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https://math.stackexchange.com/questions/793289/plotting-an-ellipse-after-an-ellipse-fit | # Plotting an Ellipse after an Ellipse Fit
I wonder if someone can assist my understanding as I'm a bit stumped with this...
I have taken the following (x,y) data which lies roughly on an ellipse:
$$\begin{pmatrix} 0.000234491 & 6855810 \\ 0.341848914 & 6856102 \\ 0.640414035 & 6874479 \\ 0.863239913 & 6908917 \\ 0.985101853 & 6955917 \\ 0.984332848 & 7006108 \\ 0.867029832 & 7056389 \\ 0.639589281 & 7100398 \\ 0.333714725 & 7134165 \\ 0.002441713 & 7147290 \\ -0.342779385 & 7146184 \\ -0.655455534 & 7137322 \\ -0.641777617 & 7136216 \\ -0.860267224 & 7116067 \\ -0.983690351 & 7072538 \\ -0.983008472 & 7021338 \\ -0.869967818 & 6973290 \\ -0.630288354 & 6923542 \\ -0.348927005 & 6889049 \\ \end{pmatrix}$$
If you were to plot this in Mathcad you will get the following:
Now, it is my intention to perform a best fit on these points to give me an equation of an ellipse. What I have done is to implement the algorithm based upon this paper - http://autotrace.sourceforge.net/WSCG98.pdf
The fit is against the following equation:
$ax^2 + bxy + cy^2 +dx + ey + f = 0$
Now, the fitting algorithm gives me the following co-effiecients:
\begin{align} & a=0.99999999999513789 \\ & b=0.0000031183817557930131 \\ & c=0.000000000045507950324787355 \\ & d=-21.87186231583247 \\ & e=-0.00063773270848852459 \\ & f=2233.2983593954009 \\ \end{align}
With a center of $(0.0116,7006431)$ which looks ok to me.
Now, here is the bit I'm stumped...
I just expected to be able to plot an ellipse centered on $(0.0116,7006431)$ that best fits the data points above. I assumed I could feed my original (x,y) data into :
$F(x,y) = ax^2 + bxy + cy^2 +dx + ey + f$
And this would give me data points lying on an ellipse which I could superimpose onto the original data.
What I get is as follows:
I guess my question is - what am I misunderstanding here? How do I use the output from the fit (a, b, ... f) to plot an ellipse which is centered on $(0.0116,7006431)$ and can be superimposed onto the original data?
Many thanks in advanced for any pointers/assistance.
You don't have a function $F(x,y)$, that would imply that you have some 3d object. you have an expression $ax^2+bxy+cy^2+dx+ey+f=0$.
You need to plug in $a,b,c,d,e$ and $f$ for $ax^2+bxy+cy^2+dx+ey+f$, and then, for each coordinate pair you want plug in either x or y, and then solve for the variable you didn't plug in.
What you did was make a form of error plot.
• Ahh ok... I was being daft!
– Mike
May 14, 2014 at 9:07
• happens to the best of us :) May 14, 2014 at 9:24
Ok, what I have done is to determine the y value from the x:
I have then determined y from x and plotted on top of the original data (blue points are the fitted values, red are the original):
• I have a problem with the reconstruction of the procedure in the mentioned article. In (eq 28) there is the explicite condition that $a_1^T \cdot C_1 \cdot a_1 = 1$ which requires a correction by division by about $288443.686$ and then the first resulting coefficient $a \ne 1$. However, to reproduce your data/function, it seems to be required that $a=1$ instead, such that the condition involving $C_1$ does not hold. What's going on here? Aug 11, 2014 at 12:19 | 2023-03-27 06:30:07 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 1, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9905574917793274, "perplexity": 232.1214339746418}, "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-2023-14/segments/1679296948609.41/warc/CC-MAIN-20230327060940-20230327090940-00392.warc.gz"} |
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Persuasive speeches to buy Latex Documentclass Phd Thesis essay help 10 per page phd thesis wur. LaTeX documentclass. The picture below compares the three LaTeX standard font sizes. \documentclass. i used \documentclass[chap, 12pt, nocenter]{thesis}. Doing Purdue University Theses Using LaTeX Mark Senn. (Purdue University thesis) LaTeX typesetting system documentclass to format. , or 12pt document class. Annotated bibliography on stress management Latex Documentclass Master Thesis online dissertations and theses center schollarship essay help. | 2017-09-21 19:33:02 | {"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": 0.8967308402061462, "perplexity": 8337.101099050225}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-39/segments/1505818687837.85/warc/CC-MAIN-20170921191047-20170921211047-00231.warc.gz"} |
https://zbmath.org/?q=an%3A0625.90056 | zbMATH — the first resource for mathematics
On the complexity of cutting-plane proofs. (English) Zbl 0625.90056
As introduced by V. Chvatal [Discrete Math. 4, 305-337 (1973; Zbl 0253.05131)] cutting planes provide a canonical way of proving that every integral solution of a given system of linear inequalities satisfies another specified inequality. In this note we make several observations on the complexity of such proofs in general and when restricted to provide the unsatisfiability of formulae in the propositional calculus.
MSC:
90C10 Integer programming 90C05 Linear programming 68Q25 Analysis of algorithms and problem complexity
Full Text:
References:
[1] L. Babai, On Lovás” lattice reduction and the nearest lattice point, Combinatorica, to appear. · Zbl 0593.68030 [2] Bell, D.E., A theorem concerning the integer lattice, Stud. appl. math., 56, 187-188, (1977) · Zbl 0388.90051 [3] S.C. Boyd and W.R. Pulleyblank, Facet generating techniques, in preparation. · Zbl 1359.90113 [4] Chang, C.L.; Lee, R.C.T., Symbolic logic and mechanical theorem proving, (1973), Academic Press New York · Zbl 0241.68039 [5] Chvátal, V., Edmonds polytopes and a hierarchy of combinatorial problems, Discrete math., 4, 305-337, (1973) · Zbl 0253.05131 [6] Chvátal, V., Edmonds polytopes and weakly Hamiltonian graphs, Math. programming, 5, 29-40, (1973) · Zbl 0267.05118 [7] Chvátal, V., Cutting-plane proofs and the stability number of a graph, () · Zbl 0395.05047 [8] Chvátal, V., Cutting planes in combinatorics, () · Zbl 0581.05015 [9] Cook, S.A., Feasibly constructive proofs and the prop ositional calculus, (), 83-97 [10] Cook, S.A.; Reckhow, R.A., The relative efficiency of propositional proof systems, J. symbolic logic, 44, 36-50, (1977) · Zbl 0408.03044 [11] W. Cook, A.M.H. Gerards, A. Schrijver and É. Tardos, Sensitivity theorems in integer linear programming, Math. Programming, to appear. · Zbl 0648.90055 [12] Dowd, M., Propositional representation of arithmetic proofs, () · Zbl 1282.68112 [13] Gomory, R.E., An algorithm for integer solutions to linear programs, (), 269-302 [14] Goodstein, R.L., Recursive number theory, (1957), North-Holland Amsterdam · Zbl 0077.01401 [15] M. Grötschel, L. Lovász and A. Schrijver, The Ellipsoid Method and Combinatorial Optimization (Springer, Berlin, to appear). [16] Haken, A., The intractability of resolution, Theor. comput. sci., 39, 297-308, (1985) · Zbl 0586.03010 [17] J. Hastad, communicated by R. Kannan. [18] Helfrich, B., Eine beziehung zwischen konvexen mengen $$P ⊂ R\^{}\{2\}$$ und den gitterbasen von $$Z$$^{2}, (1985), Manuscript, Frankfurt, W. Germany [19] Hoffman, A.J., Binding constraints and Helly numbers, () · Zbl 0403.90056 [20] Kannan, R., Imkproved algorithms for integer programming and related lattice problems, (), 193-206 [21] R. Kannan and L. Lovász, to appear. [22] Lenstra, H.W., Integer programming with a fixed number of variables, Math. oper. res., 8, 538-548, (1983) · Zbl 0524.90067 [23] H.W. Lenstra, Jr. and C.P. Schnorr, On the successive minima of a pair of polar lattices, to appear. [24] Lovász, L., An algorithmic theory of numbers, graphs and convexity, () · Zbl 0606.68039 [25] Robinson, J.A., A machine-oriented logic based on the resolution principle, Jacm, 23-41, (1965) · Zbl 0139.12303 [26] H.E. Scarf, An observation on the structure of production sets with indivisibilities, Proc. Nat. Acad. Sci. (USA) 74, 3637-3641. · Zbl 0381.90081 [27] Schrijver, A., On cutting planes, Ann. discrete math., 9, 291-296, (1980) · Zbl 0441.90070 [28] Schrijver, A., Theory of linear and integer programming, (1986), Wiley Chichester · Zbl 0665.90063 [29] Todd, M.J., The number of necessary constraints in an integer program: a new proof of Scarf’s theorem, () [30] Tseitin, G.S., On the complexity of derivations in the propositional calculus, (), 115-125, translated from Russian
This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching. | 2021-07-29 23:07:21 | {"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": 0.7684997320175171, "perplexity": 6035.679522042484}, "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-2021-31/segments/1627046153897.89/warc/CC-MAIN-20210729203133-20210729233133-00057.warc.gz"} |
http://math.stackexchange.com/questions/258451/some-irreducible-characters-of-the-symmetric-group-s-n | # Some irreducible characters of the Symmetric group $S_n$
I want to have characters of some irreducible $S_n$-modules corresponding to certain partitions $\lambda$ of $n$, the computations using Frobenius formula get complicated and I am unable to find in standard text of representation theory of symmetric groups. Where I can find these in Litrature? More precisely I am more intereseted in the irreducible modules corresponding to partitions $(n-4,2,2)$ and $(n-4,2,1,1)$. Thank you in advance.
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For general reference of the representation theory of symmetric groups, you can take a look at James, "Representation theory of the symmetric groups" or James, Kerber "Representation theory of the symmetric group". (Not the same book) – Hans Giebenrath Dec 14 '12 at 6:57
I have already seen the books you mentioned, but they do not help me much. – Jacob Dec 14 '12 at 7:26
You want to use character polynomials. Let $\mu$ be a partition of $k$. In your case, $k$ will be $4$. The character polynomial is a polynomial $$q_{\mu}(a_1, a_2, \ldots, a_k)$$ with the following property:
For any $n$ such that $n-k \geq \mu_1$, and any $\sigma \in S_n$, the character of $\sigma$ acting on the Specht module $\mathrm{Specht}(n-k, \mu_1, \mu_2, \ldots, \mu_r)$ is $q_{\mu}(a_1, \ldots, a_k)$, where $a_i$ is the number of $i$ cycles of $\sigma$.
The simplest example is $q_1 = a_1-1$. In other words, for any permutation $\sigma \in S_n$, the trace of $\sigma$ acting on $\mathrm{Specht}(n-1,1)$ is $\#(\mbox{$1$-cycles of$\sigma$})-1$.
What you want to know are $q_{22}$ and $q_{211}$. I don't know this theory well enough to compute them for you in a reasonable amount of time, but I would suggest looking at Garsia and Goupil and at Examples 1.7.13 and 1.7.14 in MacDonald's book. | 2015-07-28 03:46:24 | {"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": 0.9010093212127686, "perplexity": 107.47195108189736}, "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-2015-32/segments/1438042981525.10/warc/CC-MAIN-20150728002301-00100-ip-10-236-191-2.ec2.internal.warc.gz"} |
https://www.emathhelp.net/calculators/pre-algebra/scientific-notation-calculator/ | # Scientific Notation Calculator
The calculator will convert the given number into scientific notation.
Enter a number:
If the calculator did not compute something or you have identified an error, or you have a suggestion/feedback, please write it in the comments below.
Your input: convert $$12345.6789$$$into scientific notation. Answer Scientific notation: $$12345.6789=1.23456789\times 10^{4}$$$
Scientific e-notation: $$12345.6789=1.23456789 E{4}$$\$ | 2021-05-13 02:52:02 | {"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": 0.4945333003997803, "perplexity": 4988.085517895959}, "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-2021-21/segments/1620243992721.31/warc/CC-MAIN-20210513014954-20210513044954-00062.warc.gz"} |
https://mathoverflow.net/questions/247848/recurrence-equation-and-matrix-convergence | # Recurrence Equation and Matrix Convergence
To begin with, let us give the conceptual background needed to expose the problem. First of all, we shall consider the set $\mathbb{L}^{n} = \mathbb{R}^{n}_{\geq0} = \{\overrightarrow{x}\in\mathbb{R}^{n}_{\geq0}\mid\sum x_{i} = 1\}$. More precisely, $\overrightarrow{x} = (x_{1},x_{2},\ldots,x_{n})$ belongs to $\mathbb{L}^{n}$ iff its coordinates are non-negative and its sum totals one. From this convention, it is associated to each vector $\overrightarrow{x}\in\mathbb{L}^{n}\backslash\{(1/n,1/n,\ldots,1/n)\}$ the following function $$P(x_{i},x_{j}) = P(x_{j},x_{i}) := \frac{1}{2}\frac{|x_{i}-x_{j}|}{\displaystyle\sum_{\delta=1}^{n-1}|x_{k}-x_{l}|}\quad\text{where}\quad\delta = k - l$$ Moreover, we are going to associate to this function the following matrix $A_{0} = [a^{0}_{ij}] := [P(x_{i},x_{j})]$ with the additional restriction that $x_{1}\leq x_{2}\leq\ldots\leq x_{n}$. Given that this matrix is self-adjoint, we may associate to it the matrix $\Lambda_{0}:= [\lambda^{0}_{ij}]$ which is obtained from $A_{0}$ by diagonalizing it and taking the norm of its coordinates. We are going to impose now a restriction over $\Lambda_{0}$: $\lambda^{0}_{11}\leq\lambda^{0}_{22}\leq\ldots\leq\lambda^{0}_{nn}$. Then it is possible to propose the problem itself. Given the recurrence equation \begin{align*} A_{k+1} & = \frac{A_{k} + \Lambda_{k}}{1+\displaystyle\sum_{i=1}^{n}\lambda^{k}_{ii}} \end{align*} Where $\Lambda_{k}:= [\lambda^{k}_{ij}]$ is the diagonalization of $A_{k}$ whose coordinates have been normed and which satisfy $\lambda^{k}_{11}\leq\lambda^{k}_{22}\leq\ldots\leq\lambda^{k}_{nn}$, the question is: does this sequence converge? If so, could anyone provide me its limit? It seems to me that the non-diagonal elements converge to zero. Nonetheless I am not able to conjecture what happens to the diagonal. In order to avoid misunderstandings, here it comes an example. Let's take the vector $\overrightarrow{x} = (0.1,0.3,0.6)$ whose associated matrix is: $$A_{0} = \left[ \begin{array}{ccc} 0 & 0.1 & 0.25 \\ 0.1 & 0 & 0.15 \\ 0.25 & 0.15 & 0 \end{array} \right]$$ And its corresponding eigenvalues are $\lambda_{1} \cong -0.26,\,\lambda_{2} \cong -0.09,\,\lambda_{3} \cong 0.34$. Therefore: $$\Lambda_{0} = \left[ \begin{array}{ccc} 0.09 & 0 & 0 \\ 0 & 0.26 & 0 \\ 0 & 0 & 0.34 \end{array} \right]$$ We may now give the expression of $A_{1}$. Indeed, here it is \begin{align*} \sum_{i=1}^{3}\lambda^{0}_{ii} & = 0.09 + 0.26 + 0.34 = 0.69\Rightarrow A_{1} = \frac{1}{1.69}\left[ \begin{array}{ccc} 0.09 & 0.1 & 0.25 \\ 0.1 & 0.26 & 0.15 \\ 0.25 & 0.15 & 0.34 \end{array} \right] \end{align*} I think this is it. Thank you in advance for any contribution.
PS: the same question has been asked at MSE
https://math.stackexchange.com/questions/1897558/recurrence-equation-and-matrix-convergence
$\textbf{Remark}$. After one hundred computer-aided iterations, the above sequence seems to converge: \begin{align*} A_{100} = \left[ \begin{array}{ccc} 0.064202 & 2.1000\cdot10^{-31} & 5.2499\cdot10^{-31} \\ 2.1000\cdot10^{-31} & 0.26237 & 3.1499\cdot10^{-31} \\ 5.2499\cdot10^{-31} & 3.1499\cdot10^{-31} & 0.67343 \end{array} \right] \end{align*} From then on, the results are extremely alike.
• I'm not quite sure I understand the notation $\sum_{\delta=1}^{n-1} |x_k-x_l|$ "where $\delta=k-l$", but what is $P$ if $x=(1/n,1/n,\dots,1/n)$? – Pietro Majer Aug 21 '16 at 7:33
• Let us fix $\delta = 1$. Hence the associated sum is $|x_{2} - x_{1}| + |x_{3} - x_{2}| + \ldots + |x_{n} - x_{n-1}|$. Once again, let us fix $\delta = 2$. Thus the associated sum corresponds to $|x_{3} - x_{1}| + |x_{4} - x_{2}| + \ldots + |x_{n} - x_{n-2}|$. The same reasoning applies to the other values of $\delta$. As to the case when $\overrightarrow{x} = (1/n,1/n,\ldots,1/n)$, it does not make sense to associate to it the function $P$. The justification is based on the theoretical context where it came from. Hopefully it helps. – APC89 Aug 21 '16 at 21:29
• OK --So to be precise the domain of $P$ is not the whole set $\mathbb{L}^n$ – Pietro Majer Aug 21 '16 at 22:24
Your example does not quite fit the problem description, as the matrix $\Lambda_0$ you are adding has the absolute values of the eigenvalues of $A_0$ on its diagonal. If this variation is the problem you are interested in, then the convergence is fairly trivial.
Your initial condition is a matrix $A_0$ with nonnegative entries with the property that the sum of all entries is $1$. This property is clearly invariant under the iteration.
The iteration \begin{align*} A_{k+1} & = \frac{A_{k} + \Lambda_{k}}{1+\displaystyle\sum_{i=1}^{n}\lambda^{k}_{ii}} \end{align*} only adds to the diagonal and divides all entries by a constant greater than one. Thus the sequence of off-diagonal entries is strictly decreasing. The off-diagonal entries converge to zero as eventually $A_k$ is strictly diagonally dominant and from this point on the eigenvalues are indeed all positive. And then $\sum_{i=1}^{n}\lambda^{k}_{ii}$ is bounded away from zero. So there is even an exponential convergence of the off-diagonal entries to zero. | 2019-06-25 06:33:21 | {"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": 0.9939472675323486, "perplexity": 173.02263339234642}, "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/1560627999800.5/warc/CC-MAIN-20190625051950-20190625073950-00095.warc.gz"} |
https://hackage.haskell.org/package/gi-gtk-3.0.26/docs/GI-Gtk-Objects-PlacesSidebar.html | gi-gtk-3.0.26: Gtk bindings
Copyright Will Thompson Iñaki García Etxebarria and Jonas Platte LGPL-2.1 Iñaki García Etxebarria (garetxe@gmail.com) None Haskell2010
GI.Gtk.Objects.PlacesSidebar
Description
PlacesSidebar is a widget that displays a list of frequently-used places in the file system: the user’s home directory, the user’s bookmarks, and volumes and drives. This widget is used as a sidebar in FileChooser and may be used by file managers and similar programs.
