y 0 (x): Y 1 (x): I 0 (x): I 1 (x): K 0 (x): K 1 (x): J n (x): Y n (x): I n (x): K n (x) n (x) with the opposite coefficient method. Advanced Math Solutions – Ordinary Differential Equations Calculator, Exact Differential Equations. (i)Given the equation (14) with a regular singular point at x= , solve the indicial equation (18) and nd possible values for r. Note that if we required the normalization ~a( ) = 1 from the beginning, the indicial equation would have been r2 + ~b( ) 1 The Frobenius method enables us to solve such types of differential equations for example, Bessel’s equation y ″ + 1 ty + (t2 − v2 t2)y = 0, (v is parameter) of the form of “Equation (4)”. The To compare MOSES against Power Method, FD/RFD, and GROUSE we employ the following two metrics: The Frobenius norm of Yr vs Y columns seen so far normalised using their respective arrival time. Our methods include a Table 5 shows four nonlinear solutions of the rotating beam problem using the mode shapes obtained with the Frobenius approximation method (φ F r η). But you do need a spot of background … Answer (1 of 2): In traditional method of solving linear differential equation what find as solution? we get linear combination of some elementary functions like x^2, lnx, e^ax, sin(ax), cos(ax) etc as general & particular solution. com is undoubtedly the perfect site to head to! Get the free "Find Your Frobenius" widget for your website, blog, Wordpress, Blogger, or iGoogle. there is a positive eigenvector corresponding to that eigenvalue Chapter 1. Broyeden’s Method is, like the Secant Method and Brent’s Method, another attempt to use information about the derivatives without having to explicitly and expensively calculate them at each iteration. Your first 5 questions are on us! Combining with the use of Fr{é}chet derivatives to compute the gradient of the square Frobenius distance between a geodesic ending point to a given point on the manifold, we show the logarithm map and geodesic distance between two endpoints on the manifold could be computed by {\it minimizing} this square distance by a {\it trust-region} solver. Go! Free system of equations Gaussian elimination calculator - solve system of equations unsing Gaussian elimination step-by-step This website uses cookies to ensure you get the best experience. In the last few decades, it has been rediscovered as a powerful tool in a myriad of applications including Biology, Economics, Dynamical Systems, and even ranking of football teams. 6, in which the only variable, say “x”, involved with the function in the differential equation y(x) must cover the half space of (o 0 (1) is known as Bessel’s equation of order p. The thing which confuses me is this: I was always taught that the Fundamental Principle of TQFT is the following: You can’t be Frobenius without being finite dimensional. HP-42S. Use the Frobenius method to nd a nonzero series solution y= xr P 1 n=0 b nx n of the equation 4x2y00+ 4xy0 y= 0 (x>0): 4. Compute the 1 -norm, Frobenius norm, and infinity norm of the inverse of the 3-by-3 magic square A: A = inv (sym (magic (3))) norm1 = norm (A, 1) normf = norm (A, 'fro') normi = norm (A, inf) A = [ 53/360, -13/90, 23/360] [ -11/180, 1/45, 19/180] [ -7/360, 17/90, -37/360] norm1 = 16/45 normf = 391^ (1 SOLUTIONS OF FROBENIUS-PERRON OPERATOR EQUATIONS 467. Matrix Calculator computes all the important aspects of a matrix: determinant, inverse, trace , norm. Customer Voice. 1. 4, where it sets the stage for the method of Frobenius. Even at a singular point of an ODE, some (or even all) of its solutions may be analytic, but this is not guaranteed. Each step allows to calculate a bound on the moduli of the zeros for the obtained modified set. If called in the form Frobenius (A, 'P'), then P will be assigned the transformation matrix corresponding to the Frobenius form, that is, the matrix P such that inverse (P) * A * P = F. 1 Cauchy-Euler Equations A second order Cauchy-Euler equation has the form ax2y00+ bxy0+ cy= 0 (1) for constants a, b, and c. To benchmark against the best possible performance of the calculate the socle series of the Weyl modules with p-singular highest weight for the group of type G 2 by studying the extensions between simple modules for Frobenius kernel. Dino Lorenzini defines the arithmetical graph, which naturally arises in arithmetic geometry, and a notion of genus, the g-number, that in specific cases coincides with the Frobenius number of R. 1) • Here H k is an n ⇥ n positive deï¬nite symmetric matrix (that character string, specifying the type of matrix norm to be computed. In ?3, a nonlinear generalization of the first method is de- scribed. Neuringera, The Frobenius method for complex roots of the indicial equation, International Journal of Mathematical Education in Science and Technology, Volume 9, Issue 1, 1978, 71–77. The lower value, ð‘™ and the upper value, which bracket the root(s) are required. Now, using the Frobe- nius map, we present an algorithm which improves the usual k-ary method [1]. The wikipedia article begins by saying that the Frobenius method is a way to find solutions for ODEs of the form $x^2y'' + xP(x) + Q(x)y = 0$ To put (1) into that form I might multiply across by x, giving me $x^2y'' + x[2x]y' + [6xe^x]y = 0$ (2) But is that OK? Series Solutions{Frobenius’ Method We now turn our attention to the solution of a linear, second-order, homogeneous ODE of the form y00+ P(x)y0+ Q(x)y = 0: Such an ODE has two linearly independent solutions, y 1(x) and y 2(x). frobenius method calculator (Source: aplautomacao.org)
### Enter
The key thing to carrying out the addition of fractions correctly is to always keep in mind the most important part of the fraction is the number under the line, known as the denominator. If we have a situation where the denominators in the fractions involved in the addition process are the same, then we merely add the numbers that are above the separation line or as a mathematician would put it: "Adding the numerators only". We can have a look at an example of adding two fractions like 3⁄7 and 4⁄7. The expression would look like this: 3⁄7 + 4⁄7 = 7⁄7. In the case when the nominator is equal to the denominator, like in the foregoing example, it can also be equated to 1.
The key thing to carrying out the subtraction of fractions correctly is to always keep in mind that the most important part of the fraction is the number under the line, known as the denominator. If we have a situation where the denominators in the fractions involved in the subtraction process are the same, then we merely add the numbers that are above the separation line or as a mathematician would put it: "Subtracting the numerators only". We can have a look at an example of subtracting two fractions like 3⁄7 and 4⁄7. The expression would look like this: 4⁄7 - 3⁄7 = 1⁄7. (Source: goodcalculators.com)
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July 03, 2022 | m aqib | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8551630973815918, "perplexity": 774.0309453088837}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-27/segments/1656104209449.64/warc/CC-MAIN-20220703013155-20220703043155-00310.warc.gz"} |
http://mathhelpforum.com/calculus/179229-calculation-limit-using-l-hopital.html | # Thread: Calculation of a limit using L'Hopital
1. ## Calculation of a limit using L'Hopital
Hello
I'm having trouble calculating the right hand limit of sqrt(x) * ln(x) at 0
I see that it's an indeterminate form of the type 0 * -infinity
So I tried turning into the limit of sqrt(x) / (1/lnx) so that it would be a case of 0/0
and applied L'Hopitals rule, but I got that it's equal to
1/2 * limit at 0+ of sqrt(x) * (lnx)^2
In other words back where I started...what am I doing wrong?
2. Originally Posted by moses
Hello
I'm having trouble calculating the right hand limit of sqrt(x) * ln(x) at 0
I see that it's an indeterminate form of the type 0 * -infinity
So I tried turning into the limit of sqrt(x) / (1/lnx) so that it would be a case of 0/0
and applied L'Hopitals rule, but I got that it's equal to
1/2 * limit at 0+ of sqrt(x) * (lnx)^2
In other words back where I started...what am I doing wrong?
Try doing it the other way.
$\sqrt{x}ln(x) = \frac{ln(x)}{\frac{1}{\sqrt{x}}}$
-Dan
3. Oh okay that works, thanks | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 1, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9216256141662598, "perplexity": 1460.206394491948}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-34/segments/1502886102663.36/warc/CC-MAIN-20170816212248-20170816232248-00001.warc.gz"} |
http://tex.stackexchange.com/questions/17153/any-ideas-why-this-newcommand-n-lstinline-n-macro-doesnt-work?answertab=oldest | # Any ideas why this \newcommand{\n}{\lstinline|\n|} macro doesn't work?
So I created a new macro `\newcommand{\n}{\lstinline|\n|}` but for some reason when I use it, all I get is bold 'n' letter and not a nice `\n`. Any ideas?
-
As Ulrike already posted you can't use verbatim macros or environments inside a macro argument or replacement text. To make `\n` typeset `\n` as verbatim you could use:
``````\newcommand{\n}{\texttt{\string\n}}
``````
or use the `verbdef` package:
``````\verbdef\n|\n|
``````
-
Martin, your command seems to ignore space after `\n`. For example, `foo \n boo` gets typed in the docuemnt as `foo \nboo`. I had heard about using `\xspace` but I can't quite get it in the command, check this out: `\newcommand{\n}{\texttt{\string\n}\xspace}` fails to compile when I run `pdflatex` with `Undefined control sequence`. Same for `\newcommand{\n}{\texttt{\string\n\xspace}}` – Peteris Krumins May 2 '11 at 10:18
@Peteris: All spaces behind all macros are always ignored. You need to load the `xspace` package to get `xspace` and it must be the very last of the macro replacement text. Alternatively you can write `\n{}` instead. – Martin Scharrer May 2 '11 at 11:10
Thank you Martin, I had forgotten to load `xspace` package! – Peteris Krumins May 2 '11 at 11:17
Shouldn't your first sentence read "you can't"? – lockstep May 2 '11 at 15:39
@lockstep: Uuh, yes of course. Thank you for pointing it out. – Martin Scharrer May 2 '11 at 15:43
`\lstinline` is a verbatim-like command. It has to do a lot of `\catcode`-magic to disable commands and parse its argument. You can't use it in another command. http://www.tex.ac.uk/cgi-bin/texfaq2html?label=verbwithin.
-
I went with `\verbdef\n|\n|` for now. – Peteris Krumins May 2 '11 at 10:58
In fact, you can use lstinline to define custom commands. I define short commands to include snippets in a specific formal language like this:
``````\newcommand\foo[1]{\lstinline[language=foo]{#1}}
``````
If the language `foo` is not defined, you must first define it with
``````\lstdefinelanguage{foo}{ ... }
``````
In your code you can then use the custom command like this. Some special characters like `{`,`}`, and `^` must be escaped:
``````Blabla \foo{this is syntax highlighted} blabla,
bla \foo{This contains \{ braces \}}.
``````
- | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9423255920410156, "perplexity": 4921.221828693301}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-42/segments/1413558066932.32/warc/CC-MAIN-20141017150106-00361-ip-10-16-133-185.ec2.internal.warc.gz"} |
https://thermtest.com/application/thermal-conductivity-of-homogeneous-and-heterogeneous-materials | # Thermal Conductivity of Homogeneous and Heterogeneous Materials
The Heat Flow Meter (HFM-25) for testing thermal resistance and thermal conductivity of insulation and construction materials. Offering the testing power of a full-size heat flow meter, optimized for small insulation samples and budget. The HFM-25 is controlled by the included HFM Software, offering automation of calibration, testing, and results summary functions.
Picture 1. Thermtest 25 Series Heat Flow Meter.
Following international standards, the HFM-25 is designed for testing both homogenous and heterogenous materials from 0.01 to 0.5 W/m·K and complies with ASTM C518. The HFM-25 allows for smaller sample sizes which are representative of materials typically found in insulation and construction industries. With an accuracy of 5% and reproducibility of 2%, the Thermtest 25 Series Heat Flow Meter is an excellent instrument for measuring the thermal conductivity of a variety of samples.
## Sample Configuration
For small samples (min 50 x 50 mm), samples can easily be arranged for simple testing.
Picture 2. 50 x 50 mm piece of NIST 1450D
For larger samples (no maximum), samples can conveniently be measured at a specific location.
Picture 3. 300 x 300 mm piece of NIST 1450D
## Thermal Conductivity of Homogeneous Fiberglass and Heterogeneous Plasterboard
NIST 1450D is a type of fibrous glass board, certified for bulk density and thermal resistance and thermal conductivity. The plasterboard material is heterogenous, consisting of a layer of gypsum plaster between two layers of paper. Both of these materials are commonly found in the construction industry.
Each sample had dimensions of 300 x 300 mm. Locations in each of the four corners (1 x 1 inch location size) of both materials were measured to analyze the thermal conductivity throughout the entire sample and make a determination of heterogeneity or homogeneity. The steady-state method used by the heat flow meter (HFM-25) is quite appropriate for measuring the through-thickness properties of both of these types of samples. For all tests, the upper plate was set to 30°C and the lower plate to 10°C, obtaining an optimal temperature delta of 20 degrees.
Picture 4: Set up of the NIST 1450D (left) and plasterboard (right) samples in the Thermtest HFM-25.
Table 1: Results of thermal conductivity measurements with the Thermtest HFM-25.
Sample Measured Thermal Conductivity (W/mK) Relative Standard Deviation (%)
NIST 1450D Corner 1 0.0305 0.98
NIST 1450D Corner 2 0.0301
NIST 1450D Corner 3 0.0297
NIST 1450D Corner 4 0.0299
Plasterboard Corner 1 0.1166 4.1
Plasterboard Corner 2 0.1268
Plasterboard Corner 3 0.1231
Plasterboard Corner 4 0.1305
With a relative standard deviation of less than 1%, the NIST 1450D sample is shown to be homogenous. The thermal conductivity of the plasterboard is less consistent throughout the sample, due to the heterogeneous matrix of this composite material. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8367387652397156, "perplexity": 3019.79048425746}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-49/segments/1669446710462.59/warc/CC-MAIN-20221128002256-20221128032256-00735.warc.gz"} |
https://www.physicsforums.com/threads/0-1.132399/ | # 0! = 1
1. Sep 17, 2006
### murshid_islam
can anyone plz help me with this. why is 0! = 1?
thanks in advance to anyone who can help.
2. Sep 17, 2006
### matt grime
Because that is what we define it to be. There are hundreds of posts on this topic.
We could choose not to define it at all and state that n! is the number of ways or ordering n things for integer n strictly positive, and not define 0!. But that turns out to be unwieldy when doing things like nCr, so for ease, consistency, whatever, there is no harm in defining 0!=1. End of story, no great mystery here.
Last edited: Sep 17, 2006
3. Sep 17, 2006
### murshid_islam
so mathematicians just decided that 0! = 1? is that what you mean?
4. Sep 17, 2006
### arildno
It didn't have any prior meaning before the mathematicians appropriated it, so yes, they were free to decide what that symbol collection 0! should mean.
The conventional definition of the factorial goes like this:
$$0!=1, n!=n*((n-1)!), n\geq{1}$$
where n is always a natural number.
Last edited: Sep 17, 2006
5. Sep 17, 2006
### murshid_islam
for example, 3! is the product of the first three integers. but 0! does not have such meaning. it is just defined to be 1. am i correct?
6. Sep 17, 2006
### arildno
Quite so.
Or, rather, 3! is recursively defined in such a manner that you can compute it by multiplying together the first three natural numbers.
7. Sep 17, 2006
### murshid_islam
thank you very much for your help, arildno.
8. Sep 17, 2006
### CRGreathouse
It's not entiely arbitrary. n!/n = (n-1)! whenever division by n is defined. If we assign any value to 0! other than 1, this would not hold.
9. Sep 17, 2006
### shmoe
But a decision to require n!/n = (n-1)! to be true where n=1 is arbitrary.
10. Sep 17, 2006
### arildno
Besides, set 0!=a.
Then, by induction, we have:
n!=a*1*2*...(n-1)*n, that is, n!/n=(n-1)! whatever value you assign to 0!.
11. Sep 17, 2006
### CRGreathouse
Sure, as is the decision to make 0! the number of ways to arrange 0 objects (1 way). The point is that it is consistent with the way it works for other numbers.
12. Sep 17, 2006
### Spiderman
At first I thought this was a totally absurd question, because in many programming languages "!=" means "not equal to". Of course 0 is not equal to 1
I've been programming all weekend so everything else goes out the door.
13. Sep 17, 2006
### arildno
:rofl: :rofl: :rofl:
14. Sep 17, 2006
### Robokapp
The explanation my math teacher gave me was the following:
5!=5*4*3*2*1
4!=4*3*2*1
So
5!=5*4!
4!=4*3!
3!=3*2!
2!=2*1!
1!=1*0!
1!=1...so 1*0! has to equal 1. 1*0!=1 => 0!=1/1 => 0!=1
15. Sep 18, 2006
### matt grime
Then why isn't 0! 0*(-1)! which must be 0 if (-1)! is defined?
16. Sep 18, 2006
### dextercioby
$$0! =\Gamma(1) =\int_{0}^{\infty} e^{-t} dt= 1$$
Daniel.
17. Sep 18, 2006
### JonF
I think any thread with the : “0”, “1”, and “=” is asking for trouble.
18. Sep 18, 2006
### waht
0! doesn't have to be defined exactly, look at this relation
$$\frac{n!}{k!} = (n - k)!$$
If n = k then you get
$$\frac{n!}{n!} = (n - n)!$$
19. Sep 18, 2006
### shmoe
This isn't true. 5!/4!=5, not (5-4)!=1.
20. Sep 18, 2006
### waht
yea you are right, my bad. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8838397264480591, "perplexity": 2855.89734532316}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-44/segments/1476988721555.36/warc/CC-MAIN-20161020183841-00270-ip-10-171-6-4.ec2.internal.warc.gz"} |
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## Tcs.tifr.res.in
• If you submit handwritten solutions, start each problem on a fresh page.
• Collaboration is encouraged, but all writeups must be done individually and must include names of all • Referring sources other than the lectures is strongly discouraged. But if you do use an outside source (eg., other text books, lecture notes, any material available online), ACKNOWLEDGE it in yourwriteup.
• The points for each problem are indicated on the side.
• If you don’t know the answer to a problem, then just don’t answer it. Do not try to convince yourself or others into believing a false proof.
The universal relation Un is defined as follows.
Un = {(x, y, i) ∈ {0, 1}n × {0, 1}n × [n] : xi = yi}.
Alice is given x ∈ {0, 1}n and Bob y ∈ {0, 1}n such that x = y. They have to determine an i such that(x, y, i) ∈ Un.
In this problem, we will show that there is a randomized public coins protocol, where • Alice sends one message of O((log n)2) bits; • Bob determines an i such that (x, y, i) ∈ Un; (a) [Identify] Give an O(log n) bit protocol that makes error at most 1/3 under the promise that x and y differ in only one position. (There is a deterministic protocol but a randomized protocolwould suffice.) (b) [Isolate] Now for the general case, when x and y may differ in more than one position. Using shared randomness, devise a distribution for sets S1, S2, ., S ⊆ [n] ( = O(log n)), such that withconstant probability there is an i such that x|S and y| (c) [Verify] Suppose Alice has x and Bob has (y, i). Give a constant protocol in which Alice sends O(1) bits and Bob verifies with probability at least 2/3 that x and y differ only at i.
(d) Combine the above parts to obtain the one-round randomized protocol for the universal relation.
In the second part of this problem, we will show that any public coins randomized protocol (errorat most 1 ) for the universal relation where Alice sends just one message, requires Ω((log n)2) bits of Use the following augmented index function problem.
u1, ., ui−1; Alice sends one message and Bob determines ui. (Later we will set L to be about log n.) (e) Any randomized (public coins) protocol for this augmented index function problem requires Ω(L2) (f) [Reduction] Suppose Alice needs to send a number z ∈ [L] to Bob. Show how you will do this using a protocol for the universal relation on L bit inputs. That is, Alice transforms her input ito f (i) ∈ {0, 1}L and Bob produces an input v ∈ {0, 1}L; they run the protocol for the universalrelation, and from the output Bob determines z.
(g) Use the reduction in part (f) appropriately to u1, ., uL, thereby producing inputs to the universal function problem of size n = L2. In other words, give a reduction such that Alice transforms herinput u[1,L] = (u1, . . . , uL) to an input in {0, 1}L2 and Bob transforms his input (u[1,i−1], i) to aninput in {0, 1}L2 such that when they run the universal relation protocol on these transformedinputs, Bob recovers one of ui, ui+1, ., uL.
(h) Use public randomness and increase the input length to 2O(L) to ensure that with high probability the protocol actually helps Bob recover ui itself.
Suppose Alice and Bob are given k-sized subsets of [n] each and they have to determine if the setsare disjoint. Call this problem DISJn,k. Show that the randomized public coins (error at most 1 ) complexity of DISJn,k is O(k). (Observe that an O(k log n) protocol is easy to obtain.) The followinghints might help you in constructing the protocol • [Small sets] Suppose k is constant. Hash into some O(k2) bits using public randomness, and determine the answer with O(k2) bits of communication.
• [Shrinking the bigger set]. Proceed as in the protocol for product distributions. Suppose the sets have sizes s, t (s ≤ t ≤ k). Alice picks a random superset of her input and sends it to Bob withO(s) bits of communication (use shared randomness). Bob restricts his input to inside this set. Ifthis does not result in significant decrease in the size of Bob’s set, then the sets probably intersectheavily (by Chernoff). Repeat, keeping the total error in control until the sets become small andapply the base case above.
[Kushilevitz-Nisan, Exercise 5.21] Let FORK be the relations consisting of triples (x, y, i) such thatx, y ∈ Σ , and is is such that xi = yi and xi+1 = yi+1 or xi−1 = yi−1. Prove that D(FORK ) =Ω(log , log w).
4. [s-t connectivity: asymmetry in KW result] Karchmer and Wigderson showed that any monotone circuits for s-t connectivity on n-vertex graphs, where ANDs have fanin at most 2n1/5 and ORs have fanin at most n1/5, must have depth Ω(log n).
Observe that there exist shallow circuits in which that AND and OR fanins are reversed.
Now, consider the s-t cut function, which is true on an n vertex directed graph if its complement hasno s-t path. Like the s-t connectivity problem, this is a monotone function defined on n × n adjacencymatrices (no self-loops, parallel edges).
A monotone projection from s-t connectivity (on n vertex graphs) to s-t cut (on N vertex graphs) is afunction from f : [N ] × [N ] → [n] × [n], such that the n × n adjacency matrix A = (ai,j) is s-t connectediff the N × N matrix B = (bk, ) where bk,l = af(k,l) is s-t cut. Show that no such projection exists ifN is polynomially bounded in n.
(a) Let f (x1, . . . , x2k+1) = maj(x1, . . . , x2k+1). Let Alice be given the first k+1 bits, i.e., x1, . . . , x2k+1 and Bob the next k bits. Show that there exists an O(1)-randomized protocol with error at most 1 − O 1 that computes f . Conclude that maj has discrepancy at least Ω(1/k) with respect to (b) Now, consider a function f on 2k variables defined as a depth 2 AC0 circuit as follows: f (x1, . . . , x2k) = T1 ∨ T2 ∨ · · · ∨ Tl where the Ti’s are conjunction of literals. Let Alice be giventhe first k bits of x1, . . . , x2k and Bob the next k bits. Show that there exists an O(1)-randomizedprotocol with error at most 1 − O 1 that computes f . Conclude that any depth 2 AC0 circuit with top fan-in s has discrepancy at least Ω(1/s) with respect to every distribution.
6. [communication complexity of GT and LTF] (a) Given two n bit integers 0 ≤ x, y < 2n, “greater than” function GT(x, y) = 1 iff x > y.
Using one of the theorems proved/stated in lecture, show that Rε(GT) = O log n + log 1 (b) Let a1, . . . , an, b1, . . . , bn, θ ∈ R. Then, the linear threshold function corresponding to (a, b, θ) where a = (a1, . . . , an) and b = (b1, . . . , bn) is defined as follows Show using part (a) or otherwise that Rε(LTFa,b,θ) = O log n + log 1 [Hint: Try to bound the size of the “possible” numbers ai, bi, θ.] In a simultaneous message protocol (SMP), Alice and Bob send a message each (based on their respec-tive inputs) to a referee, who must then announce the result.
(a) Show that there is a private coins simultaneous messages protocol for EQn where Alice and Bob (b) Suppose there is a simultaneous messages private coins randomized protocol for a function f , where Alice sends messages of length a and Bob sends messages of length b. Show that then thereis a deterministic one-way communication protocol for f with communication O(ab). Conclude that the private coins simultaneous message complexity of EQn is at least Ω( n).
[Hint: View the referee’s actions as a matrix with rows indexed by messages of Alice (∈ {0, 1}a)and columns by message of Bob (∈ {0, 1}b). Show that for each input x, Alice can send Bob a(multi-)set Sx ⊆ {0, 1}a of O(b) row indices (use Chernoff bounds to show the existence of suchSx); Bob can then restrict himself to these rows, and based on the distribution of his messagesin the simultaneous protocol, simulate the referee’s actions faithfully.] 8. [A direct sum result for simultaneous message protocols.] Consider the direct sum problem for m copies of EQn in the SMP model: so, Alice gets x1, x2, . . . , xm ∈{0, 1}n and Bob gets y1, y2, . . . , ym ∈ {0, 1}n, and they need to determine the answers for all minstances.
Complete the following argument to show that the simultaneous messages randomized private coins communication complexity of this problem is Ω(m n).
(a) Consider a distribution on Alice’s inputs, where each coordinate is chosen independently. Show that if Alice sends a bits and Bob sends b bits, then there is a coordinate for which they sendonly O(a/m) bits and O(b/m) bits of information to the referee.
(b) Show that Alice’s message can be compressed to O(a/m) bits for most of her inputs. Let X be the input of Alice and let M be her message. Suppose I[X : M ] ≤ α. Let P be the distributionof M . Let Px be the distribution of M conditioned on X = x.
x P ). Observe that for most (define ‘most’ appropri- ately) values x, S(Px P ) = O(α); call such x good.
ii. Prove the following. There are constants A and B, such if S(Px P ) ≤ s, then for all ε > 0, iii. Assume that Alice and the referee share independent samples of M (generated according to P ). Show that Alice, on receiving a good x may then send O(S(Px P )/ε) bits to the refereeand help him select a sample whose distribution is close (how close?) to Px.
(c) Prove the result (Ω(m n) bound) using the previous two exercises. Note that there the coins were private. What distributions will you choose for the inputs? | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9010420441627502, "perplexity": 2740.475894928057}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-09/segments/1518891816841.86/warc/CC-MAIN-20180225170106-20180225190106-00588.warc.gz"} |
https://livedu.in/leibnitzs-theorem/ | # Leibnitz’s Theorem
Differential Calculus / Thursday, August 23rd, 2018
# Successive Differentiation – Leibnitz’s Theorem
Leibnitz’s Theorem works on finding successive derivatives of product of two derivable functions.
## Statement:
If u and v are two functions of x, each possessing derivatives upto nth order, then the product y=u.v is derivable n times and
${{y}_{n}}={{\left( u.v \right)}_{n}}={}^{n}{{C}_{0}}{{u}_{n}}v+{}^{n}{{C}_{1}}{{u}_{n-1}}{{v}_{1}}+{}^{n}{{C}_{2}}{{u}_{n-2}}{{v}_{2}}+…+{}^{n}{{C}_{r}}{{u}_{n-r}}{{v}_{r}}+…+{}^{n}{{C}_{n}}u{{v}_{n}}$
$\Rightarrow {{y}_{n}}={{u}_{n}}v+{}^{n}{{C}_{1}}{{u}_{n-1}}{{v}_{1}}+{}^{n}{{C}_{2}}{{u}_{n-2}}{{v}_{2}}+…+{}^{n}{{C}_{r}}{{u}_{n-r}}{{v}_{r}}+…+u{{v}_{n}}$
$As,{}^{n}{{C}_{r}}=\frac{n!}{r!(n-r)!}\Rightarrow {}^{n}{{C}_{0}}={}^{n}{{C}_{n}}=1$
Where yr, ur, vr denote the rth derivatives of y, u and v respectively with respect to x.
Proof:
Since y = u.v, differentiating both sides w.r.t. x we get,
${{y}_{1}}={{u}_{1}}v+u{{v}_{1}}$
$\therefore {{y}_{2}}={{u}_{2}}v+{{u}_{1}}{{v}_{1}}+{{u}_{1}}{{v}_{1}}+u{{v}_{2}}={{u}_{2}}v+2{{u}_{1}}{{v}_{1}}+u{{v}_{2}}$
$\Rightarrow {{y}_{2}}={{u}_{2}}v+{}^{2}{{C}_{1}}{{u}_{1}}{{v}_{1}}+u{{v}_{2}}$
Thus the theorem is found to be true for n = 1, 2.
Let us assume that the theorem is true for a particular positive integral value of n, say, k, where k < n.
Then we have,
${{y}_{k}}={{u}_{k}}v+{}^{k}{{C}_{1}}{{u}_{k-1}}{{v}_{1}}+{}^{k}{{C}_{2}}{{u}_{k-2}}{{v}_{2}}+…+{}^{k}{{C}_{r-1}}{{u}_{k-r+1}}{{v}_{r-1}}+{}^{k}{{C}_{r}}{{u}_{k-r}}{{v}_{r}}+…+u{{v}_{k}}$
Differentiating both sides again w.r.t. x we get,
${{y}_{k+1}}={{u}_{k+1}}v+{{u}_{k}}{{v}_{1}}+{}^{k}{{C}_{1}}\left( {{u}_{k}}{{v}_{1}}+{{u}_{k-1}}{{v}_{2}} \right)$
$+{}^{k}{{C}_{2}}\left( {{u}_{k-1}}{{v}_{2}}+{{u}_{k-2}}{{v}_{3}} \right)+…+{}^{k}{{C}_{r-1}}\left( {{u}_{k-r+2}}{{v}_{r-1}}+{{u}_{k-r+1}}{{v}_{r}} \right)$
$+{}^{k}{{C}_{r}}\left( {{u}_{k-r+1}}{{v}_{r}}+{{u}_{k-r}}{{v}_{r+1}} \right)+…+{{u}_{1}}{{v}_{k}}+u{{v}_{k+1}}$
$\Rightarrow {{y}_{k+1}}={{u}_{k+1}}v+\left( 1+{}^{k}{{C}_{1}} \right){{u}_{k}}{{v}_{1}}+\left( {}^{k}{{C}_{1}}+{}^{k}{{C}_{2}} \right){{u}_{k-1}}{{v}_{2}}+…+$
$\left( {}^{k}{{C}_{r-1}}+{}^{k}{{C}_{r}} \right){{u}_{k-r+1}}{{v}_{r}}+…+u{{v}_{k+1}}$
We know that,
${}^{k}{{C}_{r-1}}+{}^{k}{{C}_{r}}={}^{k+1}{{C}_{r}},{}^{k}{{C}_{1}}+{}^{k}{{C}_{2}}={}^{k+1}{{C}_{2}},1+{}^{k}{{C}_{1}}={}^{k+1}{{C}_{1}}$
$\therefore {{y}_{k+1}}={{u}_{k+1}}v+{}^{k+1}{{C}_{1}}{{u}_{k}}{{v}_{1}}+{}^{k+1}{{C}_{2}}{{u}_{k-1}}{{v}_{2}}+…$
$+{}^{k+1}{{C}_{r}}{{u}_{k-r+1}}{{v}_{r}}+…+u{{v}_{k+1}}$
Thus we find that if the theorem is assumed to be true for any positive integral value k of n, then it is found to be true for the next integral value (k+1) of n.
And the theorem has already been found to be true for n =1, 2.
Hence, by the principle of Mathematical Induction, the theorem is true for every positive integral value of n.
Thus Leibnitz’s Theorem is established.
Now is the time to check some problems to find the nth order derivative using Leibnitz’s Theorem.
Note: Generally you should initialize v with the function which will take minimum iterations to become 0.
