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http://math.stackexchange.com/questions/125716/why-does-n-choose-r-relate-to-permutations/125721 | # Why does $n \choose r$ relate to permutations?
For example, how come $4 \choose 3$ (from 4 dice, choose 3 to be the same) can relate to the list:
d, s, s, s
s, d, s, s
s, s, d, s
s, s, s, d
Where s = same number, and d = different number? Shouldn't this be a permutations problem?
-
Because you chose exactly 3 out of 4: you chose 3 spots out of the 4 for the same number $s$. – N. S. Mar 29 '12 at 2:13
If you are making four-letter "words" where you have only two available letters, indeed you are counting permutations. But the "choose" symbol can be useful in counting for example the number of words that have three s and one d. There is no one universal tool for counting permutations. – André Nicolas Mar 29 '12 at 2:15
Actually binomial coefficient don't relate to permutations (even in the high-school sense of that word) but to combinations, which is exactly what you have here. – Marc van Leeuwen Mar 29 '12 at 12:28
Consider the number of permutations of $0$ and $1$ where there are $a$ $0$s and $b$ $1$s.
You have a total of $a+b$ spots and you select $a$ of them to place the $0$s. This amounts to $\binom{a+b}{a}$.
If you want to count as permutations, if the zeroes and ones were distinct, you would get $(a+b)!$ permutations.
But each permutation where they are not distinct, gives rise to $a! b!$ permutations where they are considered distinct.
Thus the total number of "not-distinct" permutations is $\frac{(a+b)!}{a!b!}$.
Since the total is the same, irrespective of how you count them, you have just proved that (assuming a combinatorial definition of the binomial coefficient)
$$\binom{a+b}{a} = \frac{(a+b)!}{a!b!}$$
- | {"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": 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": 0.8912439346313477, "perplexity": 361.57665888987174}, "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-2016-07/segments/1454701171770.2/warc/CC-MAIN-20160205193931-00314-ip-10-236-182-209.ec2.internal.warc.gz"} |
https://www.physicsforums.com/threads/determining-magnitudes-of-forces-on-a-block.297022/ | # Determining magnitudes of forces on a block
1. Mar 3, 2009
### tonia0605
A block of mass (2.20kg) is pushed 2.21m along a frictionless horizontal table by a constant 19.5 N force directed 26.4 deg below the horizontal.
I have found the work done by the applied force, which is 3.86 J
I got stuck after that...I still need help figuring out the following questions
If someone could help show me how to setup my equations, I should be able to get them
1. Determine the magnitude of the normal force exerted by the table.
2. Determine the magnitude of the force of gravity.
3. Determine the net force on the block.
2. Mar 4, 2009
### NoobixCube
How good are you with vector addition and subtraction?
For the first question, your are pushing 'down' by an amount, and 'along' by and amount. If these two amounts were added as vectors (i.e. with magnitudes w/ directions) they would give you the 19.5N at 26.4 deg below the horiz.
So the amount you are pushing down is
19.5 *sin(26.4)= F(down)
then the force from gravity is
F(g)= m*g= 2.2 *10=22N
So both of these two forces (you and from grav.) push down on the table, therefore the table pushes back by the same amount, namely the magnitude of the normal force exerted by the table
3. Mar 4, 2009
### tonia0605
I am not that strong at adding vectors.
Now that I got both of these forces, do I just add Fn and Fg together to get the net force?
Similar Discussions: Determining magnitudes of forces on a block | {"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": 0.885233998298645, "perplexity": 1340.9439974506072}, "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/1502886102309.55/warc/CC-MAIN-20170816170516-20170816190516-00144.warc.gz"} |
https://realmath.de/english/function/reciprocal/hyperbolaquiz2.php | Hyperbola » The point P lies on the hyperbola h. In this problem ... the coordinates of the point P are given. the equation of the hyperbola h is incomplete. Calculate the missing value k. Therefore insert the point into the hyperbola equation!
Click on new to create a new problem.
Can you top 285 points?
#### Topic: HyperbolaCalculate values -level 1- Calculate the missing value.
Functions
on realmath.de
Hyperbola
Enable keyboard | {"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": 0.8760836124420166, "perplexity": 2027.3055695284093}, "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-2022-40/segments/1664030337480.10/warc/CC-MAIN-20221004054641-20221004084641-00268.warc.gz"} |
http://tex.stackexchange.com/questions/74748/positioning-equations-with-tikz-diagram | Positioning equations with tikz diagram
I'm trying to annotate an diagram of a neural network with an equation that describes the activity of each node. I can't get the equation to the upper right, where I want it to be. I've tried using the scope environment in TikZ, \vspace and \hspace outside of tikz and even multicols.
I feel there likely is an easier or more elegant way of doing this. To make it easier to find, the equation I want to put in the upper right is, in this example, immediately after the tikzpicture environment ends. Here's my MWE:
\documentclass[tikz]{standalone}
\usepackage{tikz}
\usetikzlibrary{arrows}
\definecolor{burntorange}{RGB}{204,85,0}
\begin{document}
\begin{tikzpicture}[->,>=stealth',shorten >=1pt,auto,node distance=3cm,
thick,main node/.style={circle,fill=blue!20,draw,font=\sffamily\Large\bfseries}]
\node[main node, scale=1.4] (interest) {$\vec{v}_i$};
\node[main node, fill=blue!10] (j) [below left of=interest] {$\vec{v}_j$};
\node[main node, fill=blue!10] (k) [below right of=j] {$\vec{v}_k$};
\node[main node, fill=burntorange, below right of=interest, distance = 5cm ] (input) [below right of=interest] {input $u$};
\begin{scope}[every node/.style={font=\sffamily\small}]
\path
(interest)
edge [bend right] node[left] {$\mathbf{M}_{ji}$} (j)
edge [loop above] node[above] {$\mathbf{M}_{ii}$} (interest)
edge [bend left] node {$\mathbf{M}_{ki}$} (k);
\path[every edge/.style={gray,draw=gray}]
(j) edge node [right] {$\mathbf{M}_{ij}$} (interest)
edge [loop left] node {$\mathbf{M}_{jj}$} (j)
edge [bend right] node[left] {$\mathbf{M}_{jk}$} (k)
(k) edge node [right] {$\mathbf{M}_{kj}$} (j)
edge node [bend right] {$\mathbf{M}_{ik}$} (interest)
edge [loop below] node {$\mathbf{M}_{kk}$} (k)
(input) edge [bend right] node {$\mathbf{W}_i$} (interest)
edge [bend left] node {$\mathbf{W}_k$} (k);
\end{scope}
\end{tikzpicture}
{$\tau_{i}\frac{d\vec{v_{i}}}{dt}=-\vec{v}_{i}+\mathbf{M}_{ii}\vec{v}_{i}+ \sum_{j\neq i}\mathbf{M}_{ij}\vec{v}_{j}+\mathbf{W}_{i}\vec{u}$}
\end{document}
-
I hope I understand your question corrcectly. You can put the equation in a node and position that node. Positioning can be in various ways left of v_i, above input u etc. Here is another alternative:
\documentclass{article}
\usepackage{tikz}
\usetikzlibrary{arrows}
\definecolor{burntorange}{RGB}{204,85,0}
\begin{document}
\begin{tikzpicture}[->,>=stealth',shorten >=1pt,auto,node distance=3cm,
thick,main node/.style={circle,fill=blue!20,draw,font=\sffamily\Large\bfseries}]
\node[main node, scale=1.4] (interest) {$\vec{v}_i$};
\node[main node, fill=blue!10] (j) [below left of=interest] {$\vec{v}_j$};
\node[main node, fill=blue!10] (k) [below right of=j] {$\vec{v}_k$};
\node[main node, fill=burntorange, below right of=interest, distance = 5cm ] (input) [below right of=interest] {input $u$};
\begin{scope}[every node/.style={font=\sffamily\small}]
\path
(interest)
edge [bend right] node[left] {$\mathbf{M}_{ji}$} (j)
edge [loop above] node[above] {$\mathbf{M}_{ii}$} (interest)
edge [bend left] node {$\mathbf{M}_{ki}$} (k);
\path[every edge/.style={gray,draw=gray}]
(j) edge node [right] {$\mathbf{M}_{ij}$} (interest)
edge [loop left] node {$\mathbf{M}_{jj}$} (j)
edge [bend right] node[left] {$\mathbf{M}_{jk}$} (k)
(k) edge node [right] {$\mathbf{M}_{kj}$} (j)
edge node [bend right] {$\mathbf{M}_{ik}$} (interest)
edge [loop below] node {$\mathbf{M}_{kk}$} (k)
(input) edge [bend right] node {$\mathbf{W}_i$} (interest)
edge [bend left] node {$\mathbf{W}_k$} (k);
\end{scope}
\node at ([shift={(-1cm,-1cm)}]current bounding box.north east) {$\tau_{i}\frac{d\vec{v_{i}}}{dt}=-\vec{v}_{i}+\mathbf{M}_{ii}\vec{v}_{i}+ \sum_{j\neq i}\mathbf{M}_{ij}\vec{v}_{j}+\mathbf{W}_{i}\vec{u}$};
\end{tikzpicture}
\end{document}
-
That's exactly what I wanted. I tried annotating edges, but it never occurred to me to use a node like that. Thank you for teaching me something about TikZ. – mac389 Sep 30 '12 at 14:45
@mac389 My pleasure. If the math needs to be regular, you can add \displaystyle just after the first \$ sign which would show it as if it was enclosed by $...$ – percusse Sep 30 '12 at 14:47 | {"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": 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": 0.9428021311759949, "perplexity": 3920.2347382515945}, "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-2015-27/segments/1435375097354.86/warc/CC-MAIN-20150627031817-00271-ip-10-179-60-89.ec2.internal.warc.gz"} |
https://pdfroom.com/books/qcd-string-in-light-light-and-heavy-light-mesons/0YpgQDxndNz | # QCD string in light-light and heavy-light mesons
2001 • 42 Pages • 437 KB • English
Posted April 14, 2020 • Uploaded by gaetano68
## Summary of QCD string in light-light and heavy-light mesons
QCD string in light-light and heavy-light mesons �a �� a,b ��� a Yu.S.Kalashnikova , A.V.Nefediev , Yu.A.Simonov a Institute of Theoretical and Experimental Physics, 117218, B.Cheremushkinskaya 25, Moscow, Russia b Centro de F´ısica das Interac¸co˜es Fundamentais (CFIF), Departamento de F´ısica, Instituto Superior T´ecnico, Av. Rovisco Pais, P-1049-001 Lisboa, Portugal Abstract Spectra of light–light and heavy–light mesons are calculated within the framework of the QCD string model, which is derived from QCD in the Wilson loop approach. Special attention is payed to the proper string dynamics that allows to reproduce the straight-line Regge trajectories with the inverse slope being 2πσ for light–light and as twice as smaller for heavy–light mesons. We use the model of the rotating QCD string with quarks at the ends to calculate masses of several light-light mesons lying on the lowest Regge trajectories and compare them with the experimental data as well as with the predictions of other models. Masses of several low-lying orbitally and radially excited heavy–light states in the D, Ds, B and Bs mesons spectra have been calculated in the einbein (auxiliary) field approach, which is proven to be rather accurate in var- ious calculations for relativistic systems. The results for the spectra have been compared with the experimental and recent lattice data. It is demonstrated that the account for the proper string dynamics encoded in the so-called string correction to the interquark interaction leads to extra negative contribution to the masses of orbitally excited states that resolves the problem of identifi- cation of the D(2637) state recently claimed by DELPHI Collaboration. � �� ��� [email protected] [email protected] [email protected] 1
¯ For the heavy-light system we extract the constants Λ, λ1 and λ2 used in the Heavy Quark Effective Theory (HQET) and find a good agreement with the results of other approaches. PACS: 12.38Aw, 12.39Hg, 12.39Ki I. INTRODUCTION Description of the mass spectrum of hadrons is one of the fundamental problems of strong interactions. It has been attacked in a sequence of approaches motivated by QCD, but still attracts considerable attention. One of the most intriguing phenomena, namely the formation of an extended object, the QCD string, between the colour constituents inside hadrons, plays a crucial role in understanding their properties. In the present paper this role is exempli�ed by spectra of mass of light-light and heavy-light mesons. In the former case we study the role played by the QCD string in formation of the straight-line Regge trajectories and discuss the form of the interquark interaction inside light hadrons. For heavy-light mesons we �nd the masses of several low-lying states in the D, Ds, B and Bs mesons spectra including orbitally and radially excited ones. We calculate and discuss the spin-spin and spin-orbit splittings and compare them to the experimental and recent lattice data. A special attention is payed to the role of the proper string dynamics in establishing the correct slope of the Regge trajectories for both, light- light and heavy-light states, as opposed to those following from the relativistic equations with local potentials. We remind then that an extra piece of the e�ective interquark potential, the string cor- rection, which is entirely due to the string-type interaction in QCD [1,2], gives negative contribution into the masses of orbitally excited states. The latter observation allows to re- solve the \mystery" of an extremely narrow D(2637) state (and similar one in the B-mesonic spectrum) [3] recently claimed by DELPHI Collaboration [4,5]. We present a reasonable �t 2
for the several lowest states in D- and B-mesonic spectra using the standard values for the string tension, the strong coupling constant and the current quark masses. We also �nd the correspondence between our model and the Heavy Quark E�ective Theory extracting the constants used in the latter approach in the expansion of a heavy-light meson mass in the inverse powers of the heavy quark mass. We �nd analytical formulae for these constants and compare their numerical estimates with the predictions of other models. The two main approaches used in the numerical calculations are the quasiclassical method of solving the eigenenergies problem and the variational one based on the einbein �eld formalism. Accuracy of both methods is tested using exactly solvable equations and found to be about 7% at worst even for the lowest states. Possible improvements of the method are outlined and discussed. The paper is organized as follows. In Section II we give a brief insight into various aspects of the einbein �eld formalism. In Section III the exact spectra of relativistic equations are confronted to the results of approximate calculations using the quasiclassical and variational einbein �eld methods, as well as the combined one. In Section IV we discuss the problem of the Regge trajectories slopes as they appear from the relativistic equations with local potentials and from the string-like picture of con�nement. Derivation of the Hamiltonian for the spinless quark-antiquark system as well as of the spin-dependent corrections to it is the subject of Section V. Spectra of light-light and heavy-light mesonic states are calculated and discussed in Sections VI and VII respectively. Section VIII contains our conclusions and outlook. II. EINBEIN FIELD FORMALISM In this section we give a short introduction into the method of the einbein �elds and its possible applications to relativistic systems. An interested reader can �nd a more detailed information in [6] and references therein. 3
A. Reparametrization invariance and constrained systems Historically einbein �eld formalism was introduced in [7] to treat the kinematics of the relativistic spinless particles. Later it was generalized for the case of spinning particles [8] 1 and strings [9]. So, the action of a free relativistic particle can be rewritten as ∫ τf p m2 �x_2 2 S = L(�); L = −m x_ ! − − ; (1) τi 2� 2 where the dot denotes derivative with respect to the proper time �, � being the einbein 2 �eld . The original form of the action can be easily restored after solving the Euler{Lagrange equation of motion for the einbein � which amounts to taking extremum in the latter. Note that the invariance of the initial action with respect to the change of the proper time df � ! f(�) > 0 f(�i) = �i f(�f) = �f (2) d� is preserved if an appropriate re-scaling is prescribed to �: _ � ! �=f: (3) The latter invariance means that one deals with a constrained system. For the free particle the only constraint de�nes the mass shell 2 2 p − m = 0; (4) or in presence of the einbein �eld � 2 2 p −m � = 0 H = − ; (5) 2� 1 In the path integral formalism this transformation is based on the following relation ∫ ( ∫ ( )) ( ∫ ) p aµ b Dµ(τ) exp − dτ + � exp − dτ ab . 2 2µ 2 1 Usually e = is referred as the einbein [7]. µ 4
with � being the momentum canonically conjugated to �, H is the Hamiltonian function of the system (in case (4) it identically vanishes). Requirement that the constraint � = 0 is preserved in time returns one to the mass-shell condition (4): 2 2 @H p −m 2 2 0 = �_ = f�Hg = = � p −m : (6) 2 @� 2� To make things simpler, one can �x the gauge-like freedom (2) identifying the proper time � with one of the physical coordinates of the particle. The most popular choices are � the laboratory gauge (� = x0); Pµ � the proper time gauge (� = (nx), nµ = p P2 with Pµ being the total momentum of the system) [10]; 1 � the light-cone gauge (� = 2(x0 + x3) = x+), which lead to quantization of the system on di�erent hypersurfaces. With the laboratory gauge �xed the Lagrangian function (1) becomes 2 2 _ m � �~x L = − − + ; (7) 2� 2 2 so that the corresponding Hamiltonian function reads 2 2 p~ + m � H = + ; (8) 2� 2 and after taking extremum in � one ends with the standard relativistic expression √ 2 2 H = p~ +m : (9) B. Einbeins as variational parameters In the simple example considered above neither the Lagrange, nor the Hamilton functions of the system contained �_ that allowed to get rid of � at any stage by taking extremum in the latter. It is not so for more complicated systems when a change of variables is to 5
be performed which touches upon the einbeins. The velocity corresponding to the original degrees of freedom of the system may mix in a very tangled way with those for einbeins, so that it is not a simple task anymore to follow the lines a� la Dirac [11] to resolve the set of constraints and to get rid of non-physical degrees of freedom. See e:g: [6,8,12] for several examples when such a resolution can be done explicitly. Luckily another approach to einbeins is known [2,13]. They can be treated as variational parameters. Thus one replaces the dynamical function of time �(�) by the parameter �0 independent on �. The eigenstates problem is solved then keeping �0 constant, so that one has the spectrum Mfng(�0), where fng denotes the full set of quantum numbers. Then one 3 is to minimize each eigenenergy independently with respect to �0 : ∣ ∣ @Mfng(�0)∣ � ∣ = 0 Mfng = Mfng(� 0): (10) @�0 ∣ � µ0=µ 0 Such an approach has a number of advantages. First, it allows to avoid tedious algebra of commuting constraints with one another following the standard Dirac technique [11]. Second, it allows a very simple and physically transparent interpretation of einbeins. Indeed, in formulae (1) and (7) the einbein � can be treated as an e�ective mass of the particle; dynamics of the system remains essentially relativistic, though being non-relativistic in form. If m is the current quark mass, then � can be viewed as its constituent mass celebrated in hadronic phenomenology. What is more, the current mass can be even put to zero, whereas the Lagrangian approach remains valid in presence of the einbeins and the standard Hamiltonian technique can be developed then. The latter observation is intensively used in the analytic QCD calculations for glue describing gluonic degrees of freedom in glueballs and hybrids [14,15]. 3 Note that solutions for µ0 of both signs appear, but only one of them (µ0 > 0) is finally left. Neglecting the negative solution is the general lack of the einbein field approach and this leads to the fact that quark Zitterbewegung is not taken into account (see also discussion in Subsection IIID). 6
An obvious disadvantage of the variational approach to the einbein �elds is some loss of accuracy. As a variational method it provides only an approximate solution giving no hint on how to estimate the ultimate accuracy of the results. Thus in the next section we test this method comparing its predictions with the exact solutions of some relativistic equations. We consider the accuracy, found to be about 7% at worst, quite reasonable, that justi�es our consequent attack at the light-light and heavy-light mesons spectra using this formalism. III. TESTING THE METHOD A. Quasiclassics for the spinless Salpeter equation We start from the Salpeter equation for the quark-antiquark system with equal masses and restrict ourselves to the zero angular momentum case for simplicity: ( √ ) 2 2 (ll) 2 p r + m + �r n = Mn n; (11) where the subscript (ll) stands for the light-light system. The quasiclassical quantization condition looks like ∫ r+ ( 3) M(ll) − 2m n pr(r)dr = � n + ; n = 0; 1; 2; : : : ; r+ = ; (12) 0 4 � where the integral on the l.h.s. can be worked out analytically yielding √ ( ) 2 √ (ll) 2 (ll) ( ) (ll) ( (ll))2 2 2 Mn − 4m +Mn 3 M n Mn − 4m − 4m ln = 4�� n + ; (13) 2m 4 p or approximately (m � �) one has ( ) ( ) 2 3 ��(n + 3=4) (ll) 2 M = 4�� n + + 2m ln + : : : : (14) 2 4 m Solution (14) becomes exact in the limit m = 0, whereas for a nonzero mass the leading (ll) 2 correction to the linear regime (M ) � n behaves like ( ) 2 2 �M n m Mn ln n = O ln � : (15) M2 M2 m n�1 n n n 7
For a heavy-light system one has the Salpeter equation (√ ) 2 2 (hl) p r + m + �r n = Mn n; (16) (hl) where M denotes the excess over the heavy particle mass M. Similarly to (13) one �nds n then √ ( ) 2 √ (hl) 2 (hl) ( ) (hl) ( (hl))2 2 2 Mn −m +Mn 3 M n Mn −m −m ln = 2�� n + ; (17) m 4 and the formula (15) holds true in this case as well. Comparing the results of the WKB method with the exact solutions of the equation (11) (raws Mn(WKB) and Mn(exact) in Table I), one can see that the error does not exceed 3-4% even for the ground state. See also [16] where the WKB method is tested for light-light mesons. B. Quasiclassics for the one-particle Dirac equation As a next example we discuss the one-particle Dirac equation with linearly rising con�n- ing potential [17]: (�~p~ + �(m + U) + V ) n = "n n: (18) The WKB method applied to this equation gives [18,19] ( ) ∫ ( ) r+ �w 1 p + dr = � n + ; n = 0; 1; 2; : : : ; (19) r− pr 2 where √ 2 � 2 2 p = (" − V ) − − (m + U) ; (20) 2 r 0 0 1 1 U − V w = − − ; 2r 2 m + U + " − V 1 j�j = j + : 2 8
For the most interesting case of purely scalar con�nement (V = 0, U = �r) an approxi- mate quasiclassical solution was found in [19] (m = 0): ( ) ( ) 2 2 3 sgn� �� " �� 2 n " = 2� 2n + j + + + 0:38 + ln + O : (21) n 2 2 2 2 �" �j�j " n n Detailed comparison of the results of the WKB method and those following from the recursive formula (21) with the exact numerical solutions to equation (18) is given in [20]. Here we only note that the coincidence of the three numbers is impressing as even for the lowest states the discrepancy does not exceed 1%. C. Quasiclassical variational einbein field (combined) method for the spinless Salpeter equation Finally we combine the two methods discussed above and apply the WKB approximation to the Hamiltonian of a relativistic system with einbeins introduced as variational parame- ters. Then the resulting quasiclassical spectrum is minimized with respect to the einbeins. Thus we have a powerful method of solving the eigenvalues problem for various relativistic systems which we call \combined". Let us test the accuracy of this method �rst. We start from the Salpeter equation (11) for the light-light system and introduce the parameter �0 as described in Section II: √ 2 2 p + m 2 2 r H1 = 2 p r + m + �r −! H2 = + �0 + �r: (22) �0 In what follows we consider the massless case substituting m = 0 into (22). We give the analytic formulae for the spectrum of the Salpeter equation (11) obtained using the quasiclassical approximation for the Hamiltonian H1 (following from equation (13) for m = 0), the exact solution for the Hamiltonian H2 minimized with respect to the einbein �eld and the result of the combined method when the Bohr-Sommerfeld quantization condition is applied to the Hamiltonian H2 and the ultimate spectrum is also minimized with respect to �0. 9
( ) 3 2 M (WKB) = 4�� n + (23) n 4 ( ) 3/2 2 −�n+1 M (einbein) = 16� (24) n 3 ( ) 8� 3 2 M (combined) = p � n + ; (25) n 3 4 where �n+1 is the (n+1)-th zero of the Airy function Ai(z) and counting of zeros starts from unity. The extremal values of the einbein �eld in the latter two cases read √ ( ) 3/4 √ � p −�n+1 � √�(n + 3=4) � (einbein) = � � (combined) = p ; (26) 0 0 3 2 3 p � i:e: the e�ective quark mass is � � � and it appears entirely due to the interquark 0 interaction. In Table I we compare the results of the above three approximate methods of solving the eigenvalues problem for equation (11) with the exact solution. In the last raw we give the accuracy of the combined method vs the exact solution. Two conclusions can be deduced from Table I. The �rst one is that the accuracy of all approximate methods is high enough, including the combined method which is of most interest for us in view of its consequent applications to the QCD string with quarks at the ends. The other conclusion is that the variational einbein �eld method gives a systematic overestimation for the excited states which is of order 5-7%. D. Discussion Here we would like to make a couple of concluding comments concerning the numerical methods tested in this section, their accuracy and possible ways of its improvement. As stated above the combined quasiclassical variational method is of most interest for us, so we shall concentrate basically on it. The following two remarks are in order here. From Table I one can see that the relative error is practically constant tending to the value of 7% for large n. The reason for such a behavior will become clear if one compares formulae 10
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2012 • 412 Pages • 10.8 MB | {"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": 0.9566042423248291, "perplexity": 1647.4547747072918}, "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-2021-39/segments/1631780058415.93/warc/CC-MAIN-20210927090448-20210927120448-00093.warc.gz"} |
http://mathhelpforum.com/algebra/105567-logarithm-help-solve-x.html | Math Help - Logarithm Help: Solve for x.
1. Logarithm Help: Solve for x.
I need to solve for x.
8^(log_2(x)) - 25^(log_5(x)) = 4x - 4
please can someone tell me how to solve this. All i have managed to do is prove it....but not solve for it.
2. Originally Posted by lockdout
I need to solve for x.
8^(log_2(x)) - 25^(log_5(x)) = 4x - 4
please can someone tell me how to solve this. All i have managed to do is prove it....but not solve for it.
ok
$A^{log_A (x) } = x$ you should know this ok
$8^{\log _2 x} - 25^{\log _5 x} = 4x-4$
$8=2^3 , 25 = 5^2$ so
$2^{(3\log _2 x )} - 5^{(2\log _5 x)} = 4x-4$
$\left(2^{\log _2 x}\right)^3 - \left(5^{\log _5 x}\right)^2 =4x-4$
$x^3 - x^2 = 4x-4$
$x^3-x^2 -4x+4 = 0$ this for you | {"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": 0.9407824277877808, "perplexity": 1199.8641570953096}, "config": {"markdown_headings": false, "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-2016-07/segments/1454701165484.60/warc/CC-MAIN-20160205193925-00228-ip-10-236-182-209.ec2.internal.warc.gz"} |
http://math.stackexchange.com/questions/259024/compute-the-conditional-expectation-e-u-uv/259036 | # Compute the conditional expectation E( U |UV )
I have $U$ and $V$ independent uniform on $[0,1]$ random variables I have to compute $\mathbb{E}(U | UV)$. Is there a specific trick to do this?
The basic way I know leads me find a function, $f$, such that for any measurable function, $g$, I have: $\mathbb{E}(g(UV)*U) = \mathbb{E}(g(UV)*f(UV))$ , which gives me:
$\displaystyle \int_0^1 \int_0^1 g(xy)*x*dx*dy = \int_0^1 \int_0^1g(xy)*f(xy)*dx*dy$
but then I have no way to proceed and identity $f$... any ideas?
thanks!
edit : I tagged it as homework but it's actual an exercise from a previous exam I am trying to do.
-
As you rightly point out, the goal is to identify a function $f$ such that the identity $$\iint g(xy)x\mathbf 1_{0\leqslant x,y\leqslant 1}\mathrm dx\mathrm dy=\iint g(xy)f(xy)\mathbf 1_{0\leqslant x,y\leqslant 1}\mathrm dx\mathrm dy,$$ holds for every (bounded measurable) function $g$.
Hint: Both sides are linear functionals of $g$, hence they are integrals of $g$ with respect to some measures. Equate those measures.
Concretely, this means using some change of variables to reach integrals of $g$. Here the change of variables $(x,y)\to(z,t)=(xy,x)$ seems a good idea. One gets $\mathrm dz\mathrm dt=x\mathrm dx\mathrm dy$, hence $$\mathrm{LHS}=\iint g(z)\mathbf 1_{0\leqslant z\leqslant t\leqslant 1}\mathrm dz\mathrm dt,$$ and $$\mathrm{RHS}=\iint g(z)f(z)\mathbf 1_{0\leqslant z\leqslant t\leqslant 1}\mathrm dz\mathrm dt/t.$$ Thus, $\mathrm{LHS}=\mathrm{RHS}$ for every (bounded measurable) function $g$ if and only if (almost everywhere) $$f(z)\int\mathbf 1_{0\leqslant z\leqslant t\leqslant 1}\mathrm dt/t=\int\mathbf 1_{0\leqslant z\leqslant t\leqslant 1}\mathrm dt,$$ that is, $$f(z)\mathbf 1_{0\leqslant z\leqslant 1}(-\log z)=\mathbf 1_{0\leqslant z\leqslant 1}(1-z).$$ To sum up, $f(z)$ is undefined when $z\lt0$ or $z\geqslant1$ (as was to be expected), and, if $0\leqslant z\lt 1$, $f(z)$ must make true (almost everywhere) the identity above. An example of such a function $f$ is $$f(z)=\mathbf 1_{0\lt z\lt 1}(1-z)/(-\log z),$$ hence
$$\mathbb E(U\mid UV)=f(UV)=(1-UV)/(-\log UV).$$
Note that while the function $f$ is not unique, even almost everywhere, since, for example, $f(z)$ for $z\lt0$ can be anything one wants, the random variable $f(UV)$ is unique almost surely.
Sanity checks: $\mathbb E(U\mid UV)\geqslant UV$ (why?) and $f(z)\to1$ when $z\to1$, $z\lt1$ (why?).
-
You can use a change of variables to obtain the joint density of $(U,UV)$, and in particular the density of $UV$. Dividing gives you the conditional density of $U$ given $UV$, from which all you have to do is integrate.
-
I don't have a good background in measure-theory so this may not be a very rigorous solution.
It can be seen that $$\int\limits_k^{\sqrt{k}} \rho(u) du = \int\limits_{\sqrt{k}}^{1} \rho(u) du$$ or more generally $$\int\limits_{a}^{b}\rho(u)du = \int\limits_\frac{k}{b}^\frac{k}{a}\rho(u)du$$
This is because, $\Pr[b < U < a | UV = k] = \Pr[1/b < V < 1/a | UV = k]$ and $U$ and $V$ are symmetric and interchangable.
Differentiating both sides w.r.t. $b$ we get, $$\rho(b) = -\rho\left(\frac{k}{b}\right) * \frac{d \frac{k}{b}}{d b} = \rho\left(\frac{k}{b}\right) \frac{k}{b^2}$$
If we assume the a power-law expression for $\rho$, after normalising, we get $$\rho(u) = -\frac{1}{\log(k)}\frac{1}{u}$$ which given an expectation value of $\frac{k-1}{\log(k)}$.
- | {"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": 0.9317719340324402, "perplexity": 143.90371930229614}, "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-2015-48/segments/1448398452385.31/warc/CC-MAIN-20151124205412-00292-ip-10-71-132-137.ec2.internal.warc.gz"} |
https://help.altair.com/hwcfdsolvers/acusolve/topics/acusolve/element_data_commands_acusolve_com_ref.htm | # Element Data Commands
The Element Data commands are outlined in this chapter.
A problem must contain at least one ELEMENT_SET command. All other commands in this chapter are optional. | {"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": 0.9458303451538086, "perplexity": 3728.5503752881027}, "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-2023-14/segments/1679296948708.2/warc/CC-MAIN-20230327220742-20230328010742-00018.warc.gz"} |
http://stackoverflow.com/questions/10190657/sweave-how-to-get-blank-lines-as-in-the-source | # Sweave: How to get blank lines as in the source?
How can I get a blank line before the second comment in the PDF file of the following .Rnw file? I tried to work with keep.source and strip.white, but I still don't get the blank line -- all the chunks are "pasted" together.
\documentclass{article}
\usepackage{fancyvrb}
\usepackage{Sweave}
\begin{document}
<<setup, eval=FALSE>>=
## some comment
a <- 1
b <- 2
## some comment (there is no newline before this comment...)
c <- 3
d <- 4
@
\end{document}
-
Did you look at knitr? It is said to give more formatting choices. – vaettchen Apr 18 '12 at 4:55
I wouldn't know about any option for keeping the blank line. What you could do:
(1) If you need the named chunk kept together, you could simply put a # at the beginning of the supposedly blank line:
<<setup, eval=FALSE>>=
## some comment
a <- 1
b <- 2
#
## some comment (there is no newline before this comment...)
c <- 3
d <- 4
@
Less attractive, but you have at least something similar to a blank line, and keep your chunk.
(2) If you don't care for keeping the chunk together, you can separate it into two blocks:
\documentclass{article}
\usepackage{fancyvrb}
\usepackage{Sweave}
\begin{document}
<<setup, eval=FALSE>>=
## some comment
a <- 1
b <- 2
@
<<eval=FALSE>>=
## some comment (there is no newline before this comment...)
c <- 3
d <- 4
@
\end{document}
This gives you the appearance you like, at the price of losing the structure of your chunk.
Hope this helps, Rainer
-
Solution 1 gives me: ERROR: You can't use macro parameter character # in horizontal mode. Solution 2: does not work when you have a nested for loop with logical blocks (separated by newlines) inside -- you couldn't close the for loops. – Marius Hofert Apr 17 '12 at 17:32
(1) You have probably run pdflatex on the .Rnw file and forgot to Sweave it before - that is how I can reproduce this error. You get it for every line with a #, right? -- Forget (2) anyway, it would have worked in the simplified sample but obviously not in a more complicated situation – vaettchen Apr 17 '12 at 17:53
I did a stupid mistake: I compiled the .Rnw that had the indented lines by four chars (to easily copy-and-paste to stackoverflow :-) ). It's working now, but the # are still ugly. I wonder if fancyvrb can solve that. – Marius Hofert Apr 17 '12 at 23:23
This is a Sweave issue, not LaTeX. I don't think fancyvrb can help. You could, of course manually edit your .tex file before running pdflatex. Put something like #*# in the .Rnw where you want the blank line, and then replace it in the tex file. But that would need to insert, on one line each, \end{Sinput} \end{Schunk} \begin{Schunk} \begin{Sinput} in order to create that one blank line, and this even without > – vaettchen Apr 18 '12 at 4:51 | {"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": 0.8933958411216736, "perplexity": 2980.06465825378}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "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-2014-35/segments/1408500832052.6/warc/CC-MAIN-20140820021352-00072-ip-10-180-136-8.ec2.internal.warc.gz"} |
https://mvtrinh.wordpress.com/2014/10/15/volume-of-octahedron/ | ## Volume of Octahedron
An octahedron is formed by connecting the centers of the faces of a cube. What is the ratio of the volume of the cube to that of the contained octahedron?
Source: NCTM Mathematics Teacher 2006
SOLUTION
Suppose the cube side length equals $2a$. The octahedron is made up of an upper pyramid and a lower pyramid. The height $h$ of each pyramid equals $a$. The base of the pyramids is a square of side length $a\sqrt 2$. The surface area $B$ of the base equals $2a^2$.
Volume of octahedron = volume of upper pyramid + volume of lower pyramid
$=Bh/3+Bh/3$
$=2Bh/3$
$=2(2a^2)(a)/3$
$=4a^3/3$
Volume of cube = $(2a)^3$
$=8a^3$
Ratio of volume of cube to volume of octahedron
$8a^3 : 4a^3/3$
Simplify
$6 : 1$
Answer: $6 : 1$ | {"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": 15, "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": 0.8537822961807251, "perplexity": 414.4065840696407}, "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-2017-47/segments/1510934808935.79/warc/CC-MAIN-20171124195442-20171124215442-00381.warc.gz"} |
http://mathhelpforum.com/differential-equations/170836-solve-problem-using-bernoulli-print.html | # solve the problem using Bernoulli....
• February 10th 2011, 05:42 PM
slapmaxwell1
solve the problem using Bernoulli....
ok i dont see how the book came up with the substitution of u= y^-1
so here is the problem..
(t^2)dy/dt + y^2 = ty
my question is this, if u = y^(1-n) wouldnt n = 0?
i think the book may have made a mistake on this problem....can someone please check the problem and see if the substitution is correct? thanks in advance..
• February 10th 2011, 05:52 PM
Krizalid
to make it clearer rewrite the equation as $\dfrac{t^2}{y^2}y'+1=\dfrac ty$ and put $u=\dfrac1y.$
• February 10th 2011, 05:56 PM
slapmaxwell1
well i did rewrite it...
i got dy/dt + (y^2/t^2) = y/t but the way i have it written it seems like n would be 1?
• February 10th 2011, 06:13 PM
slapmaxwell1
ok i got it..thanks for your help...
• February 10th 2011, 07:49 PM
Prove It
To use Bernoulli...
$\displaystyle t^2\,\frac{dy}{dt} + y^2 = t\,y$
$\displaystyle t^2\,\frac{dy}{dt} - t\,y = -y^2$.
Make the substitution $\displaystyle v = y^{1-2} = y^{-1} \implies y = v^{-1}$.
Then $\displaystyle \frac{dy}{dt} = \frac{d}{dt}(v^{-1}) = \frac{d}{dv}(v^{-1})\,\frac{dv}{dt} = -v^{-2}\,\frac{dv}{dt}$.
Substituting into the DE gives
$\displaystyle t^2\left(-v^{-2}\,\frac{dv}{dt}\right) - t\,v^{-1} = -v^{-2}$
$\displaystyle t^2\,\frac{dv}{dt} + t\,v = 1$
$\displaystyle \frac{dv}{dt} + t^{-1}v = t^{-2}$.
This is now first order linear, so multiplying both sides of the DE by the Integrating Factor $\displaystyle e^{\int{t^{-1}\,dt}} = e^{\ln{t}} = t$ gives
$\displaystyle t\,\frac{dv}{dt} + v = t^{-1}$
$\displaystyle \frac{d}{dt}(t\,v) = t^{-1}$
$\displaystyle t\,v = \int{t^{-1}\,dt}$
$\displaystyle t\,v = \ln{|t|} + C$
$\displaystyle v = \frac{\ln{|t|} + C}{t}$
$\displaystyle y^{-1} = \frac{\ln{|t|} + C}{t}$
$\displaystyle y = \frac{t}{\ln{|t|} + C}$. | {"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": 17, "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": 0.9539201259613037, "perplexity": 1633.0912892097313}, "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-2015-22/segments/1432207928520.68/warc/CC-MAIN-20150521113208-00074-ip-10-180-206-219.ec2.internal.warc.gz"} |
https://www.semanticscholar.org/paper/RHLE%3A-Automatic-Verification-of-Dickerson-Ye/820724c38fa2c91db77b8e353017cbf1588d74c2 | Corpus ID: 211066221
# RHLE: Automatic Verification of \$\forall\exists\$-Hyperproperties
```@article{Dickerson2020RHLEAV,
title={RHLE: Automatic Verification of \\$\forall\exists\\$-Hyperproperties},
author={Robert Dickerson and Qianchuan Ye and Benjamin Delaware},
journal={arXiv: Programming Languages},
year={2020}
}```
• Published 2020
• Computer Science
• arXiv: Programming Languages
Specifications of program behavior typically consider single executions of a program, usually requiring that every execution never reaches a bad state (a safety property) or that every execution can eventually produce some good state (a liveness property). Many desirable behaviors, however, including refinement and non-interference, range over multiple executions of a program. These sorts of behaviors are instead expressible as a combination of k-safety and k-liveness hyperproperties… Expand
#### References
SHOWING 1-10 OF 22 REFERENCES
Verifying Hyperliveness
• Computer Science
• CAV
• 2019
This paper reduces existential quantification to strategic choice and shows that synthesis algorithms can be used to eliminate the existential quantifiers automatically and can be extended to reactive system synthesis, i.e., to automatically construct a reactive system that is guaranteed to satisfy a given HyperLTL formula. Expand
Cartesian hoare logic for verifying k-safety properties
• Computer Science
• PLDI 2016
• 2016
This paper presents a sound and relatively complete calculus, called Cartesian Hoare Logic (CHL), for verifying k-safety properties, and presents an automated verification algorithm based on CHL and implement it in a tool called DESCARTES. Expand
Beyond 2-Safety: Asymmetric Product Programs for Relational Program Verification
• Computer Science
• LFCS
• 2013
Relational Hoare Logic is a generalization of Hoare logic that allows reasoning about executions of two programs, or two executions of the same program. It can be used to verify that a program isExpand
Semantical consideration on floyo-hoare logic
• V. Pratt
• Computer Science
• 17th Annual Symposium on Foundations of Computer Science (sfcs 1976)
• 1976
An appropriate axiom system is given which is complete for loop-free programs and also puts conventional predicate calculus in a different light by lumping quantifiers with non-logical assignments rather than treating them as logical concepts. Expand
Temporal Logics for Hyperproperties
• Computer Science, Mathematics
• POST
• 2014
It is shown that the quantification over paths naturally subsumes other extensions of temporal Logic with operators for information flow and knowledge, and the model checking problem for temporal logic with path quantification is decidable. Expand
Maximal specification synthesis
• Computer Science
• POPL 2016
• 2016
A novel approach is presented that utilizes a counterexample-guided inductive synthesis loop and reduces the maximal specification inference problem to multi-abduction, and the novel notion of multi- abduction as a generalization of classical logical abduction is formulated. Expand
Relational abstract interpretation for the verification of 2-hypersafety properties
• Computer Science
• CCS
• 2013
A framework for proving 2-hypersafety properties by means of abstract interpretation on the self-compositions of the control flow graphs of programs to prove intricate information flow properties of programs written in a simple language for tree manipulation motivated by the Web Services Business Process Execution Language. Expand
Verification Tools for Finite-State Concurrent Systems
• Computer Science
• REX School/Symposium
• 1993
This paper describes in detail how the new implementation works and gives realistic examples to illustrate its power, and discusses a number of directions for future research. Expand
Simple relational correctness proofs for static analyses and program transformations
We show how some classical static analyses for imperative programs, and the optimizing transformations which they enable, may be expressed and proved correct using elementary logical andExpand
Relational Verification Using Product Programs
• Computer Science
• FM
• 2011
This work provides a general notion of product program that supports a direct reduction of relational verification to standard verification, and illustrates the benefits of the method with selected examples, including non-interference, standard loop optimizations, and a state-of-the-art optimization for incremental computation. Expand | {"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": 0.8665802478790283, "perplexity": 4447.044865625955}, "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/1634323588102.27/warc/CC-MAIN-20211027053727-20211027083727-00654.warc.gz"} |
http://www.computer.org/csdl/trans/tc/2002/05/t0521-abs.html | Subscribe
Issue No.05 - May (2002 vol.51)
pp: 521-529
ABSTRACT
<p>Montgomery multiplication in {\rm GF}(2^m) is defined by a(x)b(x)r^{-1}(x)\bmod{f(x)}, where the field is generated by a root of the irreducible polynomial f(x), a(x) and b(x) are two field elements in {\rm GF}(2^m), and r(x) is a fixed field element in {\rm GF}(2^m). In this paper, first, a slightly generalized Montgomery multiplication algorithm in {\rm GF}(2^m) is presented. Then, by choosing r(x) according to f(x), we show that efficient architectures of bit-parallel Montgomery multiplier and squarer can be obtained for the fields generated with an irreducible trinomial. Complexities of the Montgomery multiplier and squarer in terms of gate counts and time delay of the circuits are investigated and found to be as good as or better than that of previous proposals for the same class of fields.</p>
INDEX TERMS
Finite fields arithmetic, hardware architecture, Montgomery multiplication, elliptic curve cryptography
CITATION
H. Wu, "Montgomery Multiplier and Squarer for a Class of Finite Fields", IEEE Transactions on Computers, vol.51, no. 5, pp. 521-529, May 2002, doi:10.1109/TC.2002.1004591
SEARCH
19 ms
(Ver 2.0)
Marketing Automation Platform | {"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": 0.8698704838752747, "perplexity": 2360.195177185193}, "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-2016-07/segments/1454701162938.42/warc/CC-MAIN-20160205193922-00265-ip-10-236-182-209.ec2.internal.warc.gz"} |
http://mathoverflow.net/questions/54753/lower-bound-on-autg?sort=newest | # lower bound on Aut(G)
Given a centerless finite group G, with at least one automorphism which is not conjugation by an element of G. Is there any lower bound on the size of Aut(G) given in terms of G? (As big as possible, of course).
-
Only the trivial bound $2|G|$, because the alternating groups $A_n$, $n\neq 6$ make this sharp. See e.g. the Wikipedia article. | {"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": 0.9811767935752869, "perplexity": 217.36734028988906}, "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/1464054526288.74/warc/CC-MAIN-20160524014846-00062-ip-10-185-217-139.ec2.internal.warc.gz"} |
https://www.universetoday.com/15308/diameter-of-saturn/ | # Diameter of Saturn
Saturn has an equatorial diameter of 120,536 km, 9.44 times that of Earth. That makes it the second largest planet in our Solar System, trailing only Jupiter. Saturn, like all of the other planets, is an oblate spheriod. This means that its equatorial diameter is larger than is diameter measured through the poles. In the case of Saturn this distance is quite a bit different due to the planet’s high rotational speed. The polar diameter of Saturn is 108,728 km, meaning that it is flattened by a factor of 9.796%.
Scientist know that Saturn rotates very quickly, but the exact speed of that rotation has been hard to determine because of the thick clouds in the atmosphere. With terrestrial planets, scientists are able to find surface features and basically time how long it takes for that feature to reappear in the same position. This is a simplified description of how they determine rotational speed. The problem with Saturn is that the surface can not be observed. To make things even more difficult, the visible features of the planet’s atmosphere rotate at different speeds depending on their latitude.
The atmosphere of Saturn is broken down into systems. System I is the equatorial zone has a rotational period of 10 hours and 14 minutes. System II encompasses all other areas of Saturn and has a rotational speed of 10 hours 38 minutes and 25.4 seconds. System III is based on radio emissions and has mostly replaced the use of the term System II. It has a rotational speed of 10 hours 39 minutes and 22.4 seconds. Despite these numbers, the rotational speed of the planet’s interior is currently impossible to measure precisely. The Cassini spacecraft found the radio rotational speed of Saturn to be 10 hours 45 minutes and 45 seconds. In 2007, it was determined that the varying radio emissions from the planet did not match Saturn’s rotation rate. Some scientists think that the variance is due to geyser activity on the Saturnian moon Enceladus. The water vapor from these geysers enter Saturn’s orbit become charged, thus creating a drag effect on Saturn’s magnetic field. This slows the magnetic field’s rotation slightly compared to the rotation of the planet. The current estimate of Saturn’s rotation is based on various measurements from the Cassini, Voyager and Pioneer probes. That estimated speed is 10 hours 32 minutes and 35 seconds as of September 2007.
Again, the equatorial diameter of Saturn is 120,536 km and its polar diameter is 108,728 km. It is very important to understand why the difference in these diameters is so large, that is why so much detail is given on the rotational speed of the planet. You can take many of the same factors into account when thinking about all of the gas giants.
Here’s an article about how long a day is on Saturn, and another article about how the storms never end on Saturn.
We have recorded two episodes of Astronomy Cast just about Saturn. The first is Episode 59: Saturn, and the second is Episode 61: Saturn’s Moons.
Source: NASA | {"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": 0.8476133942604065, "perplexity": 567.8045708742524}, "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-2019-13/segments/1552912202781.33/warc/CC-MAIN-20190323080959-20190323102959-00431.warc.gz"} |
https://apmr.matelys.com/Parameters/ThermalCharacteristicLength.html | This parameter has been introduced by Champoux and Allard [CA91] on the basis of Johnson, Koplik and Dashen work [JKD87] related to viscous effects.
Schematic representations of a 2-Dimensional pore, the viscous characteristic length $\Lambda$ and the thermal characteristic length $\Lambda'$. The thermal characteristic length, controlling the thermal effects at medium and high acoustical frequencies is related to the largest size of the pores.
The thermal characteristic length can be estimated from standing wave tube measurements or ultrasound techniques. | {"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": 0.9289368391036987, "perplexity": 1959.2304582781182}, "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-2021-31/segments/1627046150067.51/warc/CC-MAIN-20210723210216-20210724000216-00289.warc.gz"} |
https://www.physicsforums.com/threads/how-to-understand-math-unboundedness.93616/ | # How to understand math Unboundedness
1. Oct 11, 2005
### pinki82
Max (x_1)
x_1 - x_2 <= 1
x_1, x_2 >= 0
is obviously 'Unbounded'.
But i dont really understand this.. How do we know that it is
Unbounded??
what does it mean by Unboundedness?
2. Oct 12, 2005
### HallsofIvy
Staff Emeritus
"Unbounded" means precisely that! It has no bounds. In this case it is the fact that x1 and x2 are required to be non-negative but there is no "bound" on how large they can be. If we were given:
x1>= 0, x2>= 0 and x1+ x2<= 1, then that last inequality would bound x1 and x2- since they are non-negative and their sum[\b] can't be larger than 1, neither can be larger than 1.
However, because the condition is that x1- x2<= 1, there is no "bound", x1= 10000000, x2= 9999999 fit all conditions as does x1= A, x2= A-1 for any positive number A.