The places sidebar displays drives and volumes, and will automatically mount or unmount them when the user selects them.
Applications can hook to various signals in the places sidebar to customize its behavior. For example, they can add extra commands to the context menu of the sidebar.
While bookmarks are completely in control of the user, the places sidebar also allows individual applications to provide extra shortcut folders that are unique to each application. For example, a Paint program may want to add a shortcut for a Clipart folder. You can do this with placesSidebarAddShortcut.
To make use of the places sidebar, an application at least needs to connect to the PlacesSidebar::open-location signal. This is emitted when the user selects in the sidebar a location to open. The application should also call placesSidebarSetLocation when it changes the currently-viewed location.
# CSS nodes
GtkPlacesSidebar uses a single CSS node with name placessidebar and style class .sidebar.
Among the children of the places sidebar, the following style classes can be used:
• .sidebar-new-bookmark-row for the 'Add new bookmark' row
• .sidebar-placeholder-row for a row that is a placeholder
• .has-open-popup when a popup is open for a row
Synopsis
# Exported types
newtype PlacesSidebar Source #
Memory-managed wrapper type.
Constructors
PlacesSidebar (ManagedPtr PlacesSidebar)
Instances
Source # Instance detailsDefined in GI.Gtk.Objects.PlacesSidebar Methods Source # Instance detailsDefined in GI.Gtk.Objects.PlacesSidebar Source # Instance detailsDefined in GI.Gtk.Objects.PlacesSidebar Source # Instance detailsDefined in GI.Gtk.Objects.PlacesSidebar Source # Instance detailsDefined in GI.Gtk.Objects.PlacesSidebar Source # Instance detailsDefined in GI.Gtk.Objects.PlacesSidebar Source # Instance detailsDefined in GI.Gtk.Objects.PlacesSidebar Source # Instance detailsDefined in GI.Gtk.Objects.PlacesSidebar Source # Instance detailsDefined in GI.Gtk.Objects.PlacesSidebar
class GObject o => IsPlacesSidebar o Source #
Type class for types which can be safely cast to PlacesSidebar, for instance with toPlacesSidebar.
Instances
Source # Instance detailsDefined in GI.Gtk.Objects.PlacesSidebar Source # Instance detailsDefined in GI.Gtk.Objects.PlacesSidebar
toPlacesSidebar :: (MonadIO m, IsPlacesSidebar o) => o -> m PlacesSidebar Source #
Cast to PlacesSidebar, for types for which this is known to be safe. For general casts, use castTo.
A convenience alias for Nothing :: Maybe PlacesSidebar.
# Methods
## addShortcut
Arguments
:: (HasCallStack, MonadIO m, IsPlacesSidebar a, IsFile b) => a sidebar: a places sidebar -> b location: location to add as an application-specific shortcut -> m ()
Applications may want to present some folders in the places sidebar if they could be immediately useful to users. For example, a drawing program could add a “/usr/share/clipart” location when the sidebar is being used in an “Insert Clipart” dialog box.
This function adds the specified location to a special place for immutable shortcuts. The shortcuts are application-specific; they are not shared across applications, and they are not persistent. If this function is called multiple times with different locations, then they are added to the sidebar’s list in the same order as the function is called.
Since: 3.10
## getLocalOnly
Arguments
:: (HasCallStack, MonadIO m, IsPlacesSidebar a) => a sidebar: a places sidebar -> m Bool Returns: True if the sidebar will only show local files.
Returns the value previously set with placesSidebarSetLocalOnly.
Since: 3.12
## getLocation
Arguments
:: (HasCallStack, MonadIO m, IsPlacesSidebar a) => a sidebar: a places sidebar -> m (Maybe File) Returns: a File with the selected location, or Nothing if nothing is visually selected.
Gets the currently selected location in the sidebar. This can be Nothing when nothing is selected, for example, when placesSidebarSetLocation has been called with a location that is not among the sidebar’s list of places to show.
You can use this function to get the selection in the sidebar. Also, if you connect to the PlacesSidebar::populate-popup signal, you can use this function to get the location that is being referred to during the callbacks for your menu items.
Since: 3.10
## getNthBookmark
Arguments
:: (HasCallStack, MonadIO m, IsPlacesSidebar a) => a sidebar: a places sidebar -> Int32 n: index of the bookmark to query -> m (Maybe File) Returns: The bookmark specified by the index n, or Nothing if no such index exist. Note that the indices start at 0, even though the file chooser starts them with the keyboard shortcut "Alt-1".
This function queries the bookmarks added by the user to the places sidebar, and returns one of them. This function is used by FileChooser to implement the “Alt-1”, “Alt-2”, etc. shortcuts, which activate the cooresponding bookmark.
Since: 3.10
## getOpenFlags
Arguments
:: (HasCallStack, MonadIO m, IsPlacesSidebar a) => a sidebar: a PlacesSidebar -> m [PlacesOpenFlags] Returns: the PlacesOpenFlags of sidebar
Gets the open flags.
Since: 3.10
## getShowConnectToServer
Arguments
:: (HasCallStack, MonadIO m, IsPlacesSidebar a) => a sidebar: a places sidebar -> m Bool Returns: True if the sidebar will display a “Connect to Server” item.
Deprecated: (Since version 3.18)It is recommended to group this functionality with the drives and network location under the new 'Other Location' item
Returns the value previously set with placesSidebarSetShowConnectToServer
## getShowDesktop
Arguments
:: (HasCallStack, MonadIO m, IsPlacesSidebar a) => a sidebar: a places sidebar -> m Bool Returns: True if the sidebar will display a builtin shortcut to the desktop folder.
Returns the value previously set with placesSidebarSetShowDesktop
Since: 3.10
## getShowEnterLocation
Arguments
:: (HasCallStack, MonadIO m, IsPlacesSidebar a) => a sidebar: a places sidebar -> m Bool Returns: True if the sidebar will display an “Enter Location” item.
Returns the value previously set with placesSidebarSetShowEnterLocation
Since: 3.14
## getShowOtherLocations
Arguments
:: (HasCallStack, MonadIO m, IsPlacesSidebar a) => a sidebar: a places sidebar -> m Bool Returns: True if the sidebar will display an “Other Locations” item.
Returns the value previously set with placesSidebarSetShowOtherLocations
Since: 3.18
## getShowRecent
Arguments
:: (HasCallStack, MonadIO m, IsPlacesSidebar a) => a sidebar: a places sidebar -> m Bool Returns: True if the sidebar will display a builtin shortcut for recent files
Returns the value previously set with placesSidebarSetShowRecent
Since: 3.18
## getShowStarredLocation
Arguments
:: (HasCallStack, MonadIO m, IsPlacesSidebar a) => a sidebar: a places sidebar -> m Bool Returns: True if the sidebar will display a Starred item.
Returns the value previously set with placesSidebarSetShowStarredLocation
Since: 3.22.26
## getShowTrash
Arguments
:: (HasCallStack, MonadIO m, IsPlacesSidebar a) => a sidebar: a places sidebar -> m Bool Returns: True if the sidebar will display a “Trash” item.
Returns the value previously set with placesSidebarSetShowTrash
Since: 3.18
## listShortcuts
Arguments
:: (HasCallStack, MonadIO m, IsPlacesSidebar a)
=> a
sidebar: a places sidebar
-> m [File]
Returns: A SList of File of the locations that have been added as application-specific shortcuts with placesSidebarAddShortcut. To free this list, you can use
### C code
g_slist_free_full (list, (GDestroyNotify) g_object_unref);
Gets the list of shortcuts.
Since: 3.10
## new
Arguments
:: (HasCallStack, MonadIO m) => m PlacesSidebar Returns: a newly created PlacesSidebar
Creates a new PlacesSidebar widget.
The application should connect to at least the PlacesSidebar::open-location signal to be notified when the user makes a selection in the sidebar.
Since: 3.10
## removeShortcut
Arguments
:: (HasCallStack, MonadIO m, IsPlacesSidebar a, IsFile b) => a sidebar: a places sidebar -> b location: location to remove -> m ()
Removes an application-specific shortcut that has been previously been inserted with placesSidebarAddShortcut. If the location is not a shortcut in the sidebar, then nothing is done.
Since: 3.10
## setDropTargetsVisible
Arguments
:: (HasCallStack, MonadIO m, IsPlacesSidebar a, IsDragContext b) => a sidebar: a places sidebar. -> Bool visible: whether to show the valid targets or not. -> b context: drag context used to ask the source about the action that wants to perform, so hints are more accurate. -> m ()
Make the GtkPlacesSidebar show drop targets, so it can show the available drop targets and a "new bookmark" row. This improves the Drag-and-Drop experience of the user and allows applications to show all available drop targets at once.
This needs to be called when the application is aware of an ongoing drag that might target the sidebar. The drop-targets-visible state will be unset automatically if the drag finishes in the GtkPlacesSidebar. You only need to unset the state when the drag ends on some other widget on your application.
Since: 3.18
## setLocalOnly
Arguments
:: (HasCallStack, MonadIO m, IsPlacesSidebar a) => a sidebar: a places sidebar -> Bool localOnly: whether to show only local files -> m ()
Sets whether the sidebar should only show local files.
Since: 3.12
## setLocation
Arguments
:: (HasCallStack, MonadIO m, IsPlacesSidebar a, IsFile b) => a sidebar: a places sidebar -> Maybe b location: location to select, or Nothing for no current path -> m ()
Sets the location that is being shown in the widgets surrounding the sidebar, for example, in a folder view in a file manager. In turn, the sidebar will highlight that location if it is being shown in the list of places, or it will unhighlight everything if the location is not among the places in the list.
Since: 3.10
## setOpenFlags
Arguments
:: (HasCallStack, MonadIO m, IsPlacesSidebar a) => a sidebar: a places sidebar -> [PlacesOpenFlags] flags: Bitmask of modes in which the calling application can open locations -> m ()
Sets the way in which the calling application can open new locations from the places sidebar. For example, some applications only open locations “directly” into their main view, while others may support opening locations in a new notebook tab or a new window.
This function is used to tell the places sidebar about the ways in which the application can open new locations, so that the sidebar can display (or not) the “Open in new tab” and “Open in new window” menu items as appropriate.
When the PlacesSidebar::open-location signal is emitted, its flags argument will be set to one of the flags that was passed in placesSidebarSetOpenFlags.
Passing 0 for flags will cause GTK_PLACES_OPEN_NORMAL to always be sent to callbacks for the “open-location” signal.
Since: 3.10
## setShowConnectToServer
Arguments
:: (HasCallStack, MonadIO m, IsPlacesSidebar a) => a sidebar: a places sidebar -> Bool showConnectToServer: whether to show an item for the Connect to Server command -> m ()
Deprecated: (Since version 3.18)It is recommended to group this functionality with the drives and network location under the new 'Other Location' item
Sets whether the sidebar should show an item for connecting to a network server; this is off by default. An application may want to turn this on if it implements a way for the user to connect to network servers directly.
If you enable this, you should connect to the PlacesSidebar::show-connect-to-server signal.
Since: 3.10
## setShowDesktop
Arguments
:: (HasCallStack, MonadIO m, IsPlacesSidebar a) => a sidebar: a places sidebar -> Bool showDesktop: whether to show an item for the Desktop folder -> m ()
Sets whether the sidebar should show an item for the Desktop folder. The default value for this option is determined by the desktop environment and the user’s configuration, but this function can be used to override it on a per-application basis.
Since: 3.10
## setShowEnterLocation
Arguments
:: (HasCallStack, MonadIO m, IsPlacesSidebar a) => a sidebar: a places sidebar -> Bool showEnterLocation: whether to show an item to enter a location -> m ()
Sets whether the sidebar should show an item for entering a location; this is off by default. An application may want to turn this on if manually entering URLs is an expected user action.
If you enable this, you should connect to the PlacesSidebar::show-enter-location signal.
Since: 3.14
## setShowOtherLocations
Arguments
:: (HasCallStack, MonadIO m, IsPlacesSidebar a) => a sidebar: a places sidebar -> Bool showOtherLocations: whether to show an item for the Other Locations view -> m ()
Sets whether the sidebar should show an item for the application to show an Other Locations view; this is off by default. When set to True, persistent devices such as hard drives are hidden, otherwise they are shown in the sidebar. An application may want to turn this on if it implements a way for the user to see and interact with drives and network servers directly.
If you enable this, you should connect to the PlacesSidebar::show-other-locations signal.
Since: 3.18
## setShowRecent
Arguments
:: (HasCallStack, MonadIO m, IsPlacesSidebar a) => a sidebar: a places sidebar -> Bool showRecent: whether to show an item for recent files -> m ()
Sets whether the sidebar should show an item for recent files. The default value for this option is determined by the desktop environment, but this function can be used to override it on a per-application basis.
Since: 3.18
## setShowStarredLocation
Arguments
:: (HasCallStack, MonadIO m, IsPlacesSidebar a) => a sidebar: a places sidebar -> Bool showStarredLocation: whether to show an item for Starred files -> m ()
If you enable this, you should connect to the PlacesSidebar::show-starred-location signal.
Since: 3.22.26
## setShowTrash
Arguments
:: (HasCallStack, MonadIO m, IsPlacesSidebar a) => a sidebar: a places sidebar -> Bool showTrash: whether to show an item for the Trash location -> m ()
Sets whether the sidebar should show an item for the Trash location.
Since: 3.18
# Properties
## localOnly
No description available in the introspection data.
Construct a GValueConstruct with valid value for the “local-only” property. This is rarely needed directly, but it is used by new.
getPlacesSidebarLocalOnly :: (MonadIO m, IsPlacesSidebar o) => o -> m Bool Source #
Get the value of the “local-only” property. When overloading is enabled, this is equivalent to
get placesSidebar #localOnly
setPlacesSidebarLocalOnly :: (MonadIO m, IsPlacesSidebar o) => o -> Bool -> m () Source #
Set the value of the “local-only” property. When overloading is enabled, this is equivalent to
set placesSidebar [ #localOnly := value ]
## location
No description available in the introspection data.
clearPlacesSidebarLocation :: (MonadIO m, IsPlacesSidebar o) => o -> m () Source #
Set the value of the “location” property to Nothing. When overloading is enabled, this is equivalent to
clear #location
constructPlacesSidebarLocation :: (IsPlacesSidebar o, IsFile a) => a -> IO (GValueConstruct o) Source #
Construct a GValueConstruct with valid value for the “location” property. This is rarely needed directly, but it is used by new.
getPlacesSidebarLocation :: (MonadIO m, IsPlacesSidebar o) => o -> m (Maybe File) Source #
Get the value of the “location” property. When overloading is enabled, this is equivalent to
get placesSidebar #location
setPlacesSidebarLocation :: (MonadIO m, IsPlacesSidebar o, IsFile a) => o -> a -> m () Source #
Set the value of the “location” property. When overloading is enabled, this is equivalent to
set placesSidebar [ #location := value ]
## openFlags
No description available in the introspection data.
Construct a GValueConstruct with valid value for the “open-flags” property. This is rarely needed directly, but it is used by new.
getPlacesSidebarOpenFlags :: (MonadIO m, IsPlacesSidebar o) => o -> m [PlacesOpenFlags] Source #
Get the value of the “open-flags” property. When overloading is enabled, this is equivalent to
get placesSidebar #openFlags
setPlacesSidebarOpenFlags :: (MonadIO m, IsPlacesSidebar o) => o -> [PlacesOpenFlags] -> m () Source #
Set the value of the “open-flags” property. When overloading is enabled, this is equivalent to
set placesSidebar [ #openFlags := value ]
## populateAll
If :populate-all is True, the PlacesSidebar::populate-popup signal is also emitted for popovers.
Since: 3.18
Construct a GValueConstruct with valid value for the “populate-all” property. This is rarely needed directly, but it is used by new.
getPlacesSidebarPopulateAll :: (MonadIO m, IsPlacesSidebar o) => o -> m Bool Source #
Get the value of the “populate-all” property. When overloading is enabled, this is equivalent to
get placesSidebar #populateAll
setPlacesSidebarPopulateAll :: (MonadIO m, IsPlacesSidebar o) => o -> Bool -> m () Source #
Set the value of the “populate-all” property. When overloading is enabled, this is equivalent to
set placesSidebar [ #populateAll := value ]
## showConnectToServer
No description available in the introspection data.
Construct a GValueConstruct with valid value for the “show-connect-to-server” property. This is rarely needed directly, but it is used by new.
getPlacesSidebarShowConnectToServer :: (MonadIO m, IsPlacesSidebar o) => o -> m Bool Source #
Get the value of the “show-connect-to-server” property. When overloading is enabled, this is equivalent to
get placesSidebar #showConnectToServer
setPlacesSidebarShowConnectToServer :: (MonadIO m, IsPlacesSidebar o) => o -> Bool -> m () Source #
Set the value of the “show-connect-to-server” property. When overloading is enabled, this is equivalent to
set placesSidebar [ #showConnectToServer := value ]
## showDesktop
No description available in the introspection data.
Construct a GValueConstruct with valid value for the “show-desktop” property. This is rarely needed directly, but it is used by new.
getPlacesSidebarShowDesktop :: (MonadIO m, IsPlacesSidebar o) => o -> m Bool Source #
Get the value of the “show-desktop” property. When overloading is enabled, this is equivalent to
get placesSidebar #showDesktop
setPlacesSidebarShowDesktop :: (MonadIO m, IsPlacesSidebar o) => o -> Bool -> m () Source #
Set the value of the “show-desktop” property. When overloading is enabled, this is equivalent to
set placesSidebar [ #showDesktop := value ]
## showEnterLocation
No description available in the introspection data.
Construct a GValueConstruct with valid value for the “show-enter-location” property. This is rarely needed directly, but it is used by new.
getPlacesSidebarShowEnterLocation :: (MonadIO m, IsPlacesSidebar o) => o -> m Bool Source #
Get the value of the “show-enter-location” property. When overloading is enabled, this is equivalent to
get placesSidebar #showEnterLocation
setPlacesSidebarShowEnterLocation :: (MonadIO m, IsPlacesSidebar o) => o -> Bool -> m () Source #
Set the value of the “show-enter-location” property. When overloading is enabled, this is equivalent to
set placesSidebar [ #showEnterLocation := value ]
## showOtherLocations
No description available in the introspection data.