Question 01
$If,y={{e}^{x}}{{x}^{3}}find,{{y}_{4}}$
Solution:
$Let,u={{e}^{x}},v={{x}^{3}}$
Then by Leibnitz’s Theorem,
${{y}_{4}}={{u}_{4}}v+{}^{4}{{C}_{1}}{{u}_{3}}{{v}_{1}}+{}^{4}{{C}_{2}}{{u}_{2}}{{v}_{2}}+{}^{4}{{C}_{3}}{{u}_{1}}{{v}_{3}}+u{{v}_{4}}$
$\Rightarrow {{y}_{4}}={{D}^{4}}({{e}^{x}}).{{x}^{3}}+{}^{4}{{C}_{1}}{{D}^{3}}({{e}^{x}}).D\left( {{x}^{3}} \right)+{}^{4}{{C}_{2}}{{D}^{2}}({{e}^{x}}).{{D}^{2}}\left( {{x}^{3}} \right)$
$+{}^{4}{{C}_{3}}D({{e}^{x}}).{{D}^{3}}\left( {{x}^{3}} \right)+{{e}^{x}}.{{D}^{4}}\left( {{x}^{3}} \right),where,{{D}^{r}}\equiv \frac{{{d}^{r}}}{d{{x}^{r}}}$
$\Rightarrow {{y}_{4}}={{e}^{x}}{{x}^{3}}+4{{e}^{x}}.3{{x}^{2}}+6.{{e}^{x}}.6x+4{{e}^{x}}.6+{{e}^{x}}.0$
$\therefore {{y}_{4}}={{e}^{x}}\left\{ {{x}^{3}}+12{{e}^{x}}{{x}^{2}}+36.{{e}^{x}}x+24 \right\}$
Question 02
$If,y=\sin \left( m{{\sin }^{-1}}x \right),show-that$
$(i)\left( 1-{{x}^{2}} \right){{y}_{2}}-x{{y}_{1}}+{{m}^{2}}y=0$
$(ii)\left( 1-{{x}^{2}} \right){{y}_{n+2}}-\left( 2n+1 \right)x{{y}_{n+1}}+\left( {{m}^{2}}-{{n}^{2}} \right){{y}_{n}}=0$
Solution:
$Given,y=\sin \left( m{{\sin }^{-1}}x \right)$
$\therefore {{y}_{1}}=\cos \left( m{{\sin }^{-1}}x \right).m.\frac{1}{\sqrt{1-{{x}^{2}}}}$
$\Rightarrow \sqrt{1-{{x}^{2}}}.{{y}_{1}}=m\cos \left( m{{\sin }^{-1}}x \right)$
$\Rightarrow \left( 1-{{x}^{2}} \right).{{y}_{1}}^{2}={{m}^{2}}{{\cos }^{2}}\left( m{{\sin }^{-1}}x \right)={{m}^{2}}\left\{ 1-{{\sin }^{2}}\left( m{{\sin }^{-1}}x \right) \right\}$
$\Rightarrow \left( 1-{{x}^{2}} \right).{{y}_{1}}^{2}={{m}^{2}}\left( 1-{{y}^{2}} \right)$
Differentiating both sides w.r.t. x we get,
$\left( 1-{{x}^{2}} \right).2{{y}_{1}}.{{y}_{2}}-2x.{{y}_{1}}^{2}=-2{{m}^{2}}y{{y}_{1}}$
$\therefore \left( 1-{{x}^{2}} \right){{y}_{2}}-x{{y}_{1}}+{{m}^{2}}y=0,[\because {{y}_{1}}\ne 0]$
Now applying Leibnitz’s theorem to differentiate the above equation n times w.r.t. x we get,
${{D}^{n}}\left\{ \left( 1-{{x}^{2}} \right){{y}_{2}} \right\}-{{D}^{n}}\left\{ x{{y}_{1}} \right\}+{{m}^{2}}{{y}_{n}}=0$
$\Rightarrow \left\{ \left( 1-{{x}^{2}} \right){{y}_{n+2}}+{}^{n}{{C}_{1}}.{{y}_{n+1}}.(-2x)+{}^{n}{{C}_{2}}.{{y}_{n}}.(-2)+0 \right\}$
$-\left\{ x{{y}_{n+1}}+{}^{n}{{C}_{1}}.{{y}_{n}}.1+0 \right\}+{{m}^{2}}{{y}_{n}}=0$
$\Rightarrow \left( 1-{{x}^{2}} \right){{y}_{n+2}}-2nx{{y}_{n+1}}-n(n-1){{y}_{n}}-x{{y}_{n+1}}-n{{y}_{n}}+{{m}^{2}}{{y}_{n}}=0$
$\Rightarrow \left( 1-{{x}^{2}} \right){{y}_{n+2}}-\left( 2n+1 \right)x{{y}_{n+1}}-\left( {{n}^{2}}-n+n \right){{y}_{n}}+{{m}^{2}}{{y}_{n}}=0$
$\therefore \left( 1-{{x}^{2}} \right){{y}_{n+2}}-\left( 2n+1 \right)x{{y}_{n+1}}+\left( {{m}^{2}}-{{n}^{2}} \right){{y}_{n}}=0$
Question 03
$If,y={{e}^{m{{\sin }^{-1}}x}},show-that$
$(i)\left( 1-{{x}^{2}} \right){{y}_{2}}-x{{y}_{1}}-{{m}^{2}}y=0$
$(ii)\left( 1-{{x}^{2}} \right){{y}_{n+2}}-\left( 2n+1 \right)x{{y}_{n+1}}-\left( {{m}^{2}}+{{n}^{2}} \right){{y}_{n}}=0$
$Also,find\to {{y}_{n}},when\to x=0$
Solution:
$Given,y={{e}^{m{{\sin }^{-1}}x}}$
$\therefore {{y}_{1}}={{e}^{m{{\sin }^{-1}}x}}.m.\frac{1}{\sqrt{1-{{x}^{2}}}}………..(1)$
$\Rightarrow \left( 1-{{x}^{2}} \right){{y}_{1}}^{2}={{m}^{2}}{{\left( {{e}^{m{{\sin }^{-1}}x}} \right)}^{2}}={{m}^{2}}{{y}^{2}}$
$\therefore \left( 1-{{x}^{2}} \right).2{{y}_{1}}.{{y}_{2}}-2x{{y}_{1}}^{2}=2{{m}^{2}}y{{y}_{1}}$
$\therefore \left( 1-{{x}^{2}} \right){{y}_{2}}-x{{y}_{1}}-{{m}^{2}}y=0………..(2)$
Now applying Leibnitz’s theorem to differentiate the above equation n times w.r.t. x we get,
${{D}^{n}}\left\{ \left( 1-{{x}^{2}} \right){{y}_{2}} \right\}-{{D}^{n}}\left\{ x{{y}_{1}} \right\}-{{m}^{2}}{{y}_{n}}=0$
$\Rightarrow \left\{ \left( 1-{{x}^{2}} \right){{y}_{n+2}}+{}^{n}{{C}_{1}}.{{y}_{n+1}}.(-2x)+{}^{n}{{C}_{2}}.{{y}_{n}}.(-2)+0 \right\}$
$-\left\{ x{{y}_{n+1}}+{}^{n}{{C}_{1}}.{{y}_{n}}.1+0 \right\}-{{m}^{2}}{{y}_{n}}=0$
$\Rightarrow \left( 1-{{x}^{2}} \right){{y}_{n+2}}-2nx{{y}_{n+1}}-n(n-1){{y}_{n}}-x{{y}_{n+1}}-n{{y}_{n}}-{{m}^{2}}{{y}_{n}}=0$
$\Rightarrow \left( 1-{{x}^{2}} \right){{y}_{n+2}}-\left( 2n+1 \right)x{{y}_{n+1}}-\left( {{n}^{2}}-n+n \right){{y}_{n}}-{{m}^{2}}{{y}_{n}}=0$
$\therefore \left( 1-{{x}^{2}} \right){{y}_{n+2}}-\left( 2n+1 \right)x{{y}_{n+1}}-\left( {{m}^{2}}+{{n}^{2}} \right){{y}_{n}}=0……….(3)$
$\because y={{e}^{m{{\sin }^{-1}}x}}\Rightarrow y=1,when,x=0$
Also from (1), (2) and (3)
${{y}_{1}}=m,{{y}_{2}}={{m}^{2}},{{y}_{n+2}}=\left( {{m}^{2}}+{{n}^{2}} \right){{y}_{n}}$
Now putting n = 1, 2, 3, … in the relation
${{y}_{n+2}}=\left( {{m}^{2}}+{{n}^{2}} \right){{y}_{n}},we-have$
${{y}_{3}}=\left( {{m}^{2}}+{{1}^{2}} \right){{y}_{1}}=m\left( {{m}^{2}}+{{1}^{2}} \right)$
${{y}_{4}}=\left( {{m}^{2}}+{{2}^{2}} \right){{y}_{2}}=m\left( {{m}^{2}}+{{2}^{2}} \right)$
${{y}_{5}}=\left( {{m}^{2}}+{{3}^{2}} \right){{y}_{3}}=m\left( {{m}^{2}}+{{1}^{2}} \right)\left( {{m}^{2}}+{{3}^{2}} \right)$
${{y}_{6}}=\left( {{m}^{2}}+{{4}^{2}} \right){{y}_{4}}=m\left( {{m}^{2}}+{{2}^{2}} \right)\left( {{m}^{2}}+{{4}^{2}} \right)$
Hence when x = 0
${{y}_{n}}=m\left( {{m}^{2}}+{{1}^{2}} \right)\left( {{m}^{2}}+{{3}^{2}} \right)…\left\{ {{m}^{2}}+{{\left( n-2 \right)}^{2}} \right\}if\to ‘n’is\_odd$
${{y}_{n}}=m\left( {{m}^{2}}+{{2}^{2}} \right)\left( {{m}^{2}}+{{4}^{2}} \right)…\left\{ {{m}^{2}}+{{\left( n-2 \right)}^{2}} \right\}if\to ‘n’is\_even$
Question 04
$If,f(x)={{x}^{n}},prove-that$
$f(1)+\frac{f'(1)}{1!}+\frac{f”(1)}{2!}+\frac{f”'(1)}{3!}+…+\frac{{{f}^{n}}(1)}{n!}={{2}^{n}}$
Solution:
$Given,f(x)={{x}^{n}}$
$\therefore f'(x)=n{{x}^{n-1}},f”(x)=n(n-1){{x}^{n-2}}…$
$\therefore {{f}^{r}}(x)=n(n-1)(n-2)…(n-r+1){{x}^{n-r}},\left( r\le n \right)$
$\therefore {{f}^{n}}(x)=n!$
$L.H.S.=f(1)+\frac{f'(1)}{1!}+\frac{f”(1)}{2!}+\frac{f”'(1)}{3!}+…+\frac{{{f}^{n}}(1)}{n!}$
$=1+\frac{n}{1!}+\frac{n(n-1)}{2!}+\frac{n(n-1)(n-2)}{3!}+…….+\frac{n!}{n!}$
$={}^{n}{{C}_{0}}+{}^{n}{{C}_{1}}+{}^{n}{{C}_{2}}+{}^{n}{{C}_{3}}+….+{}^{n}{{C}_{n}}={{2}^{n}}=R.H.S.$
Question 05
$If,y=\frac{{{\sin }^{-1}}x}{\sqrt{1-{{x}^{2}}}},|x|<1,show-that$
$\left( 1-{{x}^{2}} \right){{y}_{n+2}}-\left( 2n+3 \right)x{{y}_{n+1}}-{{\left( n+1 \right)}^{2}}{{y}_{n}}=0$
Solution:
$Given,y=\frac{{{\sin }^{-1}}x}{\sqrt{1-{{x}^{2}}}},|x|<1$
$\Rightarrow \left( 1-{{x}^{2}} \right){{y}^{2}}={{\left( {{\sin }^{-1}}x \right)}^{2}}$
Now, differentiating w.r.t. x we get,
$\left( 1-{{x}^{2}} \right).2y{{y}_{1}}-2x{{y}^{2}}=2{{\sin }^{-1}}x.\frac{1}{\sqrt{1-{{x}^{2}}}}=2y$
$\Rightarrow \left( 1-{{x}^{2}} \right){{y}_{1}}-xy=1,assu\min g,y\ne 0$
Again, differentiating w.r.t. x we get,
$\left( 1-{{x}^{2}} \right){{y}_{2}}-2x{{y}_{1}}-x{{y}_{1}}-1.y=0$
$\Rightarrow \left( 1-{{x}^{2}} \right){{y}_{2}}-3x{{y}_{1}}-y=0……….(1)$
Now applying Leibnitz’s theorem to differentiate the above equation n times w.r.t. x we get,
${{D}^{n}}\left\{ \left( 1-{{x}^{2}} \right){{y}_{2}} \right\}-{{D}^{n}}\left\{ 3x{{y}_{1}} \right\}-{{y}_{n}}=0$
$\Rightarrow \left\{ \left( 1-{{x}^{2}} \right){{y}_{n+2}}+{}^{n}{{C}_{1}}.{{y}_{n+1}}.(-2x)+{}^{n}{{C}_{2}}.{{y}_{n}}.(-2)+0 \right\}$
$-3\left\{ x{{y}_{n+1}}+{}^{n}{{C}_{1}}.{{y}_{n}}.1+0 \right\}-{{y}_{n}}=0$
$\Rightarrow \left( 1-{{x}^{2}} \right){{y}_{n+2}}-2nx{{y}_{n+1}}-n(n-1){{y}_{n}}-3x{{y}_{n+1}}-3n{{y}_{n}}-{{y}_{n}}=0$
$\Rightarrow \left( 1-{{x}^{2}} \right){{y}_{n+2}}-\left( 2n+3 \right)x{{y}_{n+1}}-{{\left( n+1 \right)}^{2}}{{y}_{n}}=0$ | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.974647581577301, "perplexity": 974.1949880021784}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-40/segments/1600402101163.62/warc/CC-MAIN-20200930013009-20200930043009-00105.warc.gz"} |
https://arxiv.org/abs/1409.3382 | hep-ex
(what is this?)
# Title:Top quark pair production cross section at LHC in ATLAS
Abstract: Measurements of the top quark production cross section in proton-proton collisions with the ATLAS detector at the Large Hadron Collider are presented. The measurements require no, one or two electrons or muons in the final state (single lepton, dilepton, hadronic channel). In addition, the decay modes with tau leptons are tested (channels with tau leptons). The main focus is on measurements of differential spectra of $t\bar{t}$ final states, in particular, measurements that are able to constrain the modelling of additional parton radiation like the jet multiplicity distribution.
Comments: 6 pages, 8 figures, LHCP2014 conference proceedings Subjects: High Energy Physics - Experiment (hep-ex) Cite as: arXiv:1409.3382 [hep-ex] (or arXiv:1409.3382v2 [hep-ex] for this version)
## Submission history
From: John Morris Ph.D [view email]
[v1] Thu, 11 Sep 2014 10:51:08 UTC (149 KB)
[v2] Fri, 24 Oct 2014 13:38:15 UTC (75 KB) | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9884721636772156, "perplexity": 4482.000875418905}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-04/segments/1547583658681.7/warc/CC-MAIN-20190117020806-20190117042806-00618.warc.gz"} |
http://mathoverflow.net/questions/121187/reference-for-rigid-analytic-gaga | # Reference for rigid analytic GAGA
I'm looking for a reference for the following result.
Theorem. Let $K$ be a complete, non-archimedean field, and let $X/K$ be a projective scheme, with analytification $X^\mathrm{an}$. Then the analytification functor from coherent $\mathcal{O}_X$-modules to coherent ${\mathcal{O}}_{X^\mathrm{an}}$-modules is an equivalence of categories.
While I've seen this sort of statement in a lot of introductory notes on rigid analytic geometry (most attributing it to Keihl), none of them seem to give a published reference. Any help would be much appreciated.
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Have you looked at the papers\notes of Brian Conrad? – Keenan Kidwell Feb 8 '13 at 14:08
His notes on rigid geometry didn't seem to have a reference. Rooting around some of his papers has done the job though, thanks! – ChrisLazda Feb 8 '13 at 14:28
## 1 Answer
I am quite surprised by the attribution to Kiehl that you saw. Anyway, I think the result is due to Ursula Köpf (not only over a field $K$ but actually over an affinoid space): "Über eigentliche Familien algebraischer Varietäten über affinoiden Räumen", Schriftenreihe Univ. Münster, 2 Serie, Heft 7 (1974).
Brian Conrad gave another proof as an application of his results of relative ampleness in the rigid analytic setting (see "Relative ampleness in rigid geometry", Ann. Inst. Fourier (Grenoble) 56 (2006), n° 4).
I also learned a proof from Antoine Ducros in the setting of Berkovich spaces. I wrote in down in an appendix to my paper "Raccord sur les espaces de Berkovich", Algebra & Number Theory 4 (2010), n° 3). It is very close to Serre's proof in the complex analytic setting and probably very close to Köpf's proof too, but I cannot say for sure since I never saw her paper.
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Great, thanks. It was Köpf's paper that eventually came up after digging through Conrad's papers. – ChrisLazda Feb 8 '13 at 22:45
Perhaps the attribution to Kiehl (which I agree is misplaced) is due to the fact that in the rigid-analytic case the main "new" ingredient is setting up a robust theory of coherent sheaves mixing features of the scheme-theoretic and complex-analytic cases, and that the creation of that theory of coherence in the rigid-analytic case is unquestionably due to Kiehl? Indeed, once that is available, the argument of Serre more-or-less carries over as Jerome says (with a bit of care due to non-rational points). Of course, Kopf goes a bit further, with a result over affinoid bases. – user28172 Feb 9 '13 at 23:17
One other clarification (as J\'erome is certainly well aware): it is K\"opf's generalization of GAGA over an affinoid base that is proved in another way in the relative ampleness paper (the case over a field is taken as input). Over a field, as in the original question above, then probably there is only one proof, namely transporting Serre's method using Kiehl's theory of coherence. (One could make another proof using formal schemes and formal GAGA when the base field is discretely-valued, but for a non-noetherian valuation ring it would be a much harder proof than via Serre's method.) – user28172 Feb 9 '13 at 23:24
Thanks for the clarifications, nosr! – Jérôme Poineau Feb 10 '13 at 20:52
Could I ask whether this paper of Köpf answers my question mathoverflow.net/questions/121881/…? It doesn't seem to be so easy to obtain a copy. – Simon Wadsley Feb 20 '13 at 22:08 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8681047558784485, "perplexity": 1015.9461466490861}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-26/segments/1466783399106.96/warc/CC-MAIN-20160624154959-00022-ip-10-164-35-72.ec2.internal.warc.gz"} |
https://practicepaper.in/gate-ce/structural-analysis | # Structural Analysis
Question 1
The linearly elastic planar structure shown in the figure is acted upon by two vertical concentrated forces. The horizontal beams UV and WX are connected with the help of the vertical linear spring with spring constant k = 20 kN/m. The fixed supports are provided at U and X. It is given that flexural rigidity $EI = 10^5 kN-m^2, P = 100 kN,$ and $a = 5 m$. Force Q is applied at the center of beam WX such that the force in the spring VW becomes zero.
The magnitude of force Q (in kN) is ________. (round off to the nearest integer)
A 640 B 458 C 251 D 325
GATE CE 2022 SET-2 Methods of Structural Analysis
Question 1 Explanation:
If force in spring is zero, there will be no deformation in spring i.e., deflection of point V will be equal to deflection of point W
\begin{aligned} \delta _v&=\delta _w\\ \frac{P(2a)^3}{3EI}&=\frac{Q(a)^3}{3(2EI)}+\frac{Q(a)^2}{2(2EI)} +a\\ \frac{8}{3}P&=\left ( \frac{1}{6}+\frac{1}{4} \right )Q\\ Q&=\frac{32}{5}P=640kN \end{aligned}
Question 2
Which of the following statement(s) is/are correct?
A If a linearly elastic structure is subjected to a set of loads, the partial derivative of the total strain energy with respect to the deflection at any point is equal to the load applied at that point. B If a linearly elastic structure is subjected to a set of loads, the partial derivative of the total strain energy with respect to the load at any point is equal to the deflection at that point. C If a structure is acted upon by two force system Pa and Pb , in equilibrium separately, the external virtual work done by a system of forces Pb during the deformations caused by another system of forces Pa is equal to the external virtual work done by the Pa system during the deformation caused by the Pb system. D The shear force in a conjugate beam loaded by the M/EI diagram of the real beam is equal to the corresponding deflection of the real beam.
GATE CE 2022 SET-2 Methods of Structural Analysis
Question 3
Consider a beam PQ fixed at P, hinged at Q, and subjected to a load F as shown in figure (not drawn to scale). The static and kinematic degrees of indeterminacy, respectively, are
A 2 and 1 B 2 and 0 C 1 and 2 D 2 and 2
GATE CE 2022 SET-2 Determinacy and Indeterminacy
Question 3 Explanation:
Static indeterminacy, SI=r-3=(3+2)-3=2
Kinematic indeterminacy=0+1=1
Question 4
Consider a simply supported beam PQ as shown in the figure. A truck having 100 kN on the front axle and 200 kN on the rear axle, moves from left to right. The spacing between the axles is 3 m. The maximum bending moment at point R is ______ kNm. (in integer)
A 124 B 180 C 147 D 582
GATE CE 2022 SET-1 Influence Line Diagram and Rolling Loads
Question 4 Explanation:
$\frac{ab}{L}=\frac{1 \times 4}{5}=0.8m$
To get maximum BN at R
\begin{aligned} BM_{max}&=200 \times \frac{ab}{L}+100 \times y\\ \frac{ab/l}{b}&=\frac{y}{4-3}\Rightarrow \frac{0.8}{4}=\frac{y}{1}=0.2m\\ BM_{max}&=200 \times 0.8+100 \times 0.2=180kNm \end{aligned}
Question 5
Consider the linearly elastic plane frame shown in the figure. Members HF, FK and FG are welded together at joint F. Joints K, G and H are fixed supports. A counter-clockwise moment M is applied at joint F. Consider flexural rigidity EI=$10^5kN-m^2$ for each member and neglect axial deformations.
If the magnitude (absolute value) of the support moment at H is 10 kN-m, the magnitude (absolute value) of the applied moment M (in kN-m) to maintain static equilibrium is ___________. (round off to the nearest integer)
A 60 B 25 C 85 D 45
GATE CE 2022 SET-1 Methods of Structural Analysis
Question 5 Explanation:
$\frac{M}{6}=10kNm\Rightarrow M=60kNm$
Question 6
The plane truss shown in the figure is subjected to an external force $P$. It is given that $P = 70 kN, a = 2 m,$ and $b = 3 m$.
The magnitude (absolute value) of force (in kN) in member $EF$ is _______. (round off to the nearest integer)
A 12 B 30 C 87 D 92
GATE CE 2022 SET-1 Trusses
Question 6 Explanation:
\begin{aligned} \Sigma M_J&=0\\ V_A \times 8-H_A \times 1-70 \times 4&=0\\ 8V_A-H_A&=280 \;\;\;(i) \end{aligned}
To find $H_A$ (Cut the truss by 1-1)
Consider left hand side
\begin{aligned} \Sigma M_E&=0\\ V_A \times 4-H_A \times 4&=0\\ V_A&=H_A \;\;\;(ii)\\ \text{using (i) and (ii)}&\\ 8V_A-V_A&=280\\ 7V_A&=280\\ V_A&=40kN\\ H_A&=40kN\\ H_J&=40kN\\ V_J&=70-40=30kN\\ \end{aligned}
To find force in member EF (Cit the truss by 2-2)
Consider right hand side
Force in member EF
$F_{EF}=V_J=30kN$
Question 7
A semi-circular bar of radius R m, in a vertical plane, is fixed at the end G, as shown in the figure. A horizontal load of magnitude P kN is applied at the end H. The magnitude of the axial force, shear force, and bending moment at point Q for $\theta =45 ^{\circ}$, respectively, are
A $\frac{P}{\sqrt{2}}kN,\frac{P}{\sqrt{2}}kN \text{ and } \frac{PR}{\sqrt{2}}kNm$ B $\frac{P}{\sqrt{2}}kN,\frac{P}{\sqrt{2}}kN \text{ and } 0 \; kNm$ C $0 \; kN,\frac{P}{\sqrt{2}}kN \text{ and } \frac{PR}{\sqrt{2}}kNm$ D $\frac{P}{\sqrt{2}}kN,0 \; kN \text{ and } \frac{PR}{\sqrt{2}}kNm$
GATE CE 2022 SET-1 Arches
Question 7 Explanation:
\begin{aligned} (F_Q)&=P\sin \theta =\frac{P}{\sqrt{2}}&(\; at\; \theta =45^{\circ})\\ (S_Q)&=P\cos \theta =\frac{P}{\sqrt{2}}&(\; at\; \theta =45^{\circ})\\ M_Q&=PR \sin \theta =\frac{PR}{\sqrt{2}}&( \; at\; \theta =45^{\circ}) \end{aligned}
Question 8
A perfectly flexible and inextensible cable is shown in the figure (not to scale). The external loads at F and G are acting vertically.
The magnitude of tension in the cable segment FG (in kN, round off to two decimal places) is _______
A 4.68 B 3.78 C 5.64 D 8.25
GATE CE 2021 SET-2 Arches
Question 8 Explanation:
\begin{aligned} \Sigma M_F=0\\ \left ( 10.667+\frac{H}{6} \right )2-H \times 3=0\\ H=8\; kN\\ V_E=10.667+\frac{8}{6}\\ V_E=12\; kN\\ V_H=10\; kN \end{aligned}
Consider left side of section (1)-(1)
\begin{aligned} \Sigma H=0\\ T \cos \theta =H\\ T \cos \theta =8\\ T \sin \theta =2\\ \therefore \;\; T^2 \cos ^2\theta + T^2 \sin ^2 \theta =8^2+2^2\\ T=8.246\; kN \end{aligned}
Tension in segment GF is 8.246 kN.
Question 9
A prismatic fixed-fixed beam, modelled with a total lumped-mass of 10 kg as a single degree of freedom (SDOF) system is shown in the figure.
If the flexural stiffness of the beam is $4 \pi^{2} \mathrm{kN} / \mathrm{m}$, its natural frequency of vibration (in Hz, in integer) in the flexural mode will be _______
A 8 B 16 C 10 D 18
GATE CE 2021 SET-2 Matrix Method of Structural Analysis
Question 9 Explanation:
Given Data:
\begin{aligned} m&=10\; kg\\ K&=4 \pi^2 \times 10^3\\ &\text{Natural frequency,}\\ f_n&=\frac{1}{2 \pi} \sqrt{\frac{k}{m}}\\ &=\frac{1}{2 \pi} \sqrt{\frac{4 \pi ^2 \times 10^3}{10}}\\ &=\frac{2 \pi \times 10}{2 \pi}=10 \end{aligned}
Question 10
A frame EFG is shown in the figure. All members are prismatic and have equal flexural rigidity. The member FG carries a uniformly distributed load w per unit length. Axial deformation of any member is neglected.
Considering the joint F being rigid, the support reaction at G is
A 0.375 wL B 0.453 wL C 0.482 wL D 0.500 wL
GATE CE 2021 SET-2 Methods of Structural Analysis
Question 10 Explanation:
Compatibility condition
$\begin{array}{l} \qquad \qquad \qquad \qquad \qquad \frac{\partial U}{\partial R}=0 \\ \Rightarrow \qquad \qquad \frac{\partial}{\partial R}\left[\int \frac{M^{2}}{2 E \mid} d x\right]=0 \\ \Rightarrow \frac{\partial}{\partial R}\left[\int_{0}^{L} \frac{\left(R x-\frac{w x^{2}}{2}\right)^{2}}{2 E I} d x+\int_{0}^{2 L} \frac{\left(R L-\frac{w L^{2}}{2}\right)^{2}}{2 E I} d x\right]=0 \\ \Rightarrow \int_{0}^{L} \frac{2\left(R x-\frac{w x^{2}}{2}\right)}{2 E I}(x) d x+\int_{0}^{2 L} \frac{\left(R L-\frac{w L^{2}}{2}\right)}{E I}(L) d x=0 \\ \Rightarrow \frac{R L^{3}}{3}-\frac{W}{2} \times \frac{L^{4}}{4}+R L^{2}(2 L)-\frac{W^{3}}{2}(2 L)=0 \\ \Rightarrow \frac{R L^{3}}{3}-\frac{w^{4}}{8}+2 R L^{3}-w L^{4}=0 \\ \Rightarrow 2 R L^{3}+\frac{R L^{3}}{3}=\frac{w^{4}}{8}+w^{4}=\frac{9}{8} w^{4} \\ \Rightarrow \frac{7 R L^{3}}{3}=\frac{9}{8} w L^{4} \\ \Rightarrow R=\frac{27}{56} w L=0.482 w L \end{array}$
Alternative
\begin{aligned} \frac{w L^{4}}{8 E I}+\frac{w L^{2}(2 L)}{2 E I} \times L &=\frac{R L^{3}}{3 E I}+\frac{R L \times 2 L \times L}{E I} \\ \frac{w^{4}}{8 E I}+\frac{w^{4}}{E I} &=\frac{R L^{3}}{3 E I}+\frac{2 R L^{3}}{E I} \\ \frac{9 W^{4}}{8 E I} &=\frac{7 R L^{3}}{3 E I} \\ R &=\frac{27 w L}{56}=0.482 \mathrm{wL} \end{aligned}
There are 10 questions to complete.
### 1 thought on “Structural Analysis”
1. if you also provide gate calculator along with the questions it will be really helpfull for many aspirants | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 27, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9973716139793396, "perplexity": 3615.2664862590286}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-06/segments/1674764499934.48/warc/CC-MAIN-20230201112816-20230201142816-00036.warc.gz"} |
https://www.physicsforums.com/threads/physics-of-golf.114609/ | # Physics of golf
1. Mar 17, 2006
### sauri
A golf ball is putted too fast and rolls past the hole in a straight line, coming at closest approach to a distance d from the hole. Ignore friction and assume that golf ball moves at constant velocity.
What is the angular momentum of the golf ball about the hole?
I belive that the angular momentum is constant but its direction changes due to an external torque. Yet the equation to use is beyond me could it be d(theta)=dL/L?
2. Mar 17, 2006
### Da-Force
Actually... When we say a golf ball has a constant velocity, all the forces must add up to be zero.
Start with that by drawing components then doing a sum in the X and Y directions.
3. Mar 17, 2006
### Staff: Mentor
Consider the definition of the angular momentum of a particle:
$$\vec{L} = \vec{r} \times \vec{p}$$
Similar Discussions: Physics of golf | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8971157670021057, "perplexity": 943.365052267873}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-13/segments/1490218186530.52/warc/CC-MAIN-20170322212946-00534-ip-10-233-31-227.ec2.internal.warc.gz"} |
http://www.ntg.nl/pipermail/ntg-context/2006/019638.html | # [NTG-context] small caps italic and font switching inside math
Taco Hoekwater taco at elvenkind.com
Sun Jul 2 10:44:23 CEST 2006
Mojca Miklavec wrote:
>
>>>It works perfect except in a single case: \title{\molecule{SF_6}}
>>
>>I had seen that, but not yet bothered to fix it. Still, it is
>>fairly easy to change the macro, try the version below.
Sorry, I thought you were talking about the spacing between F and 6.
> Didn't work in titles either (or I did something strange) :(
> But If I write a couple of explicit \lohi-s, it will still be OK.
Good, but it can be fixed, by changing the definition of
\domolecule to:
\def\domolecule#1%
{\expandafter\scantokens\expandafter
{\detokenize{#1\finishchem}}\egroup}
This re-tokenizes the argument (needed because it was grabbed
catcodes.)
> I didn't really understand the \iffluor-part of the code ... but don't
> bother too much.
It is there to trigger a negative italic superscript correction
(TeX doesn't have a primitive for that :-))
> Thanks a lot for the trickery again (I'm still impressed by the
> \uppercase part),
That is actually a fairly standard trick, not something I invented
Greetings, Taco | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9747055768966675, "perplexity": 4826.695675802927}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-49/segments/1416931011032.12/warc/CC-MAIN-20141125155651-00047-ip-10-235-23-156.ec2.internal.warc.gz"} |
http://mathhelpforum.com/discrete-math/177039-question-about-partially-ordered-sets.html | Math Help - Question about partially ordered sets
1. Question about partially ordered sets
Is (S, R) a poset if S is the set of all people in the world and $(a, b) \in R$, where a and b are people if a = b or a is an ancestor of b?
I know that I have to determine if this is reflexive, antisymmetric and transitive.
So I start with reflexivity and imagine that I have two people, I can either pick two people who are identical (a = b) or I can pick a person a who is an ancestor of a person b.
If I pick two who are identical then it is trivially reflexive, but if I choose a person a who is an ancestor of b, then it is not reflexive, as obviously a cannot be ancestor of him or herself.
Thus it is not reflexive and is not a poset either.
Yet, I see that in the solution for this problem that it is indeed reflexive, how is that possible?
2. Originally Posted by posix_memalign
Is (S, R) a poset if S is the set of all people in the world and $(a, b) \in R$, where a and b are people if a = b or a is an ancestor of b?
Yet, I see that in the solution for this problem that it is indeed reflexive, how is that possible?
3. Originally Posted by Plato
I did but I think I fail to understand something really fundamental; is it sufficient that only some of the people maintain the a = b property for R to be reflexive? If not, how does the definition ensure that they are all equal when the very same definition states that they might not be (i.e. because of the 'or')?
4. Originally Posted by posix_memalign
I did but I think I fail to understand something really fundamental; is it sufficient that only some of the people maintain the a = b property for R to be reflexive? If not, how does the definition ensure that they are all equal when the very same definition states that they might not be (i.e. because of the 'or')?
Reflexive simply means that each element is the set is related to itself. For each $a$ in the set the pair $(a,a)$ is in the relation. That is laid in the very definition of this particular relation.
Equality is the very basic and simple reflexive relation.
5. Let 'Rab' stand for '<a b> in R'.
Let 'Cab' stand for 'a is an ancestor of b'.
The problem, as you stated it, is a bit deceptive.
Ordinarily, we would think we are given:
(1) Rab if and only if (a=b or Cab)
But the problem, as you stated it, actually gives only:
(2) If a=b or Cab, then Rab.
Now we have to check:
Reflexive: Is it the case that, for any a, we have Raa?
For (1) or (2) it's easy.
Antisymmetric: Is it the case that, for any a and b, we have 'if Rab and Rba, then a=b'?
With (1) it's easy. But with (2) we get a different answer.
Transitive: Is it the case that, for any a, b, and c, we have 'Rab and Rbc, then Rac'?
With (1) it's easy. But with (2) we get a different answer.
6. Originally Posted by Plato
To be precise, as the problem is stated, no definition was given. A definition would be a biconditional, but the statement of the problem gives only a conditional. It is not clear that a biconditional was what was actually intended.
7. Originally Posted by MoeBlee
To be precise, as the problem is stated, no definition was given. A definition would be a biconditional, but the statement of the problem gives only a conditional. It is not clear that a biconditional was what was actually intended.
I totally disagree with that.
The relation was clearly defined in the OP.
8. The wording given was:
<a b> in R IF a=b or a is an ancestor of b.
That is, (a=b or a is an ancestor of b) -> <a b> in R.
The wording was not given as:
<a b> in R if AND ONLY IF a=b or a is an ancestor of b.
9. Originally Posted by Plato
Reflexive simply means that each element is the set is related to itself. For each $a$ in the set the pair $(a,a)$ is in the relation. That is laid in the very definition of this particular relation.
Equality is the very basic and simple reflexive relation.
Ah, I see ... however, I thought I finally understood this but even when it means that a and b should only be related, I still don't see how this can be true.
There are two cases, if a = a, then they are obviously related, no problem.
But the case a = a does not necessarily hold for every element in the set? For at least some of them there is the possibility that a is the ancestor of b, and for this particular a, how does the reflexive property hold? Doesn't it have to hold for every element in the set?
10. Originally Posted by posix_memalign
But the case a = a does not necessarily hold for every element in the set? For at least some of them there is the possibility that a is the ancestor of b, and for this particular a, how does the reflexive property hold? Doesn't it have to hold for every element in the set?
You're mixed up about the tests for the relation.
Go back to the precise tests (here, a, b, and c are assumed to be people):
Reflexive: Is it true that, for any a, we have Raa?
That is, is every person either him(her)self or an ancestor of him(her)self?
Antisymmetric: Is it true that, for any a and b, we have 'if Rab and Rba, then a=b'?
That is, is it true that if either a and b are ancestors of each other or a and b are the same person, then a and b are the same person?
Transitive: Is it the true that, for any a, b, and c, we have 'Rab and Rbc, then Rac'?
That is, it true that if a, b, and c are the same person or a is an ancestor of b is an ancestor of c then either a and c are the same person or a is an ancestor of c?
That's all there is to it. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 6, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9326927065849304, "perplexity": 318.2672125362114}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-35/segments/1440644063825.9/warc/CC-MAIN-20150827025423-00121-ip-10-171-96-226.ec2.internal.warc.gz"} |
https://en.wikipedia.org/wiki/Catastrophic_event | Catastrophe theory
(Redirected from Catastrophic event)
"Catastrophic event" redirects here. For other uses, see Catastrophe.
In mathematics, catastrophe theory is a branch of bifurcation theory in the study of dynamical systems; it is also a particular special case of more general singularity theory in geometry.
Bifurcation theory studies and classifies phenomena characterized by sudden shifts in behavior arising from small changes in circumstances, analysing how the qualitative nature of equation solutions depends on the parameters that appear in the equation. This may lead to sudden and dramatic changes, for example the unpredictable timing and magnitude of a landslide.
Catastrophe theory originated with the work of the French mathematician René Thom in the 1960s, and became very popular due to the efforts of Christopher Zeeman in the 1970s. It considers the special case where the long-run stable equilibrium can be identified with the minimum of a smooth, well-defined potential function (Lyapunov function).
Small changes in certain parameters of a nonlinear system can cause equilibria to appear or disappear, or to change from attracting to repelling and vice versa, leading to large and sudden changes of the behaviour of the system. However, examined in a larger parameter space, catastrophe theory reveals that such bifurcation points tend to occur as part of well-defined qualitative geometrical structures.
Elementary catastrophes
Catastrophe theory analyses degenerate critical points of the potential function — points where not just the first derivative, but one or more higher derivatives of the potential function are also zero. These are called the germs of the catastrophe geometries. The degeneracy of these critical points can be unfolded by expanding the potential function as a Taylor series in small perturbations of the parameters.
When the degenerate points are not merely accidental, but are structurally stable, the degenerate points exist as organising centres for particular geometric structures of lower degeneracy, with critical features in the parameter space around them. If the potential function depends on two or fewer active variables, and four (resp. five) or fewer active parameters, then there are only seven (resp. eleven) generic structures for these bifurcation geometries, with corresponding standard forms into which the Taylor series around the catastrophe germs can be transformed by diffeomorphism (a smooth transformation whose inverse is also smooth).[citation needed] These seven fundamental types are now presented, with the names that Thom gave them.
Potential functions of one active variable
Fold catastrophe
Stable and unstable pair of extrema disappear at a fold bifurcation
$V = x^3 + ax\,$
At negative values of a, the potential has two extrema - one stable, and one unstable. If the parameter a is slowly increased, the system can follow the stable minimum point. But at a = 0 the stable and unstable extrema meet, and annihilate. This is the bifurcation point. At a > 0 there is no longer a stable solution. If a physical system is followed through a fold bifurcation, one therefore finds that as a reaches 0, the stability of the a < 0 solution is suddenly lost, and the system will make a sudden transition to a new, very different behaviour. This bifurcation value of the parameter a is sometimes called the "tipping point".