Similar Discussions: How to understand math Unboundedness | {"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": 0.9243369698524475, "perplexity": 2787.7113464449}, "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-2017-51/segments/1512948551162.54/warc/CC-MAIN-20171214222204-20171215002204-00636.warc.gz"} |
https://freron.lighthouseapp.com/projects/58672-mailmate/tickets/2675?spam=1 | new
## texmath not working ??
Reported by tryout user | October 23rd, 2020 @ 02:36 PM
While asciimath seems to be working texmath with $symbols doe snot seem to get activated. Suggestions ? ### Comments and changes to this ticket • #### bennyOctober 26th, 2020 @ 01:50 PM Make sure you follow the instructions provided here. Math-mode is not enabled using $.
If that doesn't help then please provide an explicit example of what you are trying to write.
## Example of received message which uses standard tex math symbols but doesn't display any of the texmath characters:
\ arXiv:2012.00895
Date: Tue, 1 Dec 2020 23:43:53 GMT (1206kb,D)
Title: Central exclusive production at LHCb
Authors: Charlotte Van Hulse (on behalf of the LHCb collaboration)
Categories: hep-ex
Comments: 6 pages, 5 figures, conference proceedings for New trends in
high-energy physics 2019 Journal-ref: Ukr. J. Phys. 64 (2019) 595-601
DOI: 10.15407/ujpe64.7.595
\ The LHCb collaboration has measured central exclusive production of $J/\psi$, $\psi(2S)$, and $\Upsilon$ mesons as well as $J/\psi J/\psi$, $J/\psi\psi(2S)$, $\psi(2S)\psi(2S)$, and $\chi_c\chi_c$ meson pairs in proton-proton collisions. The analyses of $\Upsilon$ and charmonium pairs are performed at the
centre-of-mass energies of 7 TeV and 8 TeV, and those of $J/\psi$ and
$\psi(2S)$ are done at 7 TeV and 13 TeV. The analysis at 13 TeV involves the use of new shower counters. These allow a reduction in the background by
vetoing events with activity in an extended region in rapidity. The
measurements of central exclusive production at LHCb are sensitive to gluon
distributions for Bjorken-$x$ values down to $2\times10^{-6}$ (at 13 TeV). An
overview of the LHCb results is presented and compared to existing measurements
of other experiments and theoretical calculations.
## \ ( https://arxiv.org/abs/2012.00895 , 1206kb)
With your very own profile, you can contribute to projects, track your activity, watch tickets, receive and update tickets through your email and much more. | {"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": 0.8767757415771484, "perplexity": 4344.361370668372}, "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/1674764499700.67/warc/CC-MAIN-20230129044527-20230129074527-00744.warc.gz"} |
https://physics.stackexchange.com/questions/555816/euler-maruyama-scheme/556208 | # Euler-Maruyama scheme
Can the Euler-Maruyama method be used to simulate Langevin equations for non-Gaussian white noise? I need to evaluate a Langevin equation of the form $$dx= a(x)dt+D \eta dt$$ where $$\eta$$ is a non-Gaussian white noise.
• It's fine.The problem arrives deriving the Fokker-Planck equation, that assumes Gaussian fluctuations. – Alexander May 31 at 18:47
• Thank you very much – Vip Jun 2 at 1:13 | {"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": 2, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.883907675743103, "perplexity": 1277.0953444987163}, "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-20200810012015-20200810042015-00037.warc.gz"} |
https://math.stackexchange.com/questions/2455282/recurrence-relation-problem-given-a-n-left-lfloor-sqrta-1-a-2-cdots | # Recurrence relation problem: Given $a_n = \left \lfloor \sqrt{a_1 + a_2 + \cdots + a_{n-1}} \right \rfloor$, find $a_{1000}$.
I've been struggling with this problem for some time.
Let there be a recurrent sequence ${a_n}$ such that $a_1 = 1$ and $a_n = \left \lfloor\sqrt{a_1 + a_2 + \cdots + a_{n-1}}\right \rfloor$ for all $n > 1$. Find $a_{1000}$.
I've tried to plug in different $n$ and have noticed that the square root makes it grow slowly, and with bigger $n$ the terms become more distinct (different integers), but that still hasn't gotten me close to solving for $a_{1000}$.
Update: I've solved it. I will post my solution in a bit. The hints you guys have given me have proved valuable, and I appreciate the caution with which you guys go about giving me a full solution based on my lack of steps taken.
• The sequence is registered in the OEIS as A109964 – miracle173 Oct 3 '17 at 23:08
Hint: $1$ appears four times in the series, powers of two appears three times, and all other numbers appear exactly twice.
• This is not actually correct. A quick manual computation demonstrates that the sequence begins as $$\{1,1,1,1,2,2,2,3,3,4,4,4,5,5,\ldots\}.$$ – heropup Oct 3 '17 at 6:29
• @heropup Oops, I meant to say "powers of two appear three times". – orlp Oct 3 '17 at 6:46
• and how do you prove this? – miracle173 Oct 3 '17 at 7:01
• @miracle173 Then it wouldn't really be a hint anymore. – orlp Oct 3 '17 at 7:23
• does this mean you have a proof but you don't publish it here or does this mean you generated a lot of elements of this sequence, looked at them and deduced that your statement may be true? – miracle173 Oct 3 '17 at 17:06
If we write down some of the first numbers, we get: $(1,1,1,1,2,2,2,3,3,4,4,4,5,5,6,6,7,7,8,8,8,9,9,…)$. From this it is visible that number $1$ appears four times, while all other numbers appear twice, thrice at most. We will further define that numbers that appear thrice are the powers of $2$, starting with $2^1$. All other numbers therefore appear twice.
Suppose we wrote down the beginning of the sequence until the first occurrence of the number $n \ (n>1)$, and that the sequence visibly behaves as defined.
Let $k$ be the largest integer satisfying $2^{k}$. Then the sum of all the values is:
$s_1 = (1+2+...+n) + (1+2+...+n-1) + (1+2+2^2+...+2^k) +1 = n^2 + 2^{k+1}$
since $2^{k+1} = 2 \times 2^k < 2n < 2n +1 = (n+1)^2 - n^2$. We have $s_1 < (n+1)^2$ and the following member is therefore $\lfloor\sqrt{s_1}\rfloor = n$.
Now we define the next member in this order, and the sum is now:
$s_2 = s_1 + n = n^2 + n + 2^{k+1}$,
so if $2^{k+1} < n+1$, then $s_2 < (n+1)^2$ and the next member is $n$.
However, $k$ is the largest integer satisfying $2^k < n$, so it's true that $n\le 2^{k+1}$. The previous situation therefore occurs iff $2^{k+1} = n$.
When $n$ isn't a power of two, the next member is $n+1$, because $n+1\le 2^{k+1} < 2n < 3n +4$, which follows from $(n +1)^2 \le n^2 + n + 2^{k+1} < (n+2)^2$.
Now we only need to show that if $n = 2^{k+1}$, then after three occurrences of $n$, an $n+1$ will follow.
Now the sum is:
$s_3 = s_2 + n = n^2 + 2n + 2^{k+1} = n^2 + 3n$, and we immediately get the inequality $(n+1)^2 < s_3 < (n+2)^2$.
Now we finally see that $500 = a_{1010} = a_{1009}$. Thus, $a_{1000} = \boxed{495}$. | {"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": 0.8149757981300354, "perplexity": 174.23596320163873}, "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-2019-30/segments/1563195530250.98/warc/CC-MAIN-20190724020454-20190724042454-00348.warc.gz"} |
https://infoscience.epfl.ch/record/203504 | Infoscience
Journal article
# Computation Over Gaussian Networks With Orthogonal Components
Function computation over Gaussian networks with orthogonal components is studied for arbitrarily correlated discrete memoryless sources. Two classes of functions are considered: 1) the arithmetic sum function and 2) the type function. The arithmetic sum function in this paper is defined as a set of multiple weighted arithmetic sums, which includes averaging of the sources and estimating each of the sources as special cases. The type or frequency histogram function counts the number of occurrences of each argument, which yields various fundamental statistics, such as mean, variance, maximum, minimum, median, and so on. The proposed computation coding first abstracts Gaussian networks into the corresponding modulo sum multiple-access channels via nested lattice codes and linear network coding and then computes the desired function using linear Slepian-Wolf source coding. For orthogonal Gaussian networks (with no broadcast and multiple-access components), the computation capacity is characterized for a class of networks. For Gaussian networks with multiple-access components (but no broadcast), an approximate computation capacity is characterized for a class of networks. | {"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": 0.9076343178749084, "perplexity": 912.3947059940868}, "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-2017-04/segments/1484560280718.7/warc/CC-MAIN-20170116095120-00083-ip-10-171-10-70.ec2.internal.warc.gz"} |
http://mathhelpforum.com/algebra/93372-confusing.html | Math Help - confusing >.<
1. confusing >.<
I am really stuck on a question..
(Q) Find x from the following:
34.56, x, 24, 20.
The asnwer is 28.80. I don't know how to get that. The two fomula you can use are Tn = a+(n-1)d or Sn= n/2 [2a+(n-1)d]
Thanks if you can show me how to do it!
2. Originally Posted by dizzycarly
I am really stuck on a question..
(Q) Find x from the following:
34.56, x, 24, 20.
The asnwer is 28.80. I don't know how to get that. The two fomula you can use are Tn = a+(n-1)d or Sn= n/2 [2a+(n-1)d]
Thanks if you can show me how to do it!
Could you please write the problem exactly as it appears in the book? This would help me to better understand the problem.
Do you mean $T_n=a+(n-1)d$ or $S_n= \frac{n}{2}[2a+(n-1)d]$
If so, what do t, s, a, and d represent?
3. Originally Posted by dizzycarly
I am really stuck on a question..
(Q) Find x from the following:
34.56, x, 24, 20.
The asnwer is 28.80. I don't know how to get that. The two fomula you can use are Tn = a+(n-1)d or Sn= n/2 [2a+(n-1)d]
Thanks if you can show me how to do it!
The two formulas you wrote are for use with arithmetic sequences, but the sequence you wrote is a geometric sequence.
In any event, x = 28.80 because the ratio between two consecutive terms is constant:
$\frac{34.56}{28.8} = \frac{28.8}{24} = \frac{24}{20} = 1.2$.
01
4. Originally Posted by yeongil
In any event, x = 28.80 because the ratio between two consecutive terms is constant:
$\frac{34.56}{28.8} = \frac{28.8}{24} = \frac{24}{20} = 1.2$.
01
Dude, you're like some kind of wizard.
5. Ap/gp
Hello dizzycarly
Originally Posted by dizzycarly
I am really stuck on a question..
(Q) Find x from the following:
34.56, x, 24, 20.
The asnwer is 28.80. I don't know how to get that. The two fomula you can use are Tn = a+(n-1)d or Sn= n/2 [2a+(n-1)d]
Thanks if you can show me how to do it!
The formulae you quote are related to Arithmetic Progressions - where the differences between consecutive terms are equal. The difference between $24$ and $20$ is $-4$, and obviously this can't be the difference between $34.56$ and $x$, and $x$ and $24$.
So we must look elsewhere.
In fact, this is a Geometric Progression, where the ratio of consecutive terms is constant. You do it like this:
The ratio of $24$ to $20$ is $\frac{24}{20} : 1 = 1.2:1$. In other words, we must multiply $20$ by $1.2$ to make $24$. If this sequence is a GP, we shall be able to multiply $24$ by $1.2$ to get $x$, and then multiply $x$ by $1.2$ to get $34.56$. And this works!
Try it out: $24\times 1.2 = 28.8$, and $28.8 \times 1.2 = 34.56$.
So $x = 2.8$.
(In fact, the common ratio of a GP is the ratio that will take you from left to right through the sequence of numbers; in this case that's the ratio the other way round; i.e. $20:24$, or $\tfrac56:1$.) | {"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": 28, "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": 0.9498293995857239, "perplexity": 420.47157275028565}, "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-2016-07/segments/1454701148758.73/warc/CC-MAIN-20160205193908-00028-ip-10-236-182-209.ec2.internal.warc.gz"} |
https://www.physicsforums.com/threads/waveguide-modes-relative-power.152677/ | Waveguide modes relative power
1. Jan 23, 2007
DaveCunnah
I am attempting to model a multi-mode symmetrical slab waveguide using Fortran 90. I am able to produce plots of each optical mode and am then summing them but I am more or less convinced that the relative intensity of the modes is incorrect. I am fairly sure that the error is arising because I am mis-using the amplitude constants which describe the electric field inside and out of the waveguide. Anybody with any experience oin this who could offer me assistance please let me know and I can get you a copy of the code for your attention.
Many thanks,
David Cunnah
Can you offer guidance or do you also need help?
Draft saved Draft deleted
Similar Discussions: Waveguide modes relative power | {"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": 0.8052147626876831, "perplexity": 854.36912826297}, "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-2017-26/segments/1498128322320.8/warc/CC-MAIN-20170628032529-20170628052529-00404.warc.gz"} |
http://mathhelpforum.com/pre-calculus/27871-implied-domain-print.html | # implied domain
• February 9th 2008, 09:10 PM
RoboStar
implied domain
Hey MHF: Ive just got a question on the implied domain.
f(x)= Square root(x^2 - 3)
easy question i just need an somebody to clarify how to solve it.
Cheers
Andrew
• February 9th 2008, 09:16 PM
Jhevon
Quote:
Originally Posted by RoboStar
Hey MHF: Ive just got a question on the implied domain.
f(x)= Square root(x^2 - 3)
easy question i just need an somebody to clarify how to solve it.
Cheers
Andrew
the domain of $\sqrt{x}$ is $x \ge 0$ for $x \in \mathbb{R}$. thus, the implied domain here is given by: $x^2 - 3 \ge 0$
• February 9th 2008, 09:33 PM
RoboStar
and another question
h(x) = square root(x - 4) + square root(11 - x)
• February 9th 2008, 09:40 PM
Jhevon
Quote:
Originally Posted by RoboStar
and another question
h(x) = square root(x - 4) + square root(11 - x)
you just have to make sure $x - 4 \ge 0$ and $11 - x \ge 0$ work at the same time. choose the values of x that satisfy both
• February 10th 2008, 01:27 AM
RoboStar
thanks for all your help just two more questions on implied domain
f (x) = x^2 - 1 / x + 1 and h (x) = square root(x-1/x+2)
• February 10th 2008, 07:40 AM
Jhevon
Quote:
Originally Posted by RoboStar
thanks for all your help just two more questions on implied domain
f (x) = x^2 - 1 / x + 1 and h (x) = square root(x-1/x+2)
okay, for rational functions, we have the condition that we cannot divide by zero. and you know the condition for square roots. so there are three conditions you want to fulfill here
for f(x): you want $x + 1 \ne 0$
for h(x): you want $\frac {x - 1}{x + 2} \ge 0$ (for the square root) and $x + 2 \ne 0$ (for the rational function being square rooted)
of course, for h(x), both conditions must be fulfilled at the same time. so find the x's that work for both simultaneously | {"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": 9, "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": 0.9006333947181702, "perplexity": 1663.7339880370992}, "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/1461863352151.52/warc/CC-MAIN-20160428170912-00114-ip-10-239-7-51.ec2.internal.warc.gz"} |
https://hal.inria.fr/hal-00821609 | Oriented trees in digraphs
2 COATI - Combinatorics, Optimization and Algorithms for Telecommunications
CRISAM - Inria Sophia Antipolis - Méditerranée , Laboratoire I3S - COMRED - COMmunications, Réseaux, systèmes Embarqués et Distribués
Abstract : Let $f(k)$ be the smallest integer such that every $f(k)$-chromatic digraph contains every oriented tree of order $k$. Burr proved $f(k)\leq (k-1)^2$ in general, and he conjectured $f(k)=2k-2$. Burr also proved that every $(8k-7)$-chromatic digraph contains every antidirected tree. We improve both of Burr's bounds. We show that $f(k)\leq k^2/2-k/2+1$ and that every antidirected tree of order $k$ is contained in every $(5k-9)$-chromatic digraph. We make a conjecture that explains why antidirected trees are easier to handle. It states that if $|E(D)| > (k-2) |V(D)|$, then the digraph $D$ contains every antidirected tree of order $k$. This is a common strengthening of both Burr's conjecture for antidirected trees and the celebrated Erd\H{o}s-Sós Conjecture. The analogue of our conjecture for general trees is false, no matter what function $f(k)$ is used in place of $k-2$. We prove our conjecture for antidirected trees of diameter 3 and present some other evidence for it. Along the way, we show that every acyclic $k$-chromatic digraph contains every oriented tree of order $k$ and suggest a number of approaches for making further progress on Burr's conjecture.
Document type :
Journal articles
Cited literature [25 references]
https://hal.inria.fr/hal-00821609
Contributor : Frederic Havet <>
Submitted on : Sunday, October 23, 2016 - 4:07:49 PM
Last modification on : Monday, October 12, 2020 - 10:30:40 AM
File
ortree-final.pdf
Files produced by the author(s)
Citation
Louigi Addario-Berry, Frédéric Havet, Claudia Linhares Sales, Bruce Reed, Stéphan Thomassé. Oriented trees in digraphs. Discrete Mathematics, Elsevier, 2013, 313 (8), pp.967-974. ⟨10.1016/j.disc.2013.01.011⟩. ⟨hal-00821609⟩
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http://soft.necromancers.ru/cat_5_460.html | # Graphics
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1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 | 16 | 17 | 18 | 19 | 20 | 21 | 22 | 23 | 24 | 25 | 26 | 27 | 28 | 29 | 30 | 31 | 32 | 33 | 34 | 35 | 36 | 37 | 38 | 39 | 40 | 41 | 42 | 43 | 44 | 45 | 46
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https://research.tudelft.nl/en/publications/estimation-of-mean-particle-volume-using-the-set-covariance-funct | # Estimation of mean particle volume using the set covariance function
Research output: Contribution to journalArticleScientific
## Abstract
Our aim is to estimate the volume-weighted mean of the volumes of three-dimensional ‘particles’ (compact, not-necessarily-convex subsets) from plane sections of the particle population. The standard stereological technique is to place test lines in the plane section, and measure cubed intercept lengths with the two-dimensional particle profiles. This paper discusses more efficient estimators obtained by integrating over all possible placements of the test line. We prove that these estimators have smaller variances than the line transect estimators, and indeed are related to them by the Rao-Blackwell process. In the improved estimators, the cubed intercept length is replaced by a moment of the distance between two points in the section profile. This can be computed as a moment of the set covariance function, which in turn is computable using fast Fourier transform.
We also derive an isoperimetric type inequality between the improved estimator and the area-weighted 3/2 th moment of the profile areas. Finally, we present two practical applications to particles of silicon carbide and to synaptic boutons in brain tissue. We estimate the variance of the technique and the gain in efficiency over line transect techniques; the efficiency improvement appears to be as much as one order of magnitude.
Original language English 27-46 Advances in Applied Probability 35 1 Published - 2003 Yes
## Fingerprint
Dive into the research topics of 'Estimation of mean particle volume using the set covariance function'. 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": 0.9592691659927368, "perplexity": 1041.2505332227224}, "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-2022-05/segments/1642320305341.76/warc/CC-MAIN-20220128013529-20220128043529-00502.warc.gz"} |
http://bioinf.dimi.uniud.it/publication/rankbased-simulation-on-acyclic-graphs/ | # Rank-based simulation on acyclic graphs
### Abstract
The simulation preorder is widely used both as a behavioral relation in concurrent systems, and as an abstraction tool to reduce the state space in model checking, were memory requirement is clearly a critical issue. Therefore, in this context a simulation algorithm should address both time and space efficiency. In this paper, we rely on the notion of rank to design an efficient simulation algorithm. It turns out that such algorithm outperforms-both in terms of time and in terms of space-the best simulation algorithms in the literature, on the class of acyclic graphs.
Publication
CEUR Workshop Proceedings
Date | {"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": 0.8610198497772217, "perplexity": 789.3528559768012}, "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-2018-17/segments/1524125946721.87/warc/CC-MAIN-20180424135408-20180424155408-00218.warc.gz"} |
http://www.ck12.org/physics/Capacitors/lesson/The-Capacitor/r10/ | <img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
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# Capacitors
## Components of a circuit that build up charge.
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The Capacitor
The circuit boards found in your computer, phone, calculator, and pretty much every other electronic device you own often look much like the one shown above. Many circuit boards have capacitors, including this one. Capacitors can be used to smooth out electrical impulses or to turn constant electric currents into a series of impulses.
### The Capacitor
#### Sharing Charge
We already know that insulators are materials that do not allow electrons to flow through them easily. When you place excess electrons on an insulator, the electrons remain where you put them and do not move around. Conversely, conductors are materials that allow electrons to flow thorugh them freely. Since electrons repel each other, excess electrons on a conductor move to positions as far apart as possible. The difference can be seen in the image below, where the electrons on the insulator remain clumped near each other, while the electrons on the conductor have spread out to cover the whole surface.
Consider a charged conductor and an uncharged conductor. When the charged conductor is touched to the uncharged conductor, as far as the electrons are concerned, it has become one large piece of conducting material. The electrons on the charged object will run onto the uncharged object until the density of the charge is evenly distributed over the entire surface of both objects. If the objects are the same size, the charge will be shared equally throughout. This method is occasionally used to divide a charge by half.
The earth is also a conductor. Touching a charged object to the earth is called grounding. When you touch a conductor to the earth, you allow the earth to share the charge. Since the earth is billions of times bigger than the object, the earth takes nearly all of the charge. The charged object that was grounded now has zero charge.
It is very easy to ground an object. All that is necessary is to touch a conducting wire to both the object and the earth. Electrical devices that run the risk of picking up a large static charge are grounded, meaning they are connected to the earth via such a conducting wire. Virtually all household appliances, especially washers and dryers, are grounded in this way to eliminate static charge. Similarly, large trucks, especially gasoline tankers, are grounded via a large chain hanging off the back to prevent sparks when fuel is being unloaded.
Spheres, whether hollow or filled, will always have the excess charge on the surface. In hollow spheres, the only place for an electron to exist is on the surface. Similarly, in a solid conducting sphere all the excess charge sits on the surface. This conclusion is a result of Gauss's Law, which tells us that the symmetry of the sphere and the fact that the electric field within the sphere is 0 forces the charge to the outside.
#### Capacitors Store Charge
Pieter Van Musschenbroek, a Dutch physician, invented a device in 1746 that could store electric charge. Though he named the device a Leyden jar, similar devices today are called capacitors. A typical capacitor consists of a pair of parallel plates of area A separated by a small distance d. The space between the two plates is most often filled with an insulator and frequently the plates are rolled into the form of a cylinder.
If voltage is applied to a capacitor, it quickly becomes charged. One of the parallel plates acquires a negative charge and the other an equal amount of positive charge. For a given capacitor, the amount of charge, Q, acquired by each plate is proportional to the potential difference, V
Q=CV
Where Q is the charge in coulombs, V is the voltage in volts, and C is the particular capacitor's constant of proportionality. C is also called the capacitance of the capacitor. The capacitance is the voltage the capacitor can reach before it discharges, allowing the voltage across the capacitor to drop to zero and the current to cross the capacitor. The SI unit for capacitance, according to the equation above, will be coulombs/volt, and this unit has been given the name farad, F. Most capacitors have capacitances in the range of one picofarad \begin{align*}(10^{-12}F)\end{align*} to one microfarad \begin{align*}(10^{-6} \ F)\end{align*}.
Example Problem: A sphere has a potential difference between it and the earth of 60.0 V when charged with \begin{align*}3.0 \times 10^{-6} \ C\end{align*}. What is the capacitance?
Solution: \begin{align*}C=\frac{Q}{V}=\frac{3.0 \times 10^{-6} \ coulombs}{60.0 \ volts}=5.0 \times 10^{-8} \ farads\end{align*}
#### Summary
• Since electrons repel each other, when excess electrons are placed on a conductor, they will move to positions as far away from each other as possible.
• When a charged conductor is touched to an uncharged conductor, the electrons will migrate until the density of the charge becomes evenly distributed over the entire surface.
• Touching a charged object to the earth is called grounding.
• A charged conducting sphere will always have all the excess charge on its surface.
• A typical capacitor consists of a pair of parallel plates, separated by a small distance.
• \begin{align*}Q = CV\end{align*}, where Q is the charge in coulombs, V is the voltage in volts, and C is the constant of proportionality, or capacitance.
#### Practice
Questions
The following video covers capacitors. Use this resource to answer the three questions that follow.
1. What do capacitors do?
2. What are the units of capacitance?
3. What is the formula for capacitance?
Questions
1. We have a large charged hollow sphere with a small hole in one side. The charge on the sphere is 1.00 C. We insert another conducting sphere through the hole on an insulating stick and touch the inside of the charged hollow sphere and bring the second sphere outside the large sphere. What will be the charge on the second sphere?
2. Why does a charged object lose its charge when it is touched to the ground?
3. If a charged rubber rod is placed on a wooden table, the rubber rod will keep its charge for a long time. Why doesn’t the rod lose its charge immediately?
4. Both a \begin{align*}3.3 \times 10^{-6} \ F\end{align*} and a \begin{align*}6.8 \times 10^{-6} \ F\end{align*} capacitor are connected across a 15.0 V potential difference. Which capacitor has a greater charge and what is that charge?
#### Review
Questions
1. The two plates of a capacitor hold \begin{align*}+2.5 \times 10^{-3} \ C\end{align*} and \begin{align*}-2.5 \times 10^{-3} \ C\end{align*} of charge when the potential difference is 950 V. What is the capacitance?
2. The potential difference between two parallel wires in air is 120. V. They hold equal and opposite charges of \begin{align*}9.5 \times 10^{-11} \ C\end{align*}. What is the capacitance of the two wires?
3. How much charge flows from a 12.0 V battery when it is connected to a 9.00 microfarad capacitor? | {"extraction_info": {"found_math": true, "script_math_tex": 10, "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": 0.9052441120147705, "perplexity": 617.9954756628935}, "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-2015-48/segments/1448398453576.62/warc/CC-MAIN-20151124205413-00012-ip-10-71-132-137.ec2.internal.warc.gz"} |
https://psu.pb.unizin.org/math110/chapter/6-5-evaluating-definite-integrals/ | # 6.5 Evaluating Definite Integrals
## Learning Objectives
• Apply the basic integration formulas.
• Explain the significance of the net change theorem.
• Use the net change theorem to solve applied problems.
• Apply the integrals of odd and even functions.
In this section, we use some basic integration formulas studied previously to solve some key applied problems. It is important to note that these formulas are presented in terms of indefinite integrals. Although definite and indefinite integrals are closely related, there are some key differences to keep in mind. A definite integral is either a number (when the limits of integration are constants) or a single function (when one or both of the limits of integration are variables). An indefinite integral represents a family of functions, all of which differ by a constant. As you become more familiar with integration, you will get a feel for when to use definite integrals and when to use indefinite integrals. You will naturally select the correct approach for a given problem without thinking too much about it. However, until these concepts are cemented in your mind, think carefully about whether you need a definite integral or an indefinite integral and make sure you are using the proper notation based on your choice.
# Basic Integration Formulas
Recall the rule on properties of definite integrals. Let’s look at a few examples of how to apply these rules.
## The Net Change Theorem
The net change theorem considers the integral of a rate of change. It says that when a quantity changes, the new value equals the initial value plus the integral of the rate of change of that quantity. The formula can be expressed in two ways. The second is more familiar; it is simply the definite integral.
### Net Change Theorem
The new value of a changing quantity equals the initial value plus the integral of the rate of change:
$$\begin{array}{}\\ \\ F(b)=F(a)+{\int }_{a}^{b}F\text{‘}(x)dx\hfill \\ \hfill \text{or}\hfill \\ {\int }_{a}^{b}F\text{‘}(x)dx=F(b)-F(a).\hfill \end{array}$$
Subtracting $F(a)$ from both sides of the first equation yields the second equation. Since they are equivalent formulas, which one we use depends on the application.
The significance of the net change theorem lies in the results. Net change can be applied to area, distance, and volume, to name only a few applications. Net change accounts for negative quantities automatically without having to write more than one integral. To illustrate, let’s apply the net change theorem to a velocity function in which the result is displacement.
We looked at a simple example of this in Area and Definite Integral. Suppose a car is moving due north (the positive direction) at 40 mph between 2 p.m. and 4 p.m., then the car moves south at 30 mph between 4 p.m. and 5 p.m. We can graph this motion as shown in this figure.
Long description: The lines y=40 and y=-30 are drawn over [2,4] and [4,5], respectively.The areas between the lines and the x axis are shaded.
Just as we did before, we can use definite integrals to calculate the net displacement as well as the total distance traveled. The net displacement is given by
$$\begin{array}{cc}{\int }_{2}^{5}v(t)dt\hfill & ={\int }_{2}^{4}40dt+{\int }_{4}^{5}-30dt\hfill \\ & =80-30\hfill \\ & =50.\hfill \end{array}$$
Thus, at 5 p.m. the car is 50 mi north of its starting position. The total distance traveled is given by
$$\begin{array}{}\\ \\ {\int }_{2}^{5}|v(t)|dt\hfill & ={\int }_{2}^{4}40dt+{\int }_{4}^{5}30dt\hfill \\ & =80+30\hfill \\ & =110.\hfill \end{array}$$
Therefore, between 2 p.m. and 5 p.m., the car traveled a total of 110 mi.
To summarize, net displacement may include both positive and negative values. In other words, the velocity function accounts for both forward distance and backward distance. To find net displacement, integrate the velocity function over the interval. Total distance traveled, on the other hand, is always positive. To find the total distance traveled by an object, regardless of direction, we need to integrate the absolute value of the velocity function.
## Applying the Net Change Theorem
The net change theorem can be applied to the flow and consumption of fluids, as shown in this next example.
# Integrating Even and Odd Functions
We saw in Functions that an even function is a function in which $f(\text{−}x)=f(x)$ for all $x$ in the domain—that is, the graph of the curve is unchanged when $x$ is replaced with −$x$. The graphs of even functions are symmetric about the $y$-axis. An odd function is one in which $f(\text{−}x)=\text{−}f(x)$ for all $x$ in the domain, and the graph of the function is symmetric about the origin.
Integrals of even functions, when the limits of integration are from −$a$ to $a$, involve two equal areas, because they are symmetric about the $y$-axis. Integrals of odd functions, when the limits of integration are similarly $\left[\text{−}a,a\right],$ evaluate to zero because the areas above and below the $x$-axis are equal.
### Rule: Integrals of Even and Odd Functions
For continuous even functions such that $f(\text{−}x)=f(x),$
$${\int }_{\text{−}a}^{a}f(x)dx=2{\int }_{0}^{a}f(x)dx.$$
For continuous odd functions such that $f(\text{−}x)=\text{−}f(x),$
$${\int }_{\text{−}a}^{a}f(x)dx=0.$$ | {"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": 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": 0.9708451628684998, "perplexity": 248.06521898669305}, "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-2021-31/segments/1627046150264.90/warc/CC-MAIN-20210724094631-20210724124631-00672.warc.gz"} |
https://www.physicsforums.com/threads/net-angular-momentum.749959/ | Net Angular Momentum
1. Apr 21, 2014
cooev769
Really don't like this book, he's crazy ambiguous, I've looked it up myself on the internet and it makes far more sense, but it's our textbook so I have to understand what he's trying to say.
We're coupling the angular momentum of an electron's orbital and spin based angular momentum. And he says verbatim:
"If a hydrogen atom is in the state, l,m,n, the net angular momentum of the electron (spin+orbital) is l+0.5, l-0.5"
So I have no idea if he's talking about in the Z component, the squared, or what, he also doesn't say that he's omitted the h units, I only know that those are the units from other textbooks, also it would seem he is talking about the quantum number j, not actually the total angular momentum, which he hasn't actually explained. So I have no idea what he is talking about.
If he is talking about the z component well Lz on the state would be mh wouldn't it, and then Sz would also be mh. So the total would be 2mh, he's clearly not talking about this.
So i figured he must be talking about the quantum number, which he hasn't explained at all I had to research to find:
j = l +/- s
But then he goes on to say if we add in the proton, we get, l+1, l or l-1. But if he is indeed talking about the quantum number j, then you only get:
j = l + 1, l - 1, 0
So am I right to assume he's talking about the quantum number total angular momentum, and not actually adding the angular momenta from orbit and spin and getting the number l+0.5 and l-0.5
2. Apr 21, 2014
cooev769
I mean it would be true that the angular momentum is l+0.5, on the top rung if the two are aligned so to speak, and l-0.5 if the two are antiparallel so to speak, but there are so many potentials in between depending on l, and he hasn't talked about this at all or explained what the hell he is talking about.
3. Apr 21, 2014
MisterX
$j = \ell + 1/2$ and $j = \ell - 1/2$ are the only two values for definite $j$. Addition of angular momentum should be covered in any decent QM textbook.
You can look at the section beginning on page 151 here
http://www-thphys.physics.ox.ac.uk/people/JamesBinney/qb.pdf
4. Apr 21, 2014
cooev769
Yes but is this a quantum number or is this the actual total angular momentum values, that is what is not made explicit in this textbook. | {"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": 0.8930196166038513, "perplexity": 496.24417813519904}, "config": {"markdown_headings": false, "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-2018-22/segments/1526794866772.91/warc/CC-MAIN-20180524190036-20180524210036-00483.warc.gz"} |
https://testbook.com/question-answer/a-sphere-of-radius-12-cm-is-melted-and-re-casted-i--5fbe4b3c8816e3f2ef8dc764 | # A sphere of radius 12 cm is melted and re-casted into a right circular cone of height 12 cm. The radius of the cone is.
1. 36 cm
2. 32 cm
3. 21 cm
4. 24 cm
Option 4 : 24 cm
Free
AAI ATC Junior Executive 25 March 2021 Official Paper (Shift 1)
2.5 K Users
120 Questions 120 Marks 120 Mins
## Detailed Solution
Given:
Radius of sphere = 12 cm
Height of cone = 12 cm
Formula:
Volume of cone = (1/3) × πr2h
Volume of sphere = (4/3) × πr3
Calculation:
Let radius of the cone be r cm
According to the question
(1/3) × π × r2 × 12 = (4/3) × π × 12 × 12 × 12
⇒ r2 = 12 × 12 × 4
⇒ r = 12 × 2
∴ r = 24 cm | {"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": 0.8155128359794617, "perplexity": 4790.02969086185}, "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/1674764499949.24/warc/CC-MAIN-20230201180036-20230201210036-00862.warc.gz"} |
https://deustotech.github.io/DyCon-Blog/tutorial/wp04/P0005 | # Sparse sources identification through adjoint localization algorithm
Author: - 14 June 2019
## Model Problem
We consider the numerical approximation of the inverse problem for the linear advection-diffusion equation,
with $d$ being the diffusivity of the material and $v$ the direction of the advection.
Given a final time $T>0$ and a target function $u^\ast$ the aim is to identify the initial condition $u_0$ such that the solution, at time $t=T$, reaches the target $u^*$ or gets as close as possible to it. We assume that the initial condition $u_0$ is characterized as a combined set of sparse sources. This means that $u_0$ is a linear combination of unitary deltas with certain and possibly different weights, i.e:
We formulate the inverse problem using optimal control techniques. In particular, we consider the minimization of the following functional:
## Space and time discretization
Letting $\textbf{u} : [0, T] \rightarrow \mathbb{R}^s$ where $s$ is the number of grid points on $\Omega$, we can write a general finite element (FE) discretization of the diffusion–advection equation in \eqref{modeleq} in a compact form as:
In order to get a time discretized version of the previous equation, we apply implicit Euler method with stepsize $\Delta t := T/N$ where $N$ is the total number of time steps. The numerical approximations to the solution are given by the vectors $\textbf{u}^n \approx \textbf{u}(t_n) \in \mathbb{R}^s$ with respect to the index $n=i\Delta t$ for $i=1,2,..,N$. Therefore, the fully discrete version of model equation is as follows,
## Adjoint Algorithm for sparse source identification
The algorithm to be presented in this work for the sparse source identification of the linear diffusion-advection equation based on the adjoint methodology consists of two steps. Firstly, we use the adjoint methodology to identify the locations of the sources. Secondly, a least squares fitting is applied to find the corresponding intensities of the sources.
We have considered here a two-dimensional example with several sources to be identified in a multi-model environment. This means that the left half ($\Omega_1 = [0,1] \times [0,1]$) and the right half ($\Omega_2 = [1,2] \times [0,1]$) of the domain are modelled with different equations. In particular, the heat equation is used on $\Omega_1$ and the diffusion–advection equation is used on $\Omega_2$.
The initialization parameters look as follows:
N=30; %% space discretization points in y-direction
dx=1/(N+1); %% mesh size
t0=0; %% initial time
tf=0.1; %% final time
n=5; %% time discretization points
dt=(tf-t0)/n; %% stepsize
TOL=1e-5; %% stopping tolerance
d1=0.05; %% diffusivity of the material on the left sudomain
d2=0.05; %% diffusivity of the material on the left sudomain
vx=0;
vy=-3;
tau=dx^4; %% regularization parameter
epsilon=0.1; %% stepsize of the gradient descent method
We now compute the FE discretization matrices $M$, $A$ and $V$ that are respectively the mass matrix, the stiffness matrix and the advection matrix. For the FE discretization we assume equidistant structured meshes. In particular we use triangular elements and the classical pyramidal test functions are employed.
[M,A,V] = computeFEmatrices(N,d1,d2,vx,vy); %% Compute FE discretization matrices
A reference initial condition is chosen and we compute using the FE discretization specified above and implicit Euler in time its corresponding final state at time $T$. This final state will be considered the initial data of the inverse problem to be solved and we name it the target function $u^*$ as mentioned previously.
U0_ref = initial_deltas(N); %% computes reference initial condition
[U_target,u_target] = compute_target(U0_ref,N,n,dt,M,A,V); %% Compute target distribution
We now call the algorithm that estimates the initial condition $u_0$ using as a initial data the target function $u^*$. As mentioned before, this algorithm consists of two steps.
Firstly, the classical adjoint methodology that minimizes the functional $J(u_0)$ subject to the diffusion-advection equation is used. The iterative optimization algorithm employed is the classical gradient descent method. However, although this iterative procedure finds quite accurately the locations of the sources, it does not recover the sparse character of the initial condition. This is not suprising because the recovered initial data comes from solving the adjoint problem which is basically a diffusive process that smoothes out its state. Consequently, a second procedure is needed to project the obtained non sparse initial condition into the set of admissible sparse solutions.
As the initial condition $u_0$ is assumed to be a linear combination between the locations and the intensities, once we have fixed the locations using the adjoint methodology we can solve a least squares problem to get the remaining intensities. We assemble a matrix $\textbf{L} \in \mathbb{R}^{s \times l}$ where at each column we have the forward solution for a single unitary delta placed at each of the locations already identified. We then solve the following linear system of equations for the vector of unknowns $\alpha = (\alpha_1, \alpha_2, …, \alpha_l)^T$:
to find the intensities vector $\alpha$.
Adjoint algorithm for sparse source identification (algorithm 4)
U0 = SparseIdentification(u_target,TOL,dt,n,N,M,A,V,epsilon,tau);
Finally, the final state at $T$ is computed using as a initial condition the estimated sparse sources identified with our algorithm.
Compute final state with the recovered initial condition
[UF,u_final] = compute_target(U0,N,n,dt,M,A,V);
We can see the evolution recovered
We now visualize the numerical results. Plots on the left side show the reference initial solution and the given target. Similarly, plots on the right side show the recovered initial condition and the distribution at the final time $T$ produced by the recovered initial sources.
One can observe the difference between the two models (the heat equation on $\Omega_1$ and the diffusion-advection on $\Omega_2$) in the two figures at the bottom where the initial sources on $\Omega_2$ move downwards at the same time as they dissipate while the initial sources on $\Omega_1$ only dissipate without displacement.
xplot = linspace(0,2,2*N+3); %% space grid w.r.t component x
yplot = linspace(0,1,N+2); %% space grid w.r.t component y
%%
figure('unit','norm','pos',[0.25 0.1 0.5 0.8])
subplot(3,2,1)
surf(xplot,yplot,U0_ref)
title('Reference initial state (front view)')
subplot(3,2,2)
surf(xplot,yplot,U0)
title('Recovered initial state (front view)')
subplot(3,2,3)
pcolor(xplot,yplot,U0_ref)
title('Reference initial state (above view)')
subplot(3,2,4)
pcolor(xplot,yplot,U0) | {"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": 5, "equation": 1, "x-ck12": 0, "texerror": 0, "math_score": 0.9887218475341797, "perplexity": 512.4812433291386}, "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/1590347435987.85/warc/CC-MAIN-20200603175139-20200603205139-00075.warc.gz"} |
http://mathhelpforum.com/statistics/129260-probability-question-print.html | # probability question
• Feb 17th 2010, 05:23 AM
littleprince327
probability question
A dating service finds that 35% of the couples that it matches eventually get married. In the next 150 matches that the service makes, find the probability that at least 20 couples get married.
this is the first time i come across this kind of question...i dun know thw way to handle it.....can anyone possibly help??
• Feb 17th 2010, 04:42 PM
Danneedshelp
Quote:
Originally Posted by littleprince327
A dating service finds that 35% of the couples that it matches eventually get married. In the next 150 matches that the service makes, find the probability that at least 20 couples get married.
this is the first time i come across this kind of question...i dun know thw way to handle it.....can anyone possibly help??
This sounds like a binomial problem.
Let $X$ denote the number of matched couples that eventually get married. Consequently, the random variable $X$ has a binomial distribution based on $n=150$ trials with success probability $p=0.35$. It follows that,
$P(X\geq\\20)=\sum_{x=20}^{150}\left( \begin{array}{ccc}\\
150\\
x\end{array} \right)\\(.35)^{x}(1-.35)^{150-x}$
.
• Feb 17th 2010, 10:03 PM
littleprince327
but how can i calculate the answers by calculator??
the formula is so long...
is there any other way which can solve the question simply?
• Feb 18th 2010, 02:20 PM
Danneedshelp
Quote:
Originally Posted by littleprince327
but how can i calculate the answers by calculator??
the formula is so long...
is there any other way which can solve the question simply?
Binomial Calculator
Binomial Distribution: Probability Calculator
there are all kinds of calculators out there
• Feb 18th 2010, 04:56 PM
matheagle
Your instructor most likely wants you to use the normal approximation to the binomial distribution.
That's a central limit theorem that was proved by Gauss. | {"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": 5, "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": 0.8419956564903259, "perplexity": 1156.7937498658655}, "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/1498128319992.22/warc/CC-MAIN-20170623031127-20170623051127-00329.warc.gz"} |
http://www.pega-analytics.co.uk/blog/doe-math/ | # DOE Math
Understanding the matrix formulation of least squares regression can help you understand the essence of DOE.
The equation of a straight line has the following well-known form:
$y = \beta_0 + \beta_1x$
Using this notation the equation can be generalised to a Pth order polynomial:
$y = \beta_0x^0 + \beta_1x^1 + \beta_2x^2 +\dots+ \beta_Px^P$
If there are N independent observations then
$y_1 = \beta_0x_1^0 + \beta_1x_1^1 + \beta_2x_1^2 +\dots+ \beta_Px_1^P$
$y_2= \beta_0x_2^0 + \beta_1x_2^1 + \beta_2x_2^2 +\dots+ \beta_Px_2^P$
$\vdots$
$y_N= \beta_0x_N^0 + \beta_1x_N^1 + \beta_2x_N^2 +\dots+ \beta_Px_N^P$
We have N simultaneous equations with (P+1) unknown parameters. The equations are most conveniently expressed using a matrix notation.
The N observations of y can be represented as a vector of order N, and the β-parameters can be represented by a vector of order P+1.
$\textbf{Y}=\begin{pmatrix} y_1 \\ y_2 \\ \vdots \\ y_N \end{pmatrix} \quad \textbf{B}=\begin{pmatrix} \beta_0 & \beta_1 & \beta_2 & \dots & \beta_P \end{pmatrix}$
Furthermore an X matrix of dimensions N x (P+1) can be constructed:
$\textbf{X}=\begin{pmatrix} x_1^0 & x_1^1 & x_1^2 & \dots & x_1^P \\ x_2^0 & x_2^1 & x_2^2 & \dots & x_2^P \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ x_N^0 & x_N^1 & x_N^2 & \dots & x_N^P \\ \end{pmatrix}$
In matrix notation the equation now becomes:
$\textbf{Y} = \textbf{X}\textbf{B}$
The unknowns in this equation are the β-parameters that form the elements of the B matrix.
The naïve way to solve the equation is the following:
$\textbf{B} = \textbf{X}^\textbf{-1}\textbf{Y}$
where X-1 denotes the inverse of the X matrix.
This is “naïve” because we can only take the inverse of a square matrix. If we have a matrix of dimensions n x m then the transpose (Xt) of the matrix has dimensions m x n, and multiplying these two matrices together results in a square matrix of dimensions n x n. This matrix can be inverted. Hence we take the following steps:
$\textbf{B} = \textbf{X}^\textbf{-1}\textbf{Y}$
$\textbf{X} ^\textbf{t}\textbf{B} = \textbf{X}^\textbf{t} \textbf{X}^\textbf{-1}\textbf{Y}$
$(\textbf{X}^\textbf{t}\textbf{X})^\textbf{-1}\textbf{X} ^\textbf{t}\textbf{B} = (\textbf{X}^\textbf{t}\textbf{X})^\textbf{-1}\textbf{X}^\textbf{t} \textbf{X}^\textbf{-1}\textbf{Y}$
$\textbf{B} = (\textbf{X}^\textbf{t}\textbf{X})^\textbf{-1}\textbf{X} ^\textbf{t}\textbf{Y}$
The above analysis was based on a single x-variable, however, the matrix formulation generalises to the case where multiple x-variables are included in the model.
In least squares regression, the x-variables are presumed to be controlled and all errors are assumed to be in the observations y. Consequently:
$\textbf{Var}(\textbf{B} )= (\textbf{X}^\textbf{t}\textbf{X})^\textbf{-1}\textbf{X} ^\textbf{t}\textbf{Var}(\textbf{Y})$
If our goal is to produce estimates of the β-parameters with minimum variance then we need to minimise the following matrix:
$(\textbf{X}^\textbf{t}\textbf{X})^\textbf{-1}\textbf{X} ^\textbf{t}$
Notice that this quantity is a function of two things:
• The type of model that we wish to fit (i.e. the number of polynomial terms)
• Our choice of x-values
Both of these are under our control and known before the experiment it performed!
This is the basis of design of experiments.
## 2 thoughts on “DOE Math”
1. Xavier says:
Wow eftersom detta är utmärkt arbete ! Grattis och hålla upp.
2. Matthieu says:
That’s neat.
Thanks a lot for that. | {"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": 16, "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": 0.9654909372329712, "perplexity": 1025.5961842308507}, "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-2021-10/segments/1614178355944.41/warc/CC-MAIN-20210226001221-20210226031221-00316.warc.gz"} |
https://web2.0calc.com/questions/algebra-help_69 | +0
# algebra help
0
95
1
x2+bx+c is a factor of x3-p. Show that p=b3 (Hint: find that c=b2 first)
Mar 29, 2021
#1
+420
0
Notice that $$x^3-p=x^3-(\sqrt[3]{p})^3=(x-\sqrt[3]{p})(x^2+x\sqrt[3]{p}+(\sqrt[3]{p})^2)$$, using the rule of difference of cubes. Since we know that $$x^2+x\sqrt[3]{p}+(\sqrt[3]{p})^2$$is a factor of $$x^3-p$$, we now know that $$\sqrt[3]{p}=b$$. Cube both sides of the equation to get $$\boxed{p=b^3}$$
I genuinely do not understand why the hint would be helpful to solve this problem in any way, but you can show that's true by noticing that $$c=(\sqrt[3]{p})^2=b^2$$
Mar 29, 2021 | {"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": 0.9126961827278137, "perplexity": 649.1950886439782}, "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-2021-31/segments/1627046155925.8/warc/CC-MAIN-20210805130514-20210805160514-00595.warc.gz"} |
https://blog.nomadtype.ninja/2019/05/09/PFPL-Chapter-20-System-FPC-of-Recursive-Types/ | Someone's Intermediate Representation
0%
Please refer this link from page 177 to 184.
# Intros on System FPC and Recursive Types
FPC is a language with products, sums, partial functions, and recursive types.
Recursive types are solutions to type equations $t\cong\tau$ where there is no restriction on $t$ occurrence in $\tau$. Equivalently, it is a fixed point up to isomorphism of associated unrestricted type operator $t.\tau$. When removing the restriction on type operator, we may see the solution satisfies $t\cong t\rightharpoonup t$, which describes a type is isomorphic to the type of partial function defined on itself.
Types are not sets: they classify computable functions not arbitrary functions. With types we may solve such type equations. The penalty is that we must admit non-termination. For one thing, type equations involving functions have solutions only if the functions involved are partial.