Construct a GValueConstruct with valid value for the “show-other-locations” property. This is rarely needed directly, but it is used by new.
getPlacesSidebarShowOtherLocations :: (MonadIO m, IsPlacesSidebar o) => o -> m Bool Source #
Get the value of the “show-other-locations” property. When overloading is enabled, this is equivalent to
get placesSidebar #showOtherLocations
setPlacesSidebarShowOtherLocations :: (MonadIO m, IsPlacesSidebar o) => o -> Bool -> m () Source #
Set the value of the “show-other-locations” property. When overloading is enabled, this is equivalent to
set placesSidebar [ #showOtherLocations := value ]
## showRecent
No description available in the introspection data.
Construct a GValueConstruct with valid value for the “show-recent” property. This is rarely needed directly, but it is used by new.
getPlacesSidebarShowRecent :: (MonadIO m, IsPlacesSidebar o) => o -> m Bool Source #
Get the value of the “show-recent” property. When overloading is enabled, this is equivalent to
get placesSidebar #showRecent
setPlacesSidebarShowRecent :: (MonadIO m, IsPlacesSidebar o) => o -> Bool -> m () Source #
Set the value of the “show-recent” property. When overloading is enabled, this is equivalent to
set placesSidebar [ #showRecent := value ]
## showStarredLocation
No description available in the introspection data.
Construct a GValueConstruct with valid value for the “show-starred-location” property. This is rarely needed directly, but it is used by new.
getPlacesSidebarShowStarredLocation :: (MonadIO m, IsPlacesSidebar o) => o -> m Bool Source #
Get the value of the “show-starred-location” property. When overloading is enabled, this is equivalent to
get placesSidebar #showStarredLocation
setPlacesSidebarShowStarredLocation :: (MonadIO m, IsPlacesSidebar o) => o -> Bool -> m () Source #
Set the value of the “show-starred-location” property. When overloading is enabled, this is equivalent to
set placesSidebar [ #showStarredLocation := value ]
## showTrash
No description available in the introspection data.
Construct a GValueConstruct with valid value for the “show-trash” property. This is rarely needed directly, but it is used by new.
getPlacesSidebarShowTrash :: (MonadIO m, IsPlacesSidebar o) => o -> m Bool Source #
Get the value of the “show-trash” property. When overloading is enabled, this is equivalent to
get placesSidebar #showTrash
setPlacesSidebarShowTrash :: (MonadIO m, IsPlacesSidebar o) => o -> Bool -> m () Source #
Set the value of the “show-trash” property. When overloading is enabled, this is equivalent to
set placesSidebar [ #showTrash := value ]
# Signals
## dragActionAsk
type C_PlacesSidebarDragActionAskCallback = Ptr () -> Int32 -> Ptr () -> IO Int32 Source #
Type for the callback on the (unwrapped) C side.
Arguments
= Int32 actions: Possible drag actions that need to be asked for. -> IO Int32 Returns: the final drag action that the sidebar should pass to the drag side of the drag-and-drop operation.
The places sidebar emits this signal when it needs to ask the application to pop up a menu to ask the user for which drag action to perform.
Since: 3.10
Connect a signal handler for the “drag-action-ask” signal, to be run after the default handler. When overloading is enabled, this is equivalent to
after placesSidebar #dragActionAsk callback
Wrap the callback into a Closure.
Generate a function pointer callable from C code, from a C_PlacesSidebarDragActionAskCallback.
A convenience synonym for Nothing :: Maybe PlacesSidebarDragActionAskCallback.
Connect a signal handler for the “drag-action-ask” signal, to be run before the default handler. When overloading is enabled, this is equivalent to
on placesSidebar #dragActionAsk callback
Wrap a PlacesSidebarDragActionAskCallback into a C_PlacesSidebarDragActionAskCallback.
## dragActionRequested
type C_PlacesSidebarDragActionRequestedCallback = Ptr () -> Ptr DragContext -> Ptr File -> Ptr (GList (Ptr File)) -> Ptr () -> IO Int32 Source #
Type for the callback on the (unwrapped) C side.
Arguments
= DragContext context: DragContext with information about the drag operation -> File destFile: File with the tentative location that is being hovered for a drop -> [File] sourceFileList: List of File that are being dragged -> IO Int32 Returns: The drag action to use, for example, GDK_ACTION_COPY or GDK_ACTION_MOVE, or 0 if no action is allowed here (i.e. drops are not allowed in the specified destFile).
When the user starts a drag-and-drop operation and the sidebar needs to ask the application for which drag action to perform, then the sidebar will emit this signal.
The application can evaluate the context for customary actions, or it can check the type of the files indicated by sourceFileList against the possible actions for the destination destFile.
The drag action to use must be the return value of the signal handler.
Since: 3.10
Connect a signal handler for the “drag-action-requested” signal, to be run after the default handler. When overloading is enabled, this is equivalent to
after placesSidebar #dragActionRequested callback
Wrap the callback into a Closure.
Generate a function pointer callable from C code, from a C_PlacesSidebarDragActionRequestedCallback.
A convenience synonym for Nothing :: Maybe PlacesSidebarDragActionRequestedCallback.
Connect a signal handler for the “drag-action-requested” signal, to be run before the default handler. When overloading is enabled, this is equivalent to
on placesSidebar #dragActionRequested callback
Wrap a PlacesSidebarDragActionRequestedCallback into a C_PlacesSidebarDragActionRequestedCallback.
## dragPerformDrop
type C_PlacesSidebarDragPerformDropCallback = Ptr () -> Ptr File -> Ptr (GList (Ptr File)) -> Int32 -> Ptr () -> IO () Source #
Type for the callback on the (unwrapped) C side.
Arguments
= File destFile: Destination File. -> [File] sourceFileList: List of File that got dropped. -> Int32 action: Drop action to perform. -> IO ()
The places sidebar emits this signal when the user completes a drag-and-drop operation and one of the sidebar's items is the destination. This item is in the destFile, and the sourceFileList has the list of files that are dropped into it and which should be copied/moved/etc. based on the specified action.
Since: 3.10
Connect a signal handler for the “drag-perform-drop” signal, to be run after the default handler. When overloading is enabled, this is equivalent to
after placesSidebar #dragPerformDrop callback
Wrap the callback into a Closure.
Generate a function pointer callable from C code, from a C_PlacesSidebarDragPerformDropCallback.
A convenience synonym for Nothing :: Maybe PlacesSidebarDragPerformDropCallback.
Connect a signal handler for the “drag-perform-drop” signal, to be run before the default handler. When overloading is enabled, this is equivalent to
on placesSidebar #dragPerformDrop callback
Wrap a PlacesSidebarDragPerformDropCallback into a C_PlacesSidebarDragPerformDropCallback.
## mount
type C_PlacesSidebarMountCallback = Ptr () -> Ptr MountOperation -> Ptr () -> IO () Source #
Type for the callback on the (unwrapped) C side.
Arguments
= MountOperation mountOperation: the MountOperation that is going to start. -> IO ()
The places sidebar emits this signal when it starts a new operation because the user clicked on some location that needs mounting. In this way the application using the PlacesSidebar can track the progress of the operation and, for example, show a notification.
Since: 3.20
Connect a signal handler for the “mount” signal, to be run after the default handler. When overloading is enabled, this is equivalent to
after placesSidebar #mount callback
Wrap the callback into a Closure.
Generate a function pointer callable from C code, from a C_PlacesSidebarMountCallback.
A convenience synonym for Nothing :: Maybe PlacesSidebarMountCallback.
Connect a signal handler for the “mount” signal, to be run before the default handler. When overloading is enabled, this is equivalent to
on placesSidebar #mount callback
Wrap a PlacesSidebarMountCallback into a C_PlacesSidebarMountCallback.
## openLocation
type C_PlacesSidebarOpenLocationCallback = Ptr () -> Ptr File -> CUInt -> Ptr () -> IO () Source #
Type for the callback on the (unwrapped) C side.
Arguments
= File location: File to which the caller should switch. -> [PlacesOpenFlags] openFlags: a single value from PlacesOpenFlags specifying how the location should be opened. -> IO ()
The places sidebar emits this signal when the user selects a location in it. The calling application should display the contents of that location; for example, a file manager should show a list of files in the specified location.
Since: 3.10
Connect a signal handler for the “open-location” signal, to be run after the default handler. When overloading is enabled, this is equivalent to
after placesSidebar #openLocation callback
Wrap the callback into a Closure.
Generate a function pointer callable from C code, from a C_PlacesSidebarOpenLocationCallback.
A convenience synonym for Nothing :: Maybe PlacesSidebarOpenLocationCallback.
Connect a signal handler for the “open-location” signal, to be run before the default handler. When overloading is enabled, this is equivalent to
on placesSidebar #openLocation callback
Wrap a PlacesSidebarOpenLocationCallback into a C_PlacesSidebarOpenLocationCallback.
## populatePopup
type C_PlacesSidebarPopulatePopupCallback = Ptr () -> Ptr Widget -> Ptr File -> Ptr Volume -> Ptr () -> IO () Source #
Type for the callback on the (unwrapped) C side.
Arguments
= Widget container: a Menu or another Container -> Maybe File selectedItem: File with the item to which the popup should refer, or Nothing in the case of a selectedVolume. -> Maybe Volume selectedVolume: Volume if the selected item is a volume, or Nothing if it is a file. -> IO ()
The places sidebar emits this signal when the user invokes a contextual popup on one of its items. In the signal handler, the application may add extra items to the menu as appropriate. For example, a file manager may want to add a "Properties" command to the menu.
It is not necessary to store the selectedItem for each menu item; during their callbacks, the application can use placesSidebarGetLocation to get the file to which the item refers.
The selectedItem argument may be Nothing in case the selection refers to a volume. In this case, selectedVolume will be non-Nothing. In this case, the calling application will have to objectRef the selectedVolume and keep it around to use it in the callback.
The container and all its contents are destroyed after the user dismisses the popup. The popup is re-created (and thus, this signal is emitted) every time the user activates the contextual menu.
Before 3.18, the container always was a Menu, and you were expected to add your items as GtkMenuItems. Since 3.18, the popup may be implemented as a Popover, in which case container will be something else, e.g. a Box, to which you may add GtkModelButtons or other widgets, such as GtkEntries, GtkSpinButtons, etc. If your application can deal with this situation, you can set PlacesSidebar::populate-all to True to request that this signal is emitted for populating popovers as well.
Since: 3.10
Connect a signal handler for the “populate-popup” signal, to be run after the default handler. When overloading is enabled, this is equivalent to
after placesSidebar #populatePopup callback
Wrap the callback into a Closure.
Generate a function pointer callable from C code, from a C_PlacesSidebarPopulatePopupCallback.
A convenience synonym for Nothing :: Maybe PlacesSidebarPopulatePopupCallback.
Connect a signal handler for the “populate-popup” signal, to be run before the default handler. When overloading is enabled, this is equivalent to
on placesSidebar #populatePopup callback
Wrap a PlacesSidebarPopulatePopupCallback into a C_PlacesSidebarPopulatePopupCallback.
## showConnectToServer
type C_PlacesSidebarShowConnectToServerCallback = Ptr () -> Ptr () -> IO () Source #
Type for the callback on the (unwrapped) C side.
Deprecated: (Since version 3.18)use the PlacesSidebar::show-other-locations signal to connect to network servers.
The places sidebar emits this signal when it needs the calling application to present an way to connect directly to a network server. For example, the application may bring up a dialog box asking for a URL like "sftp://ftp.example.com". It is up to the application to create the corresponding mount by using, for example, fileMountEnclosingVolume.
Connect a signal handler for the “show-connect-to-server” signal, to be run after the default handler. When overloading is enabled, this is equivalent to
after placesSidebar #showConnectToServer callback
Wrap the callback into a Closure.
Generate a function pointer callable from C code, from a C_PlacesSidebarShowConnectToServerCallback.
A convenience synonym for Nothing :: Maybe PlacesSidebarShowConnectToServerCallback.
Connect a signal handler for the “show-connect-to-server” signal, to be run before the default handler. When overloading is enabled, this is equivalent to
on placesSidebar #showConnectToServer callback
Wrap a PlacesSidebarShowConnectToServerCallback into a C_PlacesSidebarShowConnectToServerCallback.
## showEnterLocation
type C_PlacesSidebarShowEnterLocationCallback = Ptr () -> Ptr () -> IO () Source #
Type for the callback on the (unwrapped) C side.
The places sidebar emits this signal when it needs the calling application to present an way to directly enter a location. For example, the application may bring up a dialog box asking for a URL like "http://http.example.com".
Since: 3.14
Connect a signal handler for the “show-enter-location” signal, to be run after the default handler. When overloading is enabled, this is equivalent to
after placesSidebar #showEnterLocation callback
Wrap the callback into a Closure.
Generate a function pointer callable from C code, from a C_PlacesSidebarShowEnterLocationCallback.
A convenience synonym for Nothing :: Maybe PlacesSidebarShowEnterLocationCallback.
Connect a signal handler for the “show-enter-location” signal, to be run before the default handler. When overloading is enabled, this is equivalent to
on placesSidebar #showEnterLocation callback
Wrap a PlacesSidebarShowEnterLocationCallback into a C_PlacesSidebarShowEnterLocationCallback.
## showErrorMessage
type C_PlacesSidebarShowErrorMessageCallback = Ptr () -> CString -> CString -> Ptr () -> IO () Source #
Type for the callback on the (unwrapped) C side.
Arguments
= Text primary: primary message with a summary of the error to show. -> Text secondary: secondary message with details of the error to show. -> IO ()
The places sidebar emits this signal when it needs the calling application to present an error message. Most of these messages refer to mounting or unmounting media, for example, when a drive cannot be started for some reason.
Since: 3.10
Connect a signal handler for the “show-error-message” signal, to be run after the default handler. When overloading is enabled, this is equivalent to
after placesSidebar #showErrorMessage callback
Wrap the callback into a Closure.
Generate a function pointer callable from C code, from a C_PlacesSidebarShowErrorMessageCallback.
A convenience synonym for Nothing :: Maybe PlacesSidebarShowErrorMessageCallback.
Connect a signal handler for the “show-error-message” signal, to be run before the default handler. When overloading is enabled, this is equivalent to
on placesSidebar #showErrorMessage callback
Wrap a PlacesSidebarShowErrorMessageCallback into a C_PlacesSidebarShowErrorMessageCallback.
## showOtherLocations
type C_PlacesSidebarShowOtherLocationsCallback = Ptr () -> Ptr () -> IO () Source #
Type for the callback on the (unwrapped) C side.
Deprecated: (Since version 3.20)use the PlacesSidebar::show-other-locations-with-flagswhich includes the open flags in order to allow the user to specify to openin a new tab or window, in a similar way than PlacesSidebar::open-location
The places sidebar emits this signal when it needs the calling application to present a way to show other locations e.g. drives and network access points. For example, the application may bring up a page showing persistent volumes and discovered network addresses.
Since: 3.18
Connect a signal handler for the “show-other-locations” signal, to be run after the default handler. When overloading is enabled, this is equivalent to
after placesSidebar #showOtherLocations callback
Wrap the callback into a Closure.
Generate a function pointer callable from C code, from a C_PlacesSidebarShowOtherLocationsCallback.
A convenience synonym for Nothing :: Maybe PlacesSidebarShowOtherLocationsCallback.
Connect a signal handler for the “show-other-locations” signal, to be run before the default handler. When overloading is enabled, this is equivalent to
on placesSidebar #showOtherLocations callback
Wrap a PlacesSidebarShowOtherLocationsCallback into a C_PlacesSidebarShowOtherLocationsCallback.
## showOtherLocationsWithFlags
type C_PlacesSidebarShowOtherLocationsWithFlagsCallback = Ptr () -> CUInt -> Ptr () -> IO () Source #
Type for the callback on the (unwrapped) C side.
Arguments
= [PlacesOpenFlags] openFlags: a single value from PlacesOpenFlags specifying how it should be opened. -> IO ()
The places sidebar emits this signal when it needs the calling application to present a way to show other locations e.g. drives and network access points. For example, the application may bring up a page showing persistent volumes and discovered network addresses.
Since: 3.20
Connect a signal handler for the “show-other-locations-with-flags” signal, to be run after the default handler. When overloading is enabled, this is equivalent to
after placesSidebar #showOtherLocationsWithFlags callback
Wrap the callback into a Closure.
Generate a function pointer callable from C code, from a C_PlacesSidebarShowOtherLocationsWithFlagsCallback.
A convenience synonym for Nothing :: Maybe PlacesSidebarShowOtherLocationsWithFlagsCallback.
Connect a signal handler for the “show-other-locations-with-flags” signal, to be run before the default handler. When overloading is enabled, this is equivalent to
on placesSidebar #showOtherLocationsWithFlags callback
Wrap a PlacesSidebarShowOtherLocationsWithFlagsCallback into a C_PlacesSidebarShowOtherLocationsWithFlagsCallback.
## showStarredLocation
type C_PlacesSidebarShowStarredLocationCallback = Ptr () -> CUInt -> Ptr () -> IO () Source #
Type for the callback on the (unwrapped) C side.
Arguments
= [PlacesOpenFlags] openFlags: a single value from PlacesOpenFlags specifying how the starred file should be opened. -> IO ()
The places sidebar emits this signal when it needs the calling application to present a way to show the starred files. In GNOME, starred files are implemented by setting the nao:predefined-tag-favorite tag in the tracker database.
Since: 3.22.26
Connect a signal handler for the “show-starred-location” signal, to be run after the default handler. When overloading is enabled, this is equivalent to
after placesSidebar #showStarredLocation callback
Wrap the callback into a Closure.
Generate a function pointer callable from C code, from a C_PlacesSidebarShowStarredLocationCallback.
A convenience synonym for Nothing :: Maybe PlacesSidebarShowStarredLocationCallback.
Connect a signal handler for the “show-starred-location” signal, to be run before the default handler. When overloading is enabled, this is equivalent to
on placesSidebar #showStarredLocation callback
Wrap a PlacesSidebarShowStarredLocationCallback into a C_PlacesSidebarShowStarredLocationCallback.
## unmount
type C_PlacesSidebarUnmountCallback = Ptr () -> Ptr MountOperation -> Ptr () -> IO () Source #
Type for the callback on the (unwrapped) C side.
Arguments
= MountOperation mountOperation: the MountOperation that is going to start. -> IO ()
The places sidebar emits this signal when it starts a new operation because the user for example ejected some drive or unmounted a mount. In this way the application using the PlacesSidebar can track the progress of the operation and, for example, show a notification.
Since: 3.20
Connect a signal handler for the “unmount” signal, to be run after the default handler. When overloading is enabled, this is equivalent to
after placesSidebar #unmount callback
Wrap the callback into a Closure.
Generate a function pointer callable from C code, from a C_PlacesSidebarUnmountCallback.
A convenience synonym for Nothing :: Maybe PlacesSidebarUnmountCallback.