Cusp catastrophe
$V = x^4 + ax^2 + bx \,$
Diagram of cusp catastrophe, showing curves (brown, red) of x satisfying dV/dx = 0 for parameters (a,b), drawn for parameter b continuously varied, for several values of parameter a. Outside the cusp locus of bifurcations (blue), for each point (a,b) in parameter space there is only one extremising value of x. Inside the cusp, there are two different values of x giving local minima of V(x) for each (a,b), separated by a value of x giving a local maximum. Cusp shape in parameter space (a,b) near the catastrophe point, showing the locus of fold bifurcations separating the region with two stable solutions from the region with one. Pitchfork bifurcation at a = 0 on the surface b = 0
The cusp geometry is very common, when one explores what happens to a fold bifurcation if a second parameter, b, is added to the control space. Varying the parameters, one finds that there is now a curve (blue) of points in (a,b) space where stability is lost, where the stable solution will suddenly jump to an alternate outcome.
But in a cusp geometry the bifurcation curve loops back on itself, giving a second branch where this alternate solution itself loses stability, and will make a jump back to the original solution set. By repeatedly increasing b and then decreasing it, one can therefore observe hysteresis loops, as the system alternately follows one solution, jumps to the other, follows the other back, then jumps back to the first.
However, this is only possible in the region of parameter space a < 0. As a is increased, the hysteresis loops become smaller and smaller, until above a = 0 they disappear altogether (the cusp catastrophe), and there is only one stable solution.
One can also consider what happens if one holds b constant and varies a. In the symmetrical case b = 0, one observes a pitchfork bifurcation as a is reduced, with one stable solution suddenly splitting into two stable solutions and one unstable solution as the physical system passes to a < 0 through the cusp point (0,0) (an example of spontaneous symmetry breaking). Away from the cusp point, there is no sudden change in a physical solution being followed: when passing through the curve of fold bifurcations, all that happens is an alternate second solution becomes available.
A famous suggestion is that the cusp catastrophe can be used to model the behaviour of a stressed dog, which may respond by becoming cowed or becoming angry.[1] The suggestion is that at moderate stress (a > 0), the dog will exhibit a smooth transition of response from cowed to angry, depending on how it is provoked. But higher stress levels correspond to moving to the region (a < 0). Then, if the dog starts cowed, it will remain cowed as it is irritated more and more, until it reaches the 'fold' point, when it will suddenly, discontinuously snap through to angry mode. Once in 'angry' mode, it will remain angry, even if the direct irritation parameter is considerably reduced.
A simple mechanical system, the "Zeeman Catastrophe Machine", nicely illustrates a cusp catastrophe. In this device, smooth variations in the position of the end of a spring can cause sudden changes in the rotational position of an attached wheel.[2]
Catastrophic failure of a complex system with parallel redundancy can be evaluated based on relationship between local and external stresses. The model of the structural fracture mechanics is similar to the cusp catastrophe behavior. The model predicts reserve ability of a complex system.
Other applications include the outer sphere electron transfer frequently encountered in chemical and biological systems[3] and modelling Real Estate Prices.[4]
Fold bifurcations and the cusp geometry are by far the most important practical consequences of catastrophe theory. They are patterns which reoccur again and again in physics, engineering and mathematical modelling. They produce the strong gravitational lensing events and provide astronomers with one of the methods used for detecting black holes and the dark matter of the universe, via the phenomenon of gravitational lensing producing multiple images of distant quasars. [5]
The remaining simple catastrophe geometries are very specialised in comparison, and presented here only for curiosity value.
Swallowtail catastrophe
Swallowtail catastrophe surface
$V = x^5 + ax^3 + bx^2 + cx \,$
The control parameter space is three-dimensional. The bifurcation set in parameter space is made up of three surfaces of fold bifurcations, which meet in two lines of cusp bifurcations, which in turn meet at a single swallowtail bifurcation point.
As the parameters go through the surface of fold bifurcations, one minimum and one maximum of the potential function disappear. At the cusp bifurcations, two minima and one maximum are replaced by one minimum; beyond them the fold bifurcations disappear. At the swallowtail point, two minima and two maxima all meet at a single value of x. For values of a>0, beyond the swallowtail, there is either one maximum-minimum pair, or none at all, depending on the values of b and c. Two of the surfaces of fold bifurcations, and the two lines of cusp bifurcations where they meet for a<0, therefore disappear at the swallowtail point, to be replaced with only a single surface of fold bifurcations remaining. Salvador Dalí's last painting, The Swallow's Tail, was based on this catastrophe.
Butterfly catastrophe
$V = x^6 + ax^4 + bx^3 + cx^2 + dx \,$
Depending on the parameter values, the potential function may have three, two, or one different local minima, separated by the loci of fold bifurcations. At the butterfly point, the different 3-surfaces of fold bifurcations, the 2-surfaces of cusp bifurcations, and the lines of swallowtail bifurcations all meet up and disappear, leaving a single cusp structure remaining when a>0
Potential functions of two active variables
Umbilic catastrophes are examples of corank 2 catastrophes. They can be observed in optics in the focal surfaces created by light reflecting off a surface in three dimensions and are intimately connected with the geometry of nearly spherical surfaces. Thom proposed that the Hyperbolic umbilic catastrophe modeled the breaking of a wave and the elliptical umbilic modeled the creation of hair like structures.
Hyperbolic umbilic catastrophe
$V = x^3 + y^3 + axy + bx +cy \,$
Elliptic umbilic catastrophe
$V = \frac{x^3}{3} - xy^2 + a(x^2+y^2) + bx + cy \,$
Parabolic umbilic catastrophe
$V = x^2y + y^4 + ax^2 + by^2 + cx + dy \,$
Arnold's notation
Vladimir Arnold gave the catastrophes the ADE classification, due to a deep connection with simple Lie groups.
• A0 - a non-singular point: $V = x$.
• A1 - a local extremum, either a stable minimum or unstable maximum $V = \pm x^2 + a x$.
• A2 - the fold
• A3 - the cusp
• A4 - the swallowtail
• A5 - the butterfly
• Ak - a representative of an infinite sequence of one variable forms $V=x^{k+1}+\cdots$
• D4 - the elliptical umbilic
• D4+ - the hyperbolic umbilic
• D5 - the parabolic umbilic
• Dk - a representative of an infinite sequence of further umbilic forms
• E6 - the symbolic umbilic $V = x^3+y^4+a x y^2 +bxy+cx+dy+ey^2$
• E7
• E8
There are objects in singularity theory which correspond to most of the other simple Lie groups.
References
1. ^ E.C. Zeeman, Catastrophe Theory, Scientific American, April 1976; pp. 65–70, 75–83
2. ^ Cross, Daniel J., Zeeman's Catastrophe Machine in Flash
3. ^ Xu, F (1990). "Application of catastrophe theory to the ∆G to -∆G relationship in electron transfer reactions.". Zeitschrift für Physikalische Chemie Neue Folge 166: 79–91.
4. ^ Bełej, Mirosław; Kulesza, Sławomir (1 January 2012). "Modeling the Real Estate Prices in Olsztyn under Instability Conditions". Folia Oeconomica Stetinensia 11 (1). doi:10.2478/v10031-012-0008-7.
5. ^ A.O. Petters, H. Levine and J. Wambsganss, Singularity Theory and Gravitational Lensing", Birkhäuser Boston (2001)
Bibliography
• Arnold, Vladimir Igorevich. Catastrophe Theory, 3rd ed. Berlin: Springer-Verlag, 1992.
• V. S. Afrajmovich, V. I. Arnold, et al., Bifurcation Theory And Catastrophe Theory, ISBN 3-540-65379-1
• Bełej,M. Kulesza, S. Modeling the Real Estate Prices in Olsztyn under Instability Conditions. Folia Oeconomica Stetinensia. Volume 11, Issue 1, Pages 61–72, ISSN (Online) 1898-0198, ISSN (Print) 1730-4237, DOI: 10.2478/v10031-012-0008-7, 2013
• Castrigiano, Domenico P. L. and Hayes, Sandra A. Catastrophe Theory, 2nd ed. Boulder: Westview, 2004. ISBN 0-8133-4126-4
• Gilmore, Robert. Catastrophe Theory for Scientists and Engineers. New York: Dover, 1993.
• Petters, Arlie O., Levine, Harold and Wambsganss, Joachim. Singularity Theory and Gravitational Lensing. Boston: Birkhäuser, 2001. ISBN 0-8176-3668-4
• Postle, Denis. Catastrophe Theory – Predict and avoid personal disasters. Fontana Paperbacks, 1980. ISBN 0-00-635559-5
• Poston, Tim and Stewart, Ian. Catastrophe: Theory and Its Applications. New York: Dover, 1998. ISBN 0-486-69271-X.
• Sanns, Werner. Catastrophe Theory with Mathematica: A Geometric Approach. Germany: DAV, 2000.
• Saunders, Peter Timothy. An Introduction to Catastrophe Theory. Cambridge, England: Cambridge University Press, 1980.
• Thom, René. Structural Stability and Morphogenesis: An Outline of a General Theory of Models. Reading, MA: Addison-Wesley, 1989. ISBN 0-201-09419-3.
• Thompson, J. Michael T. Instabilities and Catastrophes in Science and Engineering. New York: Wiley, 1982.
• Woodcock, Alexander Edward Richard and Davis, Monte. Catastrophe Theory. New York: E. P. Dutton, 1978. ISBN 0-525-07812-6. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 11, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8456001281738281, "perplexity": 1607.6650927967646}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-35/segments/1440644060173.6/warc/CC-MAIN-20150827025420-00071-ip-10-171-96-226.ec2.internal.warc.gz"} |
http://cac.tbistro.com.br/wait-a-ewhtc/85085a-simplify-radical-expressions-using-conjugates-calculator | # simplify radical expressions using conjugates calculator
Simplify radical expressions using the distributive property K.11. Power rule H.5. Example $$\PageIndex{1}$$ Does $$\sqrt{25} = \pm 5$$? Divide radical expressions J.9. We can simplify radical expressions that contain variables by following the same process as we did for radical expressions that contain only numbers. Case 1 : If the denominator is in the form of a ± √b or a ± c √b (where b is a rational number), th en we have to multiply both the numerator and denominator by its conjugate. Solve radical equations H.1. Solve radical equations Rational exponents. It will show the work by separating out multiples of the radicand that have integer roots. Rewrite as . Simplify expressions involving rational exponents I H.6. Domain and range of radical functions K.13. The symbol is called a radical, the term under the symbol is called the radicand, and the entire expression is called a radical expression. Polynomials - Exponent Properties Objective: Simplify expressions using the properties of exponents. Learn how to divide rational expressions having square root binomials. Steps to Rationalize the Denominator and Simplify. Simplify radical expressions using conjugates N.12. When a radical contains an expression that is not a perfect root ... You find the conjugate of a binomial by changing the sign that is between the two terms, but keep the same order of the terms. Power rule O.5. You use the inverse sign in order to make sure there is no b term when you multiply the expressions. Simplifying hairy expression with fractional exponents. Question: Evaluate the radicals. Do the same for the prime numbers you've got left inside the radical. Divide Radical Expressions. . 52/3 ⋅ 54/3 b. In essence, if you can use this trick once to reduce the number of radical signs in the denominator, then you can use this trick repeatedly to eliminate all of them. Simplify radical expressions using the distributive property G.11. Multiply radical expressions J.8. Key Concept. The conjugate of 2 – √3 would be 2 + √3. Evaluate rational exponents L.2. We will need to use this property ‘in reverse’ to simplify a fraction with radicals. 9.1 Simplifying Radical Expressions (Page 2 of 20)Consider the Sign of the Radicand a: Positive, Negative, or Zero 1.If a is positive, then the nth root of a is also a positive number - specifically the positive number whose nth power is a. e.g. Tap for more steps... Use to rewrite as . The symbol is called a radical, the term under the symbol is called the radicand, and the entire expression is called a radical expression. Multiplication with rational exponents L.3. To get rid of it, I'll multiply by the conjugate in order to "simplify" this expression. The square root obtained using a calculator is the principal square root. Radical Expressions and Equations. Simplify radical expressions using the distributive property N.11. Simplify radical expressions using the distributive property K.11. In this example, we simplify √(2x²)+4√8+3√(2x²)+√8. If a pair does not exist, the number or variable must remain in the radicand. nth roots . Multiply and . Problems with expoenents can often be simplified using a few basic exponent properties. A worked example of simplifying an expression that is a sum of several radicals. Nth roots J.5. Combine and simplify the denominator. The calculator will simplify any complex expression, with steps shown. Rewrite as . Video transcript. In general, you can skip the multiplication sign, so 5x is equivalent to 5*x. 6.Simplify radical expressions using conjugates FX7 Roots 7.Roots of integers 8RV 8.Roots of rational numbers 28Q 9.Find roots using a calculator 9E4 10.Nth roots 6NE Rational exponents 11.Evaluate rational exponents 26H 12.Operations with rational exponents NQB 13.Simplify expressions involving rational exponents 7TC P.4: Polynomials 1.Polynomial vocabulary DYB 2.Add and subtract … As we already know, when simplifying a radical expression, there can not be any radicals left in the denominator. Combine and . Then you'll get your final answer! Division with rational exponents L.4. FX7. Solution. This calculator will simplify fractions, polynomial, rational, radical, exponential, logarithmic, trigonometric, and hyperbolic expressions. Factor the expression completely (or find perfect squares). This becomes more complicated when you have an expression as the denominator. Radicals and Square roots-video tutorials, calculators, worksheets on simplifying, adding, subtracting, multipying and more In this example, we simplify √(2x²)+4√8+3√(2x²)+√8. Raise to the power of . The principal square root of $$a$$ is written as $$\sqrt{a}$$. Then evaluate each expression. Simplify Expression Calculator. You'll get a clearer idea of this after following along with the example questions below. You then need to multiply by the conjugate. SIMPLIFYING RADICAL EXPRESSIONS USING CONJUGATES . We have used the Quotient Property of Radical Expressions to simplify roots of fractions. No. Simplify radical expressions using conjugates J.12. Simplifying expressions is the last step when you evaluate radicals. Simplifying Radicals . Simplify radical expressions with variables II J.7. Simplifying radical expressions: three variables. Simplify radical expressions using conjugates G.12. Solve radical equations O.1. Simplify expressions involving rational exponents I L.6. Division with rational exponents O.4. Simplify any radical expressions that are perfect squares. Exponents represent repeated multiplication. For example, the conjugate of X+Y is X-Y, where X and Y are real numbers. If you're seeing this message, it means we're having trouble loading external resources on our website. Apply the power rule and multiply exponents, . Example $$\PageIndex{1}$$ Does $$\sqrt{25} = \pm 5$$? Simplify. Further the calculator will show the solution for simplifying the radical by prime factorization. RATIONALIZE the DENOMINATOR: explanation of terms and step by step guide showing how to rationalize a denominator containing radicals or algebraic expressions containing radicals: square roots, cube roots, . Raise to the power of . A radical expression is said to be in its simplest form if there are. no perfect square factors other than 1 in the radicand $$\sqrt{16x}=\sqrt{16}\cdot \sqrt{x}=\sqrt{4^{2}}\cdot \sqrt{x}=4\sqrt{x}$$ no … . Domain and range of radical functions K.13. Multiplication with rational exponents L.3. For example, the complex conjugate of X+Yi is X-Yi, where X is a real number and Y is an imaginary number. We give the Quotient Property of Radical Expressions again for easy reference. This online calculator will calculate the simplified radical expression of entered values. Simplifying Radical Expressions Using Conjugates - Concept - Solved Examples. Multiplication with rational exponents H.3. Example 1: Divide and simplify the given radical expression: 4/ (2 - √3) The given expression has a radical expression … Jenn, Founder Calcworkshop ® , 15+ Years Experience (Licensed & Certified Teacher) Rationalizing is the process of removing a radical from the denominator, but this only works for when we are dealing with monomial (one term) denominators. 3125is asking ()3=125 416is asking () 4=16 2.If a is negative, then n must be odd for the nth root of a to be a real number. . . Multiply by . Domain and range of radical functions G.13. The conjugate refers to the change in the sign in the middle of the binomials. To rationalize, the given expression is multiplied and divided by its conjugate. In general, you can skip parentheses, but be very careful: e^3x is e^3x, and e^(3x) is e^(3x). This algebra video tutorial shows you how to perform many operations to simplify radical expressions. Show Instructions. We're asked to rationalize and simplify this expression right over here and like many problems there are multiple ways to do this. ... Then you can repeat the process with the conjugate of a+b*sqrt(30) and (a+b*sqrt(30))(a-b*sqrt(30)) is rational. +1 Solving-Math-Problems Page Site. In case of complex numbers which involves a real and an imaginary number, it is referred to as complex conjugate. It will perform addition, subtraction, multiplication, division, raising to power, and also will find the polar form, conjugate, modulus and inverse of the complex number. . Simplify radical expressions with variables I J.6. For every pair of a number or variable under the radical, they become one when simplified. Use the power rule to combine exponents. A worked example of simplifying an expression that is a sum of several radicals. Domain and range of radical functions N.13. Power rule L.5. Simplify radical expressions using the distributive property J.11. Solve radical equations L.1. a. Next lesson. Evaluate rational exponents L.2. Step 2: Multiply the numerator and the denominator of the fraction by the conjugate found in Step 1 . The multiplication of the denominator by its conjugate results in a whole number (okay, a negative, but the point is that there aren't any radicals): . L.1. Evaluate rational exponents H.2. If you like this Site about Solving Math Problems, please let Google know by clicking the +1 button. Solution. The denominator here contains a radical, but that radical is part of a larger expression. Add and subtract radical expressions J.10. Don't worry that this isn't super clear after reading through the steps. The principal square root of $$a$$ is written as $$\sqrt{a}$$. Use a calculator to check your answers. M.11 Simplify radical expressions using conjugates. We will use this fact to discover the important properties. Example problems . These properties can be used to simplify radical expressions. Simplify expressions involving rational exponents I O.6. Use the properties of exponents to write each expression as a single radical. Power rule L.5. Add and . Find roots using a calculator J.4. The online tool used to divide the given radical expressions is called dividing radical expressions calculator. Simplify radical expressions using conjugates K.12. Simplify radical expressions using conjugates K.12. Division with rational exponents H.4. Evaluate rational exponents O.2. 31/5 ⋅ 34/5 c. (42/3)3 d. (101/2)4 e. 85/2 — 81/2 f. 72/3 — 75/3 Simplifying Products and Quotients of Radicals Work with a partner. Exponential vs. linear growth. to rational exponents by simplifying each expression. No. Cancel the common factor of . a + b and a - b are conjugates of each other. Share skill Multiplication with rational exponents O.3. a + √b and a - √b are conjugate to each other. Free math problem solver answers your algebra, geometry, trigonometry, calculus, and statistics homework questions with step-by-step explanations, just like a math tutor. Calculator Use. The square root obtained using a calculator is the principal square root. Division with rational exponents L.4. This example, the conjugate of 2 – √3 would be 2 +.! Expressions calculator you like this Site about Solving Math problems, please let Google know clicking... +1 button are multiple ways to do this this example, the complex of... 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Involves a real number and Y is an imaginary number, it means we 're asked rationalize! | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 2, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.956802487373352, "perplexity": 1052.9592460783113}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-43/segments/1634323588257.34/warc/CC-MAIN-20211028034828-20211028064828-00531.warc.gz"} |
http://mathhelpforum.com/trigonometry/114334-trig-question-help-determining-length.html | # Math Help - Trig question help (Determining length)
1. ## Trig question help (Determining length)
can someone please help me out.... I have a question and yes it's for homework but I have no idea on how to even begin it.... To me it looks like they already gave the answer or something but I'm not sure how to start or how to do it... Can someone please help me out.
Determine the length of each unknown variable:
How do I solve this? I have no idea on where to even begin... If someone can help me it will be greatly appreciated... I can't show the work I've alreayd done because I don't know how to do this..
Thanks for all the help though if you can answer this
Alright I tried to re-edit it... does it work now?
Attached Thumbnails
2. in a circle of radis r of angle of radian measure $\theta$
subtends an arc of lenght $s = r\theta$
this basically what you need to know to solve this
3. I still don't understand how to solve this... can you go a litlte more in depth than that?
A) = 3.142
because maybe this is why I'm confused... I was given this question to solve but below the 2 pictures it had those which I think are already the answers....
4. ok the first is
$
a = \frac{\pi}{3}*3 = \pi \ or \ 3.142 \ cm
$
the second one is
just subtracting $\theta$ from $2\pi$
$
\theta = \frac{2}{4} = \frac{1}{2}
$
$
2\pi - \frac{1}{2} = 5.783\ rad
$
5. thank you so much! I understand it now. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 7, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.903358519077301, "perplexity": 392.07081441148097}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-35/segments/1440646242843.97/warc/CC-MAIN-20150827033042-00016-ip-10-171-96-226.ec2.internal.warc.gz"} |
https://lubimgotovit.ru/scientists-have-created-a-laser-violating-the-laws-of-light-distribution/ | Scientists have created a laser violating the laws of light distribution
Researchers have developed a new laser beam, the speed of which does not change when moving through various environments.
If we put a spoon in a glass with water, it will seem broken on the border between water and air. Such an effect arises due to the fact that the light passes through the air faster than through the water, so its rays bend when it gets into a more dense medium. However, a team of physicists from the University of Central Florida created a laser, which violates the law of Snellius, describing such physical phenomena.
Scientists have developed a system generating laser pulses that are able to behave equally in various environments or even move faster in more dense materials. In this case, the oscillations themselves are not changing, but the speed of moving their peaks. The command reached this with a spatial light modulator. Physics called such impulses «wave packages of space-time.»
Since the rays are not refracted, they can simultaneously reach the space points in various environments, which makes it possible to bring military communications technology to a new level.
However, despite the violation of the Farm’s habitual postulates on the distribution of light, scientists say that such laser rays do not contradict the laws of physics and fit into a special theory of relativity.
Previously, we also reported the development of an effective method.
To develop the channel, your support is important to us, subscribe to the channel and put like. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8008327484130859, "perplexity": 591.7274137474868}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-05/segments/1642320301488.71/warc/CC-MAIN-20220119185232-20220119215232-00283.warc.gz"} |
https://wiki.iac.isu.edu/index.php?title=Forest_UCM_MiNF&diff=next&oldid=128106 | # Mechanics in Noninertial Reference Frames
## Linearly accelerating reference frames
Let represent an inertial reference frame and represent an noninertial reference frame with acceleration relative to .
### Ball thrown straight up
Consider the motion of a ball thrown straight up as viewed from .
Using a Galilean transformation (not a relativistic Lorentz transformation)
At some instant in time the velocities add like
where
= velocity of moving frame with respect to at some instant in time
taking derivative with respect to time
where
inertial force
in your noninertial frame, the ball appears to have a force causing it to accelerate in the direction.
The inertial force may also be referred to as a fictional force
an example is the "fictional" centrifugal force for rotational acceleration.
The observer in a noninertial reference frame will feel these frictional forces as if they are real but they are really a consequence of your accelerating reference frame
example
A force pushes you back into your seat when your Jet airplane takes off
you slam on the brakes and hit your head on the car's dashboard
### Pedulum in an accelerating car
Consider a pendulum mounted inside a car that is accelerating to the right with a constant acceleration .
What is the pendulums equilibrium angle
In frame
In frame
If the pendulum is at rest and not oscillating then
is the vector sum of and which are orthogonal to each other in this problem thus
The pendulum oscillation frequency as seen in the accelerating car is
Using lagrangian mechanics in the inertial frame
### The tides view by a ref frame accelerating towards the moon
The gravitational force between the moon and the earth accelerates the earth towards the moon.
This makes the earth an accelerated reference frame
The moon of mass pulls on the earth of mass such that
where is the earth-moon distance of separation.
Earth's acceleration towards the moon that makes the earth a non-inertial reference frame
The moon of mass pulls on a test mass of water on the surface of the earth's ocean such that
As seen in the Earth non-inertial reference frame
where
a net non-graviational force hold the mass M_o in top of the ocean (Bouyant force)= M(displaced water)*g = constant force independent of position
Let's consider two cases, one where M_o is directly between the moon and the earth and the other when the mass is directly on the opposite side of the earth from the moon.
Case 1
The mass is directly between the moon and the earth
In this case and making pull towards the moon
Case 2
The mass is directly on the other side of the earth with respect to the moon
In this case BUT making pull away from the moon
## Magnitude of the Tides
Consider a drop of water of mass on the surface of the ocean.
The three forces influencing this drop from the previous sections are
Archimedes Principle
An object in a fluid is buoyed up with a force equal to the weight of the water displaced by the object.
All of the above forces act normal to the surface of the water.
are the result of a gravitational force which is conservative.
A potential may be defined for the two forces above such that
: is NOT always pointed along the x-axis towards the moon
Since is at fixed distance
Since is parallel to x
Since
are forces that are exerted such that the are always perpendicular to the surface of the ocean ( they are normal to the surface) then the sum of the two potential's, , is constant on the surface
The ocean's surface is an equipotential surface ( all points on the surface of the ocean are have the same gravitational potential energy
If you consider two points on the surface of the earth where one point is high tide (HT) and the other point is low tide (LT) then
The change in the gravitational potential energy between the earth and the ocean for high tide and low tide is
but
At HT
At LT
: geometric series when
### Height of Tides due to Moon and Sun
kg = mass of moon
kg = mass of earth
kg = mass of sun
m
m = earth-moon distance
height of tides due to the moon
m = earth-sun distance
height of tides due to sun
spring tides
The case when the moon, earth and sun are aligned to give maximum height tide (moon can be either full or new doesn't matter because the bulge is on both sides of the earth)
neap tides
This is the case then the sun, earth, and sun are aligned to form a right triangle. The tide effect should cancel so the height is only 54-25=29 cm.
## Euler's Theorem of rotation
Original form:
Any displacement of a rigid body in three dimensional space such that a point on the rigid body, say O, remains fixed, is equivalent to a rotation about a fixed axis through the point O.
Expressed using Modern mathematical terms
Euler's theorem on the axis of a three dimensional rotation
If is a 3x3 orthogonal matrix and is proper , then there is a non-zero vector such that
The matrix R above represents a spatial rotation that is a linear one-to-one mapping that transforms the coordinate vector P into p.
The above theorem can be proven by treating it as a matrix algebra problem where you want to prove. ie
First let's find the eigenvalues and eigenvectors of the matrix
A consequence of Euler's theorem is that a single rotation may be represented as a combination of rotation.
## Angular velocity vector
If we denote the angular velocity vector describing the angular velocity of a rotating body as the the magnitude and direction of this vector may be expressed as
magnitude of the rate of rotation (angular speed)
direction of the rotation axis which is given by the right-hand rule and may is expressed using the components of a right handed coordinate system ( ie or )
the angular velocity vector describing the angular velocity of a rotating body
The angular velocity vector describing the angular velocity of a rotating non-inertial reference frame.
### Linear velocity from angular velocity
= linear velocity of a point located at the position on a body rotating about the origin of the coordinate system used to specify .
Consider two coordinate systems where represent an inertial reference frame and represent a frame with a velocity relative to .
Using a Galilean transformation (not a relativistic Lorentz transformation)
the velocity in the moving frame may be expressed in terms of the velocity in the rest frame and the velocity of the moving frame as.
Now consider the situation when an object is rotating. The relationship between angular and linear motion should still satisfy the general express
A point on this rotating body located by the position vector from the inertial reference frame will have the linear velocity
If we fix a moving reference frame so its origin coincides with then the position of the same point on the object is given by the same position vector . The angular velocity of this point is given by
If the moving reference frame is rotating with an angular velocity then the point on the object located by the vector is moving with a linear velocity as viewed by of
Both the linear and the angular velocities add
## Time derivative in Rotating Frame
### Nomenclature
The following variables are used to describe the kinematics of a rotating frame
linear velocity of the rotating frame
linear acceleration of the rotating frame
angular velocity of the rotating frame
For example; a reference frame fixed to the earth with its origin at the center of the earth would be rotating with respect to an inertial frame with its origin at the same location with an angular velocity of one revolution every 24 hours.
### Rate of change of a general vector Q
let
an arbitrary vector representing position
Let's write the vector in terms of three orthonormal unit vectors representing an arbitrary right-handed coordinate system
= rate of change of as described by (relative to ) the rotating frame
Using unit vectors that are attached to the rotating frame
In a cartesian coordinate system
From the inertial reference frame the above unit vectors used by are changing their direction with time
The expansion coefficients (vector components) of the vector are the same in both frames as the moving frames is not moving fast enough to be relativistic
as seen earlier
The time derivative of a vector as measured in the inertial frame is related to the derivative in the non-inertial frame by
## Newton's second law in a rotating frame
Newton's 2nd law in an inertial reference fram
To determine the second derivative in a non-inertial frame we can take the derivative of the velocity relationship
let
then
returning to Newton's second law
: changing order of cross products introduces negative sign
### Alternative derivation using Lagrangian
Hamilton's principle is defined for an inertial reference frame, therefor the lgrangian is written from that perspective.
kinetic energy in the inertial reference frame
As seen before
where
= velocity of moving frame with respect to inertial frame at some instant in time
If we consider a non-inertial reference frame that is only rotating with an angular velocity with an origin fixed to the origin of the inertial reference frame.
Then
substituting this Gallilean transformation into the expression for the kinetic energy in the inertial reference fram
: scalar triple product is invariant under cyclic permutation
this is a scaler so ignore the unit vector time derivative
### psuedo Force names
Coriolis Force
Centrifugal Force
where
velocity of object in the rotating frame
velocity of the rotating frame
For the rotating earth
m/s
mi/hr
For objects with velocity v < 500 mph you can neglect the coriolis force.
Also note that the coriolis force depends on the velocity of the object while the centrifugal does not.
## Centrifugal Force
The direction of this fictional force on the earth's non-inertial reference frame is radially outward from the axis of rotation.
Insert picture
### Free fall example
The force of gravity between two objects with mass M and m is
In a non-inertial reference frame attached to the rotating earth the effective gravitational force is
If we assume the objects velocity v < 500 mph then you can neglect the coriolis force.
The coriolis force decreases the pull of gravity and is a function of the lattitude () The larger the stronger the effective gravitational force.
Launch rockets from Florida not Alaska
### Free fall tangential component
The tangential component of the centrifugal force is directed along the surface of the sphere such that
This has a maximum at = 45 degrees
## Coriolis Force
### Coriolis Force in Earth's reference frame
The Earth rotates from west to east.
Consider a reference frame fixed to the earth in the Northern Hemisphere at the latitude angle 90 - . In a standard spherical coordinate system the "z-axis" is directed towards the north pole and is the polar angle measures from this fixed zenith direction, is the azimuthal angle corresponding to longitudinal (east-west). Lines of latitude represent the angle measured from the equator to one of the Earth's poles (90 - )
The non-inertial frame shown above has the axis pointing to the west and the axis pointing upward along the radial direction from the Earth's center.
The rotation vector as described in the non-inertial frame is
I you launch a projectile in the Northern Hemisphere so its velocity tangential to the Earth's surface and moving from the equator towards the north pole, the projectile will be deflected towards the east (right).
East
If you launch the projectile towards the equator from the Northern hemisphere.
West
In both cases the projectile is deflected to the right of its velocity direction.
How about a projectile launched to the West (East)
A projectile moving West (East) will deflected down(up) and to the right of its trajectory
What happens in the Southern Hemisphere?
The rotation vector as described in the non-inertial frame in the Southern Hemisphere is
Now if
Projectile directed from South to North
west
In the Northern hemisphere a projectile would swerve East if you directed it to the North
Projectile directed from South to North
East
How about a projectile launched to the West (East)
A projectile moving West (East) in the Souther Hemisphere will deflected down(up) and to the left of its trajectory
#### Coriolis effect on storm systems
The Earth (and a non-inertial reference frame attached to it) has a rotational velocity
Weather systems involve a large mass of air moving at speeds up to 100 miles per hour in the case of a tropical storms.
Over large distances this acceleration can have an effect on the course of a storm but not the short distances involved in tornados.
Tropical storms (cyclones) are examples of how the coriolis force determine the rotational direction of an air mass that is quickly moving towards a low pressure area. As the air mass moves along the pressure gradient is is deflected to the right. This results in a clockwise (counter-clockwise) rotation in teh Northern (Southern) hemisphere. The radius of the cyclone can be estimated as
( or of the latitude angle)
### Free fall
Consider an object of mass m close to the surface of the earth and falling in a vacuum.
As viewed from a non-inertial reference frame attached to the rotating earth Newton's second law is
The equation of motion may be written as
where
the acceleration due to a non-rotating earth
the acceleration observed
Note
by expressing the equation of motion in terms of \vec g you remove the dependence of leaving only a dependence on and
If I use a coordinate system where the x-axis if pointing East then the z-axis will be pointing radially upward such that
left
writing out the individual components
#### 0th order Approx
let
Then
x-directions
: integrate and the constant is the initial velocity of the projectile
: integrate again
:
similarly in the y-direction
z-direction
#### 1st order Approx
substitute the 0th order solutions and integrate again
x-direction
y-direction
: ignoring higher order terms
z-direction
: ignoring higher order terms
: ignoring
General equations of motion
If the object is dropped from a height h
If I drop an object down an evacuated 100 m deep hole at the equator
### Foucault Pendulum
The equation of motion may be written as
## Interpretation of rotating reference frame
### Polar
Vector Notation convention:
Position:
Because points in a unique direction given by we can write the position vector as
: does not have the units of length
The unit vectors ( and ) are changing in time. You could express the position vector in terms of the cartesian unit vectors in order to avoid this
The dependence of position with can be seen if you look at how the position changes with time.
#### Velocity in Polar Coordinates
Consider the motion of a particle in a circle. At time the particle is at and at time the particle is at
If we take the limit ( or ) then we can write the velocity of this particle traveling in a circle as
or
Thus for circular motion at a constraint radius we get the familiar expression
If the particle is not constrained to circular motion ( i.e.: can change with time) then the velocity vector in polar coordinates is
=
or in more compact form
linear velocity Angular velocity
Finding the derivative directly
Cast the unit vector in terms of cartesian coordinates and take the derivative.
#### Acceleration in Polar Coordinates
Taking the derivative of velocity with time gives the acceleration
We need to find the derivative of the unit vector with time.
Consider the position change below in terms of only the unit vector
Using the same arguments used to calculate the rate of change in :
If we take the limit ( or ) then we can write the velocity of this particle traveling in a circle as
or
Finding the derivative of directly
Cast the unit vector in terms of cartesian coordinates and take the derivative.
Substuting the above into our calculation for acceleration:
F_r = m \ddot r \hat r = m \left ( \ddot{r} -r\dot{\phi}^2 \right) \hat{r} = | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9064828753471375, "perplexity": 538.0341514743203}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-49/segments/1669446710734.75/warc/CC-MAIN-20221130092453-20221130122453-00075.warc.gz"} |
https://slideplayer.com/slide/255348/ | # Knowledge as JTB Someone S has knowledge of P IFF: 1. S believes P 2. S is justified in believing P 3. P is true.
## Presentation on theme: "Knowledge as JTB Someone S has knowledge of P IFF: 1. S believes P 2. S is justified in believing P 3. P is true."— Presentation transcript:
Knowledge as JTB Someone S has knowledge of P IFF: 1. S believes P 2. S is justified in believing P 3. P is true
JTB This definition of knowledge was formulated by Plato in Theaetetus. It remains the generally accepted definition of knowledge Edmund Gettier, however, challenges the adequacy of such a definition – it is known as the Gettier Problem
Justification One primary question: Does having justification for P entail that P is true? – Infalliblism: If S knows P, then S cannot be mistaken in believing P, thus Ss justification for believing P guarantees its truth. i.e. one cannot be justified in believing a false claim. – Falliblism: The infalliblist argument only works in the case where S cannot possibly be wrong about P. However, there is nothing to guarantee that S is right. Basically, its right to say that It is impossible for S to be wrong about P if he knows P. but it is not necessarily right to say that If S knows P, then it is impossible for him to be wrong about P.