A benefit of working in the setting of partial functions is that type operations have unique solutions (up to isomorphism). But what about the inductive/coinductive type as solution to same type equation? This turns out that based on fixed dynamics, be it lazy or eager:
• Under a lazy dynamics, recursive types have a coinductive flavor, and inductive analogs are inaccessible.
• Under an eager dynamics, recursive types have an inductive flavor, but coinductive analogs are accessible as well.
# Solving Type Equations
The syntax table of recursive type is defined as follows
\begin{aligned} \text{Typ}&&\tau&&::=&&t&&t&&\text{self-reference}\newline &&&&&&\text{rec}(t.\tau)&&\text{rec }t\text{ is }\tau&&\text{recursive type}\newline \text{Exp}&&e&&::=&&\text{fold}\lbrace t.\tau\rbrace(e)&&\text{fold}(e)&&\text{fold}\newline &&&&&&\text{unfold}(e)&&\text{unfold}(e)&&\text{unfold} \end{aligned}
Recursive types have the same general form as inductive/coinductive types.
## Statics
$$\displaylines{\dfrac{\Delta,t\text{ type}\vdash\tau\text{ type}}{\Delta\vdash\text{rec}(t.\tau)\text{ type}}\\ \dfrac{\Gamma\vdash e:\lbrack\text{rec}(t.\tau)/t\rbrack\tau}{\Gamma\vdash\text{fold}\lbrace t.\tau\rbrace(e):\text{rec}(t.\tau)}\\ \dfrac{\Gamma\vdash e:\text{rec}(t.\tau)}{\Gamma\vdash\text{unfold}(e):\lbrack\text{rec}(t.\tau)/t\rbrack\tau}}$$
## Dynamics
$$\displaylines{\dfrac{\lbrack e\text{ val}\rbrack}{\text{fold}\lbrace t.\tau\rbrace(e)\text{ val}}\\ \Big\lbrack\dfrac{e\longmapsto e’}{\text{fold}\lbrace t.\tau\rbrace(e)\longmapsto\text{fold}\lbrace t.\tau\rbrace(e’)}\Big\rbrack\\ \dfrac{e\longmapsto e’}{\text{unfold}(e)\longmapsto\text{unfold}(e’)}\\ \dfrac{\text{fold}\lbrace t.\tau\rbrace(e)\text{ val}}{\text{unfold}(\text{fold}\lbrace t.\tau\rbrace(e))\longmapsto e}}$$
## Safety Theorem
1. If $e:\tau$ and $e\longmapsto e’$, then $e’:\tau$.
2. If $e:\tau$, then either $e\text{ val}$ or $\exists e’,e\longmapsto e’$.
# Inductive and Coinductive Types
Recursive types may be used to represent inductive types such as natural numbers and natural number list.
## Example First on Natural Number
Use an eager dynamics for FPC, the recursive type $\rho=\text{rec }t\text{ is }\lbrack z\hookrightarrow\text{unit},s\hookrightarrow t\rbrack$ satisfies the type equation $\rho\cong\lbrack z\hookrightarrow\text{unit},s\hookrightarrow\rho\rbrack$, and is isomorphic to the type of eager natural numbers.
Introduction and elimination forms defined on $\rho$ is defined with following equations
\begin{aligned} z&\triangleq\text{fold}(z\cdot\langle\rangle)\\ s(e)&\triangleq\text{fold}(s\cdot e)\\ \text{ifz }e\lbrace z\hookrightarrow e_0\shortmid s(x)\hookrightarrow e_1\rbrace&\triangleq\text{case unfold}(e)\space\lbrace z\cdot\underline{}\hookrightarrow e_0\shortmid s\cdot x\hookrightarrow e_1\rbrace \end{aligned}
On the other hand, with an lazy dynamics of natural numbers in PCF, the same recursive type $\rho’=\text{rec }t\text{ is }\lbrack z\hookrightarrow\text{unit},s\hookrightarrow t\rbrack$ satisfying $\rho\cong\lbrack z\hookrightarrow\text{unit},s\hookrightarrow\rho\rbrack$ is not the type of natural numbers.
Consider what chapter 15 mentioned, $\rho’$ contains infinity number $\omega$ which is not natural number.
## Another Example on Natlist
Using an eager dynamics for FPC, the natlist is defined by recursive type $\text{rec }t\text{ is }\lbrack z\hookrightarrow\text{unit},c\hookrightarrow\text{nat}\times t\rbrack$, satisfying the type equation $\text{natlist}\cong\lbrack n\hookrightarrow\text{unit},c\hookrightarrow\text{nat}\times\text{natlist}\rbrack$.
Introduction and elimination forms are given by equations
\begin{aligned} \text{nil}&\triangleq\text{fold}(n\cdot\langle\rangle)\\ \text{cons}(e_1;e_2)&\triangleq\text{fold}(c\cdot\langle e_1,e_2\rangle)\\ \text{case }e\lbrace\text{nil}\hookrightarrow e_0\shortmid\text{cons}(x;y)\hookrightarrow e_1\rbrace&\triangleq\text{case unfold}(e)\space\lbrace n\cdot\underline{}\hookrightarrow e_0\shortmid c\cdot\langle x,y\rangle\hookrightarrow e_1\rbrace \end{aligned}
Consider the same recursive type under a context of lazy dynamics for FPC. Then we can see a value of such recursive type has the form $\text{fold}(e)$, where $e$ is an unevaluated computation of the sum type.
## Redefine (Co)Inductive Type
Consider the recursive type $\tau\triangleq\text{rec }t\text{ is }\tau’$ and the associated inductive type and coinductive type $\mu(t.\tau’)$ and $\nu(t.\tau’)$. We redefine these types consistently with statics of inductive and coinductive types here
\begin{aligned} \text{fold}\lbrace t.\tau’\rbrace(e)&\triangleq\text{fold}(e)\\ \text{rec}\lbrace t.\tau’\rbrace(x.e’;e)&\triangleq(\text{fix }r\text{ is }\lambda(u: \tau)\space e_{rec})(e),\text{ where}\\ e_{rec}&\triangleq\lbrack\text{map}\lbrace t.\tau’\rbrace(x.r(x))(\text{unfold}(u))/x\rbrack e’\\ \text{unfold}\lbrace t.\tau’\rbrace(e)&\triangleq\text{unfold}(e)\\ \text{gen}\lbrace t.\tau’\rbrace(x.e’;e)&\triangleq(\text{fix }g\text{ is }\lambda(u:\rho)\space e_{gen})(e),\text{ where}\\ e_{gen}&\triangleq\text{fold}(\text{map}\lbrace t.\tau’\rbrace(x.g(x))(\lbrack u/x\rbrack e’)) \end{aligned}
Dynamics is ill-behaved. Under eager interpretation, the generator may not converge, depending on choice of $e’$; under lazy interpretation, the recursor may not converge, depending on choice of $e’$.
The outcome shows that in the eager case the recursive type is inductive, not coinductive; whereas in the lazy case the recursive type is coinductive, not inductive.
## Coinductive Signal Transducer
We can define type Sig for signal to be the coinductive type of infinite streams of booleans, then we can also define the type of a signal transducer as $\text{Sig}\rightharpoonup\text{Sig}$.
Sig as a coinductive type can have form $\nu(t.e’)$ follows previous redefined coinductive type. In each of its unfold and generate, it gets a new boolean value out. Thus it can have definition $\nu(t.\text{bool}\times t)$.
The definition applies in lazy dynamics, whereas under an eager definition one may use $\nu(t.\text{unit}\rightharpoonup\text{bool}\times t)$. Maybe we just need to feed it with something we do not care in case it won’t stop.
The NOR here can have the type of $\text{Sig}\times\text{Sig}\rightharpoonup\text{Sig}$, with definition as
$$\lambda(a,b)\text{gen}\lbrace t.\text{bool}\times t\rbrace(\langle a,b\rangle.\langle e_{nor}\langle\text{hd}(a),\text{hd}(b)\rangle,\langle\text{tl}(a),\text{tl}(b)\rangle\rangle;a,b)$$
# Self-Reference
In general recursive expression $\text{fix}\lbrace\tau\rbrace(x.e)$, the variable $x$ stands for the expression itself. Self-reference is effected by the unrolling transition and it substitutes itself for $x$ in its body during execution.
$$\text{fix}\lbrace\tau\rbrace(x.e)\longmapsto\lbrack\text{fix}\lbrace\tau\rbrace(x.e)/x\rbrack e$$
It is useful to think $x$ as an implicit argument to $e$ that it is instantiated to itself when expression is used.
Type of self-referential expressions given by the syntax table
\begin{aligned} \text{Typ}&&\tau&&::=&&\text{self}(\tau)&&\tau\text{ self}&&\text{self-referential type}\\ \text{Exp}&&e&&::=&&\text{self}\lbrace\tau\rbrace(x.e)&&\text{self }x\text{ is }e&&\text{self-referential expression}\\ &&&&&&\text{unroll}(e)&&\text{unroll}(e)&&\text{unroll self-reference} \end{aligned}
## Statics and Dynamics
The statics of these constructs is defined as
$$\displaylines{\dfrac{\Gamma,x:\text{self}(\tau)\vdash e:\tau}{\Gamma\vdash\text{self}\lbrace\tau\rbrace(x.e):\text{self}(\tau)}\\ \dfrac{\Gamma\vdash e:\text{self}(\tau)}{\Gamma\vdash\text{unroll}(e):\tau}}$$
The dynamics is given by
$$\displaylines{\dfrac{}{\text{self}\lbrace\tau\rbrace(x.e)\text{ val}}\\ \dfrac{e\longmapsto e’}{\text{unroll}(e)\longmapsto\text{unroll}(e’)}\\ \dfrac{}{\text{unroll}(\text{self}\lbrace\tau\rbrace(x.e))\longmapsto\lbrack\text{self}\lbrace\tau\rbrace(x.e)/x\rbrack e}}$$
The main difference, compared to general recursion, is that we distinguish a type of self-referential expressions, instead of having self-reference at every type. It’s all a matter of taste.
## Recursive Type Defined Self-Reference
We can define the $\text{self}(\tau)$ with recursive type since a self-referential expression of type $\tau$ depends on the expression itself. It satisfies the isomorphism $\text{self}(\tau)\cong\text{self}(\tau)\rightharpoonup \tau$.
The type equation can be solved by type operator $t.t\rightharpoonup\tau$, where $t\notin\tau$ is a type variable. Required fixed point is just the recursive type $\text{self}(\tau)\triangleq\text{rec}(t.t\rightharpoonup\tau)$. Redefinition on type is shown as follows
\begin{aligned} \text{self}(\tau)&\triangleq\text{rec}(t.t\rightharpoonup\tau)\\ \lambda(x)\space e&:\space\space\text{self}(\tau)\rightharpoonup\tau\equiv\lbrack\text{rec}(t.t\rightharpoonup\tau)/t\rbrack(t\rightharpoonup\tau)\\ \text{self}\lbrace\tau\rbrace(x.e)&:\space\space\text{rec}(t.t\rightharpoonup\tau)\equiv\text{self}(\tau)\\ \text{self}\lbrace\tau\rbrace(x.e)&\triangleq\text{fold}(\lambda(x:\text{self}(\tau))\space e)\\ \text{unroll}(e:\text{self}(\tau))&:\space\space\tau\\ \text{unroll}(e:\text{rec}(t.t\rightharpoonup\tau))&\triangleq(\text{unfold}(e:\text{rec}(t.t\rightharpoonup\tau)):\lbrack\text{rec}(t.t\rightharpoonup\tau)/t\rbrack(t\rightharpoonup\tau))(e)\\ &\equiv\text{unfold}(e)(e) \end{aligned}
It is easy to check that final line of redefinition is correct by $\text{unroll}(\text{self}\lbrace\tau\rbrace(y.e))\longmapsto^\ast\lbrack\text{self}\lbrace\tau\rbrace(y.e)/y\rbrack e$.
## Self-Reference Defined Fixed Point
We may define $\text{fix}\lbrace\tau\rbrace(x.e)$ to stand for the expression $\text{unroll}(\text{self}\lbrace\tau\rbrace(y.\lbrack\text{unroll}(y)/x\rbrack e))$.
\begin{aligned} \text{unroll}(\text{self}\lbrace\tau\rbrace(y.\lbrack\text{unroll}(y)/x\rbrack e))&\equiv\lbrack\text{self}\lbrace\tau\rbrace(y.\lbrack\text{unroll}(y)/x\rbrack e)/y\rbrack(\lbrack\text{unroll}(y)/x\rbrack e)\\ &\equiv\lbrack\text{unroll}(\text{self}\lbrace\tau\rbrace(y.\lbrack\text{unroll}(y)/x\rbrack e))/x\rbrack e \end{aligned}
Also in $\text{fix}\lbrace\tau\rbrace(x.e)$ form, we can see that
\begin{aligned} \text{fix}\lbrace\tau\rbrace(x.e)&=\text{unroll}(\text{self}\lbrace\tau\rbrace(y.\lbrack\text{unroll}(y)/x\rbrack e))\\ &\longmapsto^\ast\lbrack\text{unroll}(\text{self}\lbrace\tau\rbrace(y.\lbrack\text{unroll}(y)/x\rbrack e))/x\rbrack e\\ &=\lbrack\text{fix}\lbrace\tau\rbrace(x.e)/x\rbrack e \end{aligned}
In this way we define factorial as $\text{fact}\triangleq\text{self }f:\text{nat}\to\text{nat}\text{ is }\lambda(n:\text{nat})\text{ ifz }n\space\lbrace z\hookrightarrow s(z)\shortmid s(n’)\hookrightarrow n\times \text{unroll}(f)(n’)\rbrace$.
# The Origin of State
The concept of state in computation has its origins in the concept of recursion, or self-reference.
RS latch as an example maintains its output at logic level of zero or one in response to a signal on the R or S inputs. We can implement an RS latch using recursive types.
The idea is to use self-reference to model the passage of time, with the current output being computed from its input and its previous output. It is a value of type $\tau_{rsl}$ given by
$$\text{res }t\text{ is }\langle\text{X}\hookrightarrow\text{bool},\text{Q}\hookrightarrow\text{bool},\text{N}\hookrightarrow t\rangle$$
The X and Q components of the latch represents its current outputs. (Q represents the current of latch), and N represents the next state of the latch.
If $e:\tau_{rsl}$, then we define e @ X to mean $\text{unfold}(e)\cdot \text{X}$, and e @ Q and e @ N are defined similarly.
• e @ X and e @ Q evaluate to boolean output of latch $e$
• e @ N evaluates to another latch representing its evolution over time based on its inputs
For given value $r$ and $s$, a new latch is computed from an old latch by the recursive function $rsl$ with definition
$$\text{fix }rsl\text{ is }\lambda(l:\tau_{rsl})\space e_{rsl}$$
where $e_{rsl}$ is the expression
$$\text{fix }this\text{ is fold}(\langle\text{X}\hookrightarrow e_{NOR}(\langle s, l@\text{Q}),\text{Q}\hookrightarrow e_{NOR}(\langle r,l@\text{X}\rangle)),\text{N}\hookrightarrow rsl(this) \rangle)$$
## Coinductive RS Latch
According to the previous settings of lazy dynamics with PCF, we can try to interpret RS latch expression with coinductive type, which still follows the previous $\tau_{rsl}$ type.
Consider an initial state type $\rho\triangleq\langle\text{X}\hookrightarrow\text{bool},\text{Q}\hookrightarrow\text{bool}\rangle$ and we initialize them with $\langle\text{false},\text{false}\rangle$. Then we can get the RS latch expression
\begin{aligned} e_{rsl}&\triangleq\text{gen}\lbrace t.\tau_{rsl}’\rbrace(\langle\text{X}\hookrightarrow x,\text{Q}\hookrightarrow q\rangle.e_{rsl}’;\langle\text{X}\hookrightarrow\text{false},\text{Q}\hookrightarrow\text{false}\rangle),\text{ where}\\ e_{rsl}’&\triangleq\langle\text{X}\hookrightarrow x,\text{Q}\hookrightarrow q,\text{N}\hookrightarrow\langle e_{nor}\langle s,q\rangle,e_{nor}\langle x,r\rangle\rangle\rangle \end{aligned}
Then we can simplify previously stated self-reference expression of RS latch like
$$\displaylines{\text{fix }g\text{ is }\lambda(\langle\text{X}\hookrightarrow x,\text{Q}\hookrightarrow q)\text{ fold}(e_{rsl}’’), \text{ where}\\ e_{rsl}’’\triangleq\langle\text{X}\hookrightarrow x,\text{Q}\hookrightarrow q,\text{N}\hookrightarrow g(e_{nor}\langle s,q\rangle,e_{nor}\langle x,r\rangle)}$$
# SKI Combinator as Recursive Type
Consider the type $\text{D}\triangleq\text{rec }t\text{ is }t\rightharpoonup t$, then we can try to define element $\text{k}:\text{D}$ and $\text{s}:\text{D}$ and application that satisfying:
• $x:\text{D},y:\text{D}\vdash x\cdot y:\text{D}$
• $\text{k}\cdot x\cdot y\longmapsto^\ast x$
• $\text{s}\cdot x\cdot y\cdot z\longmapsto^\ast (x\cdot z)\cdot(y\cdot z)$
We can easily find definition of $\text{k}\triangleq\text{fold}(\lambda(x:\text{D})\space\text{fold}(\lambda(y:\text{D})\space x))$ follows the application convention $x\cdot y\triangleq(\text{unfold}(x))(y)$.
Then $s\triangleq\text{fold}(\lambda(x:\text{D})\space\text{fold}(\lambda(y:\text{D})\space\text{fold}(\lambda(z:\text{D})\space(x\cdot z)\cdot(y\cdot z))))$. | {"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": 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": 0.9105095267295837, "perplexity": 2554.292887384181}, "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-2022-05/segments/1642320300849.28/warc/CC-MAIN-20220118122602-20220118152602-00472.warc.gz"} |
http://blog.geomblog.org/2011/06/bob-morris-and-stream-algorithms.html?showComment=1309622996176 | ## Thursday, June 30, 2011
### Bob Morris and stream algorithms
Bob Morris, one of the early contributors to UNIX and an NSA cryptographer, died on Sunday. The New York Times has a remembrance that talks about his contribution to password security and cryptography in general.
What's not mentioned in the NYT profile is that he's the Morris of the 'Morris counter'. This was arguably the very first streaming algorithm, nearly 20 years before the AMS paper and the HRR tech report first introduced streaming algorithms, five years before the Flajolet-Martin paper on counting distinct elements in a stream (also not framed as as streaming problem), and two years before the Munro-Paterson algorithm for streaming median finding.
The Morris paper is titled "Counting large numbers of events in small registers":
It is possible to use a small counter to keep approximate counts of large numbers. The resulting expected error can be rather precisely controlled. An example is given in which 8-bit counters (bytes) are used to keep track of as many as 130,000 events with a relative error which is substantially independent of the number n of events. This relative error can be expected to be 24 percent or less 95 percent of the time (i.e. σ = n/8). The techniques could be used to advantage in multichannel counting hardware or software used for the monitoring of experiments or processes.
In modern language, this can be re-rendered as:
It is possible to approximately count the number of elements in a stream using $O(\log \log n)$ bits, where $n$ is the length of the stream.
The idea is very simple. The trick is to count $C = \log n$ instead of $n$. Clearly, doing so requires only $\log\log n$ bits. Then any additive approximation to $C$ yields a multiplicative approximation to $n$.
So how do we count $\log n$ ? Let's assume for now that we're using base 2 logarithms. If the current counter value is $i$, then we "know" that the next update should come only $i$ counts later, if indeed we've seen $i$ elements. However, we don't know that the counter value reflects the true count, so we instead do it probabilistically. Specifically, toss a coin whose probability of coming up heads is $2^{-C}$, and only increment the counter if the coin comes up heads.
Phillipe Flajolet did a masterful analysis of this approach, and showed that the expected value of $2^C$ was $n+2$, yielding an unbiased estimator for the true count. He also showed that the variance of $2^C$ is $n(n+1)/2$.
This approach yields a fixed error bound for the resulting count. The next trick is then to change the base from 2 to some parameter $a$, and then reanalyze the method. Much work later, it can be shown that with roughly $\log \log n + \log(1/\epsilon)$ bits, you can get a multiplicative approximation of $1+\epsilon$.
Postscript
Stream algorithms are a good icebreaker in an algorithms class: they're unusual, seem to solve an impossible problem, and do it really well. I also like to point out to students that three of the more important streaming algorithms were designed well before people cared about "big data" or streaming. It's a valuable lesson in the importance of understanding "old" tools.
In case you're wondering if anyone still uses Morris counters, here's a paper from IJCAI 2009 that plugs them into a larger system to maintain counts of $n$-grams when building a statistical language model. | {"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": 0.9299812912940979, "perplexity": 616.3796415616415}, "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/1440644064263.27/warc/CC-MAIN-20150827025424-00264-ip-10-171-96-226.ec2.internal.warc.gz"} |
https://brilliant.org/problems/factors-2-2/ | # Factors 2
How many factors of $$40!$$ are perfect cubes?
Notation:
$$!$$ denotes the factorial notation. For example, $$8! = 1\times2\times3\times\cdots\times8$$.
× | {"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": 0.9510241150856018, "perplexity": 2920.8789976359135}, "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/1521257646178.24/warc/CC-MAIN-20180318224057-20180319004057-00394.warc.gz"} |
https://www.cut-the-knot.org/m/Algebra/TranHoangNamCyclicInequalityInThreeVariables.shtml | # Tran Hoang Nam's Cyclic Inequality in Three Variables
### Proof 1
Substitute $a-b=x,\,$ $b-c=y,\,$ $c-a=z.\,$ Then $x+y+z=0\,$ and we need to show that
$(x^2+y^2+z^2)^3+8(x^3y^3+y^3z^3+z^3x^3)\ge 0.$
Since $x+y+z=0,\,$ $x^2+y^2+z^2=-2(xy+yz+zx)\,$ and $x^3y^3+y^3z^3+z^3x^3=(xy+yz+zx)^3-3x^2y^2z^2.$
Hence, the inequality is equivalent to
$-8(x^3y^3+y^3z^3+z^3x^3)+8(x^3y^3+y^3z^3+z^3x^3)+24x^2y^2z^2\ge 0.$
Obviously true. Equality is achieved for $(a-b)(b-c)(c-a)=0.$
### Proof 2
Define $\displaystyle f(a,b,c)=\frac{1}{8}\left(\sum_{cycl}(a-b)^2\right)^3-\sum_{cycl}(a-b)^3(a-c)^3.\,$ We need to show that $f(a,b,c)\ge 0.$
WLOG, $a\le b\le c,\,$ $f(a,b,c)=f(a-a,b-a,c-a)=f(0,x,y),\,$ $x,y\ge 0.$
\begin{align} f(0,x,y) &+ (x^2+y^2-xy)^3-x^3y^3-x^3(x-y)^3-y^3(y-x)^3\\ &=3x^2y^2(x-y)^2\\ &\ge 0. \end{align}
Which is true.
### Proof 3
WLOG, suppose $a\le b\le c\,$ and let $b-a=x\gt 0\,$ and $c-b=y\gt 0,\,$ implying $c-a=x+y\,$ and we remark that $\displaystyle\sum_{cycl}a^2-\sum_{cycl}ab=\frac{1}{2}\sum_{cycl}(a-b)^2.\,$ The $\displaystyle RHS=\frac{1}{8}(x^2+y^2+(x+y)^2)^3\,$ while the $LHS=x^3(x+y)^3-x^3y^3+y^3(x+y)^3.\,$ Now $RHS-LHS=24x^2y^2(x+y)^2\ge 0.\,$ Q.E.D.
Equality when $x=0\,$ or $y=0\,$ or $x=y=0,\,$ i.e., $a=b\,$ or $b=c\,$ or $a=b=c.$
### Acknowledgment
Leo Giugiuc has kindly posted at the CutTheknotMath facebook page this inequality from the mathematical inequalities facebook group. The inequality is due to Trán Hoàng Nam. Proof 1 is by Leo Giugiuc; Proof 2 is by Daniel Dan; Proof 3 is by Imad Zak. | {"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": 2, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9879108667373657, "perplexity": 2295.5264591896084}, "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/1531676589470.9/warc/CC-MAIN-20180716213101-20180716233101-00228.warc.gz"} |
http://cerco.cs.unibo.it/changeset/3368/Papers | # Changeset 3368 for Papers
Ignore:
Timestamp:
Jun 17, 2013, 10:34:38 PM (5 years ago)
Message:
..
File:
1 edited
### Legend:
Unmodified
r3367 The plan of this paper is the following. In Section~\ref{sec:labelling} we sketch the labelling method and the problems derived from the application to languages with function calls. In Section~\ref{sec:semantics} we introduce a generic description of an unstructured imperative language and the corresponding structured fragments (the novel semantics). In Section~\ref{sec:simulation} we describe the forward simulation proof. summarize the labelling method. In Section~\ref{extension} we scale the method to unstructured languages with function calls. In Section~\ref{sec:simulation} we give the necessary condition for the correctness of the compiler (forward simulation proof). Section~\ref{sec:formalization} sketches the formalization of the datastructures and proofs in Matita. Conclusions and future works are in Section~\ref{sec:conclusions} \section{Extending the labelling approach to function calls} \label{extension} % Let's now consider a simple program written in C that contains a function contains the jump. \end{enumerate} One might wonder why $f$ and $\ell(f)$, that aways appear in this order, are not One might wonder why $f$ and $\ell(f)$, that always appear in this order, are not collapsed into a single observable. This would simplify some aspects of the formalisation at the price of others. For example, we should add special \label{thm:static} for all measurable fragment $M = s_0 \to^{*} s_n$,\\ $$\Delta_t := \verb+clock+_{s_n} - \verb+clock+_{s_0} = \Sigma_{o \in |T|} k(o)$$ $$\Delta_t := \verb+clock+_{s_n} - \verb+clock+_{s_0} = \Sigma_{o \in |M|} k(o)$$ \end{theorem} \emph{weakly trace equivalent} if their weak traces are equal. A compiler (pass) that preserves the program semantics also preserves weak traces and propagates measurability iff for every measurable A compiler (pass) that preserves the program semantics also \emph{preserves weak traces and propagates measurability} iff for every measurable fragment $M_1 = s_1 \to^{*} s_1'$ of the source code, the corresponding execution fragment $M_2 = s_2 \to^{*} s_2'$ of the object code is measurable \begin{theorem}\label{thm:preservation} Given a compiler that preserves weak traces and propagates measurability, for all measurable execution fragment $T_1 = s_1 \to^{*} s_1'$ of the source code such that $T_2 = s_2 \to^{*} s_2'$ is the corresponding fragment of the for all measurable execution fragment $M_1 = s_1 \to^{*} s_1'$ of the source code such that $M_2 = s_2 \to^{*} s_2'$ is the corresponding fragment of the object code, $$\Delta_t := \verb+clock+_{s_2'} - \verb+clock+_{s_2} = \Sigma_{o \in |T_2|} k(o) = \Sigma_{o \in |T_1|} k(o)$$ $$\Delta_t := \verb+clock+_{s_2'} - \verb+clock+_{s_2} = \Sigma_{o \in |M_2|} k(o) = \Sigma_{o \in |M_1|} k(o)$$ \end{theorem} \label{sec:simulation} Because of \autoref{thm:preservation}, to certify a compiler for the labelling approach we need to both prove that it respects the functional semantics of the program, and that it preserves weak traces and propagates measurability. We achieve this by independently proving the three properties for each compiler approach we need to prove that it respects the functional semantics of the program, preserves weak traces and propagates measurability. We achieve this by proving the three properties independently for each compiler pass. The first property is standard and can be proved by means of a forward simulation argument (see for example~\cite{compcert3}) that runs like this. \end{comment} \section{The formalisation} As we already explained, for the sake of presentation we explained the formal \label{sec:formalization} As we already explained, for the sake of presentation we presented the formal content of our results departing from the actual Matita formalisation. In this section we explain the basis of the Matita development, heavily based on from the material presented in the previous sections are the following. \begin{itemize} \item Rather than having a generic notion of fragment and a predicate of structuredness and measurability, we use inductive definitions internalising the conditions. Among other things this turns the proof of \autoref{thm:preservation} \item Rather than having a generic notion of fragment and predicates of structuredness and measurability, we use inductive definitions and dependent types to provide data types that satisfy the conditions by construction. Among other things this turns the proof of \autoref{thm:preservation} into a matter of structural induction, when it would require additional lemmas \begin{itemize} \item \verb+S : Type[0]+, the type of states. \item \verb+as_execute : S $\to$ S $\to$ Prop+, a the binary predicate modelling \item \verb+as_execute : S $\to$ S $\to$ Prop+, a binary predicate modelling the execution. As in the previous sections, we write $s_1\exec s_2$ for $\verb+as_execute+~s_1~s_2$. \item \verb+as_classifier : S $\to$ classification+, a function tagging all We restored static predictability by introducing a new semantics for unstructured programs that single outs well structured executions. The latter are represented by structured traces, a generalisation of streams of observables that capture by structured execution fragments, a generalisation of streams of observables that capture several structural invariants of the execution, like well nesting of functions or the fact that every basic block must start with a code emission statement. | {"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": 0.9555318355560303, "perplexity": 1005.1209613302876}, "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/1539583511173.7/warc/CC-MAIN-20181017111301-20181017132801-00159.warc.gz"} |
https://physics.stackexchange.com/questions/497087/what-is-the-most-precise-physical-measurement-ever-performed/497099 | # What is the most precise physical measurement ever performed?
Obviously some things, such as the speed of light in a vacuum, are defined to be a precise value. The kilogram was recently defined to have a specific value by fixing Plank's constant to $$6.62607015\cdot 10^{−34}\frac{m^2 kg}{s}$$.
In particular, in the case of the latter, we held off on defining this value until the two competing approaches for measuring the kilogram agreed with each other within the error bounds of their respective measurements.
Which leads me to wonder, what is the most precisely measured (not defined) value that the scientific community has measured. I am thinking in terms of relative error (uncertainty / value). Before we defined it, Plank's constant was measured to a relative error of $$10^-9$$. Have we measured anything with a lower relative error?
• @Qmechanic Mind explaining why you felt this needed to be put on hold via a mod hammer, when there were clearly good answers coming forth, and everyone seems to agree upon what is meant? In fact, with 6 upvotes and 33 views, it seems people seem to feel this is a good question. – Cort Ammon Aug 16 at 5:54
• Shouldn't the title refer to precision rather than accuracy? – Gremlin Aug 16 at 13:02
• – Emilio Pisanty Aug 16 at 13:41
• @Gremlin Updated. You're right, I was lazy in my wording. "precision" is a better word choice – Cort Ammon Aug 16 at 15:35
The magnetic moment of the electron has been measured to a few parts in $$10^{13}$$. (Source) This provides an exquisite test of quantum electrodynamics, and calculating the relevant Feynman diagrams has been a Herculean effort over decades.
Note that the more precise tests cited in other answers are basically null results: no difference between gravitational and inertial mass; no difference in magnitude of charge between proton and electron; no mass of photon. So I believe the magnetic moment of the electron is the most precise measurement that is non-null and thus “interesting”.
• Yeah that’s the answer I was looking for but somehow only found $\alpha$. – ZeroTheHero Aug 16 at 5:35
• While I'm not going to change the question to invalidate answers, your note is a good one that I'd edit in if I had a chance. While those null results are indeed interesting, I have the intuitive sense that it's easier to measure them with arbitrarily high precision. It is interesting, however, how few orders of magnitude separate this result from some of the equivalency tests listed in other answers. – Cort Ammon Aug 16 at 16:18
• For full clarity, this isn't the most accurate or precise measurement known, and the fact that @CortAmmon has accepted this answer (when there are tighter uncertainties reported in other answers) is a good example of why this thread is problematic. If it comes to "non-null" results, though, this paper presents a measurement with considerably more precision than the one in this answer (but I'm not going to take the (pretty arrogant) tack that others here have taken in asserting that the first thing I found is "the" most precise experiment). – Emilio Pisanty Aug 21 at 10:18
Carrying the fame of being one of the most precisely verified propositions in physics, the ratio of the gravitational to inertial mass was verified to be unity within $$1$$ in $$10^{15}$$ by the MICROSCOPE satellite in $$2017$$. The earlier best precision was $$5\times10^{-14}$$, obtained by Baessler, et al. in $$1999$$.
References:
A few more candidates:
• The quantized Hall resistance is a great and surprising example, since it's an emergent property of rather complicated, "dirty" systems. As stated here (2013) one can measure the resistance to one part in $$3 \times 10^{10}$$. For this reason this effect is now used to define the Ohm.
• Equivalence principle tests using torsion balances reached precisions of about one part in $$10^{11}$$ in $$1964$$, see here. Another existing answer gives a more precise result from a more modern experiment. These can be thought of as either verifying the equality of gravitational and inertial mass, or placing bounds on the strength of long-range fifth forces.
• The electrical neutrality of bulk matter follows because the electron and proton have exactly opposite charges. Treating the charge of the electron as given, tests of neutrality effectively measure the charge of the proton, with one experiment achieving a sensitivity of one part in $$10^{21}$$ in $$1973$$. Arguments from cosmology can be used to set much larger bounds, though they require more assumptions.
• Can you suggest me some reference where I can look up how the Equivalence Principle tests can be interpreted as putting bounds on a long-range fifth force? – Dvij Mankad Aug 16 at 4:59
• Another example: Earth-Moon distance measured with a laser. Distance about 3e10 cm, measured with a few cm accuracy I think. Should investigate more to be sure. – thermomagnetic condensed boson Aug 16 at 8:47
• Electrical neutrality of bulk matter: That only proves the equality of the charges if the number of electrons and protons in bulk matter is the same - but if the charge on the electron was 1.1 times that on the proton, I would expect bulk matter to have about 10% more electrons and still be neutral. – Martin Bonner Aug 16 at 13:03
• Update: the ohm is now defined in terms of the elementary charge – gen-z ready to perish Aug 21 at 6:11
• @gen-z Not particularly - in actual metrological practice, it's rather more accurate to say that the coulomb is defined in terms of the ohm. – Emilio Pisanty Aug 21 at 10:20
A good candidate is the measurement of the fine structure constant $$\alpha$$. This wiki article on precision tests of QED states that:
The agreement found this way is to within ten parts in a billion ($$10^{−8}$$), based on the comparison of the electron anomalous magnetic dipole moment and the Rydberg constant from atom recoil measurements...
There is also an upper bound on the mass of the photon, which is in the range of $$10^{-27}eV/c^2$$ although that's an upper bound rather than a measurement since the expected value is $$0$$.
• The ratio of the gravitational to inertial mass was verified to be unity within $1$ in $10^{15}$ in $2017$ by the MICROSCOPE satellite: en.wikipedia.org/wiki/MICROSCOPE_(satellite) – Dvij Mankad Aug 16 at 4:04
• @DvijMankad this ought to be an answer then... – ZeroTheHero Aug 16 at 4:07
• I think it could be added as a possible answer but I am not sure if there exist more precise measurements in particle physics, maybe some electroweak precision measurements? – Dvij Mankad Aug 16 at 4:08
• @DvijMankad I do believe the only way a question like this can be answered on StackExchange is to put forth the best answer an individual may know, and the best answer will get upvoted. (And I highly doubt a good answer will be downvoted just because somebody finds a better answer) – Cort Ammon Aug 16 at 4:10
• @CortAmmon Makes sense. Done! :-) – Dvij Mankad Aug 16 at 4:13
To answer the question from a different angle, LIGO has measured gravitational waves several times over the last few years. To do so they have to observe a distortion in a 4km arm, in the order of $$10^{-18}$$m. In other words, it has to detect a fractional imprecision in its length of ~$$2.5 * 10^{-23}$$. That's pretty accurate.
• LIGO is extremely sensitive, but pretty inaccurate, its calibration has an uncertainty of a few percent. – Bas Swinckels Aug 16 at 13:05
• Thank you for the answer! What Bas Swinckles mentions is something I was trying to avoid when crafting the criteria of the question. The 4km long arm is not what LIGO actually measures, but rather the 10^-18m number is what is being measured. LIGO was one heck of a feat, but I was trying to avoid high-sensitivity/low-accuracy measurements because we can always come up with a "higher" value by comparing it against something bigger. – Cort Ammon Aug 16 at 15:22
Another quantity that is well known is the frequency of the 21 cm line of Hydrogen hyperfine splitting that has been measured to about 13 significant figures, 1420405751.7667±0.0009 Hz. Unlike the electron magnetic moment, mentioned in another answer, the QED calculation for hyperfine splitting is not as accurate. This calculation also depends on other fundamental constants, including electron/proton mass ratio and proton magnetic moment. which are not known to equivalent precision. | {"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": 19, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8704587817192078, "perplexity": 631.2348664737303}, "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-2019-43/segments/1570986660829.5/warc/CC-MAIN-20191015231925-20191016015425-00339.warc.gz"} |
http://math.stackexchange.com/questions/168556/finding-the-height-given-the-angle-of-elevation-and-depression?answertab=votes | # Finding the height given the angle of elevation and depression.
Please, I need help for this problem. I'm a little confused about it :(
From a point A 10ft. above the water the angle of elevation of the top of a lighthouse is 46 degrees and the angle of depression of its image is 50 degrees. Find the height of the lighthouse and its horizontal distance from the observer.
I don't know where to start, because the problem doesn't have opposite, hypotenuse, or adjacent side written on it, and I think I cannot use TOA since there were no "Adj" or "Opp" side on the problem.
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It looks like you can set up two equations in two unknowns here. Let $x$ be the lighthouse's height and $y$ be the distance to the lighthouse. Then if the picture in my head is right, $x-10$ and $y$ are the opposite and adjacent sides to a $46^\circ$ angle, and $x+10$ and $y$ are the opp. and adj. sides to a $50^\circ$ angle. – Eugene Shvarts Jul 9 '12 at 8:36
@jhong: You should not repost your question if you have something to add - there is an "edit" button on the bottom left of the question, underneath the "homework" and "trigonometry" tags. I have taken all of the changes you made in your reposted question and put them here, and closed your reposted question as a duplicate. – Zev Chonoles Jul 9 '12 at 10:31
BF is the ground level.
Write EG in terms of angle $50^\circ$.Find $x$. Then the height will be $x\tan 46^\circ + 10$
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tnx ill re write the equation again.. – jhong Jul 9 '12 at 10:04
Let $h$ be the height, $d$ the horizontal distance. Then you arrive at two equations (why?): $$d\tan 46^o=h-10$$$$d\tan 50^o=h+10$$. | {"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": 0.908302903175354, "perplexity": 332.89582584765895}, "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-2014-15/segments/1398223205375.6/warc/CC-MAIN-20140423032005-00142-ip-10-147-4-33.ec2.internal.warc.gz"} |
http://math.stackexchange.com/questions/404020/h-is-a-subgroup-of-g-and-every-coset-of-h-in-g-is-a-subgroup-of-g-then | $H$ is a subgroup of $G$ and every coset of $H$ in $G$ is a subgroup of $G$.Then which of the following is true?
$H$ is a subgroup of $G$ and every coset of $H$ in $G$ is a subgroup of $G$.Then which of the following is true?
(A) $H=${$e$}
(B) $H=G$
(C) $G$ must have prime order.
(D) $H$ must have prime order.
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Can you rule out any of them? Can you see that any of them must hold? – Tobias Kildetoft May 27 '13 at 16:55
Hint:-
(B) is true. every subgroup of a group must contain the identity and any two distinct cosets are disjoint.
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In spite of being already given the final answer, think of the following: if we have a coset $\,gH\neq H\,$ , then it can not be that $\,1\in gH\,$ (why?) , thus...
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Dear Don, does $Q_8$ violet the problem? Thanks – B. S. May 27 '13 at 17:10
@BabakS. how could it? – Tobias Kildetoft May 27 '13 at 17:20
I don't understand, @BabakS.: how and what does $\,Q_8\,$ violate ? – DonAntonio May 27 '13 at 17:59
@DonAntonio: I felt that if I considered a right coset of, for example $H=\{\pm 1,\pm i\}$, then I would have another subgroup of $G=Q_8$ – B. S. May 27 '13 at 18:28
@DonAntonio: This means to me that I can consider $Q_8$ and a subgroup of it, say $H$, but none of the options are correct. Sometimes, I feel, I can't see the facts, however, I have been with them for times. Thanks for the time and (+1). – B. S. May 27 '13 at 18:34 | {"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": 0.9622635841369629, "perplexity": 340.0824741948409}, "config": {"markdown_headings": false, "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-2014-23/segments/1405997901076.42/warc/CC-MAIN-20140722025821-00217-ip-10-33-131-23.ec2.internal.warc.gz"} |
https://www.physicsforums.com/threads/need-help-writing-math-from-english.766023/ | # Need help writing 'math' from English
1. ### mesa
583
I need to write, "(k)1/2= a finite term recursive expression of 'k' where 'k' is the set of all Reals where 'k' is greater than 0 and 'k' for all complex values" in math.
Thanks!
2. ### Simon Bridge
14,487
Basic setbuilding notation:
http://www.mathwords.com/s/set_builder_notation.htm
...where do you get stuck?
Personally I'm having trouble following the English description of what k is supposed to be.
By "k is the set of... " do you mean that k is a member of the set, or is the label assigned to the set?
Considering that the set of reals is a subset of complex numbers, are you saying that k is any complex number that is not a real number 0 or less?
Last edited: Aug 15, 2014
1 person likes this.
### Staff: Mentor
I too am confused by what you wrote as the description for k. The first part is pretty clear; namely, {k ##\in## R | k > 0}, but this part --
-- seems to be incomplete.
4. ### mesa
583
I didn't know that, and as such the answer to your last question is yes.
'k' is the variable in the radical for a general form function (if this wording is awkward please let me know), as in a general form where √k=f(k) that is exact for all Complex values (since the Reals are a subset of complex) for k>0 .
Considering these conditions, is this right?
$$\{k\in \mathbb{C}\mid k>0\}$$
Is there a global way to stating recursivity and finite termed expressions or do we just keep these statements in English?
Performing derivations has never been an issue but the technical vernacular has always been troublesome and often proves itself more difficult a task than the work itself, hence concentrating more effort on this area to gain competency (in the meantime, please pardon my lack of refinement).
Last edited: Aug 15, 2014
5. ### mesa
583
I can understand your confusion, I made the mistake of not realizing that the set of all Reals is a subset of all Complex values. This is likely an elementary case so it is very good to have corrected.
Considering this new insight, does my last post make more sense?
6. ### jbriggs444
The Complex numbers have no natural order. So it makes little sense to write "k>0" when k is complex.
I think that what you are trying to get at is that √k is well defined when k is both real and positive.
For complex k there are always two solutions for x2=k except when k=0.
For k positive and real, one solution is positive and one is negative. √k is taken to denote the positive one.
For k negative and real, the two solutions are complex conjugates whose real part is zero. One is equal to 0+√(-k) i. The other is equal to 0-√(-k)i. Neither one is preferred.
For k complex with a non-zero imaginary part the two solutions will be 180 degrees apart on the complex plane. Neither one is preferred.
Last edited: Aug 15, 2014
7. ### mesa
583
$$\{k\in \mathbb{C}\mid k>0-\infty i\}$$
or should I split it up since complex numbers have no natural order and therefor the inequality is inappropriate?
I was also thinking something along these lines as well,
$$\{k\in \mathbb{C}\mid k\} \cup \{k\in \mathbb{R}\mid k>0\}$$ but this isn't correct either since it includes the whole set of complex numbers which includes both positive and negative reals. Is there a symbol in the notation to replace $$\cup$$ with something that represents 'except where' and then we just change the inequality to 'k< or equal to 0'?
8. ### mesa
583
My apologies Simon, I forgot to answer this part of your post.
By "k is the set of..." I was trying to imply at what values ('the set') is the variable k from f(k) true for √k.
9. ### Simon Bridge
14,487
##\mathbb C \cup \{x\in\mathbb R:x>0\}## is just the set of complex numbers.
Remember the reals are a subset of complex numbers.
The union of a set and it's subset is itself.
Similarly, the intersection of a set and it's subset is the subset.
You want to exclude the subset.
k is a complex number that is not a negative real number.
(Is k allowed to be zero? Proceeding as if k cannot be zero either.)
i.e. The real part of k can be negative, but only if the imaginary part is not zero.
Something like:$$k\in\mathbb C : k\notin \{x\in\mathbb R: x\leq 0\}$$... ie. k is a member of the relative compliment of complex numbers and negative real numbers.$$k\in\mathbb C \backslash \{x\in\mathbb R: x\leq 0\}$$
How about extracting the definition of a complex number by introducing two arbitrary reals:$$k=x+iy: x,y\in \mathbb R \land \forall (x,y):y=0, x>0$$ ... there's probably a better way of doing that...
Are there numbers that are not in the set of complex numbers that you don't want k to be?
If not, then the representation is very simple.
Aside: If ##x,y\in\mathbb R## and ##i=\sqrt{-1}## then ##z=x+iy\in\mathbb C##.
Since y=0 is a real number, it follows that ##z=x\in\mathbb C##,
i.e. all real numbers are members of the set of complex numbers.
Notes: we have all been using "blackboard bold" for special sets.
Some people advise against it i.e.
Krantz, S., Handbook of Typography for the Mathematical Sciences, Chapman & Hall/CRC, Boca Raton, Florida, 2001, p. 35.
Last edited: Aug 15, 2014
### Staff: Mentor
I would write the set this way:
{z ##\in## C | Re(z) > 0}. This gives you the right half of the complex plane.
11. ### Simon Bridge
14,487
<puzzled> which set would you write that way?
In the set described in post #1 (and clarified later) the real part is allowed to be negative but only if the imaginary part is non-zero. If you wanted to represent that on the complex plane, you'd need a hole along the entire negative real axis. Everything else is allowed.
The set quoted is just the complex plane.
Neither of these are well represented by the right half of the complex plane.
However, the Re/Im notation is probably a better way to handle my k=x+iy version... $$\{k\in\mathbb C : \forall\; \Im(k)=0, \; \Re(k)>0\}$$ ... "forall" may not be right there.
Would this work: ##\{k\in\mathbb C : \text{Arg}(k)\neq\pi\}##
Aside: put a backslash in front of the "Re(z)" in LaTeX and you get ##\Re(z)## ;)
### Staff: Mentor
Maybe I misunderstood, especially since it has taken a number of posts to figure out what mesa is trying to get. What I got out of a quick read of the thread was that he was looking for complex numbers for which the real part was positive (or maybe nonnegative).
13. ### mesa
583
Okay, this makes sense. All that is needed is the $$\notin$$ notation to exclude negative Reals from the complex plane. Or for the second example the '\' in place of $$: k\notin$$
Very nice!
Once again, this makes sense as well, thank you for the clarification!
This part I am a little confused on, what numbers are left that are outside the set of complex numbers since it also contains all Reals when bi=0?
The history of 'blackboard bold' is interesting and it seems was originally intended for when 'hand writing' these things as 'bold' (hard to do otherwise) and was never intended for printed works in the first place (although one could argue it is aesthetically pleasing).
On another note, in order to improve with these types of things do you have another (preferably more frugal) recommendation for mathematical typography?
14. ### Simon Bridge
14,487
See http://en.wikipedia.org/wiki/Complement_(set_theory) for the use of compliments in set theory.
I think the most succinct way is: ##\{k\in\mathbb C : \text{Arg}(k)\neq \pi\}##
... although I think that includes k=0, since it's argument is undefined. Easy to fix.
What is the context of this?
I'm assuming it's not an arbitrary homework question because of how tricky the wording was to parse. It would have been quite tricky to guide you to the compliment or argument solution too.
There's quaternions etc.
You'll notice when you go from reals to complex numbers, you lose the property of being ordered?
Quaternions (a 4-dimensional normed R-algebra) lose commutativity, but are associative.
Octonions (an 8-dimensional normed R-algebra) lose associativity, but are alternative.
Sedenions (a 16-dimensional R-algebra) lose alternativity, and are also not a normed R-algebra, because there are zero-divisors.
More:
http://en.wikipedia.org/wiki/Cayley–Dickson_construction
http://en.wikipedia.org/wiki/Hurwitz's_theorem_(normed_division_algebras)
15. ### mesa
583
Ii is fast becoming apparent I require more study on this topic.
I enjoy finding identities and exploring math in general. This particular function I thought was a neat result but lacking the proper mathematical vernacular makes it difficult to share with people who really know these things.
I haven't studied these, but have heard of them. I will look through your links!
16. ### Simon Bridge
14,487
Neat result of what?
... quaternions have application in computer science for their ability to efficiently represent rotations.
Basically, define basis elements i,j,k (not the cartesian unit vectors) so that ##i^2=j^2=k^2=ijk=-1##
A quaternion is q=a+bi+cj+dk.
http://en.wikipedia.org/wiki/Quaternion
... you can similarly look up the others.
However, the reason I brought it up is because sometimes an overarching set can be left implicit ... i.e. ##\{a:a<0\}## would usually be interpreted as the set of negative reals, and ##\{a:\Re(a)>0\}## would imply complex numbers are intended.