Connect a signal handler for the “unmount” signal, to be run before the default handler. When overloading is enabled, this is equivalent to
on placesSidebar #unmount callback
Wrap a PlacesSidebarUnmountCallback into a C_PlacesSidebarUnmountCallback. | 2020-08-13 15:10:42 | {"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": 0.22625894844532013, "perplexity": 11379.713546597735}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-34/segments/1596439739046.14/warc/CC-MAIN-20200813132415-20200813162415-00438.warc.gz"} |
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### Horizontal Cylindrical Tank Volume Calculator - Imperial
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### HOW TO CALCULATE THE VOLUMES OF PARTIALLY FULL
cylindrical tanks, either in horizontal or vertical configuration. Consider, for example, a cylindrical tank with length L and radius R, filling up to a height H. If you want to obtain the volume of the liquid that partially fills the tank, you should indicate if the tank is in horizontal or vertical position.Horizontal Cylindrical Tank Volume and Level CalculatorVolume calculation on a partially filled cylindrical tank Some Theory. Using the theory. Use this calculator for computing the volume of partially-filled horizontal cylinder-shaped tanks.With horizontal cylinders, volume changes are not linear and in fact are rather complex as the theory above shows. Fortunately you have this tool to do the work for you.Volume in horizontal round tanks??? - Excel Help ForumJan 16, 2013 · The following formula will calculate the volume of a fluid in a partially full tank (cylindrical and flat on sides, and set up horizontally) based on the tank dimensions and the fluid depth. Enter the length of tank in A2, diameter of the tank in B2, and fluid depth in C2, and the following formula in D2.
### Tank Volume Calculator - Oil Tanks
Mar 26, 2015 · Lets look at how to calculate the volume of both of these tanks using tank capacity calculators or a cylindrical tank calculator. In the case of the horizontal cylindrical tank, you need to calculate the area of a cross-section of the tank and then multiply this figure by the total length of the tank.Tank Volume CalculatorTotal volume of a cylinder shaped tank is the area, A, of the circular end times the length, l. A = r 2 where r is the radius which is equal to 1/2 the diameter or d/2. Therefore: V(tank) = r 2 l Calculate the filled volume of a horizontal cylinder tank by first finding the area, A, of a circular segment and multiplying it by the length, l.Cylindrical Tank CalculatorI n s t r u c t i o n s Use this calculator for computing the volume of partially-filled horizontal cylinders. If you need a vertical cylinder calculator, click here.. Besides calculating volume for any particular depth, this calculator can also produce a "dipstick chart" showing volume across the entire range of tank
### Calculation of Liquid Volume in a Horizontal Cylindrical , ghana horizontal cylindrical tank fire volume
Calculation of Liquid Volume in a Horizontal Cylindrical Container: This calculator calculates the volume of liquid inside a horizontal cylindrical container at any given height of liquid. The other required dimensions are the diameter and length of the tank. Values to be Entered Values to be Calculated; Diameter of CylinderHorizontal Tank Volume Calculations - HagraHorizontal Cylindrical Tank Volume Calculator. Horizontal Oval Tank Volume Calculator. Disclaimer. The calculations on these pages are a purely theoretical exercise! Therefore the outcomes of the calculations on these pages can only be used for indicative purposes. It might help you to estimate the content of a tank.Tank Volume Calculator - www.specialtytankandweldingSpecialty Tank & Welding Company is dedicated to bringing you tanks of the highest quality, with the most economical and competitive pricing, at the quickest convenience to you. STW specializes in custom builds and tanks for specific situations and conditions.
### Radio frequency liquid level gauging in propane tank car , ghana horizontal cylindrical tank fire volume
Selected radio frequency (rf) resonances of an empty 30,300 liter (8,000 gallon) tank car were measured to determine whether rf can be used to gauge the propane liquid levels during tank car fire safety tests. The change of resonant frequencies of a small horizontal cylindrical tank as a function of liquid volume has been tested in order to estimate the precision to which the amount of propane , ghana horizontal cylindrical tank fire volumeVolume and Wetted Area of Partially Filled Horizontal , ghana horizontal cylindrical tank fire volumeIntroduction. The calculation of the liquid volume or wetted area of a partially filled horizontal vessel is best performed in parts, by calculating the value for the cylindrical section of the vessel and the heads of the vessel and then adding the areas or volumes together.Math Forum - Ask Dr. Math Archives: Volume of a TankFinding the Volume of a Tank, a selection of answers from the Dr. Math archives. Volume of a Cylinder What is the volume of the storage tank with a diameter 6m, height 5m? Volume of Liquid in a Cylindrical Tank How can I calculate volume of liquid in a cylinder that's not full and lying horizontally? Units and Cylinder Volume
### Volume in horizontal round tanks??? - Excel Help Forum
Jan 16, 2013 · The following formula will calculate the volume of a fluid in a partially full tank (cylindrical and flat on sides, and set up horizontally) based on the tank dimensions and the fluid depth. Enter the length of tank in A2, diameter of the tank in B2, and fluid depth in C2, and the following formula in D2.Calculation of Liquid Volume in a Horizontal Cylindrical , ghana horizontal cylindrical tank fire volumeCalculation of Liquid Volume in a Horizontal Cylindrical Container: This calculator calculates the volume of liquid inside a horizontal cylindrical container at any given height of liquid. The other required dimensions are the diameter and length of the tank. Values to be Entered Values to be Calculated; Diameter of CylinderCylindrical Tank CalculatorI n s t r u c t i o n s Use this calculator for computing the volume of partially-filled horizontal cylinders. If you need a vertical cylinder calculator, click here.. Besides calculating volume for any particular depth, this calculator can also produce a "dipstick chart" showing volume across the entire range of tank
### Tank calculations - MrExcel
Dec 19, 2014 · We use what are called tank strappings for some of our additive tanks. They were built from formulas for cylindrical horizontal or vertical tanks. I pugged in the size of the tanks and a strapping was produced. From there I set up a form using vlookups to the strappings. Give me some tank measurements and I'll see if what I have at work will , ghana horizontal cylindrical tank fire volumecalculus - Volume of a horizontal cylinder using height of , ghana horizontal cylindrical tank fire volumeVolume of a horizontal cylinder using height of liquid. Ask Question , ghana horizontal cylindrical tank fire volume length, and height of liquid in the tank to find the volume of the liquid. $\endgroup$ Billybob686 Oct 13 , ghana horizontal cylindrical tank fire volume then the correction factor is 1/2 -- if the height measured was half the diameter, then you would have half the volume of a horizontal cylinder. Simple. shareTank Volume Calculator - Horizontal Elliptical - MetricPlease help promote this free service - Tell a Friend about this site! Create PDF to print diagrams on this page. Help & Settings Printing Help (new window) Copy all diagrams on this page to bottom of page - Make multiple copies to Print or Compare.
### Spill Prevention Control and Countermeasure (SPCC)
Spill Prevention Control and Countermeasure (SPCC) Plan Construct New Secondary Containment EXAMPLE D. Accounting for the displacements from other rectangular tanks to be located in dike or berm with the largest tank 1. Calculate the total displacement volume from the additional horizontal cylindrical tanks in the dike or bermalbania horizontal cylindrical tank fire volume - Oil , ghana horizontal cylindrical tank fire volumeThe general construction of fire tube boiler is as a tank of water perforated by tubes that carry the hot flue gases from the fire. The tank is usually cylindrical for the most part being the strongest practical shape for a pressurized container and this cylindrical tank may be either horizontal SPCC Plan - Calculation Guidance - AsmarkSPCC Plan - Calculation Guidance The following example compares two different design criteria: one based on the volume of the tank and one based on precipitation. Scenario: A 20,000-gallon horizontal tank is placed within an engineered secondary containment structure, such as a concrete dike. The tank is 35 feet long by 10 feet in diameter.
### Content of Horizontal - or Sloped - Cylindrical Tank and Pipe
Volume of partly filled horizontal or sloped cylindrical tanks and pipes - an online calculator Sponsored Links The online calculator below can be used to calculate the volume and mass of liquid in a partly filled horizontal or sloped cylindrical tank if you know the inside diameter and the level of the liquid the tank.Calculate volume in a horizontal cylinder or a cylindrical , ghana horizontal cylindrical tank fire volumeAbout this Horizontal Cylinder Calculator; Note 1: A fluid can fill the cylinder or cylindrical tank only partially, in order to calculate the fluid volume. Note 2: The selection of measurement unit is optional. Reference (ID: N/A)SPREADSHEET MASTER - Tank Level CalculatorSPREADSHEET MASTER. Tank Level Calculator. Cylindrical Horizontal Tank. I have created lots of Excel tool. But I was not able to publish them on a web interface so users can access it live. , ghana horizontal cylindrical tank fire volume It measures the tank volume based on the dimensions that you will enter. Also when you enter your dipstick level, it allows you to know accurately your , ghana horizontal cylindrical tank fire volume
### Calculation of Height of Liquid in a Horizontal Container , ghana horizontal cylindrical tank fire volume
Calculation of Height of Liquid in a Horizontal Cylindrical Container: Data Table Generator for Calculated Volumes of Liquid in a Horizontal Cylindrical Container: Data Table Generator for Calculated Heights of Liquid in a Horizontal Cylindrical Container: Calculator for Volume, Diameter, and Length of a Cylindrical Container or TubeHow to mathematically calculate a fire water tank capacity , ghana horizontal cylindrical tank fire volume1. You need to determine the fire water tank demand for each area or equipment or tank in the facility 2. After determination of the individual firewater demand, you design your firewater facility based on the highest firewater demand 3. Based on , ghana horizontal cylindrical tank fire volume* Sloped Bottom Tank - arachnoid, ghana horizontal cylindrical tank fire volumeThe easy part the cylindrical section above the slope, which has a volume of: (1) $\displaystyle v = \pi r^2 h$ v = volume; r = tank radius; h = cylindrical section height; More difficult the tank's sloped section, which lies between the tank's bottom and the top of the slope where the tank
### L/D Ratio of storage tank - Chemical engineering other , ghana horizontal cylindrical tank fire volume
- Desired vapor space volume and tank venting requirements - Minimum Fire protection distance requirements for combustible and flammable liquids , ghana horizontal cylindrical tank fire volume Do you mean vertical cylindrical tanks or horizontal cylindrical tanks? (L/D implies horizontal "bullet" like tanks, ghana horizontal cylindrical tank fire volume.) What sort of volume / size are you looking at?Cylindrical Tank ProblemsIn order to find the volume, we would have to find the area of the section covered by the oil at the end of the tank and then multiply by the length of the tank. But, how do you find the area of such a figure? Let's begin by examining the end view of the tank (in general so that we can do it for any size cylindrical tank Gauge Charts - Highland TankHighland Tanks on-line Tank Gauge Chart Utility generates gauge charts to help you track the volume of fluid in your storage tanks. Please note that these charts are theoretical and intended as a guide for estimating tank and vessel volumes. The accuracy of these calculations can be affected by many factors and may vary per tank/vessel.
### SPCC Plan - Calculation Guidance - Asmark
SPCC Plan - Calculation Guidance The following example compares two different design criteria: one based on the volume of the tank and one based on precipitation. Scenario: A 20,000-gallon horizontal tank is placed within an engineered secondary containment structure, such as a concrete dike. The tank is 35 feet long by 10 feet in diameter.Volume of horizontal cylindrical tank - OnlineConversion , ghana horizontal cylindrical tank fire volumeRe: Volume of horizontal cylindrical tank Folks, Re the solution below- I actually need the inverse function, whereby I can calculate the depth measured along a vertical diameter as a function of the occupied volue of the horizontal tank. (I am pumping fluid into the tank at a known rate, and I need to measure how fast the level rises)Fuel Oil - Storage Tanks - Engineering ToolBoxDimensions of fuel oil storage tanks. Related Topics . Combustion - Boiler house topics - fuels like oil, gas, coal, wood - chimneys, safety valves, tanks - combustion efficiency; Related Documents . Content of Horizontal - or Sloped - Cylindrical Tank and Pipe - Volume of partly filled horizontal or sloped cylindrical tanks and pipes - an online calculator
### Calculating Tank Wetted Area - Chemical Processing
Calculating Tank Wetted Area Saving time, increasing accuracy By Dan Jones, Ph.D., P.E. alculating wetted area in a partially-filled horizontal or vertical cylindrical or elliptical tank can be complicated, depending on fluid height and the shape of the heads (ends) of a horizontal tank or the head (bottom) of a vertical tank.Fireguard - Highland TankFireguard ® tanks are thermally protected, double-wall steel storage tanks and are the best alternative for safe storage of motor fuels and other flammable and combustible liquids aboveground. They are used where a fire-protected tank is needed because of setback limitations or regulatory requirements. Each tank is constructed with a minimum 3 interstice around the inner tank.
Tags: | 2020-06-05 23:33:16 | {"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": 0.6613344550132751, "perplexity": 2157.8898839240032}, "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/1590348504341.78/warc/CC-MAIN-20200605205507-20200605235507-00179.warc.gz"} |
https://blog.csdn.net/chenzhenyu123456/article/details/46779495 |
# Leapin' Lizards
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1568 Accepted Submission(s): 637
Problem Description
Your platoon of wandering lizards has entered a strange room in the labyrinth you are exploring. As you are looking around for hidden treasures, one of the rookies steps on an innocent-looking stone and the room's floor suddenly disappears! Each lizard in your platoon is left standing on a fragile-looking pillar, and a fire begins to rage below... Leave no lizard behind! Get as many lizards as possible out of the room, and report the number of casualties.
The pillars in the room are aligned as a grid, with each pillar one unit away from the pillars to its east, west, north and south. Pillars at the edge of the grid are one unit away from the edge of the room (safety). Not all pillars necessarily have a lizard. A lizard is able to leap onto any unoccupied pillar that is within d units of his current one. A lizard standing on a pillar within leaping distance of the edge of the room may always leap to safety... but there's a catch: each pillar becomes weakened after each jump, and will soon collapse and no longer be usable by other lizards. Leaping onto a pillar does not cause it to weaken or collapse; only leaping off of it causes it to weaken and eventually collapse. Only one lizard may be on a pillar at any given time.
Input
The input file will begin with a line containing a single integer representing the number of test cases, which is at most 25. Each test case will begin with a line containing a single positive integer n representing the number of rows in the map, followed by a single non-negative integer d representing the maximum leaping distance for the lizards. Two maps will follow, each as a map of characters with one row per line. The first map will contain a digit (0-3) in each position representing the number of jumps the pillar in that position will sustain before collapsing (0 means there is no pillar there). The second map will follow, with an 'L' for every position where a lizard is on the pillar and a '.' for every empty pillar. There will never be a lizard on a position where there is no pillar.Each input map is guaranteed to be a rectangle of size n x m, where 1 ≤ n ≤ 20 and 1 ≤ m ≤ 20. The leaping distance is
always 1 ≤ d ≤ 3.
Output
For each input case, print a single line containing the number of lizards that could not escape. The format should follow the samples provided below.
Sample Input
4 3 1 1111 1111 1111 LLLL LLLL LLLL 3 2 00000 01110 00000 ..... .LLL. ..... 3 1 00000 01110 00000 ..... .LLL. ..... 5 2 00000000 02000000 00321100 02000000 00000000 ........ ........ ..LLLL.. ........ ........
Sample Output
Case #1: 2 lizards were left behind. Case #2: no lizard was left behind. Case #3: 3 lizards were left behind. Case #4: 1 lizard was left behind.