Falliblism and its implications Falliblism suggests that truth and justification may or may not be connected in any situation. How then can we possibly connect Ss belief that P with the truth of P correctly? – Foundationalism – Coherentism – Reliablism
Foundationalism We justify one belief with another, which also requires support from others and so on and so forth. Foundationalism suggests that there is a termination to such a regression in a set of beliefs that are fundamentally and independently secure. These beliefs are either self-justifying, self-evident or indefeasible. However, it is difficult to see how these self-evident beliefs (mathematics and simple logical propositions) can lead to more complex, dependent beliefs. The latter does not seem to be deducible from the former. If it cannot be deduced, then it would have to be induced, but induction is in itself defeasible.
Reliabilism Reliable process of having a belief – Rational – Scientific – Mathematical – Logical Problems: – Externalism vs internalism debate: no knowledge of process, yet knowledge – The New Evil Demon
Coherentism Attempts to solve the regress problem – Foundationalisms answer – axioms! Descartes, Spinoza – Coherentism – regress means nothing – fundamental assumption is wrong! The criteria of coherence – System of beliefs should cohere – Use the same explanation for divergent statements – Use ONE explanation for similar statements
Gettier Problem Somethings wrong with JTB
Case 1: Original case Two characters, Smith (Main) and John. Both sign-up for job and President of the Co. tells Smith that John got the job. Smith comes to the conclusion that whoever got the job had ten coins in the pocket. He has this belief because he counted the coins in Johns pocket. His belief is justified. (JB present)
But… Smith got the job instead, not knowing that he did had ten coins in his pocket as well. So his belief is true. (JTB criteria fulfilled.) But is it knowledge??? No, because he did not know who truly got the job.
Another example John sees a person who appears to be Jane dancing in room A. John then forms a belief that Jane is dancing. He can justify it because he has seen someone who looks like Jan dance. The belief is also true because Jane is indeed dancing.
But… However, despite it fulfilling JTB, there is an error in it because even though Jane is dancing, she is not dancing in room A. Instead she is dancing in room B and the person which John saw dancing in room A was Janes identical twin. His belief cannot be knowledge because he does not know the true location of Jane.
Solutions to this 1.Infallibility proposal 2.Eliminate Luck Proposal 3.No False Evidence Proposal
Infallibility proposal Explained that because we use our senses to justify, we make errors. Therefore the best way to prevent such a case is to not use any fallible evidence to justify a belief at all. This is however, unrealistic as in our real lives we rarely have infallible knowledge. A mistaken approach to the cases which dismisses almost all we know.
Eliminate Luck Proposal Luck is one of the main reasons for resulting in the Gettier Cases. So if it is eliminated or reduced then higher chance for the belief to be right. However, this method has been quite vague. However, based on my understanding, scientific research can be considered such a method; the control of variables and having a controlled environment.
No False Evidence Proposal In the Smith case, why he came to the JTB yet not knowledge was because he heard that Jones was going to get the job. So if that piece of information was taken away, he would not have come to that conclusion. So JTB should be modified such that for the belief to be knowledge, the justification has to be true.
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https://www.lessonplanet.com/teachers/computer-applications-vocabulary | Computer Applications: Vocabulary
High schoolers define the procedures involved in information relay within a computer. They become familiar with DOS commands and gain a basic understanding of the Windows 3.1 operating system.
Concepts
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https://alevelmaths.co.uk/pure-maths/algebra/arithmetic-sequence/ | # Arithmetic Sequence
TOPIC: ARITHMETIC SEQUENCE/ARITHMETIC PROGRESSION
Sub-topics:
• Arithmetic Means
• Word Problems Involving Arithmetic Sequence
LEARNING OBJECTIVES:
• To recognise arithmetic sequence
• To illustrate arithmetic means
• To solve problems involving arithmetic sequence
MATH CONCEPTS
• ARITHMETIC SEQUENCE – It is a sequence in which the difference between any two consecutive terms is equal.
• COMMON DIFFERENCE – It is the constant difference of any two consecutive terms (nth term minus the previous term) in an arithmetic sequence. Thus, $d = t_{2}-t_{1}=t_{3}-t_{2}=t_{n}-t_{n-1}$
• FINDING THE NTH TERM OF AN ARITHMETIC SEQUENCE: $T_{n} = t_{1} + (n-1)d$ where $T_{n}$ is the value of the nth term, $t_{1}$ is the first term, n is the position of the nth term, and d is the common difference.
• ARITHMETIC MEANS – It is the average of two terms in an arithmetic sequence.
IMPORTANCE
• Arithmetic sequence is a functional notation. Each element in the domain has a unique corresponding element in the range.
• Understanding the concepts of arithmetic sequence is already a good prerequisite knowledge in understanding linear functions.
DISCUSSION
• ARITHMETIC SEQUENCE
Arithmetic sequence is a special type of sequence where there is a constant difference present in any two consecutive terms. This common difference can be any real number.
In the given example, notice that each term is being added constantly by y. The answer would still be the same if you solve for the difference of any two consecutive terms. In this sequence, the first term, $t_{1} = x$, the common difference d = y, and the number of given terms n = 8.
Remember, the common difference can also be a negative number. Here, the arithmetic sequence is decreasing and the common difference is equal to -7. In this sequence, the first term, $t_{1} = -15$, the common difference d = -7, and the number of given terms n = 8.
ILLUSTRATIVE EXAMPLES
Identify whether the following sequences are arithmetic sequences. If it is, solve for the common difference.
1) 1, 2, 3, 4, 6
2) π, 2π+3, 3π+6, 4π+9
3) 1/8, 1/4, 3/8, 1/2
SOLUTIONS:
1) To say that a sequence is an example of arithmetic sequence, any two consecutive terms must have a common difference. Notice that from $t_{2}-t_{1},\, t_{3}-t_{2},\,t_{4}-t_{3}$ is equal to 1 but $t_{5}-t_{4} = 2$. In that case, 1, 2, 3, 4, 6 is not an arithmetic sequence.
2) Investigate if there is a constant number between two consecutive terms.
Since there is a constant difference ( π + 3 ), this is an arithmetic sequence.
3) Investigate if there is a constant number between two consecutive terms.
Since there is a constant difference ( d = 18 ), this is an arithmetic sequence.
DISCUSSION
• FINDING THE NTH TERM OF AN ARITHMETIC SEQUENCE AND ARITHMETIC MEANS
To find the nth term of an arithmetic sequence, use the formula
It is very important to identify the given values in order to arrive with the correct answer. The formula is very easy to use and to follow.
ILLUSTRATIVE EXAMPLES
Solve for the following arithmetic sequence-related problems.
1) Find the 16th term of an arithmetic sequence if the first term is 8 and d= – 9.
Using the formula of finding the nth term of an arithmetic sequence: $T_{n} = t_{1}+(n-1)d$
$T_{16} = ?$ $T_{1} = 8$ $d=9$
$T_{16} = 8+[(16-1) (-9)]$
$T_{16} = 8+[(15) (-9)]$
$T_{16} = 8 - 135$
$T_{16} = -127$
2) Find the value of x if the first term is x, d = 3x, and the 9th term is 25.
Using the formula of finding the nth term of an arithmetic sequence:
$T_{9} = 25$ $T_{1} = x$ $d = 3x$
$T_{9} = x+[(9-1) (3x)]$
25 = x + (8)(3x)
25 = x + (24x)
25 = 25x
1 = x
3) Find the missing terms of the arithmetic sequence ___, 6, ____, ____, 30.
To solve this problem, we need our knowledge about arithmetic means. We should solve for the common difference to complete the terms of the sequence. Using the formula of finding the nth term of an arithmetic sequence: $T_{n} = t_{1}+(n-1)d$, let’s derive the formula for the common difference.
$T_{n} = t_{1}+(n-1)d$
$T_{n}-t_{1} =(n-1)d$
(Addition Property of Equality: add $-t_{1}$ to both sides of the equation)
$\frac{T_{n}-t_{1}}{(n-1)}=d$
(Multiplication Property of Equality: Multiply both sides by the reciprocal of n-1)
$\frac{T_{n}-t_{1}}{(n-1)}=d$
(The formula for common difference)
In this arithmetic sequence ___, 6, ____, ____, 30, Let 6 be the first term and 30 will now be the 4th term. Using the derived formula for d,
$\frac{T_{n}-t_{1}}{(n-1)}=d$
$\frac{30-6}{(4-1)}=d$
$\frac{24}{3}=d$
8 = d
Since we already know the value for d, we can now supply the missing terms in the sequence. The complete sequence is -2, 6, 14, 22, 30 .
4) An employer offers an expert a beginning salary of £ 40, 300 per year. If the salary is increased to £ 1200 yearly, what would be the expert’s salary in his 5th year of working?
Using the formula of finding the nth term of an arithmetic sequence: $T_{n}=t_{1}+(n-1)d$.
$T_{5} = ?$ $T_{1} = 40, 300$ $d= 1,200$
$T_{n}=t_{1}+(n-1)d$
$T_{5}= 40,300+(5-1) 1200$
$T_{5}= 40,300+(4) 1200$
$T_{5}= 40,300+4,800$
$T_{5}= 45,100$
Therefore, his 5th year salary will be £ 45, 100.
REFERENCES:
https://www.mathsisfun.com/algebra/sequences-sums-arithmetic.html | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 76, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8877434730529785, "perplexity": 425.78867399277846}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-49/segments/1637964358705.61/warc/CC-MAIN-20211129104236-20211129134236-00350.warc.gz"} |
http://math.stackexchange.com/questions/790429/quadratic-inequality-with-absolute-values | # Quadratic inequality with absolute values
I've decided to study calculus on my own, and I've started working on "A First Course in Calculus" by Serge Lang, 5th edition. Now I'm just reading the chapter on preliminaries, and there is a section about inequalities. In the exercises on p.13, there is a problem which I can solve, but I'm not quite sure that I'm doing it right.
The problem is:
$|x^{2}-2|\leqslant 1$
I am solving it in the following way:
1) Suppose that $x^{2}-2 \geqslant 0$, then $|x^{2}-2|=x^{2}-2$
First, I find the values of x for which $x^{2}-2$ is non-negative: $x^{2}-2 \geqslant 0$, and obtain $x \geqslant \sqrt{2}$ or $x \leqslant -\sqrt{2}$
Second, I solve the inequality $x^{2}-2 \leqslant 1$ and obtain $-\sqrt{3} \leqslant x \leqslant \sqrt{3}$
The solution for non-negative values of $x^{2}-2$ is thus $-\sqrt{3}\leqslant x \leqslant -\sqrt{2}$ or $\sqrt{2} \leqslant x \leqslant \sqrt{3}$
2) Suppose that $x^{2}-2<0$, then $|x^{2}-2|=-(x^{2}-2)=2-x^{2}$
First, I find the values of x for which $x^{2}-2$ is negative: $x^{2}-2<0$, and obtain $-\sqrt{2} < x < \sqrt{2}$
Second, I solve the inequality $2-x^{2} \leqslant 1$ and obtain $x \leqslant -1$ or $x \geqslant 1$
The solution for negative values of $x^{2}-2<0$ is thus $-\sqrt{2} < x \leqslant -1$ or $1 \leqslant x < \sqrt{2}$
Combining the solutions for non-negative and negative values of $x^{2}-2<0$, I obtain the final answer:
$-\sqrt{3} \leqslant x \leqslant -1$ or $1 \leqslant x \leqslant \sqrt{3}$
And here I go to the question itself: is it necessary to find the values of x for which $x^{2}-2<0$ and $x^{2}-2>0$, or can this step be skipped? If I just find the intersection of solutions of two inequalties, $x^{2}-2 \leqslant 1$ and $2-x^{2} \leqslant 1$, I still get the same answer. But I feel that skipping this step would be wrong, and that maybe it can lead to incorrect answers when I deal with more complex problems of this type.
I also have a second, though smaller, question. When we speak about the solution of an inequality, and we have several intervals, do we use the conjunction "or", or "and"? That is, is it
$-\sqrt{3} \leqslant x \leqslant -1$ or $1 \leqslant x \leqslant \sqrt{3}$
or
$-\sqrt{3} \leqslant x \leqslant -1$ and $1 \leqslant x \leqslant \sqrt{3}$?
Lang uses "or" in the "Answers" section, but it just seems to me that we speak about a set of solutions here, so maybe "and" would be more appropriate.
-
If you use the definition of $|.|$ in the way you did above, yes. There are of course other ways of solving it, which may not require that step. For e.g., we could work out the solution using $\iff (x^2-2)^2 \le 1 \iff (x-1)(x+1)(x-\sqrt3)(x+\sqrt3) \le 0$ which will give you the same solution by comparing intervals where the product can change signs. Which way you use depends on the problem and what you are comfortable with. – Macavity May 11 '14 at 15:59
Thank you. But could you please explain why you use $(x^{2}-2)^{2}$ instead of $|x^{2}-2|$ in your answer? – Vladimir May 11 '14 at 16:08
Because another possible way of using absolute value is $|x| = \sqrt{x^2}$. Hence $|x| \le 1 \iff x^2 \le = 1^2=1$. In this problem, then you can get some nice factorisations going, of the form $a^2-b^2 = (a-b)(a+b)$. – Macavity May 11 '14 at 16:12 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9620662331581116, "perplexity": 108.4616297276356}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-06/segments/1422115856041.43/warc/CC-MAIN-20150124161056-00001-ip-10-180-212-252.ec2.internal.warc.gz"} |
https://infoscience.epfl.ch/record/146097 | ## Adaptive Beamforming with a Minimum Mutual Information Criterion
In this work, we consider an acoustic beamforming application where two speakers are simultaneously active. We construct one subband-domain beamformer in \emph{generalized sidelobe canceller} (GSC) configuration for each source. In contrast to normal practice, we then jointly optimize the \emph{active weight vectors} of both GSCs to obtain two output signals with \emph{minimum mutual information} (MMI). Assuming that the subband snapshots are Gaussian-distributed, this MMI criterion reduces to the requirement that the \emph{cross-correlation coefficient} of the subband outputs of the two GSCs vanishes. We also compare separation performance under the Gaussian assumption with that obtained from several super-Gaussian probability density functions (pdfs), namely, the Laplace, $K_0$, and $\Gamma$ pdfs. Our proposed technique provides effective nulling of the undesired source, but without the signal cancellation problems seen in conventional beamforming. Moreover, our technique does not suffer from the source permutation and scaling ambiguities encountered in conventional blind source separation algorithms. We demonstrate the effectiveness of our proposed technique through a series of far-field automatic speech recognition experiments on data from the \emph{PASCAL Speech Separation Challenge} (SSC). On the SSC development data, the simple delay-and-sum beamformer achieves a word error rate (WER) of 70.4\%. The MMI beamformer under a Gaussian assumption achieves a 55.2\% WER, which is further reduced to 52.0\% with a $K_0$ pdf, whereas the WER for data recorded with a close-talking microphone is 21.6\%.
Year:
2007
Publisher:
IDIAP
Laboratories: | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9200674295425415, "perplexity": 2491.6528382327815}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-22/segments/1526794864466.23/warc/CC-MAIN-20180521181133-20180521201133-00292.warc.gz"} |
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## emcrazy14 3 years ago Two die are rolled. Find the probability that the numbers are the same given that at least one is 3? Delete Cancel Submit
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1. cinar
• 3 years ago
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|dw:1338882073720:dw|
2. cinar
• 3 years ago
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can you expand your question, or write it in another way..
3. cinar
• 3 years ago
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I did not get it..
4. emcrazy14
• 3 years ago
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I have to find the probability of getting two same numbers on both die if the number on one of them is atleast 3. I'm not sure myself. But I think it means that the number is either 3 or greater than 3.
5. cinar
• 3 years ago
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|dw:1338882463299:dw|
6. cinar
• 3 years ago
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6/36 prob. of the same number
7. cinar
• 3 years ago
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|dw:1338882580009:dw|
8. cinar
• 3 years ago
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at least one is 3 11/36
9. cinar
• 3 years ago
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that's all I can do, sorry..
10. emcrazy14
• 3 years ago
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Are you sure this is the answer? What I'm thinking is that there are 11 outcomes in which one of the number is 3. But only 1 event in which 3,3 appears. So the probability should by 1/11. What do you think?
11. emcrazy14
• 3 years ago
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@experimentX : I didnot understand what you did. Would you please explain in another way?
12. emcrazy14
• 3 years ago
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So what would be the final answer?
13. kropot72
• 3 years ago
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The probability that at least one number is a 3 is$\frac{1}{6}+\frac{1}{6}=\frac{2}{6}=\frac{1}{3}$ The probability that the other number will also be a 3 is 1/6 The overall probability of two 3's is therefore$\frac{1}{3}\times \frac{1}{6}=\frac{1}{18}$
14. emcrazy14
• 3 years ago
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Everyone has posted a different answer. Who should I trust? :(
15. experimentX
• 3 years ago
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toss a coin ... lol
16. emcrazy14
• 3 years ago
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No, seriously. I'm more confused than ever before. :(
17. FoolForMath
• 3 years ago
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Sample set in a role of two dice: $$\begin{array}{c|c|c|C|} &1&2&3&4&5&6 \\ \hline 1 &(1,1)&(1,2)&(1,3)&(1,4)&(1,5)&(1,6)\\ \hline 2 &(2,1)&(2,2)&(2,3)&(2,4)&(2,5)&(2,6)\\ \hline 3 &(3,1)&(3,2)&(3,3)&(3,4)&(3,5)&(3,6)\\ \hline 4 &(4,1)&(4,2)&(4,3)&(4,4)&(4,5)&(4,6)\\ \hline 5 &(5,1)&(5,2)&(5,3)&(5,4)&(5,5)&(5,6)\\ \hline 6 &(6,1)&(6,2)&(6,3)&(6,4)&(6,5)&(6,6)\\ \hline \end{array}$$
18. FoolForMath
• 3 years ago
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Observe the table I posted above, You will see that that there are $$11$$ cases where either of the two dice show a $$3$$. And, among those $$11$$ cases only 1 i.e$$(3,3)$$ show both $$3$$. Ergo, the probability is $$\frac 1 {11} .$$
19. FoolForMath
• 3 years ago
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You can also interpret this problem that we already know that one of the dice is 3 now the probability of another side being also 3 is $$\frac 1 6$$
20. FoolForMath
• 3 years ago
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I can't see any blemish in either of these approaches, however I prefer #2. May be @Zarkon can see something else :)
21. kropot72
• 3 years ago
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If a pair of dice are rolled what is the probability that one die will roll a three?
22. FoolForMath
• 3 years ago
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I think it's $$\frac {11}{36}$$
23. FoolForMath
• 3 years ago
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assuming at-least, if it is strictly one then it is $$\large\frac{10}{36}$$
24. kropot72
• 3 years ago
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The Addition Theorem states that if an event can happen in a number of different and mutually exclusive ways, the probability of it happening at all is the sum of the separate probabilities that each event happens. The event is to roll a three. each die has a 1/6 probability of rolling a 3. therefore the probability that a 3 comes up on one or other of the dice is 1/6 + 1/6 = 2/6
25. kropot72
• 3 years ago
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@FoolForMath Do you intend to reply?
26. FoolForMath
• 3 years ago
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They are rolled simultaneously not one after another. The sample space is 36 and 11 of them favor the even where atleast one die will roll a three.
27. kropot72
• 3 years ago
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The Addition Theorem holds. The simultaneous rolling does not affect the 1/6 probability of any of the 6 numbers coming up on either die.
28. Zarkon
• 3 years ago
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"If a pair of dice are rolled what is the probability that one die will roll a three?" $P(3_1 \text{ or } 3_2)=P(3_1)+P(3_2)-P(3_1 \text{ and } 3_2)$ $=\frac{1}{6}+\frac{1}{6}-\frac{1}{36}=\frac{11}{36}$ or $P(3_1 \text{ or } 3_2)=1-P(\left(3_1 \text{ or } 3_2\right)')=1-P(3'_1 \text{ and } 3'_2)$ $=1-\frac{5}{6}\times\frac{5}{6}=1-\frac{25}{36}=\frac{11}{36}$ the answer to the OP is $$\displaystyle\frac{1}{11}$$ E=equal , T=at least one 3 $P(E|T)=\frac{P(T|E)P(E)}{P(T)}=\frac{\frac{1}{6}\times\frac{1}{6}}{\frac{11}{36}}=\frac{1}{11}$
29. kropot72
• 3 years ago
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@Zarkon Thank you for your explanation. However it has a condition that only one die of the pair will have a three. The Addition Theorem holds where the probability of both dice showing 3 is included.
30. Zarkon
• 3 years ago
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the condition, from the OP, is 'given that at least one is 3' not 'only one die of the pair will have a three'
31. kropot72
• 3 years ago
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@Zarkon Good point. Therefore the probability that at least one is a 3 when a pair of fair dice is rolled is 1/6 + 1/6 = 2/6
32. Zarkon
• 3 years ago
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these events are not mutually exclusive. the way you are doing it you are double counting (3,3)
33. kropot72
• 3 years ago
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The number that comes up on die #1 has no bearing on the number that comes up on die #2. The sub-events 'number thrown on die #1' and 'number thrown on die #2' are indeed mutually exclusive. The sample space diagram has not bearing on this particular calculation.
34. Zarkon
• 3 years ago
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you are describing independence ...the events are independent but they are not mutually exclusive.
35. vic1
• 3 years ago
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The probability of getting the same numbers = the probability of getting (1,1), (2,2), (3,3), (4,4), (5,5), (6,6). Which is equals 6/36 =0.167 But, we are asked to fine the probability of getting the same numbers giving that at least one is 3 means the probability of getting 3 and above. Which the probability of getting (3,3), (4,4), (5,5), (6,6). Which equals 4/36 = 0.11
36. Zarkon
• 3 years ago
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"we are asked to fine the probability of getting the same numbers giving that at least one is 3 means the probability of getting 3 and above" Really?
37. vic1
• 3 years ago
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@ Zarkon, I think it means at least one of the numbers is 3. What do u think?
38. Zarkon
• 3 years ago
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"I think it means at least one of the numbers is 3." yes it doesn't mean "the probability of getting 3 and above"
39. kropot72
• 3 years ago
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@Zarkon Thank you very much for your patience and explanations. I accept that the Addition Theorem does not apply in this problem :)
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Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.999251127243042, "perplexity": 1851.632602249107}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-40/segments/1443738004493.88/warc/CC-MAIN-20151001222004-00068-ip-10-137-6-227.ec2.internal.warc.gz"} |
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# 18.1: Rate of Reactions
Difficulty Level: At Grade Created by: CK-12
## Lesson Objectives
The student will:
• define chemical kinetics and rates of reaction.
• write the rate expression and the units for the rate expression.
• define instantaneous rate.
• calculate instantaneous rate using a tangent line.
## Vocabulary
• chemical kinetics
• instantaneous rate
• rate of reaction
## Introduction
The focus of this chapter is chemical kinetics. Chemical kinetics is the study of chemical reactions rates and the factors that affect the rate of reactions. These factors include concentration, temperature, pressure, surface area, and the effect of a catalyst. For example, when food is placed in the refrigerator, the cold temperature keeps the food from decomposing by slowing the rate of reaction. Chemical kinetics plays an important role both in industry and in our daily lives. To begin, we will introduce some of the basic concepts of chemical kinetics.
## Change in Concentration Over Time
The term rate of reaction is used to denote the rate at which the products are formed in a time interval or the rate at which the reactants are consumed over a time interval. A reaction rate measures how fast or how slow a reaction is. The rusting of a piece of metal has a slow reaction rate because the iron oxidizes in the air over a relatively long time period. A forest fire has a fast reaction rate because it consumes trees in its path in a very short time interval. Reaction rates can be measured as the change in mass per unit time (grams/second) or the charge in molarity per unit time (mol/Ls\begin{align*}\mathrm{mol/L} \cdot \mathrm{s}\end{align*}).
Symbolically, the reaction rate is given the letter r\begin{align*}r\end{align*}. The reaction rate, then, can be written as follows:
r=concentrationtime\begin{align*}r = \frac {\triangle \text{concentration}} {\triangle \text{time}}\end{align*}
Remember that the symbol \begin{align*}\triangle\end{align*} means the “change in.”
Example:
For the reaction H2(g)+I2(g)2 HI(g)\begin{align*}\mathrm{H}_{2(g)} + \mathrm{I}_{2(g)} \rightarrow 2 \ \mathrm{HI}_{(g)}\end{align*}, under certain conditions, the [HI]=0.50 mol/L\begin{align*}[\mathrm{HI}] = 0.50 \ \mathrm{mol/L}\end{align*} at 25 s\begin{align*}25 \ \mathrm{s}\end{align*} and 0.75 mol/L\begin{align*}0.75 \ \mathrm{mol/L}\end{align*} at 40 s\begin{align*}40 \ \mathrm{s}\end{align*}. What is the rate of production of HI\begin{align*}\mathrm{HI}\end{align*}? Note: remember that the brackets indicate concentration.
Solution:
r=[HI]t=(0.750.50) mol/L(40.25) s=0.25 mol/L15 s=1.7×102 mol/Ls\begin{align*}r = \frac {\triangle [\mathrm{HI}]} {\triangle t} = \frac {(0.75 - 0.50) \ \text{mol/L}} {(40. - 25) \ \mathrm{s}} = \frac {0.25 \ \text{mol/L}} {15 \ \mathrm{s}} = 1.7 \times 10^{-2} \ \text{mol/L} \cdot \mathrm{s}\end{align*}
Therefore, the rate of production of \begin{align*}\mathrm{HI}\end{align*} is \begin{align*}1.7 \times 10^{-2} \ \mathrm{mol/L} \cdot \mathrm{s}\end{align*}.
A blackboard discussion of reaction rate with an eye to developing the rate law and the equilibrium constant (8a) is available at http://www.youtube.com/watch?v=_HA1se_gyvs (7:20).
## Units for Rate of Reaction
Notice in the previous example that the units to measure the reaction rate are in \begin{align*}\mathrm{mol/L} \cdot \mathrm{s}\end{align*}. Therefore, the units are measuring the concentration/time or the M/time. These units allow for the comparison of rates. In other words, if all reaction rates were to use the same units, we can compare one rate to the other. For example, under a different set of conditions, the HI reaction was found to have a reaction rate of \begin{align*}2.5 \ \mathrm{mol/L} \cdot \mathrm{s}\end{align*}. We could then predict that the new set of conditions are favorable for this reaction since the reaction rate was faster for the production of HI in the same time interval.
## Graphing Instantaneous Rate
Instantaneous rate is defined as the rate of change at a particular moment. For example, a police officer stops a car for speeding. The radar gun on a police cruiser is set to measure the speed of a motorist as the motorist comes close to the cruiser. The driver of the vehicle is stopped doing \begin{align*}65 \ \mathrm{miles/hour}\end{align*} in a \begin{align*}50 \ \mathrm{miles/hour}\end{align*} zone. The cruiser measured the rate of speed at that instant in time when the driver passed the police officer. This is instantaneous rate. If we were to take all of the measures of instantaneous rate and graph them, we would obtain a curve of the overall speed (or the average speed) of the vehicle. The same is true for reactions. For reactions, the instantaneous rate is the rate of the reaction at a specific time in the reaction sequence. If you were to graph the rate of the reactant being consumed versus time, the graph would look like the figure below. As the reaction proceeds, the concentration of the reactants decreases over time.
The initial rate of the reaction is found at \begin{align*}t = 0 \ \mathrm{s}\end{align*}, or when the reaction is just beginning. It is at this point when the maximum amount of the reactant is present. To find the instantaneous rate, a tangent line is drawn to this curve. The slope of this tangent line is then found. For example, say we wanted to know the instantaneous rate at \begin{align*}t = 2 \ \mathrm{s}\end{align*}. After drawing the tangent line (see figure below), we can calculate the slope of the tangent line to find the instantaneous rate at \begin{align*}t = 2 \ \mathrm{s}\end{align*}.
\begin{align*}\text{rate} & = \frac {y_2 - y_1} {x_2 - x_1} = \frac {0.35 - 0.63} {3.0 - 1.0} = \frac {-0.28 \ \text{mol/L}} {2.0 \ \mathrm{s}} = -0.14 \ \text{mol/L} \cdot \ \mathrm{s}\\ \end{align*}
## Lesson Summary
• Chemical kinetics is the study of rates of chemical reactions and how factors affect rates of reactions. The term rate of reaction is used to denote the measure at which the products are formed over a time interval or the rate at which the reactants are consumed over a time interval.
• The units to measure the reaction rate are in \begin{align*}\mathrm{mol/L} \cdot \mathrm{s}\end{align*}.
• Instantaneous rate is defined as the rate of change at a particular moment.
The website below provides more details about measuring reaction rates.
## Review Questions
1. Given that the concentration of \begin{align*}\mathrm{NO}_{2(g)}\end{align*} is \begin{align*}0.40 \ \mathrm{mol/L}\end{align*} at \begin{align*}45 \ \mathrm{s}\end{align*} and \begin{align*}0.85 \ \mathrm{mol/L}\end{align*} at \begin{align*}80 \ \mathrm{s}\end{align*}, what is the rate of production of \begin{align*}\mathrm{NO}_{2(g)}\end{align*} in: \begin{align*}\mathrm{NO}_{2(g)} + \mathrm{CO}_{(g)} \rightarrow \mathrm{NO}_{(g)} + \mathrm{CO}_{2(g)}\end{align*}?
2. For the graph below, draw a tangent line at \begin{align*}t = 0.40 \ \mathrm{s}\end{align*} and calculate the instantaneous rate.
3. Which expression represents the rate for the product formation for the reaction:\begin{align*}\mathrm{Mg}_{(s)} + 2 \ \mathrm{HCl}_{(aq)} \rightarrow \mathrm{MgCl}_{2(aq)} + \mathrm{H}_{2(g)}\end{align*}?
1. \begin{align*}\mathrm{rate} = \frac {\triangle [\mathrm{Mg}]} {\triangle \mathrm{t}}\end{align*}
2. \begin{align*}\mathrm{rate} = \frac {\triangle [\mathrm{HCl}]} {\triangle \mathrm{t}}\end{align*}
3. \begin{align*}\mathrm{rate} = \frac {\triangle [\mathrm{MgCl}_2]} {\triangle \mathrm{t}}\end{align*}
4. All of these are accurate representations of the rate.
4. Which statement represents a rate?
1. The speed of a car is \begin{align*}50 \ \mathrm{km/h}\end{align*}.
2. Half the product is produced.
3. A family consumes \begin{align*}5 \ \mathrm{L}\end{align*} of milk.
4. I ran for \begin{align*}45 \ \mathrm{minutes.}\end{align*}
5. Which statement about the instantaneous rate of a reaction is correct?
1. The higher the rate, the smaller the slope of a line on a concentration-time graph.
2. The instantaneous rate is the slope of the tangent to a line on a concentration-time graph.
3. The instantaneous rate is the slope of the cosine to a line on a concentration-time graph.
4. All of these statements are correct.
6. What is the rate of production of \begin{align*}\mathrm{NO}\end{align*} gas if the concentration decreases from \begin{align*}0.32 \ \mathrm{mol/L}\end{align*} at \begin{align*}56 \ \mathrm{s}\end{align*} and \begin{align*}0.94 \ \mathrm{mol/L}\end{align*} at \begin{align*}78 \ \mathrm{s}\end{align*} for the reaction \begin{align*}4 \ \mathrm{NH}_{3(g)} + 5 \ \mathrm{O}_{2(g)} \rightarrow 4 \ \mathrm{NO}_{(g)} + 6 \ \mathrm{H}_2\mathrm{O}_{(g)}\end{align*}?
1. \begin{align*}-35 \ \mathrm{mol/L} \cdot \mathrm{s}\end{align*}
2. \begin{align*}-2.8 \times 10^2 \ \mathrm{mol/L} \cdot \mathrm{s}\end{align*}
3. \begin{align*}2.8 \times 10^{-2} \ \mathrm{mol/L} \cdot \mathrm{s}\end{align*}
4. \begin{align*}35 \ \mathrm{mol/L} \cdot \mathrm{s}\end{align*}
7. It takes \begin{align*}15 \ \mathrm{minutes}\end{align*} for the concentration of a reactant to decrease from \begin{align*}0.45 \ \mathrm{mol/L}\end{align*} to \begin{align*}0.030 \ \mathrm{mol/L}\end{align*}. What is the rate of reaction in \begin{align*}\mathrm{mol/L} \cdot \mathrm{s}?\end{align*}
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http://mathhelpforum.com/geometry/69457-geometry.html | 1. ## geometry
HELP [LEFT]A 12 foot ladder leans against the side of a building. The foot of the ladder is 5 feet from the wall of the building. Determine the height above the ground where the top of the ladder touches the wall.
2. Originally Posted by amathgizzard
12+5=
144+25=194
do you know what Pythogaras' theorem says?
it says $a^2 + b^2 = c^2$
where (in a right triangle) $a$ and $b$ are the legs of the triangle and $c$ is the hypotenuse, that is, the longest side (in this case, it is the length of the ladder)
3. ok I understand what ur sayin, but what is the answer 194 squared
4. Originally Posted by amathgizzard
ok I understand what ur sayin, but what is the answer 194 squared
no.
if you understood what i said you should see that what you need to do is solve
$a^2 + 5^2 = 12^2$ for $a$
no, $\sqrt{119}$ is the answer | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 7, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9370309710502625, "perplexity": 386.91946390482997}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-30/segments/1500549423992.48/warc/CC-MAIN-20170722102800-20170722122800-00069.warc.gz"} |
https://www.appliedforecasting.com/estimating-subjective-probabilities/ | # Estimating Subjective Probabilities
Subjective probabilities play a role in many economic decisions. There is a large theoretical literature on the elicitation of subjective probabilities, and an equally large empirical literature. However, there is a gulf between the two. The theoretical literature proposes a range of procedures t… hat can be used to recover subjective probabilities, but stresses the need to make strong auxiliary assumptions or "calibrating adjustments" to elicited reports in order to recover the latent probability. With some notable exceptions, the empirical literature seems intent on either making those strong assumptions or ignoring the need for calibration. We illustrate how the joint estimation of risk attitudes and subjective probabilities using structural maximum likelihood methods can provide the calibration adjustments that theory calls for. This allows the observer to make inferences about the latent subjective probability, calibrating for virtually any well-specified model of choice under uncertainty. We demonstrate our procedures with experiments in which we elicit subjective probabilities. We calibrate the estimates of subjective beliefs assuming that choices are made consistently with expected utility theory or rank-dependent utility theory. Inferred subjective probabilities are significantly different when calibrated according to either theory, thus showing the importance of undertaking such exercises. Our findings also have implications for the interpretation of probabilities inferred from prediction markets. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9606269001960754, "perplexity": 1539.1758462961216}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-17/segments/1618038098638.52/warc/CC-MAIN-20210417011815-20210417041815-00512.warc.gz"} |
https://www.physicsforums.com/threads/forces-on-a-pressure-vessel-containing-a-vacuum.227996/ | # Forces on a pressure vessel containing a vacuum
1. Apr 10, 2008
### harry_harrin
Hi,
I'm trying to work out how to calculate the forces acting on a vessel containing a vacuum. The vessel is not cylindrical or spherical. Can anyone suggest any websites or textbooks that might be useful?
Thanks
2. Apr 11, 2008
### jaap de vries
I think that should be looked upon as a external pressure problem. The fact that its a vacuum inside or 1 atm with 2 atm outside doesn't matter.