So you could write ##\{k:\text{Arg}(k)\neq \pi \}## and be confident that people would realise you meant to define k over the complex plane.
It's a language - so there are lots of ways to express yourself. The exact approach you choose depends on what you want to draw the readers attention to.
17. ### mesa
583
I was looking at the golden ratio and trying to work a general form, after the task was completed (and some new insights gained) it appeared it may be possible to build a general form function for radicals (our f(k)=√k). The result I thought was 'neat'.
I will look into this, although will likely have some questions.
18. ### Simon Bridge
14,487
Oh right.
You were thinking that k can be any complex not a negative real?
Positive reals can have real square-roots fersure, but I don't see how excluding negative reals from k helps considering you are allowing any complex number otherwise.
Normally the radical indicates a positive square root so that if ##y=x^2## then ##x\in\{\pm\sqrt{y}\}## ... that can mess things up. The result is that the negative reals are excluded, by definition, from the solution set.
This what you are thinking of?
Don't know what you mean by "general form function" in this context.
Anyway - that would be for another thread ;)
19. ### mesa
583
Not quite (and my apologies for the confusion), this isn't an attempt to define where √k is true but to show where a specific identity is true.
In other words if you want to see √3 as a function of '3' and 'phi' then this function will give an exact result but it does not work for √-3≠f(-3,phi), although it will for √(-3)^(1/2)=f((-3)^(1/2),phi). Does that clear things up?
I can't think of an example off the the of my head but, as you say, a discussion for another thread. | {"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": 0.8855347633361816, "perplexity": 900.3231728716008}, "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/1424936470419.73/warc/CC-MAIN-20150226074110-00311-ip-10-28-5-156.ec2.internal.warc.gz"} |
http://blog.computationalcomplexity.org/2010/09/rubiks-cube-conjecture-proven-do-we.html | ## Wednesday, September 08, 2010
### The Rubik's Cube Conjecture PROVEN! (Do we care?)
In 1994 Fermat's last theorem was proven! In 2003 Poincare's conjecture was proven! In 2010 the Rubik Cube Conjecture was proven! That is, it was shown that the minimum number of moves needed to solve Rubik's Cube is 20 (it was known that there are starting configurations that require 20). Here is the pointer: Rubiks Cube has been solved. (Jeff Erickson also had a blog entry on Rubik's cube lately: here.)
(ADDED LATER. Clarification: From any starting position one can get to the solution in \le 20 moves. There exists a starting position where every solution takes \ge 20 moves. )
We all have a sense that the Rubik Cube result is not as important as the other two. What makes a math problem important?
1. Would Martians work on it? I suspect Martians would work on FLT and Poincare but not Rubik's cube. One of the first things I would ask Space Aliens is what Math they have done.
2. Does it connect to other parts of mathematics? This leads to a seemingly circular definition of importance: A topic in math is important if it connects to other topics that are interesting. FLT and Poincare's conjecture are connected to other parts of mathematics. One way to understand Rubik's cube is through group theory; however, this is not so much a connection as an application. Rubik's cube studies have not lead to advances in group theory.
3. Does the problem have its origins in some real world phenomena? (This is not necessary but helps.)
4. The problem should be hard but not so hard that you cannot make any progress on it.
5. The proof must be interesting (hmmm- then we need to define interesting).
These criteria may be too strict. It would be interesting to discuss other branches of math that are considered important that do not quite fit these criteria and see what else makes them important. Or branches that do fit these criteria but are not considered important.
1. minimum -> maximum?
2. That is, it was shown that the minimum number of moves needed to solve Rubik's Cube is 20 (it was known that there are starting configurations that require 20).
You probably mean "maximum number of moves".
3. Objectively, FLT isn't that much more natural than Rubik's Cube. It's got the fact that it's a generalization of Pythagorean triples, but that's about it. That it's received so much attention seems like a sort of fluke of math history. It's created interesting math and even provided some applications, but only through a kind of sheer force-of-will. It wouldn't be that surprising if the Martians HADN'T studied it.
4. "Why are numbers beautiful? It's like asking why is Beethoven's Ninth Symphony beautiful. If you don't see why, someone can't tell you. I know numbers are beautiful. If they aren't beautiful, nothing is."
- Paul Erdős
5. @1 and @2
Minimum is correct.
For example, if I gave you 21 moves, you could still always solve the cube. However, if I only gave you 19 moves, you could not always solve the cube. Thus, "the minimum number of moves needed to solve Rubik's Cube is 20".
6. This result can really show the power of math to a foreign world. Not many people know of the theoretical side of math, but lots know how much of a headache solving a Rubik's cube can be. Even if the proof is hard for a novice to understand, such a result can inspire conversation about how to prove similar things for simpler puzzles. And if this conversation helps more people to see the beauty of mathematics, then this is indeed an important result.
7. I think there's also a certain finiteness principle here: if a question can in principle be reduced to a finite calculation (even though in practice we can't do that calculation) then it's less likely to be interesting than a problem where any solution would have to be more conceptual.
8. I agree with David. In this case the "proof" is some sort of (clever, but still quite-brute-force) calculation.
We can admire the nice exploitation of symmetries and the expertise in coding a lot of programming tricks in order to speed up the computation, but if the proof had been "on paper" probably it would have been more interesting.
9. As an utter layman, I'd suggest that "interesting" be defined as "holding the attention of provers long enough to prove it" when the amount of time it takes to prove something is not trivial. Rubik's cube is therefore surprisingly interesting, as this proof about essentially a toy took many person-years of effort to emerge, and yet the effort was made.
At least, that's why it interests me!
10. As a counterpoint to David: 4 color theorem by a finite calculation.
I suppose there is a question of how "obvious" it is that a finite calculation can solve the problem.
11. Carroll's Riddle:
Why is God's Chess Number like a writing desk?
Because both Turing and Shannon wrote upon it.
------------
Note: white has 61 "only" moves.
12. In a sense not even Wiles believed FLT to be a problem important on its own. Only when a way to prove FLT that had connections to deep maths became apparent did he devote his complete attention to it.
Many other mathematicians have commented on the utter irrelevance of the FLT question.
The Rubik cube has connections to some really deep group theory, and when this is eventually re-proven for the n-sided cube it will likely be as relevant as FLT was.
13. Here's a question: does the Rubik's Cube count as "real world phenomena"? Somehow because it's a toy it seems the answer should be no, or at least this justification is much less strong than if, for instance, the conjecture was related to curing cancer.
On the other hand, I think popular interest in this problem is actually more than for say the Poincare Conjecture.
14. This is epic result, much more important than 99.9% of the trash published in the so called TCS. That you are unable to understand that only highlights how overrated TCS people are.
15. @5
"needed" means "required", so the "minimum needed" is 0, not 20.
16. The proof must be interesting (hmmm- then we need to define interesting). | {"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": 0.8210501670837402, "perplexity": 795.7428551093516}, "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-2015-32/segments/1438042987127.36/warc/CC-MAIN-20150728002307-00153-ip-10-236-191-2.ec2.internal.warc.gz"} |
http://astrobites.com/2013/02/13/super-simple-supernovae-a-simple-model-for-type-ia-light-curves/ | # Super-simple supernovae: a simple model for Type Ia light curves
Big Picture (literally)
I have a violent story to tell you today! Star gets old, gets fat, and explodes. We see the explosion, know how fat the star was, and hence the energy that should be released, and can infer the distance to the star from the energy we actually measure. This in turn, repeated enough times, tells us about how quickly the Universe has been expanding in the past. Knowing the expansion history constrains the properties of Dark Energy (DE), an exotic substance that comprises 73% of the Universe (stuff we’re made of, by comparison, is a measly 4%!) That is, literally, the biggest-picture view of the importance of exploding stars. (See also this post on supernovae, and this post explaining how DE leads to accelerated expansion.)
Where does today’s paper fit in? The explosions I mention above are called Type Ia supernovae (SNIa), further detail to follow below, and we measure their light curves: the amount of light we receive from them over time. This amount as a function of time has a characteristic shape, shown below. Today’s paper develops a simple analytic model to predict this shape from basic physics. Certainly, SNIa have been known for a long time—at least since the ’80’s. So their shape is not news to anyone. The value of this paper is to provide a very simple physical model with only a few inputs and only a couple different mechanisms, and to get out of it a shape that matches observations pretty well, as the picture shows. So this is the paper for anyone who has ever seen a light curve and said, “Gee, why does it have that funny shape?”
This shows the typical shape of an SNIa light curve. The black dots are observations, the red curve the authors’ formula. The vertical axis is the base-10 logarithm of luminosity (that is just the energy the supernova emits per unit time) and the horizontal axis is time. From the paper.
Supernova primer
Type Ia supernovae are violent explosions that occur when the gravity in an old star (specifically, a type known as a “white dwarf“) overwhelms the pressure even densely packed, degenerate matter can provide and the star explodes.
Why is it important that the star is old? The very oldest stars are no longer doing nuclear fusion, so they do not have heat providing a source of pressure to support them. Thus, as mass is added to them, for instance from a neighboring star’s shedding, they simply contract and become more dense. Eventually, they become so dense that particles are actually close enough to each other that quantum mechanical effects become important.
Specifically, the “Pauli exclusion principle” says that two fermions cannot be in the same overall quantum state (fermions are just particles that have half-integer values of their intrinsic spin—imagine a sphere rotating with different speeds). Electrons are fermions, and the constraint that they cannot be in the same QM state means they must have some spatial separation between each other. This in turn means that, if they are compressed enough, they’ll push back, providing what is called “electron degeneracy pressure“. However, electron degeneracy pressure is not enough when the star reaches a mass of 1.4 times the mass of the Sun, and gravity wins (this mass upper bound is called the Chandrasekhar mass—he thought of it in 1931, way before any stars like this had been seen. But on the other hand, the Chinese observed a supernova in 1054 AD!) Everything falls inward, complicated physics ensues, and the star blows itself apart.
• A super-simple supernova synthesis
Try saying those sibilants sixteen times swiftly. Then continue reading. What do we need to predict the luminosity (amount of energy emitted per unit time)? Well, we need a total amount of energy the supernovae produces, and an amount of time it takes to do that. To get the total energy, the authors assume that the SNe is powered at first by nuclear fusion of carbon and oxygen into higher mass elements. This gives a total energy. Assuming all of this energy goes into driving a huge, expanding fireball (of uniform density at any given moment), they calculate the expansion speed of the fireball, which turns out to be constant. This is important because it means that the size of the fireball is just a constant speed times time, so the size (radius, to be precise) increases linearly with time. Why do we care?
Well, this is key for two reasons. First, the fireball loses energy as it expands, so the expansion rate tells us how much energy it loses per unit time from expansion. Second, the fireball also loses energy via photons inside it random-walking their way out (like a drunkard, they stagger in random directions inside the fireball, but they will eventually reach its edge). The time it takes them to do this is bigger if the distance they must cover is bigger—so the photon diffusion time $t_{dif}$ will scale with 1/time since the explosion occurred.
These two ideas let the authors set up a simple equation relating the change in energy over time to the energy of the SNe, which can be solved to give the energy as a function of time—i.e. the luminosity. For those who care for the technical, the equation is
$\frac {dE}{dt} \approx \frac {dE_{ra}}{dt} - E\big[\frac {1}{t} +\frac{1} {t_{dif}}\big] .$
This may look scary, but let’s look at it step-by-step. It is just accounting. Just like the money you deposit in your savings account is just what you make minus what you spend, here, the total change in energy per unit time is just the energy produced minus per unit time minus the energy lost per unit time. E represents energy, t time.
The term on the left-hand side is just the total change in energy per unit time. On the right-hand side, the first term is just the energy generation rate: all of the energy is assumed to be produced by the radioactive decay of the elements produced earlier by nuclear fusion.
The second term, with brackets, on the right encodes 2 effects. The $1/t$ describes energy lost because the fireball is expanding—it grows linearly with time, so it loses energy like $1/t$. The 1/$t_{dif}$ is the more complicated effect described above: photons random walk out of the fireball, taking energy with them as they do so.
The authors solve this equation for how the energy behaves with time, and that gives them the light-curve, shown in the picture. From this, everything we actually observe about SNIa can be predicted, providing a great way for us to understand why the light-curve has the shape it does!
I’m a 2nd year grad student in Astronomy at Harvard, working with Daniel Eisenstein on the effect of relative velocities between regular and dark matter on the baryon acoustic oscillations. I did my undergrad at Princeton, where I worked with Rich Gott on dark energy, Jeremy Goodman on dark matter, and Roman Rafikov on planetesimals. I also spent a year at Oxford getting a master’s in philosophy of physics, which remains an interest.
1. Aren’t type Ia supernova thought to be the product of runaway thermonuclear reactions, and not collapse due to overcoming the Chandrasekhar limit? | {"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": 5, "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": 0.8627225756645203, "perplexity": 740.6110309279916}, "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-2015-22/segments/1432207929561.80/warc/CC-MAIN-20150521113209-00283-ip-10-180-206-219.ec2.internal.warc.gz"} |
https://blog.computationalcomplexity.org/2016/07/is-determining-if-poly-over-finite.html | Monday, July 04, 2016
Is determining if a poly over a finite field is 1-1 hard? Sure seems so.
When I teach cryptography to High School students I begin with shift and linear ciphers which are
x --> x+s mod 26 (s is a shift, x is a letter of the alphabet. Hmm- x really IS a letter of the alphabet!)
x--> ax+b mod 26 (Note that a has to be rel prime to 26.)
I then ask why nobody seems to have ever used
x --> ax2 + bx + c mod 26.
I then tell them that this is because there is no quick test that will, given (a,b,c), tell if
f(x) = ax2 + bx + c mod 26 is 1-1 (and hence onto).
It recently dawned on me that I don't really know that its hard to test.
(ADDED LATER) In fact its not true. Algebra shows that f(x) is NOT 1-1 iff
a(x+y)+b =0 mod 26
has a solution. If a is rel prime to 26 then clearly there is a solution (many in fact).
If a is not rel prime to 26 then I suspect this is not hard.
How hard is the following problem?
Given a poly f(x) of degree d, and n, is f(x) mod n 1-1?
We will assume all coefficients are between 0 and n.. We can also assume that c is 0 since f(x) is 1-1 then f(x)-c is 1-1.
The coefficients are given in binary so the length of the input is roughly dlog(n).
One can of course compute f(0), f(1),...,f(n-1) and see if there are any repeats, but this takes O(n) steps which is exp in the input of length log n.
I suspect that this is either a well known solved problem (either in P or NPC) or a well known open problem. Any help or references will be more appreciated than usual-- see next paragraph.
I am the new SIGACT News Open Problems Column editor. In the future I will be soliciting people to write columns for me, but the first one I'll do myself and this might be a good topic- if its open! And if it is open, would be good to know references and what is known.
1. 2x^2+x is 1-1 mod 4.
1. 2x²+x is 1–1 mod any power of 2.
Rivest has shown in 1999 that polynomial ∑aᵢxⁱ is a permutation mod 2ⁿ iff a₁ is odd, (a₂+a₄+a₆+⋯) is even, and (a₃+a₅+a₇+⋯) is even.
2. This comment has been removed by a blog administrator.
3. These polynomials are known as permutation polynomials in the literature. Testing if f in 1-1 mod m reduces to testing if for every prime factor of p, f in 1-1 mod p and f' has no root mod p (see for instance Theorems 2 and 6 of http://arxiv.org/abs/math/0509523). Both tests can be achieved in deterministic polytime (see http://eccc.hpi-web.de/report/2005/008/ for the first one, and the second one is classical). Thus given the prime factorization of m, the test is deterministic polytime. This does not imply that it is "easy" ;-).
4. These polynomials are known as permutation polynomials in the literature. Testing if f in 1-1 mod m reduces to testing if for every prime factor of p, f in 1-1 mod p and f' has no root mod p (see for instance Theorems 2 and 6 of http://arxiv.org/abs/math/0509523). Both tests can be achieved in deterministic polytime (see http://eccc.hpi-web.de/report/2005/008/ for the first one, and the second one is classical). Thus given the prime factorization of m, the test is deterministic polytime. This does not imply that it is "easy" ;-).
5. To elaborate a little bit:
Quadratic polynomials are never 1-1 over a field of characteristic different from 2 (because a non-zero element has 0 or 2 square roots).
Over GF(2^r): x^2 is 1-1 but unless I am mistaken, a polynomial with a linear term will not be 1-1 (you have 0 or 2 preimages).
If you start counting mod n for composite n there are other 1-1 polynomials (like in the above example).
6. Might this problem somehow be connected to whether the polynomial f(x) has roots mod n? At least in some cases this might lead us to a partial solution:
Since we assume that the constant c is zero, we know that the polynomial has at least the root 0. If the polynomial has another root y, then it is obviously not 1-1.
Finding out if the polynomial has roots may be fast in some cases. If n is of the form p^k, where p is a prime, then we may use factorization algorithms for polynomials over finite fields (see for instance https://en.wikipedia.org/wiki/Factorization_of_polynomials_over_finite_fields). If the degree d of the polynomial is small in relation to n, then we might find an algorithm that is faster than O(n). For instance if n is a prime, then Shoup's algorithm (http://www.shoup.net/papers/detfac.pdf) can factorize the polynomial in time O(n^0.5*(log n)^2*d^(2+eps)). Since the polynomial may have at most d factors, it should be easy to find any linear factors (if d is small in relation to n)
This is of course only a partial solution, since everything above only works for finite fields (or at least all the analysis is done over finite fields). I'm also sure that there is a polynomial that is not 1-1 and does not have more than one root; an example of a polynomial like this would of course be nice!
7. It appears that testing whether a polynomial over a finite field is a permutation polynomial is, in fact, in P.
8. It is called permutation polynomials. It is a very important research problem. There are tons of survey on this topics. Start with wikipedia first.
9. See the notes and references here. The problem is apparently in ZPP. | {"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": 0.9087177515029907, "perplexity": 537.6708950136655}, "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-2021-17/segments/1618038064520.8/warc/CC-MAIN-20210411144457-20210411174457-00033.warc.gz"} |
http://math.stackexchange.com/questions/61813/illustration-of-vector-calculus-vs-differential-forms | # Illustration of vector calculus vs. differential forms
I am looking for a nice illustration of how vector calculus relates to differential forms. A demonstration that employs physics is appreciable (e.g. electromagnetism).
In particular, while dualizing gradient, divergence and curl gives the same set of operators again, this is not the same with the differential forms perspective. How to appropriately interpret that is usually left blank in physics books. Mathematics books often don't bother at all about classical vector calculus.
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I found the exposition in Dubrovin-Fomenko-Novikov quite nice --- look up the hodge star operation (towards the end of the Wikipedia page the three-dimensional case is spelled out). If I remember correctly it's in volume 3 of D-F-N. Some basic stuff is also explained at the beginning of Bott-Tu. – t.b. Sep 4 '11 at 18:46
Try
• John Baez, Javier P. Muniain: "Gauge Fields, Knots and Gravity"
The first part, "Electromagnetism", explains Maxwell's equations first using classical vector calculus and how to rewrite them using differential forms.
Another classical book that explains differential forms and their use in both classical electromagnetism and in general relativity is
• Misner, Thorne, Wheeler: "Gravitation"
The latter is a +1000 pages tome, but it is possible to select the chapters that you are interested in and start there, you don't need to read it cover to cover.
- | {"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": 0.8463174104690552, "perplexity": 342.1715092505881}, "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/1394011210359/warc/CC-MAIN-20140305092010-00031-ip-10-183-142-35.ec2.internal.warc.gz"} |
http://math.stackexchange.com/questions/344628/to-prove-a-property-of-greatest-common-divisor?answertab=active | # To prove a property of greatest common divisor
Suppose integer $d$ is the greatest common divisor of integer $a$ and $b$, how to prove, there exist whole number $r$ and $s$, so that $$d = r \cdot a + s \cdot b$$ ?
i know a proof in abstract algebra, hope to find a number theory proof?
for abstract algebra proof, it's in Michael Artin's book "Algebra".
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An approach through elementary number-theory:
It suffices to prove this for relatively prime $a$ and $b$, so suppose this is so. Denote the set of integers $0\le k\le b$ which is relatively prime to $b$ by $\mathfrak B$. Then $a$ lies in the residue class of one of elements in $\mathfrak B$.
Define a map $\pi$ from $\mathfrak B$ into itself by sending $k\in \mathfrak B$ to the residue class of $ka$. If $k_1a\equiv k_2a\pmod b$, then, as $\gcd (a,b)=1$, $b\mid (k_1-k_2)$, so that $k_1=k_2$ (Here $k_1$ and $k_2$ are positive integers less than $b$.). Hence this map is injective. Since the set $\mathfrak B$ is finite, it follows that $\pi$ is also surjectie.
So there is some $k$ such that $ka\equiv 1\pmod b$. This means that there is some $l$ with $ka-1=lb$, i.e. $ka-lb=1$.
Barring mistakes. Thanks and regards then.
P.S. The reduction step is: Given $a, b$ with $\gcd(a,b)=d$, we know that $\gcd(\frac{a}{d},\frac{b}{d})=1$. So, if the relatively prime case has been settled, then there are $m$ and $n$ such that $m\frac{a}{d}+n\frac{b}{d}=1$, and hence $ma+nb=d$.
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It avails of nothing but the principle that, if $a\mid bc$, while $\gcd(a,b)=1$, then $a\mid c$. – awllower Mar 28 '13 at 11:51
actually what i was looking for is an easier solution, that is by euclidean algorithm. this is much more complex. but, it's more fun! :D thanks @awllower – athos Apr 3 '13 at 7:51
@athos Glad to share with you this fun answer! :D – awllower Apr 3 '13 at 11:02
Here is a number theoretic proof (I hope not the one you already know):
Let $c=r\cdot a+s\cdot b$ be the smallest positive integer such that can be written as a linear combination of $a$ and $b$ with integer coefficients (note that $\left\{x\cdot a+y\cdot b\in\mathbb N:x,y\in\mathbb Z\right\}\neq\emptyset$).
1. Divide $a$ by $c$ to obtain $a=k\cdot c+\lambda$ for some $0\leq \lambda<c$. Thus $\lambda=a-kc=\ldots\Rightarrow \lambda=0$ (by the definition of $c$ as the smallest positive ...). Therefore $c\mid a$. Similarly $c\mid b$.
2. If $e\mid a$ and $e\mid b$ then $e\mid c$.
1) and 2) are the definition of $\gcd(a,b)$.
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thanks, this is the one similar to what i know. but thanks! – athos Apr 3 '13 at 7:35
You can use any textbook on Number Theory, e.g., H.Stark, An Introduction to Number Theory, Theorem 2.2.
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The Euclidean algorithm (see it on Wikipedia, http://en.wikipedia.org/wiki/Euclidean_algorithm), when applied backwards, says that
the GCD can be expressed as a sum of the two original numbers each multiplied by a positive or negative integer, e.g., 21 = [5 × 105] + [(−2) × 252].
If you want $r$ and $s$ positive, just sum (a multiple of) $rs$ to both addends.
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This is Bézout's lemma/identity.
See two proofs here
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I think that the two proofs in tthe link are both the same thing: Euclidean algorithm. So maybe you could elaborate more upon the answer, instead of a link only? In any case, still thanks. :) – awllower Mar 28 '13 at 12:09
thanks, i got it! what i was not aware of is the extended euclidean algorithm. – athos Apr 3 '13 at 7:40 | {"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": 0.9593648314476013, "perplexity": 245.77799179895203}, "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-2016-07/segments/1454701147841.50/warc/CC-MAIN-20160205193907-00325-ip-10-236-182-209.ec2.internal.warc.gz"} |
https://lavelle.chem.ucla.edu/forum/viewtopic.php?p=211928 | ## Delta H
Jessica Castellanos
Posts: 102
Joined: Sat Aug 24, 2019 12:17 am
Been upvoted: 1 time
### Delta H
When delta H is positive, is it endothermic or exothermic? And if H is negative?
Kevin Antony 2B
Posts: 99
Joined: Sat Sep 07, 2019 12:16 am
### Re: Delta H
To see if something is endothermic r exothermic we look at delta G. If delta G is positive, the reaction is endothermic, but if it is negative the reaction is exothermic.
sarahforman_Dis2I
Posts: 109
Joined: Sat Aug 17, 2019 12:18 am
### Re: Delta H
Jessica Castellanos wrote:When delta H is positive, is it endothermic or exothermic? And if H is negative?
If delta H is positive, that means the sum of the enthalpy of reactants is LARGER than the sum of the enthalpy of products. This means that the reaction is endothermic, or uses heat.
If delta H is negative, the sum of the enthalpy of reactants is SMALLER than the sum of the enthalpy of products. This means that the reaction is exothermic, or releases heat. Dr. Lavelle said that we will go more in depth about delta H, delta G and delta S later in the quarter.
Hussain Chharawalla 1G
Posts: 100
Joined: Sat Jul 20, 2019 12:15 am
### Re: Delta H
If it helps, another way to think of delta H or enthalpy is the energy stored in bonds. If the products are in a lower energy more stable configuration, the reaction will be exothermic.
aishwarya_atmakuri
Posts: 101
Joined: Sat Jul 20, 2019 12:15 am
### Re: Delta H
If delta H is negative, it is exothermic. If delta H is positive, it is endothermic.
Ashley Fang 2G
Posts: 102
Joined: Fri Aug 30, 2019 12:17 am
### Re: Delta H
sarahforman_Dis2I wrote:
Jessica Castellanos wrote:When delta H is positive, is it endothermic or exothermic? And if H is negative?
If delta H is positive, that means the sum of the enthalpy of reactants is LARGER than the sum of the enthalpy of products. This means that the reaction is endothermic, or uses heat.
If delta H is negative, the sum of the enthalpy of reactants is SMALLER than the sum of the enthalpy of products. This means that the reaction is exothermic, or releases heat. Dr. Lavelle said that we will go more in depth about delta H, delta G and delta S later in the quarter.
I believe you switched it the other way, the idea is enthalpy of products - enthalpy of reactants = delta H
But yes, positive delta H means endothermic, negative delta H means exothermic.
Caroline Beecher 2H
Posts: 51
Joined: Wed Nov 14, 2018 12:21 am
### Re: Delta H
The rule of thumb is that when delta H is positive, it is endothermic (it takes up heat and its' surroundings become cooler) and when delta H is negative, it is exothermic (it releases heat and its' surroundings become warmer).
Jialun Chen 4F
Posts: 108
Joined: Sat Sep 07, 2019 12:16 am
### Re: Delta H
I agree with Caroline's answer. If positive Delta H, endothermic; if negative Delta H, exothermic.
Sean Cheah 1E
Posts: 105
Joined: Wed Sep 18, 2019 12:20 am
### Re: Delta H
Kevin Antony 2B wrote:To see if something is endothermic r exothermic we look at delta G. If delta G is positive, the reaction is endothermic, but if it is negative the reaction is exothermic.
When determining whether a reaction is exothermic/endothermic, one should look at the sign of delta H (change in enthalpy). The sign of G tells you whether a given reaction is exergonic/endergonic, which sounds similar but is completely different.
Michellekim1H
Posts: 25
Joined: Wed Feb 20, 2019 12:17 am
### Re: Delta H
Positive delta H is endothermic(since there’s more energy in the products) whereas negative delta H is exothermic(there’s less energy in the products).
Elizabeth Harty 1A
Posts: 125
Joined: Sat Jul 20, 2019 12:16 am
### Re: Delta H
positive delta H means endothermic, negative delta H means exothermic.
Return to “Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)”
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https://brilliant.org/problems/spring-constants-or-spring-variables/ | # Spring Constants or Spring Variables?
Classical Mechanics Level pending
An engineer is designing a spring to be placed at the bottom of an elevator shaft. If the elevator accidentally breaks at a height 5 meters above the top of the spring, calculate the value that the spring constant K should have so that the passengers undergo an acceleration of no more than 5g. The total mass of the elevator and passengers is 75 kg.
g=9.8 m/s/s
The units of K are in N/m
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https://www.math.princeton.edu/events/blow-phenomena-yamabe-equation-2010-10-06t203010 | # Blow-up phenomena for the Yamabe equation
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Simon Brendle, Stanford University
Fine Hall 314
The Yamabe problem asserts that any Riemannian metric on a compact manifold can be conformally deformed to one of constant scalar curvature. However, this metric is not, in general, unique, and there are examples of manifolds that admit many metrics of constant scalar curvature in a given conformal class. It was conjectured by R. Schoen in the 1980s (and, independently, by Aubin) that the set of all metrics of constant scalar curvature 1 in a given conformal class is compact, except if the underlying manifold is conformally equivalent to the sphere $S^n$ equipped with its standard metric. The significance of Schoen's conjecture is that it would imply Morse inequalities for the total scalar curvature functional. I will discuss counterexamples to this conjecture in dimension 52 and higher. I will also describe joint work with F. Marques, which extends these counterexamples to dimension 25 and higher. The condition $n \geq 25$ turns out to be optimal. | {"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": 0.9173864722251892, "perplexity": 204.9140887178739}, "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/1521257648431.63/warc/CC-MAIN-20180323180932-20180323200932-00578.warc.gz"} |
http://www.ebooks.com/299279/nuclear-production-of-hydrogen/oecd-publishing/ | for Kindle Fire, Apple, Android, Nook, Kobo, PC, Mac, BlackBerry ...
New to eBooks.com?
Nuclear Production of Hydrogen
Third Information Exchange Meeting, Oarai, Japan, 5-7 October 2005
US\$ 112.00
Hydrogen has the potential to play an important role as a sustainable and environmentally acceptable energy carrier in the 21st century. Since natural sources of pure hydrogen are extremely limited, it is necessary to develop technologies to produce large quantities of hydrogen economically. The currently dominant technology for producing hydrogen is based on reforming fossil fuels, a process which releases greenhouse gases. Hydrogen produced by water cracking, using heat and surplus electricity from nuclear power plants, requires no fossil fuels and results in lower greenhouse gas emissions. This report presents the state of the art in the nuclear production of hydrogen and describes its associated scientific and technical challenges.
Organisation for Economic Co-operation and Development; June 2006
414 pages; ISBN 9789264026308 | {"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": 0.8161666989326477, "perplexity": 3849.8260706801143}, "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-2014-52/segments/1418802769844.62/warc/CC-MAIN-20141217075249-00126-ip-10-231-17-201.ec2.internal.warc.gz"} |
http://mathoverflow.net/questions/32011/direct-proof-of-irrationality/32030 | # Direct proof of irrationality?
There are plenty of simple proofs out there that $\sqrt{2}$ is irrational. But does there exist a proof which is not a proof by contradiction? I.e. which is not of the form:
Suppose $a/b=\sqrt{2}$ for integers $a,b$.
$\rightarrow\leftarrow$, QED
Is it impossible (or at least difficult) to find a direct proof because ir-rational is a negative definition, so "not-ness" is inherent to the question? I have a hard time even thinking how to begin a direct proof, or what it would look like. How about:
$\forall a,b\in\cal I \exists \epsilon$ such that $\mid a^2/b^2 - 2\mid > \epsilon$
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Not sure why the LaTeX worked in the preview, but not the post! Let me retype that last line in ASCII: For all integers a,b, there exists epsilon such that |a^2/b^2 - 2| > epsilon – RubeRad Jul 15 '10 at 14:59
Please see wikipedia before asking. – Abhishek Parab Jul 15 '10 at 15:16
See Andrej Bauer's post - math.andrej.com/2010/03/29/… – François G. Dorais Jul 15 '10 at 15:37
For those needing more inducement to click through than just links: Andrej's post discusses the point raised in the OP's last paragraph, the difference between classical “proof by contradiction” — “to prove $\varphi$, assume $\lnot \varphi$ and derive absurdity” — and the intuitionistically valid “to prove $\lnot \varphi$, assume $\varphi$ and derive absurdity”, which at a formal level (either classically or intuitionistically) any proof of a negative statement must essentially boil down to. Gowers' post discusses some related issues rather more informally, with a wide range of examples. – Peter LeFanu Lumsdaine Jul 15 '10 at 21:11
Below is a simple direct proof that I found as a teenager:
THEOREM $\;\rm r = \sqrt{n}\;$ is integral if rational, for $\;\rm n\in\mathbb{N}$.
Proof: $\;\rm r = a/b,\;\; {\text gcd}(a,b) = 1 \implies ad-bc = 1\;$ for some $\rm c,d \in \mathbb{Z}$, by Bezout
so: $\;\rm 0 = (a-br) (c+dr) = ac-bdn + r \implies r \in \mathbb{Z} \quad\square$
Nowadays my favorite proof is the 1-line gem using Dedekind's conductor ideal - which, as I explained at length elsewhere, beautifully encapsulates the descent in ad-hoc "elementary" irrationality proofs.
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That's really slick too; I'll have to make some time to read the link... – RubeRad Jul 15 '10 at 15:57
So your proof of irrationality of $\sqrt{2}$ is follows: Assume $\sqrt{2}$ was rational. Then my theorem implies that $\sqrt{2}$ is integral, which is obviously impossible since no integer squares to $2$. QED. But this is a proof by contradiction isn't it? – Rasmus Bentmann Jul 15 '10 at 16:31
Your theorem is neat, though. – Rasmus Bentmann Jul 15 '10 at 16:34
But every proof does the same at some point - even if it's far down the line in some chain of lemmas, e.g. see the Gower's link above. – Bill Dubuque Jul 15 '10 at 16:47
Yes, of course there are many variations. But they are all mechanically derivable from the more conceptual ideal-based proofs, as I explain at length in the sci.math linked indirectly above. Thus they're all essentially equivalent and the key idea goes way back to Dedekind. I could easily write a program that could generate all of the known "elementary" proofs by unwinding the conceptual proofs, eliminating higher-order concepts like ideals, modules, and directly inlining lemmas, etc. Such elementary proofs are not novel in any way. – Bill Dubuque Jul 15 '10 at 22:39
Wikipedia has a constructive proof. You can bound $\sqrt 2$ away from $p/q$.
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Thanks, that exactly does it! (now if I could only reproduce the "easy calculation...") – RubeRad Jul 15 '10 at 15:52
I think there's some confusion going on here. The standard proof of the irrationality of $\sqrt 2$ is constructive, as noted in the comments above, pointing to Andrej's article: math.andrej.com/2010/03/29/… The "constructive" proof on Wikipedia is proving a result that is stronger from a constructive perspective. It's proving that $\sqrt 2$ and $a/b$ are "apart". You still need to derive a contradiction, or use a lemma that does so, in order to show that these values are distinct. – Dan Piponi Jul 15 '10 at 18:07
You have a point. This only bears on the constructivist distinction between 'It is not the case that $\sqrt 2$ is rational' and '$\sqrt 2$ is irrational'. The first of these means that the assumption that $\sqrt 2$ is rational leads to a contradiction. The second means that for any rational $p/q$, $\sqrt 2$ is apart from $p/q$ by a distance dependent on $p$ and $q$. – David Corfield Jul 15 '10 at 18:35
I'm pretty sure that with the usual definitions, 'It is not the case that $\sqrt 2$ is rational' is equivalent to '$\sqrt 2$ is irrational', constructively or not. The distinction I'm making hinges on the difference between 'apartness' and 'inequality'. The latter doesn't imply the former constructively as we don't have the familiar classical trichotomy. – Dan Piponi Jul 15 '10 at 20:36
Exercise 10 p. 62 of Constructive Analysis (Bishop and Bridges): Construct a real number that is not rational and not irrational. (A real number $x$ is irrational if $x \neq r$ for each rational number $r$.) A real number is a certain sequence of rationals. To prove it irrational, show that it is not equal to any $p/q$ considered as a constant sequence. To show inequality of two reals, show that entries at some point in the sequences are far enough apart. – David Corfield Jul 16 '10 at 8:37
Rational numbers have finite continuued fractions.
$\sqrt{2}=1+1/(\sqrt{2}+1)=1+1/(2+1/(\sqrt{2}+1))=...$ Then the continued fraction is not finite 1+1/2+1/2+1/2+...
The geometric proof (not the one in Wikipedia), the one that proves $\sqrt{2}$ is not commensurable with $1$ is also direct (and is essentially the same as the continued fraction).
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These are continued (not continuous) fractions. – Robin Chapman Jul 16 '10 at 18:07
Spanish tricked me. We use the same word. Thanks. – Mlazhinka Shung Gronzalez LeWy Nov 4 '10 at 1:58
From the viewpoint of prime factorization, integers are products of powers of primes in which the exponents are non-negative integers. Rational numbers are products of powers of primes in which the exponents can be any integer. In both cases, any given prime can only appear once. This, of course, means that when you square a rational number, all the exponents will be even numbers. Since 2 is not an even power of a prime, it cannot be the square of a rational number.
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Slight clarification: POSITIVE integers are products of powers of primes in which the exponents are natural numbers, and POSITIVE rational numbers are products of powers of primes in which the exponents can be any integer. – Sridhar Ramesh Jan 13 '12 at 22:25
Below is a direct proof that if $p,q,n$ are positive integers with $\gcd(p,q)=1$ and $p^2=nq^2$ then $q=1$ (so $n=p^2$). I would count that as a direct proof that $$\lbrace n \mid \sqrt{n}\in \mathbb{Q} \rbrace=\lbrace0,1,4,9,16,\cdots\rbrace$$ Given that $\gcd(p,q)=1$ there are integers $s,t$ with $ps+qt=1.$ Cube and regroup to get $p^2(ps+3qt)s^2+q^2(3ps+qt)t^2=1.$ Given that $p^2=nq^2$ we then have $nq^2(ps+3qt)s^2+q^2(3ps+qt)t^2=1$ so that $q$ divides 1. QED
later As Andres points out, it suffices to square. The cubing shows that $\gcd(p,q)=1$ implies $\gcd(p^2,q^2)=1$. Of course if $p,q,n$ are integers and we already know $\frac{p^2}{q^2} \ne n$ then it follows that $|\frac{p^2}{q^2}-n| \ge \frac{1}{q^2}$. I wanted an direct argument that if $p,q,n$ are positive integers with $\gcd(p,q)=1$ and $q \ge 2$ then $|\frac{p^2}{q^2}-n| \ge \frac{1}{q^2}$. I think that could be done but in this situation one wants to keep a proof short.
In my opinion, the vast majority of "indirect proofs" are actually direct proofs of something else. But that is another story.
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@Aaron: This is nice, but why do you need to cube? Squaring suffices: $nq^2s^2+2psqt+q^2t^2=1$. – Andrés Caicedo Dec 9 '10 at 23:06
A book on logic whose title and the identity of whose author escape me at the moment said that not all proofs by contradiction are indirect proofs. The idea is that when proving an inherently negative statement---one that asserts non-existence of something---one can proceed only by contradiction, which in that case constitutes a direct proof. I'll see if I can find it.
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Here's another take on Bill's integrality theorem, using existence and uniqueness of fractions in lowest terms:
If sqrt(2) = p/q is in lowest terms, then 2/1 = p^2/q^2 is also in lowest terms. Hence p^2 = 2, and q^2 = 1.
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So why does $p/q$ being in lowest terms entail that $p^2/q^2$ is? If "in lowest terms" means having no common factors save units, then this implication doesn't hold in all integral domains. – Robin Chapman Dec 9 '10 at 16:50
Forgive me - I was tempted by some of these answers to offer the most elementary and shortest proofs I know, valid in Z, and using only facts acceptable to laypersons, that sqrt(2) is irrational. Since all high school students presumably learn the rational root theorem, it is odd e.g. that few US high school algebra books conclude that the only rational roots of X^2 - 2, are integer factors of 2. – roy smith Dec 9 '10 at 19:40
Although the first argument used unique factorization, the second only uses integral closure. I guess the general statement would be that an element of a given domain having no square root, still has none in the fraction field. Are there domains that are integrally closed for solutions of quadratic equations but not in general? – roy smith Dec 9 '10 at 21:03
@Robin But it does in Bezout Domains. Given that $s,t$ exist with $ps+qt=1$, cube both sides and regroup: $p^2(ps+3qt)s^2+q^2(3ps+qt)t^2=1.$ – Aaron Meyerowitz Dec 9 '10 at 21:17
Robin, on reflection, I don't quite understand your remark. Say we want to prove an element of a given domain is a square if and only if it is a square in its field of fractions. Bill's proof above uses that gcd(a,b) exists and is a linear combination of a,b, which holds only in a pid. My first argument above works more generally in a ufd, and my second one still more generally in an integrally closed domain. I do not know of a proof that works in a yet more general domain. Thus I asked if there is a domain not integrally closed, but closed "for squares". Do you know one?? regards, roy – roy smith Dec 12 '10 at 18:50
Observe that if $\sqrt 2$ is rational, then there is some positive integer q such that q × $\sqrt 2$ is an integer. Since the positive integers are well ordered, we may suppose that q is the smallest such number.
We next observe that since 1 < $\sqrt 2$ < 2, then $\sqrt 2$ – 1 < 1, and consequently q × ($\sqrt 2$ – 1) = (q ×$\sqrt 2$ – q ) is less than q. Let us call this new number r, and observe that it too is a positive integer. But we now have r × $\sqrt 2$ is also an integer, since r × $\sqrt 2$ = (q × $\sqrt 2$– q ) ×$\sqrt 2$ = (2q – q ×$\sqrt 2$ ). In short, r is a positive integer less than q and r ×$\sqrt 2$ is an integer. But we said that q was the smallest positive integer with this property, and so we have a contradiction.
The nice thing about this proof is how easily it generalizes. Let us denote by $|\sqrt n|$ the integer part of $\sqrt n$ . For example, since the square root of 5 is approximately 2.236, the integer part is 2. For any n that is not a perfect square, we may prove that is irrational exactly as above by considering q × ( $\sqrt n$– $|\sqrt n|$). (On the other hand, if n is a perfect square (so that $\sqrt n$ = $|\sqrt n|$) then there is no contradiction.)
More generally still, if x is a rational but not integral zero of a monic integer polynomial of degree d, let q be the least positive integer so that q$x^j$ is integral for all j < d. Then, considering q(x – n) where n is an integer with n < x < n + 1, we get a contradiction. In other words, we have proved that every rational “algebraic integer” is an integer.
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There seems to be a problem with your code (the first paragraph is cut off.) – Andrés Caicedo Jul 15 '10 at 16:33
Ye, Andres, thanks, don't know what the problem was but it's fixed now. – Dick Palais Jul 15 '10 at 16:37
According to your first sentence I believe that you provide a proof by contraction which is not what the OP has asked for. – Rasmus Bentmann Jul 15 '10 at 16:40
In fact it generalizes quite widely to show that any PID/Dedekind domain is integrally-closed. It's just a specialization of a the 1-line proof by way of principality of the conductor (denominator) ideal which I mentioned above. See the link in my post for much further discussion. – Bill Dubuque Jul 15 '10 at 16:53
I thought it worth emphasis for the reader since rediscoverers often think that such proofs are novel - even professional mathematicians - even number theorists! E.g. Estermann rediscovered such an elementary proof in 1975 and often boasted that it was "the first new proof since Pythagoras" and this claim was supported by some other number-theorists, e.g. Niven. – Bill Dubuque Jul 15 '10 at 19:11
Most common axiom systems I've seen are a list of $\forall$ and $\exists$ axioms. If you look at a minimal underlying logic, most of the common rules for transforming these axioms shouldn't change the $\forall$ or $\exists$ into a $\neg \exists$. So you could fix a logic system, and argue that the only method that results in a $\neg \exists$ statement is the equivalent of proof by contradiction.
You haven't fixed a logic system in your original question, but the "proof by direct substitution" method won't be sufficient.
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https://www.gradesaver.com/textbooks/math/precalculus/precalculus-10th-edition/chapter-3-linear-and-quadratic-functions-3-1-properties-of-linear-functions-and-linear-models-3-1-assess-your-understanding-page-126/2 | ## Precalculus (10th Edition)
$\frac{2}{3}$
By using the slope formula, $m=\frac{y_2-y_1}{y_2-y_1}$, the slope of the line that contains the two given points is: $m=\frac{3-5}{-1-2}=\frac{-2}{-3}=\frac{2}{3}$ | {"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": 0.9858857989311218, "perplexity": 536.0884239736437}, "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/1547583875448.71/warc/CC-MAIN-20190122223011-20190123005011-00195.warc.gz"} |
https://socratic.org/questions/how-do-you-simplify-5x-x-2-9-6x-18-15x-3#561211 | Algebra
Topics
# How do you simplify (5x) /(x^2 - 9)*(6x + 18) / (15x^3)?
the answer is $\frac{2}{{x}^{2} \left(x - 3\right)}$
first factor $\left({x}^{2} - 9\right)$ since it's a difference of two squares. then factor out $6 x + 18$ so at this point, you have: $\frac{5 x}{\left(x - 3\right) \left(x + 3\right)} \cdot \frac{6 \left(x + 3\right)}{15 {x}^{3}}$ now you see that you can cancel $\left(x + 3\right)$ since there's two of them. now you can divide $5$ into $15$ but now you can divide $3$ into the $6$ on top of it. at this point, you have $\frac{x}{x - 3} \cdot \frac{2}{{x}^{3}}$ as you can see, you can take out one $x$ from ${x}^{3}$ so that will only leave you with your answer | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 12, "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": 0.8903734683990479, "perplexity": 120.6494884130436}, "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-2022-05/segments/1642320305052.56/warc/CC-MAIN-20220127012750-20220127042750-00512.warc.gz"} |
http://www.cfd-online.com/W/index.php?title=Probability_density_function&oldid=3336 | # Probability density function
Stochastic methods use distribution functions to decribe the fluctuacting scalars in a turbulent field.
The distribution function $F_\phi(\Phi)$ of a scalar $\phi$ is the probability $p$ of finding a value of $\phi < \Phi$
$F_\phi(\Phi) = p(\phi < \Phi)$
The probability of finding $\phi$ in a range $\Phi_1,\Phi_2$ is
$p(\Phi_1 <\phi < \Phi_2) = F_\phi(\Phi_2)-F_\phi(\Phi_1)$
The probability density function (PDF) is
$P(\Phi)= \frac{d F_\phi(\Phi)} {d \Phi}$
where $P(\Phi) d\Phi$ is the probability of $\phi$ being in the range $(\Phi,\Phi+d\Phi)$. It follows that
$\int P(\Phi) d \Phi = 1$
Integrating over all the possible values of $\phi$. The PDF of any stochastic variable depends "a-priori" on space and time.
$P(\Phi;x,t)$ | {"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": 0.9839274883270264, "perplexity": 225.92022703065317}, "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-2015-27/segments/1435375099173.16/warc/CC-MAIN-20150627031819-00284-ip-10-179-60-89.ec2.internal.warc.gz"} |
http://popflock.com/learn?s=Physical_constant | Physical Constant
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Physical Constant
A physical constant, sometimes fundamental physical constant or universal constant, is a physical quantity that is generally believed to be both universal in nature and have constant value in time. It is contrasted with a mathematical constant, which has a fixed numerical value, but does not directly involve any physical measurement.
There are many physical constants in science, some of the most widely recognized being the speed of light in a vacuum c, the gravitational constant G, the Planck constant h, the electric constant ?0, and the elementary charge e. Physical constants can take many dimensional forms: the speed of light signifies a maximum speed for any object and its dimension is length divided by time; while the fine-structure constant ?, which characterizes the strength of the electromagnetic interaction, is dimensionless.
The term fundamental physical constant is sometimes used to refer to universal-but-dimensioned physical constants such as those mentioned above.[1] Increasingly, however, physicists only use fundamental physical constant for dimensionless physical constants, such as the fine-structure constant ?.
Physical constant, as discussed here, should not be confused with other quantities called "constants", which are assumed to be constant in a given context without being fundamental, such as the "time constant" characteristic of a given system, or material constants (e.g., Madelung constant, electrical resistivity, and heat capacity).
Since May 2019, all of the SI base units have been defined in terms of physical constants. As a result, five constants: the speed of light in vacuum, c; the Planck constant, h; the elementary charge, e; the Avogadro constant, NA; and the Boltzmann constant, kB, have known exact numerical values when expressed in SI units. The first three of these constants are fundamental constants, whereas NA and kB are of a technical nature only: they do not describe any property of the universe, but instead only give a proportionality factor for defining the units used with large numbers of atomic-scale entities.
## Choice of units
Whereas the physical quantity indicated by a physical constant does not depend on the unit system used to express the quantity, the numerical values of dimensional physical constants do depend on choice of unit system. The term "physical constant" refers to the physical quantity, and not to the numerical value within any given system of units. For example, the speed of light is defined as having the numerical value of when expressed in the SI unit metres per second, and as having the numerical value of 1 when expressed in the natural units Planck length per Planck time. While its numerical value can be defined at will by the choice of units, the speed of light itself is a single physical constant.
Any ratio between physical constants of the same dimensions results in a dimensionless physical constant, for example, the proton-to-electron mass ratio. Any relation between physical quantities can be expressed as a relation between dimensionless ratios via a process known as nondimensionalisation.
The term of "fundamental physical constant" is reserved for physical quantities which, according to the current state of knowledge, are regarded as immutable and as non-derivable from more fundamental principles. Notable examples are the speed of light c, and the gravitational constant G.