AC代码：
#include <cstdio>
#include <cstring>
#include <queue>
#include <algorithm>
#define MAXN 1000+10//最多800个点
#define MAXM 40000
#define INF 100000+10
using namespace std;
struct Edge
{
int from, to, cap, flow, next;
}edge[MAXM];
int N, M, D;
int head[MAXN], cur[MAXN], top;
int dist[MAXN];
bool vis[MAXN];
char Map[21][21];
char val[21][21];//记录每个位置最多可以跳几次
int source = 900, sink = 1000;//超级汇点
int sum;//蜥蜴总数目
bool judge(int x, int y)//判断是否越界
{
return x >= 0 && x < N && y >= 0 && y < M;
}
void init()
{
top = 0;
memset(head, -1, sizeof(head));
}
void addEdge(int u, int v, int w)
{
Edge E1 = {u, v, w, 0, head[u]};
edge[top] = E1;
head[u] = top++;
Edge E2 = {v, u, 0, 0, head[v]};
edge[top] = E2;
head[v] = top++;
}
void getMap()
{
int a, b, c;
int x, y;
sum = 0;
int move1[4][2] = {0,1, 0,-1, 1,0, -1,0};//D为1 可以跳4个点
int move2[8][2] = {0,2, 0,-2, 2,0, -2,0, 1,1, 1,-1, -1,1, -1,-1};//D为2 可以跳着8个点加上D为1的4个点 下同
int move3[16][2] = {0,3, 0,-3, 3,0, -3,0, 1,2, 1,-2, -1,2, -1,-2, 2,1, 2,-1, -2,1, -2,-1, 2,2, 2-2, -2,2, -2,-2};//D为3
int move4[20][2] = {0,4, 0,-4, 4,0, -4,0, 1,3, 1,-3, -1,3, -1,-3, 2,3, 2,-3, -2,3, -2,-3, 3,1, 3,-1, -3,1, -3,-1, 3,2, 3,-2, -3,2, -3,-2};
for(int i = 0; i < N; i++)
{
scanf("%s", val[i]);
M = strlen(val[i]);
for(int j = 0; j < M; j++)
{
if(val[i][j] == '0') continue;
a = j + i * M;
b = 400 + a;
addEdge(a, b, val[i][j]-'0');//拆点建边
}
}
for(int i = 0; i < N; i++)
{
scanf("%s", Map[i]);
for(int j = 0; j < M; j++)
{
if(Map[i][j] == 'L')//有蜥蜴
{
sum++;
a = j + i * M;
addEdge(source, a, 1);//从源点引一条 容量为一的边
}
}
}
for(int i = 0; i < N; i++)
{
for(int j = 0; j < M; j++)
{
if(val[i][j] > '0')
{
a = j + i * M;
b = 400 + a;
for(int k = 0; k < 4; k++)
{
x = i + move1[k][0];
y = j + move1[k][1];
c = y + x * M;
if(judge(x, y))//没有越界
{
if(val[x][y] != '0')//当前位置必须 可跳跃至少一次
addEdge(b, c, INF);//直接连接
}
else//越界说明可以直接跳出去
addEdge(b, sink, INF);//连接汇点
}
if(D == 1)
continue;
for(int k = 0; k < 8; k++)
{
x = i + move2[k][0];
y = j + move2[k][1];
c = y + x * M;
if(judge(x, y))//没有越界
{
if(val[x][y] != '0')
addEdge(b, c, INF);//直接连接
}
else
addEdge(b, sink, INF);//连接汇点
}
if(D == 2)
continue;
for(int k = 0; k < 16; k++)
{
x = i + move3[k][0];
y = j + move3[k][1];
c = y + x * M;
if(judge(x, y))//没有越界
{
if(val[x][y] != '0')
addEdge(b, c, INF);//直接连接
}
else
addEdge(b, sink, INF);//连接汇点
}
if(D == 3)
continue;
for(int k = 0; k < 20; k++)
{
x = i + move4[k][0];
y = j + move4[k][1];
c = y + x * M;
if(judge(x, y))//没有越界
{
if(val[x][y] != '0')
addEdge(b, c, INF);//直接连接
}
else
addEdge(b, sink, INF);//连接汇点
}
}
}
}
}
bool BFS(int start, int end)//寻找是否存在增广路
{
queue<int> Q;
memset(dist, -1, sizeof(dist));
memset(vis, false, sizeof(vis));
Q.push(start);
dist[start] = 0;
vis[start] = true;
while(!Q.empty())
{
int u = Q.front();
Q.pop();
for(int i = head[u]; i != -1; i = edge[i].next)
{
Edge E = edge[i];
if(!vis[E.to] && E.cap - E.flow > 0)
{
dist[E.to] = dist[u] + 1;
vis[E.to] = true;
if(E.to == end)
return true;
Q.push(E.to);
}
}
}
return false;
}
int DFS(int x, int a, int end)//增广路
{
if(x == end || a == 0) return a;
int flow = 0, f;
for(int &i = cur[x]; i != -1; i = edge[i].next)
{
Edge &E = edge[i];
if(dist[E.to] == dist[x] + 1 && (f = DFS(E.to, min(a, E.cap - E.flow), end)) > 0)
{
E.flow += f;
edge[i^1].flow -= f;
flow += f;
a -= f;
if(a == 0)
break;
}
}
return flow;
}
int Maxflow(int start, int end)
{
int flow = 0;
while(BFS(start, end))
{
memcpy(cur, head, sizeof(head));
flow += DFS(start, INF, end);
}
return flow;
}
int main()
{
int t;
int k = 1;
int ans;
scanf("%d", &t);
while(t--)
{
scanf("%d%d", &N, &D);
init();
getMap();
ans = sum - Maxflow(source, sink);
printf("Case #%d: ", k++);
if(ans == 0)
printf("no lizard was left behind.\n");
else if(ans == 1)
printf("1 lizard was left behind.\n");
else
printf("%d lizards were left behind.\n", ans);
}
return 0;
}
#### HDU 2732 Leapin' Lizards（拆点法+最大流）
2015-08-30 17:13:39
#### HDU 2732 Leapin' Lizards(最大流)
2014-08-31 22:21:11
#### HDU 2732 Leapin' Lizards
2014-08-31 16:57:03
#### HDU 2732 Leapin' Lizards
2017-02-26 10:35:37
#### 【HDU】 2732 Leapin' Lizards
2016-05-06 16:20:45
#### HDU 2732 —— Leapin' Lizards
2016-05-08 14:51:31
#### hdu 2732 Leapin' Lizards
2015-07-05 20:12:18
#### HDU 2732 Leapin' Lizards(拆点+最大流)
2017-08-20 18:16:03
#### Leapin' Lizards (hdu 2732 最大流)
2015-04-18 21:50:37
#### HDU 2732 Leapin' Lizards(拆点+最大流）
2014-11-12 15:09:54 | 2018-07-15 19:25:14 | {"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": 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": 0.2417021244764328, "perplexity": 14169.006608031983}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "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-2018-30/segments/1531676588961.14/warc/CC-MAIN-20180715183800-20180715203800-00160.warc.gz"} |
https://xianblog.wordpress.com/tag/parallel-tempering/ | ## parallel tempering on optimised paths
Posted in Statistics with tags , , , , , , , , , , , , , , , on May 20, 2021 by xi'an
Saifuddin Syed, Vittorio Romaniello, Trevor Campbell, and Alexandre Bouchard-Côté, whom I met and discussed with on my “last” trip to UBC, on December 2019, just arXived a paper on parallel tempering (PT), making the choice of tempering path an optimisation problem. They address the touchy issue of designing a sequence of tempered targets when the starting distribution π⁰, eg the prior, and the final distribution π¹, eg the posterior, are hugely different, eg almost singular.
“…theoretical analysis of reversible variants of PT has shown that adding too many intermediate chains can actually deteriorate performance (…) [while] on non reversible regime adding more chains is guaranteed to improve performances.”
The above applies to geometric combinations of π⁰ and π¹. Which “suffers from an arbitrarily suboptimal global communication barrier“, according to the authors (although the counterexample is not completely convincing since π⁰ and π¹ share the same variance). They propose a more non-linear form of tempering with constraints on the dependence of the powers on the temperature t∈(0,1). Defining the global communication barrier as an average over temperatures of the rejection rate, the path characteristics (e.g., the coefficients of a spline function) can then be optimised in terms of this objective. And the temperature schedule is derived from the fact that the non-asymptotic round trip rate is maximized when the rejection rates are all equal. (As a side item, the technique exposed in the earlier tempering paper by Syed et al. was recently exploited for a night high resolution imaging of a black hole from the M87 galaxy.)
## ABC, anytime!
Posted in Books, pictures, Statistics, Travel, University life with tags , , , on January 18, 2021 by xi'an
Last June, Alix Marie d’Avigneau, Sumeet Singh, and Lawrence Murray arXived a paper on anytime ABC I intended to review right away but that sat till now on my virtual desk (and pile of to-cover-arXivals!). The notion of anytime MCMC was already covered in earlier ‘Og entries, but this anytime ABC version bypasses the problem of asynchronicity, namely, “randomly varying local move completion times when parallel tempering is implemented on a multi-processor computing resource”. The different temperatures are replaced by different tolerances in ABC. Since switches between tolerances are natural if a proposal for a given tolerance ε happens to be eligible for a lower tolerance ε’. And accounting for the different durations required to simulate a proposal under different tolerances to avoid the induced bias in the stationary distributions. Or the wait for other processors to complete their task. A drawback with the approach stands in calibrating the tolerance levels in advance (or via preliminary runs that may prove costly).
## general perspective on the Metropolis–Hastings kernel
Posted in Books, Statistics with tags , , , , , , , , , , , , , on January 14, 2021 by xi'an
[My Bristol friends and co-authors] Christophe Andrieu, and Anthony Lee, along with Sam Livingstone arXived a massive paper on 01 January on the Metropolis-Hastings kernel.
“Our aim is to develop a framework making establishing correctness of complex Markov chain Monte Carlo kernels a purely mechanical or algebraic exercise, while making communication of ideas simpler and unambiguous by allowing a stronger focus on essential features (…) This framework can also be used to validate kernels that do not satisfy detailed balance, i.e. which are not reversible, but a modified version thereof.”
A central notion in this highly general framework is, extending Tierney (1998), to see an MCMC kernel as a triplet involving a probability measure μ (on an extended space), an involution transform φ generalising the proposal step (i.e. þ²=id), and an associated acceptance probability ð. Then μ-reversibility occurs for
$\eth(\xi)\mu(\text{d}\xi)= \eth(\phi(\xi))\mu^{\phi}(\text{d}\xi)$
with the rhs involving the push-forward measure induced by μ and φ. And furthermore there is always a choice of an acceptance probability ð ensuring for this equality to happen. Interestingly, the new framework allows for mostly seamless handling of more complex versions of MCMC such as reversible jump and parallel tempering. But also non-reversible kernels, incl. for instance delayed rejection. And HMC, incl. NUTS. And pseudo-marginal, multiple-try, PDMPs, &c., &c. it is remarkable to see such a general theory emerging a this (late?) stage of the evolution of the field (and I will need more time and attention to understand its consequences).
## QuanTA
Posted in Books, pictures, Running, Statistics, University life with tags , , , , , , , on September 17, 2018 by xi'an
My Warwick colleagues Nick Tawn [who also is my most frequent accomplice to running, climbing and currying in Warwick!] and Gareth Robert have just arXived a paper on QuanTA, a new parallel tempering algorithm that Nick designed during his thesis at Warwick, which he defended last semester. Parallel tempering targets in parallel several powered (or power-tempered) versions of the target distribution. With proposed switches between adjacent targets. An improved version transforms the local values before operating the switches. Ideally, the transform should be the composition of the cdf and inverse cdf, but this is impossible. Linearising the transform is feasible, but does not agree with multimodality, which calls for local transforms. Which themselves call for the identification of the different modes. In QuanTA, they are identified by N parallel runs of the standard, or rather N/2 to avoid dependence issues, and K-means estimates. The paper covers the construction of an optimal scaling of temperatures, in that the difference between the temperatures is scaled [with order 1/√d] so that the acceptance rate for swaps is 0.234. Which in turns induces a practical if costly calibration of the temperatures, especially when the size of the jump is depending on the current temperature. However, this cost issue is addressed in the paper, resorting to the acceptance rate as a proxy for effective sample size and the acceptance rate over run time to run the comparison with regular parallel tempering, leading to strong improvements in the mixture examples examined in the paper. The use of machine learning techniques like K-means or more involved solutions is a promising thread in this exciting area of tempering, where intuition about high temperatures can be actually misleading. Because using the wrong scale means missing the area of interest, which is not the mode!
## love-hate Metropolis algorithm
Posted in Books, pictures, R, Statistics, Travel with tags , , , , , , , , , on January 28, 2016 by xi'an
Hyungsuk Tak, Xiao-Li Meng and David van Dyk just arXived a paper on a multiple choice proposal in Metropolis-Hastings algorithms towards dealing with multimodal targets. Called “A repulsive-attractive Metropolis algorithm for multimodality” [although I wonder why XXL did not jump at the opportunity to use the “love-hate” denomination!]. The proposal distribution includes a [forced] downward Metropolis-Hastings move that uses the inverse of the target density π as its own target, namely 1/{π(x)+ε}. Followed by a [forced] Metropolis-Hastings upward move which target is {π(x)+ε}. The +ε is just there to avoid handling ratios of zeroes (although I wonder why using the convention 0/0=1 would not work). And chosen as 10⁻³²³ by default in connection with R smallest positive number. Whether or not the “downward” move is truly downwards and the “upward” move is truly upwards obviously depends on the generating distribution: I find it rather surprising that the authors consider the same random walk density in both cases as I would have imagined relying on a more dispersed distribution for the downward move in order to reach more easily other modes. For instance, the downward move could have been based on an anti-Langevin proposal, relying on the gradient to proceed further down…
This special choice of a single proposal however simplifies the acceptance ratio (and keeps the overall proposal symmetric). The final acceptance ratio still requires a ratio of intractable normalising constants that the authors bypass by Møller et al. (2006) auxiliary variable trick. While the authors mention the alternative pseudo-marginal approach of Andrieu and Roberts (2009), they do not try to implement it, although this would be straightforward here: since the normalising constants are the probabilities of accepting a downward and an upward move, respectively. Those can easily be evaluated at a cost similar to the use of the auxiliary variables. That is,
– generate a few moves from the current value and record the proportion p of accepted downward moves;
– generate a few moves from the final proposed value and record the proportion q of accepted downward moves;
and replace the ratio of intractable normalising constants with p/q. It is not even clear that one needs those extra moves since the algorithm requires an acceptance in the downward and upward moves, hence generate Geometric variates associated with those probabilities p and q, variates that can be used for estimating them. From a theoretical perspective, I also wonder if forcing the downward and upward moves truly leads to an improved convergence speed. Considering the case when the random walk is poorly calibrated for either the downward or upward move, the number of failed attempts before an acceptance may get beyond the reasonable.
As XXL and David pointed out to me, the unusual aspect of the approach is that here the proposal density is intractable, rather than the target density itself. This makes using Andrieu and Roberts (2009) seemingly less straightforward. However, as I was reminded this afternoon at the statistics and probability seminar in Bristol, the argument for the pseudo-marginal based on an unbiased estimator is that w Q(w|x) has a marginal in x equal to π(x) when the expectation of w is π(x). In thecurrent problem, the proposal in x can extended into a proposal in (x,w), w P(w|x) whose marginal is the proposal on x.
If we complement the target π(x) with the conditional P(w|x), the acceptance probability would then involve
{π(x’) P(w’|x’) / π(x) P(w|x)} / {w’ P(w’|x’) / w P(w|x)} = {π(x’) / π(x)} {w/w’}
so it seems the pseudo-marginal (or auxiliary variable) argument also extends to the proposal. Here is a short experiment that shows no discrepancy between target and histogram:
nozero=1e-300
#love-hate move
move<-function(x){
bacwa=1;prop1=prop2=rnorm(1,x,2)
while (runif(1)>{pi(x)+nozero}/{pi(prop1)+nozero}){
prop1=rnorm(1,x,2);bacwa=bacwa+1}
while (runif(1)>{pi(prop2)+nozero}/{pi(prop1)+nozero})
prop2=rnorm(1,prop1,2)
y=x
if (runif(1)<pi(prop2)*bacwa/pi(x)/fowa){
y=prop2;assign("fowa",bacwa)}
return(y)}
#arbitrary bimodal target
pi<-function(x){.25*dnorm(x)+.75*dnorm(x,mean=5)}
#running the chain
T=1e5
x=5*rnorm(1);luv8=rep(x,T)
fowa=1;prop1=rnorm(1,x,2) #initial estimate
while (runif(1)>{pi(x)+nozero}/{pi(prop1)+nozero}){
fowa=fowa+1;prop1=rnorm(1,x,2)}
for (t in 2:T)
luv8[t]=move(luv8[t-1]) | 2021-10-20 19:27:46 | {"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": 1, "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": 0.7101938724517822, "perplexity": 1668.669337367228}, "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/1634323585348.66/warc/CC-MAIN-20211020183354-20211020213354-00590.warc.gz"} |
https://www.physicsforums.com/threads/group-theory-permutations-formula.228280/ | # Homework Help: Group Theory, permutations formula
1. Apr 12, 2008
### Coto
When proving that $$A_n$$ with $$n \geq 5$$ is simple, we require the following lemma:
If N is a normal subgroup of $$A_n$$ with $$n \geq 5$$ and N contains a 3-cycle, then $$N = A_n$$.
The proof is actually given for us in the lecture notes, however he utilizes a formula that I'm not sure how he derived:
Let $$f \in S_n$$ be a permutation and let $$x = (i_1 i_2 ... i_k)$$ be a k-cycle. Then:
$$fxf^{-1} = f(i_1 i_2 ... i_k)f^{-1} = ( f(i_1) f(i_2) ... f(i_k) )$$
where the right side is a cycle. I'm blind to how he gets from left to right.. and we were left with the ubiquitous "it is easy to check that..."
Any help to get started on seeing how this is true would be appreciated. Thanks,
Coto
Last edited: Apr 12, 2008
2. Apr 12, 2008
### tiny-tim
Hi Coto!
For any number j:
fxf^{-1}(f(i_j)) = fx(i_j) = f(i_j+1);
and that is the definition of ( f(i_1) f(i_2) ... f(i_k) ), isn't it?
3. Apr 12, 2008
### Coto
Thanks tiny-tim. I must be having a brain-dead day, I'm still having troubles seeing this.
Specifically I get lost in the reasoning from moving left to right. Could you elaborate between each equality, specifically why you've decided to add (f(i_j)) to fxf^{-1}, and why fx(i_j) = f(i_j+1) ?
The definition of the cycle on the right would be all f(i_j) move to f(i_j+1) (j < k)? But I'm having troubles making that connection with your help given above.
4. Apr 12, 2008
### tiny-tim
Hi Coto!
ok … I think what you're missing is the basic definition of f{-1}, which of course is that f{-1}f = I, the identity … so in particular f{-1}f (i_j) = (i_j);
so fxf^{-1}(f(i_j)) = fx [ f^{-1}f (i_j)) ] = fx [ (i_j) ] = f [ x(i_j) ] = f(i_j+1).
5. Apr 12, 2008
### Coto
Thanks again for your time tiny-tim.
Actually I understood where the derivation was coming from in terms of the inverse.. but I seem to be stuck more on why fxf^{-1}(f(i_j)) appears.
1) We are looking for fxf^{-1} so why do we look at fxf^{-1}(f(i_j))? Why do we append the cycle (f(i_j)) to it?
2) we have x(i_j) = (i_1 ... i_k)(i_j) = (i_j+1) ?? I guess I'm lost on this step too. I'm not sure why this is true. If we're considering (i_j) as a "cycle" of one element, then wouldn't x(i_j) = x?
I suppose these may seem basic.. my mind is having troubles wrapping around this.
Thanks again for the help. Sorry for being a bit dense right now.. I'm just not quite seeing it for some reason.
6. Apr 12, 2008
### tiny-tim
Because we want to prove that fxf^{-1} is the cycle that takes f(i_j) to f(i_j+1).
AND (f(i_j)) is not a cycle … it's an element, just as i_j is.
Ah … all is clear now … you're misreading the cycle notation. And you're confusing cycles with elements. AND there's no such thing as a "cycle" of one element … how would you define it?!
If C = (a b c d …) is the notation for a cycle C, then that means that Ca = b, Cb = c, …
So if C = (i_1 ... i_k), then that means C(i_1) = i_2, C(i_2) = i_3, … and generally C(i_j) = i_j+1.
x is a permutation, just like f, not an element … x operates on the same elements as f does.
7. Apr 12, 2008
### Coto
Gotcha. I think that actually was my problem. I didn't realize this notation at all. It makes sense now what is going on.
By mentioning (i_j).. it was sometimes used when writing a permutation as a product of disjoint cycles, i.e. (1345)(2) would be the cycle as 1>3>4>5>1 and the invariant 2>2. Superfluous given that we known when using cycle notation that all other elements remain the same, but it seems that it was the source of my confusion when interpreting x(i_j).