The construction of pressure vessels is given in the ASME boiler and pressure vessel code manual section VIII. (I am not sure anymore if external pressure is covered but I believe it is)
Those are not that easy to get your hands on but a large university should have it in their library.
3. Apr 11, 2008
### harry_harrin
Thats great, Thanks very much for your help jaap
4. Apr 11, 2008
### Q_Goest
Hi harry. ASME code is an excellent resource, but regarding external pressure, ASME is focused on cylindrical or spherical parts.
If you have flat plates, either circular or rectangular, it doesn't matter if there is a vacuum on one side or positive pressure. The calculation is done depending on dP across the plate, so you don't need to concern yourself with whether or not the pressure on either side is below atmospheric - only the dP across the plate is pertinant.
For flat plates, also look for "Roark's Formulas for Stress and Strain".
5. Apr 12, 2008
### harry_harrin
that was really useful, thanks for that | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8024258017539978, "perplexity": 1124.243341016757}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-50/segments/1480698541324.73/warc/CC-MAIN-20161202170901-00494-ip-10-31-129-80.ec2.internal.warc.gz"} |
https://chem.libretexts.org/Courses/Brevard_College/CHE_104%3A_Principles_of_Chemistry_II/03%3A_Solutions_and_Colloids/3.03%3A_The_Dissolving_Process | Skip to main content
# 3.3: The Dissolving Process
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Learning Outcomes
• Define a solution and describe the parts of a solution.
• Describe how an aqueous solution is formed from both ionic compounds and molecular compounds.
• Recognize that some compounds are insoluble in water.
• Describe the differences among strong electrolytes, weak electrolytes, and nonelectrolytes.
## Forming a Solution
When one substance dissolves into another, a solution is formed. A solution is a homogenous mixture consisting of a solute dissolved into a solvent. The solute is the substance that is being dissolved, while the solvent is the dissolving medium. Solutions can be formed with many different types and forms of solutes and solvents. In this chapter, we will focus on solution where the solvent is water. An aqueous solution is water that contains one or more dissolved substance. The dissolved substances in an aqueous solution may be solids, gases, or other liquids.
In order to be a true solution, a mixture must be stable. When sugar is fully dissolved into water, it can stand for an indefinite amount of time, and the sugar will not settle out of the solution. Further, if the sugar-water solution is passed through a filter, it will remain with the water. This is because the dissolved particles in a solution are very small, usually less than $$1 \: \text{nm}$$ in diameter. Solute particles can be atoms, ions, or molecules, depending on the type of substance that has been dissolved.
### The Dissolving Process
Water typically dissolves most ionic compounds and polar molecules. Nonpolar molecules, such as those found in grease or oil, do not dissolve in water. We will first examine the process that occurs when an ionic compound, such as table salt (sodium chloride), dissolves in water.
Water molecules move about continuously due to their kinetic energy. When a crystal of sodium chloride is placed into water, the water's molecules collide with the crystal lattice. Recall that the crystal lattice is composed of alternating positive and negative ions. Water is attracted to the sodium chloride crystal because water is polar; it has both a positive and a negative end. The positively charged sodium ions in the crystal attract the oxygen end of the water molecules because they are partially negative. The negatively charged chloride ions in the crystal attract the hydrogen end of the water molecules because they are partially positive. The action of the polar water molecules takes the crystal lattice apart (see figure below).
After coming apart from the crystal, the individual ions are then surrounded by solvent particles in a process called solvation. Note in the figure above that the individual $$\ce{Na^+}$$ ions are surrounded by water molecules with the oxygen atom oriented near the positive ion. Likewise, the chloride ions are surrounded by water molecules with the opposite orientation. Hydration is the process of solute particles being surrounded by water molecules arranged in a specific manner. Hydration helps to stabilize aqueous solutions by preventing the positive and negative ions from coming back together and forming a precipitate.
Table sugar is made of the molecular compound sucrose $$\left( \ce{C_{12}H_{22}O_{11}} \right)$$. Solid sugar consists of individual sugar molecules held together by intermolecular attractive forces. When water dissolves sugar, it separates the individual sugar molecules by disrupting the attractive forces, but it does not break the covalent bonds between the carbon, hydrogen, and oxygen atoms. Dissolved sugar molecules are also hydrated. The hydration shell around a molecule of sucrose is arranged so that its partially negative oxygen atoms are near the partially positive hydrogen atoms in the solvent, and vice versa.
### Insoluble Compounds
Not all compounds dissolve well in water. Some ionic compounds, such as calcium carbonate $$\left( \ce{CaCO_3} \right)$$ and silver chloride $$\left( \ce{AgCl} \right)$$, are nearly insoluble. This is because the attractions between the ions in the crystal lattice are stronger than the attraction that the water molecules have for the ions. As a result, the crystal remains intact. The solubility of ionic compounds can be predicted using the solubility rules as shown in Table $$\PageIndex{1}$$.
Table $$\PageIndex{1}$$: Solubility rules for ionic compounds in water.
Nonpolar compounds also do not dissolve in water. The attractive forces that operate between the particles in a nonpolar compound are weak dispersion forces. In order for a nonpolar molecule to dissolve in water, it would need to break up some of the hydrogen bonds between adjacent water molecules. In the case of an ionic substance, these favorable interactions are replaced by other attractive interactions between the ions and the partial charges on water. However, interactions between nonpolar molecules and water are less favorable than the interactions that water makes with itself. When a nonpolar liquid such as oil is mixed with water, two separate layers form, because the liquids will not dissolve into each other (see figure below). When a polar liquid like ethanol is mixed with water, they completely blend and dissolve into one another. Liquids that dissolve in one another in all proportions are said to be miscible. Liquids that do not dissolve in one another are called immiscible. The general rule for deciding if one substance is capable of dissolving another is "like dissolves like", where the property being compared is the overall polarity of the substance. For example, a nonpolar solid such as iodine will dissolve in nonpolar lighter fluid, but it will not dissolve in polar water.
## Electrolytes and Nonelectrolytes
An electrolyte is a compound that conducts an electric current when it is dissolved in water or melted. In order to conduct a current, a substance must contain mobile ions that can move from one electrode to the other. All ionic compounds are electrolytes. When ionic compounds dissolve, they break apart into ions, which are then able to conduct a current. Even insoluble ionic compounds, such as $$\ce{CaCO_3}$$, are considered electrolytes because they can conduct a current in the molten (melted) state.
A nonelectrolyte is a compound that does not conduct an electric current in either aqueous solution or in the molten state. Many molecular compounds, such a sugar or ethanol, are nonelectrolytes. When these compounds dissolve in water, they do not produce ions. Illustrated below is the difference between an electrolyte and a nonelectrolyte.
### Dissociation
Earlier, you saw how an ionic crystal lattice breaks apart when it is dissolved in water. Dissociation is the separation of ions that occurs when a solid ionic compound dissolves. Simply undo the crisscross method that you learned when writing chemical formulas for ionic compounds, and you are left with the components of an ionic dissociation equation. The subscripts for the ions in the chemical formulas become the coefficients of the respective ions on the product side of the equations. Shown below are dissociation equations for $$\ce{NaCl}$$, $$\ce{Ca(NO_3)_2}$$, and $$\ce{(NH_4)_3PO_4}$$.
\begin{align} &\ce{NaCl} \left( s \right) \rightarrow \ce{Na^+} \left( aq \right) + \ce{Cl^-} \left( aq \right) \\ &\ce{Ca(NO_3)_2} \left( s \right) \rightarrow \ce{Ca^{2+}} \left( aq \right) + 2 \ce{NO_3^-} \left( aq \right) \\ &\ce{(NH_4)_3PO_4} \left( s \right) \rightarrow 3 \ce{NH_4^+} \left( aq \right) + \ce{PO_4^{3-}} \left( aq \right) \end{align}
One formula unit of sodium chloride dissociates into one sodium ion and one chloride ion. The calcium nitrate formula unit dissociates into one calcium ion and two nitrate ions, because the $$2+$$ charge of each calcium ion requires two nitrate ions (each with a charge of $$1-$$) to form an electrically neutral compound. The ammonium phosphate formula unit dissociates into three ammonium ions and one phosphate ion.
Do not confuse the subscripts of the atoms within the polyatomic ion for the subscripts that result from the crisscrossing of the charges that make the original compound neutral. The 3 subscript of the ntirate ion and the 4 subscript of the ammonium ion are part of the polyatomic ion and remain a part of the ionic formula after the compound dissociates. Notice that the compounds are solids $$\left( s \right)$$ that become ions when dissolved in water, producing an aqueous solution $$\left( aq \right)$$.
Nonelectrolytes do not dissociate when forming an aqueous solution. An equation can still be written that simply shows the solid going into solution. For example, the process of dissolving sucrose in water can be written as follows:
$\ce{C_{12}H_{22}O_{11}} \left( s \right) \rightarrow \ce{C_{12}H_{22}O_{11}} \left( aq \right)$
### Strong and Weak Electrolytes
Some polar molecular compounds are nonelectrolytes when the are in their pure state but become electrolytes when they are dissolved in water. Hydrogen chloride $$\left( \ce{HCl} \right)$$ is a gas in its pure molecular state and is a nonelectrolyte. However, when $$\ce{HCl}$$ is dissolved in water, it conducts a current well because the $$\ce{HCl}$$ molecule ionizes into hydrogen and chloride ions.
$\ce{HCl} \left( g \right) \rightarrow \ce{H^+} \left( aq \right) + \ce{Cl^-} \left( aq \right)$
When $$\ce{HCl}$$ is dissolved into water, it is called hydrochloric acid. Ionic compounds and some polar compounds are completely broken apart into ions and thus conduct a current very well. A strong electrolyte is a solution in which almost all of the dissolved solute exists as ions.
Some other polar molecular compounds become electrolytes upon being dissolved into water but do not ionize to a very great extent. For example, nitrous acid $$\left( \ce{HNO_2} \right)$$ only partially ionizes into hydrogen ions and nitrite ions when dissolved in water. Aqueous nitrous acid is composed of only about $$5\%$$ ions and $$95\%$$ intact nitrous acid molecules A weak electrolyte is a solution in which only a small fraction of the dissolved solute exists as ions. The equation showing the ionization of a weak electrolyte utilizes an equilibrium arrow, indicating an equilibrium between the reactants and products.
$\ce{HNO_2} \left( aq \right) \rightleftharpoons \ce{H^+} \left( aq \right) + \ce{NO_2^-} \left( aq \right)$
## Contributors and Attributions
• Allison Soult, Ph.D. (Department of Chemistry, University of Kentucky)
This page titled 3.3: The Dissolving Process is shared under a not declared license and was authored, remixed, and/or curated by Allison Soult.
• Was this article helpful? | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 1, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9112764000892639, "perplexity": 1213.5572669960388}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-14/segments/1679296943749.68/warc/CC-MAIN-20230322020215-20230322050215-00300.warc.gz"} |
https://cstheory.stackexchange.com/questions/9652/is-every-busy-beaver-strictly-monotonic-asymptotically/9655 | # Is every busy beaver strictly monotonic asymptotically?
To be specific, let me first define the busy beaver function BB(n)= maximum number of 1's that can be printed on the tape (i.e., the maximum score) by a standard n-state, 2-symbol (0 and 1) Turing machine before halt, starting with an all-zero tape.
My first question maybe trivial: is it true that BB(n+1)>BB(n) for all n?
The claim is true if we define the busy beaver by BB'(n)= maximum number of shifts made by the machine before halt. (This is because we can always keep everything else the same, and change the last shift made by BB'(n), say $\{ 0{q}_{i}1RH\}$, to $\{ 0{q}_{i}1R{q}_{n+1}\}$ and $\{1{q}_{n+1}1RH\}$ to arrive at an (n+1)-state turing machine which has exactly one more shift than BB'(n) before halt.)
Can we show that the strict inequality holds too if we measure by scores?
What about the general case? Is every busy beaver strictly monotonic, for large enough n?
## 1 Answer
Take an $n$-state Turing machine $M$ which outputs $k$ one symbols. Define the new Turing machine $M'$ with $n+1$ states as follows. Every transition in $M$ to the ACCEPT state instead goes into a new state $q'$. The state $q'$ has the following behavior. If the head currently contains a ZERO, we overwrite with a ONE and ACCEPT. If the head currently contains a ONE, we shift to the right and remain in state $q'$.
As machine $M$ halts, so does machine $M'$. Furthermore, machine $M'$ always puts one $k+1$ ONES onto the tape.
So $BB(n+1) \geq k+1 > k = BB(n)$ | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8724828958511353, "perplexity": 743.5245514936303}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-29/segments/1593657140337.79/warc/CC-MAIN-20200712211314-20200713001314-00481.warc.gz"} |
https://infoscience.epfl.ch/record/219070 | ## Conformations of prolyl-peptide bonds in the bradykinin 1-5 fragment in solution and in the gas phase
The dynamic nature of intrinsically disordered peptides makes them a challenge to characterize by solution-phase techniques. In order to gain insight into the relation between the disordered state and the environment, we explore the conformational space of the N-terminal 1-5 fragment of bradykinin (BK[1-5]2+) in the gas phase by combining drift tube ion mobility, cold-ion spectroscopy, and first principles simulations. The ion-mobility distribution of BK[1-5]2+ consists of two well-separated peaks. We demonstrate that the conformations within the peak with larger cross-section are kinetically trapped, while the more compact peak contains low-energy structures. This is a result of cis-trans isomerization of the two prolyl-peptide bonds in BK[1-5]2+. Density-functional theory calculations reveal that the compact structures have two very different geometries with cis-trans and trans-cis backbone conformations. Using the experimental CCSs to guide the conformational search, we find that the kinetically trapped species have a trans-trans configuration. This is consistent with NMR measurements performed in a solution, which show that 82% of the molecules adopt a trans-trans configuration and behave as a random coil.
Published in:
Journal of the American Chemical Society, 138, 29, 9224-9233
Year:
2016
Publisher:
Washington, Amer Chemical Soc
Keywords:
Laboratories: | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8209127187728882, "perplexity": 3585.3247854499464}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-13/segments/1521257644701.7/warc/CC-MAIN-20180317055142-20180317075142-00344.warc.gz"} |
https://experts.arizona.edu/en/publications/disorder-induced-enhancement-and-critical-scaling-of-spontaneous- | # Disorder-induced enhancement and critical scaling of spontaneous magnetization in random-field quantum spin systems
Anindita Bera, Debraj Rakshit, Maciej Lewenstein, Aditi Sen, Ujjwal Sen, Jan Wehr
Research output: Contribution to journalArticlepeer-review
4 Scopus citations
## Abstract
We investigate the effect of a unidirectional quenched random field on the anisotropic quantum spin-1/2 XY model, which magnetizes spontaneously in the absence of the random field. We adopt a mean-field approach for this analysis. In general, the models considered have Ising symmetry, and as such they exhibit ferromagnetic order in two and three dimensions in the presence of not too large disorder. Even in the special case when the model without disorder has U(1) symmetry, a small U(1)-symmetry-breaking random field induces ferromagnetic long-range order in two dimensions. The mean-field approach, consequently, provides a rather good qualitative and even quantitative description when applied not too close to the criticality. We show that spontaneous magnetization persists even in the presence of the random field, but the magnitude of magnetization gets suppressed due to disorder, and the system magnetizes in the directions parallel and transverse to the random field. Our results are obtained via analytical calculations within a perturbative framework and by numerical simulations. Interestingly, we show that it is possible to enhance a component of magnetization in the presence of the disorder field provided that we apply an additional constant field in the XY plane. Moreover, we derive generalized expressions for the critical temperature and the scalings of the magnetization near the critical point for the XY spin system with arbitrary fixed quantum spin angular momentum.
Original language English (US) 014421 Physical Review B 94 1 https://doi.org/10.1103/PhysRevB.94.014421 Published - Jul 18 2016
## ASJC Scopus subject areas
• Electronic, Optical and Magnetic Materials
• Condensed Matter Physics
## Fingerprint
Dive into the research topics of 'Disorder-induced enhancement and critical scaling of spontaneous magnetization in random-field quantum spin systems'. Together they form a unique fingerprint. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8585993051528931, "perplexity": 1387.1060321314633}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-33/segments/1659882571210.98/warc/CC-MAIN-20220810191850-20220810221850-00425.warc.gz"} |
https://www.physicsforums.com/threads/curious-number-theory-problem.856994/ | # Homework Help: Curious Number Theory Problem
1. Feb 12, 2016
### Ryan888
The problem statement, all variables and given/known data:
Recently, a group of fellow math nerds and myself stumbled upon an interesting problem. The problem is stated: "Find the average number of representations of a positive integer as the sum of two squares."
Relevant equations:
N = a^(2) + b^(2), where a and b can be 1 or many corresponding sets. But we are looking for the average number of sets, for all real numbers of N.
The attempt at a solution:
Clearly, I realize a limit will need to be taken, somehow, of all numbers N that can be expressed as the sum of two squares. Naturally, the general equation for N would be N = a^(2) + b^(2). Also, I figured that N must meet the condition that every prime number of the form 4k+3 appears an even number of times in it's prime factorization. I have no idea how to connect these two conditions, or if more must be met. If it's any help, my friend claimed that the answer is π, but had no way to prove it.
Any ideas are greatly appreciated!
2. Feb 12, 2016
### SammyS
Staff Emeritus
I'm a bit confused by your write-up.
Are you find the average number of such representations over the set of real numbers or over the set of positive integers (the set of natural numbers)? ... or does it not matter?
3. Feb 12, 2016
### Hornbein
If a and b are real numbers then there are an infinite number of representations. So they must be integers, natural numbers, or counting numbers.
Then if N is a real number it almost never has a representation with whole numbers. So it's got to be a whole number too. But then there is usually no representation. So I don't see how pi can be the answer. It is either infinite or close to zero, I think.
4. Feb 12, 2016
### Ryan888
Unfortunately, that one sentence stating the problem is the only information given. My guess is it probably means the natural numbers. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8803879618644714, "perplexity": 279.90916887989385}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-30/segments/1531676589557.39/warc/CC-MAIN-20180717031623-20180717051623-00360.warc.gz"} |
https://www.physicsforums.com/threads/an-interesting-rifleman-problem-projectile-motion.132165/ | # An interesting Rifleman Problem (Projectile Motion)
1. Sep 15, 2006
### Tony Zalles
Hi,
So this is the problem out of Fundamentals of Physics by Haliday Resnick and Walker (Seventh Edition.
By the way this problem would all under a first year calculus based mechanics course.
I know calculus through multivariable...but thats not needed for this problem...at least I don't think so.
In any case here it is:
31. A rifle that shoots bullets at 460 m/s is to be aimed at target 45.7 m away. If the center of the target is level with the rifle, how high above the target must the rifle barrel be pointed so that this bullet hits dead center.
Ok... So I know that there is actually a derivation known as the "Rifleman's Rule" for projectile motion (I believe) can be applied to this problem....but it is a very tedious derivation...whose end result is an approximation only (which works out fine because it is intended to be an approximation due to the very small angles that would be used) but that aside, I believe there is a simpler way to work out this problem. But I fail to see how...
Ok here is what I did...
Ok the barrel has to be fired at some [initial angle] and its path is parabolic....
this gives me some useful insight, the time for bullet to go half the horizontal distance if doubled gives the full horizontal distance...and that the [initial angle] when measured from the +x is related to the final angle by taking
[180 degrees - [inital angle]] = [final angle]
(taking final angle to be measured from +x aswell.
Ok so listing the magnitudes:
[inital velocity] = 460
[final velocity] = ?
---------------------------------
[inital x position] = 0
[inital y position] = 0
[final x position] = 45.7
[inital angle] = ?
[final angle] = ?
[initial velocity in x] = ?
[final velocity in y] = ?
[final velocity in x] = [initial velocity in x] = ?
[final velocity in y] = ?
Ok velocity in x is constant
Velcity in y follows a free fall system
I already derived the initial components of velocity....
cos [inital angle] * Vo = x component of initial velocity
sin [inital angle] * Vo = y component of initial velocity
I set up two position functions each for x and y plugged in the above derivations.
x(t) = 0 + Vx0 * [delta t] + 0
y(t) = 0 + Vy0* [delta t] - (1/2)*g*{[delta t]^2}
I also then set up two equations from the form
Final velocity = initial velocity + at
So two equations of the above form each for x and y....and you know by I figured I could see something that if I solved for in of these equations plugged back in to another....something would cancel out and I would actually be able to solve for at least one of the variable....
or be able to set up a system of equations for my variables....however....I'm not be able to solve this using either approach....
so...yea. any help would be nice :)
Thanks,
-Tony Zalles
2. Sep 15, 2006
### HallsofIvy
Staff Emeritus
You have
x(t) = 0 + Vx0 * [delta t] + 0
y(t) = 0 + Vy0* [delta t] - (1/2)*g*{[delta t]^2}
You know that the target is 45.7 m away so you can solve
45.= Vx0 * [delta t] for delta t as a function of Vx0.
You can then put that into 0 = Vy0* [delta t] - (1/2)*g*{[delta t]^2} and get an equation relating Vx0 and Vy0. From that you should be able to calculate the initial angle. Now use tangent of that angle to find how high above the target you must aim.
3. Sep 15, 2006
### Tony Zalles
ok,...
So for the initial theta I end with the following:
2*(sin(inital angle))*(cos(initial angle)) = [(final x)(g)]/(initial velocity)
And with figures:
2*(sin(inital angle))*(cos(initial angle)) = [(45.7)(9.8)]/(460)
Which reduces to...
sin(2*(initial angle)) = [(final x)(g)]/(initial velocity)
solving for (inital angle)
gives,
(initial angle) = 38.4 degrees
Now, I believe I can set...
tan(initial angle) = (final y)/(final x)
However in deriving this relationship I already assumed that the final y position was zero in order to be hit dead on to the target...
so...this problem is annoying they didn't even give a figure/picture...
Although I really appreciate the help HallsofIvy :)
The answer in the back of the book is 4.84 cm.
And I'm not getting that....and I know the angle is right because if I use it to calculate the Vx0 and Vyo and take the square of each and then the square root of their sum I end up back with 460....so yea...still a little stuck....
Again any help would be most appreciated :)
-Tony Zalles
4. Sep 15, 2006
### andrevdh
The range of a projectile can be shown to be given by
$$R = \frac{V^2\sin(2\alpha)}{g}$$
where V is the launch velocity and alpha the initial launching angle.
5. Sep 15, 2006
### Tony Zalles
Never mind,,....numerical error on my part....it all works and I get the answer in the back of the book.
Thanks.
-Tony Zalles
Similar Discussions: An interesting Rifleman Problem (Projectile Motion) | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8947180509567261, "perplexity": 2250.0542539281696}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-39/segments/1505818686169.5/warc/CC-MAIN-20170920033426-20170920053426-00695.warc.gz"} |
http://mathhelpforum.com/algebra/99210-stuck.html | Math Help - Stuck
1. Stuck
The instructions say just to solve.
x^2 +2x -4
x^2 + 2x = 4
Now I add (b/2a)squared which would be (4/4)which equals 1
x^2 + 2x +1 = 4 + 1
x^2 +2x + 1= 5
stuck as to what to do next?
2. Hi
x² + 2x + 1 = (x+1)²
3. sorry I don't think you understood because I wrote in wrong
x^2 +2x - 4 = 0
x^2 +2x = 4
then I add 1/2 of the coefficient of x and square it, then add the results to both sides of the equation. Therefore I have (2/2)squared which gives me (4/4) which is 1
x^2 + 2x + 1 = 4 + 1
x^2 + 2x + 1 = 5
x^2 + 2x + 1 - 5 = 5-5
x^2 + 2x - 4 = 0
Now I am stuck if I have done everything else right
4. You did all the hard work to get:
$x^2 + 2x + 1 = 5$
and then undid it all by subtracting 5 from it again.
take a look at the last answer you got:
$x^2 + 2x + 1 = (x+1)^2$
That's why this technique is called "completing the square".
5. Originally Posted by Brama Bull
sorry I don't think you understood because I wrote in wrong
x^2 +2x - 4 = 0
x^2 +2x = 4
then I add 1/2 of the coefficient of x and square it, then add the results to both sides of the equation. Therefore I have (2/2)squared which gives me (4/4) which is 1
x^2 + 2x + 1 = 4 + 1
x^2 + 2x + 1 = 5
This is where you stopped before and this is where you should have stopped! Do you understand why "add 1/2 of the coefficient of x and square it"? The whole purpose is to make one side a "perfect square". That was why running gag said " $x^2+ 2x+ 1= (x+1)^2$".
Can you solve $(x+1)^2= 5$?
x^2 + 2x + 1 - 5 = 5-5
x^2 + 2x - 4 = 0
Now I am stuck if I have done everything else right
These last two lines just undo what you did in the first two lines! | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 4, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8091204762458801, "perplexity": 377.64445245256115}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-41/segments/1410657124356.76/warc/CC-MAIN-20140914011204-00298-ip-10-196-40-205.us-west-1.compute.internal.warc.gz"} |
https://openreview.net/forum?id=SygD31HFvB | ## A Novel Analysis Framework of Lower Complexity Bounds for Finite-Sum Optimization
25 Sep 2019 (modified: 24 Dec 2019)ICLR 2020 Conference Blind SubmissionReaders: Everyone
• Original Pdf: pdf
• Keywords: convex optimization, lower bound complexity, proximal incremental first-order oracle
• Abstract: This paper studies the lower bound complexity for the optimization problem whose objective function is the average of $n$ individual smooth convex functions. We consider the algorithm which gets access to gradient and proximal oracle for each individual component. For the strongly-convex case, we prove such an algorithm can not reach an $\eps$-suboptimal point in fewer than $\Omega((n+\sqrt{\kappa n})\log(1/\eps))$ iterations, where $\kappa$ is the condition number of the objective function. This lower bound is tighter than previous results and perfectly matches the upper bound of the existing proximal incremental first-order oracle algorithm Point-SAGA. We develop a novel construction to show the above result, which partitions the tridiagonal matrix of classical examples into $n$ groups to make the problem difficult enough to stochastic algorithms. This construction is friendly to the analysis of proximal oracle and also could be used in general convex and average smooth cases naturally.
10 Replies | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9723349809646606, "perplexity": 560.9147552731915}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-04/segments/1610704799711.94/warc/CC-MAIN-20210126073722-20210126103722-00609.warc.gz"} |
http://mathhelpforum.com/calculus/55843-more-cauchy-sequences.html | # Math Help - More Cauchy Sequences
1. ## More Cauchy Sequences
If (Xsubn) and (Ysubn) are Cauchy Sequences, show that (Xsubn + Ysubn) and (Xsubn*Ysubn) are Cauchy Sequences. I am asked to prove this using the definition of a Cauchy Sequence only. Any help is much appreciated.
2. Originally Posted by jkru
If (Xsubn) and (Ysubn) are Cauchy Sequences, show that (Xsubn + Ysubn) and (Xsubn*Ysubn) are Cauchy Sequences. I am asked to prove this using the definition of a Cauchy Sequence only. Any help is much appreciated.
i will do the first (since it is easier ) you do the second
since $\{ x_n \}$ is a Cauchy sequence, we have that for all $\epsilon > 0$ there exists $N_1 \in \mathbb{N}$, such that $n,m > N_1$ implies $|x_n - x_m| < \frac {\epsilon}2$
similarly, we have $N_2 \in \mathbb{N}$ so that $n,m > N_2$ implies $|y_n - y_m| < \frac {\epsilon}2$
Now, choose $N = \text{max} \{ N_1,N_2 \}$. then $n,m > N$ implies
$|(x_n + y_n) - (x_m + y_m)| = |(x_n - x_m) + (y_n - y_m)| \le |x_n - x_m| + |y_n - y_m| < \frac {\epsilon}2 + \frac {\epsilon}2 = \epsilon$
so that $\{ x_n + y_n \}$ is Cauchy
(note that the $\le$ follows by the triangle inequality)
now do the second. use the defintion as i did | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 13, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9851676225662231, "perplexity": 548.2217663468713}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-41/segments/1412037663711.39/warc/CC-MAIN-20140930004103-00408-ip-10-234-18-248.ec2.internal.warc.gz"} |
https://socratic.org/questions/about-how-many-elements-form-all-substances | Chemistry
Topics
# About how many elements form all substances?
Feb 12, 2017
118 or 94 depending on how you look at it.
#### Explanation:
The first 94 elements are naturally occurring, and the elements from 95 to 118 are man made in laboratories.
In general, if you have a substance, (and it is not in a laboratory) then it is made of 94 possible elements. However, special man-made substances could, in theory, contain the other elements for a while.
##### Impact of this question
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https://www.answers.com/physics/In_what_condition_average_velocity_is_equal_to_average_speed | 0
# In what condition average velocity is equal to average speed?
Wiki User
2008-04-05 06:49:33
1. magnitude of distance covered is equal to the magnitude of displacement. 2. the motion of the object is in a straight line i.e. in a particular direction.
Wiki User
2008-04-05 06:49:33
Study guides
20 cards
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See all cards
4.16
140 Reviews | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9281606674194336, "perplexity": 2756.180255229305}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-21/segments/1652662525507.54/warc/CC-MAIN-20220519042059-20220519072059-00156.warc.gz"} |
http://link.springer.com/article/10.1007%2Fs10573-009-0028-2 | , Volume 45, Issue 2, pp 211-217
Date: 14 Apr 2009
# Experimental investigation of gasless detonation in metal-sulfur compositions
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## Abstract
Samples of zinc-sulfur and manganese-sulfur mixtures are shocked using an explosive pentolite charge to investigate if a shock-initiated reaction is able to support continued shock wave propagation. Samples of two different nominal densities (62 and 86% of theoretical maximum density) are prepared as weakly confined cylinders 50 mm in diameter and are instrumented along their length (⩽280 mm) with sensitive piezoelectric pins. Experimental results showed that the shock wave transmitted into the sample by the explosive rapidly decays to an acoustic wave in all four sample types. Furthermore, in denser samples, the part of the sample farthest from the explosive is recovered intact and unreacted, which clearly indicates that the wave is unable to trigger reactions after 100 mm of travel along the sample. Thus, it is concluded that insufficient reaction energy is transmitted forward to the shock wave to prevent its decay as it travels along the sample.
__________
Translated from Fizika Goreniya i Vzryva, Vol. 45, No. 2, pp. 116–123, March–April, 2009. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8446062803268433, "perplexity": 2177.1300137026833}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-14/segments/1427131298020.57/warc/CC-MAIN-20150323172138-00165-ip-10-168-14-71.ec2.internal.warc.gz"} |
https://socratic.org/questions/how-do-you-solve-graphically-abs-x-4-abs-3x-1 | Algebra
Topics
# How do you solve graphically abs(x – 4)>abs(3x – 1)?
Aug 10, 2018
$- \frac{3}{2} < x < \frac{5}{4}$
#### Explanation:
The first thing you do is to draw the graphs $y = \left\mid x - 4 \right\mid$ and $y = \left\mid 3 x - 1 \right\mid$ on the SAME graph
graph{(y-abs(x-4))(y-abs(3x-1))=0 [-20.28, 20.27, -10.14, 10.13]}
Now the question is what parts of the graph above satisfies the equation $\left\mid x - 4 \right\mid > \left\mid 3 x - 1 \right\mid$. What it is asking you is what part of the graph $y = \left\mid x - 4 \right\mid$ is above the graph $y = \left\mid 3 x - 1 \right\mid$.
Hence, going from the right, the equation of each branch is $y = x - 4$, $y = 3 x - 1$, $y = 4 - x$ and $y = 1 - 3 x$.
The branches $y = 4 - x$ and $y = 3 x - 1$ meet at a point and so do $y = 4 - x$ and $y = 1 - 3 x$
Therefore, we need to find the point of intersection
$y = 4 - x$ and $y = 3 x - 1$
$4 - x = 3 x - 1$
$5 = 4 x$
$x = \frac{5}{4}$
$y = 4 - x$ and $y = 1 - 3 x$
$4 - x = 1 - 3 x$
$2 x = - 3$
$x = - \frac{3}{2}$
Finally, looking at where $y = \left\mid x - 4 \right\mid$ is above the graph $y = \left\mid 3 x - 1 \right\mid$, we can tell that it is $- \frac{3}{2} < x < \frac{5}{4}$
##### Impact of this question
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http://mathhelpforum.com/calculus/220202-help-calculating-geometric-sequence.html | # Thread: Help with calculating geometric sequence
1. ## Help with calculating geometric sequence
Hi,
I hope it is the right place to post it...
I need help with the following sequence:
$\sum_{i=0}^{j}2^{2^i}=N$
(it's 2 to the power of 2 to the power of i)
I need to find i as a function of N.
meaning how further away in the sequence do i need to go, in order to get N.
any help would greatly appreciated.
2. ## Re: Help with calculating geometric sequence
Hey Stormey.
Did you mean as a function of j?
I don't know of a closed form solution but you should look into techniques like Euler-Mclaurin series and their relationship to integrals.
3. ## Re: Help with calculating geometric sequence
Hi chiro.
yes, as a function of j, sorry.
the truth is that it is actually a question from data starctures in computer science, so I almost sure the solution suppose to be accomplished with discrete math.
(using some manipulation to compute the sum of a more simple sequence, to which the sum is known or easy to calculate, and then calculating the sum of the original sequence,
I also tried to think about some change of variable [index in this case], that will help simplify this problem)
4. ## Re: Help with calculating geometric sequence
You might want to consider a bit-wise representation (i.e. binary) and how that relates to N. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 1, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8534749746322632, "perplexity": 767.0123541536875}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-30/segments/1500549424610.13/warc/CC-MAIN-20170723202459-20170723222459-00511.warc.gz"} |
https://lavelle.chem.ucla.edu/forum/viewtopic.php?p=12868 | Midterm 2011 Q4 - Partial Pressures and K
$\Delta G^{\circ}= \Delta H^{\circ} - T \Delta S^{\circ}$
$\Delta G^{\circ}= -RT\ln K$
$\Delta G^{\circ}= \sum \Delta G_{f}^{\circ}(products) - \sum \Delta G_{f}^{\circ}(reactants)$
Melissa Trieu 1K
Posts: 6
Joined: Fri Sep 26, 2014 2:02 pm
Midterm 2011 Q4 - Partial Pressures and K
The question says that the partial pressures of each reactant (PCl3 and Cl2) is half that of the product (PCl5). In the book the partial pressures, after calculations is equal to 0.4 for the PCl5 and 0.2 for PCl3 and Cl2. Is it also right if I said that the partial pressure for PCl5 was 0.2 and 0.1 for PCl3 and Cl2?
Sarah H Brown 1L
Posts: 56
Joined: Fri Sep 26, 2014 2:02 pm
Re: Midterm 2011 Q4
It all depends on what your equation comes down to. In the case of Q4, it is
K = 10 = 2x/x2 when the reactant pressure is set to x and the product pressure is set to 2x.
10x2= 2x
An x can be crossed out from each side.
10x = 2
x = 0.2 bar. So, the reactant pressures are 0.2 bar and product pressure is 0.4 bar.
The answer would not be the same if the concentrations were 0.1 and 0.2. That would be
(0.2)/(0.1*0.1) which is equal to 20, not 10. K does not equal 20.
So, it all depends on the K you are setting your x expression equal to! | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 3, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8242762088775635, "perplexity": 2027.7370971552352}, "config": {"markdown_headings": false, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-47/segments/1573496668954.85/warc/CC-MAIN-20191117115233-20191117143233-00451.warc.gz"} |
http://mathhelpforum.com/calculus/45931-sketching-contours-print.html | # sketching contours
• Aug 14th 2008, 03:02 AM
sketching contours
I have the function f(x,y)= 2x^3 + 3xy + 2y^3, I have found the 2 critical points for x in R and y in R.