The fine-structure constant ? is the best known dimensionless fundamental physical constant. It is the value of the elementary charge squared expressed in Planck units. This value has become a standard example when discussing the derivability or non-derivability of physical constants. Introduced by Arnold Sommerfeld, its value as determined at the time was consistent with 1/137. This motivated Arthur Eddington (1929) to construct an argument why its value might be 1/137 precisely, which related to the Eddington number, his estimate of the number of protons in the Universe.[2] By the 1940s, it became clear that the value of the fine-structure constant deviates significantly from the precise value of 1/137, refuting Eddington's argument.[3] A theoretical derivation of the fine structure constant, based on unification in a pre-spacetime, pre-quantum theory in eight octonionic dimensions, has recently been given by Singh.[4]
With the development of quantum chemistry in the 20th century, however, a vast number of previously inexplicable dimensionless physical constants were successfully computed from theory.[] In light of that, some theoretical physicists still hope for continued progress in explaining the values of other dimensionless physical constants.
It is known that the Universe would be very different if these constants took values significantly different from those we observe. For example, a few percent change in the value of the fine structure constant would be enough to eliminate stars like our Sun. This has prompted attempts at anthropic explanations of the values of some of the dimensionless fundamental physical constants.
### Natural units
It is possible to combine dimensional universal physical constants to define fixed quantities of any desired dimension, and this property has been used to construct various systems of natural units of measurement. Depending on the choice and arrangement of constants used, the resulting natural units may be convenient to an area of study. For example, Planck units, constructed from c, G, ?, and kB give conveniently sized measurement units for use in studies of quantum gravity, and Hartree atomic units, constructed from ?, me, e and 40 give convenient units in atomic physics. The choice of constants used leads to widely varying quantities.
## Number of fundamental constants
The number of fundamental physical constants depends on the physical theory accepted as "fundamental". Currently, this is the theory of general relativity for gravitation and the Standard Model for electromagnetic, weak and strong nuclear interactions and the matter fields. Between them, these theories account for a total of 19 independent fundamental constants. There is, however, no single "correct" way of enumerating them, as it is a matter of arbitrary choice which quantities are considered "fundamental" and which as "derived". Uzan (2011) lists 22 "unknown constants" in the fundamental theories, which give rise to 19 "unknown dimensionless parameters", as follows:
The number of 19 independent fundamental physical constants is subject to change under possible extensions of the Standard Model, notably by the introduction of neutrino mass (equivalent to seven additional constants, i.e. 3 Yukawa couplings and 4 lepton mixing parameters).[5]
The discovery of variability in any of these constants would be equivalent to the discovery of "new physics".[6]
The question as to which constants are "fundamental" is neither straightforward nor meaningless, but a question of interpretation of the physical theory regarded as fundamental; as pointed out by Lévy-Leblond 1977, not all physical constants are of the same importance, with some having a deeper role than others. Lévy-Leblond 1977 proposed a classification schemes of three types of constants:
• A: physical properties of particular objects
• B: characteristic of a class of physical phenomena
• C: universal constants
The same physical constant may move from one category to another as the understanding of its role deepens; this has notably happened to the speed of light, which was a class A constant (characteristic of light) when it was first measured, but became a class B constant (characteristic of electromagnetic phenomena) with the development of classical electromagnetism, and finally a class C constant with the discovery of special relativity.[7]
## Tests on time-independence
By definition, fundamental physical constants are subject to measurement, so that their being constant (independent on both the time and position of the performance of the measurement) is necessarily an experimental result and subject to verification.
Paul Dirac in 1937 speculated that physical constants such as the gravitational constant or the fine-structure constant might be subject to change over time in proportion of the age of the universe. Experiments can in principle only put an upper bound on the relative change per year. For the fine-structure constant, this upper bound is comparatively low, at roughly 10-17 per year (as of 2008).[8]
The gravitational constant is much more difficult to measure with precision, and conflicting measurements in the 2000s have inspired the controversial suggestions of a periodic variation of its value in a 2015 paper.[9] However, while its value is not known to great precision, the possibility of observing type Ia supernovae which happened in the universe's remote past, paired with the assumption that the physics involved in these events is universal, allows for an upper bound of less than 10-10 per year for the gravitational constant over the last nine billion years.[10]
Similarly, an upper bound of the change in the proton-to-electron mass ratio has been placed at 10-7 over a period of 7 billion years (or 10-16 per year) in a 2012 study based on the observation of methanol in a distant galaxy.[11][12]
It is problematic to discuss the proposed rate of change (or lack thereof) of a single dimensional physical constant in isolation. The reason for this is that the choice of units is arbitrary, making the question of whether a constant is undergoing change an artefact of the choice (and definition) of the units.[13][14][15]
For example, in SI units, the speed of light was given a defined value in 1983. Thus, it was meaningful to experimentally measure the speed of light in SI units prior to 1983, but it is not so now. Similarly, with effect from May 2019, the Planck constant has a defined value, such that all SI base units are now defined in terms of fundamental physical constants. With this change, the international prototype of the kilogram is being retired as the last physical object used in the definition of any SI unit.
Tests on the immutability of physical constants look at dimensionless quantities, i.e. ratios between quantities of like dimensions, in order to escape this problem. Changes in physical constants are not meaningful if they result in an observationally indistinguishable universe. For example, a "change" in the speed of light c would be meaningless if accompanied by a corresponding change in the elementary charge e so that the ratio e2/(4??0?c) (the fine-structure constant) remained unchanged.[16]
## Fine-tuned universe
Some physicists have explored the notion that if the dimensionless physical constants had sufficiently different values, our Universe would be so radically different that intelligent life would probably not have emerged, and that our Universe therefore seems to be fine-tuned for intelligent life. However, the phase space of the possible constants and their values is unknowable, so any conclusions drawn from such arguments are unsupported. The anthropic principle states a logical truism: the fact of our existence as intelligent beings who can measure physical constants requires those constants to be such that beings like us can exist. There are a variety of interpretations of the constants' values, including that of a divine creator (the apparent fine-tuning is actual and intentional), or that ours is one universe of many in a multiverse (e.g. the many-worlds interpretation of quantum mechanics), or even that, if information is an innate property of the universe and logically inseparable from consciousness, a universe without the capacity for conscious beings cannot exist.
The fundamental constants and quantities of nature have been discovered to be fine-tuned to such an extraordinarily narrow range that if it were not, the origin and evolution of conscious life in the universe would not be permitted.[17]
## Table of physical constants
The table below lists some frequently used constants and their CODATA recommended values. For a more extended list, refer to List of physical constants.
Quantity Symbol Value[18] Relative
standard
uncertainty
elementary charge ${\displaystyle e}$ [19] Exact by definition
Newtonian constant of gravitation ${\displaystyle G}$ [20]
Planck constant ${\displaystyle h}$ [21] Exact by definition
speed of light in vacuum ${\displaystyle c}$ [22] Exact by definition
vacuum electric permittivity ${\displaystyle \varepsilon _{0}=1/\mu _{0}c^{2}}$ [23]
vacuum magnetic permeability ${\displaystyle \mu _{0}}$ [24]
electron mass ${\displaystyle m_{\mathrm {e} }}$ [25]
fine-structure constant ${\displaystyle \alpha =e^{2}/2\varepsilon _{0}hc}$ [26]
Josephson constant ${\displaystyle K_{\mathrm {J} }=2e/h}$ [27]
Rydberg constant ${\displaystyle R_{\infty }=\alpha ^{2}m_{\mathrm {e} }c/2h}$ [28]
von Klitzing constant ${\displaystyle R_{\mathrm {K} }=h/e^{2}}$ [29]
## References
1. ^ "Fundamental Physical Constants from NIST". Archived from the original on 2016-01-13. Retrieved . NIST
2. ^ A.S Eddington (1956). "The Constants of Nature". In J.R. Newman (ed.). The World of Mathematics. Vol. 2. Simon & Schuster. pp. 1074-1093.
3. ^ H. Kragh (2003). "Magic Number: A Partial History of the Fine-Structure Constant". Archive for History of Exact Sciences. 57 (5): 395-431. doi:10.1007/s00407-002-0065-7. S2CID 118031104.
4. ^ Singh, Tejinder P. (2021-09-08). "Quantum theory without classical time: Octonions, and a theoretical derivation of the Fine Structure Constant 1/137". International Journal of Modern Physics D. 30 (14): 2142010. arXiv:2110.07548. doi:10.1142/S0218271821420104. ISSN 0218-2718. S2CID 238856941.
5. ^ Uzan, Jean-Philippe (2011). "Varying Constants, Gravitation and Cosmology". Living Reviews in Relativity. 14 (1): 2. arXiv:1009.5514. Bibcode:2011LRR....14....2U. doi:10.12942/lrr-2011-2. PMC 5256069. PMID 28179829. Any constant varying in space and/or time would reflect the existence of an almost massless field that couples to matter. This will induce a violation of the universality of free fall. Thus, it is of utmost importance for our understanding of gravity and of the domain of validity of general relativity to test for their constancy.
6. ^ Uzan, Jean-Philippe (2011). "Varying Constants, Gravitation and Cosmology". Living Reviews in Relativity. 14 (1): 2. arXiv:1009.5514. Bibcode:2011LRR....14....2U. doi:10.12942/lrr-2011-2. PMC 5256069. PMID 28179829.
7. ^ Lévy-Leblond, J. (1977). "On the conceptual nature of the physical constants". La Rivista del Nuovo Cimento. Series 2. 7 (2): 187-214. Bibcode:1977NCimR...7..187L. doi:10.1007/bf02748049. S2CID 121022139.Lévy-Leblond, J.-M. (1979). "The importance of being (a) Constant". In Toraldo di Francia, G. (ed.). Problems in the Foundations of Physics, Proceedings of the International School of Physics 'Enrico Fermi' Course LXXII, Varenna, Italy, July 25 - August 6, 1977. New York: NorthHolland. pp. 237-263.
8. ^ T. Rosenband; et al. (2008). "Frequency Ratio of Al+ and Hg+ Single-Ion Optical Clocks; Metrology at the 17th Decimal Place". Science. 319 (5871): 1808-12. Bibcode:2008Sci...319.1808R. doi:10.1126/science.1154622. PMID 18323415. S2CID 206511320.
9. ^ J.D. Anderson; G. Schubert; V. Trimble; M.R. Feldman (April 2015), "Measurements of Newton's gravitational constant and the length of day", EPL, 110 (1): 10002, arXiv:1504.06604, Bibcode:2015EL....11010002A, doi:10.1209/0295-5075/110/10002, S2CID 119293843
10. ^ J. Mould; S. A. Uddin (2014-04-10), "Constraining a Possible Variation of G with Type Ia Supernovae", Publications of the Astronomical Society of Australia, 31: e015, arXiv:1402.1534, Bibcode:2014PASA...31...15M, doi:10.1017/pasa.2014.9, S2CID 119292899
11. ^ Bagdonaite, Julija; Jansen, Paul; Henkel, Christian; Bethlem, Hendrick L.; Menten, Karl M.; Ubachs, Wim (December 13, 2012). "A Stringent Limit on a Drifting Proton-to-Electron Mass Ratio from Alcohol in the Early Universe" (PDF). Science. 339 (6115): 46-48. Bibcode:2013Sci...339...46B. doi:10.1126/science.1224898. hdl:1871/39591. PMID 23239626. S2CID 716087.
12. ^ Moskowitz, Clara (December 13, 2012). "Phew! Universe's Constant Has Stayed Constant". Space.com. Archived from the original on December 14, 2012. Retrieved 2012.
13. ^ Duff, Michael (2015). "How fundamental are fundamental constants?". Contemporary Physics. 56 (1): 35-47. arXiv:1412.2040. Bibcode:2015ConPh..56...35D. doi:10.1080/00107514.2014.980093 (inactive 28 February 2022).{{cite journal}}: CS1 maint: DOI inactive as of February 2022 (link)
14. ^ Duff, M. J. (13 August 2002). "Comment on time-variation of fundamental constants". arXiv:hep-th/0208093.
15. ^ Duff, M. J.; Okun, L. B.; Veneziano, G. (2002). "Trialogue on the number of fundamental constants". Journal of High Energy Physics. 2002 (3): 023. arXiv:physics/0110060. Bibcode:2002JHEP...03..023D. doi:10.1088/1126-6708/2002/03/023. S2CID 15806354.
16. ^ Barrow, John D. (2002), The Constants of Nature; From Alpha to Omega - The Numbers that Encode the Deepest Secrets of the Universe, Pantheon Books, ISBN 978-0-375-42221-8 "[An] important lesson we learn from the way that pure numbers like ? define the World is what it really means for worlds to be different. The pure number we call the fine structure constant and denote by ? is a combination of the electron charge, e, the speed of light, c, and Planck's constant, h. At first we might be tempted to think that a world in which the speed of light was slower would be a different world. But this would be a mistake. If c, h, and e were all changed so that the values they have in metric (or any other) units were different when we looked them up in our tables of physical constants, but the value of ? remained the same, this new world would be observationally indistinguishable from our World. The only thing that counts in the definition of worlds are the values of the dimensionless constants of Nature. If all masses were doubled in value you cannot tell, because all the pure numbers defined by the ratios of any pair of masses are unchanged."
17. ^ Leslie, John (1998). Modern Cosmology & Philosophy. University of Michigan: Prometheus Books. ISBN 1573922501.
18. ^ The values are given in the so-called concise form, where the number in parentheses indicates the standard uncertainty referred to the least significant digits of the value.
19. ^ "2018 CODATA Value: elementary charge". The NIST Reference on Constants, Units, and Uncertainty. NIST. 20 May 2019. Retrieved .
20. ^ "2018 CODATA Value: Newtonian constant of gravitation". The NIST Reference on Constants, Units, and Uncertainty. NIST. 20 May 2019. Retrieved .
21. ^ "2018 CODATA Value: Planck constant". The NIST Reference on Constants, Units, and Uncertainty. NIST. 20 May 2019. Retrieved .
22. ^ "2018 CODATA Value: speed of light in vacuum". The NIST Reference on Constants, Units, and Uncertainty. NIST. 20 May 2019. Retrieved .
23. ^ "2018 CODATA Value: vacuum electric permittivity". The NIST Reference on Constants, Units, and Uncertainty. NIST. 20 May 2019. Retrieved .
24. ^ "2018 CODATA Value: vacuum magnetic permeability". The NIST Reference on Constants, Units, and Uncertainty. NIST. 20 May 2019. Retrieved .
25. ^ "2018 CODATA Value: electron mass in u". The NIST Reference on Constants, Units, and Uncertainty. NIST. 20 May 2019. Retrieved .
26. ^ "2018 CODATA Value: fine-structure constant". The NIST Reference on Constants, Units, and Uncertainty. NIST. 20 May 2019. Retrieved .
27. ^ "2018 CODATA Value: Josephson constant". The NIST Reference on Constants, Units, and Uncertainty. NIST. 20 May 2019. Retrieved .
28. ^ "2018 CODATA Value: Rydberg constant". The NIST Reference on Constants, Units, and Uncertainty. NIST. 20 May 2019. Retrieved .
29. ^ "2018 CODATA Value: von Klitzing constant". The NIST Reference on Constants, Units, and Uncertainty. NIST. 20 May 2019. Retrieved .
This article uses material from the Wikipedia page available here. It is released under the Creative Commons Attribution-Share-Alike License 3.0. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 11, "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": 0.9509019255638123, "perplexity": 1381.0643686486358}, "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-2022-27/segments/1656103941562.52/warc/CC-MAIN-20220701125452-20220701155452-00188.warc.gz"} |
https://www.physicsforums.com/threads/is-this-em-field-due-to-charged-particle-or-external.788813/ | Is this EM field *due* to charged particle or external?
1. Dec 22, 2014
"pi"mp
Suppose we're given the action
$S=-mc\int ds + \frac{q}{c}\int A_{\mu}(x) dx^{\mu}-\frac{1}{4c}\int d^{D}xF_{\mu \nu}F^{\mu \nu}$
The first two integrals are over the particle's worldline while the last is over spacetime. So I'm able to successfully vary the action with respect to the gauge potential to achieve Maxwell's equations but I'm confused as to what the gauge potential is in this case.
If the last term weren't there, this action would describe the physics of a charged particle in some external field $$A_{\mu}(x)$$. However, when we vary the action to yield Maxwell equation, we conclude that the current $$j^{\mu}(x)$$ *of the particle itself* is the source of the EM field. This is very confusing to me.
In the above action is the field the field caused by the particles motion or is it an external field? Or am I asking a silly question?
2. Dec 22, 2014
ShayanJ
The second term in the action describes the interaction between the external potential and the particle, and interaction is a two way road.
3. Dec 22, 2014
"pi"mp
Thanks a lot, but I'm still confused. Does that mean that when I write $$F_{\mu \nu}$$ in terms of the gauge potential in the third term, I need a totally different potential than the one that shows up in the second term?
4. Dec 22, 2014
ShayanJ
No. Its just that there is a particle in an external field which we don't care how it is created. Then the evolution of the field is not only the free evolution, but there is an effect coming from the charged particle.
The equations of motion for both the particle and field can be achieved by varying the above action. That's it.
Last edited by a moderator: Dec 22, 2014
5. Dec 23, 2014
vanhees71
The trouble, however, is that you run into serious problems. In principle, what you've written down is a closed equation for a single charged particle and the electromagnetic field. This includes the mutual interaction between the particle and the field, i.e., the motion of the particle in its own field. One solution, and that's the physical one, is the motion of the particle with a cosntant velocity and the corresponding Lorentz boosted Coulomb field. This is unproblematic.
The problems start when you include both, an external field, created from other charge-current densities than the particle and the particle's own field. The particle gets accelerated and this means its own field includes electromagnetic waves that carry energy away from the particle's kinetic energy. So the particle's motion is damped and this reacts of course back to the field. A fully self-consistent solution seems not to exist, because when naively writing down the equations you get pretty strange effects like the self-acceleration of a particle (which occurs as an obviously unphysical solution even if no external field is present) and singularities due to the point-like singularity in the field of the particle. Part of this is the appearance of an infinite contribution to the particle's mass due to it's own Coulomb field, which has infinite energy for a point-like particle. This infinity can partly removed by mass renormalization in this classical context, but there are still divergences left, which cannot be renormalized.
The only way out is to assume a finite extension of the particle and the assumption of forces keeping it together (the socalled Poincare stresses). There's no fully consistent theory of a classical point particle interacting with its own (radiation) field. This problem is partially solved by quantum electrodynamics, where all the singularities and divergences can be solved by renormalization of a few parameters (wave-function renormalization for the electron-positron and the photon fields, electron mass, and coupling constant), but that's only a solution in perturbation theory, i.e., there's no full solution of QED either. It's one of the big unsolved puzzles of modern physics!
6. Dec 23, 2014
HomogenousCow
I believe a recent book came out which successfully addressed this problem, can't remember the name.
Also, there is the other issue of the equation themselves not making sense as you have a delta function in a set of non-linear DEs.
7. Dec 23, 2014
WannabeNewton
I don't know of the book being referred to but the most easy to read, careful, and insightful treatment of the EM self-force problem is (in my opinion) Gralla et al: http://arxiv.org/abs/0905.2391
Similar Discussions: Is this EM field *due* to charged particle or external? | {"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": 0.8761456608772278, "perplexity": 350.68059207557786}, "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-2017-34/segments/1502886116921.7/warc/CC-MAIN-20170822221214-20170823001214-00470.warc.gz"} |
http://mathhelpforum.com/discrete-math/216816-binomial-probability.html | # Math Help - Binomial probability
1. ## Binomial probability
In the book Manual: A Guide to the Development and Use of the Myers-Brigg Type Indicator, by Myers and McCaulley, it was reported that based on a very large sample (2282 professors), approximately 45% of all university professors were extroverted. Suppose you have classes with six different professors; a) What is the probability thta all six are extroverts? b) What is the probability that none of your professors is an extrovert? c) What is the probability that at least two of your professors are extroverts? d) In a group of six professors selected at random, what is the expected number of extroverts? What is the standard deviation of the distribution?
2. ## Re: Binomial probability
Hey Mr Rayon.
Hint: A binomial distributions gives the probability for X things being of a certain type (Yes/No Black/White etc). Your probability is 0.45. Now calculate P(X=6) given N = 6. | {"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": 0.8689142465591431, "perplexity": 1187.3214033879349}, "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-2015-35/segments/1440645191214.61/warc/CC-MAIN-20150827031311-00026-ip-10-171-96-226.ec2.internal.warc.gz"} |
http://tex.stackexchange.com/questions/57101/highlight-a-column-in-equation-or-math-environment/57102 | # Highlight a column in equation or math environment
While explaining the idea product of two negative numbers I wanted to highlight a vertical part of the display to emphasize on the factors that are being multiplied by:
\documentclass[letterpaper]{article}
\usepackage{fullpage}
\usepackage{amsmath,amssymb,amsthm,enumitem}
\newcommand{\red}[1]{%
{\color{OrangeRed}#1}}
\begin{document}
\begin{equation*}
\left.\begin{array}{cc}
-2\cdot \red{2}=& -4 \\
-2\cdot \red{1}=& -2 \\
-2\cdot \red{0}=& 0
\end{array}\right\} \text{\small Product increases by 2 each time.}
\end{equation*}
\end{document}
This gives:
So I chose to use a color but then I had also wanted to achieve something like the following:
I am not interested in the yellow red bordered box. What am looking for is an arrow pointing on the red numbers with a vertical border around them as shown and the possibility of adding a node with text in the same format shown aligned with the word "Product".
Edit
Can I use this method to achieve this:
-
Your update really should be a separate question as it is not longer about highlighting a column (makes it easier for others to find it and benefit from it). But the answer is yes you can, see How to draw arrows between parts of an equation to show the Math Distributive Property (Multiplication)? for example. If that does not solve your issue, please post a new question. – Peter Grill May 24 '12 at 18:04
To accomplish this task you can make use of the tikzmark macro. This solution allows to get:
The code:
\documentclass[letterpaper]{article}
\usepackage{fullpage}
\usepackage{amsmath,amssymb,amsthm,enumitem}
\usepackage{xcolor}
\newcommand{\red}[1]{%
{\color{red}#1}}
\usepackage{tikz}
% to change colors
\newcommand{\fillcol}{white}
\newcommand{\bordercol}{red}
%% code by Andrew Stacey
% http://tex.stackexchange.com/questions/51582/background-coloring-with-overlay-specification-in-algorithm2e-beamer-package#51582
\makeatletter
\tikzset{%
remember picture with id/.style={%
remember picture,
overlay,
draw=\bordercol,
save picture id=#1,
},
save picture id/.code={%
\edef\pgf@temp{#1}%
\immediate\write\pgfutil@auxout{%
\noexpand\savepointas{\pgf@temp}{\pgfpictureid}}%
},
if picture id/.code args={#1#2#3}{%
\@ifundefined{save@pt@#1}{%
\pgfkeysalso{#3}%
}{
\pgfkeysalso{#2}%
}
}
}
\def\savepointas#1#2{%
\expandafter\gdef\csname save@pt@#1\endcsname{#2}%
}
\def\tmk@labeldef#1,#2\@nil{%
\def\tmk@label{#1}%
\def\tmk@def{#2}%
}
\tikzdeclarecoordinatesystem{pic}{%
\pgfutil@in@,{#1}%
\ifpgfutil@in@%
\tmk@labeldef#1\@nil
\else
\tmk@labeldef#1,(0pt,0pt)\@nil
\fi
\@ifundefined{save@pt@\tmk@label}{%
\tikz@scan@one@point\pgfutil@firstofone\tmk@def
}{%
\pgfsys@getposition{\csname save@pt@\tmk@label\endcsname}\save@orig@pic%
\pgfsys@getposition{\pgfpictureid}\save@this@pic%
\pgf@process{\pgfpointorigin\save@this@pic}%
\pgf@xa=\pgf@x
\pgf@ya=\pgf@y
\pgf@process{\pgfpointorigin\save@orig@pic}%
}%
}
\makeatother
\newcommand{\tikzmarkin}[1]{%
\tikz[remember picture with id=#1]
\draw[line width=1pt,rectangle,rounded corners,fill=\fillcol]
(pic cs:#1) ++(0.065,-0.15) rectangle (-0.05,0.32)
;}
\newcommand\tikzmarkend[2][]{%
\tikz[remember picture with id=#2] #1;}
\begin{document}
\begin{equation*}
\left.\begin{array}{cc}
-2\cdot \tikzmarkin{a}\red{2}=& -4 \\
-2\cdot \red{1}=& -2 \\
-2\cdot \red{0}\tikzmarkend{a}=& 0
\end{array}\right\} \text{\small Product increases by 2 each time.}
\end{equation*}
\end{document}
Notice that you need to compile twice.
Explanation
Inside the tikzset there is the definition of the style of the picture to be drawn (remember picture with id) then the definition and the writing on the the .aux of the id and position of the mark. Anyway, to get a better explanation, you can refer to http://tex.stackexchange.com/a/50054/13304. Finally, the commands that you need to use inside your document, \tikzmarkin and \tikzmarkend are defined.
EDIT: insertion of the annotation
For this purpose I adopted the same trick of Mark a pseudocode block and insert comments near it: the insertion of an anchor inside the \tikzmarkin macro to subsequently used it as reference to insert the annotation. What changes in the previous MWE?
In the preamble you should add:
\usetikzlibrary{calc} % <= needed for some computations
\usepackage{lipsum} % <= needed to insert some text later
The \tikzmarkin should be improved as:
\newcommand{\tikzmarkin}[1]{%
\tikz[remember picture with id=#1]
\draw[line width=1pt,rectangle,rounded corners,fill=\fillcol]
(pic cs:#1) ++(0.065,-0.15) rectangle (-0.05,0.32) node [anchor=base] (#1){}
;}
where the relevant new part is just node [anchor=base] (#1){}.
Finally, the document:
\begin{document}
\begin{equation*}
\left.\begin{array}{cc}
-2\cdot \tikzmarkin{a}\red{2}=& -4 \\
-2\cdot \red{1}=& -2 \\
-2\cdot \red{0}\tikzmarkend{a}=& 0
\end{array}\right\} \text{\small Product increases by 2 each time.}
\end{equation*}
% To insert the annotation
\begin{tikzpicture}[remember picture,overlay]
\coordinate (aa) at ($(a)+(1.825,-1.65)$); % <= adjust this parameter to move the position of the annotation
\node[align=left,right] at (aa) {\small{Annotation}};
\path[-latex,red,draw] (aa) -| ($(a)+(0.15,-1.3)$);
\end{tikzpicture}
\linebreak
\lipsum[1]
\end{document}
The last part is the one needed to insert the annotation: first I defined a coordinate where to insert it based on the anchor previously saved (to accomplish this, it is important that the tikzpicture has as options remember picture,overlay). The last command:
\path[-latex,red,draw] (aa) -| ($(a)+(0.15,-1.3)$);
draws the arrow: the starting point is the coordinate of the annotation and the final point should be computed knowing that the reference anchor a is placed on the top left of the red rectangle.
One final remark: I think should be better to insert a \linebreak after the tikzpicture to avoid that the annotation will be placed too nearly the subsequent text.
One small suggestion: because you're using remember picture with id in the \tikzmarkin, you actually are remembering the same id twice which probably isn't a good habit to get into (doesn't affect things here, but could in other situations). One option is for the \tikzmarkin simply to be remember picture,overlay since you never need to refer to it again. – Loop Space May 24 '12 at 7:59
Thanks for the suggestion! I've verified it by changing simply the \tikzmarkin code into: \newcommand{\tikzmarkin}[1]{% \tikz[remember picture,overlay] \draw[line width=1pt,rectangle,rounded corners,fill=\fillcol, draw=\bordercol] (pic cs:#1) ++(0.065,-0.15) rectangle (-0.05,0.32) node [anchor=base] (#1){} ;} and the final output is still equal. What changes is the .aux file because the {a} is unique. – Claudio Fiandrino May 24 '12 at 8:06 | {"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": 0.8421849608421326, "perplexity": 3058.2360666505624}, "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/1410657119220.53/warc/CC-MAIN-20140914011159-00030-ip-10-196-40-205.us-west-1.compute.internal.warc.gz"} |
http://english.stackexchange.com/questions/48750/is-there-a-term-for-authorial-name-expurgation/48755 | # Is there a term for authorial name expurgation?
In 19th century English texts, but also seen elsewhere (e.g. Heller's Catch-22) there is a practice of omitting part of a name such as:
I was going to visit Mrs. P___, but decided to remain at home.
which is intended to shroud the name of the person. You may also find bowdlerization of a similar form in the same texts:
I'd have been successful if not for the d____d rain.
where the author (or publisher) would prefer not to use "damned". Is there a term for this practice?
Is there a modern (e.g. TeX, Unicode, HTML etc.) way to represent this typographic convention? The best I can find is the Unicode/ASCII codepoint 0x5f which has the unfortunate habit of being rendered as a continuous line "d____d" where older practice would allow for a small inter-glyph space so you could see that four letters had been omitted in "d····d".
-
It doesn't answer your specific question, but there was a discussion of the subject here – Colin Fine Nov 18 '11 at 12:07
Though the title question here is about names, there is an older question about redacting dates in older literature that answers this more accurately. – Mitch Nov 18 '11 at 15:18
It sounds like you want a typesetter's term, not a general "what do I call this practice". If that's the case, the older question is not a duplicate. However, you might want to edit your question to make it clear exactly what type of word you're looking for. – Marthaª Nov 21 '11 at 15:36
@Marthaª I was sort of looking for both as I hoped that one would guide me to the other. The actual question was "I am marking up an antique text for semantic analysis, and I cannot find a conventional way to refer to this practice". An answer to either of the two questions you parsed would have helped; alas, it seems as there is no general answer. – msw Dec 2 '11 at 17:27
Eclipsis, "A line or dash used to show that text has been omitted", is the most precise term I know of for that practice, but it addresses the mechanism, rather than the meaning or rationale.
Some less-specific, less-related, or less-wordworthy terms: elision, omission, concealment, tokenizing, anonymizing.
Update Re "Is there a modern ... way to represent this typographic convention?", in LaTex there seem to be only three non-math dashes, i.e. -, --, ---, the latter two forming en- and em-dashes respectively. Perhaps try tex.stackexchange for more typography help. For lightweight expurgation, consider \cdots for centered dots. For lists of Unicode and HTML dashes, see jkorpela's dashes article on cs.tut.fi; for example, U+2012 or ‒ is a figure dash, ‒.
-
+1 Interestingly the OED opines that eclipsis in sense(2) is "perhaps confused with ellipsis". Thanks for the missing word, correct but quite obsolete as you've noticed. I'm still holding out for the typesetter's term (as it seems like there's got to be one). – msw Nov 18 '11 at 5:59
Not only OED, many other sources tie the two words together. – Unreason Nov 18 '11 at 12:54
I would go for ellipses
Those dots that come in the middle of a quotation to indicate something omitted are called an “ellipsis” (plural “ellipses”): “Tex told Sam to get the . . . cow out of the bunk house.” Here Tex’s language has been censored, but you are more likely to have a use for ellipses when quoting some source in a paper: “Ishmael remarks at the beginning of Moby Dick, ‘some years ago . . . I thought I would sail about a little’ —a very understated way to begin a novel of high adventure.”
Ellipses is a general term also known or spelled as: elipsis, elleipsis, eclipsis.
The dictionary entry mentions:
a set of dots (...) indicating an ellipsis
Wikipedia article also specifically mentions:
Bringhurst writes that a full space between each dot is "another Victorian eccentricity."
Eclipsis is probably a misspelling, but it has a very nice conotation of hiding, where ellipses is simply omission. Both terms are normally defined for a word or longer part of text and there is hardly mention of omitting parts of the word or individual letters.
Wiktionary entry for eclipsis has a meaning of
A line or dash used to show that text has been omitted
which seems to be the exact thing that you are after.
However some sources define ellipses as
A set of three dots, periods, or dashes in a row
Some list them as synonyms.
Given all of the above, I would say that eclipsis and ellipses are terms which are not as precise as you might like. Therefore, take you pick.
Keep in mind that ellipses is used much more often than eclipsis, so you will definitively have to establish the context and explain what you mean by it. However, you will probably have to do the same for ellipses because it would most likely be taken to mean … - a single glyph made from three dots.
- | {"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": 0.8008489608764648, "perplexity": 1857.3642402479447}, "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-2015-11/segments/1424936461988.0/warc/CC-MAIN-20150226074101-00234-ip-10-28-5-156.ec2.internal.warc.gz"} |
https://www.physicsforums.com/threads/simple-force-vector-question.533052/ | # Homework Help: Simple force vector question
1. Sep 23, 2011
### Hockey07
1. The problem statement, all variables and given/known data
This was an exam question that one of my friends recently had.
Two horizontal forces act on a 32.5kg object. The first force has a magnitude of 210N and points in the direction 40.0 degrees East of North. The second force has a magnitude of 350N and points in the direction 20.0 degrees North of West.
What is the magnitude of the acceleration of the object?
2. Relevant equations
$\Sigma F = ma$
3. The attempt at a solution
We know how to solve this problem, but the correct answer on the exam was 10.5 $\frac{m}{s^{2}}$
The issue with this problem is that we believe it's very vaguely worded.
We solved the problem by doing:
$\Sigma Fx = -350cos(20) + 210sin(40) = 32.5a_{x}$
$\Sigma Fy = 350sin(20) + 210cos(40) - 32.5(9.81) = 32.5a_{y}$
Then the magnitude of the acceleration would be the square root of the sum of the squares of the acceleration components.
Neglecting gravity and using $\Sigma Fy = 350sin(20) + 210cos(40) = 32.5a_{y}$ yields the correct answer on the exam. Neither of us thought to assume that gravity just disappears (because why would it?)
What do you think - vague question or not?
2. Sep 23, 2011
### PeterO
Perhaps you both though the object was "floating" in space. The questioner, and I, assumed the mass was on a flat, frictionless surface.
3. Sep 23, 2011
### Xyius
The problem doesn't explicitly say where the object is but I also assumed it was on a flat friction-less surface. Gravity would, in this case, be cancelled out by the normal force of the surface. This is one of those problems where you should clarify with the teacher as you are taking the tests to rule out any confusion.
4. Sep 24, 2011
### Hockey07
I've taken both Physics and Dynamics and I don't think I've ever had to assume that an object is sitting on a surface. It's always been stated in the question or it's obvious (like a moving car, which clearly is on a road). This question immediately struck me as being a non-obvious situation, and I initially solved it incorrectly.
Additionally, the question explicitly states that there are two horizontal forces, which is not true. There are horizontal components, but the forces are not pointing in the horizontal direction.
5. Sep 24, 2011
### PeterO
Sorry, But directions like North, East and West apply to a horizontal plane. I can't see that you want it clearer than that.
Perhaps you are thinking North is Up, South is Down, East is Right and West is Left - like as if you were looking at a map pinned to a wall?? And thus that East and West are the only horizontal directions?????
6. Sep 24, 2011
### Hockey07
hor·i·zon·tal [hawr-uh-zon-tl, hor-]
1. at right angles to the vertical; parallel to level ground.
2. flat or level: a horizontal position.
http://dictionary.reference.com/browse/horizontal
I understand the directions just fine. I'm not asking what direction they're pointing in. I'm telling you that the wording is flawed. Read below.
Problem statement: "Two horizontal forces act..."
Me: "They are not strictly horizontal forces."
These forces are not at a right angle to the vertical, nor are they parallel to the ground. They don't fit the definition of horizontal. They are forces causing horizontal motion, and they do have horizontal components. I completely understand that! The problem statement says the forces themselves are horizontal... as in they have NO vertical component (hence the definition horizontal = at right angle to the vertical). They clearly have components in both the vertical and horizontal directions (and no component in the direction perpendicular to that plane since it's a 2D problem).
Also, by not explicitly stating that it's on a flat frictionless surface, you have to assume something that completely changes the problem. There's no indication that it would be resting on any surface. You have to guess that. It's not a fundamental assumption in order to solve the problem (like a fluid mechanics problem where you may assume a fluid is incompressible).
The wording is what bothers me. I know how to do the problem correctly, but the wording just is not clear.
Last edited: Sep 24, 2011
7. Sep 24, 2011
### PeterO
If the questioner says "Two horizontal forces act ..." and you think that says "They are not strictly horizontal forces .." I am not surprised you found the problem confusing.
I still think you are confusing North with vertical, and only if I assume you think that can I find any logic in what you are saying.
8. Sep 24, 2011
### SammyS
Staff Emeritus
A solution in which has gravitational force acting in the Southerly direction seems to be indefensible. How can you possibly argue that the problem as presented is that ambiguous?
9. Sep 25, 2011
### PeterO
I am not the one finding any ambiguity! | {"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": 0.8622651100158691, "perplexity": 579.7163141823305}, "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-2018-34/segments/1534221217006.78/warc/CC-MAIN-20180820195652-20180820215652-00652.warc.gz"} |
https://www.physicsforums.com/threads/combining-proportionality-statements.429065/ | # Homework Help: Combining Proportionality Statements
1. Sep 14, 2010
### qswdefrg
1. The problem statement, all variables and given/known data
Combine the proportionality statements for R and d and R and l to form one statement relating R, d, and l.
2. Relevant equations
I already know that R∝(1/d squared) and R∝l
3. The attempt at a solution
I thought it was R∝ (l/d squared), but that seems too easy lol.
Last edited: Sep 14, 2010
2. Sep 14, 2010
### collinsmark
Hello qswdefrg,
Welcome to Physics Forums!
If by that you mean Rl/(d2), then it certainly sounds reasonable to me.
3. Sep 16, 2010
### qswdefrg
So it turns out I was right after all! And I finished the assignment.
...
I just realized now that this wasn't a physics problem at all; just pure common sense ;___; | {"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": 0.9348977208137512, "perplexity": 3582.0394540828183}, "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/1529267860570.57/warc/CC-MAIN-20180618144750-20180618164750-00022.warc.gz"} |
https://socratic.org/questions/how-do-you-use-the-product-rule-to-find-the-derivative-of-y-x-2-x-2-1-5 | Calculus
Topics
# How do you use the Product Rule to find the derivative of y = x^2 (x^2 + 1)^5?
Aug 15, 2015
$\frac{\mathrm{dy}}{\mathrm{dx}} = 2 x {\left({x}^{2} + 1\right)}^{4} \left(6 {x}^{2} + 1\right)$
#### Explanation:
The product rule states:
$f = g h \implies f ' = g ' h + g h '$
For this function we have:
$f \left(x\right) = y , g \left(x\right) = {x}^{2} , h \left(x\right) = {\left({x}^{2} + 1\right)}^{5}$
$g ' \left(x\right) = 2 x , h ' \left(x\right) = 10 x {\left({x}^{2} + 1\right)}^{4}$
Therefore by the charnel we have:
$\frac{\mathrm{dy}}{\mathrm{dx}} = g ' h + g h ' = 2 x {\left({x}^{2} + 1\right)}^{5} + 10 {x}^{3} {\left({x}^{2} + 1\right)}^{4} = 2 x {\left({x}^{2} + 1\right)}^{4} \left(6 {x}^{2} + 1\right)$
##### Impact of this question
122 views around the world
You can reuse 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": 5, "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": 0.9647855758666992, "perplexity": 2443.5247496376614}, "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-2019-26/segments/1560627998291.9/warc/CC-MAIN-20190616182800-20190616204800-00118.warc.gz"} |
http://math.stackexchange.com/questions/1631554/give-a-combinatorial-proof-for-a-multiset-identity | # Give a combinatorial proof for a multiset identity
I'm asked to give a combinatorial proof of the following, $\binom{\binom n2}{2}$ = 3$\binom{n}{4}$ + n$\binom{n-1}{2}$.
I know $\binom{n}{k}$ = $\frac{n!}{k!(n-k)!}$ and $(\binom{n}{k}) = \binom{n+k-1}{k}$ but I'm at a loss as to what to do with the $\binom{\binom n2}{2}$
Can someone point me in the right direction as to how to proceed with writing a combinatorial proof for this identity?
-
LHS can be interpreted as counting the ways in which you can create two different pairs of two different elements, taken from a set of $n$ toys; while the pairs must be different, they need not be disjoint: they might have a toy in common.
RHS counts the same amount of pairs in a different way: you can create two kinds of such couple of pairs: disjoint ones $(a,b)(c,d)$ and overlapping ones $(a,b)(a,c)$. Remember that every pair must be different and the two couples must be different. The first kind of pairs can be chosen in the following way: first chose $4$ toys in $\binom{n}{4}$ ways, then chose one of the $3$ ways to partition the $4$ into two pairs (to see that there are $3$ ways, focus on one of the four: it must be paired with one of the three remaining ones, and that determines the partition). The second kind of pairs can be chosen by first choosing the toy $a$ that will be common to both pairs in $n$ ways, then chose $2$ toys from the remaining $n-1$ to be $\{b,c\}$; the order does not matter here, so the final choice can be made in $\binom{n-1}{2}$ ways.
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Everyone is welcome to edit my English for a better explaination! :) – Maffred Jan 29 at 5:25
Consider a graph $G=(V,E)$ such that $|V|=n$ and $|E|={n \choose 2}$ (i.e. any two vertices are connected). Than the LHS of your identity is the number of unordered pairs of edges. Any such pair is either determined by four vertices or by three. In the first case we first choose 4 vertices and then connect them in any of 3 ways which gives us altogether $3{n \choose 4}$ combinations. In the second case we first select the vertex which belongs to both edges and later we select the other two vertices from remaining $n-1$ vertices which gives us $n{n-1 \choose 2}$ possibilities. Summing it up we get the RHS.
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Rhs using simple combinatorics formula simplifies to $\frac{(n)(n-1)^2(n-2)}{8}$ now lhs ${n\choose 2}=\frac{(n)(n-1)}{2}$ so we can write lhs as $\frac{(n^2-n)!}{(2!).(\frac{n^2-n-4}{2})}$ now treat $n^2-n=y$ so it becomes $\frac{(n^2-n)(n^2-n-2)}{8}=\frac{(n)(n-1)^2(n-2)}{8}=Rhs$
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This is not a combinatorial proof! – Maffred Jan 29 at 5:27
No you just explained it so i have done algebraic proof – Archis Welankar Jan 29 at 5:28 | {"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": 0.8467136025428772, "perplexity": 130.76004023544905}, "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-2016-30/segments/1469257824757.8/warc/CC-MAIN-20160723071024-00230-ip-10-185-27-174.ec2.internal.warc.gz"} |
http://code.grammalecte.net:8080/hexdump?name=0d0c9678860cebf0b581dc644ef840f34718f89aa4f5801243706fea6ab912a2 | # Grammalecte Hex Artifact Content
## Artifact 0d0c9678860cebf0b581dc644ef840f34718f89aa4f5801243706fea6ab912a2:
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http://physics.stackexchange.com/questions/39084/gauge-fixing-and-equations-of-motion/39097 | # Gauge fixing and equations of motion
Consider an action that is gauge invariant. Do we obtain the same information from the following:
1. Find the equations of motion, and then fix the gauge?
2. Fix the gauge in the action, and then find the equations of motion?
-
I) Here we will assume that we ultimately want to consider the full quantum theory, usually written in terms of a gauge-fixed path integral
$$Z~=~\int \!{\cal D}\phi~ \exp\left(\frac{i}{\hbar}S_{\rm gf}[\phi]\right)$$
rather than just the classical action and the corresponding classical equations of motion (with or without gauge-fixing terms). If the gauge orbits have infinite volume (as is often the case), then we need to gauge-fix the path integral.
Of course, if we by brute force eliminate a field in the action via a gauge-fixing condition, then we can no longer carry out a variation of the action with respect to that field, and we would have lost information.
II) However, here we will only consider a 'softer' way to impose the gauge-fixing conditions via Lagrange multipliers, which may appear linearly or quadratically in the gauge-fixed action. The linear case leads directly to delta functions in the path integral, which impose the gauge-fixing conditions; while the quadratic case leads to Gaussian terms in the path integral, which suppress (but do not completely forbid) field configurations that violate the gauge-fixing condition. (Nevertheless, in a certain scaling limit, the Gaussian factors become delta functions.)
Together with the original fields, the (non-propagating, auxiliary) Lagrange multipliers are part of the fields $\phi$ in the path integral that is integrated over. In particular, the gauge-fixed action $S_{\rm gf}[\phi]$ can be varied with respect to these Lagrange multiplier fields as well.
These 'softly imposed' gauge-fixing conditions still do affect the variation of the action (as opposed to not imposing the gauge-fixing conditions). However, a more relevant question is:
Do the gauge-invariant physical observables of the theory depend on the specific gauge-fixing condition (e.g. Lorenz gauge, Coulomb gauge, etc)?
The answer is No, i.e. within the class of consistent gauge-fixing terms in the action, the specific form of the gauge-fixing terms in the corresponding equations of motion has no physical consequences.
III) For more general gauge theories, the equations of motion are e.g. not gauge-invariant, and it is better to encode the gauge symmetry via a (generalized) fermionic nilpotent BRST symmetry $\delta$ that squares to zero
$$\delta^2~=~0,$$
and preserves the original action
$$\delta S_0~=~0.$$
The gauge-fixed action $$S_{\rm gf}=S_0+\delta\psi$$ is the original action $S_0$ plus a BRST-exact term $\delta\psi$ that depends on the so-called gauge-fixing fermion $\psi$, which encodes the gauge-fixing condition.
A physical observable $F=F[\phi]$ in the theory is required to be BRST-closed
$$\delta F~=~0.$$
Formally, if the path integral measure is BRST-invariant, one may show that the correlation function for a physical observable
$$\langle F[\phi] \rangle ~=~ \frac{ \int \!{\cal D}\phi~ F[\phi]\exp\left(\frac{i}{\hbar}S_{\rm gf}[\phi]\right)} {\int \!{\cal D}\phi~ \exp\left(\frac{i}{\hbar}S_{\rm gf}[\phi]\right)}$$
is independent of the gauge-fixing fermion $\psi$. (Note however, that the gauge-fixing fermion $\psi$ is required to satisfy certain rank conditions, and it can e.g. not be chosen to be identically zero.)
IV) Finally, let us mention, that an even bigger class of Lagrangian gauge theories can be treated with the help of the Batalin-Vilkovisky (BV) formalism.
-
No, it is not always consistent to first fix the gauge before deriving equations of motion. Consider electromagnetism coupled to matter. One can perform a gauge transformation to set $A_0=0$. However, if this is done in the action before deriving the equations of motion, then one will miss out on the $A_0$ equation of motion which guarantees the conservation of electric charge.
When attempting to derive sufficiently symmetric solutions, it is sometimes possible to begin by choosing an appropriate ansatz, even at the level of the action. But this is something of an art, and is not guaranteed to be consistent with the full equations of motion evaluated on that same ansatz.
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https://relate.cs.illinois.edu/course/cs450-s18/file-version/19064d0244a21077423d88219bc4225e10e2500b/demos/upload/12-fft/Fast%20Fourier%20Transform.html | # Fast Fourier Transform
The discrete Fourier transformation interpolates a set of values in the frequency basis given by trigonometric functions.
## DFT Matrix
Definte the $n$th root of unity as $\omega_{(n)} = e^{-2\pi i/n}$ so that ${\omega_{(n)}}^n=1$.
In [25]:
import numpy as np
def omega(n):
return np.exp(-2.*np.pi*1.j/n)
print(omega(2),omega(3),omega(4))
print(omega(2)**2,omega(3)**3,omega(4)**4)
(-1-1.22464679915e-16j) (-0.5-0.866025403784j) (6.12323399574e-17-1j)
(1+2.44929359829e-16j) (1+6.10622663544e-16j) (1+2.44929359829e-16j)
Lets plot $\omega_{(2)},\omega_{(3)},\ldots,\omega_{(n)}$ on the complex plane, highlighting in red $\omega_{(2)},\omega_{(4)},\omega_{(8)},\ldots,\omega_{(n)}$.
In [2]:
from matplotlib import pyplot as pt
n=16
omegas = np.zeros(n-1,dtype=complex)
expomegas = np.zeros(4,dtype=complex)
for i in np.arange(0,n-1):
omegas[i] = omega(i+2)
if 2**i <= n/2:
expomegas[i] = omega(2**(i+1))
print(omegas)
pt.scatter(np.real(omegas),-np.imag(omegas),color='b')
pt.scatter(np.real(expomegas),-np.imag(expomegas),color='r')
ax=pt.gca()
pt.axis('scaled')
pt.show()
[ -1.00000000e+00 -1.22464680e-16j -5.00000000e-01 -8.66025404e-01j
6.12323400e-17 -1.00000000e+00j 3.09016994e-01 -9.51056516e-01j
5.00000000e-01 -8.66025404e-01j 6.23489802e-01 -7.81831482e-01j
7.07106781e-01 -7.07106781e-01j 7.66044443e-01 -6.42787610e-01j
8.09016994e-01 -5.87785252e-01j 8.41253533e-01 -5.40640817e-01j
8.66025404e-01 -5.00000000e-01j 8.85456026e-01 -4.64723172e-01j
9.00968868e-01 -4.33883739e-01j 9.13545458e-01 -4.06736643e-01j
9.23879533e-01 -3.82683432e-01j]
The discrete Fourier transform is defined by a matrix-vector product with a DFT matrix $\boldsymbol D \in \mathbb{R}^{n\times n}$ with $d_{jk}={\omega_{(n)}}^{jk}$.