Combine that with not understanding the meaning behind Ca = b and you have a confused student! ;) | 2018-06-18 02:39:39 | {"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": 0.8063600659370422, "perplexity": 1404.5801944330472}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-26/segments/1529267859923.59/warc/CC-MAIN-20180618012148-20180618032148-00543.warc.gz"} |
https://newspaperarchive.com/us/oklahoma/ada/ada-evening-news/1919/09-03/ | # Ada Evening News Newspaper Archives
## September 03, 1919
Issue Date:
Pages Available: 8
NewspaperARCHIVE.com - Used by the World's Finest Libraries and Institutions
Pages Available: 241,891
Years Available: 1904 - 1978
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Ada Evening News (Newspaper) - September 3, 1919, Ada, Oklahoma "The Homebreaker"Comes Bere Today—And We Warn You to Look Out for Her. But Miss Dalton Isn’t the Homebreaker. LibertyWit Stoa Cerning i^etos VOLUME XVI. NUMBER ADA, OKLAHOMA, WEDNESDAY, SEPTEMBER 3, 1919 Carnegie Medal for Ada Negro Opinion of Agent for Hero Fund Commission |U Carranza, In Message to Congress, Takes Slap at U. S. and Monroe Doctrine PUK UALU* FOR THI u n SKT speeches in principal CITIES OF THK WEST. By thv AMocMWd Pm^> WASHINGTON, Sept. 3, President Wilson will leave the national capitol tonight on his speaking tour of the*country, for the purpose 01 urging an early ratification of the peace treaty and the league of nations covenant b> the senate without reservations. The first stop of the president’s special will he at Columbus, Ohio, where he will make his first address on Thursday. The will be accompanied on the trip by Mrs. Wilson, Admiral Cary Grayson. Secretary Tumulty and a score of stenographers, secretaries and press men. and a number of private detectives. Although the trip calls for only thirty set speeches in the principal cities of the west, it is expected that the president will deliver many short speeches* from the platform of his train along the route. ADA COMPANY NUE SELL INVENTION Stater* Have Herndon. The four daughters of H. K. Nettles had their first reunion in thirteen years yesterday at the home of C. M Gwaltney near Jones Chapel. The four sisters were Mrs. Etta Ward of Quay, New Mexico. Mrs. D. W. Shilling of Ada. Mrs. L. R Luckey of Kus*, and Mrs. Gwalt ney. Another guest at the reunion was Mrs. Kila Buster, sister-in-law of Mrs. Ward. A splendid dinner was served and all enjoyed the day to the utmost. particularly Mrs. Buster, who has been an invalid for quite a while. Mr. Ward and wife left Ada thirteen years ago and settled in New Mexico. They live in the heart of the prospective oil territory of that state and can see many drills now boring for oil. They will return to H. H. Gladwell has Just received the finished model of a machine that he patented some time ago. It is a machine for stamping bank checks and operates by being attached to an adding machine and is run with the same mechanism that president j operates the *iddin*’ machine. Mr. Glad well invented this machine some two years ago and or-i .misted a company composed of Ada citizens to manufacture and sell his invention. A contract has been made with an eastern company for rnanu-: actu ring the invention, but the opening of the war prevented the carrying out of the contract. The Patent Holding Company of Ada will make arrangements in the immediate future for manufacturing ihis invention in quantities aud putting it on the market. The simplicity of Mr. Gladwell's invention makes it possible to offer the machine to the public at a nominal charge. It works with accuracy aud rapidity and will be a great labor saver to banks and other institutions. There is every reason for believing that this invention will om be a pronounced success. Watchful Waiting Mr. Baker's Plan Regarding Mexico their home soon. in the west some time That Thelma Cannon, a negro boy of Ada, is eat tied to the Carnegie Medal for heroism, is the opinion of Mr. Leonard C. Snell of Pittsburgh, Pennsylvania, who has been in Ada since yesterday investigating the story of Cannon’s rescue of five negroes from drowning in Little Sandy more than three years ago. Citizens of Ada will remember the flood of rain that fell here January 20, 1916. For four day* snow and rain had been falling and Little Sandy was already swollen. On the 20th a perfect cloudburst came, the rainfall of the afternoon amount-to 6.5 inches. All the lowlands of the city were flooded, the water ran thru the postoffice on 12th Street, even swamping the Palm Garden thru its back door on Main. Little Sandy became an angry torrent, sweeping away many houses from their foundation til the northern part of the city. Among the families living in the colored section was that of Joe Col-, lins, h section ham! on the Katy. The water rose to a depth of morel than five feet in his house, preventing the escape of his family. Several unsuccessul attempts were' made to rescue the marooned ne-groes, men on horseback trying to reach tile house, bul the current was too strong for even a horse to swim. Express At that time Thelma Cannon wasj one at only seventeen .'ears old, but he had!company. Thus. been raised on the North Canadian been stolen by at El Keno and was a powerful swimmer. He went into Little Sandy, swam fifty .'aids thru the icy cur-r* in, and rescued the five negroes, each trip. Three of them were children, one a woman, and the other a man. The danger of Ills « \pl<>iI was doubly great because of the fact that the water was mixed with snow and ice. From swimming in the cold water Cannon contracted rheumatism and came near dying. He was placed on a stretcher and conveyed phur Springs. Texas, where tnained for year after "FAT” BF RNN* ARR ENTER NEV I'KAL HAYN AGO, NOW CX>N-FRHNKN TO BEING AUTO THIEF. By Ne**- S*<«St*r>ic*- MUSKOGEE. Okla.. Sept. 3.— Onej of the most notorious safe cracksmen in Oklahoma is believed to be in the county jail in the person of “Fat” Burns, who was arrested uev-j t rill days ago at the Eagle flats here by Al Bartlett ot th* Fulk National; detective bureau, and Deputy Sheriff' Ed MaJoney, and who is alleged to | have made a confession this morning latter being held on investigation. He was arrested with George Wilson and Mrs. George Wilson, alias Buster Cody, who formerly served a term in the penitentiary on charge of a violation of the narcotic act. A quantity of nitro glycerine and a complete set of cracksman’s tools Were found at their rooms at the Eagle flats. The arrest was the re>ult of tracing the recent theft of adding machines which cases were in the hands of Mur*! Grady, former police officer. and now captain |f detectives for the Fulk bureau. Three adding machines were recovered by Grad}, one being located at Eufaula, one at the American company’s depot here, and Harry Bowman’s Printing are alleged to have the three prisoners. “Fat'' Burna, according to Bartlett. mach a confession this morning that he Lad planned a big “haul” Jus' bt to • U lug captured and would have made it if his arrest had been delayed. He als* admitted being an automobile thief, according to Bartlett. . ad becoming boastful, explained to Bartlett how easy it was for him to cover up his theft whenever lit* stole an automobile. One of ti t adding' machines recover* d was the one stolen from MEXICO CITY, Sept. I.— A de lense of Mexico against foreign rep- I resentativei, particularly from the United States, regarding the lives I of ioreignera and their property ' was contained in a message of Pres- j ident Carranza, read at the opening I session of congress last night. Particular reference was made in the message to alleged injustices practiced against Mexicans in the United States. A long list of the alleged offenses was given under the section of the address devoted to foreign relations. Th* message also defended Mexico’s neutrality during the war and protested against charges that the Mexican government was incapable and unwilling to protect foreign lives and property. In the message President Carranza asserted that Mexica City did not ask admission to the league of nations, because, he said, the league did not establish equality for all nations and races. He reiterated that Mexico had not and would not recognize the Monroe Doctrine. Regarding oil legislation, the message said the government was willing to conciliate, but would not sacrifice its national sovereignty. FLORIDA PROSPECTORS ARRIVING ONE BY ONE GRAND ENC AMPMENT AT COLUMBUS, OHIO, NEXT WEEK MAY BE LAST ONE HELD. PETKOGKAD TO BE ATTACKED IS REPORT fly the Assaulted Press HELSINGFORS, Fin . Sept. 3-— Admiral Cowen, commander of the British squadron in the Baltic sea, has arrived here in connection, it is understood, with preparation for an attack on Fetrograd. A Berlin dispatch Tuesday carried German reports that Gen. Gouth of the British army, had issued a proc-the imitation to the people of Petrograd, to Sui- O’Bannoi, Grain company of Clare-; declaring that an .attack was about lie re- more. Okla., where also a safe was to be made on the city, adding that manv months. It was a blown about a month ago. tux .SH EY IK FORCES OCCUPY SOUTHERN PART OF KIRT By the Associated Piw LONDON, Sept. 2. -The entire Bolshevik forces are occupying the southern portion of Kirt today, according to a wiieless dispatch sent by the Soviet government at Moscow. and picked up here. The dispatch states that the fighting is still going on. he contracted the disease I before he was able to walk again. By Few** s* rvicc ^e still suffers from rheumatism du- WASHINGTON, Sept. 3.- -The war! ring cold or damp weather, department will await offical re-; Thelma Cannon is a cousin of Bill port from Major-General Dtckman, Cannon, the well known porter of commander of the southern depart-; the Elks club. When only a small meat, Secretary Baker said, before lad he distinguished himself by considering the question of whether an unsuccessful attempt to save from some action shall be taken against drowning the son of Judge Frank the Mexicans who fired on an Anteri- Gillette at El Keno. Mr. Snell, who can army airplane on the Rio la the special agent of the Carnegie Grande, wounding one officer. Until Hero Fund Commission, will make some report "as received Mr. Baker his report to the commission im- declined to speculate on what line mediately, the department’s action might take, j ------------- The former dispatch while it does not refer to any particular district in Russia probably has reference to the Lithuanian front where the Bolshy k forces were reported Tuesday to be surrounded and to be offering to make peace. BKI/GI AN KING AND QUEEN COMING EARLY IN «MTOHER In military codes the firing upon the army fliers was considered far more serious than the recent detention of two aviators for ransom. All army machines are plainly marked, it was said, and the reports from the border would indicate that Captain McNabb’* airplane was either on the American side or was following the river, the international boundary at this po.nt. In cither case it should have be*n minnie from hostile attack. The fact that press disptaches referred to repeated “volleys” from the Mexican side also was considered ! significant as indicating that the as-! s&ilants were under some sort of I military command. Reports persist here that General Dickman has authority to meet such First Baptist Church. situations without waiting on his Regular praver meeting tonight superiors by sending a column across 8:15. All members are urged to! to capture or scatter the offenders HE GIRLS Sn HERI Oldest Man in America Wants Life Insurance By ti*** As--Axiate*! Pr«* LEXINGTON, My., Sept. 3.—, ’Uncle Johnny” Shell, probably the oldest man in the country, celebrated his 13 Isl birthday here yesterday by sending for a life insurance agent. “You can’t tell what will happen.” he told the agent, "and I want lo be prepared for the worst.” The agent looked over his rate card and found that it covered the first one hundred years of a man’s life. He is now busy with his headquarters trying to learn if ii will be possible for him to insure Mr. Shell. as soon as Petrograd had been freed from Bolshevik rule food supplies', would be sent in. HONDURAN REVOLUTIONISTS lUHnrER BY (JOVT. TROOPS By th* Ak-KHA khI Pre sr SAN SALVADOR, Sept. 3.—The newspapers today say that about one thousand Honduran revolutionists have routed a superior force of government troops, killing and wounding many soldiers and capturing much war material. By the Associated Pret-s COLUMBUS, O.. Sept. 3.—Their ranks reduced to approximately 135,000, members of the Grand Army of the Republic are about ready to retire from the stage of action, and bequeath their traditions to 11heil sons, veterans of the European w ar. now organizing into the American Legion. The fifty-third annual encampment of the G. A. R . which opens here I Sunday, September 7, for one week, probably will be the last great encampment of the old veterans, according to James E. Campbell, former governor of Ohio and chairman of the local committee on arrangements. “Of course there will be other encampments, but in ail probability this will be the last big one. This was indicated at the last encampment when thousands of veterans in the far West voted to hold this year’s encampment in Columbus in preference to St. Paul. Most of them were born in the Central and Eastern states and they wanted to return, as they expressed it, to pay a last visit. to their old homes and bid a final farewell to old and dear friends, Governor Campbell said. Governor Campbell expressed the belief that the G. A. R. after this year’s encampment, though maintaining its organization, will gradually turn over to veterans of the European war their activities. “This is indicated,” he said, “by the fact that for the first time in its history, the G. A. R. has invitetd veterans of another war, the world conflict, to participate in its annual parade and other exercises this year. The G. A.R has been a very jealous organization, never permitting any organization except the Sons of Veterans to participate in any of its activities, then I only in the parade. Now our own By News** .special Service | sons are standing ready to take up PARIS, Sept. 2.—A distressing the work and they no doubt finally era of speculation in foodstuffs in assume it.” the United States and throughout Besides the G. A. R. seven a iilithe world’s primary food markets> ated organizations will meet is largely responsible for high food at the same time. They are; ine costs, in the opinion of Herbert • ^ omans Relief Corpt, Ladies of the Hoover, chairman of the inter-allied! tL A. R., relief organization, who, on the eve * Veterans of his departure for the United V eterans, States, granted the Associated Press Civil War Nurses. an interview today. Wharves and One of warehouses in northern European annual parade which will be J*8-2’" ports are overflowing with food- ticipated in by the G. A. R., Sons stuffs, principally meats, fats and dairy products, sent by merchants About two weeks ago John McKinley, Joe Cole, N. B. Haney, Jr., and Bart Smith left Ada on a “wild goose chase” to Florida, where they were supposed to look into the proposition of investments in grapefruit orchards. That they had a go-jJ time on the trip is attested by the numerous; letters received from them while on the trip, but the News has refrained from publishing more than one of them to* the reason that ic detected ihat at least some of them weie forgeries. It seems that each tried to play a practical joke or, one of the others and for fear of getting a libel suit on its hands the News called a halt after publisning one letter which bore the name of Bart Smith at the bottom but which, in all likelihood was a forgery. Ar any rate the soys had a fine trip, and one by one are “coming into camp.” John McKinley is home, having taken the shortest cut to get here, while the other three of the party boarded a boat as Jacksonville and are making the trip home via New York City. This route is a little 'round about” way to get to Ada, but the experience by boat from Jacksonville to the world’s metropolis will no doubt be worth the time and expense. Speculation Has Caused H. C. L. is Hoover's Belier Sons of Veterans, Sons of Auxiliary, Daughters of Ex-Prisoners of War, and featuers besides the I By the Pmw WASHINGTON, Sept. 3.--King Alben and Queen Elizabeth of Belgium, will arrive in Washington about October I st, and will be the guest of President and Mrs. Wilson at the White House, remaining, probably, three days. at ______ be present. This will be the first and to rlean up the district in which class in Bible study. We will take up they committed the objectionable ac-the first chapter of Genesis. Let Hon. everybody read and study the chap- It would not come as a surprise to , . ter before coming.—Clyde C. Morris,| many officers if dispatches related toit,le larS** fllnd “aa raised. SIN THOUSAND TELEPHONE I SKIIS WITHOUT SERVICE * AS RESULT OF STRIKE OF OPERATORS. SHAWNEE. Okla., Sept. 3. Entrenched behind a fund of $6,800 the Shawnee striking telephone girls are pursuing a policy of watchful waiting and six thousand telephone users are without any sort of phone service or protection, the different unions of the city ha\e subscribed I to the telephone girls cause and HftluKlist I*ra)it Meeting Lt t all Methodist people attend pray* r meeting this evening at eight-thirtj We will pray and offer thanks to (Too toi>-th«“r. We have a great deal for which to be thankful nobody has more than we. Wallace M. Crutchfield. Pastor. FIVE THOUSAND PEOPLE SEE THE FIREWORKS EXHIBIT AT THE ITTY LAKE LAST NIGHT. MICKIE SAYS Pastor. (heir Meeting Tills Evening. The choir of the First Christian church will meet this evening at 9:30 o'clock immediately after prayer meeting. Anyone who can read music or carry* a tune is invited to come. a southward movement by cavalry.! CORPORATION OOMMISSION CHAIRMAN IS COMING; From this fund, the girls will draw weekly benefits. The Southwestern standing by their they will close the entirely unless they SPECIAL TRAIN LOOM AOA TO SEE PRESIDENT Hon. K. L. Echols, chairman of protection from the the state corporation commission, | wht,n rHCejve has notified the city authorities here } t hat he will be in the city Friday to hear the complaints against the Southern Ice & Utilities Co., of this city. The county attorney filed complaint against this company with the corporation commission several days ago, and the hearing is set for Friday at the district court chamber in this city. President J M. Gordon of the East Central State Normal, sent a message to President Wilson, through Congressman Tom D. Mc* Keown, stating that a special train of Ada citizens would be in Oklahoma City on the 26th of to pay their tribute of respect to President Wilson. Tom I). McKeown sent a wire in response to this message as follows: “President Gordon, “Ada, Oklahoma. **I conveyed your message to the president and he requested me to express bis cordial acknowledgement and appreciation of your message. “TOM D. MCKEOWN.” ADMIRAL KORCHAK RETREATS AND MOVES HEADQUARTERS By the AMoriated Pre** LONDON, Sept. 3.-—A wire from | Moscow received here reports that ; Admiral Kolchak, head of all Rub- ; October • Man government, has evacuated Omsk and has moved hi sheadquart-ers to Irkutsk. Firwt Christian Church. Ever eat red bananas? It At the prayer meeting at the First Christian church those present will tell why and how they became Christians. This will be a very interesting meeting. A large number is expected to be present. The meeting will open at 8:30 and close at 9:30. All are welcome. Bidi people are statement that exchange here, are guaranteed; authorities and! such guarantee they state they will continue their service. The strike here is paralyz-j lug many lines of business, grocery men report their sales almost cut iii half. Spot cotton men report that they will have to leave this city and go to Oklahoma City or some oth**r nc»ar by market lf the strike goes into many days. Police and fire protection is ruined and telephone shopping has been stopped. There is no long distance service in or out of the city. Western Union wires I are alone available and after nine o’clock each night the town is cut off from the world. The business men are hacking the i telephone girls, however, and at a j large mass meeting last night many i speeches were given by local busl-' iness and professional men. At that 1 meeting over$700 was donated to I the girls. Miss Anna Bell Glenn enjoyed the exhilirating sensations of an aeroplane flight yesterday. Her physical condition does not seem to be impaired\in the least from having soared in gteiherial regions. And she says it was a wonderful ride. at The two-days picnic of the Woodman of the World ended last night at the city lake in a blaze of glory. The exhibition of fire works that had been arranged for the closing hours of the picnic was carried out according to schedule, to the great delight of all who witnessed it. Fully 5,000 people were present to witness The display and take part in the festivities. The attendance Monday was seriously interferred with by the ball games a* the fair grounds, which attracted many people who otherwise would have gone to the city lake. A pleasing feature of the first day’s program was the patriotic and Woodmen songs rendered by fifty little girls under the direction of Mrs. Dowd. This feature is highly complimented by all who heard the rendition. Diving and swimming contests were numerous and * njoyed immense ^ ly by both participants and wit* j nesses, dam Scheinberg of The Model Had charge of the aquatics and carried it through in fine form. The only speech of the occasion was delivered by Judge J. W. Bolen, who spoke to the delight of the large audience. A fist fight last evening was the only thing that marred the whole affair, but it was of little consequence and no one was seriously hurt. all over the world. Mr. Hoover said. These merchants, he declared, had “gambled” on sales in Poland, Czecho Slovakia, the Baltic states and Germany at high prices, but these states have only a depreciated local currency and many commodities are in danger of spoiling as the central European market for foodstuffs is limited to the ability of the peoples to buy on credit. “This year’s speculations,” Mr. Hoover said, “are due, in my opinion, chiefly to the belief of food merchants and manufacturers that when the blockade was removed there would be an enormous demand for foodstuffs and other commodities in Central and Eastern Europe. This speculation was not due to any important shortage, at present, of actual supply. This speculative fever which was not confined to foods, was greatly stimulated by the long of Veterans and recent war, will be service on Sunday mortal Hall when soldiers of the a union religious afternoon at Me-ministers of all religious denominations will speak and take part in the exercises. In the evening a musical program will be rendered at the hall. In a letter addressed to world war soldiers, sailors and marines, especially those joining the American Legion, Governor Campbell after speaking of the general order of Clarendon E. Adams of Omaha, commander in chief of the G. A. R.. inviting world war and Spanish-American war veterans to march in the parade, says: “The reason for this reversal of the former policy is that the members of the Grand Army of the Republic exult in the existence in this nation of 4,000,000 young men. largely the sons and grandsons of old veterans, who are able and ready to take up the welcome of another half century of keeping alive delay in the removal of the blockade until the peace treaty was signed!the ot Patriotism.” Versailles.” NKW SUPREME tXHJNCHi SENDS ULTIMATUM TO BOUMAN IA ZEALAND RATIFIES THK PEACE TREATY By th** Associated Press PARIS. Sept. 3.—The supreme council today decided to send an ultimatum to the Roumanian government regarding her acts in Hungary. The ultimatum, in drastic terms and with a time limit, will be delivered by special envoy to the council. Should Rou man ta refuse to comply with the terms of the ultimatum a given time diplomatic relations will cease and the allied; envoy will bring away with him' from Bucharest the belongings cfi I he government. By the Associated Press WELLINGTON, N. Z., Sept. 3.-The peace treaty with Germany was unanimously ratified by parliament j her \ although the entire labor forces continues to criticise it. Ever eat red bananas? It JOHN RAWL’S BROTHER KILLED IK TEXAS MEX IU AN FEDERAL SOLDIERS ATT AUK AMERICAN AIRPLANE John Rawls, the city chief of po- - I lice, received a message last night By the Associated Press RUSSIAN BOLSHEVIKS WANT PEACE SINCE THK ROUT WEATHER FORECAST. Weather fair tonight and tomorrow. Not much change, is the advice >f the weather man today. By the Associated Presa BERNE, Switzerland, Sept. 3.— The Ruslan Bolshevik forces have proposed peace negotiations following a rout of their armies which are surrounded, according to an official announcement. Let a Want Att get It for you. LAREDO. Tex., Sept. 3.—Mexican federal soldiers made the attack yesterday upon an American army airplane while on American patrol duty, but the machine was in Mexican territory, according to Mexican consul Garcia of Iaredo. The firing was unwarranted. the consul said, and was done in the absence of the detachment commander. The case has been referred to superior authority for instructions in regard to the punishment of those guilty. 1 that his brother had been instantly killed at Gorman, Tex. Mr. Rawls left at once on the southbound Frisco for Gorman. As no message has yet been received from Mr. Rawls it is not known how his brother met his death. Until last week the brothers had not met in twenty-seven years, and Mr. Rawls had returned from a visit at Gorman only last Sunday. His brother owned some valuable oil lands in the Texas fields. Moved next door to Gwin -C E. Sprague, Jeweler. & Mays. 9-3-tf Let’s have one hundred people in the Bible Study class tonight at prayer meeting at First Baptist Church. 9-3-lt ; | 2017-02-22 08:23:24 | {"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": 0.1731334924697876, "perplexity": 11374.044066251223}, "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-2017-09/segments/1487501170925.44/warc/CC-MAIN-20170219104610-00133-ip-10-171-10-108.ec2.internal.warc.gz"} |
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# Convergence in total variation of the Euler-Maruyama scheme applied to diffusion processes with measurable drift coefficient and additive noise
2 MATHRISK - Mathematical Risk Handling
Inria de Paris, ENPC - École des Ponts ParisTech, UPEM - Université Paris-Est Marne-la-Vallée
Abstract : We are interested in the Euler-Maruyama discretization of a stochastic differential equation in dimension $d$ with constant diffusion coefficient and bounded measurable drift coefficient. In the scheme, a randomization of the time variable is used to get rid of any regularity assumption of the drift in this variable. We prove weak convergence with order $1/2$ in total variation distance. When the drift has a spatial divergence in the sense of distributions with $\rho$-th power integrable with respect to the Lebesgue measure in space uniformly in time for some $\rho \ge d$, the order of convergence at the terminal time improves to $1$ up to some logarithmic factor. In dimension $d=1$, this result is preserved when the spatial derivative of the drift is a measure in space with total mass bounded uniformly in time. We confirm our theoretical analysis by numerical experiments.