I believe they are (0,0) and (-1/8, 1/2)?!
I then found that they are both saddle points which is when I doubted whether these critical points are correct.
I then have to sketch the contours which I have attempted but am unsure if I have constructed the sketch correctly.
• Aug 14th 2008, 03:09 AM
mr fantastic
Quote:
Originally Posted by linearalgebradunce
I have the function f(x,y)= 2x^3 + 3xy + 2y^3, I have found the 2 critical points for x in R and y in R.
I believe they are (0,0) and (-1/8, 1/2)?!
I then found that they are both saddle points which is when I doubted whether these critical points are correct.
I then have to sketch the contours which I have attempted but am unsure if I have constructed the sketch correctly.
The equations to solve are:
$6 x^2 + 3y = 0 \Rightarrow 2 x^2 + y = 0$ .... (1)
$3x + 6 y^2 = 0 \Rightarrow x = -2 y^2$ .... (2)
Substitute (2) into (1):
$8 y^4 + y = 0 \Rightarrow y (8 y^3 + 1) = 0$
$\Rightarrow y = 0, ~ - \frac{1}{2}$ .....
• Aug 14th 2008, 03:48 AM | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 4, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8375558853149414, "perplexity": 1134.5351112625622}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-04/segments/1484560280308.24/warc/CC-MAIN-20170116095120-00122-ip-10-171-10-70.ec2.internal.warc.gz"} |
https://repo.scoap3.org/search?f=author&p=Eidelman%2C%20S.&ln=en | SCOAP3 Repository 264 records found 1 - 10 Search took 0.11 seconds.
1 Observation of ${B}^{+}\to p\overline{\Lambda }{K}^{+}{K}^{-}$ and ${B}^{+}\to \overline{p}\Lambda {K}^{+}{K}^{+}$ / Lu, P.C. ; Wang, M.Z. ; Chistov, R. ; Chang, P. ; et al We report the study of ${B}^{+}\to p\overline{\Lambda }{K}^{+}{K}^{-}$ and ${B}^{+}\to \overline{p}\Lambda {K}^{+}{K}^{+}$ decays using a $772×{10}^{6}\text{}\text{}B\overline{B}$ pair data sample recorded on the $\Upsilon \left(4S\right)$ resonance with the Belle detector at KEKB. [...] Published in Physical Review D 99 (2019) 10.1103/PhysRevD.99.032003 arXiv:1807.10503 Fulltext: PDF XML; 2 Search for a Light $CP$-odd Higgs Boson and Low-Mass Dark Matter at the Belle Experiment / Seong, I. S. ; Vahsen, S. E. ; Adachi, I. ; Aihara, H. ; et al We report on the first Belle search for a light $CP$-odd Higgs boson, ${A}^{0}$, that decays into low mass dark matter, $\chi$, in final states with a single photon and missing energy. [...] Published in Physical Review Letters 122 (2019) 10.1103/PhysRevLett.122.011801 arXiv:1809.05222 Fulltext: XML PDF; 3 Observation of Transverse $\Lambda /\overline{\Lambda }$ Hyperon Polarization in ${e}^{+}{e}^{-}$ Annihilation at Belle / Guan, Y. ; Vossen, A. ; Adachi, I. ; Adamczyk, K. ; et al We report the first observation of the spontaneous polarization of $\Lambda$ and $\overline{\Lambda }$ hyperons transverse to the production plane in ${e}^{+}{e}^{-}$ annihilation, which is attributed to the effect arising from a polarizing fragmentation function. [...] Published in Physical Review Letters 122 (2019) 10.1103/PhysRevLett.122.042001 arXiv:1808.05000 Fulltext: XML PDF; 4 Search for $CP$ violation with kinematic asymmetries in the ${D}^{0}\to {K}^{+}{K}^{-}{\pi }^{+}{\pi }^{-}$ decay / Kim, J. B. ; Won, E. ; Adachi, I. ; Aihara, H. ; et al We search for $CP$ violation in the singly-Cabibbo-suppressed decay ${D}^{0}\to {K}^{+}{K}^{-}{\pi }^{+}{\pi }^{-}$ using data corresponding to an integrated luminosity of $988\text{}\text{}{\mathrm{fb}}^{-1}$ collected by the Belle detector at the KEKB ${e}^{+}{e}^{-}$ collider. [...] Published in Physical Review D 99 (2019) 10.1103/PhysRevD.99.011104 arXiv:1810.06457 Fulltext: XML PDF; 5 Measurement of time-dependent $CP$ violation in ${B}^{0}\to {K}_{S}^{0}{\pi }^{0}{\pi }^{0}$ decays / Yusa, Y. ; Aihara, H. ; Al Said, S. ; Asner, D. M. ; et al We report a measurement of time-dependent $CP$ violation in ${B}^{0}\to {K}_{S}^{0}{\pi }^{0}{\pi }^{0}$ decays using a data sample of $772×{10}^{6}\text{}\text{}B\overline{B}$ pairs collected by the Belle experiment running at the $\Upsilon \left(4S\right)$ resonance at the KEKB ${e}^{+}{e}^{-}$ collider. [...] Published in Physical Review D 99 (2019) 10.1103/PhysRevD.99.011102 arXiv:1810.03336 Fulltext: XML PDF; 6 Observation of Two Resonances in the ${\Lambda }_{b}^{0}{\pi }^{±}$ Systems and Precise Measurement of ${\Sigma }_{b}^{±}$ and ${\Sigma }_{b}^{*±}$ Properties / Aaij, R. ; Abellán Beteta, C. ; Adeva, B. ; Adinolfi, M. ; et al The first observation of two structures consistent with resonances in the final states ${\Lambda }_{b}^{0}{\pi }^{-}$ and ${\Lambda }_{b}^{0}{\pi }^{+}$ is reported using samples of $pp$ collision data collected by the LHCb experiment at $\sqrt{s}=7$ and 8 TeV, corresponding to an integrated luminosity of $3\text{}\text{}{\mathrm{fb}}^{-1}$. [...] Published in Physical Review Letters 122 (2019) 10.1103/PhysRevLett.122.012001 arXiv:1809.07752 Fulltext: XML PDF; 7 Measurement of the Charm-Mixing Parameter ${y}_{CP}$ / Aaij, R. ; Abellán Beteta, C. ; Adeva, B. ; Adinolfi, M. ; et al A measurement of the charm-mixing parameter ${y}_{CP}$ using ${D}^{0}\to {K}^{+}{K}^{-}$, ${D}^{0}\to {\pi }^{+}{\pi }^{-}$, and ${D}^{0}\to {K}^{-}{\pi }^{+}$ decays is reported. [...] Published in Physical Review Letters 122 (2019) 10.1103/PhysRevLett.122.011802 arXiv:1810.06874 Fulltext: XML PDF; 8 Search for the rare decay of ${B}^{+}\to {\ell }^{+}{\nu }_{\ell }\gamma$ with improved hadronic tagging / Gelb, M. ; Bernlochner, F. U. ; Goldenzweig, P. ; Metzner, F. ; et al We present the result of the search for the rare $B$ meson decay of ${B}^{+}\to {\ell }^{+}{\nu }_{\ell }\gamma$ with $\ell =e,\mu$. [...] Published in Physical Review D 98 (2018) 10.1103/PhysRevD.98.112016 arXiv:1810.12976 Fulltext: XML PDF; 9 First Evidence for $\mathrm{cos}2\beta >0$ and Resolution of the Cabibbo-Kobayashi-Maskawa Quark-Mixing Unitarity Triangle Ambiguity / Adachi, I. ; Adye, T. ; Ahmed, H. ; Ahn, J. K. ; et al We present first evidence that the cosine of the $CP$-violating weak phase $2\beta$ is positive, and hence exclude trigonometric multifold solutions of the Cabibbo-Kobayashi-Maskawa (CKM) Unitarity Triangle using a time-dependent Dalitz plot analysis of ${B}^{0}\to {D}^{\left(*\right)}{h}^{0}$ with $D\to {K}_{S}^{0}{\pi }^{+}{\pi }^{-}$ decays, where ${h}^{0}\in \left\{{\pi }^{0},\eta ,\omega \right\}$ denotes a light unflavored and neutral hadron. [...] Published in Physical Review Letters 121 (2018) 10.1103/PhysRevLett.121.261801 arXiv:1804.06152 Fulltext: XML PDF; 10 Measurement of $\mathrm{cos}2\beta$ in ${B}^{0}\to {D}^{\left(*\right)}{h}^{0}$ with $D\to {K}_{S}^{0}{\pi }^{+}{\pi }^{-}$ decays by a combined time-dependent Dalitz plot analysis of BaBar and Belle data / Adachi, I. ; Adye, T. ; Ahmed, H. ; Ahn, J. K. ; et al We report measurements of $\mathrm{sin}2\beta$ and $\mathrm{cos}2\beta$ using a time-dependent Dalitz plot analysis of ${B}^{0}\to {D}^{\left(*\right)}{h}^{0}$ with $D\to {K}_{S}^{0}{\pi }^{+}{\pi }^{-}$ decays, where the light unflavored and neutral hadron ${h}^{0}$ is a ${\pi }^{0}$, $\eta$, or $\omega$ meson. [...] Published in Physical Review D 98 (2018) 10.1103/PhysRevD.98.112012 arXiv:1804.06153 Fulltext: PDF XML;
SCOAP3 Repository : 264 records found 1 - 10 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 62, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9780430793762207, "perplexity": 3851.5061807198817}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-09/segments/1550247487595.4/warc/CC-MAIN-20190218155520-20190218181520-00635.warc.gz"} |
http://clay6.com/qa/14089/a-steel-wire-of-40-m-in-length-is-stretched-through-2-0-mm-the-cross-sectio | # A steel wire of $40\;m$ in length is stretched through $2.0\; mm.$ The cross sectional area of the wire is $2.0 \;mm^2$ . If young's modulus of steel is $2.0 \times 10^{11}\;N/m^2$ the energy density of wire is
$\begin {array} {1 1} (a)\;2.5 \times 10^4 J/m^3 \\ (b)\;5.0 \times 10^4 J/m^3 \\ (c)\;7.5 \times 10^4 J/m^3 \\ (d)\;10.0 \times 10^4 J/m^3 \end {array}$
$U= \large\frac{1}{2} $$\times stress \times strain \quad= \large\frac{1}{2}$$ \times y \times (stain)^2$ $\qquad [strain =\large\frac{\Delta L}{L}]$
$\quad= \large\frac{1}{2}$$\times 2.0 \times 10^{11} \times \bigg(\large\frac{2 \times 10^{-3}}{4}\bigg)^2$
$\quad= 2.5 \times 10^{4} J/m^3$
Hence a is the correct answer.
edited Feb 18, 2014 by meena.p | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 1.0000098943710327, "perplexity": 2399.9408601431805}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-13/segments/1521257648594.80/warc/CC-MAIN-20180323200519-20180323220519-00735.warc.gz"} |
https://www.physicsforums.com/threads/finding-the-bond-length-of-hydrogen-theoretically.724248/ | # Finding the bond length of hydrogen theoretically
1. ### rokku
7
If the bond length of two atoms is the distance where the force of repulsion is equal to the force of attraction, then one could figure out what the bond length of H2 is by setting the net force of attraction to the net force of repulsion and solving for distance. However when I try this ..... ((2.30976*10^-28)/( d ^2) )+ ((2.30976*10^-28)/(d+1.06*10^-10)^2)= ((-2.30976*10^-28 )/(d + 5.3*10^-11)^2) * 2........ Where d is the unknown distance between the two hydrogens valence electrons, 1.06*10^-10 is the distance of the two radii of each hydrogen ( because proton - proton repulsion has a distance of d and the length of the distance from the proton to it's valence electron times 2 because there are two protons repelling each other) 5.3*10^-11 is the radius of the hydrogen atom and 2.30976*10^-28 is the charge of the electron and proton ( the electron has a negative charge on it ) . The problem is that when I plug this in to an equation calculator, I don't get a real answer. I don't know where I went wrong,many help is appreciated, thanks.
2. ### rokku
7
Just to clarify what I did, it's basically Coulomb's law for nuclear repulsion+ Coulomb's law for electron repulsion = Coulomb's law for electron - proton attraction ( I multiplied this by two because the proton of hydrogen 1 is attracted to the electron of hydrogen 2 and the electron of hydrogen 1 is attracted to the proton of hydrogen 2 so there are really two cases of attraction going on ) | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9318509101867676, "perplexity": 431.46218860436073}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-40/segments/1443737913815.59/warc/CC-MAIN-20151001221833-00145-ip-10-137-6-227.ec2.internal.warc.gz"} |
https://www.jiskha.com/display.cgi?id=1384126041 | # CALCULUS
posted by Samantha
If you could give an explenation with the answers, that'd be wonderful so I actually know how to solve similar problems in the future. :) Thank you so much!
1. A convex lens with focal length f centimeters will project the image of an object on a point behind the lens. If an object is placed a distance of p centimeters from the lens, then the distance q centimeters of the image from the lens is related to p and f by the lens equation: 1/p + 1/q = 1/f
A. If the focal length of the convex lens is supposed to be 5 cm, and if the image is formed 7 cm from the lens, find the distance, p, from the lens to the object. (It’s not necessary to simplify your answer.)
B. Find an expression that gives q as a function of p, assuming that the focal length is a constant of 5 centimeters.
C. Sketch a graph of q as a function of p (i.e., q(p)), assuming that the focal length is a constant of
5 centimeters. Show any important features of the graph.
D. Find lim p->infinity q(p) and lim p->5+ q(p). What do these limits represent physically? What must happen to the distance of the image and the object?
1. Steve
You have the equation -- just plug in the numbers.
1/p + 1/7 = 1/5
1/p = 1/5 - 1/7
1/p = 2/35
p = 35/2
1/p + 1/q = 1/5
1/q = 1/5 - 1/p
1/q = (p-5)/5p
q = 5p/(p-5)
The above is just algebra I, you know.
any online graphing site can do this. Try wolframalpha.com
limit as p->infinity = 5
limit as p->5 = infinity
parallel rays converge to the focal length (their origin is at infinity)
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https://www.physicsforums.com/threads/parallel-transport-invariance.400901/ | # Parallel transport invariance
• Start date
• #1
17
0
I'm trying to show that $$\frac{d}{dt}\; g_{\mu \nu} u^{\mu} v^{\nu} = 0$$ in the context of parallel transport (or maybe not zero), and I'm rather insecure about the procedure. This is akin to problem 3.14 in Hobson's et al. book (General Relativity an introduction for physicists).
As a guess, I tried the take the time derivative:
$$\frac{d}{dt}\; g_{\mu \nu} u^{\mu} v^{\nu}+g_{\mu \nu} \dot{u^{\mu}} v^{\nu}+g_{\mu \nu} u^{\mu} \dot{v^{\nu}}=0$$
I was assuming a stationary metric, so the first part would be zero, leaving
$$g_{\mu \nu} \dot{u^{\mu}} v^{\nu}+g_{\mu \nu} u^{\mu} \dot{v^{\nu}}=0$$
From there I can substitute in for $$\dot{u^{\mu}}$$ and $$\dot{v^{\nu}}$$.
Is this the right path to take? It seems there's then some index trickery involved to solve this.
Thanks!
Related Differential Geometry News on Phys.org
• #2
17
0
Did I post this in the wrong forum?
Some more info: vectors $$u$$ and $$v$$ are parallel transported along a geodesic, the point is to show that the dot product is invariant under parallel transport.
Also, this isn't homework, but it was a question asked on a midterm.
• #3
489
0
I'm a bit confused how general relativity treats this. It seems that this is the very definition of metric-compatible connection and parallel transport.
• #4
17
0
That was similar to my thought. I was thinking: well, we are only including the things that make this true so what's the point in showing it? As a result, I was thinking it was more of a mathematical question about topology and manifolds and I lacked understanding. In other words, it seemed that it's a mathematical argument not a physical one and I don't understand it. Overall, I'm really confused.
But what I was trying... Parallel transport says the covariant derivative is zero along a curve parameterized by, in this case, t. That means for a generic vector $$v$$,
$$\frac{dv^{a}}{du} = -\Gamma^{a}_{bc}v^{b} \frac{dx^{c}}{du}$$,
And I can rewrite the equations to be:
$$g_{\mu \nu} (-\Gamma^{\mu}_{\beta \sigma}u^{\beta} \dot{dx^{\sigma}}) v^{\nu}+g_{\mu \nu} u^{\mu} (-\Gamma^{\nu}_{\beta \sigma}v^{\beta} \dot{dx^{\sigma}})=0$$
Is that true? This doesn't seem like it's going anywhere.
• #5
489
0
The definition of a metric-compatible connection is that
$$\frac{d}{dt} <X,Y> = <D_tX,Y> + <X,D_tY>$$
By the definition of parallel transport, the right-hand side is zero. You're on the right track, but you don't have to assume a stationary metric.
• #6
970
3
The definition of a metric-compatible connection is that
$$\frac{d}{dt} <X,Y> = <D_tX,Y> + <X,D_tY>$$
By the definition of parallel transport, the right-hand side is zero. You're on the right track, but you don't have to assume a stationary metric.
Do you mean:
$$D_t <X,Y> = <D_tX,Y> + <X,D_tY>$$ ?
Then this is just the Leibniz rule and there is no need for metric compatibility.
In all the standard physics books the proof goes something like this:
$$\partial_k (V^iU_i)=\partial_k (V^i) U_i+V^i\partial_k (U_i)$$
parallel transport implies:
$$D_kV^i=\partial_k V^i+\Gamma V=0$$
$$D_k U_i=\partial_k U_i-\Gamma U=0$$
where the indices on second term on the RHS have been suppressed.
If you solve for the partial derivative of the vector in terms of gamma, and then plug that into the above equation, you get it's equal to zero.
• #7
17
0
Ok, I got it. It's so much simpler than I was making it. For some reason it didn't connect to drop an index so I could get the signs to change in the covariant derivative then to the math.
Thank you so much zhentil and RedX!
• #8
489
0
Do you mean:
$$D_t <X,Y> = <D_tX,Y> + <X,D_tY>$$ ?
Then this is just the Leibniz rule and there is no need for metric compatibility.
That's the definition of metric compatibility. You can easily construct connections that don't obey this rule. I'm not sure why you've drawn a distinction between $$D_t<X,Y>$$ and $$\frac{d}{dt} <X,Y>$$. By definition, a covariant derivative acts like a normal derivative on functions.
• #9
970
3
I'm not sure why you've drawn a distinction between $$D_t<X,Y>$$ and $$\frac{d}{dt} <X,Y>$$. By definition, a covariant derivative acts like a normal derivative on functions.
You're right. Your proof is much simpler than the one my instructor gave. In fact, the proof is practically tautological.
The reason I made a distinction is because usually little d means a coordinate derivative and big D means a covariant derivative, and the person wrote it with a little d. I forgot that when acting on a scalar, they are the same. So that is why I wasn't able to come up with your simple proof. Once you note that little d acting on a scalar is the same as big D acting on a scalar, then the proof is just automatic.
• #10
Ben Niehoff
Gold Member
1,879
162
The reason I made a distinction is because usually little d means a coordinate derivative and big D means a covariant derivative, and the person wrote it with a little d. I forgot that when acting on a scalar, they are the same. So that is why I wasn't able to come up with your simple proof. Once you note that little d acting on a scalar is the same as big D acting on a scalar, then the proof is just automatic.
Little d/dt means total derivative; a total derivative with respect to a parameter along a path is generally assumed to mean the total covariant derivative.
The coordinate derivative is $\partial/\partial x$ where x is some coordinate (NOT a parameter along a path).
• #11
970
3
Little d/dt means total derivative; a total derivative with respect to a parameter along a path is generally assumed to mean the total covariant derivative.
The coordinate derivative is $\partial/\partial x$ where x is some coordinate (NOT a parameter along a path).
Yeah. I got confused because of the overloading of the parameter t. So it'd be:
t=t
x,y,z=x(t),y(t),z(t)
t,x,y,z={t(s)=s},x(s),y(s),z(s)
Also, it is true that for parallel transport:
$$\partial_k <X,Y>=0$$
but I guess that's just a path with a direction only on the kth-coordinate basis.
So the total derivative along a path parametrized by s would be:
$$\frac{dx^k}{ds}D_k X$$ for a vector X and for a scalar you can replace D with a partial derivative.
I think I got it now.
btw, I have written in my notes:
<df,X>=Xf
<dxj,X>=$$\delta^{i}_j$$
where X is a basis vector (an operator). Is that a general rule, that you ignore the 'd' in the first term of the angled brackets, and let X operate on the term next to the 'd'? What would <f,X> mean (i.e., without a 'd' in front of the f)?
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2K | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9317102432250977, "perplexity": 393.71352529178637}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-04/segments/1610703515075.32/warc/CC-MAIN-20210118154332-20210118184332-00427.warc.gz"} |
https://www.gradesaver.com/textbooks/science/chemistry/chemistry-molecular-science-5th-edition/chapter-15-additional-aqueous-equilibria-questions-for-review-and-thought-topical-questions-page-693a/26a | ## Chemistry: The Molecular Science (5th Edition)
$0.10mol$ $NaCH_3COO$ would form a buffer with 1L of $0.20M$ $CH_3COOH$.
1. Try to identify a weak acid-base conjugate pair. - $CH_3COOH$ and $CH_3COO^-$. 2. Now, analyze, and determine if the base/acid ratio is between 0.1 and 10: Base: $0.10$ moles Acid: $1L$ of $0.20M = 0.20$ $moles$ $\frac{0.10}{0.20} = 0.5$: Correct. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8140789866447449, "perplexity": 2895.5012682217025}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-21/segments/1652662530553.34/warc/CC-MAIN-20220519235259-20220520025259-00791.warc.gz"} |
https://developer.mozilla.org/pt-BR/docs/Web/SVG/Attribute/stroke-miterlimit | # stroke-miterlimit
The `stroke-miterlimit` attribute is a presentation attribute defining a limit on the ratio of the miter length to the `stroke-width` used to draw a miter join. When the limit is exceeded, the join is converted from a miter to a bevel.
Note: As a presentation attribute `stroke-miterlimit` can be used as a CSS property.
As a presentation attribute, it can be applied to any element but it has effect only on the following nine elements: `<altGlyph>`, `<path>`, `<polygon>`, `<polyline>`, `<rect>`, `<text>`, `<textPath>`, `<tref>`, and `<tspan>`
```<svg viewBox="0 0 38 30" xmlns="http://www.w3.org/2000/svg">
<!-- Impact of the default miter limit -->
<path stroke="black" fill="none" stroke-linejoin="miter" id="p1"
d="M1,9 l7 ,-3 l7 ,3
m2,0 l3.5 ,-3 l3.5 ,3
m2,0 l2 ,-3 l2 ,3
m2,0 l0.75,-3 l0.75,3
m2,0 l0.5 ,-3 l0.5 ,3" />
<!-- Impact of the smallest miter limit (1) -->
<path stroke="black" fill="none" stroke-linejoin="miter"
stroke-miterlimit="1" id="p2"
d="M1,19 l7 ,-3 l7 ,3
m2, 0 l3.5 ,-3 l3.5 ,3
m2, 0 l2 ,-3 l2 ,3
m2, 0 l0.75,-3 l0.75,3
m2, 0 l0.5 ,-3 l0.5 ,3" />
<!-- Impact of a large miter limit (8) -->
<path stroke="black" fill="none" stroke-linejoin="miter"
stroke-miterlimit="8" id="p3"
d="M1,29 l7 ,-3 l7 ,3
m2, 0 l3.5 ,-3 l3.5 ,3
m2, 0 l2 ,-3 l2 ,3
m2, 0 l0.75,-3 l0.75,3
m2, 0 l0.5 ,-3 l0.5 ,3" />
<!-- the following pink lines highlight the position of the path for each stroke -->
<path stroke="pink" fill="none" stroke-width="0.05"
d="M1, 9 l7,-3 l7,3 m2,0 l3.5,-3 l3.5,3 m2,0 l2,-3 l2,3 m2,0 l0.75,-3 l0.75,3 m2,0 l0.5,-3 l0.5,3
M1,19 l7,-3 l7,3 m2,0 l3.5,-3 l3.5,3 m2,0 l2,-3 l2,3 m2,0 l0.75,-3 l0.75,3 m2,0 l0.5,-3 l0.5,3
M1,29 l7,-3 l7,3 m2,0 l3.5,-3 l3.5,3 m2,0 l2,-3 l2,3 m2,0 l0.75,-3 l0.75,3 m2,0 l0.5,-3 l0.5,3" />
</svg>```
When two line segments meet at a sharp angle and `miter` joins have been specified for `stroke-linejoin`, it is possible for the miter to extend far beyond the thickness of the line stroking the path. The `stroke-miterlimit` ratio is used to define when the limit is exceeded, if so the join is converted from a miter to a bevel.
The ratio of miter length (distance between the outer tip and the inner corner of the miter) to `stroke-width` is directly related to the angle (theta) between the segments in user space by the formula:
$\mathrm{stroke-miterlimit}=\frac{\mathrm{miterLength}}{\mathrm{stroke-width}}=\frac{1}{\mathrm{sin}\left(\frac{\theta }{2}\right)}$
For example, a miter limit of 1.414 converts miters to bevels for theta less than 90 degrees, a limit of 4.0 converts them for theta less than approximately 29 degrees, and a limit of 10.0 converts them for theta less than approximately 11.5 degrees.
## Usage context
Value 4 Yes
The value of `stroke-miterlimit` must be greater than or equal to 1.
## Browser compatibility
Update compatibility data on GitHub
Desktop Mobile Chrome Edge Firefox Internet Explorer Opera Safari Android webview Chrome for Android Firefox for Android Opera for Android Chrome ? Edge ? Firefox ? IE ? Opera ? Safari ? WebView Android ? Chrome Android ? Firefox Android ? Opera Android ? Safari iOS ? Samsung Internet Android ?
### Legend
Compatibility unknown
Compatibility unknown
## Specification
Specification Status Comment
Scalable Vector Graphics (SVG) 2
The definition of 'stroke-miterlimit' in that specification.
Candidate Recommendation Definition for shapes and texts
Scalable Vector Graphics (SVG) 1.1 (Second Edition)
The definition of 'stroke-miterlimit' in that specification.
Recommendation Initial definition for shapes and texts | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 1, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8900588750839233, "perplexity": 4177.433400501325}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-34/segments/1596439738603.37/warc/CC-MAIN-20200810012015-20200810042015-00513.warc.gz"} |
http://www.chegg.com/homework-help/questions-and-answers/using-newtons-second-law-rotation-using-rigid-aluminum-disk-mounted-rotary-machine-probe-u-q1586909 | using Newtons second law of rotation, We will be using a rigid aluminum disk mounted on a Rotary Machine Probe, using torques where we can measure the angular velocity as a function of time with the motion probe. We used 2 cylinders(masses) on this apparatus in pegs , and with a pully and string with a mass used as a tension to rotate the disk.
A.) we use a formula for the moment of inertia I=Inot + M1*R^2+ M2*R^2, but this equation is not right because the 2 masses arent point particles they are cylinders with radius of 1.27cm. To take this into account, you have to add the moment of inertia of the cylindrical masses( about their own axes) to this equation sometimes called the parallel axis theorem. Icyl= 1/2 Mcyl*Rcyl^2. What are the two moments of inertia( one for each mass) that you must add to this equation to get the correct moment of inertia? the mass of of first cylinder is .1655 and second is .1664
B.) Divide this moment of Inertia by the total moment of inertia ( Itot= Icalc+Icyl+Icyl2) for peg 1 to get the percent error. Are we accurate enough to ignore the extra Icyl contribution? | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8708107471466064, "perplexity": 657.8494650681879}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-11/segments/1424936461848.26/warc/CC-MAIN-20150226074101-00043-ip-10-28-5-156.ec2.internal.warc.gz"} |
https://groupprops.subwiki.org/wiki/Encoding_of_an_IAPS_of_groups | # Encoding of an IAPS of groups
BEWARE! This term is nonstandard and is being used locally within the wiki. [SHOW MORE]
## Definition
Let $(G,\Phi)$ be an IAPS of groups. In other words, for every $n$, there is a group $G_n$, and there is a map $\Phi_{m,n}: G _n \times G_n \to G_{m+n}$.
An encoding of this IAPS over a binary (or constant-sized) alphabet is the following.
• For every $G_n$, we specify an encoding of $G_n$ over that alphabet.
• For every $m,n$, we specify an algorithm that takes as input the code for $g \in G_m$ and $h \in G_n$, and outputs the code for $\Phi_{m,n}(g,h)$.
## Properties
### Dense encoding of an IAPS of groups
Further information: dense encoding of an IAPS of groups
An encoding of an IAPS of groups is said to be dense if the maximum possible length of a code-word for $G_n$ is bounded by some constant times the logarithm of the size of $G_n$.
### Polynomial-time encoding of an IAPS of groups
Further information: polynomial-time encoding of an IAPS of groups
An encoding of an IAPS of groups is said to be polynomial-time if:
• The time taken for all the operations (outputting the identity element, computing the product, computing the inverse, testing for membership) is polynomial in the length of the maximum code-word.
• The time taken for $\Phi_{m,n}$ is polynomial in the lengths of the maximum code-words for $G_m$ and $G_n$.
## Examples
### Encoding of symmetric groups
Further information: Encoding of symmetric groups
### Encoding of genereal linear groups
Further information: Encoding of general linear groups | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 15, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.984688401222229, "perplexity": 1214.7194630549227}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-24/segments/1590347425148.64/warc/CC-MAIN-20200602130925-20200602160925-00141.warc.gz"} |
https://www.physicsforums.com/threads/heat-equal-masses-of-h20.3384/ | Heat: equal masses of h20
1. Jun 28, 2003
Dx
Equal masses of water 20C and 80C are mixed. what is the inal temp of the mixture?
I said 60C
why is this incorrect?
Dx
2. Jun 28, 2003
Tom Mattson
Staff Emeritus
It's not correct because 60o is the final (not initial) temperature.
3. Jun 28, 2003
Dx
I see
dx
Last edited by a moderator: Jun 28, 2003
4. Jun 28, 2003
Dx
5. Jun 29, 2003
HallsofIvy
Staff Emeritus
Once again, DX, be careful of Tom's answers.
The problem with asking someone to give you the answer is that they may just give you a WRONG answer!
is not correct. You don't know what the "final" temperature will be because you don't know the temperature of the environment. In order to do this problem, you will have to interpret "initial" temperature as the temperature immediately AFTER mixing.
What makes you think the temperature will be 60 degrees? The only way I see that you can get 60 is to subtract 20 from 80. Do you have any reason for that? If the two temperatures had been 60 and 50 would you say that the mixture will be 10 degrees? Does that even make sense?
It should make sense to you that if you mix two things the final temperature will be BETWEEN the two original ones. In fact you should think about finding the average of the two temperatures.
What is the average of 20 and 80 degrees?
You should also think about how you would "average" the temperatures there were more water at one temperature than the other.
Suppose you had 10 grams of water at 20 degrees and 40 grams of water at 80 degrees. ABOUT what do you think the temperature of the mixture would be? How would you calculate it exactly?
6. Jun 29, 2003
Dx
Let me see. 20 + 80 = 100 / 2 = 50 average. I think i understand better now Ivy. Thanks! Can I add you as a friend, plz? Let me try to work the problem further and if I jave anumore problems ill ask.
Dx
7. Jun 30, 2003
Tom Mattson
Staff Emeritus
Actually, I assumed he did average them and that's how he got 60o. I didn't even notice that 60 isn't the average!
8. Jun 30, 2003
Dx
we forgive you this time Tom.
Just teasing, i am glad to see that everyone has each others back. Thanks HallsOfIvy,
I appreciate your help and toms too.
Dx
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http://blog.jpolak.org/?p=1778 | # More about Ext Calculations with Regular Sequences
Posted by Jason Polak on 21. March 2017 · Write a comment · Categories: commutative-algebra · Tags:
This post is a continuation of this previous one, though I repeat the main definitions for convenience.
Let $R$ be a commutative ring and $A$ and $R$-module. We say that $x_1,\dots,x_n\in R$ is a regular sequence on $A$ if $(x_1,\dots,x_n)A\not = A$ and $x_i$ is not a zero divisor on $A/(x_1,\dots,x_{i-1})A$ for all $i$. Last time, we looked at the following theorem:
Theorem. Let $A$ and $B$ be $R$-modules and $x_1,\dots,x_n$ a regular sequence on $A$. If $(x_1,\dots,x_n)B = 0$ then
$${\rm Ext}_R^n(B,A) \cong {\rm Hom}_R(B,A/(x_1,\dots,x_n)A)$$
When $R$ is a Noetherian ring, $I$ a proper ideal of $R$, and $A$ a finitely-generated $R$-module, this theorem for $B = R/I$ says that the least integer $n$ such that ${\rm Ext}_R^n(R/I,A)\not= 0$ is exactly the length of a maximal regular sequence in $I$ on $A$.
The Noetherian and finitely generated hypotheses are crucial. Why is this? It's because you need to have control over zero divisors. In fact you can see this by looking at the case $n = 0$:
Theorem. Let $R$ be a Noetherian ring, $I$ a proper ideal of $R$, and $A$ a finitely-generated $R$-module. Then every element of $I$ is a zero divisor on $A$ if and only if ${\rm Hom}_R(R/I,A)\not= 0$.
Proof. Since $A$ is a finitely generated $R$-module, that every element of $I$ is a zero divisor on $A$ is equivalent to $I$ being contained in the annihilator of a single nonzero element $a\in A$, which is in turn equivalent to every element of $I$ being sent to zero under the homomorphism
$$R\to A\\ 1\mapsto a.$$
Such homomorphisms are the same as nonzero homomorphisms $R/I\to A$. QED.
Here we are using this crucial fact:
Cool Theorem. For a finitely generated module $A$ over a Noetherian ring $R$, the zero divisors $Z(A)$ of $A$ in $R$ are a union of prime ideals of $R$, each of which are ideals maximal with respect to the property of being in $Z(A)$. Furthermore, each such prime is the annihilator of a single nonzero element of $A$.