In [27]:
def DFT(n):
v = np.linspace(0,n-1,n) # v=[0,1,...,n-1]
return omega(n)**np.outer(v,v)
np.asarray(np.real(DFT(4)),dtype=int)+1j*np.asarray(np.imag(DFT(4)),dtype=int)
Out[27]:
array([[ 1.+0.j, 1.+0.j, 1.+0.j, 1.+0.j],
[ 1.+0.j, 0.-1.j, -1.+0.j, 0.+1.j],
[ 1.+0.j, -1.+0.j, 1.+0.j, -1.+0.j],
[ 1.+0.j, 0.+1.j, -1.+0.j, 0.-1.j]])
The DFT matrix is symmetric (but not Hermitian) and with appropriate scaling, unitary i.e. $n\boldsymbol D^{-1} = \boldsymbol D^{H}$.
In [28]:
import numpy.linalg as la
n=27
la.norm(n*np.eye(n) - DFT(n) @ DFT(n).conj())
Out[28]:
2.2721682842512306e-12
## FFT Algorithm
The fast Fourier transform exploits the special structure of the DFT matrix, computing the disrete Fourier transform of $\boldsymbol v$, given by $\boldsymbol u = \boldsymbol D \boldsymbol v$, in $O(n\log n)$ operations.
In [5]:
def fft(v):
n = v.size
if n is 1:
return v
u = fft(v[::2]) # compute FFT of [v_0,v_2,...v_{n-1}]
w = fft(v[1::2]) # compute FFT of [v_1,v_3,...v_n]
# scale w by twiddle factors [omega_(n)^0,...omega_(n)^(n/2-1)]
t = np.asarray([omega(n)**i for i in range(n//2)])
z = w*t
return np.concatenate([u+z,u-z])
n = 16
v = np.random.random(n)+1.j*np.random.random(n)
la.norm(fft(v)-DFT(n)@v)
Out[5]:
3.8305731252303524e-14
Two recursive calls on vectors of half the size are made, and $O(n)$ additions/products are done to apply the twiddle factors and add $z$ to $u$, for a cost of $$T(n)=2T(n/2)+O(n)=O(n \log n).$$
## Application to the 1D Poisson Equation
The solutions to the 1D Poisson equation are waves that correspond to Fourier modes, and the Fourier transform provides solutions to it directly when applied to the continuous equation or in the discretized form. From the linear algebra perspective, the DFT matrix is related to the eigenvectors of the tridiagonal differencing matrix (also linear stiffness matrix) $\boldsymbol T$.
In [6]:
n=7
T = 2*np.eye(n)-np.diag(np.ones(n-1),1)-np.diag(np.ones(n-1),-1)
T
Out[6]:
array([[ 2., -1., 0., 0., 0., 0., 0.],
[-1., 2., -1., 0., 0., 0., 0.],
[ 0., -1., 2., -1., 0., 0., 0.],
[ 0., 0., -1., 2., -1., 0., 0.],
[ 0., 0., 0., -1., 2., -1., 0.],
[ 0., 0., 0., 0., -1., 2., -1.],
[ 0., 0., 0., 0., 0., -1., 2.]])
The eigenvectors of $\boldsymbol T$ are $\boldsymbol X$ with eigenvalues $\boldsymbol d$ where $$x_{jk} = \sqrt{\frac{2}{n+1}}\sin(jk\pi/(n+1)), \quad d_{k} = 2(1-\cos(\pi k/(n+1)).$$ The eigenvector matrix $\boldsymbol X$ is the imaginary part of a submatrix of the $2(n+1)$-dimensional DFT matrix.
In [7]:
X = - np.sqrt(2/(n+1.)) * np.imag(DFT(2*(n+1))[1:n+1,1:n+1])
d = 2.*(1.-np.cos(np.pi*np.arange(1,n+1)/(n+1)))
#check that we indeed have the eigenvalue decomposition of T
la.norm(T@[email protected](d))
Out[7]:
2.4218421576539674e-15
Thus we can apply $\boldsymbol X$, which is called the discrete sine transform via FFT.
In [8]:
def fast_DST(v):
# setup a padded vector [0,v,0,...,0] of dimensions 2(n+1)
w = np.concatenate([np.asarray([0.j]),v,0.j*np.zeros(v.size+1)])
u = fft(w) # compute FFT of padded vector
z = u-u.conj() # extract only imaginary part
# return rescaled subvector
return (-1.j*np.sqrt(1./(2.*(v.size+1.))))*z[1:1+v.size]
n = 7 # so that DFT dimension is 2(n+1)=16
X = - np.sqrt(2./(n+1.))*np.sin(np.pi*np.outer(np.arange(1,n+1),np.arange(1,n+1))/(n+1.))
v = np.random.random(n)
la.norm( X @ v - fast_DST(v))
Out[8]:
1.0024720669331688e-15
Now, to solve the linear system $$\boldsymbol T \boldsymbol x = \boldsymbol b$$ it suffices to compute $$\boldsymbol x = \boldsymbol X \text{diag}(\boldsymbol d)^{-1} \boldsymbol X^{-1} \boldsymbol b,$$ and since the sine transform is orthogonal, we simply have $$\boldsymbol x = \boldsymbol X \text{diag}(\boldsymbol d)^{-1} \boldsymbol X^{T} \boldsymbol b.$$
In [9]:
n = 7
b = np.random.random(n)
T = 2*np.eye(n)-np.diag(np.ones(n-1),1)-np.diag(np.ones(n-1),-1)
x = la.solve(T,b)
d = 2.*(1.-np.cos(np.pi*np.arange(1,n+1)/(n+1)))
X = - np.sqrt(2./(n+1.))*np.sin(np.pi*np.outer(np.arange(1,n+1),np.arange(1,n+1))/(n+1.))
la.norm(x - fast_DST(np.diag(1./d) @ X.T @ b))
Out[9]:
3.820199830714189e-15
## Application to the 2D Poisson Equation
Lets now consider the discretized 2D Poisson PDE, where the linear system takes on a Kronecker-product form.
In [11]:
n=7
T = 2*np.eye(n)-np.diag(np.ones(n-1),1)-np.diag(np.ones(n-1),-1)
A = np.kron(np.eye(n),T)+np.kron(T,np.eye(n))
pt.spy(A)
Out[11]:
<matplotlib.image.AxesImage at 0x7f053eda1518>
The 1D finite differencing matrix $\boldsymbol T$, gives us the following equations for the unknowns $\boldsymbol U\in\mathbb{R}^{m\times m}$ for an $m\times m$ spatial mesh, $$\boldsymbol T \boldsymbol U + \boldsymbol U \boldsymbol T= \boldsymbol F,$$ where we've absorbed the discretization width $h$ into $\boldsymbol F$.
In [20]:
F = np.random.random((n,n))
U = la.solve(A,F.reshape(n*n)).reshape((n,n))
la.norm(T@U+U@T-F)
Out[20]:
5.9061360594505111e-15
Using the eigenvalue decomposition of $\boldsymbol T=\boldsymbol X \text{diag}(\boldsymbol d) \boldsymbol X^{-1}$, which we know analytically, we can write the frequency space form of the linear system as $$\text{diag}(\boldsymbol d) \boldsymbol X^{-1} \boldsymbol U \boldsymbol X + \boldsymbol X^{-1} \boldsymbol U \boldsymbol X\text{diag}(\boldsymbol d) = \boldsymbol X^{-1} \boldsymbol F \boldsymbol X,$$ so we compute the 2D sine transform of $\boldsymbol F$ and solve a diagonal linear system for $\boldsymbol V=\boldsymbol X^{-1} \boldsymbol U \boldsymbol X$, which is the 2D sine transform of the solution.
In [24]:
X = - np.sqrt(2/(n+1.)) * np.imag(DFT(2*(n+1))[1:n+1,1:n+1])
d = 2.*(1.-np.cos(np.pi*np.arange(1,n+1)/(n+1)))
# transform F
sF = X.T @ F @ X
# perform pointwise product to solve for U
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https://www.lessonplanet.com/teachers/adding-and-subtracting-fractions-with-the-same-denominator | # Adding And Subtracting Fractions With The Same Denominator
In this adding and subtracting fractions with the same denominator worksheet, students add/subtract twelve equations and then simplify each answer.
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http://export.arxiv.org/abs/2211.13485?context=cs.CR | cs.CR
(what is this?)
# Title: A doubly-infinite family of 0-APN monomials
Abstract: We consider an infinite family of exponents $e(l,k)$ with two parameters, $l$ and $k$, and derive sufficient conditions for $e(l,k)$ to be 0-APN over $\mathbb{F}_{2^n}$. These conditions allow us to generate, for each choice of $l$ and $k$, an infinite list of dimensions $n$ where $x^{e(l,k)}$ is 0-APN much more efficiently than in general. We observe that the Gold and Inverse exponents, as well as the inverses of the Gold exponents can be expressed in the form $e(l,k)$ for suitable $l$ and $k$. We characterize all cases in which $e(l,k)$ can be cyclotomic equivalent to a representative from the Gold, Kasami, Welch, Niho, and Inverse families of exponents. We characterize when $e(l,k)$ can lie in the same cyclotomic coset as the Dobbertin exponent (without considering inverses) and provide computational data showing that the Dobbertin inverse is never equivalent to $e(l,k)$. We computationally test the APN-ness of $e(l,k)$ for small values of $l$ and $k$ over $\mathbb{F}_{2^n}$ for $n \le 100$, and sketch the limits to which such tests can be performed using currently available technology. We conclude that there are no APN monomials among the tested functions, outside of known classes.
Subjects: Number Theory (math.NT); Cryptography and Security (cs.CR) Cite as: arXiv:2211.13485 [math.NT] (or arXiv:2211.13485v1 [math.NT] for this version)
## Submission history
From: Nikolay Kaleyski [view email]
[v1] Thu, 24 Nov 2022 09:23:47 GMT (33kb)
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http://math.stackexchange.com/questions/135770/determine-which-of-the-following-functions-are-uniformly-continuous-on-the-open | # Determine which of the following functions are uniformly continuous on the open unit interval (0,1)
Could someone help me through this problem? Determine which of the following functions are uniformly continuous on the open unit interval $(0,1)$ :
a) $1/(1-x)$
b) $1/(2-x)$
c) $\sin{x}$
d) $\sin{1/x}$
e) $x^3$
-
A nice exercise: $f : (0,1) \to \mathbb{R}$ is uniformly continuous if and only if it is continuous on $(0,1)$ and $\lim_{x \downarrow 0} f(x)$ and $\lim_{x \uparrow 1}f(x)$ both exist. – Nate Eldredge Apr 24 '12 at 13:46
One way in which I check uniform continuity of function is the following: Given an interval $[a,b]$ look at points where the function increases rapidly, for example consider $f(x)=\frac{1}{x}$ on $(0,1)$. This function can't be uniformly continuous because as $x \to 0, \frac{1}{x} \to \infty$ and hence you can't have $|f(x)-f(y)| < \epsilon$ for all $x \in (0,1)$.
Another useful way of checking uniform continuity of differentiable functions to look at:"$\text{their derivative and see if they are bounded in that given interval.}$" For example, $f(x)=x^{3}$ on $(0,1)$ has derivative $3x^{2}$ and attains a maximum of $3$ as $x \to 1$. So $f$ here is bounded by $3$. So this function has to be uniformly continuous. Similarly functions such as $\sin{x},\cos{x}$ are uniformly continuous on any given interval. But functions such as $f(x)=x^{2},x^{3}$ aren't uniformly continuous over $\mathbb{R}$ but they are $\text{uniformly continuous}$ over any interval.
-
The test of derivatives is actually a test of "lipschitzianity". A lipschitz map is always uniformly continuous. – Siminore Apr 23 '12 at 13:18
@Siminore: Well I am not actually familiar with such terms. – user9413 Apr 23 '12 at 13:19
A function $f$ is liptschiz when there exists a universal constant $L>0$ such that $|f(x)-f(y)|\leq L |x-y|$ for every $x$ and $y$ (in the domain of $f$, of course). This happens, as you suggest, if the first derivative $f'$ is bounded. It is immediate to check that such a function is uniformly continuous, by choosing $\delta=\varepsilon/L$ in the definition. – Siminore Apr 23 '12 at 13:24
@Siminore. I'm pretty sure that Chandrasekhar is familiar with the definition of a Lipschitz map. Your terminology, in particular the word "lipschitzianity", was likely what caused the confusion. – T. Eskin Apr 23 '12 at 13:49
Well, $x \mapsto \sin x$ can be extended continuously to $[0,1]$, and therefore it is uniformly continuous on $(0,1)$. The same for $x \mapsto x^3$. The function in (a) has a vertical asymptote $x=1$, and it is very easy to check that it cannot be uniformy continuous. Formally, choose $x_n=1-\frac{1}{n}$ and $y_n=1-\frac{1}{n+1}$. Then $x_n \to 1$, $y_n \to 1$, but $$\frac{1}{1-x_n}-\frac{1}{1-y_n}=n-(n+1)=-1.$$ Try to understand what happens in case (b). More generally, it is a nice exercise to prove the following: let $f \colon [a,b) \to \mathbb{R}$ a continuous function. If $\lim_{x \to b-} f(x) = \pm \infty$, then $f$ can't be uniformly continuous. Hint: otherwise, there would exist a continuous extension $\tilde{f} \colon [a,b] \to \mathbb{R}$ of $f$, namely $\tilde{f}(x)=f(x)$ for every $x \in [a,b)$.
- | {"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": 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": 0.970910370349884, "perplexity": 109.78277107831194}, "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-2016-26/segments/1466783402516.86/warc/CC-MAIN-20160624155002-00193-ip-10-164-35-72.ec2.internal.warc.gz"} |
https://www.physicsforums.com/threads/compressed-air-bottle-rocket-variable-mass.264773/ | # Homework Help: Compressed Air Bottle Rocket (Variable Mass)
1. Oct 15, 2008
### Aftermarth
1. The problem statement, all variables and given/known data
I need to find the maximum distance a rocket (which is actually a bottle filled with compressed air to 654kPa) can travel when the exhuast nozzle is 22cm (and can be changed to any smaller size if needed). Neglect air resistance. The bottles original mass is 50g
2. Relevant equations
Ideal Gas Law: PV = nRT
Bernoullis equation: $$\rho$$gh + (1/2)$$\rho$$V^2 + P = constant
Thrust Equation: T = mv
$$\rho_{1}$$$$A_{1}$$$$V_{1}$$=$$rho_{2}A_{2}V_{2}$$=m
R = $$\dot{m}$$v
= $$\rho$$AV(v-$$v_{0}$$
Density of Air = 1.2062kg/m^3
Pressure of Air = 101.3kPa
3. The attempt at a solution
Well i dont know.
I cant seem to figure out anything here. Bernoulli's equation wont work cause i end up with unknowns on both sides.
m doesnt make sense. I dont know if i just just sub in a number
veloctity of the gases - i dont know how to find them
im just confused how to do this :x
(ps - i dont want the thing solved. can i just get some help in formulating expressions which can be integrated between 1. Angles of launch and 2. Area of the nozzle)
2. Oct 16, 2008
### Andrew Mason
This is a complicated question, even if you ignore air resistance. The mass of the rocket is continually changing as the gas is expelled. If you ignore this change in mass, the problem becomes more manageable but is still complicated because the thrust force is not constant - it decreases as the gas is expelled.
When the rocket reaches maximum height, it has kinetic and potential energy. The kinetic energy due to its horizontal speed (vertical speed = 0) and its potential energy at maximum height a function of that height. This energy has to come from the work done by rocket propulsion. Determine the expression for that work (ie. thrust x distance) and that will give you the total energy imparted to the rocket. It is not a trivial calculation.
AM
3. Oct 16, 2008
### Aftermarth
ok. here is an update on my progress....
Thrust = $$\dot{m}$$v
where v = (v - $$v_{0}$$)
and $$\dot{m}$$ = $$\rho$$AV
using the gas laws (PV = nRT) and Boyle's Law ($$\rho$$ = (MP)/(RT)
along with Pressure/ Temperature = constant
T(inside bottle) = 654/T = 101.325/293.15 (NOTE - i have assumed the atmosphere outside the bottle is STP P = 101.325kPa, T = 293.15K)
T = 1892K (this seems very high can someone please verify this!!)
Subbing average molar mass of air = 28.97g/mol and T into boyle's law gives
$$\rho$$ = 1.2062
(so density of the air does not change inside the bottle??)
Now i have for INSIDE the bottle: P = 654kPa, T = 1892K $$\rho$$=1.2062kg/m^3
for OUTSDIE the bottle P = 101.325kPa T = 293.15K $$\rho$$= 1.2062kg/m^3
subbing into Bernoulli's Equation and rearranging gives
($$V^{2}_{1}$$ - $$V^{2}_{2}$$ = 916.39 (u haven't worked out units for this value yet)
Also i now have the equation
mv + pA -mg - R = m$$v^{.}$$
ignoring R as the resistance force gives
T + pA - mg = m$$v^{.}$$ (as thrust is m`v)
and i dont know how to work out p which is supposed to be nozzle exit pressure or something like that :s
4. Oct 22, 2008
### Paul McHugh
See me tomorrow
- paul
5. Oct 22, 2008
### Paul McHugh
lol jks
6. Oct 22, 2008 | {"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": 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": 0.8537308573722839, "perplexity": 1506.6544445797474}, "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/1526794868316.29/warc/CC-MAIN-20180527131037-20180527151037-00466.warc.gz"} |
http://mathhelpforum.com/algebra/4130-word-problem-involving-conversion-2.html | # Thread: Word Problem involving conversion
1. Originally Posted by Quick
Your method is strange but correct.
Thanks
umm...why do you keep substituting different measurements into your equation. Why not just do ratios?
~ $\displaystyle Q\!u\!i\!c\!k$
Which is bigger--mile or meter? mile, of course.
Hence, $\displaystyle \frac{mile}{meter}=\frac{1609}{1}$
Keep Smiling
Malay
2. Originally Posted by malaygoel
Thanks
Which is bigger--mile or meter? mile, of course.
Hence, $\displaystyle \frac{mile}{meter}=\frac{1609}{1}$
Keep Smiling
Malay
In measurements, ratios are equal to 1,
an example: 1 mile = 1609 meters, therefore you write the ratio as $\displaystyle \frac{1mile}{1609meters}$ such as if I wanted to convert meters to miles I would do this...
$\displaystyle 1609m_e=1609m_e$
$\displaystyle 1609m_e\times1=1609m_e\times\frac{1m_i}{1609m_e}$
$\displaystyle 1609m_e=\frac{1609m_im_e}{1609m_e}$
$\displaystyle 1609m_e=\frac{ \not{1609}m_i\not{m_e}}{\not{1609} \not{m_e}}$
$\displaystyle 1609m_e=1m_i$
admitedly I can just substitute 1 mile for 1609 meters, but that is not what the way the person who asked this question wants.
~ $\displaystyle Q\!u\!i\!c\!k$
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https://math.stackexchange.com/questions/795501/verifying-the-cofinite-topology | Verifying the Cofinite topology
I'm trying to provee that the cofinite topology is a valid topological space. I've defined it as
$$C=\{\emptyset\}\cup\{S\subseteq X: X-S \quad\text{is finite }\}.$$
Now, clearly $\emptyset$ and $X$ are in $C$. I just have to prove that it is closed under infinite union and finite intersection:
1) We need to show that $$V=\bigcup_{i\in I}V_i\in C \quad\forall i\in I$$ where $I$ is the indexing set. Proving that $V$ is open is the same as showing that the complement is closed; now with the help of De Morgan's laws we have:
$$X-V = X-\left(\bigcup_{i\in I}V_i\right)=\bigcap_{i\in I}\left(X-V_i\right)$$
it's obvious that if one of the $V_i$, say $V_0$ is non-empty, we have
$$\bigcap_{i\in I}\left(X-V_i\right)\subseteq X-V_0$$ which is finite. Otherwise, every $V_i$ is empty and the union is empty therefore open
2) We need to show that $$V=\bigcap_{i=1}^nV_i\in C \quad$$ where $I$ is the indexing set. Proceeding in a similar manner as before: $$X-V = X-\left(\bigcap_{i=1}^nV_i\right)=\bigcup_{i=1}^n\left(X-V_i\right)$$
Which shows that we have finite unions of finite sets which will be again finite. Therefore, $\bigcap_{i=1}^nV_i\in C$
Is my logic correct? Have I missed something?
• Nope. Looks good. – Sammy Black May 15 '14 at 5:14 | {"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": 0.9601243138313293, "perplexity": 126.98759026203106}, "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-2019-35/segments/1566027315132.71/warc/CC-MAIN-20190819221806-20190820003806-00286.warc.gz"} |
http://www.ck12.org/physics/Archimedes-Law/postread/Archimedes-Principle-and-Buoyancy/ | <img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# Archimedes' Law
## The force exerted by a fluid on a floating or submerged object is equal to the weight of the water displaced.
%
Progress
Progress
%
Archimedes Principle and Buoyancy
Teacher Contributed
This is an activity for students to complete after reading the Archimedes Principle and Buoyancy Concept | {"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": 0.8622079491615295, "perplexity": 3478.2414501309436}, "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-2016-36/segments/1471982295192.28/warc/CC-MAIN-20160823195815-00002-ip-10-153-172-175.ec2.internal.warc.gz"} |
https://www.physicsforums.com/threads/superposition-question.10269/ | # Superposition Question
1. Dec 4, 2003
### timtng
A transverse wave of frequency 40 Hz
propagates down a string. Two points 5 cm apart are
out of phase by p/6. (a) What is the wavelength of the
wave? (b) At a given point, what is the phase
difference between two displacements for times 5 ms
apart? (c) What is the wave velocity?
for a.) I use theta=(sπx)/λ
solving for λ I get λ=.6m
Please help me on part b and c, and check to see if I did part a correctly.
2. Dec 4, 2003
### chroot
Staff Emeritus
For part b, you can figure out the phase difference by looking at the frequency of the wave. The frequency is 40 Hz, so the wave has a period T = 1/40 Hz = 0.025 s. The time period 50 ms is thus 0.050 / 0.025 = 2 periods, exactly. If the point on the string executes exactly an integer number of periods in 50 ms, then its phase difference between the beginning and end of that 50 ms period is zero.
For part c, you know the frequency, 40 Hz, and the wavelength, 0.6 m. You can find the velocity with
$$v = \lambda \cdot \nu$$
Does this make sense?
- Warren | {"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": 0.9012659192085266, "perplexity": 1001.7671687418734}, "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-2017-47/segments/1510934805114.42/warc/CC-MAIN-20171118225302-20171119005302-00456.warc.gz"} |
http://math.stackexchange.com/questions/49424/probability-metric-spaces-for-which-r-leq-r-implies-fracprbx-rprbx | # Probability metric spaces for which $r \leq R$ implies $\frac{Pr(B(x,r))}{Pr(B(x,R))} \geq \frac{r}{R}$
I'm looking for examples of spaces $X$ such that:
1. $X$ is a probability space.
2. $X$ is a metric space.
3. If $x \in X$ and $0 < r \leq R$ then $\frac{Pr(B(x,r))}{Pr(B(x,R))} \geq \frac{r}{R}$.
1. A segment $[a,b] \subset \mathbb{R}$ with the natural probability and metric.
2. Generalization of $1$: A finite tree with positive weights on the edges (where the weights sum up to 1, each edge $e$ is equivalent to a segment $[0,weight(e)]$ and the entire space is a quotient of the disjoint union of the edges which identifies the ends of the edges according to the structure of the tree). See Didier's comment.
Sorry but I fail to see how example 2. works except when the tree is in fact linear. When one crosses the distance $d$ between $x$ and its nearest neighbor(s), for small positive $t$, it seems that $P(B(x,d-t))/P(B(x,d+t))$ becomes too small to exceed $(d-t)/(d+t)$ because of the new branch(es) appearing at distance $d$. – Did Jul 4 '11 at 15:22
The first thing that comes to mind is that if you're working in the $d$-dimensional unit cube, then (except near the edges) $Pr(B(x,r))/Pr(B(x,R)) = (r/R)^d$. So in some sense you want spaces which are of dimension one or less. – Michael Lugo Jul 4 '11 at 17:25
For a very slightly different example, let $X$ be the unit circle with uniform probability distribution. Also $d(x,y)$ is the length of the smaller arc between $x$ and $y$. – Srivatsan Aug 1 '11 at 21:38
Here's a fancier version of the above example. (I hope this works; I cannot trust my calculations yet.) Take a unit interval, and a circle of circumference $2$. Identify one endpoint of the interval with one of the points in the circle. The metric is the natural metric. The probability density is uniform over the interval and over the circle individually, and the total probability of the interval (or the circle) is $1/2$. (Note that since the circumference of the circle is twice the length of the interval, the density over the interval is double that over the circle.) – Srivatsan Aug 1 '11 at 21:52 | {"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": 0.9789822101593018, "perplexity": 192.2726222092397}, "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-2015-22/segments/1432207930866.20/warc/CC-MAIN-20150521113210-00154-ip-10-180-206-219.ec2.internal.warc.gz"} |
https://quics.umd.edu/research/publications?s=year&o=desc&f%5Bauthor%5D=71 | # Publications
Export 1 results:
Author Title Type [ Year]
Filters: Author is David Jao [Clear All Filters]
2014
, Constructing elliptic curve isogenies in quantum subexponential time, Journal of Mathematical Cryptology, vol. 8, no. 1, pp. 1 - 29, 2014. | {"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": 0.8099740743637085, "perplexity": 4485.247164781118}, "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/1637964358953.29/warc/CC-MAIN-20211130050047-20211130080047-00490.warc.gz"} |
http://mathonline.wikidot.com/divisors-of-elements-in-commutative-rings | Divisors of Elements in Commutative Rings
# Divisors of Elements in Commutative Rings
So far we have discussed the term "divisor" with regards to integers and polynomials over a field $F$. We will now extend the notion of a divisor to a general commutative ring.
Definition: Let $(R, +, \cdot)$ be a commutative ring and let $a, b \in R$. Then $b$ is said to be a Divisor of $a$ denoted $b | a$ if there exists an element $q \in R$ such that $a = bq$.
Many of the proofs regarding divisors of integers and polynomials also applies to divisors of elements in a commutative ring.
Theorem 1: Let $(R, +, \cdot)$ be a commutative ring. Then: a) If $a | b$ and $b | c$ then $a | c$. b) If $a | b$ then $a | bc$. c) If $a | b$ and $a | c$ then for all $x, y \in R$, $a | (bx + cy)$.
• Proof of a) Since $a | b$ and $b | c$ there exists $q_1, q_2 \in R$ such that $aq_1 = b$ and $bq_2 = c$. Plugging the second equation into the first yields:
(1)
\begin{align} \quad a(q_1q_2) = c \end{align}
• So $a | c$. $\blacksquare$
• Proof of b) Since $a | b$ there exists a $q \in R$ such that $aq = b$. Multiply both sides of this equation by $c$ to get $a(cq) = bc$. So $a | bc$. $\blacksquare$
• Proof of c) Since $a | b$ and $a | c$ there exists $q_1, q_2 \in R$ such that $aq_1 = b$ and $aq_2 = c$. So for any $x, y \in R$ we have that $a(q_1x) = bx$ and $a(q_2y) = cy$. Adding these equations yields:
(2)
\begin{align} \quad a(q_1x + q_2y) = bx + cy \end{align}
• So $a | (bx + cy)$. $\blacksquare$
Theorem 2: Let $(R, +, \cdot)$ be a commutative ring and let $a, b \in R$. Then $aR \subseteq bR$ if and only if $b | a$.
• Proof: $\Rightarrow$ Suppose that $aR \subseteq bR$. Then since $a \in aR$ we have that $a \in bR$. Since $bR = \{ bq : q \in R \}$ we see that $a = bq$ for some $q \in R$. So $b | a$.
• $\Leftarrow$ Suppose that $b | a$. Then there exists an element $q \in R$ such that $a = bq$.
• Let $x \in aR$. Then $x = ay$ for some $y \in R$. Thus $x = b(qy)$ which shows that $x \in bR$. Therefore $aR \subseteq bR$. $\blacksquare$ | {"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": 2, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.999542236328125, "perplexity": 63.864604518390806}, "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-2018-51/segments/1544376832559.95/warc/CC-MAIN-20181219151124-20181219173124-00475.warc.gz"} |
http://www.geekblog.eu/mad-cow-jljiyu/what-is-division-in-math-429b74 | Division is often shown in algebra and science by placing the dividend over the divisor with a horizontal line, also called a fraction bar, between them. Definitions vary regarding integer division when the dividend or the divisor is negative: rounding may be toward zero (so called T-division) or toward −∞ (F-division); rarer styles can occur – see Modulo operation for the details. The user is responsible, however, for mentally keeping track of the decimal point. K-8 Math. If you like this Site about Solving Math Problems, please let Google know by clicking the +1 button. {\displaystyle b{\overline {)a}}} × This usage, though widespread in anglophone countries, is neither universal nor recommended: the ISO 80000-2 standard for mathematical notation recommends only the solidus / or fraction bar for division, or the colon for ratios; it says that this symbol "should not be used" for division. Sometimes this remainder is added to the quotient as a fractional part, so 10 / 3 is equal to 3+1/3 or 3.33..., but in the context of integer division, where numbers have no fractional part, the remainder is kept separately (exceptionally, discarded or rounded). If the dividend has a fractional part (expressed as a decimal fraction), one can continue the algorithm past the ones place as far as desired. Ask Question Asked 8 years ago. One can define a division operation for matrices. This is tedious and doesn't quite look right. There are four terms which describe the four numbers in a division problem. Quotient: A quotient is a result obtained in the division process. If the divisor has a fractional part, one can restate the problem by moving the decimal to the right in both numbers until the divisor has no fraction. ( Other Stuff. Division Sign. Division (logical fallacy), when one reasons logically that something true of a thing must also be true of all or some of its parts Tom Clancy's The Division , a multiplayer … Some mathematical software, such as MATLAB and GNU Octave, allows the operands to be written in the reverse order by using the backslash as the division operator: A typographical variation halfway between these two forms uses a solidus (fraction slash), but elevates the dividend and lowers the divisor: Any of these forms can be used to display a fraction. b Letters and symbols in multiplication and division equations Find missing divisors and dividends (1-digit division) CCSS.Math: 3.OA.A.4 , 3.OA.C.7 means 6 divided by 2, which is the division … If you're seeing this message, it means we're having trouble loading external resources on our website. 6 / 2 = 3 . This division sign is also used alone to represent the division operation itself, as for instance as a label on a key of a calculator. The dividend is the number that is being divided. Math Help for Division: Easy-to-understand lessons for kids, parents and teachers. Apart from division by zero being undefined, the quotient is not an integer unless the dividend is an integer multiple of the divisor. One can define the division operation for polynomials in one variable over a field. For example, 26 cannot be divided by 11 to give an integer. Help With Your Math Homework. × All four quantities p, q, r, s are real numbers, and r and s may not both be 0. Pu… Division definition is - the act or process of dividing : the state of being divided. The obelus was introduced by Swiss mathematician Johann Rahn in 1659 in Teutsche Algebra. a Just like all division problems, a large number, which is the dividend , is divided by another number, which is called the divisor , to give a result called the quotient and sometimes a remainder . Don’t worry though, you've come to the right place for help - and the good news is that although it sounds complicated, it really is quite simple. Divide Into Equal Groups. The other operations are addition, subtraction, and multiplication (which can be viewed as the inverse of division). If a number is BOTH divisible by three (see the three rule) AND an even number (ending in 0,2,4, 6 or 8) then it is divisible by six too. Divisor:The number by which dividend is being divided by is called divisor. Divide Two Numbers. to denote a divided by b, especially when discussing long division. For example; when 41 is divided by 7, the quotient is 5 and the remainder is 6. [citation needed]. The other operations are addition, subtraction, and multiplication (which can be viewed as the inverse of division). Copyright © 2020 Studypad Inc. All Rights Reserved. Division Sign. Definition of Quotative Division explained with real life illustrated examples. The result is placed under the number divided into. This is denoted as 20 / 5 = 4, or .mw-parser-output .sr-only{border:0;clip:rect(0,0,0,0);height:1px;margin:-1px;overflow:hidden;padding:0;position:absolute;width:1px;white-space:nowrap}20/5 = 4. The divisor is the number that the dividend will be divided by. Quotient: Up… Long division helps in breaking the division problem into a sequence of easier steps. A person can calculate division with a slide rule by aligning the divisor on the C scale with the dividend on the D scale. + b Take another piece and put it in a second pile. 70 synonyms of division from the Merriam-Webster Thesaurus, plus 104 related words, definitions, and antonyms. c The division sign ÷, a symbol consisting of a short horizontal line with a dot above and another dot below, is often used to indicate mathematical division. − How many will be in each group? This notation was introduced by Gottfried Wilhelm Leibniz in his 1684 Acta eruditorum. Therefore, 4 balls should be kept in each box. In abstract algebra, given a magma with binary operation ∗ (which could nominally be termed multiplication), left division of b by a (written a \ b) is typically defined as the solution x to the equation a ∗ x = b, if this exists and is unique. or The result of dividing two rational numbers is another rational number when the divisor is not 0. Division. is called 'realisation' or (by analogy) rationalisation. [7] Division is also not, in general, associative, meaning that when dividing multiple times, the order of division can change the result. In Year 2 they may be asked to solve a word problem like this one: I have 20 sweets. Some programming languages, such as C, treat integer division as in case 5 above, so the answer is an integer. Dividing two complex numbers (when the divisor is nonzero) results in another complex number, which is found using the conjugate of the denominator: This process of multiplying and dividing by Note: Division is a stepping stone of math! more... Division is splitting into equal parts or groups. Here's the problem I have.. : You want to buy a pair of new shoes that costs 56. divide, division, dividend, divisor • to divide or division is sharing or grouping a number into equal parts. The division with remainder or Euclidean division of two natural numbers provides a quotient, which is the number of times the second one is contained in the first one, and a remainder, which is the part of the first number that remains, when in the course of computing the quotient, no further full chunk of the size of the second number can be allocated. Students can also use repeated addition to find the product. In some non-English-speaking countries, a colon is used to denote division:[12]. To learn about when algebras (in the technical sense) have a division operation, refer to the page on division algebras. This page will show you a complete "long division" solution for the division of two numbers. Distributing the objects several at a time in each round of sharing to each portion leads to the idea of 'chunking' – a form of division where one repeatedly subtracts multiples of the divisor from the dividend itself. [2]:7 This number of times is not always an integer (a number that can be obtained using the other arithmetic operations on the natural numbers), which led to two different concepts. 12 Dogs total, placed into 3 groups of 4: Group 1: Group 2: Group 3: 4 Dogs : 4 Dogs: 4 Dogs . When dividing something by 1, the answer will always be the original number. Search for courses, skills, and videos. The division is a method of distributing a group of things into equal parts. The remainder is the portion of the dividend that is left over after division. Division: something that divides, separates, or marks off. Unlike multiplication and addition, division is not commutative, meaning that a / b is not always equal to b / a. The division is an operation inverse of multiplication. Algebra. In an integral domain, where not every element need have an inverse, division by a cancellative element a can still be performed on elements of the form ab or ca by left or right cancellation, respectively. Some symbols for division can be a slash, a line, or the division sign, as in: 6 / 3 {\displaystyle 6/3\,} or 6 3 {\displaystyle {\frac {6}{3}}} or 6 ÷ 3. In these cases, a division by x may be computed as the product by the multiplicative inverse of x. Division is one of the four basic operations of arithmetic, the ways that numbers are combined to make new numbers. {\displaystyle b)a} A fraction is a division expression where both dividend and divisor are integers (typically called the numerator and denominator), and there is no implication that the division must be evaluated further. The division is a method of distributing a group of things into equal parts. A division is made up of different numbers, and each of these numbers has a special name. In this tutorial, you'll see how cookies can help you find the answer to division problems! It includes unlimited math lessons on number counting, addition, subtraction etc. • divisor or factor: a number that will divide the dividend … Any remainders are ignored at this point. Take the next piece and put it in a third pile. In modular arithmetic (modulo a prime number) and for real numbers, nonzero numbers have a multiplicative inverse. Dividing by 2 If the remainder is greater than the divisor, it means that the division is incomplete. La division décimale Méthode : Poser une division décimale 1) Poser 45 : 8 et 32,12 : 4 The first number is the dividend, and … The division of the same dividend and divisor is always 1. Division by 0 1 ÷ 0 = not allowed 2 ÷ 0 = not allowed 3 ÷ 0 = not allowed Another generalization of division to algebraic structures is the quotient group, in which the result of 'division' is a group rather than a number. [11]:295 Leibniz disliked having separate symbols for ratio and division. 17 ] in these cases, a colon is used throughout mathematics, the word division '' solution the... At any level and professionals in related fields lot of math called the dividend is being divided in case! There is no remainder ) unlimited math lessons on number counting,,. De calculs our first division equation: 12 ÷ 3 = 4 says that there are four terms describe! Kids for fun math worksheet online at SplashLearn languages also provide functions to get the results of of... 'S the problem I have 20 sweets below, we need your age to give an!. [ 15 ] and addition, subtraction, and r and s may not both be 0 it a. You like this one: I have..: you want to know how many in each group in process... Integers in a second pile Math®, SplashLearn™ & Springboard™ are Trademarks of,! About math division, dividend, 5 is the number entirely and leaves remainder. Reste D ’ une division well defined, B−1 need not exist division! Remainder is the number that the division operation for polynomials in one variable over field. Product, students can build a model of three groups with five items each! The special case of dividing polynomials people on Pinterest a divisor that divides an integer winning math learning program for. Your child to divide a number up into an equal number of pineapples that we have give! Hand-Written computation, polynomial long division helps in breaking the division is a method used integer... May be computed as the answer from the Merriam-Webster Thesaurus, plus 104 related,! The Merriam-Webster Thesaurus, plus 104 related words, definitions, and 4 is the entirely. Also define a left division or slash-division in this basic introduction to division problems 'd it of the product! Divisor is not 0 slide rule by aligning the divisor = 4, 8 is divisible by,. Not both be 0 words, definitions, and % single result ) by all nonzero is. Below, we want to share them, how to put 16 balls and 4 is the expression! 3 five times: 3 + 3 + 3 + 3 = 4 says that there are 12,! Another rational number when the remainder is greater than or lesser than the quotient is... / b is not entirely clear because it evolved over time. [ ]. Aligning the divisor is not commutative, one can also be defined as multiplication by the pseudoinverse quickly determine one... It includes unlimited math lessons on number counting, addition, division as in the case in ;. You 're behind a web filter, please make sure that the dividend on the scale... By is called divisor 's in 8 associated with the dividend on the D scale where it is one the! The original number in case 5 above, so the answer is award... Notation was introduced by de Morgan in 1845 mathematician Johann Rahn in 1659 in Algebra. First division equation: 12 ÷ 3 = 15 that numbers are combined to make new numbers of case above. Every computer Algebra system return a rational number rational numbers is another rational number as the answer for example 20. European countries, a division operation, refer to the page on algebras... Ways that numbers are combined to make new numbers costs 56 how you teach your to. Practice online or download as pdf to print rédaction et de calculs undefined, the ways that numbers are to... A quotient is a structure in which a division operation, refer to the page on division algebras when... By 13923 people on Pinterest undefined, the answer, as in case 5,! Not commutative, one can also use repeated addition to find the product by the.! The colon is restricted to expressing the related concept of ratios in terms of four... The three expressions means 6 divided by is called division 4 make 12 in multiplication, gives... As there is no remainder over after division a factor, however A−1 does need exist! Learning in mathematics, the quotient is the number of times the divisor is 1, the quotient signs can. Special case of integers, and thus distributive this approach is often associated with the cancellation property computed.! C if and only if a +1 button, too not 0 numbers, and (. Involved in a division by x may be 0 additional subscription based content in quantity two. And the remainder is kept as a fraction, it leads to a rational number the! Notation was introduced by Swiss mathematician Johann Rahn in 1659 in Teutsche Algebra to division... Or division is not commutative, meaning that twenty divided by 3 '', by! Between two numbers four numbers in a division ( with a slide rule by the! Computed as the set of all rational numbers p/q and r/s can be viewed the! Online or download as pdf to print is possible by the multiplicative inverse x... Kept as a sign in division process is sometimes called right division or slash-division in this basic introduction division! The problem I have 20 sweets an identity element and hence inverses, on dividing 83 by 2 divisor! Is followed in the division process was introduced by Gottfried Wilhelm Leibniz in his 1684 Acta.. Mathantics.Comvisit http: //www.mathantics.com for more Free math videos and additional subscription based!! Tête avec reste and divisor is 1, the quotient can be found on C. To share them, how to do division grouping of Objects: division is sharing or a!, parents and teachers students can also use repeated addition to find answer... Without an identity element and hence inverses hence inverses is placed under the number is. Is very important that children are taught division in the example, 26 can not be divided by is the... That children are taught division in the special case of integers r, s are real numbers, numbers. Of case 3: I have..: you want to know many. And does n't quite look right and teachers without an identity element and hence inverses Teutsche.. It refers to the 12 dogs in multiplication ; 12 divided into 3 equal groups polynomials in one over! Obtained in the context of problem-solving make sure that the division of two real numbers results in another real (! Quotient: a quotient is a shortcut method of distributing a group of things into equal parts groups... Mathematician Johann Rahn in 1659 in Teutsche Algebra times: 3 + 3 + 3 + 3 + +... Video for kids, learning division is different from what is division in math definitions is to... Math, math classroom 4 can we get from 36 his 1684 Acta eruditorum synthétique euclidienne de avec! 16 balls into four equal sized boxes studypad®, Splash Math®, SplashLearn™ Springboard™... Math®, SplashLearn™ & Springboard™ are Trademarks of StudyPad, Inc and got 4 cookies in each in!
Spodobał Ci się ten wpis? Podziel się nim: | {"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": 0.8418316841125488, "perplexity": 860.0424588349899}, "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-2022-05/segments/1642320304528.78/warc/CC-MAIN-20220124094120-20220124124120-00515.warc.gz"} |
http://mathhelpforum.com/algebra/24812-elp-semester-exams-please.html | # Math Help - elp for semester exams! please! :)
1. ## elp for semester exams! please! :)
Classify each system as idependent, dependent or inconsistant:
x-3y=2
4x-12y=8
any help would be greatly appreciated! whenever i do it (through elimination) i get 0 = 10... what category does that come under?? but then again i might have that wrong aswell :P
2. 0 = 10?? If you had to pick "consistent" or "inconsistent", which would you choose? Words still mean things in mathematics. You do not have to check your vocabulary at the door.
3. Originally Posted by mazer05
Classify each system as idependent, dependent or inconsistant:
x-3y=2
4x-12y=8
any help would be greatly appreciated! whenever i do it (through elimination) i get 0 = 10... what category does that come under?? but then again i might have that wrong aswell :P
Both equations are the same. Multiply the first one by four. I'm not sure what makes a system of equations independent, dependent, or consistent, but that tidbit of information should help you. | {"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": 0.9045346975326538, "perplexity": 2244.8125845584095}, "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/1424936462035.86/warc/CC-MAIN-20150226074102-00021-ip-10-28-5-156.ec2.internal.warc.gz"} |
http://tex.stackexchange.com/questions/64282/leaders-and-evaluation-order | I made this simple latex file :
\documentclass{minimal}
\usepackage{tikz}
\newcommand\randdot{%
\tikz{\pgfmathsetmacro{\r}{0.01+0.05*random()}\fill (0,0) circle (\r);}\ }
\begin{document}
\randdot\randdot\randdot\randdot\randdot\randdot
Test \leaders\hbox{\randdot\ }\hfill 3 \hfill \
\randdot\randdot\randdot\randdot\randdot\randdot
\end{document}
The new command randdot typeset a dot of random size. The command works good, BUT when I use it in a leader, the dots are no more random. The leaders command must evaluate only once the randdot, and copy it...
How can I obtain a random leader, with each dot being evaluated differently ?
Thanks !!
-
Leaders simply repeat copies of the box, they don't rebuild it at every instance. – egreg Jul 22 '12 at 20:10
Would it be possible to know how much will TeX expand a hfill, perhaps from lua? If that were possible, the problem would be solved writting a TikZ loop which fill that given dimension with random dots. – JLDiaz Jul 22 '12 at 20:43
The TikZ way David mentions is the now-famous (thanks to Andrew Stacey and Peter Grill) \tikzmark macro. Basically you leave TikZ coordinates at places you like and then refer to them in a later TikZ picture.
\documentclass{article}
\usepackage{tikz}
\usetikzlibrary{decorations.markings}
markings,% switch on markings
between positions 0 and 1 step 5mm
with
{
\pgfmathsetmacro{\r}{0.01+0.05*random()}\fill (0,0) circle (\r);
}
},
decorate
}
}
\newcommand{\tikzmark}[1]{\tikz[overlay,remember picture,baseline=-0.5ex] \node (#1) {};}
\begin{document}
Test\tikzmark{a1} \hfill \tikzmark{a2}3 \hfill
Test some text and then Test \tikzmark{a3} \hfill \tikzmark{a4}165 \hfill
\begin{tikzpicture}[overlay,remember picture]
\end{tikzpicture}
\end{document}
Further tweaks are possible but take this as a proof of concept.
-
This is very elegant. Why do you need to set the baseline to -0.5ex in the definition of the tikzmark ? – Xoff Jul 22 '12 at 22:00
@Xoff Well just cosmetics. It starts a little higher on the text line. Also you need two runs with this solution too. – percusse Jul 22 '12 at 22:01
There's probably a tikz way of getting the coordinates but I use the pdftex primitive here. Use glue instead of leaders, but measure the distance and overlay some boxes on a second run.
\documentclass{minimal}
\usepackage{tikz}
\newcommand\randdot{%
\tikz{\pgfmathsetmacro{\r}{0.01+0.05*random()}\fill (0,0) circle (\r);}\ }
\begin{document}
\makeatletter
\randdot\randdot\randdot\randdot\randdot\randdot
Test \leaders\hbox{\randdot\ }\hfill 3 \hfill \
Test %
\rlap{%
\loop
\setbox0\hbox{\randdot\ }%
\ifdim\dimen@>\z@
\box\z@
\repeat
}%
\fi
\hfill | {"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": 0.9935833811759949, "perplexity": 4242.035345123396}, "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-2016-07/segments/1454701148758.73/warc/CC-MAIN-20160205193908-00019-ip-10-236-182-209.ec2.internal.warc.gz"} |
http://www.mathportal.org/calculators/analytic-geometry/graphing-lines-calculator.php | Math Lessons, Calculators and Homework Help
It is time to solve your math problem
You are here:
# Calculators :: Analytic Geometry :: Graphing Lines Calculator
This calculator will plot lines given in following forms:
1.Slope y-intercept form - this is a line of the form $y = mx + b$ where $m$ is the slope of the line and $b$ is the y-intercept.
2.Standard form - this is the line of the form $Ax + By = C$, where $A, B,$ and $C$ are real numbers and A and B are both not zero
The calculator will generate a step-by-step explanation how to graph lines.
## Graphing lines calculator ( line is given in slope y-intercept form )
You can enter either integers (10), decimal numbers(10.12) or FRACTIONS (10/3).
Important: The form will NOT let you enter wrong characters (like *, (, ), y, p, ...).
## Graphing Lines Calculator ( line is given in standard form )
You can enter either integers (10), decimal numbers(10.12) or FRACTIONS (10/3).
Important: The form will NOT let you enter wrong characters (like *, (, ), y, 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": 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": 0.9840112924575806, "perplexity": 1844.6920975933594}, "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-2014-23/segments/1405997894865.50/warc/CC-MAIN-20140722025814-00063-ip-10-33-131-23.ec2.internal.warc.gz"} |
https://math.libretexts.org/Bookshelves/Analysis/Book%3A_Mathematical_Analysis_(Zakon)/3%3A_Vector_Spaces_and_Metric_Spaces/3.6%3A_Normed_Linear_Spaces | $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$
# 3.6: Normed Linear Spaces
$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$
$$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$
By a normed linear space (briefly normed space) is meant a real or complex vector space $$E$$ in which every vector $$x$$ is associated with a real number $$|x|$$, called its absolute value or norm, in such a manner that the properties $$\left(\mathrm{a}^{\prime}\right)-\left(\mathrm{c}^{\prime}\right)$$ of §9 hold. That is, for any vectors $$x, y \in E$$ and scalar $$a,$$ we have
$$\left(\mathrm{i}\right) |x| \geq 0;$$
$$\left(\mathrm{i}^{\prime}\right)|x|=0$$ iff $$x=\overrightarrow{0};$$
$$\left(\mathrm{ii}\right) |a x|=|a||x| ;$$ and
$$\left(\mathrm{iii}\right)|x+y| \leq|x|+|y|\text{ (triangle inequality). }$$
Mathematically, the existence of absolute values in $$E$$ amounts to that of a map (called a norm map) $$x \rightarrow|x|$$ on $$E,$$ i.e., a map $$\varphi : E \rightarrow E^{1},$$ with function values $$\varphi(x)$$ written as $$|x|,$$ satisfying the laws (i)-(iii) above. Often such a map can be chosen in many ways (not necessarily via dot products, which may not exist in $$E$$ , thus giving rise to different norms on $$E .$$ Sometimes we write $$\|x\|$$ for $$|x|$$ or use other similar symbols.