Keywords :
Type de document :
Pré-publication, Document de travail
Domaine :
https://hal-enpc.archives-ouvertes.fr/hal-02613774
Contributeur : Oumaima Bencheikh Connectez-vous pour contacter le contributeur
Soumis le : mercredi 20 mai 2020 - 13:01:41
Dernière modification le : vendredi 4 février 2022 - 03:13:41
### Identifiants
• HAL Id : hal-02613774, version 1
• ARXIV : 2005.09354
### Citation
Oumaima Bencheikh, Benjamin Jourdain. Convergence in total variation of the Euler-Maruyama scheme applied to diffusion processes with measurable drift coefficient and additive noise. 2020. ⟨hal-02613774⟩
### Métriques
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https://socratic.org/questions/54aade32581e2a444beda09a | Question da09a
Jan 5, 2015
Actually, Avogadro's number was not selected per se, it was calculated after experimental data on the charge of an electron was gathered.
An accurate calculation of Avogadro's number became possible after American physicist Robert Millikan successfully measured the charge on a single electron in his famous oil drop experiment.
This value was used with the charge on a mole of electrons, which had been known at that point, to calculate Avogadro's number. The charge on a mole of electrons is known as the Faraday constant, and equal to
$96485.3383$ $\text{Coulombs/electron}$
The charge on a single electron was determined to be
$1.60217653 \cdot {10}^{- 19}$ $\text{Coulombs}$
Avogadro's number was calculated by dividing the charge on a mole of electrons by the charge of a single electron:
N_A = (96485.3383 C/(mol))/(1.60217653 * 10^(-19)"C") = 6.0221 * 10^23 "mol"^-1# | 2020-05-26 01:33:57 | {"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": 0.6375678181648254, "perplexity": 747.5120405497984}, "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-2020-24/segments/1590347390437.8/warc/CC-MAIN-20200525223929-20200526013929-00244.warc.gz"} |
https://chem.libretexts.org/Ancillary_Materials/Laboratory_Experiments/Wet_Lab_Experiments/Organic_Chemistry_Labs/Experiments/9%3A_Multistep_Synthesis_(Experiment) | Skip to main content
# 9: Multistep Synthesis (Experiment)
When synthesizing complex organic molecules, it is common to have at least a dozen individual transformations whereby the product of one reaction is then used as the starting material for the next reaction. You will have an opportunity to do a multistep synthesis starting with inexpensive, readily available benzaldehyde. The sequence you will attempt is first the conversion of benzaldehyde to benzoin using the vitamin, thiamin, as a catalyst. In the second step, the benzoin is oxidized to benzil through the use of an oxidizing agent. The third step is a condensation reaction of benzil with dibenzyl ketone (1,3-diphenyl-2-propanone) to produce tetraphenylcyclopentadienone. An alternative third step is the reduction of benzil to dihydrobenzoin with a reducing agent, sodium borohydride. An additional fourth step is possible converting the tetraphenylcyclopenta-dienone to a substituted naphthalene via a Diels-Alder reaction (followed by decarbonylation) using microwaves as the energy source.
One problem which becomes apparent is that the yield of the overall final product will be limited by the lowest yielding individual reaction. Therefore, each reaction in the sequence must be a high yielding reaction. Second, the overall final product yield is the product of each individual percentage yield. Therefore, if each step is a 90% yield and there are 10 steps, the overall final product yield is (0.90)10 or 35%. For a twenty step reaction, the overall yield would only be 12%. In a two step reaction, if one step had a yield of 50%, the highest overall yield possible is 50%. It is not necessary, or desirable, to use all of your material in each step.
## Step 1: Synthesis of Benzoin
This reaction is a classic--the conversion of two molecules of an aldehyde to an alpha-hydroxy ketone. The reaction is known as a benzoin condensation ("condensation" because two molecules become condensed to one molecule). This reaction, which requires a catalyst, if often performed with cyanide ion. We will use thiamine as a catalyst. It is heat-sensitive and may decompose if heated too vigorously. Instead of running this reaction at elevated temperatures for a few hours, we will allow the reaction to proceed closer to room temperature for 24 hours or more. Benzaldehyde is easily oxidized to benzoic acid which can impede the desired reaction so freshly distilled benzaldehyde is used. The concentration of reactants and temperatures of solutions are critical to obtaining a good yield so procedures must be followed carefully. Too much water will force benzaldehyde out of solution preventing an efficient reaction. Too little water prevents the thiamine hydrochloride from dissolving. Some of the base reacts with the thiamine hydrochloride to produce thiamine which is the active catalyst.
Procedure: Place 1.5 mL of 5M NaOH (CAUTION: extremely caustic) in a 10 mL Erlenmeyer flask and cool in an ice bath. In a 50 mL Erlenmeyer flask dissolve 0.80 g of thiamine hydrochloride (MW=337) in 2.5 mL of water. Add 7.5 mL of 95% ethanol to the thiamine and cool the solution for several minutes in an ice bath. While keeping both flasks in the ice bath, add the 1.5 mL of previously cooled 5M sodium hydroxide dropwise (3-5 minutes) to the thiamine solution with swirling so that the solution stays below room temp. Remove the 50 mL flask from the ice bath, add 5.0 mL of benzaldehyde (d=1.044 g/mL) at one time, swirling the flask so that the benzaldehyde mixes with the yellow, aqueous, basic layer. The solution becomes milky but then clears*. Seal the flask with Parafilm and place it in your drawer until the next lab period.
*If the mixture does not go to solution (eg, if two layers are obvious), place the flask in a warm water bath at approx. 50 oC until the solution clears or for a maximum of 10 minutes. You can use hot water from the faucets at the front of the lab. The mixture should become homogeneous in the water bath but it may not stay homogeneous once it cools.
The following lab period: Filter the crystals, wash them free of mother liquor with 10-15 mL of a cold 2:1 mixture of water and 95% ethanol, and air dry the solid for 15 min. Weigh your crude yield, break up any clumps of solid and recrystallize from hot 95% ethanol (8 mL per gram). You should not have to filter the hot solution. After cooling, the recrystallized benzoin should be filtered, washed with a minimum of a cold 2:1 mixture of water and 95% ethanol, and air dried for 15 min or left until the next lab period. Obtain the mp of the recrystallized benzoin (lit mp listed as 133 and 137 oC for d,l-benzoin, most students will see a mp of 133 oC) .
When you are satisfied that you have the product you want, you may dispose of the first filtrate by neutralizing with dilute HCl, then flushing the aqueous layer down the drain with plenty of water. The second filtrate (from recrystallization) can be flushed down the drain with water.
## Step 2: Oxidation of Benzoin to Benzil
CAUTION: Concentrated Nitric Acid is extremely caustic and will burn exposed skin.
Work in a hood! Into a 125 mL Erlenmeyer flask, place 2.0 g* of benzoin (weighed to the nearest tenth of a g) and carefully add 7 mL of concentrated nitric acid. Heat the mixture on a steam bath with occasional slow swirling for 30 minutes or until the brown-red nitric oxide gases are no longer evolved. The fumes are toxic and noxious so be certain that the fume hood safety shield is pulled down.
Carefully cool the flask and contents using tap water (keep the flask covered with a plastic seal or a cork), then pour into 35 mL of cool water and swirl to coagulate the precipitated product. Collect the yellow solid using suction filtration and wash twice with 5 mL of cool water to remove some of the nitric acid present. Press the crystals to remove more water by placing another piece of filter paper over the crystals and pushing with a beaker or cork; the suction flask MUST be supported and sitting flat on the desktop. This crude product can be recrystallized from 95% ethanol while it is still slightly wet (4 mL/g). Dissolve it in hot ethanol, add water dropwise to reach the cloud point, and allow it to slowly crystallize. Filter, dry, record the yield, and take the mp.
* Do not use all of the benzoin you have synthesized. If you do not have more than 2.0 g, save 100 mg of the benzoin for a mp, IR, and to hand in and use the rest for this next step, altering the procedure to scale.
If your instructor requests, run a tlc of the recrystallized product with known samples of benzoin and benzil for comparison of Rf values.
When you are satisfied that you have the product you want, you may dispose of the filtrate by first neutralizing with sodium carbonate, diluting with water, and flushing down the drain. Ethanol from the recrystallization goes to the non-halogenated waste container.
## Step 3: Preparation of Tetraphenylcyclopentadienone
Cyclopentadienone is a relatively unstable compound which will dimerize even at low temperature. However, the corresponding tetraphenyl compound is quite stable.
Procedure: Into a 50 mL rbf, place 0.7 g* of benzil, 0.7 g of dibenzyl ketone, and 5 mL of absolute ethanol. Attach a reflux condenser and heat the mixture on a steam bath, water bath or sand bath until the solids dissolve. It is critical to prevent water (moisture) from coming into contact with the reactants. Raise the temperature to provide a slow reflux and add a solution of 0.1 g of potassium hydroxide (CAUSTIC) in 1 mL of absolute ethanol (this solution may be already prepared) dropwise through the top of the condenser. The reaction is very fast and a purple color will appear.
After addition of the base, allow the mixture to reflux for 15 minutes while periodically shaking the flask. Cool the reaction flask to room temperature, then in an ice bath. Filter using a Buchner funnel, wash twice with 5-mL portions of cold 95% ethanol, and air dry for an hour, if possible. When the crystals are dry (which may take until the next lab period), weigh, record the yield and the percentage yield . You may recrystallize a portion of the purple product using a 1:1 mixture of 95% ethanol and toluene (12 mL/0.5 g). Record the mp; literature mp 219-220 oC).
When you are satisfied that you have the product you want, the filtrate can be neutralized with dilute aqueous HCl and flushed down the drain. The recrystallizing solvent should be placed in the non-halogenated waste container.
*Do not use all of the benzil you have synthesized. If you do not have more than 0.7 g, save 100 mg of the benzil for a mp, IR, and to hand in and use the rest for this next step, altering the procedure to scale or obtaining additional benzil from your instructor.
## Alternative Step 3: Reduction of Benzil with Sodium Borohydride
There are a wide variety of hydride reducing agents that convert carbonyl compounds into alcohols. One of the least reactive of these agents is sodium borohydride. Although it reduces aldehydes and ketones, it is fairly stable in aqueous and alcoholic solutions. The more reactive hydride reducing agents can reduce other functional groups such as carboxylic acids, esters, epoxides, and nitriles. Such hydrides react violently with water releasing hydrogen gas and must be handled very carefully. You will reduce the diketone, benzil, using sodium borohydride. Three stereoisomers of hydrobenzoin can be formed in this reaction, RR, SS, and RS which is the meso isomer. It is the meso isomer that predominates. The stoichiometry of the typical borohydride reaction is:
$\ce{4 R_2C=O + NaBH_4} \rightarrow \ce{(R_2CHO)_4B}^-\ce{Na+}$
hydrolysis of the borate ester
$\ce{(R_2CHO)_4B}^-\ce{Na+} + \ce{H_2O} \rightarrow \ce{4 R_2CHOH}$
Your starting compound is a diketone so you need one mmol of borohydride for every two mmol of ketone .
Procedure: Using a 25 mL or a 50 mL Erlenmeyer flask, dissolve 0.50 g of benzil (weigh to the nearest hundredth of a gram) in 5 mL of warm 95% ethanol. Cool the solution in a water bath which will produce a fine suspension of benzil particles. Add 0.10 g of sodium borohydride (weight to the nearest hundredth of a g) which will cause the solution to warm and dissolve the suspended benzil. As the reduction reaction proceeds in the next few minutes, the yellow color of benzil will disappear. After a total of 10 minutes, add 5 mL of water, heat to boiling on a steam bath, filter or decant if the solution is not clear. When the solution cools, dilute to the saturation point with as much as 10 mL of water and set the solution aside for crystallization to occur. In your discussion, mention that three stereoisomers are possible and suggest why the meso isomer (lit mp 136-7 oC) predominates.
When you are satisfied that you have the product you want, you may dispose of the filtrate by diluting with water, neutralizing the excess, unreacted borohydride with acetic acid, and flushing down the drain.
## Step 4: Dimethyltetraphenylphthalate
Into a 10 mL beaker, place 39 mg of tetraphenylcyclopentadienone (0.10 mmol), 3 drops of dimethyl acetylenedicarboxylate (an excess) and 0.3 mL of triethylene glycol. Mix the ingrediants by swirling, cover the beaker with a thin watch glass or appropriate microwave safe film (thick watch glasses often break in the microwave) and place it in the microwave oven. Set the oven at power level=6 for 5 minutes. After irradiating for 5 min, the beaker will be hot. Let it cool for a minute. The reaction mixture should be a golden color which cools to colorless crystals. It may take a few hours to see crystals. If necessary, leave the material to crystallize until the next class. Collect the crystals in a micro Hirsch funnel and wash with a few drops of cold 95% ethanol. Recrystallize using 95% ethanol. Record the product mp; literature mp 255-257oC.
## Contributors and Attributions
• Was this article helpful? | 2021-09-18 01:38:21 | {"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": 0.5116620063781738, "perplexity": 3899.674076474373}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-39/segments/1631780056120.36/warc/CC-MAIN-20210918002951-20210918032951-00685.warc.gz"} |
http://math.stackexchange.com/questions/124365/self-intersections-in-a-product-of-two-curves | self-intersections in a product of two curves
Let $X$ be a smooth projective curve of genus $g$ over an algebraically closed field and consider the intersection pairing on the surface $X \times X$. I remember hearing that $\Delta^2 = 2-2g$: how does one prove this? I understand that the self-intersection is defined by intersecting $\Delta$ with a general divisor linearly equivalent to $\Delta$, but I'm in the dark about how to compute such things.
Also, what are $(X \times \{ * \})^2$ and $(\{ * \}\times X)^2$? My intuition suggests that these are both $0$, but again I don't know how to compute.
-
I'm not an algebraist, but over $\mathbb{C}$, $\Delta^2=2-2g=\chi(X)$ follows since the normal bundle of $\Delta$ in $X\times X$ is isomorphic to the tangent bundle of $X$. The second follow since $X\times\{*_1\}$ and $X\times \{*_2\}$ are disjoint and homologous for any $*_1\neq *_2\in X$. – Adam Mar 25 '12 at 20:17
1) Use the adjunction formula. For the surface $S=X\times X$ it reads:
$$\Delta^2+\Delta \cdot K_S=2g-2$$ Now, denoting by $p_i$ the two projections projections $S\to X$, we have $K_S=p_1^*K_X+ p_2^*K_X$
$$2g-2= \Delta^2+ \Delta \cdot (p_1^*K_X)+ \Delta \cdot (p_1^*K_X)= \Delta^2+2g-2+2g-2$$ from which follows the required formula $\Delta^2=2-2g \:$.