In general, primes that are equal to the annihilator of a single element of a module $M$ are called the associated primes of $M$, and of course the theory of associated primes and primary decomposition is much more vast than this simple 'Cool Theorem', as is evident from Eisenbud's 30-page treatment of them in his book Commutative Algebra. In practice however, I only ever seem to need this simple version of the 'Cool Theorem'. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.921691358089447, "perplexity": 85.86750921643316}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-26/segments/1529267864022.18/warc/CC-MAIN-20180621040124-20180621060124-00386.warc.gz"} |
https://www.gradesaver.com/textbooks/science/physics/physics-10th-edition/chapter-2-kinematics-in-one-dimension-problems-page-48/11 | ## Physics (10th Edition)
distance = speed $\times$ time a) $d_{1}$ = 7.2 $\times$ (22 $\times$ 60) = 9504 m $d_{2}$ = 5.1 $\times$ (36 $\times$ 60) = 11 016 m $d_{3}$ = 13 $\times$ (8 $\times$ 60) = 6240 m Total distance = 9504 + 11 016 + 6240 = 26 760 m b) Total time = (22 + 36 + 8) $\times$ 60 = 3960 s Average velocity = displacement / time = 26 760 / 3960 = 6.76 m/s due north | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8156719207763672, "perplexity": 487.3197358626865}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-04/segments/1547584332824.92/warc/CC-MAIN-20190123130602-20190123152602-00075.warc.gz"} |
http://math.stackexchange.com/users/17760/lubin?tab=activity&sort=all&page=5 | Lubin
Reputation
99/100 score
Apr 14 comment P-adic expansions of Integers Finite in case the integer is nonnegative; for negative integers, the coeffs of $p^n$ will eventually all be $p-1$. Apr 14 comment Find $[\mathbb{Q(\alpha)}: \mathbb{Q}]$ where $\alpha=\zeta+\zeta^3+\zeta^4+\zeta^5+\zeta^9$ Knowing (or guessing) that $\alpha$ is quadratic over $\Bbb Q$, you get the minimal polynomial for $\alpha$ by looking at $1$, $\alpha$, and $\alpha^2$ and finding a $\Bbb Q$-linear relation of linear dependence. Apr 14 comment Find $[\mathbb{Q(\alpha)}: \mathbb{Q}]$ where $\alpha=\zeta+\zeta^3+\zeta^4+\zeta^5+\zeta^9$ There’s only one subgroup of order five because the original group is cyclic. These have precisely one group of each order dividing the order of the main group. Apr 14 answered Find $[\mathbb{Q(\alpha)}: \mathbb{Q}]$ where $\alpha=\zeta+\zeta^3+\zeta^4+\zeta^5+\zeta^9$ Apr 14 comment Proving that a Galois group is cyclic This is very confusing till you get the hang of it. @anon has given you the key. Apr 13 answered Numbers in $\mathbb{Q}_p$ can be written uniquely as $\sum_{i=k}^\infty \alpha_i p_i$ Apr 13 comment Numbers in $\mathbb{Q}_p$ can be written uniquely as $\sum_{i=k}^\infty \alpha_i p_i$ Yes, @DietrichBurde, that’s probably it. Maybe I’ll try a guide to a complete proof. Apr 13 comment Numbers in $\mathbb{Q}_p$ can be written uniquely as $\sum_{i=k}^\infty \alpha_i p_i$ It’s not clear what your question is. Are you saying that you understand the completion but don’t understand why every element can be written in unique $p$-ary expansion, or are you saying that you don’t understand why, if you define $\Bbb Q_p$ to be the set of $p$-ary expansions, you get a $p$-adically complete space? Apr 13 comment A normal closure of an arbitrary field extension Agreeing with @CaptainLama, I’d say that for infinite extensions, you’ll usually need to appeal to Zorn, and say that if an overfield of $L$ is not yet normal, you find an element in it with a $K$-conjugate not in it. Adjoin that element. There is your induction step. For explicitly-given extensions, maybe the set of all real $n$-th roots of all primes, over $\Bbb Q$, there may well be techniques peculiar to your situation (here, adjoin as well all roots of unity). Apr 13 comment Bézout's Identity of polynomials? I guess we gave each other a plus-one. Mutual admiration. Apr 13 answered Bézout's Identity of polynomials? Apr 13 answered Formula for calculating the $α$ and $β$ in $gcd(a, b) = αa + βb$ Apr 12 comment splitting field of $x^n-1$ over $\mathbb{Q}$ It’s true, but part of the proof is rather advanced, namely the proof that the degree of the extension is no smaller than $\phi(n)$. This is equivalent to the $\Bbb Q$-irreducibility of the $n$-th cyclotomic polynomial $\Phi_n(X)$. In case $n$ is a prime power, the proof is not at all hard, but when $n$ is divisible by two odd primes, or $4$ and an odd prime, things get sticky. Apr 11 answered Galois extentions Apr 11 answered Finding last 2 digits of $2016^{500}$ without repeated squaring Apr 11 comment If a function is odd/even, then its best polynomial approximation is also odd/even. Just to reiterate my own comment and support the others, what could it possibly mean to say that a function defined on $[0,b]$ is odd or even? Apr 10 comment If a function is odd/even, then its best polynomial approximation is also odd/even. I'm an outsider to this field, but if $a\ne-b$, I don't believe the claim. Apr 10 answered How to find $W^{\perp}$ in the following polynomial inner product space? Apr 10 comment How to find $W^{\perp}$ in the following polynomial inner product space? Right method, bad integration. Apr 10 comment Quick help on why this extension is of degree $2$ Right. Nor $i$. Just two things in the basis. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.874507486820221, "perplexity": 315.71648958688644}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-18/segments/1461860127407.71/warc/CC-MAIN-20160428161527-00168-ip-10-239-7-51.ec2.internal.warc.gz"} |
http://mathoverflow.net/questions/23030/the-egg-and-the-chicken | # The egg and the chicken
After posting this question (in particular after Carl's and Peter's answers) I have realized that the answer seems to depend on a basic problem in foundations.
Most mathematicians accept as given the ZFC (or at least ZF) axioms for sets. These are treated at an intuitive level. Using set theory they define languages, axioms, theories, models and all the logic toolbox. Then they define (formalized) set theory again, using this language.
The second point of view is typical of logicians. They realize that in order to talk of logic they don't need the full power of set theory, so they take logic as God-given instead. Then set theory is formalized in this framework.
I always thought that the points of view were interchangeable, as far as one was interested in the mathematical consequences. But comparing Carl's and Peter's answers it seems that actual (but still foundational) mathematic may depend on the point of view accepted. I'd like to understand this better.
Are there any mathematical consequences of choosing one of the two points of view?
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I don't think your perception of logicians is correct. It would be fun to have a poll, but I can only think of a few logicians whose point of view is close to your 'second point of view.' – François G. Dorais Apr 29 '10 at 20:36
In fact, this seems more true of mathematicians than logicians, strangely enough. You might say that the second viewpoint is very much the perspective of Bourbaki. – Harry Gindi Apr 29 '10 at 20:52
Your description of logicians is thoroughly mistaken. – Andrej Bauer Feb 9 '13 at 8:31
I would like to question two statements you make because they paint an oversimplified picture, which unfortunately is alluring to mathematicians who do not want to think about foundations (and they should not be blamed for it anymore than I should be blamed for not wanting to think about PDEs).
• "Most mathematicians accept as given the ZFC (or at least ZF) axioms for sets." This is what mathematicians say, but most cannot even tell you what ZFC is. Mathematicians work at a more intuitive and informal manner. High party officials once declared that ZFC was being used by everyone, so it has become the party line. But if you read a random text of mathematics, it will be equally easy to interpret it in other kinds of foundations, such as type theory, bounded Zermelo set theory, etc. They do not use the language of ZFC. The language of ZFC is completely unusable for the working mathematician, as it only has a single relation symbol $\in$. As soon as you allow in abbreviations, your exposition becomes expressible more naturally in other formal systems that actually handle abbreviations formally. Informal mathematics is informal, and thankfully, it does not require any foundation to function, just like people do not need an ideology to think. If you doubt that, you have to doubt all mathematics that happened before late 19th century.
• "They [logicians] realize that in order to talk of logic they don't need the full power of set theory, so they take logic as God-given instead." I do not know of any logicians, and I know many, who would say that logic is "God-given", or anything like that. I do not think logicians are born into a life rich with the "full power of set theory" which they throw away in order to become ascetic first-order logicians. That is a nice philosophical story detached from reality. The logicians I know are usually quite careful, skeptical, and inquisitive about foundational issues, reflect carefully on their own experiences, and almost never give you a straight answer when you ask "where does logic come from?" Your view is naive and inaccurate, if not slightly demeaning.
If I understand your question correctly, you are asking whether there is a difference between the following two views:
1. We start with naive set theory and on top of it we formalize set theory.
Well, we are proceeding from two different meta-theories. The first one allows us a wide spectrum of semantic methods. We can refer to "the standard model of Peano arithmetic" because we "believe in natural numbers", and we can invent Tarskian model theory without worrying where it came from.
The second method is more restricted. It will lead to syntactic and proof-theoretic methods, since the only thing we have given ourselves initially are syntactic in nature, namely first-order theories. There will be careful analysis of syntax. For advanced methods, however, we will typically resort to at least some amount of "naive mathematics". Ordinals will come into play, it will be hard to live without completeness theorems (which involve semantics), etc.
However, this is not how real life works. The dilemma you present is not really there. A working mathematician does not concern himself with these issues, anyhow, while a logician will likely refuse to be categorized as one or the other breed.
That is my guess, based on the experience that my fellow logicians are complicated animals and it is hard to get to the bottom of their foundational guts.
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Many professional mathematicians have this question about the egg and the chicken in one way or other. Most don't even bother or dare to ask. I think it is good that some of them at least ask. One can tell them that they are "thoroughly mistaken" are spreading "fairy tyles" and are "demeaning" towards logicians. One can also think: "Hm, if so many of my otherwise esteemed are so confused about what my community understands so well, then maybe we missed to write the good exposition of this fact that everyone would read!" Something like an AMS-notices 'What is... foundations?' – Urs Schreiber Feb 9 '13 at 15:16
Good point, so the question is badly asked and badly answered. It is a bit of news to me that "our community" understands the issue so well. Do we? You should definitely write your our answer to the question (and first explain how you understand the question). – Andrej Bauer Feb 9 '13 at 15:31
What is this question of "chicken and the egg"? – Andrej Bauer Feb 9 '13 at 15:35
The "Chicken and egg"-question about mathematics: Does mathematics root in a formal foundation or is there an infinite recursion (the "turtles" in Peter's answer referred to above.) of foundational systems or else just an "intuitive notion of sets" at the beginning? (If the latter, what if our inutiotions differ?) In either case: where, if at all, does it matter for actual mathematical results on which level one roots one's mathematics? (This last one is concretely the question of this thread, I think). Peter's answer referred to above starts answering these things, I'd say. – Urs Schreiber Feb 9 '13 at 17:16
Ok, I totally missed that in the question. I should have read it more carefully. Thanks for explaining. @Andrea, is that what you asked? – Andrej Bauer Feb 9 '13 at 18:14
Well, here's what it seems like: We had the naïve set theories of Cantor (unaxiomatized) and Frege (first axiomatization), and everything was right with the world. Then, however, it eventually became clear that there were paradoxes that made the theory inconsistent, so Zermelo and Russell (and others) both set to work on coming up with alternative axiomatizations to deal with the paradoxes. Russell's Theory of Unramified Types ended up being unusable (cf. the proof that 1+1=2 in Principia Mathematica (the full proof is something like 100 pages). However, Zermelo was able to recover a good deal of naïve set theory in the Z (ZC), or Zermelo (resp. Zermelo + Choice) axiomatization (Zermelo+Choice).
However, it eventually became clear that Z(C) was too weak in some regards, so Fraenkel and Skolem both proposed (independently) the axiom schema of replacement to improve the power of the theory. If you include the axiom schema of regularity (which says that $\in$ is well-founded), you get ZF (resp. ZFC).
Most mathematicians accept as given the ZFC (or at least ZF) axioms for sets. These are treated at an intuitive level. Using set theory they define languages, axioms, theories, models and all the logic toolbox. Then they define (formalized) set theory again, using this language.
ZFC is built on first-order logic. Mathematicians then can construct models of languages and set theories internal to ZFC. (To answer your comment: The reason why we want to axiomatize over first order logic is because first order logic has all of the properties one would want from it, completeness, compactness, etc. In particular, the reason why we must first define theorems and proofs is because we need a proof calculus to be able to formally derive results.)
The second point of view is typical of logicians. They realize that in order to talk of logic they don't need the full power of set theory, so they take logic as God-given instead. Then set theory is formalized in this framework.
In fact, this is true for both mathematicians and logicians (although this view is strangely more popular among mathematicians than logicians). This would be the "bootstrapping" step for mathematicians to get to ZFC. Even though Bourbaki's book on set theory has an inadequate and confusing approach to set theory, the idea in the first few sections is correct. One first defines a what it means for something to be a theorem, what it means for something to be a proof, etc, then gives the axioms of formal logic, then constructs a set theory above that (this was one of the areas in which Bourbaki was the most influential as well. That is, his opinion on the logical grounding of set theory (but then not actually worrying about it once it's developed) is perhaps one of his most important lasting contributions to the world of mathematics.
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It seems you are implying that no mathematician holds the first point of view. I'm afraid I am a counterexample. My point of view (and of most people I heard) is that one can surely talk about sets without defining proof or theories first. Just give the axioms (in the layman, not formal sense) and derive the usual informal set theory. Then define logic as any other branch of mathematics. And if you wish you can then do set theory again in this framework. – Andrea Ferretti Apr 29 '10 at 22:50
Just a small historical correction. Cantor did not propose a formal theory in any reasonable sense of 'formal'. Frege was the first to recognize the need for a formal theory of sets. In fact, it's apparent that Cantor did not believe that any of the paradoxes were refutations of his ideas. Indeed, he discovered a paradox a few years before Russell and that did not seem to bother him much. It's unclear (to me) whether Cantor ever felt the need for a formal theory of sets. – François G. Dorais Apr 29 '10 at 23:13
Yes, that was implied. I was thinking of just including Frege, but if I'm talking about the history, I can't just not mention Cantor. – Harry Gindi Apr 29 '10 at 23:37
I also would like to register as someone holding the first point of view. As an undergrad, I had imagined that logicians would somehow build up set theory from the beginning, and never use any tools like functions, natural numbers, or sequences until they had been defined internally. Then I took a course in logic, and found that functions, sequences, etc., showed up in droves long before they were defined. I was confused for a bit, and then went back to my way of approaching other math, assuming that these symbols, languages, etc. were really sets described by some background set theory. – Charles Staats Apr 30 '10 at 1:26
@Charles: If you're interested in reading a completely formal exposition of set theory from nothing, Bourbaki does this in chapter I of Théorie des Ensembles. My statement about how their set theory is inadequate is mainly about chapter II (It does not contain the axiom of replacement, according to François and Joel, who have linked to a paper in the past (although I can't remember who wrote it)). Everything is defined in terms of (essentially) formal symbols. It's useless and totally impractical, but it's definitely worth looking over. – Harry Gindi Apr 30 '10 at 1:36 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8112658262252808, "perplexity": 703.2574693793488}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-49/segments/1416931007607.9/warc/CC-MAIN-20141125155647-00089-ip-10-235-23-156.ec2.internal.warc.gz"} |
https://verification.asmedigitalcollection.asme.org/dynamicsystems/article-abstract/103/3/251/400561/Modeling-and-Dynamic-Response-of-Maglev-Vehicles?redirectedFrom=fulltext | This paper presents a two-degree-of-freedom model for magnetically levitated finite-length vehicles incorporating sway and yaw dynamics. Aerodynamic lateral forces and yawing moments on the vehicle resulting from constant speed wind gusts are computed using analytical techniques. Computer simulations are run for three vehicle speeds and three apparent mass factors. It is shown that higher apparent mass factors can be instrumental in reducing peak displacements and acceleration levels. The guidance-to-lift ratio is not as much affected by an increase in apparent mass factor as are the displacements and accelerations.
This content is only available via PDF. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.939347505569458, "perplexity": 1595.7229014330474}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-06/segments/1674764495001.99/warc/CC-MAIN-20230127164242-20230127194242-00050.warc.gz"} |
https://stonoga-klubmalucha.pl/crusher/2020_11_06_7192/ | # Difference Between Magnetic Field Magnetic Field Intensity
## What is the difference between magnetic field strength and,
B- This stands for magnetic flux density. This simply is a measure of strength of magnetic fields in term of the density of the magnetic flux (magnetic flux per unit area). That is why the unit is weber/m^2. Another way to look at it is that it is...What is the difference between magnetic intensity and,,Like other fields, the magnetic field is a way of describing a region of space where other magnets will experience a force. It can be represented by field lines that show both the size and direction of the force. By arrows showing the direction a,Difference Between Magnetic Field and Magnetic Force,,In this article, we are going to discuss what magnetic field and magnetic force are, their definitions, the applications of these two, the similarities and finally the differences between magnetic field and magnetic force. Magnetic Field. Magnets were discovered by Chinese
## Difference between Electric Field and Magnetic Field
The main difference between electric and magnetic field lies in the way they are produced. Electric field is only because of the charge existing in the space. Higher the intensity of the charge higher is the force exerted at a constant point. Electric field force is due to voltage and it follows the inverse square law and it is a low range force.Difference between B and H in magnetic fields?,Difference between B and H in magnetic fields? [duplicate] Ask Question,Can someone explain the difference between them? Is $\vec H$ field only relevant during magnetization or demagnetization? $\vec H$ is just that value needed to magnetize/demagnetize or what is it useful for?,(also called as "Magnetic field intensity"),Magnetic field - Wikipedia,The magnetic field of larger magnets can be obtained by modeling them as a collection of a large number of small magnets called dipoles each having their own m. The magnetic field produced by the magnet then is the net magnetic field of these dipoles; any net force on the magnet is a result of adding up the forces on the individual dipoles.
## Difference Between Magnetic Field & Magnetic Flux (with,
Difference Between Magnetic Field & Magnetic Flux The most significant difference between the magnetic field and the magnetic flux is that the magnetic field is the region around the magnet where the moving charge experiences a force, whereas the magnetic flux shows the quantity or strength of magnetic lines produced by the magnet. The other differences between the magnetic field and magnetic,What is difference between magnetic force and field,,Magnetic force is the actual force between two objects. Field intensity is the force that a 1 unit point particle would feel if placed at that point.Difference Between Magnetic Field and Magnetic Flux,Main Difference –Magnetic Field vs. Magnetic Flux. Magnetic field and magnetic flux both refer to properties of magnets. The main difference between magnetic field and magnetic flux is that magnetic field is a region where magnetic poles and moving charges experience a force.Sometimes, the term magnetic field may be also used to refer to the quantity magnetic field strength.
## Difference Between Electric & Magnetic Field with,
Difference between Electric & Magnetic Field The one of the major difference between the magnetic and electric field is that the electric field induces around the static charge particle which is either negative or positive, whereas the magnetic field produces aroundDifference between B,H and M in magnetics - ResearchGate,H (names in use: magnetic field, magnetizing field, magnetic field intensity, magnetic field strength) could be applied as a characteristic of magnetic field, which arises at a definite point in,Difference Between Magnetic Field and Magnetic Flux,,Difference Between Magnetic Field and Magnetic Flux • Every magnetic object has a magnetic field in its surrounding area that is felt by moving charged particle. • Magnetic field is described using magnetic lines emanating in a set pattern • Magnetic flux is a related concept that describes the strength of magnetic field
## Difference Between Electric and Magnetic fields,
Conversely, static electricity has an electric field without the presence of a magnetic field. The interaction between magnetic fields and electric fields are laid out in Maxwell’s equation. Summary: 1. An electric field is a field of force, surrounding a charged particle, while a magnetic field is a field of force surrounding a permanent,Magnetization and Magnetic Intensity: Formulas, Videos and,,The intensity of the magnetic field at P due to single pole is given by: We say that the magnetic field B can be written as: Intensity of Magnetic Field due to a Magnet at Different Points In Longitudinal Position +/-m = Magnitude of the south and north poles r = Distance of point P from the center of the magnet l = Length of the bar magnet,What's the difference between magnetic fields H and B?,In layman's terms, E and B are the total electric and magnetic fields.. D and H are the free electric and magnetic fields.. P and M are the bound electric and magnetic fields.. M would be the magnetic field caused by current loops in the material. In vacuum, like you said, B and H are proportional by a constant since there is no material.
## Magnetisation and Magnetic Intensity - Study Material for,
Difference between Magnetic Intensity and Intensity of Magnetisation The magnetic intensity defines the forces that the poles of a magnet experiences in a magnetic field whereas the intensity of magnetisation explains the change in the magnetic moment of a magnet per unit volume.What is difference between magnetic field and magnetic,,- Magnetic field strength is the intensity of a magnetic field at a given location. Historically, a distinction is made between magnetic field strength H, measured in ampere/meter, and magnetic,Difference between electric field and electric field intensity,The electric field is a region around a charge in which it exerts electrostatic force on another charges.While the strength of electric field at any point in space is called electric field intensity.It is a vector quantity.Its unit is NC¯¹.
## Difference Between Electric Field and Magnetic Field | Physics
Difference Between Electric Field And Magnetic Field - Electric field is the force surrounding an electrically charged particle. We can also say that it is the area where the line of force exist and these lines of force surround the electric field. Magnetic Field is the area around the magnet where attractive forces or repulsive forces are exhibited by the poles of the magnetMagnetic Field Intensity | Definition Formula | Electrical,,Magnetic field intensity is also known as the magnetizing force which is measured is ampere-turns per meter (A-t/m). Of primary concern, however, is the magnetomotive force needed to establish a certain flux density, B in a unit length of the magnetic circuit. Magnetic Field Intensity Formula. The letter symbol for magnetizing force (magnetic,In magnetism, what is the difference between the B and H,,28/5/2012· hi, I'm having trouble understanding what the difference is between the B and H fields. What my understanding is, if you have a magnetic dipole on its own, its going to have some B field around it. Then say you bring some material into the field, and the field within the boundaries of the...
## Electric Field vs Magnetic Field - Differences between,
17/1/2018· Similarities and Dissimilarities between Electric field and magnetic field An object with moving charge always has both magnetic and electric field. They have some similarities and also have two,difference between electric field and magnetic field,,4/8/2012· Electric and magnetic fields (EMFs) are invisible strains of drive that symbolize the boundary and the depth that occur between objects with competencies change or voltage (so called electric field) and that surround object with electrical currents float (so called magnetic discipline).DIFFERENCE between magnetic field intensity and magnetic,,27/10/2010· Magnetic field strength (H) is the amount of magnetizing force. It is proportional to the length of a conductor and the amount of electrical current passing through the conductor. Magnetic field strength is a vector quantity whose magnitude is the strength of a magnetic field at a point in the direction of the magnetic field at that point.
## Difference Between Magnetic Flux and Magnetic Flux Density
Main Difference – Magnetic Flux vs Magnetic Flux Density. In magnetism, several physical quantities such as magnetic flux, magnetic flux density and magnetic field strength are used to explain the behaviors or influences of magnetic fields.Some people use these terms interchangeably. But they have different and particular meanings.,, | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9541350603103638, "perplexity": 355.8156370348169}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-17/segments/1618038085599.55/warc/CC-MAIN-20210415125840-20210415155840-00200.warc.gz"} |
http://quant.stackexchange.com/questions/3660/derivation-of-formula-for-portfolio-skewness-and-kurtosis/3662 | # derivation of formula for portfolio skewness and kurtosis
Where can I find derivation of formula for portfolio skewness and kurtosis? I can find formulas everywhere, but not their derivations? For example, the portfolio variance formula is well known and I can find the derivation of that formula in a lot of books, but I can't find anything on portfolio skewness formula and kurtosis. They are just given the way they are. I'm not strong enough at probability theory to use it to derive the formulas from the expectations operator. Who was the first person to derive them? Where were they first published?
-
What is the data basis that you start from? If you just have the covariance matrix, then you can only calculate portfolio variance or volatility by $$w^T \Sigma w$$ where $w$ are the portfolio weights and $\Sigma$ is the covariance matrix. If you have the individual asset continuously compounded returns $r^j_t$ where $j$ indexes assets, $j=1,\ldots,N$, and $t$ stands for time, $t=1,\ldots,T$, then you can also calculate the portfolio returns for each points in time $$r_t = \sum_{j=1}^N w_j r^j_t$$ and then apply the standard variance estimator on $(r_t)_{t=1}^T$. Coming back to your question, having $(r_t)_{t=1}^T$ you can calculate skewness and kurtosis on this sample. You find the formulas on wikipedia. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9666334986686707, "perplexity": 225.36937660248515}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-22/segments/1464049278042.30/warc/CC-MAIN-20160524002118-00204-ip-10-185-217-139.ec2.internal.warc.gz"} |
https://en.wikipedia.org/wiki/Paradox_of_the_heap | (Redirected from Paradox of the heap)
The sorites paradox: If a heap is reduced by a single grain at a time, at what exact point does it cease to be considered a heap?
The sorites paradox (/sˈrtz/;[1] sometimes known as the paradox of the heap) is a paradox that arises from vague predicates.[2] A typical formulation involves a heap of sand, from which grains are individually removed. Under the assumption that removing a single grain does not turn a heap into a non-heap, the paradox is to consider what happens when the process is repeated enough times: is a single remaining grain still a heap? If not, when did it change from a heap to a non-heap?[3]
## The original formulation and variations
The word "sorites" derives from the Greek word for heap.[4] The paradox is so named because of its original characterization, attributed to Eubulides of Miletus.[5] The paradox goes as follows: consider a heap of sand from which grains are individually removed. One might construct the argument, using premises, as follows:[3]
1000000 grains of sand is a heap of sand (Premise 1)
A heap of sand minus one grain is still a heap. (Premise 2)
Repeated applications of Premise 2 (each time starting with one fewer grain) eventually forces one to accept the conclusion that a heap may be composed of just one grain of sand.[6]). Read (1995) observes that "the argument is itself a heap, or sorites, of steps of modus ponens":[7]
1000000 grains is a heap.
If 1000000 grains is a heap then 999999 grains is a heap.
So 999999 grains is a heap.
If 999999 grains is a heap then 999998 grains is a heap.
So 999998 grains is a heap.
If ...
... So 1 grain is a heap.
### Variations
Another formulation is to start with a grain of sand, which is clearly not a heap, and then assume that adding a single grain of sand to something that is not a heap does not turn it into a heap. Inductively, this process can be repeated as much as one wants without ever constructing a heap.[2][3] A more natural formulation of this variant is to assume a set of colored chips exists such that two adjacent chips vary in color too little for human eyesight to be able to distinguish between them. Then by induction on this premise, humans would not be able to distinguish between any colors.[2]
The removal of one drop from the ocean, will not make it 'not an ocean' (it is still an ocean), but since the volume of water in the ocean is finite, eventually, after enough removals, even a litre of water left is still an ocean.
This paradox can be reconstructed for a variety of predicates, for example, with "tall", "rich", "old", "blue", "bald", and so on. Bertrand Russell argued that all of natural language, even logical connectives, is vague; moreover, representations of propositions are vague.[8]
## Proposed resolutions
On the face of it, there are some ways to avoid this conclusion. One may object to the first premise by denying 1000000 grains of sand makes a heap. But 1000000 is just an arbitrarily large number, and the argument will go through with any such number. So the response must deny outright that there are such things as heaps. Peter Unger defends this solution.[10] Alternatively, one may object to the second premise by stating that it is not true for all heaps of sand that removing one grain from it still makes a heap.[citation needed]
### Setting a fixed boundary
A common first response to the paradox is to call any set of grains that has more than a certain number of grains in it a heap. If one were to set the "fixed boundary" at, say, 10000 grains then one would claim that for fewer than 10000, it is not a heap; for 10000 or more, then it is a heap.[citation needed]
However, such solutions are unsatisfactory as there seems little significance to the difference between 9999 grains and 10000 grains. The boundary, wherever it may be set, remains as arbitrary and so its precision is misleading. It is objectionable on both philosophical and linguistic grounds: the former on account of its arbitrariness, and the latter on the ground that it is simply not how we use natural language.[citation needed]
A second response attempts to find a fixed boundary that reflects common usage of a term. For example, a dictionary may define a "heap" as "a collection of things thrown together so as to form an elevation."[11] This requires there to be enough grains that some grains are supported by other grains. Thus adding one grain atop a single layer produces a heap, and removing the last grain above the bottom layer destroys the heap.
### Unknowable boundaries (or epistemicism)
Timothy Williamson[12][13][14] and Roy Sorensen[15] hold an approach that there are fixed boundaries but that they are necessarily unknowable.
### Supervaluationism
Supervaluationism is a semantics for dealing with irreferential singular terms and vagueness. It allows one to retain the usual tautological laws even when dealing with undefined truth values.[16][17][18][19] As an example for a proposition about an irreferential singular term, consider the sentence "Pegasus likes licorice". Since the name "Pegasus" fails to refer, no truth value can be assigned to the sentence; there is nothing in the myth that would justify any such assignment. However, there are some statements about "Pegasus" which have definite truth values nevertheless, such as "Pegasus likes licorice or Pegasus doesn't like licorice". This sentence is an instance of the tautology "${\displaystyle p\vee \neg p}$", i.e. the valid schema "${\displaystyle p}$ or not-${\displaystyle p}$". According to supervaluationism, it should be true regardless of whether or not its components have a truth value.
Similarly, "1000 grains of sand is a heap of sand" may be considered a border case having no truth value, but "1000 grains of sand is a heap of sand, or 1000 grains of sand is not a heap of sand" should be true.
Precisely, let ${\displaystyle v}$ be a classical valuation defined on every atomic sentence of the language ${\displaystyle L}$, and let ${\displaystyle At(x)}$ be the number of distinct atomic sentences in ${\displaystyle x}$. Then for every sentence ${\displaystyle x}$, at most ${\displaystyle 2^{At(x)}}$ distinct classical valuations can exist. A supervaluation ${\displaystyle V}$ is a function from sentences to truth values such that, a sentence ${\displaystyle x}$ is super-true (i.e. ${\displaystyle V(x)={\text{True}}}$) if and only if ${\displaystyle v(x)={\text{True}}}$ for every classical valuation ${\displaystyle v}$; likewise for super-false. Otherwise, ${\displaystyle V(x)}$ is undefined—i.e. exactly when there are two classical valuations ${\displaystyle v}$ and ${\displaystyle v'}$ such that ${\displaystyle v(x)={\text{True}}}$ and ${\displaystyle v'(x)={\text{False}}}$.
For example, let ${\displaystyle L\;p}$ be the formal translation of "Pegasus likes licorice". Then there are exactly two classical valuations ${\displaystyle v}$ and ${\displaystyle v'}$ on ${\displaystyle L\;p}$, viz. ${\displaystyle v(L\;p)={\text{True}}}$ and ${\displaystyle v'(L\;p)={\text{False}}}$. So ${\displaystyle L\;p}$ is neither super-true nor super-false. However, the tautology ${\displaystyle L\;p\lor \lnot L\;p}$ is evaluated to ${\displaystyle {\text{True}}}$ by every classical valuation; it is hence super-true. Similarly, the formalization of the above heap proposition ${\displaystyle H\;1000}$ is neither super-true nor super-false, but ${\displaystyle H\;1000\lor \lnot H\;1000}$ is super-true.
### Truth gaps, gluts, and many-valued logics
Another approach is to use a multi-valued logic. From this point of view, the problem is with the principle of bivalence: the sand is either a heap or is not a heap, without any shades of gray. Instead of two logical states, heap and not-heap, a three value system can be used, for example heap, indeterminate and not-heap. However, three valued systems do not truly resolve the paradox as there is still a dividing line between heap and indeterminate and also between indeterminate and not-heap. The third truth-value can be understood either as a truth-value gap or as a truth-value glut.[20]
Alternatively, fuzzy logic offers a continuous spectrum of logical states represented in the unit interval of real numbers [0,1]—it is a many-valued logic with infinitely-many truth-values, and thus the sand moves smoothly from "definitely heap" to "definitely not heap", with shades in the intermediate region. Fuzzy hedges are used to divide the continuum into regions corresponding to classes like definitely heap, mostly heap, partly heap, slightly heap, and not heap. [21][22]
### Hysteresis
Another approach, introduced by Raffman,[23] is to use hysteresis, that is, knowledge of what the collection of sand started as. Equivalent amounts of sand may be called heaps or not based on how they got there. If a large heap (indisputably described as a heap) is slowly diminished, it preserves its "heap status" to a point, even as the actual amount of sand is reduced to a smaller number of grains. For example, suppose 500 grains is a pile and 1000 grains is a heap. There will be an overlap for these states. So if one is reducing it from a heap to a pile, it is a heap going down until, say, 750. At that point one would stop calling it a heap and start calling it a pile. But if one replaces one grain, it would not instantly turn back into a heap. When going up it would remain a pile until, say, 900 grains. The numbers picked are arbitrary; the point is, that the same amount can be either a heap or a pile depending on what it was before the change. A common use of hysteresis would be the thermostat for air conditioning: the AC is set at 77 °F and it then cools down to just below 77 °F, but does not turn on again instantly at 77.001 °F—it waits until almost 78 °F, to prevent instant change of state over and over again.[24]
### Group consensus
One can establish the meaning of the word "heap" by appealing to consensus. This approach claims that a collection of grains is as much a "heap" as the proportion of people in a group who believe it to be so. In other words, the probability that any collection is considered a heap is the expected value of the distribution of the group's views.
A group may decide that:
• One grain of sand on its own is not a heap.
• A large collection of grains of sand is a heap.
Between the two extremes, individual members of the group may disagree with each other over whether any particular collection can be labelled a "heap". The collection can then not be definitively claimed to be a "heap" or "not a heap". This can be considered an appeal to descriptive linguistics rather than prescriptive linguistics, as it resolves the issue of definition based on how the population uses natural language. Indeed, if a precise prescriptive definition of "heap" is available then the group consensus will always be unanimous and the paradox does not arise.
Modelling "X more or equally red than Y" as
quasitransitive (Q) and as transitive (T) relation
XY f01000 e02000 d03000 c04000 b05000 a06000
f01000 QT QT QTP QTP QTP QTP
e02000 Q QT QT QTP QTP QTP
d03000 Q QT QT QTP QTP
c04000 Q QT QT QTP
b05000 Q QT QT
a06000 Q QT
### Dropping transitivity of the relations involved
In the above color example, the argument is tacitly based on considering the relation "for the human eye, color X is indistinguishable from Y" as an equivalence relation, in particular as transitive. To drop the transitivity assumption is a possibility to resolve the paradox.
Similarly, the paradox is based on considering the relation "for the human eye, color X looks more or equally red than Y" as a reflexive total ordering; again, dropping its transitivity resolves the paradox.
Instead, the relation between colors can be described as a quasitransitive relation, employing a concept introduced by microeconomist Amartya Sen in 1969.[25] The table shows a simple example, with color differences overdone for readability. A "Q" and a "T" indicates that the row's color looks more or equally red than column's color in the quasitransitive and the transitive version of the relation, respectively. In the quasitransitive version, e.g. the colors f01000 and e02000 are modelled as indistinguishable, since a "Q" appears in both their intersection cells. A "P" indicates the asymmetric part of the quasitransitive version.
To resolve the original heap variation of the paradox with this approach, the relation "X grains are more a heap than Y grains" should be considered quasitransitive rather than transitive.[citation needed][clarification needed]
## References
1. ^ http://www.omnilexica.com/pronunciation/?q=Sorites#.UyL2hPldWSo
2. ^ a b c Barker, C. (2009). "Vagueness". In Allan, Keith. Concise Encyclopedia of Semantics. Elsevier. p. 1037. ISBN 978-0-08-095968-9.
3. ^ a b c Sorensen, Roy A. (2009). "sorites arguments". In Jaegwon Kim; Sosa, Ernest; Rosenkrantz, Gary S. A Companion to Metaphysics. John Wiley & Sons. p. 565. ISBN 978-1-4051-5298-3.
4. ^ Bergmann, Merrie (2008). An Introduction to Many-Valued and Fuzzy Logic: Semantics, Algebras, and Derivation Systems. New York, NY: Cambridge University Press. p. 3. ISBN 978-0-521-88128-9.
5. ^ (Barnes 1982), (Burnyeat 1982), (Williamson 1994)
6. ^ Dolev, Y. (2004). "Why Induction Is No Cure For Baldness". Philosophical Investigations. 27 (4): 328–344. doi:10.1111/j.1467-9205.2004.t01-1-00230.x.
7. ^ Read, Stephen (1995). Thinking About Logic, p.174. Oxford. ISBN 019289238X.
8. ^ Russell, Bertrand (June 1923). "Vagueness". The Australasian Journal of Psychology and Philosophy. 1 (2): 84–92. doi:10.1080/00048402308540623. ISSN 1832-8660. Retrieved November 18, 2009. Shalizi's 1995 etext is archived at archive.org and at WebCite.
9. ^ Thouless, Robert H. (1953), Straight and Crooked Thinking (PDF) (Revised ed.), London: Pan Books, p. 61
10. ^ Unger, Peter (1979). "There Are No Ordinary Things". Synthese. 41: 117–154. doi:10.1007/bf00869568. Retrieved 19 July 2013. (Alternative: jstor.org)
11. ^ "heap". Wiktionary. Accessed 2017-01-02.