Note 1. From (iii), we also obtain $$|x-y| \geq| | x|-| y| |$$ exactly as in $$E^{n}.$$
Example $$\PageIndex{1}$$
(A) Each Euclidean space (§9) such as $$E^{n}$$ or $$C^{n},$$ is a normed space, with norm defined by
$|x|=\sqrt{x \cdot x},$
as follows from formulas (a')-(c') in §9. In $$E^{n}$$ and $$C^{n},$$ one can also equivalently define
$|x|=\sqrt{\sum_{k=1}^{n}\left|x_{k}\right|^{2}},$
where $$x=\left(x_{1}, \ldots, x_{n}\right) .$$ This is the so-called standard norm, usually presupposed in $$E^{n}\left(C^{n}\right) .$$
(B) One can also define other, "nonstandard," norms on $$E^{n}$$ and $$C^{n} .$$ For example, fix some real $$p \geq 1$$ and put
$|x|_{p}=\left(\sum_{k=1}^{n}\left|x_{k}\right|^{p}\right)^{\frac{1}{p}}.$
One can show that $$|x|_{p}$$ so defined satisfies $$(\mathrm{i})-($$ iii) and thus is a norm (see Problems 5-7 below).
(C) Let $$W$$ be the set of all bounded maps
$f : A \rightarrow E$
from a set $$A \neq \emptyset$$ into a normed space $$E,$$ i.e., such that
$(\forall t \in A) \quad|f(t)| \leq c$
for some real constant $$c>0$$ (dependent on $$f$$ but not on $$t ) .$$ Define $$f+g$$ and $$a f$$ as in Example (d) of §9 so that $$W$$ becomes a vector space. Also, put
$\|f\|=\sup _{t \in A}|f(t)|,$
i.e., the supremum of all $$|f(t)|,$$ with $$t \in A .$$ Due to boundedness, this supremum exists in $$E^{1},$$ so $$\|f\| \in E^{1}.$$
It is easy to show that $$\|f\|$$ is a norm on $$W .$$ For example, we verify (iii) as follows.
By definition, we have for $$f, g \in W$$ and $$x \in A,$$
\begin{aligned}|(f+g)(x)| &=|f(x)+g(x)| \\ & \leq|f(x)|+|g(x)| \\ & \leq \sup _{t \in A}|f(t)|+\sup _{t \in A}|g(t)| \\ &=\|f\|+\|g\|. \end{aligned}
(The first inequality is true because (iii) holds in the normed space $$E$$ to which $$f(x)$$ and $$g(x)$$ belong.) By (1), $$\|f\|+\|g\|+\|g\|$$ is an upper bound of all expressions $$|(f+g)(x)|, x \in A .$$ Thus
$\|f\|+\|g\| \geq \sup _{x \in A}|(f+g)(x)|=\|f+g\|.$
Note 2. Formula (1) also shows that the map $$f+g$$ is bounded and hence is a member of $$W .$$ Quite similarly we see that $$a f \in W$$ for any scalar $$a$$ and $$f \in W .$$ Thus we have the closure laws for $$W .$$ The rest is easy.
In every normed (in particular, in each Euclidean) space $$E,$$ we define distances by
$\rho(x, y)=|x-y| \quad\text{ for all x, y } \in E.$
Such distances depend, of course, on the norm chosen for $$E ;$$ thus we call them norm-induced distances. In particular, using the standard norm in $$E^{n}$$ and $$C^{n}$$ (Example (A)), we have
$\rho(x, y)=\sqrt{\sum_{k=1}^{n}\left|x_{k}-y_{k}\right|^{2}}.$
Using the norm of Example (B), we get
$\rho(x, y)=\left(\sum_{k=1}^{n}\left|x_{k}-y_{k}\right|^{p}\right)^{\frac{1}{p}}$
instead. In the space $$W$$ of Example (C) we have
$\rho(f, g)=\|f-g\|=\sup _{x \in A}|f(x)-g(x)|.$
Proceeding exactly as in the proof of Theorem 5 in §§1-3, we see that norm- induced distances obey the three laws stated there. (Verify!) Moreover, by definition,
$\rho(x+u, y+u)=|(x+u)-(y+u)|=|x-y|=\rho(x, y).$
Thus we have
$\rho(x, y)=\rho(x+u, y+u)\text{ for norm-induced distances;}$
i.e., the distance $$\rho(x, y)$$ does not change if both $$x$$ and $$y$$ are "translated" by one and the same vector $$u .$$ We call such distances translation-invariant.
A more general theory of distances will be given in §§11ff. | {"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": 0.9997214674949646, "perplexity": 254.54923603475908}, "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-2019-51/segments/1575541301014.74/warc/CC-MAIN-20191215015215-20191215043215-00282.warc.gz"} |
https://www.physicsforums.com/threads/find-a-formula-for-this-sequence.643259/ | # Find a formula for this sequence!
1. Oct 11, 2012
### nate9228
1. The problem statement, all variables and given/known data
A sequence (xj) where j can go from 0 to infinity satisfies the following:
(1) x1= 1 and
(2) for all m≥n≥0, xm+n+ xm-n= (1/2)(x2m+x2n)
Find a formula for xj and prove that the formula is correct
2. Relevant equations
3. The attempt at a solution
All I have done so far is is play around with m and n values so far and I think a proper formula:
xj= j2. Now I would have to prove this (for which I plan to use induction most likely) but I wanted to see if I was on the write track in my thinking.
2. Oct 11, 2012
### Dick
Looks fine to me. Now try and check it. Induction isn't really necessary. Just check that your guess works.
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https://mathoverflow.net/questions/68009/downgrading-from-zfc-with-universes-to-zfc/68013 | Downgrading from ZFC with universes to ZFC
Is the following correct?
• If something is provable for small sets in ZFC with Axiom of Universes ("For any set x, there exists a universe U such that x∈U.") then it is provable for any sets in ZFC (without Axiom of Universes).
I want to work in ZFC with Axiom of Universes, but I wish my results be downgradable to plain ZFC. Help?
-
Can you please define "small set" in ZFC? – Goldstern Jun 16 '11 at 22:20
Have you seen the following MO question? It deals exactly with this topic mathoverflow.net/questions/12804/… – Ali Enayat Jun 16 '11 at 22:24
– Joel David Hamkins Jun 16 '11 at 23:54
@goldstern - small set in ZFC+U is an element of U (and perhaps one isomorphic to an element of U, depending on one's proclivities) – David Roberts Jun 17 '11 at 0:45
@david roberts: I thought the question was not about "ZFC plus one additional constant (or predicate) U", but about an additional axiom postulating many universes. If every x is in some universe, which x should be called "small"? Perhaps those in the smallest universe? – Goldstern Jun 17 '11 at 7:43
No can do. ZFC+AU proves con(ZFC), ZFC doesn't.
-
What is "con"? Consistency? – porton Jun 17 '11 at 10:29
yes (char min) – Ricky Demer Jun 17 '11 at 11:20
As I understand, consistency of ZFC is a statement about small (even finite) sets and thus this gives a counter-example to my claim that we can move statements about small sets from ZFC+AU to ZFC. Right? – porton Jun 17 '11 at 12:52
Porton, Yes, that is right. Con(ZFC) is an assertion of arithmetic that is provable in ZFC+AU, but not in ZFC, if ZFC is consistent. – Joel David Hamkins Jun 17 '11 at 18:03
If you want a universe-like theory that is conservative over ZFC, that is, which proves no additional facts about sets that ZFC cannot prove alone, then the thing to do is to work in the following theory, which is also described in the answers to this MO question.
The theory consists of ZFC plus the assertion that there is a hierarchy of universe-like sets, namely, $V_\theta$ for all $\theta\in C$, a closed unbounded proper class of cardinals, and furthermore that truth in these $V_\theta$ cohere with each other and with the full set-theoretic universe $V$, so that they form an elementary chain. Specifically, the theory has ZFC, the assertion that $C$, a new class predicate, is a closed unbounded proper class of cardinals, and the scheme asserting that $V_\theta$ is an elementary substructure of $V$ for every $\theta\in C$, namely, the scheme expressing of each formula $\varphi$ in the language of set theory that
• $\forall x\ \forall\theta\in C\text{ above the rank of }x \ (\varphi(x)\iff V_\theta\models\varphi[x])$.
It follows from this theory that the models $V_\theta$ for $\theta\in C$ form an elementary chain, all agreeing with each other and with the full set-theoretic background universe on what is true as you ascend to higher models. It follows that every $\theta$ in $C$ will be a strong limit cardinal, a beth fixed point and so on, and so these cardinal exhibit very strong closure properties. In particular, I could have written $H_\theta$ instead of $V_\theta$---these are essentially the $\theta$-small universes, the collection of sets of hereditary size less than $\theta$. The only difference between these $V_\theta$ and an actual Grothendieck universe is that in this theory, you may not assume that $\theta$ is regular. But otherwise, they function just like universes in many ways, and indeed, every $V_\theta$ for $\theta\in C$ is a model of ZFC. Because of the coherence in the theories, these weak universes can be more useful than Grothendieck universes for certain purposes. For example, any statement true about an object in the full background universe will also be true about that object in every weak universe $V_\theta$ for $\theta\in C$ in which it resides. Thus, it one takes care, one can use the $V_\theta$ much like Grothendieck universes, and this was the point of my linked answer above (as well as Andreas's).
Meanwhile, the theory is conservative over ZFC, since in fact every model of ZFC can be elementary embedded into (a reduct of) a model of this theory. This can be proved by a simple compactness argument, using the reflection theory. If $M\models ZFC$, then add constants for every element of $M$, add the full elementary diagram of $M$, add a new predicate symbol for $C$ and all the axioms of the new theory. Every finite subtheory of this theory is consistent, by the reflection theorem, and so we get a model of the new theory, which elementary embeds $M$ since it satisfies the elementary diagram of $M$.
(Although it seems counterintuitive at first to some set-theorists, this theory does not prove Con(ZFC), if ZFC is consistent, even though it asserts in a sense that $V_\theta$ is elementary in $V$ for all $\theta\in C$ and hence that $V_\theta$ is a model of ZFC. The explanation is that the theory only makes the assertion that $V_\theta$ is elementary in $V$ as a scheme, and not as a single assertion (which is not expressible anyway by Tarski's theorem), and thus the theory does not actually prove that $V_\theta\models ZFC$ for $\theta\in C$, even though they do model ZFC, since the theory only proves every finite instance of this, rather than the universal assertion that every axiom of ZFC is satisfied in every $V_\theta$.)
-
Does this theory have any wellfounded models? – François G. Dorais Jun 17 '11 at 5:54
If $\delta$ is an inaccessible cardinal, then by a Lowenheim-Skolem argument you can find a club $C\subset\delta$ with $V_\theta$ elementary in $C_\delta$ for $\theta\in C$, and so $\langle V_\delta,{\in},C\rangle$ is a model of the theory. – Joel David Hamkins Jun 17 '11 at 10:25
In the displayed formulation of the reflection schema, $\theta$ should be big enough so that $x\in V_\theta$. – Andreas Blass Jun 17 '11 at 13:23
Yes, I have edited. – Joel David Hamkins Jun 17 '11 at 18:02
The theory described by Joel has the curious feature that it provides a definition of " $\phi$ is true in $(V, \in)$ for standard sentences of set theory [as opposed to those with nonstandard length] with via "$\phi$ holds on a TAIL of the structures of the form $V_{\alpha}$', where $\alpha\in C$". This does not contradict Tarski's theorem on undefinability of truth since $\phi$ is in the language using only $\in$, and is not allowed to mention $C$. – Ali Enayat Jun 17 '11 at 20:50 | {"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": 0.9238967299461365, "perplexity": 361.61813001579947}, "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-2015-48/segments/1448398465089.29/warc/CC-MAIN-20151124205425-00053-ip-10-71-132-137.ec2.internal.warc.gz"} |
https://stacks.math.columbia.edu/tag/00X1 | Definition 7.14.1. Let $\mathcal{C}$ and $\mathcal{D}$ be sites. A morphism of sites $f : \mathcal{D} \to \mathcal{C}$ is given by a continuous functor $u : \mathcal{C} \to \mathcal{D}$ such that the functor $u_ s$ is exact.
There are also:
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An uninformed prior has a distribution that would be considered to add no information to the Bayesian inference. For example, a Uniform(0,1) distribution could be considered an uninformed prior when estimating a binomial probability because it states that prior to collection of any data we consider every possible value for the true probability to be as likely as every other. An uninformed prior is often desirable in the development of public policy to demonstrate impartiality. Laplace (1812), who also independently stated Bayes' Theorem in Laplace (1774) 11 years after Bayes' essay was published (he apparently had not seen Bayes' essay), proposed that public policy priors should assume all allowable values to have equal likelihood (i.e. Uniform or Duniform distributions).
At first glance then, it might seem that uninformed priors will just be Uniform distributions running across the entire range of possible values for the parameter:
Figure 1: These intuitively seem to be logical uninformed priors,
except for setting ranges, but are they?
That this is not true can be easily demonstrated from the following example. Consider the task of estimating the intensity l of a Poisson process (this is equivalent to fitting a Poisson distribution to data). We have observed a certain number of events within a certain period, which we can use to give us a likelihood function. It might seem reasonable to assign a Uniform(0, z) prior to l, where z is some large number. However, we could just have easily parameterised the problem in terms of b, the mean exposure between events. Since b = 1/l, we can quickly check what a Uniform(0, z) prior for l would look like as a prior for b by running a simulation on the formula: =1/Uniform(0, z):
The result is actually impossible to show as a density plot because it has an infinite peak at zero, but this plot compares the Uniform(0.01,100) with a 1/Uniform(0.01,100) on a cumulative scale:
Figure 2: Cumulative prior distributions:
p(l)
= Uniform(0.01,100) and b = 1/l
The prior for b is alarmingly far from being uninformed with respect to b! Of course, the reverse equally applies: if we had performed a Bayesian inference on b with a Uniform prior, the prior for l would be just as far from being uninformed. The probability density function for the prior distribution of a parameter must be known in order to perform a Bayesian inference calculation. However, one can often chose between a number of different parameterizations that would equally well describe the same stochastic process. For example, one could describe a Poisson process by l, the mean number events per unit exposure, by b, the mean exposure between events as above, or by P(x>0), the probability of at least one event in a unit of exposure.
The Jacobian transformation let's us formally calculate the prior distribution for a Bayesian inference problem after reparameterizing, but it is easy to see why the two priors are not the same with an example:
If l is given a Uniform(0,z) prior then the fraction 1/z of the distribution is below 1 and the rest between 1 and z. The prior for b would then be 1/Uniform(0,z) and that has the fraction (z-1)/z below 1 because it is <1 when the Uniform(0,z) is >1.
There is no all-embracing solution to the problem of finding an uninformed prior that doesn't become "informed" under some reparameterizing of the problem, but one approach is to select a prior that is the same under a transformation, so that it is at least not affecting the estimation depending on the parameterisation chosen.
For example, consider a prior such that log10(q) (or loge(q)) is Uniform( , ) distributed:
The parameter 1/q would have a prior log10(1/q) = -log10(q) = Uniform( , )
The parameter c*q would have a prior log10(c*q) = log10(c) + log10(q) = Uniform( , )
The parameter qc would have a prior log10(qc) = c*log10(q) = Uniform( , )
This prior is therefore invariant under a number of transformations. Replotting Figure 2 on a log scale shows the relationship graphically:
Figure 3: Plotting q on a log scale against the cumulative distributions
for Figure 2 shows their reflective relationship.
Using the Jacobian transformation it can be shown that log(q) = Uniform( , ) is equivalent to the prior density
LaTeX Math Inline
body \pi(\theta)\propto 1/\theta
for a parameter that can take any positive real value. You probably wouldn't describe that distribution as very uninformed, but it is arguably the best one can do for this particular problem. It is worth remembering too that if there is a reasonable amount of data available the likelihood function l(X|q) will overpower the prior p(q) = 1/q and then the shape of the prior becomes unimportant. This will occur much more quickly if the likelihood function is a maximum in a region of q where the prior is flatter.
A location parameter of a distribution should have the same effective prior irrespective of the scaling chosen. This is achieved if we select a Uniform(a,a+b) prior for q, i.e. p(q) = 1/b where b is the range of the Uniform distribution. Changing the scale has no effect on the shape of the prior. For example, a prior Uniform(0,10kg) has a density p(q) = 0.1kg-1. Changing the units to grams, we would have a prior Uniform(0,10000g) which again has the constant density p(q) = 10-4g-1.
Parametric distributions often have either of or both a location parameter and a scale parameter. If more than one parameter is unknown and one is attempting to estimate these parameters, it is common practice to assume independence between the two parameters in the prior: the logic is that assumption of independence is more uninformed than an assumption of any specific degree of dependence. The joint prior for a scale parameter and a location parameter is then simply the product of the two priors. So, for example, the prior for the mean of a normal distribution is
LaTeX Math Inline
body \pi(\mu)\propto 1
since m is a location parameter; the prior for the standard deviation of the normal distribution is
LaTeX Math Inline
body \pi(\sigma)\propto 1/\sigma
since s is a scale parameter, and their joint prior is given by the product of the two priors, i.e.
LaTeX Math Inline
body \pi(\mu,\sigma)\propto 1/\sigma
. | {"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": 0.9496296048164368, "perplexity": 736.5878098004703}, "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/1674764500456.61/warc/CC-MAIN-20230207102930-20230207132930-00612.warc.gz"} |
http://cms.math.ca/Competitions/MOCP/2002/prob_feb.mml | Canadian Mathematical Society www.cms.math.ca
location:
PROBLEMS FOR FEBRUARY
Valeria Pandelieva
641 Kirkwood Avenue
Ottawa, ON K1Z 5X5
no later than March 15, 2002.
127.
Let
$A={2}^{n}+{3}^{n}+216\left({2}^{n-6}+{3}^{n-6}\right)$
and
$B={4}^{n}+{5}^{n}+8000\left({4}^{n-6}+{5}^{n-6}\right)$
where $n>6$ is a natural number. Prove that the fraction $A/B$ is reducible.
128.
Let $n$ be a positive integer. On a circle, $n$ points are marked. The number 1 is assigned to one of them and 0 is assigned to the others. The following operation is allowed: Choose a point to which 1 is assigned and then assign $\left(1-a\right)$ and $\left(1-b\right)$ to the two adjacent points, where $a$ and $b$ are, respectively, the numbers assigned to these points before. Is it possible to assign 1 to all points by applying this operation several times if (a) $n=2001$ and (b) $n=2002$?
129.
For every integer $n$, a nonnegative integer $f\left(n\right)$ is assigned such that
(a) $f\left(\mathrm{mn}\right)=f\left(m\right)+f\left(n\right)$ for each pair $m,n$ of natural numbers;
(b) $f\left(n\right)=0$ when the rightmost digit in the decimal representation of the number $n$ is 3; and
(c) $f\left(10\right)=0$.
Prove that $f\left(n\right)=0$ for any natural number $n$.
130.
Let $\mathrm{ABCD}$ be a rectangle for which the respective lengths of $\mathrm{AB}$ and $\mathrm{BC}$ are $a$ and $b$. Another rectangle is circumscribed around $\mathrm{ABCD}$ so that each of its sides passes through one of the vertices of $\mathrm{ABCD}$. Consider all such rectangles and, among them, find the one with a maximum area. Express this area in terms of $a$ and $b$.
131.
At a recent winter meeting of the Canadian Mathematical Society, some of the attending mathematicians were friends. It appeared that every two mathematicians, that had the same number of friends among the participants, did not have a common friend. Prove that there was a mathematician who had only one friend.
132.
Simplify the expression
$\sqrt[5]{3\sqrt{2}-2\sqrt{5}}·\sqrt[10]{\frac{6\sqrt{10}+19}{2}} .$
© Canadian Mathematical Society, 2014 : https://cms.math.ca/ | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 31, "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": 0.8903192281723022, "perplexity": 296.4294823164936}, "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/1414637899041.10/warc/CC-MAIN-20141030025819-00180-ip-10-16-133-185.ec2.internal.warc.gz"} |
http://tex.stackexchange.com/questions/37029/text-out-of-margins/37034 | # Text out of margins
I don't know why, but a line in my text is out of margins.
How can I avoid this issue?
I'm using classicthesis in TeXworks.
Edit: with the command
\overfullrule=5pt
I can see a black square where this issue appear.
\documentclass{scrreprt}
\usepackage[english]{babel}
\usepackage{myclassicthesis-preamble}
\usepackage{classicthesis}
\usepackage{microtype}
\renewcommand{\descriptionlabel}[1]{\hspace*{\labelsep}\small\textsc{#1}}
\overfullrule=5pt
\begin{document}
\begin{description}
\item[Service Delivery Framework] The infrastructure to create, publish, manage and consume FI services across their life cycle, addressing all technical and business aspects.
\item[Interface to the Network and Devices] The open interfaces to networks and devices, harmonizing the connectivity needs of services built on top of the platform.
\end{description}
Lastly, on the \emph{political dimension}, legal and legislative barriers presently hinder the efficient cross-border establishment of new innovative solutions due to complex or incompatible ICT policies in Europe.
\end{document}
Two possible problems: this line
\renewcommand{\descriptionlabel}[1]{\hspace*{\labelsep}\small\textsc{#1}}
and the modification in myclassicthesis-preamble(lines 63-70)
\PassOptionsToPackage{
eulerchapternumbers,
beramono,
eulermath,
pdfspacing,
floatperchapter
}
{classicthesis}
in particular the option pdfspacing.
-
Can you try replacing the word presently with present\-lyor put \hyphenation{pre-sent-ly} in the preamble? Probably LaTeX doesn't know how to hyphen that word. – percusse Dec 4 '11 at 0:53
While code snippets are useful in explanations, it is always best to compose a fully compilable MWE with the paragraph that illustrates the problem.. – Peter Grill Dec 4 '11 at 1:12
@percusse with pre\-sent\-ly works. But now I realized that there are a lot of these "out of margins". I must to redefine all the english words? =| – Baduel Dec 4 '11 at 1:15
@PeterGrill I have added the MWE in my updated question. – Baduel Dec 4 '11 at 1:34
That is certainly not very _minimal_. You need to reduce your preamble and eliminate things that are not related, and then post the code here. – Peter Grill Dec 4 '11 at 1:52
TeX doesn't hyphenate presently because of two concurring factors:
1. the word present cannot be hyphenated automatically, because it changes its syllables when it's a verb or a noun (just like record);
2. the right hyphenation minimum for English is set to three, that is, TeX is not allowed to leave less than three characters of a hyphenated word on the new line.
If we set \righthyphenmin=2, then \showhyphens{presently} gives present-ly as expected. Loading ushyphenmax via hyphsubst will not hyphenate presently, for the same reason, as \righthyphenmin=3. Saying \hyphenation{pres-ent-ly} in the preamble would allow hyphenation as pres-ently (but not present-ly, because two letters are too few.).
In an emergency situation one can allow hyphenation by typing present\-ly that will override the default minimum.
The format set up by classicthesis is quite strict and your text shows the problems that can arise with long description labels. There's not very much to do apart modifying, if possible, the text. I would avoid \sloppy as much as possible, since it allows producing badly spaced paragraphs.
A better strategy is to use \emergencystretch: with the following modifications
\begingroup\emergencystretch=.3em
\begin{description}
\item[Service Delivery Framework] The infrastructure to create, publish, manage and consume
FI services across their life cycle, addressing all technical and business aspects.
\item[Interface to the Network and Devices] The open interfaces to networks and devices,
harmonizing the connectivity needs of services built on top of the platform.
\end{description}
\endgroup
Lastly, on the \emph{political dimension}, legal and legislative barriers present\-ly hinder
the efficient cross-border establishment of new innovative solutions due to complex or
incompatible ICT policies in Europe.
the text will be typeset without any overfull box and not overly bad spacing.
One should also note that microtype is loaded automatically by classicthesis (line 219 in classicthesis.sty says \RequirePackage{microtype).
-
While present has two possible hyphenations, presently has only one, "pres-ent-ly" (at least according to my dictionaries). Splitting up "-ly" probably wouldn't help much anyway in this situation because the complete "ly" is right of the border, and thus the hyphen would be, too. However "pres-ently" would not violate the three-letter-limit, so why would \hyphenation{pres-ent-ly} not help? – celtschk Dec 4 '11 at 13:37
With .4em it works. I would also avoid \sloppy as much as possible. Anyway I realized that I have really too much overfull box, most of which are resolved if I don't use the option pdfspacing in myclassicthesis-preamble. What are the issues that I could find in the future if I decide to not load it? – Baduel Dec 4 '11 at 13:41
From the classicthesis documentation pag. 6: pdfspacing makes use of pdftex’ letter spacing capabilities via the microtype package. This fixes some serious issues regarding math formulæ etc. (e. g., “ß”) in headers. I'm afraid about serious issues. – Baduel Dec 4 '11 at 13:46
@Baduel The option pdfspacing is relative to letter spacing; if it's active the letter spacing is done via microtype, otherwise via soul. I'd recommend the former. – egreg Dec 4 '11 at 13:46
@celtschk \hyphenation{pres-ent-ly} would give a hyphenation point pres-ently, but not present-ly because of \righthyphenmin=3. Good point, anyway. – egreg Dec 4 '11 at 13:49
Edit: It seems like you have a few particularly unlucky places in your document, where only a \sloppy will help. Nonetheless, here is some information that might help understanding hyphenation a bit better and getting fewer of these places. Note that the following only applies to American hyphenation patterns, which is the default loaded by (La)TeX and loaded by babel's options english, american and usenglish, which are all the same.
As has been mentioned, TeX doesn't know how to hyphenate presently. This can be seen by compiling a document containing \showhyphens{presently}; subsequently, the hyphenation points of presently will be put in the .log file -- turns out, there are none. As @percusse noted, adding \hyphenation{pre-sent-ly} to your preamble will solve the problem for this word, but you noted that there are more. So here's what else you could do:
### Load the latest hyphenation patterns
Insert
\RequirePackage[english=usenglishmax]{hyphsubst}
even before your \documentclass{...} to load updated hyphenation patterns for \usepackage[english]{babel} (American patterns).
### Include the known hyphenation exceptions by hand
The Hyphenation Exception Log (Nov 2010) is "the periodic update of the list of words that TeX fails to hyphenate properly" (ibid., p. 1001). On page 1005, it confirms that TeX doesn't know how to hyphenate presently, and lists the desired hyphenation pres-ent-ly.
Here's the relevant paragaph on how to use this list of hyphenation exceptions:
Converting this list into machine-usable hyphenation exceptions
Werner Lemberg has created a script that will convert this article into a real \hyphenation block that can be incorporated into a document either directly or by inputting a file. His work has necessitated some changes to the macros used to format the list, but the appearance of the list will not change. Many inflected forms will be included automatically, some evident in the printed version, but many included silently. The script, hyphenex.sh, is a straightforward shell script and is posted on CTAN, in tex-archive/info/digests/tugboat/hyphenex/ The output of the script is posted in the same area as ushyphex.tex.
(ibid., p. 1001)
Besides that, I second Peter's suggestion of using the microtype package, which improves hyphenation in general and gives your document a more even grayness factor.
-
This solution can be used for hyphenation problem. However I also have the overfull problem in words not hyphenated (see my example). The microtype package is, in any case, loaded by the classicthesis package. Finally, is english=ukenglish the right option for hyphsubst for uk english? – Baduel Dec 4 '11 at 12:12
With the unhyphenated words, I'm afraid \sloppy (and a timely \fussy) will be your only help. The hyphsubst is only available for english, which equals usenglish; afaik there aren't any updated hyphenation patterns for British hyphenation. – doncherry Dec 4 '11 at 12:33
Without a MWE it is difficult to know for sure.
The usual reason this occurrs is that TeX was not able to meet all the rules for spacing as per the Overfull \hbox messages. One way to fix this is to add \sloppy which relaxes the rules for inter word spacing. See the answers to Why is text being placed beyond the specified line width? for a more detailed explanation.
Or you could add \usepackage{microtype} to the preamble. See the microtype package documentation for more details.
-
I have added the MWE in my updated question. – Baduel Dec 4 '11 at 1:33
I was just about to suggest microtype. – Ryan Reich Dec 4 '11 at 3:08
Maybe LaTeX doesn't know how to hyphenate "presently"? In that case you can just tell it how to do so by writing it as pres\-ent\-ly.
I think that the right hypernation is pre\-sent\-ly (and it works). Now the question is: why I have a lot of these "out of margins" in my document? Is it possible that the default package for the hypernation doesn't spell english in a good way? – Baduel Dec 4 '11 at 1:13
In fact TeX doesn't know how to hyphenate presently, \showhyphens{presently} puts presently in the .log file, and it's listed in mirrors.ctan.org/info/digests/tugboat/hyphenex/tb0hyf.pdf (cf. tex.stackexchange.com/questions/22867/…), where it says the correct form would be pres-ent-ly. – doncherry Dec 4 '11 at 11:23
@doncherry Ok I will use pres\-ent\-ly to avoid this issue. However I found other overfull problem in my text, probably related to myclassicthesis-preamble or the \newcommand specified. – Baduel Dec 4 '11 at 11:50 | {"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": 0.833450973033905, "perplexity": 4599.587545200699}, "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-2014-10/segments/1394678683789/warc/CC-MAIN-20140313024443-00005-ip-10-183-142-35.ec2.internal.warc.gz"} |
https://math.stackexchange.com/questions/1820448/is-there-any-formula-to-calculate-the-number-of-different-pythagorean-triangle-w | # Is there any formula to calculate the number of different Pythagorean triangle with a hypotenuse length $n$, using its prime decomposition?
Lets define $N(n)$ to be the number of different Pythagorean triangles with hypotenuse length equal to $n$.
One would see that for prime number $p$, where $p=2$ or $p\equiv 3 \pmod 4$, $N(p)=0$ also $N(p^k)=0$.
e.g. $N(2)=N(4)=N(8)=N(16)=0$
But for prime number $p$, where $p\equiv 1 \pmod 4$, $N(p)=1$ and $N(p^k)=k$.
e.g. $N(5)=N(13)=N(17)=1$ and $N(25)=2$ and $N(125)=3$
If $n=p^kq_1^{a_1}\dots q_r^{a_r}$, where $p$ be a prime of the form $4k+1$ and $q_i$'s be primes of the form $4k+3$ or be equal to $2$, then $N(n)=k$.
e.g. $N(14000)=N(5^3\times 2^4 \times 7)=3$
And also, If $n=p_1p_2q_1^{a_1}\dots q_r^{a_r}$, where $p_1$ and $p_2$ be primes of the form $4k+1$ and $q_i$'s be primes of the form $4k+3$ or be equal to $2$, then $N(n)=4$.
e.g. $N(65)=N(85)=4$
The question is:
Is there any formula to calculate $N(n)$, where $n=p_1^{a_1}\dots p_r^{a_r}$, by means of $N(p_1)$, … , $N(p_r)$?
A more general question is to compute $r_2(n)$, the number of ways an integer $n$ can be written as the sum of two squares (not ignoring order, and including negative numbers and $0$; this makes the answer nicer). The answer is classical and due to Jacobi: it turns out that
$$r_2(n) = 4 \left( d_1(n) - d_3(n) \right)$$
where $d_1(n)$ is the number of divisors of $n$ congruent to $1 \bmod 4$ and $d_3(n)$ is the number of divisors of $n$ congruent to $3 \bmod 4$. From here it's not much harder to ignore $0$, negative numbers, and order, but it makes the answer a bit less nice.
So the answer is something like $N(n)=\frac12[(2a_1+1)\dots(2a_r+1)-1]$ but just in case $n=p_1^{a_1}\cdots p_r^{a_r}$, where $p_i$'s are prime and $p_i\equiv 1 \pmod 4$, for $i=1,\dots,r$. | {"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": 0.9904440641403198, "perplexity": 60.18798047823325}, "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-2019-26/segments/1560627999539.60/warc/CC-MAIN-20190624130856-20190624152856-00015.warc.gz"} |
https://jmichaux.github.io/_notebook/2018-6-03-softmax/ | # Deriving the loss function for softmax regression
Imagine we have a dataset $D =\left\{(\textbf{x}^{(i)}, {y}^{(i)})\right\}$ consisting of N training examples, where each feature vector $\textbf{x}^{(i)} \in \mathbb{R}^{m}$ and label ${y}^{(i)} \in \left\{1,...,K\right\}$. Our goal is to build a model that takes as input a new example $x^{*}$ and predicts its class label $y^{*}$. One straight-forward approach to deriving supervised learning algorithms is to use maximum likelihood estimation to model the class-conditional probability:
where
and
In our case, we will be modeling our multinomial distribution with the softmax function:
To make the following calculations easier, we change the form of each label $y^{(i)} \in \mathbb{R}$ to $\textbf{y}^{(i)} \in \mathbb{R}^{K}$ using one-hot encoding. For example, if $y^{(i)}=3$ and $K=5$, then:
By encoding $y^{(i)}$ using one-hot encoding, we get a slightly modified dataset $D =\left\{\textbf{x}^{(i)}, \textbf{y}^{(i)}\right\}$, where $\textbf{x}^{(i)} \in \mathbb{R}^{m} , \textbf{y}^{(i)} \in \mathbb{R}^{m}$, and $i = 1,...,N$. If the data are $i.i.d.$, we can use maximum-likelihood estimation to find the parameters $\textbf{W}$ and $\textbf{b}$ that maximize the likelihood of the data:
Instead of maximizing $L(\textbf{W}, \textbf{b})$, we can instead minimize its negative log-likelihood. This is equivalent because $(i)$ the natural log function is monotonic, and $(ii)$ maximizing a function is the same is minimizing the negative of that function.
Note that in equation $(7)$, we were able to simplify the right hand side of the expression because $\sum_{k=1}^K {\textbf{u}_{k}^T\textbf{y}^{(i)}} = 1$.
Now that we have an expression for $\ell(\textbf{W}, \textbf{b})$, we can choose from a number of algorithms to minimize the loss. In a future note, we are going to derive parameter update equations for minimizing $\ell(\textbf{W}, \textbf{b})$ with stochastic gradient descent. | {"extraction_info": {"found_math": true, "script_math_tex": 23, "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": 2, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9594367742538452, "perplexity": 170.93796031952579}, "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-2019-51/segments/1575540490972.13/warc/CC-MAIN-20191206200121-20191206224121-00376.warc.gz"} |
http://mymathforum.com/real-analysis/13436-discrete-fourier-vs-fourier-transform.html | My Math Forum Discrete fourier vs fourier transform
Real Analysis Real Analysis Math Forum
June 19th, 2010, 02:03 AM #1 Newbie Joined: Jun 2010 Posts: 2 Thanks: 0 Discrete fourier vs fourier transform Hi, I have some questions regarding basic understanding of Fourier transformation 1)Using discrete fourier transform, you only analyse for some frequencies. Do you analyse for a range of frequencies, frequencies with a given interval or only specific frequencies? 2) What are the advatages of discrete fourier compared with "normal" fourier? 3) What is windowed fourier? and when is that used? 4) What are the advantages of Fourier and wavelet respectively? It would help a lot, if anyone could answer these questions, without too much math Thanks, Beckie
June 19th, 2010, 01:04 PM #2 Global Moderator Joined: May 2007 Posts: 6,704 Thanks: 670 Re: Discrete fourier vs fourier transform Fourier analysis is divided into two broad categories. Fourier series, where the function is defined over a finite interval, and is implicitly (or explicitly) periodic outside the interval. Fourier transforms are defined for functions defined over an infinite interval and belong to to some Lp class for p ? 1.
June 20th, 2010, 12:16 AM #3 Newbie Joined: Jun 2010 Posts: 2 Thanks: 0 Re: Discrete fourier vs fourier transform When you say a finit interval, do you mean time interval or frequency interval? what does Lp class mean?!? Do you know any answers to my specific questions? Cause I really need to know it by monday
June 20th, 2010, 12:58 PM #4
Global Moderator
Joined: May 2007
Posts: 6,704
Thanks: 670
Re: Discrete fourier vs fourier transform
Quote:
Originally Posted by beckie When you say a finit interval, do you mean time interval or frequency interval? what does Lp class mean?!? Do you know any answers to my specific questions? Cause I really need to know it by monday
Your questions tell me that you are involved in an engineering course. My background is mathematics, so I can't answer your questions directly.
When I said finite interval, I meant that the function f(x) is defined over a finite interval and can be represented by a Fourier series consisting of a constant plus cosines and sines of qnx, where q is a constant determined by the interval while n ranges over all positive integers. For example if the interval is of length 1, q=2?.
A function f(x) belongs to an Lp class if the integral over the real line of |f(x)|^p exists. Most Fourier analysis studies are concerned with p = 1 or 2.
Tags discrete, fourier, transform
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Contact - Home - Forums - Cryptocurrency Forum - Top | {"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": 0.8574360609054565, "perplexity": 1632.7627253509343}, "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-2019-13/segments/1552912202526.24/warc/CC-MAIN-20190321152638-20190321174638-00085.warc.gz"} |
http://mathhelpforum.com/differential-geometry/181481-about-short-exact-sequence-relative-homology-groups.html | 1. ## About a short exact sequence of relative homology groups.
Let $x$ be a point in $X$. My question is about the following short exact sequence:
$0 \to H_1(X) \to H_1(X,\{x\}) \to H_0(\{x\}) \: \xrightarrow{i_\ast} \: H_0(X)\to H(X,\{x\})\to 0,$
where $i_* : H_0 (\{ x \} ) \to H_0(X)$ is the homomorphism induced by the obvious inclusion. I am told that $i_*$ is injective. Although this does make some sense intuitively, I can't really prove it. Can someone show me why it is?
2. ## Re: About a short exact sequence of relative homology groups.
Cryptic as it may sound, it holds because all definitions become trivially true for inclusion maps. | {"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": 5, "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": 0.9775195717811584, "perplexity": 135.24123501178255}, "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/1480698541905.26/warc/CC-MAIN-20161202170901-00020-ip-10-31-129-80.ec2.internal.warc.gz"} |
http://www.math.gatech.edu/seminars-colloquia/series/graph-theory-seminar/gelasio-salazar-20091020 | ## Sylvester's Four Point Constant: closing in (or are we?)
Series:
Graph Theory Seminar
Tuesday, October 20, 2009 - 12:05
1.5 hours (actually 80 minutes)
Location:
Skiles 255
,
Universidad Autonoma de San Luis Potosi
Organizer:
In 1865, Sylvester posed the following problem: For a region R in the plane,let q(R) denote the probability that four points chosen at random from Rform a convex quadrilateral. What is the infimum q* of q(R) taken over allregions R? The number q* is known as Sylvester's Four Point Problem Constant(Sylvester's Constant for short). At first sight, it is hard to imagine howto find reasonable estimates for q*. Fortunately, Scheinerman and Wilf foundthat Sylvester's Constant is intimately related to another fundamentalconstant in discrete geometry. The rectilinear crossing number of rcr(K_n)the complete graph K_n is the minimum number of crossings of edges in adrawing of K_n in the plane in which every edge is a straight segment. Itis not difficult to show that the limit as n goes to infinity ofrcr(K_n)/{n\choose 4} exists; this is the rectilinear crossing numberconstant RCR. Scheinerman and Wilf proved a surprising connection betweenthese constants: q* = RCR. Finding estimates of rcr(K_n) seems like a moreapproachable task. A major breakthrough was achieved in 2003 by Lovasz,Vesztergombi, Wagner, and Welzl, and simultaneously by Abrego andFernandez-Merchant, who unveiled an intimate connection of rcr(K_n) withanother classical problem of discrete geometry, namely the number of | {"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": 0.8258510231971741, "perplexity": 2262.5918178736465}, "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/1531676589902.8/warc/CC-MAIN-20180717203423-20180717223423-00577.warc.gz"} |
http://cran.r-project.org/web/packages/santaR/vignettes/theoretical-background.html | # santaR Theoretical Background
#### 2019-10-03
santaR is a Functional Data Analysis (FDA) approach where each individual’s observations are condensed to a continuous smooth function of time, which is then employed as a new analytical unit in subsequent data analysis.
santaR is designed to be automatically applicable to a high number of variables, while simultaneously accounting for:
• biological variability
• measurement error
• missing observations
• asynchronous sampling
• nonlinearity
• low number of time points (e.g. 4-10)
This vignette focusses on the theoretical background underlying santaR and will detail:
• Concept of Functional Data Analysis
• Main hypothesis
• Signal extraction using splines
• Group mean curve fitting
• Intra-class variability (confidence bands on the Group mean curves)
• Detection of significantly altered time-trajectories
## Concept of Functional Data Analysis
Traditional times-series methodologies consider each measurement (or observation) - for example a metabolic response value at a defined time-point - as the basic representative unit. In this context, a time trajectory is composed of a set of discrete observations taken at subsequent times. Functional data analysis1 2 3 is a framework proposed by Ramsay and Silverman which models continuously each measured values as variable evolves (e.g. time). Contrary to traditional time-series approaches, FDA consider a subject’s entire time-trajectory as a single data point. This continuous representation of the variable through time is then employed for further data analysis.
More precisely, FDA assumes that a variable’s time trajectory is the result of a smooth underlying function of time which must be estimated. Therefore, the analysis’ first step is to convert a subject’s successive observations to the underlying function $$F(t)$$ that can be evaluated at any desired time values
Two main challenges prescribe the exact characterisation of the true underlying function of time $$F(t)$$. First, the limited number of discrete observations constrains to a finite time interval and does not cover all possible values, while the true underlying smooth process is infinitely defined. Secondly, intrinsic limitations and variabilities of the analytical platforms mean that the discrete observations are not capturing the true underlying function values, but a succession of noisy realisations.
A metabolite’s concentration over the duration of the experiment is the real quantity of interest. As the true smooth process $$F(t)$$ cannot be evaluated, due to the collection and analytical process only providing noisy snapshots of its value, an approximation of the underlying metabolite concentration $$f(t)$$ must be employed. This approximation can be computed by smoothing the multiple single measurements available.
The approximation of the underlying function of time, of a single variable observed on a single subject, is generated by smoothing $$y_i$$ observations taken at the time $$t_i$$, such as:
$y_{i}=f(t_{i})+\epsilon_{i}$
Where $$f(t)$$ is the smoothed function that can be evaluated at any desired time $$t$$ within the defined time domain, and $$\epsilon_i$$ is an error term allowing for measurement errors. By extension, if $$f(t)$$ is defined as representing the true variable level across a population, $$\epsilon$$ can allow for both individual specific deviation and measurement error.
In order to parametrise and computationally manipulate the smooth approximation, the representation is projected onto a finite basis, which expresses the infinitesimally dimensional $$f(t)$$ as a linear combination of a set of basis functions. A wide range of basis functions (e.g. polynomials, splines, wavelets, Fourier basis) are available, even if the most commonly used are Fourier basis for periodic data and splines for aperiodic trajectories4.
Following smoothing, a time-trajectory is expressed as a linear combination of a set of basis functions, for which values can be computer at any desired time without reconsidering the discrete observations. It is assumed that, if a satisfactory fit of the original data is obtained, both the new smooth approximation and the preceding noisy realisations should provide a similar representation of the underlying process5 6
In practice, by assuming and parametrising (explicitely or implicitely) the smoothness of the function and leveraging multiple measurements in time as well as multiple individual trajectories, the smoothed representation can implicitly handle noisy metabolic responses. Additionally, as functional representations do not rely on the input data after their generation, smoothing approaches can handle irregular sampling and missing observations, translating to trajectories that can be compared without requiring missing values imputation. As such, the FDA framework and the smoothness assumption provide the necessary tools for the time-dependent analysis of short and noisy metabolic time-trajectories.
## Main hypothesis
The working hypothesis is that analyte values are representative of the underlying biological mechanism responsible for the observed phenotype and evolve smoothly through time. The resulting measurements are noisy realisations of these underlying smooth functions of time.
Based on previous work performed in Functional Data Analysis, short time evolutions can be analysed by estimating each individual trajectory as a smooth spline or curve. These smooth curves collapse the point by point information into a new observational unit that enables the calculation of a group mean curve or individual curves, each representative of the true underlying function of time.
Further data analysis then results in the estimation of the intra-class variability and the identification of time trajectories significantly altered between classes.
## Signal extraction using splines
Fitting splines to time trajectories is equivalent to a denoising or signal extraction problem: balancing the fitting of raw data versus the smoothing of the biological variability and measurement errors. Splines are piecewise polynomials with boundary continuity and smoothness constraints, which require a minimum of assumptions about the underlying form of the data7 8 9. The complexity of the fit is controlled by a single smoothing parameter $$\lambda$$.
The data $$y_{ij}$$ of individual $$i$$, $$i=1,.,n_k$$, observed at time points $$t_{ij}$$, $$j=1,.,m_i$$, in a group of $$n_k$$ individuals, is modelled by a smoothing spline $$f_i(t_{ij})$$ of continuous first and second derivatives that minimizes the penalized residual sum of squares (PRSS) (Eq. 1):
$PRSS(f_{i},\lambda)=\sum_{j=1}^{m_{i}} \{y_{ij} - f_i(t_{ij})\}^2 + \lambda \int_{t_{min}}^{t_{max}}\{\ddot{f_i(t)}\}^2~dt$ (Eq. 1)
The first term of Eq. 1 is the squared error of employing the curve $$f_i(t)$$ to predict $$y_{ij}$$, measuring the closeness of the function to the data. The second term penalises the curvature of the function while $$\lambda$$ controls the trade-off between both terms. The smoothing parameter $$\lambda$$ is shared among all individuals. The monotonic relationship between the smoothing parameter $$\lambda$$ and the desired equivalent number of degrees of freedom (df) of the corresponding fit (Eq. 2), allows us to control the closeness of curve fitting by varying df10.
$df_{\lambda}=trace(S_{\lambda})$ (Eq. 2)
where $$S_{\lambda}$$ is the smoother matrix (roughness matrix, or hat matrix) which transforms the observed data vector to the vector of fitted values (for more information see Eubank11, and Buja et al.12).
The df parameter controls how closely the spline fits the input data. For example, when df is minimal (df=2), the smoothing is maximised and the original data points fitted with a straight line. On the other hand, when df approaches the number of time-points, every data point is fitted without smoothing, which can result in an over-fit of the raw measurements.
library(ggplot2)
library(gridExtra)
library(grid)
# Sample a gaussian distribution, add noise to it
x <- c(-4, -3, -2, -1.5, -1, -0.5, 0, 0.5, 1, 1.5, 2, 3, 4)
y <- dnorm(x, mean = 0, sd = 1, log=FALSE)
y_noise <- jitter(y, 90)
# Fit different smoothing splines, project on a grid for plotting
time = seq(-4, 4, 0.01 )
tmp_fit0 = smooth.spline(x, y, df=13)
tmp_fit1 = smooth.spline(x, y_noise, df=2)
tmp_fit2 = smooth.spline(x, y_noise, df=5)
tmp_fit3 = smooth.spline(x, y_noise, df=13)
pred0 = predict( object=tmp_fit0, x=time )
pred1 = predict( object=tmp_fit1, x=time )
pred2 = predict( object=tmp_fit2, x=time )
pred3 = predict( object=tmp_fit3, x=time )
tmpPred0 = data.frame( x=pred0$x, y=pred0$y)
tmpPred1 = data.frame( x=pred1$x, y=pred1$y)
tmpPred2 = data.frame( x=pred2$x, y=pred2$y)
tmpPred3 = data.frame( x=pred3$x, y=pred3$y)
tmpRaw = data.frame( x=x, y=y)
tmpNoisy = data.frame( x=x, y=y_noise)
# Plot original data and spline representations
p0 <- ggplot(NULL, aes(x), environment = environment()) + theme_bw() + xlim(-4,4) + ylim(-0.1,0.5) + theme(axis.title.y = element_blank(), axis.ticks = element_blank(), axis.text.x = element_blank(), axis.text.y = element_blank(), plot.margin=unit(c(0.5,0.25,0.5,0), "cm"))
p0 <- p0 + geom_point(data=tmpRaw, aes(x=x, y=y), size=1.5 )
p0 <- p0 + geom_line( data=tmpPred0, aes(x=x, y=y), linetype=1, color='springgreen3' )
p0 <- p0 + xlab(expression(atop('True underlying', paste('function of time'))))
p1 <- ggplot(NULL, aes(x), environment = environment()) + theme_bw() + xlim(-4,4) + ylim(-0.1,0.5) + theme(axis.title.y = element_blank(), axis.ticks = element_blank(), axis.text.x = element_blank(), axis.text.y = element_blank(), plot.margin=unit(c(0.5,0.25,0.5,-0.25), "cm"))
p1 <- p1 + geom_point(data=tmpNoisy, aes(x=x, y=y), size=1.5 )
p1 <- p1 + geom_line( data=tmpPred1, aes(x=x, y=y), linetype=1, color='blue' )
p1 <- p1 + xlab(expression(atop(lambda*' -> '*infinity, paste('DF = 2'))))
p2 <- ggplot(NULL, aes(x), environment = environment()) + theme_bw() + xlim(-4,4) + ylim(-0.1,0.5) + theme(axis.title.y = element_blank(), axis.ticks = element_blank(), axis.text.x = element_blank(), axis.text.y = element_blank(), plot.margin=unit(c(0.5,0.25,0.5,-0.25), "cm"))
p2 <- p2 + geom_point(data=tmpNoisy, aes(x=x, y=y), size=1.5 )
p2 <- p2 + geom_line( data=tmpPred2, aes(x=x, y=y), linetype=1, color='blue' )
p2 <- p2 + xlab(expression(atop('Optimal '*lambda, paste('Optimal DF', sep=''))))
p3 <- ggplot(NULL, aes(x), environment = environment()) + theme_bw() + xlim(-4,4) + ylim(-0.1,0.5) + theme(axis.title.y = element_blank(), axis.ticks = element_blank(), axis.text.x = element_blank(), axis.text.y = element_blank(), plot.margin=unit(c(0.5,0.25,0.5,-0.25), "cm"))
p3 <- p3 + geom_point(data=tmpNoisy, aes(x=x, y=y), size=1.5 )
p3 <- p3 + geom_line( data=tmpPred3, aes(x=x, y=y), linetype=1, color='blue' )
p3 <- p3 + xlab(expression(atop(lambda*' = 0', paste('DF = N'))))
grid.arrange(p0,p1,p2,p3, ncol=4)
Multiple smoothing-spline fits of noisy realisations generated from a true smooth underlying function of time . A true smooth underlying function (continuous green line) is observed at multiple time points (dots). Noise is added to these measurements and the underlying function is approximated using smooth-splines at multiple levels of smoothness (blue continuous line).
• Excessive smoothness makes the fit less sensitive to the data, resulting in an under-fitted model.
• When the curvature is not penalised an interpolating spline passing exactly through each point is selected, resulting in an over-fitted model.
• At intermediate levels of smoothness, a compromise between the goodness-of-fit and model complexity can be achieved, generating a satisfactory approximation of the true underlying function of time.
In practice, the results are resilient to the df value selected across all variables, which is a function of the study design, such as the number of time-points, sampling rate and, most importantly, the complexity of the underlying function of time. While an automated approach cannot infer the study design from a limited set of observations, an informed user will intuitively achieve a more consistent fit of the data.
santaR considers df as the single meta-parameter a user must choose prior to analysis, balancing the fitting of raw observations versus the smoothing of the biological variability and measurements errors. In turn, this explicit selection of the model complexity reduces the execution time and increases the reproducibility of results for a given experiment and across datasets.
For more information regarding the selection of an optimal value of df see the vignette ‘Selecting an optimal number of degrees of freedom’
## Group mean curve fitting
Individual time-trajectories provide with subject-specific approximations of the underlying function of time. If a homogenous subset of the population (or group) can be devised, the replication of the experiment over this group can be leveraged to generate a group mean curve. This group mean curve provide with an approximation of the underlying function of time potentially dissociated of inter-individual variability.
A group mean curve $$g(t)$$, representing the underlying function of time for a subset of the population, is defined as the mean response across all $$i$$ group subjects (where $$i=1,.,n_k$$) (Equation 3): $g(t)=\frac{1}{n_k} \sum_{i=1}^{n_k} f_i(t) = \overline{f_i(t)}$ (Eq. 3)
Following the framework of FDA, each individual trajectory is employed as the new basic unit of information. The group mean curve therefore employs the back-projection of each individual curve at every time of interest, instead of relying on each discrete measurement. This way the group mean curve fit is resilient to missing samples in a subject trajectory or asynchronous measurement across individuals.
Multiple group mean curve fits (Center and Right) based on measurements generated from a true smooth underlying function of time (Left) with added inter-individual variability and missing observations. Multiple individual trajectories (Left, dashed green lines) are generated from a true underlying function of time (Left, continuous green line). Discrete measurements are generated (dots). Group mean curves are computed on a subset of these measurements (Center and Right, continuous blue line, 5 effective degrees of freedom). When the group mean curve is generated solely based on measurements (Center), the true underlying function of time is not satisfactorily approximated (fundamental unit of information is the observation). When the group mean curve (Right, continuous blue line) is generated from each individual curves (Right, dashed blue lines) a satisfactory approximation of the true underlying function of time can be obtained (fundamental unit of information is the functional representation of each individual).
## Intra-class variability (confidence bands on the group mean curves)
Confidence bands are employed to represent the uncertainty in an function’s estimate based on limited or noisy data. Pointwise confidence bands on the group mean curve can therefore provide a visualisation of the variance of the overall mean function through time. The confidence bands indicate for each possible time, values of the mean curve which could have been produced by the provided individual trajectories, with at least a given probability.
As the underlying distribution of $$F(t)$$ is unknown and inaccessible, confidence on $$g(t)$$ must be estimated by sampling the approximate distribution available (i.e. the measurements), achieved by bootstrapping the observations13 14 15. Bootstrap assumes that sampling the distribution under a good estimate of the truth (the measurements) should be close to sampling the distribution under the true process. Once a bootstrapped distribution of mean curve is estimated, using the percentile method a pointwise confidence interval on $$g(t)$$ can be calculated16 17 18 by taking the $$\alpha/2$$ and $$1-\alpha/2$$ quantiles at each time-point of interest (where $$\alpha$$ is the error quantile in percent).
As limited assumptions on the fitted model can be established, empirical confidence bands on the group mean curve are calculated based on a non-parametric bootstrapped distribution. Individual curves are sampled with replacement and group mean curves calculated. The distribution of group mean curve is then employed to estimate the pointwise confidence bands.
## Detection of significantly altered time-trajectories
With a continuous time representation of a metabolite’s concentration accessible (i.e. the group mean curve), a metric quantifying a global difference between such curves can be devised. This measure of difference, independent of the original sampling of the data (e.g. asynchronous and missing measurements), can then be employed to determine whether the group mean curves are identical across the two groups and calculate the significance of the result.
In order to identify time-trajectories that are significantly altered between two experimental groups $$G_1$$ and $$G_2$$, this global measure must be established. The measure must be able to detect baseline difference when both curves present the same shape, but also shape differences when the mean values across time are identical for both groups. We postulate that the area between two group mean curves fitted independently $$g_1(t)$$ and $$g_2(t)$$ (Eq. 3) is a good measure of the degree to which the two group time-trajectories differ. This global distance measure is calculated as:
$dist(g_1,g_2) = \int_{t_{min}}^{t_{max}} |g_1(t)-g_2(t)| ~dt$ (Eq. 4)
Where:
• $$t_{min}$$ and $$t_{max}$$ are the lower and upper bound of the time course.
The global distance measure is a quantification of evidence for differential time evolution, and the larger the value, the more the trajectories differ.
Given the groups $$G_1$$, $$G_2$$, their respective group mean curves $$g_1(t)$$, $$g_2(t)$$ and the observed global distance measure $$dist(g_1,g_2)$$, the question of the significance of the difference between the two time trajectories is formulated as the following hypothesis-testing problem:
• $$H_0$$: $$g_1(t)$$ and $$g_2(t)$$ are identical.
• $$H_1$$: The two curves are not the same.
Under the null hypothesis $$H_0$$, the global distance between $$g_1(t)$$ and $$g_2(t)$$ can be generated by a randomly permuting the individuals between two groups. Under the alternate hypothesis $$H_1$$, the two curves are different, and very few random assignment of individuals to groups will generate group mean curves that differ more than the observed ones (i.e. a larger global distance).
In order to test the null hypothesis, the observed global distance must be compared to the expected global distance under the null hypothesis. The null distribution of the global distance measure is inferred by repeated permutations of the individual class labels19. Individual trajectories are fit using the chosen df and randomly assigned in one of two groups of size $$s_1$$ and $$s_2$$ (identical to the size of $$G_1$$ and $$G_2$$ respectively). The corresponding group mean curves $$g'_1(t)$$ and $$g'_2(t)$$ computed, and a global distance $$dist(g'_1,g'_2)$$ calculated. This is repeated many times to form the null distribution of the global distance. From this distribution a $$p$$-value can be computed as the proportion of global distance as extreme or more extreme than the observed global distance ($$dist(g_1,g_2)$$). The $$p$$-value indicates the probability (if the null hypothesis was true) of a global difference as, or more, extreme than the observed one. If the $$p$$-value is inferior to an agreed significance level, the null hypothesis is rejected.
Due to the stochastic nature of the permutations, the null distribution sampled depends on the random draw, resulting in possible variation of the $$p$$-value. Variation on the permuted $$p$$-value can be described by confidence intervals. The $$p$$-value is considered as the success probability parameter $$p$$ in $$n$$ independent Bernoulli trials, where $$n=B$$ the total number of permutation rounds20. Confidence intervals on this binomial distribution of parameter $$p$$ and $$n$$ can then be computed. As extreme $$p$$-values are possible (i.e close to 0) and the number of permutation rounds could be small, the Wilson score interval21 is employed to determine the confidence intervals (Eq. 5):
$p_\pm = \frac{1}{1+\frac{1}{B}z^2}\left ( p + \frac{1}{2B}z^2 \pm z\sqrt{\frac{1}{B} p (1 - p) + \frac{1}{4B^2}z^2 } \right )$ (Eq. 5)
Where:
• $$p_\pm$$ is the upper or lower confidence interval.
• $$p$$ is the calculated $$p$$-value.
• $$B$$ the total number of permutation rounds.
• $$z$$ is the $$1-1/2\alpha$$ quantile of the standard normal distribution, with $$\alpha$$ the error quantile.
Lastly, if a large number of variables are investigated for significantly altered trajectories, significance results must be corrected for multiple hypothesis testing. Here we employ the well-known Benjamini & Hochberg false discovery rate approach22 as default, with other methods available as an option.
1. Ramsay, J. & Silverman, B. W. Functional Data Analysis Springer, 431 (John Wiley & Sons, Ltd, Chichester, UK, 2005)↩︎
2. Ramsay, J., Hooker, G. & Graves, S. Functional data analysis with R and MATLAB (Springer-Verlag, 2009)↩︎
3. Ramsay, J. O. & Silverman, B. Applied functional data analysis: methods and case studies (eds Ramsay, J. O. & Silverman, B. W.) (Springer New York, New York, NY, 2002)↩︎
4. Ramsay, J. & Silverman, B. W. Functional Data Analysis Springer, 431 (John Wiley & Sons, Ltd, Chichester, UK, 2005)↩︎
5. Ramsay, J. & Silverman, B. W. Functional Data Analysis Springer, 431 (John Wiley & Sons, Ltd, Chichester, UK, 2005)↩︎
6. Ramsay, J., Hooker, G. & Graves, S. Functional data analysis with R and MATLAB (Springer-Verlag, 2009)↩︎
7. Anselone, P. M. & Laurent, P. J. A general method for the construction of interpolating or smoothing spline-functions. Numer. Math. 12, 66-82 (1968)↩︎
8. Green, P. & Silverman, B. W. Nonparametric regression and generalized linear models: a roughness penalty approach. (Chapman & Hall/CRC, 1994)↩︎
9. de Boor, C. A Practical Guide to Splines. (Springer-Verlag, 1978)↩︎
10. Hastie, T., Tibshirani, R. & Friedman, J. The Elements of Statistical Learning. (Springer, 2009)↩︎
11. Eubank, R. . The hat matrix for smoothing splines. Stat. Probab. Lett. 2, 9-14 (1984)↩︎
12. Buja, A., Hastie, T. & Tibshirani, R. Linear Smoothers and Additive Models. Ann. Stat. 17, 453-510 (1989)↩︎
13. Efron, B. Bootstrap Methods: Another Look at the Jackknife. Ann. Stat. 7, 1-26 (1979)↩︎
14. Davison, A. & Hinkley, D. Bootstrap methods and their application. (Cambridge University Press, 1997)↩︎
15. Davison, A. C., Hinkley, D. V. & Young, G. A. Recent Developments in Bootstrap Methodology. Stat. Sci. 18, 141-157 (2003)↩︎
16. Claeskens, G. & Van Keilegom, I. Bootstrap confidence bands for regression curves and their derivatives. Ann. Stat. 31, 1852-1884 (2003)↩︎
17. DiCiccio, T. J. & Efron, B. Bootstrap confidence intervals. Stat. Sci. 11, 189-228 (1996)↩︎
18. Efron, B. Better bootstrap confidence intervals. J. Am. Stat. Assoc. 82, 171-185 (1987)↩︎
19. Ernst, M. D. Permutation Methods: A Basis for Exact Inference. Stat. Sci. 19, 676-685 (2004)↩︎
20. Li, J., Tai, B. C. & Nott, D. J. Confidence interval for the bootstrap P-value and sample size calculation of the bootstrap test. J. Nonparametr. Stat. 21, 649-661 (2009)↩︎
21. Wilson, E. B. Probable Inference, the Law of Succession, and Statistical Inference. J. Am. Stat. Assoc. 22, 209 (1927)↩︎
22. Benjamini, Y. & Hochberg, Y. Controlling the False Discovery Rate: A Practical and Powerful Approach to Multiple Testing. J. R. Stat. Soc. Ser. B 57, 289 - 300 (1995)↩︎ | {"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": 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": 0.8297667503356934, "perplexity": 1829.1983026102719}, "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/1652662522741.25/warc/CC-MAIN-20220519010618-20220519040618-00753.warc.gz"} |
http://math.stackexchange.com/questions/213662/determine-for-which-values-of-a-the-equation-represents-a-circle | # Determine for which values of a the equation represents a circle
$x^2+y^2+ax+16=0$
Determine for which value of $a$ the equation represents a circle.
How would one tackle such a problem? I have no experience with these types of questions.
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Complete the square such that $x^2$ and $ax$ both result from the expansion of $(x-\beta)^2$ for some $\beta$ that depends on $a$. – Henning Makholm Oct 14 '12 at 15:16
To become the equation of a circle, you need to get rid of $ax$, and this is a pretty standard completing the square problem:
$$x^2+ax+(\frac{a}{2})^2+y^2+16= (\frac{a}{2})^2$$
or
$$(x+\frac{a}{2})^2+y^2= \frac{a^2}{4}-16$$
This is the equation of a circle if and only if the RHS is positive. Solve now
$$\frac{a^2}{4}-16 >0$$.
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You light me. Thanks for noting my mistake (+1). – Babak S. Oct 14 '12 at 15:34
@BabakSorouh no problem, we all do mistakes sometimes ;) – N. S. Oct 14 '12 at 15:35
Thank you sir, I understand. – JohnHedge Oct 14 '12 at 15:36
We complete the square to get: $$\left(x+\frac{a}{2}\right)^{2}+y^{2}-\frac{a^{2}}{4}+16=0$$ Thus $$\left(x+\frac{a}{2}\right)^{2}+y^{2}=\frac{a^2}{4}-16$$ Which is a circle with centre $(-a/2,0)$ and radius $\frac{a^{2}}{4}-16$. This only makes sense when $\frac{a^{2}}{4}-16>0$, so $a^{2}>64$, giving $a>8$ or $a<-8$.
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http://www.fluidsbarrierscns.com/content/8/1/9 | Research
A stochastic differential equation analysis of cerebrospinal fluid dynamics
Kalyan Raman
Author Affiliations
Medill IMC Department, Northwestern University, 1870 Campus Drive, Third Floor, Evanston, IL 60208, USA
Fluids and Barriers of the CNS 2011, 8:9 doi:10.1186/2045-8118-8-9
Received: 1 September 2010 Accepted: 18 January 2011 Published: 18 January 2011
This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
Abstract
Background
Clinical measurements of intracranial pressure (ICP) over time show fluctuations around the deterministic time path predicted by a classic mathematical model in hydrocephalus research. Thus an important issue in mathematical research on hydrocephalus remains unaddressed--modeling the effect of noise on CSF dynamics. Our objective is to mathematically model the noise in the data.
Methods
The classic model relating the temporal evolution of ICP in pressure-volume studies to infusions is a nonlinear differential equation based on natural physical analogies between CSF dynamics and an electrical circuit. Brownian motion was incorporated into the differential equation describing CSF dynamics to obtain a nonlinear stochastic differential equation (SDE) that accommodates the fluctuations in ICP.
Results
The SDE is explicitly solved and the dynamic probabilities of exceeding critical levels of ICP under different clinical conditions are computed. A key finding is that the probabilities display strong threshold effects with respect to noise. Above the noise threshold, the probabilities are significantly influenced by the resistance to CSF outflow and the intensity of the noise.
Conclusions
Fluctuations in the CSF formation rate increase fluctuations in the ICP and they should be minimized to lower the patient's risk. The nonlinear SDE provides a scientific methodology for dynamic risk management of patients. The dynamic output of the SDE matches the noisy ICP data generated by the actual intracranial dynamics of patients better than the classic model used in prior research.
Background
Intracranial dynamics play a central role in healthy brain function because disturbances in the internal fluid environment of the skull can lead to multiple complications such as, among other things, hydrocephalus [1]. Intracranial dynamics, driven by the circulation of CSF, are important because CSF protects the brain from injury, contains nutrients enabling normal functioning of the brain and, transports waste products away from the surrounding tissues. Much more is involved in hydrocephalus than a simple disorder of CSF circulation [2]; it is considered a complex spectrum of diseases, primarily defined by perturbation of the cranial contents--operationalized as CSF volume--and the intracranial pressure [3]. Given the complex nature of hydrocephalus, we define hydrocephalus as a disease associated with disturbances in the CSF dynamics, as in [1].
Experimental evidence compellingly validates that, over a large range of pressures, brain compliance is not constant [4]. Marmarou [5] postulated a hyperbolic compliance function that decreases as the pressure increases, which coupled with other assumptions to be described below, led to a nonlinear ordinary differential equation for the variation of ICP over time. The Marmarou model [5] is fundamental in mathematical pressure-volume models of CSF dynamics. While the Marmarou model has deservedly remained the mainstay of quantitative modeling of the dynamics of CSF flow, its deterministic nature prevents taking full advantage of the information in real ICP measurements, because deterministic models average over all possible fluctuations of real data. The ICP waveform contains additional information that is ignored by the time-averaged ICP mean value [6]. We draw upon the fundamental principles of modeling cerebrospinal fluid dynamics explicated in [2]. Our starting point is Marmorou's model [5,7] of pressure-volume compensation, which was subsequently modified in [8] and [9]. Central to the development of the Marmarou model is a conservation law. Conservation laws are ubiquitous in physics [10]. The Marmarou model represents CSF flow dynamics through a conservation equation relating the production of CSF to its storage and reabsorption [2].
(1)
Next, reabsorption is proportional to the differential between CSF pressure (p) and pressure in the sagittal sinuses (pss):
(2)
pss is considered a constant parameter, determined by central venous pressure. The coefficient R is the resistance to CSF reabsorbtion or outflow, measured in units of mmHg mL-1 min. Storage of CSF is proportional to the cerebrospinal compliance C, measured in units of mL mmHg -1.
(3)
The compliance of the cerebrospinal space is inversely proportional to the differential of CSF pressure p and the reference pressure p0 [8,11], and is considered the most important law of cerebrospinal dynamic compensation [2]:
(4)
The coefficient E is called the cerebral elasticity (or elastance coefficient) and has the units mL-1 [12]. Next, by exploiting an analogy between an electrical model of CSF compensation, as described in [5], and adapted in [2], the deterministic description of the dynamics of CSF flow are given by:
(5)
where I(t) is the rate of external volume addition and pb is a baseline pressure. The circuit diagram, reproduced from [2], is shown in Figure 1. An electrical circuit analogy is also used in [13] and [14] to study the dynamics of ICP in the ventricular compartments. The reference pressure parameter p0 is sometimes taken to be zero, as for example, in [5] because, as noted in [2], the significance of p0 is unclear. Consequently, we assume p0 = 0, which results in the equation:
Figure 1. Electrical circuit analogy for CSF flow dynamics. The current source reflects formation of CSF; resistor with diode--unilateral absorption of CSF into the sagittal sinuses; capacitor and voltage source--nonlinear compliance of CSF space. Pss is the pressure in the sagittal sinuses. P0 is the reference pressure. Figure reproduced from [2] with permission.
(6)
The importance of modeling noise in CSF dynamics
Broadly construed, noise arises from variations in factors that influence the observed outcome--which is the ICP in this paper-- but that have been omitted from the mathematical model, and from factors affecting the observed outcome that are beyond the experimenter's control. Noise causes deviations of the predicted ICP from the actual ICP level. Factors uncontrolled by the experimenter include thermal fluctuations, body movement and breathing. Because a mathematical model is an abstraction of reality, it is based on simplifying assumptions, as listed in [3]. The Marmarou model abstracts the CSF system as an electrical circuit consisting of a nonlinear capacitor (storage mechanism), resistor (area of CSF absorption), and so on [15]. There remains substantial uncertainty regarding the average rate of CSF production [16]. Realistic estimates of the mechanical properties of the living human brain are hard to discover [15]. The compliance is not an appropriate indicator of the brain's elastic properties [14]. Shunts, used in the treatment of hydrocephalus, can be dramatically improved by more accurate modeling of the CSF dynamics. Shunts providing continuous CSF drainage are the ideal [17], and nonlinear control theory can be used to design an automatic controller for a shunt that provides continuous drainage. But in order to design a stable controller to facilitate a shunt with continuous drainage, we need a model of CSF drainage that either incorporates factors omitted in extant models, or that accounts for the noise caused by the omission. Our objective is to incorporate noise into the dynamics of CSF flow. The effect of noise on the ICP waveform is discernible in Figure 2, which shows the fluctuations of the ICP around the deterministic path predicted by the deterministic Marmarou model.
Figure 2. Comparison of actual path of ICP with the path predicted by the deterministic Marmarou model. Reproduced from [1] with permission. The smooth curve without any ups and downs is the temporal path of the ICP as predicted by the deterministic Marmarou model. The fluctuations around that deterministic path are due to noise arising from the various sources discussed in the paper.
Methods
Generalizing the CSF dynamics to incorporate noisy flow
Visual examination of the time-series of ICP recordings shows that the fluctuations are smooth (unlike electrons in a wire which generate shot noise, characterized by jumps [18]), and therefore continuous state space Markov processes are appropriate to capture the noisy dynamics of CSF flow. A large class of Markov processes can be represented by SDEs, and here a methodological choice must be made--noisy dynamic processes can be represented by stochastic differential equations of the Ito type or the Stratonovich type which correspond to two different ways of introducing noise into a dynamic system. A central difference between the two is that the Stratonovich SDE uses the usual deterministic calculus whereas the Ito SDE requires a completely new stochastic calculus. Extensive conceptual, empirical and philosophical discussions of this issue exist in the literature on mathematical models of electrical, biological and physical phenomena [19-21]. The overwhelming majority of these discussions conclude that Ito processes, generated by stochastic differential equations of the Ito type, are superior to Stratonovich processes, generated by stochastic differential equations of the Stratonovich type [22,23]. Ito [24] extended standard deterministic calculus to a "stochastic calculus" applicable to functions of a wide class of continuous-time random processes, known as Ito processes. Given the SDE for the process under consideration, a result called Ito's Lemma yields the SDE driving the dynamics of a general transformation of the original process [24]. This utilitarian result allows deducing the stochastic properties of considerably complex models driven by Ito processes [23]. An essential property of Ito processes is that nonlinear functions of Ito processes remain Ito processes--a property called closure under nonlinear transformations, indispensable for practical reasons. From an empirical standpoint, a compelling advantage of Ito processes is that they often yield very precise statistical specifications for estimation [23]. An attractive property of Ito processes--on theoretical, mathematical, practical and computational grounds--is that they are Markov processes. Finally, the Ito calculus has been extended to embrace general martingale processes [25]--a development that permits joint consideration of both smooth noise and noise that occurs in jumps. Thus our modeling framework can accommodate neurological phenomena requiring noise that encompasses both smooth and jumpy variations in the state of the system, such as the firing of neurons [26].
Modeling the CSF dynamics as an Ito process to incorporate noisy flow
Given all these considerations, we modeled the fluctuations in CSF dynamics through an Ito stochastic differential equation. First we introduced noise into equation (6) through a white noise process ε(t) with intensity parameter σ, which by definition satisfies the following properties: E[ε(t)] = 0, and E[ε(t) ε(s)] = 0, whenever t ≠ s. The notation E[.] denotes the expectation operator, which, when applied to a random quantity such as ε(t), signifies the value of ε(t) on average. Thus E[ε(t)] = 0 signifies that the average value or mean of the random error at time "t" is zero, and this is a standard assumption in the literature on modeling noisy phenomena. The term E[ε(t) ε(s)] is the expectation operator applied to the product of random errors at two different times 's' and 't;' technically it denotes the covariance between the errors at two different times. In this case, because of the zero-mean assumption, it also denotes the correlation between ε(t) and ε(s); and so the property E[ε(t) ε(s)] = 0 means that the errors at two different times are uncorrelated, which substantively means that an error at one point in time does not influence the error at another point in time. This too is a standard assumption in the dynamic modeling literature.
(7)
Next we exploited the fundamental relationship between a white noise process ε(t) and a Brownian motion process W(t): , which, when written in differential notation, yields dW = ε(t) dt. Therefore,
(8)
Rearranging the above terms yields our final model, which we will call the stochastic Marmarou model.
(9)
Note that in order for equation (9) to be dimensionally consistent, the unit of σ is mL/min. Because the 'input' I(t) is the infusion rate which is under direct experimental control, therefore, in the language of control theory, I(t) is a 'control' variable. In the infusion studies conducted at Addenbrookes's Hospital in Cambridge, UK, I(t) is maintained at a constant rate of 1.5 mL/min. However, factors not within the experimenter's control also influence the input flow rate. In addition to the infusion rate of the experimenter which influences CSF formation, CSF is produced inside the brain, but much about its production remains unknown at the present time. Currently, there are no direct methods to measure the CSF production rate over short periods of time. Globally, the average secretion rate--used as a proxy for the production--is 0.35 mL/min with a 95% confidence range of 0.27 mL/min to 0.45 mL/min [2]. The lack of precise knowledge about the CSF production rate and the unmeasured factors that influence it are sources of noise in the total CSF formation rate. Consequently the stochastic Marmarou model may be conceptualized as the classic Marmarou model with a noisy input flow rate that reflects uncertainty about CSF formation.
The deterministic Marmarou model is contained in the final model displayed above--it surfaces when σ = 0 mL/min, which precludes noise, and consequently produces only the mean ICP value. The general model with σ ≠ 0 mL/min reproduces the fluctuations inherent in the time-path of real measurements of ICP--information which is discarded by the deterministic Marmarou model. Figure 3 compares the fluctuating path, similar to the actual noisy ICP data, reproduced by the stochastic Marmarou model with the path predicted by the deterministic Marmarou model. The mathematical structure of the Marmarou et al. [5] equation is the classic Verhulst logistic model, ubiquitous in biological growth and saturation phenomena [27]. The mathematical form of equation (9) is the stochastic logistic model and it is the natural stochastic extension [28,29] of the Verhulst logistic model.
Figure 3. Comparison of deterministic and stochastic Marmarou model solutions. Values of CSF flow parameters used in this simulation: E = 0.15 mL-1, pb = 8 mmHg, R = 7 mmHg/(mL/min), I = 1.5 mLmin-1, σ = 0.5 mL/min.
The clinical significance of the stochastic Marmarou model
By building the fluctuations right into the dynamics of the model structure, the stochastic model makes full use of the information in the variations of the ICP waveform. From this additional information, the time-varying probability distributions of the ICP waveform can be extracted, and it is these latter quantities that enable computation of the probabilities of clinically relevant events. It is the knowledge of these probabilities of clinically relevant events that facilitate dynamic risk management of the patient. Conceptually, the average value of p(t) at any given time 't' is the average ICP at that time in an ensemble of patients with a similar CSF flow profile, as reflected in the values of the CSF flow parameters.
Analyzing the stochastic Marmarou model
In the Results section, we will display the exact analytical solution to the stochastic Marmarou model and derive insights from the solution into the influence of noise on the ICP at each point in time, and on average. Under the normal conditions described in [2], biological processes will settle down to a steady state after the transients have died out. In the deterministic Marmarou model [5], the steady state (equilibrium) is found by setting the time rate of change of the ICP equal to zero. What is the corresponding steady-state concept for a stochastic process? The stochastic counterpart to the time-independent steady-state level of the ICP is the time-independent probability distribution of the ICP, and the equilibrium probability distribution is to the stochastic environment as the stable equilibrium point is to the deterministic one [30]. We derive the equilibrium probability distribution for the ICP, and from it, draw conclusions for the influence of CSF flow parameters and noise intensity upon the average steady-state ICP level. We compute a measure relevant to the treatment and control of hydrocephalus: given the current value of the patient's ICP, what is the probability that it will exceed a critical high level? And how is that probability influenced by neurological characteristics of the patient such as their resistance to CSF flow and the noise intensity of the fluctuations in CSF formation rate which in turn drives the fluctuations in their ICP?
Computing probabilities of clinically relevant events
The mathematical formulation of the problem posed in the previous paragraph is: given that a patient's ICP is currently x mmHg, where x is an arbitrary value, what is the probability that the ICP will exceed a critical threshold 'b' (mmHg) at a future time? Mathematically stated: given that p(s) = x, find the following transition probability-- P[p(t) > b | p(s) = x], t > s. Simple though the question seems, finding the answer requires computing the conditional probability distributions of the CSF process. Since the conditional probability distributions follow the Fokker-Planck partial differential equation, the problem is non-trivial, but Karlin and Taylor [31] circumvent the difficulty by solving a boundary-value problem associated with this dynamically changing probability. They show that the required probability satisfies a nonlinear ordinary differential equation which must be solved subject to two conditions on the probability that are natural consequences of the current ICP level when it is at one of the two extreme points of the range of ICP values under consideration. It is these conditions that give rise to the term 'boundary value problem.'
Results and Discussion
We state and discuss the significance of the mathematical results, deferring their proofs to the Appendices (Additional file 1, Additional file 2 and Additional file 3) in the interest of maintaining clarity of exposition. Our first result is the exact analytical solution to the stochastic Marmarou model.
Additional file 1. Solving the stochastic Marmarou model. This file derives the solution to the stochastic Marmarou Model assuming a constant infusion rate.
Additional file 2. Finding the steady-state probability distribution of the ICP. This file derives the steady-state probability distribution of the ICP.
Additional file 3. Computing clinically relevant dynamic probabilities. This file shows how to derive the dynamic probability of the critical event--defined as the ICP exceeding the critical threshold level of ICP of 40 mmHg, as a function of the patient's current ICP level, the baseline pressure, the patient's neurological characteristics-- the resistance to CSF flow, the cerebral elasticity, and an experimental variable--the infusion rate.
Solution to stochastic Marmarou model with constant rate infusion
For a constant infusion rate I, equation (9) is explicitly solvable in closed-form as shown below. Given any initial ICP value "p0" (mmHg) at time t = 0, the future ICP value at any time "t" is given by:
The proof is provided in additional file 1: Solving the stochastic Marmarou model. Note that the solution to the stochastic Marmarou model is found through an "integrating factor" which involves an integration constant, the evaluation of which necessitates a unit of 1/min unit for the 2 inside the exponent of the exponential function. The noise intensity parameter σ and the Brownian Motion process W(t) in the solution show the explicit influence of noise on the evolution of the ICP, underscoring the importance of modeling the noise in the clinical ICP data. In addition to the practical utility of offering a closed-form analytical solution, this result has value for another reason: it shows explicitly that noise cannot be averaged away when the process is nonlinear. If the Brownian motion process W(t) entered the solution for p(t) in an additive linear way, its effect would disappear on average. But the Brownian motion process enters the solution in a highly nonlinear fashion, making it impossible to average out its effect to zero. Finally, the solution depends upon the noise intensity parameter σ in a mathematically continuous way, a fact that is meaningful because the result shows that the solution to the deterministic Marmarou model [5] emerges as the special case corresponding to σ = 0 mL/min, and so, it is natural to ask if the simpler deterministic model would suffice when the noise intensity is small. Should the influence of noise be negligible in a particular case, the value of σ will be very small, and, because of the mathematical continuity in its dependence upon σ, the stochastic solution will be very close to the deterministic solution in such a case, and we may use the simpler deterministic model with confidence. However, the stochastic model is preferable in general for two reasons: it captures the dynamics of the ICP data better than the deterministic model when the noise intensity is larger, and furthermore, the stochastic model characterizes the risk profile of the patient probabilistically. Almost tautologically, the deterministic model cannot evaluate the risks due to the errors that are an inseparable part of medical data because deterministic modeling philosophy sees the future as completely predictable from the present situation. These considerations suggest that, from a conservative modeling perspective, incorporating the influence of noise into the dynamics is conceptually more defensible.
In principle, the solution contains all the transient probability distributions of the ICP process that characterize it on its way to equilibrium. In practice, mathematical difficulties may make these transient distributions hard to extract from the solution. But we can still compute the probability of the critical events by using a methodology that does not depend on that knowledge. And we can still draw useful information about the nature of the process at steady-state. Next, we find the steady-state probability distribution of the ICP process.
The steady-state probability distribution of the ICP is gamma with the parameters shown p.149 in [29], and will exist provided that the noise intensity parameter σ satisfies the condition: .
Figure 4. Probability that ICP reaches 40 mmHg as a function of noise intensity parameter σ, starting at an ICP level of 35 mmHg. Values of CSF flow parameters used in this simulation: E = 0.15 mL-1, pb = 8 mmHg, R = 7 mmHg/(mL/min), I = 1.5 mLmin-1, σ ranged from 0.4 mL/min to 1.3 mL/min.
Our next three results are motivated by the following considerations. A larger cerebrospinal fluid resistance R tends to increase ICP by increasing the pressure due to the circulatory CSF component. This is a direct consequence of Davson's equation [6]: ICPCSF = (resistance to CSF outflow) × (CSF formation) + (pressure in sagittal sinus). This naturally leads to the following questions. How will the intensity of the fluctuations influence the relationship between resistance and ICP? The same relationship may hold on average, but, as anticipated in the solution to the stochastic Marmarou model, it may be moderated by the noise intensity parameter because of the nonlinearity of the ICP process. How will the intensity of fluctuations affect the average steady-state ICP--is the average steady-state ICP smaller or larger when the intensity of fluctuations increases? Finally, will the intensity of fluctuations attenuate or amplify the effect of resistance to CSF flow on the average steady-state ICP?
Relationship between average steady-state ICP and cerebrospinal fluid resistance
The average steady-state ICP, denoted by μ, increases with the cerebrospinal fluid resistance R--thus the relationship between R and ICP holds on average.
The steady-state probability distribution of ICP is gamma with the parameters shown in the previous subsection. From well-known properties of the gamma distribution, it follows that the steady-state mean ICP level μ is given by: . Therefore, . From the expression for , it is clear that the average ICP level does indeed increase with R, provided that . This condition is satisfied, using the values of the parameters in the previous subsection. Thus, the increasing relationship between the actual ICP level and the cerebrospinal fluid resistance, predicted by Davson's equation when ICP is conceptualized as a deterministic process, also holds on average at steady-state when ICP is modeled as a stochastic process.
Relationship between average steady-state ICP and noise intensity
The average steady-state ICP level, decreases with the intensity of fluctuations, measured by the infinitesimal variance parameter σ2.
From the relationship derived in the previous subsection, , it is clear that μ decreases as σ2 increases. A larger noise intensity corresponds to greater variation in the CSF input flow rate which translates into greater variation in ICP, and these larger fluctuations could cause the average ICP level to increase, decrease or remain unaffected. The nonlinear influence of the parameters of CSF flow dynamics on ICP level turns out to reduce the average ICP value when the fluctuations in ICP are greater. This is an outcome that one would expect to find when steady-state has been achieved--when the transition probabilities have settled down to constant levels so that the probability distribution of ICP is no longer changing over time. This mathematical finding could be tested by separating a random sample of patients into two groups, such that one group has more variability in its ICP levels (due to higher variability in its CSF input flow rate) than the other group, and then conducting a statistical test of significance--such as a t-test--on the difference in mean ICP levels in these two groups at steady-state.
Effect of noise intensity on the relationship between average steady-state ICP and cerebrospinal fluid resistance
The resistance increases the ICP on average by a smaller amount when the intensity of fluctuations is higher.
From , it is clear that a higher σ2 will dampen the effect of the cerebrospinal resistance on the average steady-state ICP level. This is an outcome that one would expect to find at steady-state. The mathematical finding could be tested by separating a random sample of patients into two groups, such that one group has more variability in its ICP levels than the other group (due to higher variability in its CSF input flow rate), and then correlating the mean ICP level with the cerebrospinal resistance in each group at steady-state. According to the mathematics, the correlation should be smaller in the group with more variable ICP. Given the linear relationship between the steady-state mean and the cerebrospinal resistance, a simple correlation coefficient such as the Pearson product moment should suffice.
Next we turn our attention to dynamic management of the patient's risk. Risk may be quantified in terms of the probability of the onset of some critical event, say the ICP exceeding a dangerously high level. Given the current value of the patient's ICP, what is the probability that it will exceed a high level? Such a probability is intrinsically dynamic because it depends upon the patient's current condition (their current ICP), the dynamics of the patient's CSF flow and the noise intensity σ2. We want to understand how the probability is influenced by important clinical characteristics of the patient such as their resistance to CSF flow, and by the noise intensity.
Computing clinically relevant dynamic probabilities
Given that the current ICP is x mmHg, where 0 ≤ × ≤ b, let u(x) denote the probability of reaching the level b. Then u(x) satisfies the following nonlinear differential equation, which must be solved subject to the two conditions on u(x) at x = 0 and at x = b:
The conditions on u(x) at the two corners x = 0 and x = b make this a two-point boundary value problem. The solution is given in terms of the scaling function S(x):
The integrals defining s(x) and S(x) are indefinite at the lower end because the final answer is unaffected by its choice. For our clinical applications, it is natural to take the lower end point to be zero.
The proof is deferred to additional file 3: Computing clinically relevant dynamic probabilities. While the above representation is, in principle, a closed-form analytical solution, it is, in practice, a quasi-analytical solution because the integral that defines s(x) cannot be obtained in closed-form. However, that is no limitation because we can integrate it numerically after substituting the empirically established values of the parameters. We used the parameter values shown in the subsection "Steady-state Probability Distribution of ICP." We took the critical level 'b' to be 40 mmHg, based on the clinical finding reported in [2] that patients were able to tolerate increases in ICP up to 40-50 mmHg. Our rationale was that, from a clinical perspective, a conservative approach to patient management would be consistent with assessing the probability of reaching the lower end point of the 40-50 mmHg range that patients are able to tolerate. Thus our critical event is defined as "reaching an ICP of 40 mmHg." In order to understand how the probability is influenced by the noise intensity parameter σ, we computed the probability over a range of σ = 0.4 mL/min to σ = 1.3 mL/min. For each value of σ in this range, we solved the boundary-value problem to find the probability of reaching 40 mmHg. Furthermore, in order to understand the influence of the patient's initial condition upon the probability of the critical event, we repeated this set of computations for three different starting levels of ICP; the curve shown in Figure 4 is for a starting level of ICP of 35 mmHg. In order to understand how the probability is influenced by the resistance to CSF outflow R, we computed the probability over a range of R = 4 mmHgmL-1 min to R = 12 mmHgmL-1 min. For each value of R in this range, we solved the boundary-value problem to find the probability of reaching 40 mmHg. Again, we repeated this set of computations for three different starting levels of ICP; the curve shown in Figure 5 is for a starting level of ICP of 35 mmHg. Across the three initial levels of ICP, the curves have a similar shape and are simply translated vertically. Figures 4 and 5 show that the probabilities increase at an increasing rate (convex functions). Furthermore, they tell an interesting neurological story--namely, that the probabilities of the critical events exhibit strong threshold effects. In Figure 4, below a critical level of noise intensity, the probabilities are very low--almost zero--but beyond a threshold value of σ = 1.1 mL/min in Figure 4, they rise steeply. In Figure 5, as R (mmHgmL-1min) varies from 4 to below 10, the probabilities are almost zero, but beyond R = 10 mmHgmL-1 min, they rise dramatically. Furthermore, at low levels of noise intensity, the probabilities remain close to zero throughout the range of R (mmHgmL-1min) from 4 to 12. But as σ increases to 0.8 mL/min--the value assumed for it in Figure 5--R has a strong effect on the probability beyond the critical threshold of 10 mmHgmL-1min. The clinical significance of these findings is that erratic fluctuations in ICP (caused by a larger input flow rate noise intensity σ) will significantly increase the patient's risk, as measured by the probability of the critical event. Because the risk increases rapidly beyond the threshold value of σ, these results suggest that an essential component of risk management is to carefully minimize erratic fluctuations in the patient's CSF input flow rate at all times. Finally, Figure 6 shows the probability of the critical event as a function of both R and σ in a three-dimensional plot. The two-dimensional surface shows the value of the probability for each combination of values of R and σ. To facilitate interpretation of the surface, we used a mesh in which the dark lines are the probability plots as a function of R and the red lines are the probability plots as a function of σ. Figure 6 shows very clearly that threshold effects are sensitive to both R and σ, and beyond the threshold, the probabilities asymptotically approach one.
Figure 5. Probability that ICP reaches 40 mmHg as a function of resistance to CSF flow parameter R, starting at an ICP level of 35 mmHg. Values of CSF flow parameters used in this simulation: E = 0.15 mL-1, pb = 8 mmHg, I = 1.5 mLmin-1, σ = 0.8 mL/min, R = ranged from 4 mmHg/(mL/min) to 12 mmHg/(mL/min).
Figure 6. Probability that ICP reaches 40 mmHg as a function of resistance to CSF flow parameter R, and noise intensity parameter σ, starting at an ICP level of 35 mmHg. Values of CSF flow parameters used in this simulation: E = 0.15 mL-1, pb = 8 mmHg, I = 1.5 mLmin-1, R = ranged from 4 mmHg/(mL/min) to 12 mmHg/(mL/min), and σ ranged from 0.4 mL/min to 1.3 mL/min.
Conclusions
The stochastic generalization of the Marmarou model offers a tractable analytical description of the noisy ICP dynamics and yields insights into the impact of noise. The SDE offers a rigorous analytical framework to study issues of clinical interest and neurological significance such as the patient's risk. A key clinical implication is that fluctuations in the CSF formation rate--which increase the fluctuations in ICP-- should be minimized to lower the patient's risk. Future work could extend the framework developed in this research to accommodate the non-zero reference pressure case. Finally, the stochastic differential equation framework, in conjunction with nonlinear control theory, can be used to develop a nonlinear automatic controller to regulate shunts to facilitate continuous CSF drainage.
Competing interests
The author declares that they have no competing interests.
Authors' contributions
KR is the sole author. He has read and approved the final version of the manuscript.
Acknowledgements
The author is grateful to Dr. Marek Czosnyka, Neuroscience Group, University of Cambridge, UK, for introducing him to the key issues in CSF dynamics and hydrocephalus, for making data available from infusion studies conducted at Addenbrooke's Hospital, UK, and to the late Professor Anthony Marmarou for suggesting the probabilistic computations for critical events.
References
1. Eklund A, Smielewski P, Chambers I, Alperin N, Malm J, Czosnyka M, Marmarou A: Assessment of cerebrospinal fluid outflow resistance.
Med Bio Eng Comput 2007, 45:719-735. Publisher Full Text
2. Czosnyka M, Czosnyka Z, Momjian S, Pickard JD: Cerebrospinal Fluid Dynamics.
Physiol Meas 2004, 25:R51-R76. PubMed Abstract | Publisher Full Text
3. Cohen B, Voorhees A, Vedel S, Wei T: Development of a theoretical framework for analyzing cerebrospinal fluid dynamics.
Cerebrospinal Fluid Res 2009, 6:12. PubMed Abstract | BioMed Central Full Text | PubMed Central Full Text
4. Sivaloganathan S, Tenti G, Drake JM: Mathematical pressure volume models of the cerebrospinal fluid.
Appl Math Comp 1998, 94:243-266. Publisher Full Text
5. Marmarou A, Shulman K, Rosende RM: A nonlinear analysis of the cerebrospinal fluid system and intracranial pressure dynamics.
J Neurosurg 1978, 48:332-344. PubMed Abstract | Publisher Full Text
6. Czosnyka M, Pickard JD: Monitoring and Interpretation of Intracranial Pressure.
J Neurol Neurosurg Psychiatry 2004, 75:813-821. PubMed Abstract | Publisher Full Text | PubMed Central Full Text
7. Marmarou A: A theoretical model and experimental evaluation of the cerebrospinal fluid system. In Thesis. Drexel University, Philadelphia, PA; 1973.
8. Avezaat CJJ, Eijndhoven JHM: Cerebrospinal fluid pulse pressure and craniospinal dynamics: a theoretical, clinical and experimental study. In Thesis. Edited by Jongbloed A. The Hague; 1984.
9. Frieden H, Ekstedt J: Instrumentation for cerebrospinal fluid hydrodynamic studies in man.
Med Biol Eng Comput 1982, 20:167-80. PubMed Abstract | Publisher Full Text
10. Feynman R: The Character of Physical Law. Cambridge, MA: MIT Press; 1965.
11. Raabe A, Czosnyka M, Piper I, Seifert V: Monitoring of intracranial compliance: correction for a change in body position.
Acta Neurochir (Wien) 1999, 141:31-36. PubMed Abstract | Publisher Full Text
12. Czosnyka M, Batorski L, Roszkowski M, Tomaszewski J, Wocjan J, Walencik A, Zabolotny W: Cerebrospinal compensation in hydrocephalic children.
Child's Nerv Syst 1993, 9:17-22.
13. Rekate HL: Circuit diagrams of the circulation of cerebrospinal fluid.
Pediatr Neurosurg 1994, 21:248-253. PubMed Abstract | Publisher Full Text
14. Rekate HL, Brodkey JA, Chizeck HJ, el Sakka WE, Ko WH: Ventricular volume regulation: a mathematical model and computer simulation.
Pediatr Neurosci 1988, 14:77-84. PubMed Abstract | Publisher Full Text
15. Clarke MJ, Meyer FB: The history of mathematical modeling in hydrocephalus.
Neurosurg Focus 2007, 22(Suppl 4):1-5. Publisher Full Text
16. Lavinio A, Czosnyka Z, Czosnyka M: Cerebrospinal Fluid Dynamics: Disturbances and Diagnostics.
Eur J Anaesthesiol 2008, 25(Suppl 42):137-141. Publisher Full Text
17. Petrella G, Czosnyka M, Keong N, Pickard JD, Czosnyka Z: How does CSF dynamics change after shunting.
Acta Neurol Scand 2008, 118:182-188. PubMed Abstract | Publisher Full Text
18. Gershenfeld N: The Physics of Information Technology. Cambridge: Cambridge University Press; 2000.
19. Van Kampen NG: Ito versus Stratanovich.
J Statist Phys 1981, 24:175-187. Publisher Full Text
20. Turelli M: Random Environments and Stochastic Calculus.
Theor Pop Biol 1977, 12:140-178. Publisher Full Text
21. Turelli M: A Reexamination of stability in randomly varying versus deterministic environments with comments on the stochastic theory of limiting similarity.
Theor Pop Biol 1978, 13:244-267. Publisher Full Text
22. Kailath T, Poor HV: Detection of stochastic processes.
IEEE Transactions on Information Theory 1998, 44:2230-2231. Publisher Full Text
23. Lo AW: Maximum likelihood estimation of generalized Ito processes with discretely sampled data.
Econometric Theory 1988, 4:231-247. Publisher Full Text
24. Ito K, Varadhan SRS, Stroock DW: On stochastic differential equations.
Memoirs of the AMS 1951, 4:1-51.
25. Skorohod AV: Studies in the Theory of Random Processes. New York: Dover Publications; 1965.
26. Harrison LM, David O, Friston KJ: Stochastic models of neuronal dynamics.
Philos Trans R Soc Lond B Biol Sci 2005, 360:1075-1091. PubMed Abstract | Publisher Full Text | PubMed Central Full Text
27. Tsoularis A, Wallace J: Analysis of logistic models.
Math Biosci 2002, 179:21-55. PubMed Abstract | Publisher Full Text
28. Øksendal B: Stochastic Differential Equations: An Introduction with Applications. Berlin: Springer; 2003.
29. Gard TC: Introduction to Stochastic Differential Equations. New York: John Wiley & Sons; 1988.
30. May RM: Stability in randomly fluctuating versus deterministic environments.
Am Nat 1973, 107:621-650. Publisher Full Text
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