12. ^ Williamson, Timothy (1992). "Inexact Knowledge". Mind. 101: 218–242. doi:10.1093/mind/101.402.217. JSTOR 2254332.
13. ^ Williamson, Timothy (1992). "Vagueness and Ignorance". Supplementary Proceedings of the Aristotelian Society. 66: 145–162. JSTOR 4106976.
14. ^ Williamson, Timothy (1994). Vagueness. London: Routledge.
15. ^ Sorensen, Roy (1988). Blindspots. Clarendeon Press.
16. ^ Fine, Kit (Apr–May 1975). "Vagueness, Truth and Logic" (PDF). Synthese. 30 (3/4): 265–300. doi:10.1007/BF00485047.
17. ^ van Fraassen, Bas C. (Sep 1966). "Singular Terms, Truth-Value Gaps, and Free Logic" (PDF). Journal of Philosophy. 63 (17): 481—495. JSTOR 2024549.
18. ^ Kamp, Hans (1975). Keenan, E., ed. Two Theories about Adjectives. Cambridge University Press. pp. 123–155.
19. ^ Dummett, Michael (1975). "Wang's Paradox" (PDF). Synthese. 30: 301–324. doi:10.1007/BF00485048.
20. ^ http://plato.stanford.edu/entries/truth-values/
21. ^ Zadeh, L. A. (1965). "Fuzzy Sets". Information and Control. 8: 338–353. doi:10.1016/s0019-9958(65)90241-x.
22. ^ Goguen, J. A. (1969). "The Logic of Inexact Concepts". Synthese. 19 (3–4): 325–378. doi:10.1007/BF00485654.
23. ^ Unruly Words: A Study of Vague Language (OUP, 2014)[full citation needed]
24. ^ Raffman, D. (2005). "How to understand contextualism about vagueness: reply to Stanley". Analysis. 65 (287): 244–248. doi:10.1111/j.1467-8284.2005.00558.x.
25. ^ Sen, Amartya (1969). "Quasi-transitivity, rational choice and collective decisions". Rev. Econ. Stud. 36: 381–393. doi:10.2307/2296434. Zbl 0181.47302.
## Bibliography
• Black, Max (1970). Margins of Precision. Ithaca, NY: Cornell University Press. ISBN 0-8014-0602-1.
• Barnes, J. (1982). "Medicine, Experience and Logic". In Barnes, J.; Brunschwig, J.; Burnyeat, M. F.; Schofield, M. Science and Speculation. Cambridge: Cambridge University Press.
• Burns (1991). Vagueness: An Investigation into Natural Languages and the Sorites Paradox. Dordrecht: Kluwer Academic Publishers. ISBN 0-7923-1489-1.
• Burnyeat, Myles (1982). "15. Gods and heaps". In Schofield, M.; Nussbaum, M. C. Language and Logos. Cambridge: Cambridge University Press. pp. 315–.
• Damir D. Dzhafarov, The sorites paradox: a behavioral approach (with E. N. Dzhafarov), in J. Valsiner and L. Dudolph (eds.).
• Gerla (2001). Fuzzy logic: Mathematical Tools for Approximate Reasoning. Dordrecht, Netherlands: Kluwer Academic Publishers. ISBN 0-7923-6941-6.
• Kirk Ludwig & Greg Ray, "Vagueness and the Sorites Paradox", Philosophical Perspectives 16, 2002.
• Nouwen, Rick; Rooij, Robert van; Sauerland, Uli; Schmitz, Hans-Christian (2009). International Workshop on Vagueness in Communication (ViC; held as part of ESSLLI). LNAI. 6517. Springer. ISBN 978-3-642-18445-1.
• Sainsbury, R. M. (2009). Paradoxes (3rd ed.). Cambridge University Press.; Sect.3 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 30, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8076545596122742, "perplexity": 2547.895578632077}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-04/segments/1484560282932.75/warc/CC-MAIN-20170116095122-00110-ip-10-171-10-70.ec2.internal.warc.gz"} |
https://cdsweb.cern.ch/collection/PH-EP%20Preprints?ln=pl | # PH-EP Preprints
Ostatnio dodane:
2016-12-23
07:36
Measurement of the $J/\psi$ pair production cross-section in $pp$ collisions at $\sqrt{s} = 13 \,{\mathrm{TeV}}$ / LHCb Collaboration The production cross-section of $J/\psi$ pairs is measured using a data sample of $pp$ collisions collected by the LHCb experiment at a centre-of-mass energy of $\sqrt{s} = 13 \,{\mathrm{TeV}}$, corresponding to an integrated luminosity of $279 \pm 11 \,{\mathrm{pb^{-1}}}$. [...] arXiv:1612.07451 ; CERN-EP-2016-307 ; LHCB-PAPER-2016-057. - 2016. - 40 p. Full text - Related data file(s) - Related supplementary data file(s)
2016-12-17
08:13
Measurements of top-quark pair differential cross-sections in the $e\mu$ channel in $pp$ collisions at $\sqrt{s} = 13$ TeV using the ATLAS detector / ATLAS Collaboration This article presents measurements of $t\bar{t}$ differential cross-sections in a fiducial phase-space region, using an integrated luminosity of 3.2 fb$^{-1}$ of proton--proton data at a centre-of-mass energy of $\sqrt{s} = 13$ TeV recorded by the ATLAS experiment at the LHC in 2015. [...] CERN-EP-2016-220 ; arXiv:1612.05220. - 2016. - 43 p. Full text - Previous draft version
2016-12-07
07:18
Search for massive long-lived particles decaying semileptonically in the LHCb detector / LHCb Collaboration A search is presented for massive long-lived particles decaying into a muon and two quarks. [...] arXiv:1612.00945 ; LHCB-PAPER-2016-047 ; CERN-EP-2016-283. - 2016. - 25. Full text - Related data file(s) - Related supplementary data file(s)
2016-12-07
07:18
Final COMPASS results on the deuteron spin-dependent structure function $g_1^{\rm d}$ and the Bjorken sum rule / COMPASS Collaboration Final results are presented from the inclusive measurement of deep-inelastic polarised-muon scattering on longitudinally polarised deuterons using a $^6$LiD target. [...] arXiv:1612.00620 ; CERN-EP-2016-299. - 2016. - 16 p. Full text
2016-11-29
17:25
Reconstruction of primary vertices at the ATLAS experiment in Run 1 proton--proton collisions at the LHC / ATLAS Collaboration This paper presents the method and performance of primary vertex reconstruction in proton--proton collision data recorded by the ATLAS experiment during Run 1 of the LHC. [...] arXiv:1611.10235 ; CERN-EP-2016-150. - 2016. - 52 p. Previous draft version - Preprint - Full text
2016-11-29
10:12
Performance of the ATLAS Trigger System in 2015 / ATLAS Collaboration During 2015 the ATLAS experiment recorded $3.8 \mathrm{fb}^{-1}$ of proton--proton collision data at a centre-of-mass energy of $13 \mathrm{TeV}$. [...] arXiv:1611.09661 ; CERN-EP-2016-241. - 2016. - 76 p. Previous draft version - Preprint - Full text
2016-11-28
15:00
Search for CP violation in $\mathrm{ t \bar{t} }$ production and decay in proton-proton collisions at $\sqrt{s} =$ 8 TeV / CMS Collaboration The results of a first search for CP violation in the production and decay of top quark-antiquark ($\mathrm{ t \bar{t} }$) pairs are presented. [...] arXiv:1611.08931 ; CMS-TOP-16-001 ; CERN-EP-2016-266. - 2016. - 32 p. Preprint - Full text
2016-11-24
09:31
Evidence for the two-body charmless baryonic decay $B^+ \to p \kern 0.1em\overline{\kern -0.1em\Lambda}$ / LHCb Collaboration A search for the rare two-body charmless baryonic decay $B^+ \to p \kern 0.1em\overline{\kern -0.1em\Lambda}$ is performed with $pp$ collision data, corresponding to an integrated luminosity of $3\mbox{fb}^{-1}$, collected by the LHCb experiment at centre-of-mass energies of $7$ and $8\mathrm{\,Te\kern -0.1em V}$. [...] arXiv:1611.07805 ; LHCB-PAPER-2016-048 ; CERN-EP-2016-275. - 2016. - 18 p. Related data file(s) - Preprint - Preprint
2016-11-21
09:33
Search for supersymmetry in events with photons and missing transverse energy in pp collisions at 13 TeV / CMS Collaboration The results of a search for new physics in final states with photons and missing transverse energy are reported. [...] arXiv:1611.06604 ; CMS-SUS-15-012 ; CERN-EP-2016-269. - 2016. - 30 p. Preprint - Full text
2016-11-21
07:41
Measurements of charm mixing and $C\!P$ violation using $D^0 \to K^\pm \pi^\mp$ decays / LHCb collaboration Measurements of charm mixing and $C\!P$ violation parameters from the decay-time-dependent ratio of $D^0 \to K^+ \pi^-$ to $D^0 \to K^- \pi^+$ decay rates and the charge-conjugate ratio are reported. [...] arXiv:1611.06143 ; CERN-EP-2016-280 ; LHCB-PAPER-2016-033. - 2016. - 24 p. Preprint - Preprint - Related data file(s) - Related supplementary data file(s) | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9938520193099976, "perplexity": 3589.047009985378}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-04/segments/1484560281419.3/warc/CC-MAIN-20170116095121-00143-ip-10-171-10-70.ec2.internal.warc.gz"} |
http://tex.stackexchange.com/questions/11287/how-to-draw-inside-a-tikz-node-using-node-style | # How to draw inside a TikZ node, using node style?
I would like to define a node style, so that the node will look like the LaTeX symbol \oplus, i.e. a circle with lines from north to south, and east to west. (I also have more complicated examples in mind.)
I think that I should draw the lines in the execute at end node, or after mode path arguments, but nothing I have tried seems to work. The code I have in mind is something like this:
\tikzstyle{foo}=[circle,draw,execute at end node={\draw ??????}]
To rephrase the question: How can one do arbitrary drawing within a node, and define this as a style? In particular, how do you get the coordinates of the current node in a style?
-
a tip: you can use backticks to mark your inline code as I did in my edit. – Hendrik Vogt Feb 17 '11 at 11:07
All TikZ node shapes are defined using lower-level PGF code. This is described in the pgfmanual in section 75.5 Declaring New Shapes, on page 625 of the v2.1 manual.
You can use the existing code of the circle shape as a base and add the lines to it. The code can be found in the file ${TEXMF}/tex/generic/pgf/modules/pgfmoduleshapes.code.tex. Here the way I would do it: ## Shape declaration: \pgfdeclareshape{oplus} % % Shaped like '\oplus' math symbol. Based on 'circle' shape % {% % All anchors are taken from the 'circle' shape: \inheritsavedanchors[from={circle}]% \inheritanchor[from={circle}]{center}% \inheritanchor[from={circle}]{mid}% \inheritanchor[from={circle}]{base}% \inheritanchor[from={circle}]{north}% \inheritanchor[from={circle}]{south}% \inheritanchor[from={circle}]{west}% \inheritanchor[from={circle}]{east}% \inheritanchor[from={circle}]{mid west}% \inheritanchor[from={circle}]{mid east}% \inheritanchor[from={circle}]{base west}% \inheritanchor[from={circle}]{base east}% \inheritanchor[from={circle}]{north west}% \inheritanchor[from={circle}]{south west}% \inheritanchor[from={circle}]{north east}% \inheritanchor[from={circle}]{south east}% \inheritanchorborder[from={circle}]% % % Only the background path is different % \backgroundpath{% % First the existing 'circle' code: \pgfutil@tempdima=\radius% \pgfmathsetlength{\pgf@xb}{\pgfkeysvalueof{/pgf/outer xsep}}% \pgfmathsetlength{\pgf@yb}{\pgfkeysvalueof{/pgf/outer ysep}}% \ifdim\pgf@xb<\pgf@yb% \advance\pgfutil@tempdima by-\pgf@yb% \else% \advance\pgfutil@tempdima by-\pgf@xb% \fi% \pgfpathcircle{\centerpoint}{\pgfutil@tempdima}% % % Now the | and -- lines: \pgfmoveto{\pgfpointadd{\centerpoint}{\pgfpoint{0pt}{\pgfutil@tempdima}}}% \pgflineto{\pgfpointadd{\centerpoint}{\pgfpoint{0pt}{-\pgfutil@tempdima}}}% \pgfmoveto{\pgfpointadd{\centerpoint}{\pgfpoint{\pgfutil@tempdima}{0pt}}}% \pgflineto{\pgfpointadd{\centerpoint}{\pgfpoint{-\pgfutil@tempdima}{0pt}}}% }% } ## Usage example: \documentclass{standalone} \usepackage{tikz} \usepackage{pgfshape_oplus} \begin{document} \begin{tikzpicture} \node [draw=blue,shape=oplus] {Test}; \end{tikzpicture} \end{document} ## Result: You might also want to draw the additional lines in the \foregroundpath instead. For this see the forbidden sign shape in the file pgflibraryshapes.symbols.code.tex is an good example. - very good answer: the right method, the right code. But note the forbidden sign shape is backwards! – Matthew Leingang Feb 17 '11 at 13:05 @Matthew: You mean the line is drawn in the wrong angle, do you? That's something which should be reported to the PGF/TikZ author. Nevertheless the code of it is a good example for having a circle as background path and the other lines in the foreground. – Martin Scharrer Feb 17 '11 at 13:29 Yes, that's what I meant. I did report it to the mailing list a while back (google tells me it was Feb 2009). Maybe I should follow up on that since it's just a one-line patch—unless now you want it to be an option to have it backwards! – Matthew Leingang Feb 17 '11 at 13:35 @Matthew: Yeah, backwards compatibility (wow what a pun! :-) ) is always an issue even with smaller stuff like this. – Martin Scharrer Feb 17 '11 at 13:48 @Martin: Tell me about it. The new pgfmath engine has introduced bugs in lots of the graphs that I use for teaching. :-( – Matthew Leingang Feb 17 '11 at 14:17 A possibility : \begin{tikzpicture} \tikzset{oplus/.style={path picture={% \draw[black] (path picture bounding box.south) -- (path picture bounding box.north) (path picture bounding box.west) -- (path picture bounding box.east); }}} \node[oplus,fill=blue!10,draw=blue,thick,circle] {}; \end{tikzpicture} It's possible to send the cross in the background and to change the color \begin{tikzpicture} \tikzset{oplus/.style={path picture={% \begin{pgfonlayer}{background} \draw[red,opacity=.5] (path picture bounding box.south) -- (path picture bounding box.north) (path picture bounding box.west) -- (path picture bounding box.east); \end{pgfonlayer}}} } \node[oplus,draw=blue,thick,circle](a) {TEST}; \end{tikzpicture} Now your code execute at end node is not good in this case. This code modifies the content of the node. You can try the next code : \begin{tikzpicture} \tikzset{foo/.style ={circle,draw, execute at end node={\ #1}}} \node[foo=bad](A) at (2,3){good}; \end{tikzpicture} Interesting is execute at begin node=$ and execute at end node=\$
Now if my answer is not what you want, you can define new shape like shape =oplus` like in the other answer from Martin. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8734897971153259, "perplexity": 711.2154084167033}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-35/segments/1440646249598.96/warc/CC-MAIN-20150827033049-00073-ip-10-171-96-226.ec2.internal.warc.gz"} |
http://www.princeton.edu/~naomi/tac94.html | ## Motion Control of Drift-Free, Left-Invariant Systems on Lie Groups
### N.E. Leonard and P.S. Krishnaprasad
IEEE Transactions on Automatic Control, Vol. 40, No. 9, September, 1995, p. 1539-1554.
#### Abstract
In this paper we address the constructive controllability problem for drift-free, left-invariant systems on finite-dimensional Lie groups with fewer controls than state dimension. We consider small ($\epsilon$) amplitude, low-frequency, periodically time-varying controls and derive average solutions for system behavior. We show how the $p$th-order average formula can be used to construct open-loop controls for point-to-point maneuvering of systems that require up to $(p-1)$ iterations of Lie brackets to satisfy the Lie algebra controllability rank condition. In the cases $p = 2,3$, we give algorithms for constructing these controls as a function of structure constants that define the control authority, i.e., the actuator capability, of the system. The algorithms are based on a geometric interpretation of the average formulas and produce sinusoidal controls that solve the constructive controllability problem with $O(\epsilon^p)$ accuracy in general (exactly if the Lie algebra is nilpotent). The methodology is applicable to a variety of control problems and is illustrated for the motion control problem of an autonomous underwater vehicle with as few as three control inputs. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8491902351379395, "perplexity": 706.4198197935417}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-50/segments/1480698541773.2/warc/CC-MAIN-20161202170901-00248-ip-10-31-129-80.ec2.internal.warc.gz"} |
https://graph.subwiki.org/wiki/Radius_of_a_graph | ## Definition
The radius of a graph is defined for any connected graph as the radius of the metric space induced by it. Explicitly, for a graph with vertex set , it is:
where denotes the distance between two vertices. In words, the radius of a graph is the minimum, over all vertices, of the eccentricity of that vertex.
Note that for a finite graph, the radius is finite. For an infinite graph, the radius may be finite or .
For a graph that is not connected, we can consider the radius to be either or undefined.
## Related invarants
Diameter of a graph maximum of distances between pairs of vertices Radius Diameter Twice the radius | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 8, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.993790328502655, "perplexity": 280.2417217417259}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-27/segments/1656103941562.52/warc/CC-MAIN-20220701125452-20220701155452-00207.warc.gz"} |
https://brilliant.org/problems/what-is-it-5/ | # What is it?
Calculus Level 3
$$\int _{ 0 }^{ 1 }{ \left| x \right| \left| 1-x \right| dx}$$
can be expressed as the form $$\frac { a }{ b }$$ where $$a$$ and $$b$$ are co-prime integers what is the the value of $$a+b$$.
× | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8929600119590759, "perplexity": 520.8366148535189}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-30/segments/1500549426133.16/warc/CC-MAIN-20170726102230-20170726122230-00633.warc.gz"} |
https://brilliant.org/problems/can-we-treat-it-as-an-integral/ | # Can We Treat It As An Integral?
Calculus Level 4
Find the number of positive integers $$m$$ less than 100 such that $$\dfrac1{(\ln 2)^m} + \dfrac1{(\ln 3)^m} + \dfrac1{(\ln 4)^m} + \dfrac1{(\ln 5)^m} + \cdots$$ converges.
× | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8888789415359497, "perplexity": 562.6457186351458}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-44/segments/1476990033880.51/warc/CC-MAIN-20161020190033-00447-ip-10-142-188-19.ec2.internal.warc.gz"} |
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Sep 24 revised $C_c(X)$ dense in $L_1(X)$ added 15 characters in body Sep 24 comment $C_c(X)$ dense in $L_1(X)$ Did you use the Tietze theorem anywhere? Sep 24 comment $C_c(X)$ dense in $L_1(X)$ Thanks! Where do you get the increasing sequence from? $\sigma$-finite doesn't give you that: en.wikipedia.org/wiki/%CE%A3-finite_measure Sep 23 accepted Homology groups of unit square with parts removed — revisited Sep 23 accepted Why is an integral domain a commutative ring with unity? Sep 23 asked $C_c(X)$ dense in $L_1(X)$ Sep 23 accepted Lie algebra of $GL_n(\mathbb{C})$ Sep 19 comment Lie algebra of $GL_n(\mathbb{C})$ I know nothing. I started to read about Lie groups yesterday. I read that the exponential map maps elements in the Lie algebra to elements in the Lie group. I wonder if that is useful. Probably not because the exponential map is not surjective, I suppose.... Sep 19 asked Lie algebra of $GL_n(\mathbb{C})$ Sep 19 comment Definition of tangent space Thanks, actually anon's read it right. I'm just generally confused at the moment by the new subject of lie algebras and groups and I didn't see that it was so obvious. Sep 19 comment Definition of tangent space Thanks! And thanks for the book recommendation. As for the bonus: I'm still struggling with definitions so I'm not quite ready to answer that. Sep 19 accepted Definition of tangent space Sep 19 revised Definition of tangent space added 135 characters in body Sep 19 asked Definition of tangent space Sep 19 accepted Exact meaning of homology Sep 17 comment Exact meaning of homology I picture a one dimensional hole as something you can not get $S^1$ passed. I'm not sure it's right but in a cell complex such as the torus I can consider the one skeleton and then I see that it contains $2$ one dimensional holes. Now going to look at the other questions you gave me... thank you! Sep 17 asked Exact meaning of homology Sep 8 comment $X$ $n$-connected $\iff$ any continuous map $f:K \rightarrow X$ is null-homotopic @Shaun Ault: But $X$ is not necessarily a CW complex. It's just a topological space. Sep 8 comment $X$ $n$-connected $\iff$ any continuous map $f:K \rightarrow X$ is null-homotopic @Michael: I thought that too, several times even. But then I always thought I'm just not advanced enough yet... Sep 7 accepted $X$ a CW complex is contractible if it's the union of an increasing sequence | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8853076696395874, "perplexity": 683.0566454889331}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-35/segments/1440645281325.84/warc/CC-MAIN-20150827031441-00317-ip-10-171-96-226.ec2.internal.warc.gz"} |
https://communitycrypto.network/article/9e35e9-formal-letter-format-icse-class-9-2020 | # formal letter format icse class 9 2020
So this is a "prepare the way" video about symmetric matrices and complex matrices. Thus, the diagonal of a Hermitian matrix must be real. Eigenvalues of real symmetric matrices. And I guess the title of this lecture tells you what those properties are. I'll have to tell you about orthogonality for complex vectors. Eigenvalues of a triangular matrix. Are eigenvectors of real symmetric matrix all orthogonal? Similarly, show that A is positive definite if and ony if its eigenvalues are positive. Can you hire a cosigner online? A real symmetric matrix is a special case of Hermitian matrices, so it too has orthogonal eigenvectors and real eigenvalues, but could it ever have complex eigenvectors? B is just A plus 3 times the identity-- to put 3's on the diagonal. Orthogonality of the degenerate eigenvectors of a real symmetric matrix, Complex symmetric matrix orthogonal eigenvectors, Finding real eigenvectors of non symmetric real matrix. The length of x squared-- the length of the vector squared-- will be the vector. That's why I've got the square root of 2 in there. So I'll just have an example of every one. We give a real matrix whose eigenvalues are pure imaginary numbers. Here are the results that you are probably looking for. In engineering, sometimes S with a star tells me, take the conjugate when you transpose a matrix. Even if and have the same eigenvalues, they do not necessarily have the same eigenvectors. (Mutually orthogonal and of length 1.) Then prove the following statements. So these are the special matrices here. I must remember to take the complex conjugate. Then prove the following statements. the complex eigenvector $z$ is merely a combination of other real eigenvectors. The row vector is called a left eigenvector of . Real lambda, orthogonal x. But it's always true if the matrix is symmetric. Download the video from iTunes U or the Internet Archive. If I transpose it, it changes sign. Moreover, the eigenvalues of a symmetric matrix are always real numbers. Square root of 2 brings it down there. If $A$ is a matrix with real entries, then "the eigenvectors of $A$" is ambiguous. Namely, the observation that such a matrix has at least one (real) eigenvalue. OK. What about complex vectors? What about A? But recall that we the eigenvectors of a matrix are not determined, we have quite freedom to choose them: in particular, if $\mathbf{p}$ is eigenvector of $\mathbf{A}$, then also is $\mathbf{q} = \alpha \, \mathbf{p}$ , where $\alpha \ne 0$ is any scalar: real or complex. (In fact, the eigenvalues are the entries in the diagonal matrix (above), and therefore is uniquely determined by up to the order of its entries.) When we have antisymmetric matrices, we get into complex numbers. So that's really what "orthogonal" would mean. Here that symmetric matrix has lambda as 2 and 4. And those columns have length 1. So I take the square root, and this is what I would call the "magnitude" of lambda. All hermitian matrices are symmetric but all symmetric matrices are not hermitian. And now I've got a division by square root of 2, square root of 2. MATLAB does that automatically. And does it work? And eigenvectors are perpendicular when it's a symmetric matrix. 1 plus i over square root of 2. So if I have a symmetric matrix-- S transpose S. I know what that means. As the eigenvalues of are , . Real, from symmetric-- imaginary, from antisymmetric-- magnitude 1, from orthogonal. Distinct Eigenvalues of Submatrix of Real Symmetric Matrix. Again, real eigenvalues and real eigenvectors-- no problem. So if a matrix is symmetric--and I'll use capital S for a symmetric matrix--the first point is the eigenvalues are real, which is not automatic. Sponsored Links Formal definition. Namely, the observation that such a matrix has at least one (real) eigenvalue. Even if you combine two eigenvectors $\mathbf v_1$ and $\mathbf v_2$ with corresponding eigenvectors $\lambda_1$ and $\lambda_2$ as $\mathbf v_c = \mathbf v_1 + i\mathbf v_2$, $\mathbf A \mathbf v_c$ yields $\lambda_1\mathbf v_1 + i\lambda_2\mathbf v_2$ which is clearly not an eigenvector unless $\lambda_1 = \lambda_2$. So if a matrix is symmetric-- and I'll use capital S for a symmetric matrix-- the first point is the eigenvalues are real, which is not automatic. True or False: Eigenvalues of a real matrix are real numbers. The matrix A, it has to be square, or this doesn't make sense. I have a shorter argument, that does not even use that the matrix $A\in\mathbf{R}^{n\times n}$ is symmetric, but only that its eigenvalue $\lambda$ is real. Sponsored Links Measure/dimension line (line parallel to a line). The rst step of the proof is to show that all the roots of the characteristic polynomial of A(i.e. Thus, the diagonal of a Hermitian matrix must be real. Has anyone tried it. A Hermitian matrix always has real eigenvalues and real or complex orthogonal eigenvectors. As for the proof: the $\lambda$-eigenspace is the kernel of the (linear transformation given by the) matrix $\lambda I_n - A$. For n x n matrices A and B, prove AB and BA always have the same eigenvalues if B is invertible. Made for sharing. It's the fact that you want to remember. 1, 2, i, and minus i. If T is a linear transformation from a vector space V over a field F into itself and v is a nonzero vector in V, then v is an eigenvector of T if T(v) is a scalar multiple of v.This can be written as =,where λ is a scalar in F, known as the eigenvalue, characteristic value, or characteristic root associated with v.. Real symmetric matrices (or more generally, complex Hermitian matrices) always have real eigenvalues, and they are never defective. @Joel, I do not believe that linear combinations of eigenvectors are eigenvectors as they span the entire space. ), Learn more at Get Started with MIT OpenCourseWare, MIT OpenCourseWare makes the materials used in the teaching of almost all of MIT's subjects available on the Web, free of charge. Here the transpose is minus the matrix. Does for instance the identity matrix have complex eigenvectors? A full rank square symmetric matrix will have only non-zero eigenvalues It is illuminating to see this work when the square symmetric matrix is or. Do you have references that define PD matrix as something other than strictly positive for all vectors in quadratic form? I'll have 3 plus i and 3 minus i. And sometimes I would write it as SH in his honor. If I multiply a plus ib times a minus ib-- so I have lambda-- that's a plus ib-- times lambda conjugate-- that's a minus ib-- if I multiply those, that gives me a squared plus b squared. Prove that the matrix Ahas at least one real eigenvalue. How is length contraction on rigid bodies possible in special relativity since definition of rigid body states they are not deformable? That matrix was not perfectly antisymmetric. The transpose is minus the matrix. Here are the results that you are probably looking for. On the circle. The fact that real symmetric matrix is ortogonally diagonalizable can be proved by induction. Prove that the eigenvalues of a real symmetric matrix are real. Download files for later. So you can always pass to eigenvectors with real entries. Now for the general case: if $A$ is any real matrix with real eigenvalue $\lambda$, then we have a choice of looking for real eigenvectors or complex eigenvectors. A matrix is said to be symmetric if AT = A. There's no signup, and no start or end dates. Real symmetric matrices not only have real eigenvalues, they are always diagonalizable. So if a matrix is symmetric-- and I'll use capital S for a symmetric matrix-- the first point is the eigenvalues are real, which is not automatic. A matrix is said to be symmetric if AT = A. They pay off. Let . Symmetric Matrices There is a very important class of matrices called symmetric matrices that have quite nice properties concerning eigenvalues and eigenvectors. observation #4: since the eigenvalues of A (a real symmetric matrix) are real, the eigenvectors are likewise real. Every matrix will have eigenvalues, and they can take any other value, besides zero. This is the great family of real, imaginary, and unit circle for the eigenvalues. How can ultrasound hurt human ears if it is above audible range? All its eigenvalues must be non-negative i.e. Alternatively, we can say, non-zero eigenvalues of A are non-real. Real symmetric matrices have only real eigenvalues. site design / logo © 2020 Stack Exchange Inc; user contributions licensed under cc by-sa. A real symmetric n×n matrix A is called positive definite if xTAx>0for all nonzero vectors x in Rn. That gives you a squared plus b squared, and then take the square root. Well, that's an easy one. Real … Here is the lambda, the complex number. Feel I 've done is add 3 times the identity, so a real-valued Hermitian must... Get 0 and real eigenvalues, and they 're on the axis or that axis or axis! They span the entire space eigenvectors with real entries, symmetric and Hermitian have different meanings guess that that is... -- 3 plus I squared of $a$ is odd and not I have a one-way mirror layer! Gave me a 3 plus I and minus 1, 1 minus times. Tell you about orthogonality for complex matrices sponsored Links real symmetric matrix a is square! Matrix is ortogonally diagonalizable can be proved by induction OCW to guide your own life-long learning, or this n't! Ah = at, so I would call the magnitude '' of that any level and in! If Ais an n nsymmetric matrix with the property of being Hermitian for complex matrices the MIT OpenCourseWare site materials... Use the top silk layer symmetric matrix choose a game for a 3 plus I over square root Mn C. The eigenvalues of a real symmetric matrix, that is, AT=−A special relativity since definition rigid! References that define PD matrix as something other than strictly positive for all I and 3 I. That symmetric matrix are real do symmetric matrices always have real eigenvalues? that case, we get is only the. Looking for me give an example a Hermitian matrix is symmetric are all real and positive eigenvalues squares! Cite OCW as the transpose, it has to be square, this! S was a complex eigenvector $z=u+ v\cdot I$ with $,... If its eigenvalues are pure imaginary numbers is 0 get 0 and real eigenvalues, go! Antisymmetric, but still a good matrix multiply real eigenvectors -- no problem their eigenvectors can, and then the... N'T change when we view it as a complex number times its conjugate what if matrix... & open publication of material from thousands of MIT courses, covering the entire space let n an! Not necessarily have the same eigenvectors every I to a line ) should... To choose a game for a real skew-symmetric matrix then its eigenvalue will be the vector squared -- be. B is invertible and ony if its eigenvalues answer site for people studying math at any level and professionals related. Following fact: eigenvalues of Hermitian ( real or complex ) matrices are always real complex entries then... Did George Orr have in his coffee in the novel the Lathe of Heaven involves a choice that for! Row vector is not possible to diagonalize one by a real symmetric matrix, that is the! Slightly over my head... what is Mn ( C ) you what those properties are Spectral states... A question and answer site for people studying math at any level and professionals in related fields single senator passing. Transpose Q is the family of real, the diagonal of a ( a ) prove that the of! Sorry, that 's the fact that you are probably looking for he studied complex... 2 and 4 Pythagoras lived, or this does n't make sense positive for all vectors in form. I$ with $U, v\in do symmetric matrices always have real eigenvalues? { R } ^n$ when... Go along a, up B are imaginary, it satisfies by both... Should say -- I would have 1 plus the identity matrix have complex eigenvectors nonetheless ( by taking linear. A do symmetric matrices always have real eigenvalues? complex entries, then the eigenvectors certainly are determined '' they! Licensed under cc by-sa A_ij=A_ji for all I and 1 minus i. I flip across the real matrix. A single senator from passing a bill they want with a star tells me, take the conjugate well. 2001–2018 Massachusetts Institute of Technology same eigenvalues a square matrix with real entries, symmetric and Hermitian different... I know is becoming head of department, do I mean by orthogonal eigenvectors leads. ( line parallel to a minus i. Oh I send congratulations or condolences of MIT courses, covering the space. That gives me lambda is I and minus I iTunes U or the circle thank Pythagoras... If $n$ minus the rank of a symmetric matrix, and 's... About the diagonalization be an n×n real matrix I over square do symmetric matrices always have real eigenvalues? of two a. Mean I am long hair '' and not I have a complex eigenvector $z=u+ I. Real matrix are equal to its eigenvalues the magnitude of lambda S, an orthogonal one there. Thus we may take U to be 1 plus I squared would 0. In Rn passing a bill they want with a star tells me, take the conjugate as as. Long hair '' and not I am long hair '' and not I... ; user contributions licensed under cc by-sa do symmetric matrices always have real eigenvalues? j special properties, in! And professionals in related fields right, I do not necessarily have the same eigenvalues, they!, be taken orthonormal matrices, we get not believe that linear combinations ) other! If S was a complex eigenvector$ z=u+ v\cdot I $with$ U, v\in \mathbf { R ^n... An odd integer and let a be a real symmetric positive-definite matrix Aare all.... Need to be a real matrix does n't make sense systems of differential.... And 4 both sides of the matrix Ahas at least one real.! Eigenvectors -- no problem has lambda as 2 and 4 symmetric too other than positive. Be proved by induction their eigenvectors can, and I guess that that matrix is symmetric when we it... Positive length -- how do I prove that a symmetric matrix are always real same eigenvalues orthogonal if.... Eigenvectors like $z$ is ortogonally diagonalizable can be found -- recognize. And n real symmetric matrices a and B, prove AB and BA always have the same eigenvectors copy paste. We do n't offer credit or certification for using OCW antisymmetric -- magnitude 1 2! Real and positive I go along a, up B Internet Archive take x transpose x, I not... Real eigenvalues, and he understood to take -- I should have written linear combination of eigenvectors perpendicular... Would be 0 division by square root of 2 Theorem states that if eigenvalues of a real matrices! Guide your own pace property of being symmetric for real matrices corresponds to property. For 2 do not necessarily have the same eigenvalues vectors '' mean that finding a of. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8766297101974487, "perplexity": 443.81482770945235}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-49/segments/1669446710462.59/warc/CC-MAIN-20221128002256-20221128032256-00598.warc.gz"} |
https://crypto.stackexchange.com/questions/25471/the-non-interactive-proof-of-verifiable-computation-pinocchio | # The non-interactive proof of verifiable computation: Pinocchio
I am reading the Pinocchio paper. The paper says, in paragraph "polynomial asymptotics" of section 4.2.1, a worker, in order to include $h(s)$ into the proof, has to interpolate $p(x)$, and then divide it with $t(x)$ to get $h(x)$, then calculate $h(s)$.
As a worker, I would calculate $p(s)$, then $t(s)$, and let $h(s)=p(s)*(t(s))^{-1}$. Suppose this $h(s)$ is good, this approach is much less complicated, compared to the approach described in the paper, which dictates at least $O(n (log(n))^2)$ complexity.
I am wondering why this simpler approach to get $h(s)$ for the worker does not work. Any help is appreciated.
The worker does not know $s, t(s), v_k(s), w_k(s), y_k(s)$, the workers knows only the content of the public evaluation key and the public verification key. That's to say, the worker knows only $g^s, g^{t(s)}, g^{v_k(s)}$ etc. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9468845725059509, "perplexity": 464.0579628296105}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-05/segments/1579250593994.14/warc/CC-MAIN-20200118221909-20200119005909-00470.warc.gz"} |
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