2) If $P, Q$ are any points of $X$, the fibers $p_1^*(P)$ and $p_1^*(Q)$ are algebraically equivalent so that $(\lbrace P\rbrace\times X)^2 =(p_1^*(P))^2 = p_1^*(P)\cdot p_1^*(Q) =0$ since algebraic equivalence implies numerical equivalence. | 2015-08-01 10:21:23 | {"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": 0.9756990075111389, "perplexity": 72.44331873485102}, "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-2015-32/segments/1438042988650.53/warc/CC-MAIN-20150728002308-00064-ip-10-236-191-2.ec2.internal.warc.gz"} |
https://code.kx.com/platform/stream/dw_log_replay/ | # Log replay¶
The log replay process is responsible for the intraday writedowns from the RDB to the IHDB. It also does the EOD writedowns from the RDB and IHDB to the HDB. Writedown frequency is determined by setting the intraday frequency parameters in the LR and the TP.
• To create an LR, right-click on the Navigation panel in the KxWarehouse Package and select New > Service Class from the context menu.
• In Service Details, provide a description for the instance.
• Configure the Service Parameters. The tables in the intradayTableList should match the ones configured in the TP section.
parameter value | 2021-04-11 13:17:34 | {"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": 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": 0.2019493281841278, "perplexity": 6328.421902551402}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-17/segments/1618038062492.5/warc/CC-MAIN-20210411115126-20210411145126-00466.warc.gz"} |
https://math.stackexchange.com/questions/2469821/foundations-of-mathematical-analysis-johnsonbaugh-20-22 | # Foundations of Mathematical Analysis [Johnsonbaugh 20.22]
Question: Let $\{a_n\}$ and $\{b_n\}$ be sequences such that:
$$b_{n+1} = a_n + a_{n+1}$$
for every positive integer $n$, and suppose that $\{b_n\}$ is convergent. Prove that $\lim_{n\to\infty} \left(\frac{a_n}{n}\right)=0$.
Things I know:
• Since $\{b_n\}$ is convergent with limit $L$ then : $$\forall\epsilon>0, \exists N\in\mathbb{N}:\forall n\ge N, |b_n-L|<\epsilon$$
• $b_{n+1}=a_n+a_{n+1} \implies a_n = b_{n+1}-a_{n+1}$
• $\lim_{n\to\infty} \left(\frac{a_n}{n}\right)=0 \implies \lim_{n\to\infty} \left(\frac{ b_{n+1}-a_{n+1}}{n}\right) = 0$
I'm not necessarily sure how to go about proving this using the above information. It's clear that I have to show that the difference between $\{b_n\}$ and $\{a_n\}$ converges to a point, and this might have something to do with knowing that a convergent sequences is always Cauchy.
Any help or direction would be appreciated.
Thank you
Edit: I would like to use an approach that involves the following topics: Convergent Sequences, Cauchy Sequences, Supremums and Infimums, Limits, Epsilon Characterizations, Bounded Sequences, Bolzano-Weierstrass Theorem.
## 1 Answer
First we do the case that $(b_n)$ converges to $0$, the general case will follow from this. Define $(c_n)$ by $c_1=a_1$ and $c_n=b_n$ for $n \geq 2$. Then I claim we have $$a_n=(-1)^n\sum_{k=1}^n (-1)^kc_k,$$ for all $n$. This can be seen by induction, it clearly holds for $n=1$ and if it holds for some $n \geq 1$ then $$a_{n+1}=c_{n+1}-a_n=c_{n+1}-(-1)^n\sum_{k=1}^n (-1)^kc_k=\sum_{k=1}^{n+1}(-1)^{n+1+k}c_k=(-1)^{n+1}\sum_{k=1}^{n+1} (-1)^kc_k,$$ so it holds for $n+1$. Now that we have this, we need to show $\frac{a_n}{n}$ goes to zero, which is equivalent to showing $|\frac{a_n}{n}|$ goes to zero. Notice the following $$|\frac{a_n}{n}| \leq \frac{1}{n}\sum_{k=1}^n |c_k|,$$ so we see that $|\frac{a_n}{n}|$ is bounded above by the average of the first $n$ terms of a sequence converging to $0$ since $c_k \rightarrow 0$. The average of the first $n$ elements of a sequence that converges to $0$, converges to $0$ (see the lemma beneith), that is we have $$\lim_{n \rightarrow \infty} \frac{1}{n} \sum_{k=1}^n |c_k| = 0,$$ which establishes the result for the case that $(b_n)$ converges to $0$. If $(b_n)$ does not converge to zero but to some $L \in \mathbb{R}$, then the sequence $(b'_n)$ defined by $b'_n=b_n - L$ does converge to $0$ and if we define $a'_n=a_n-\frac{L}{2}$, we have $$a'_{n+1}+a'_{n}=b'_{n+1}$$ so by the preceding argument we have that $\lim \frac{a'_n}{n}=0$. But from this we have $$\lim \frac{a_n}{n}=\lim (\frac{a'_n}{n}+\frac{L/2}{n})=0,$$ which completes the proof.
EDIT: Now for the lemma:
Let $(s_n)$ be a sequence of real numbers converging to $0$. Then the sequence $(t_n)$ defined by $$t_n=\frac{1}{n}\sum_{k=1}^n s_k$$ converges to $0$ as well.
Proof Let $\epsilon>0$. There is some $N_1 \in \mathbb{N}$ such that if $n>N_1$ then $|s_n|<\frac{\epsilon}{2}$. We can now find $N_2>N_1$ such that if $n>N_2$, then $$|\frac{\sum_{k=1}^{N_1} s_k}{n}|<\frac{\epsilon}{2}.$$ Now by the triangle inequality, if $n>N_2$ we also have $$|t_n|<\frac{\sum_{k=1}^{N_1} |s_k| + \sum_{k=N_1+1}^n |s_k|}{n} < \frac{\epsilon}{2}+\frac{n-N_1}{n} \frac{\epsilon}{2}<\epsilon.$$
• Firstly, thank you for the detailed answer. Although I do understand your method and approach, the book does not touch series or summations until after this chapter, and so I don't know how I would arrive at your claim for the claim of $a_n$ in the form of a summation.Is there anyway to go about this only using things related to convergent sequences, limsups liminfs, and their properties? I would also appreciate a proof in regards to the average converging to 0. – Hushus46 Oct 13 '17 at 0:18
• Sure, shall I edit it into my answer? – M. Van Oct 13 '17 at 7:48
• That would be fine, thank you very much again! – Hushus46 Oct 13 '17 at 7:49
• And to answer your questions: I arrived at the conclusion about $(a_n)$ when I started realizing that if I'd treat the $(b_n)$ as a "given" sequence and $(a_n)$ as a sequence inductively constructed by $a_{n+1}=b_{n+1}-a_n$, I could get a closed form for $(a_n)$ in terms of $(b_n)$ and $a_1$. (write out some terms and you can see it happening, $a_2=b_2-a_1$, $a_3=b_3-a_2=b_3-b_2+a_1$, etc). Moreover, I claim I only use things related to sequences and no theory about series! Maybe it is confusing since I use sigma notation but I could use other notation as well. – M. Van Oct 13 '17 at 8:49 | 2019-08-24 00:31:49 | {"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": 0.9694512486457825, "perplexity": 74.48650761965372}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-35/segments/1566027319155.91/warc/CC-MAIN-20190823235136-20190824021136-00305.warc.gz"} |
http://mathonline.wikidot.com/criterion-for-g-to-be-isomorphic-to-h-k-when-h-k-are-normal | Criterion for G to be Isomorphic to H × K When H, K ◁ G
# Criterion for G to be Isomorphic to H × K When H, K are Normal Subgroups of G
Theorem 1: Let $G$ be a group and let $H$ and $K$ be normal subgroups of $G$. If $G = HK$ and $H \cap K = \{ e \}$ then $G \cong H \times K$.
• Proof: Let $G$ be a group and let $H$, $K$ be normal subgroups of $G$ with $G = HK$ and $H \cap K = \{ e \}$. Let $\phi : H \times K \to G$ be defined for all $(h, k) \in H \times K$ by:
(1)
\begin{align} \quad \phi((h, k)) = hk \end{align}
• We first show that $\phi$ is a homomorphism from $H \times K$ to $G$. Observe that for all $(h_1, k_1), (h_2, k_2) \in H \times K$ we have that:
(2)
\begin{align} \quad \phi((h_1,k_1)(h_2,k_2)) = \phi((h_1h_2, k_1k_2)) = h_1h_2k_1k_2 \end{align}
(3)
\begin{align} \quad \phi((h_1, k_1)) \phi((h_2, k_2)) = h_1k_1h_2k_2 \end{align}
• We want to show that $h_1h_2k_1k_2 = h_1k_1h_2k_2$ for all $h_1, h_2 \in H$ and for all $k_1, k_2 \in K$. To do this, it is sufficient to show that $h_2k_1 = k_1h_2$ for all $h_2 \in H$ and for all $k_1 \in K$, or equivalently, show that $hk = kh$ for all $h \in H$ and for all $k \in K$.
• Let $h' \in H$ and let $k' \in K$. Consider the element $hkh^{-1}k^{-1}$.
• Since $H$ is a normal subgroup of $G$ we have that $ghg^{-1} \in H$ for all $g \in G$ and for all $h \in H$. Since $K \subseteq G$, we have in particular that $khk^{-1} \in H$ for all $k \in K$ and for all $h \in H$. Since $khk^{-1} \in H$ and $h^{-1} \in H$ (and since $H$ is a group) we have that $khk^{-1}h^{-1} \in H$. So $hkh^{-1}k^{-1} \in H$ for all $h \in H$ and for all $k \in K$. In particular:
(4)
• Similarly, since $K$ is a normal subgroup of $G$ we have that $gkg^{-1} \in K$ for all $g \in G$ and for all $k \in K$. Since $H \subseteq G$, we have in particular that $hkh^{-1} \in K$ for all $h \in H$ and for all $k \in K$. Since $hkh^{-1} \in K$ and and $k^{-1} \in K$ for each $k \in K$ (and since $K$ is a group) we have that $hkh^{-1}k^{-1} \in K$ for all $h \in H$ and for all $k \in K$. In particular:
(5)
• From $(*)$ and $(**)$ we see that:
(6)
\begin{align} \quad h'k'h'^{-1}k'^{-1} \in H \cap K = \{ e \} \end{align}
• Thus $h'k'h'^{-1}k'^{-1} = e$, i.e., $h'k' = k'h'$ for all $h' \in H$ and for all $k' \in K$, so the claim is proved and indeed, $\phi$ is a homomorphism from $H \times K$ to $G$.
• Now observe that:
(7)
\begin{align} \quad \ker (\phi) &= \{ (h, k) \in H \times K : \phi((h, k)) = e \} \\ &= \{ (h, k) \in H \times K : hk = e \} \end{align}
• Observe that $hk = e$ implies that $h = k^{-1}$. So $k^{-1} = h \in H$. Thus $k \in H$. So $k \in H \cap K = \{ e \}$, i.e., $k =e$. But then $hk = e$ gives us $h = e$. So $\ker(\phi) = \{ (e, e) \}$, i.e., $\ker(\phi)$ is the trivial group. Thus $(H \times K)/\ker(\phi) \cong H \times K$. So by The First Group Isomorphism Theorem we have that:
(8)
\begin{align} \quad H \times K = (H \times K)/\{(e, e)\} = (H \times K)/\ker(\phi) \cong \phi(H \times K) \end{align}
• Lastly, observe that $\phi$ is surjective since $G = HK$. Indeed, if $g \in G$ then there exists $h \in H$ and $k \in K$ such that $g = hk$. So $(h, k) \in H \times K$ is such that $\phi(h, k) = hk = g$. Thus $\phi(H \times K) = G$, and so from above we conclude that:
(9) | 2019-06-20 01:32:32 | {"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": 9, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 1.0000028610229492, "perplexity": 92.61513461998483}, "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/1560627999130.50/warc/CC-MAIN-20190620004625-20190620030625-00542.warc.gz"} |
https://www.wyzant.com/resources/answers/287673/a_collection_of_27_coins_consists_of_0_20_and_2_00_coins_and_has_a_value_of_23_40_find_the_number_of_coins_of_each_denomination | JamesBill B.
# A collection of 27 coins consists of $0.20 and$2.00 coins, and has a value of $23.40. Find the number of coins of each denomination. A collection of 27 coins consists of$0.20 and $2.00 coins, and has a value of$23.40. Find the number of coins of each denomination.
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Choose an expert and meet online. No packages or subscriptions, pay only for the time you need. | 2021-10-22 18:30:04 | {"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": 0.18405084311962128, "perplexity": 5125.588628420276}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "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-2021-43/segments/1634323585518.54/warc/CC-MAIN-20211022181017-20211022211017-00080.warc.gz"} |
http://lt-jds.jinr.ru/record/72419 | / hep-ex CERN-EP-2017-311
Search for heavy neutral lepton production in $K^+$ decays
Abstract: A search for heavy neutral lepton production in $K^+$ decays using a data sample collected with a minimum bias trigger by the NA62 experiment at CERN in 2015 is reported. Upper limits at the $10^{-7}$ to $10^{-6}$ level are established on the elements of the extended neutrino mixing matrix $|U_{\ell 4}|^2$ ($\ell=e,\mu$) for heavy neutral lepton mass in the range $170-448~{\rm MeV}/c^2$. This improves on the results from previous production searches in $K^+$ decays, setting more stringent limits and extending the mass range.
Note: *Temporary entry*
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Numbers of unique views: 206 | 2018-10-24 05:52:41 | {"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": 0.7869168519973755, "perplexity": 2343.8988670186104}, "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/1539583519859.77/warc/CC-MAIN-20181024042701-20181024064201-00253.warc.gz"} |
https://www.biostars.org/p/474666/ | Error with gff file while building snpeff database
1
0
Entering edit mode
7 months ago
I built a snpEff database for my nuclear genome. It worked. I am now trying to build own for the chloroplast and mitochondria genomes. I've edited the config file and I have fasta files in the genome folder and their corresponding .gff files in their respective folders. I'm getting this error after I run the build command:
Reading GFF3 data file : '/home/kmmahan/snpEff/./data/scenedesmus_obliquus_doe0152z_mitochondria/genes.gff' java.lang.StackOverflowError
I checked the fasta file and the gff file and the genome names match. It's not a memory error because I was able to build the nuclear genome database. I get the error for both the chloroplast and mitochondria database. I've searched for answers on biostars but I am stuck. Something seems to be wrong when it tries to read the gff file. What changes can I make?
snpeff gff java error • 341 views
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0
Entering edit mode
7 months ago
Fatima ▴ 950
/snpEff/./data/
I'm not sure, but this "." in the path seems weird to me. Are you sure the path is correct?
ls /home/kmmahan/snpEff/./data/scenedesmus_obliquus_doe0152z_mitochondria/genes.gff
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0
Entering edit mode
It's the same path I used for the nuclear genome (except folder was scenedesmus_obliquus_doe0152z_nuclear/genes.gff). And same way that it was done here: A: Building Snpeff Database
I cd's using the path you listed (/home/kmmahan/snpEff/./data/scenedesmus_obliquus_doe0152z_mitochondria) and was taken to that directory. I think the path is correct.
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0
Entering edit mode
Oh, okay. What did you use to read the gff file? That program probably makes a lot of recursive calls, that fills up the stack.
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0
Entering edit mode
What do you mean read? I just entered the build command to build the database11. java -jar snpEff.jar build -gff3 -v scenedesmus_obliquus_doe0152z_mitochondria
Oh maybe its the -gff3? Except I used the same thing to build the database for the nuclear genome . . .
ADD REPLY
0
Entering edit mode
Yes, -gff3 probably calls a different set of functions.
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0
Entering edit mode
If I am using a gff file- I have to put -gff3 in the build command. Also - I used -gffs to build the database for the nuclear genome and it worked. Something seems to be wrong with the gff files themselves but just haven't figured it out.
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Powered by the version 2.3.6 | 2021-06-21 10:25:24 | {"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": 0.3116270899772644, "perplexity": 5631.078685492393}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-25/segments/1623488269939.53/warc/CC-MAIN-20210621085922-20210621115922-00562.warc.gz"} |
https://mathhelpboards.com/threads/problem-of-the-week-59-may-13th-2013.4760/#post-22030 | # Problem of the week #59 - May 13th, 2013
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#### Jameson
Staff member
The line $L_1$ goes through the point $(4,3,-2)$ and is parallel to the line defined by $(x=1+3t, y=2-4t, z= 3-t)$. If the point $(m,n,-5)$ is also on $L_1$ then find the values of $m$ and $n$.
--------------------
#### Jameson
Staff member
Congratulations to the following members for their correct solutions:
1) MarkFL
2) anemone
3) Sudharaka
Solution (from Sudharaka):
The line $$L_1$$ should go through the point $$(4,\,3,\,-2)$$ and should be parallel to the vector $$(3,\,-4,\,-1)$$. The equation of the line $$L_1$$ in vector form can be written as,
$L_1:\, (x,\,y,\,z)=(4,\,3,\,-2)+t(3,\,-4,\,-1)$
Since $$(m,\,n,\,-5)$$ is on $$L_1$$ we have,
$L_1:\, (m,\,n,\,-5)=(4,\,3,\,-2)+t(3,\,-4,\,-1)$
$\Rightarrow t=3$
$\therefore m=4+3t=13\mbox{ and }n=3-4t=-9$
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Not open for further replies. | 2022-07-02 19:55:55 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6716271042823792, "perplexity": 523.4065572609982}, "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-2022-27/segments/1656104204514.62/warc/CC-MAIN-20220702192528-20220702222528-00552.warc.gz"} |
https://www.journaltocs.ac.uk/index.php?action=browse&subAction=subjects&publisherID=17&journalID=38360&pageb=2&userQueryID=&sort=&local_page=1&sorType=&sorCol=5 | Subjects -> SCIENCES: COMPREHENSIVE WORKS (Total: 374 journals)
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Edinburgh, EH14 4AS, UK
Email: journaltocs@hw.ac.uk
Tel: +00 44 (0)131 4513762 | 2022-12-01 17:10:30 | {"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": 0.41156524419784546, "perplexity": 1198.3390241838936}, "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-2022-49/segments/1669446710829.5/warc/CC-MAIN-20221201153700-20221201183700-00803.warc.gz"} |
https://meijun.github.io/ | # Logic Programming Is Amazing
Ordinary programming Logic programming
Identify problem Identify problem
Assemble information Assemble information
Figure out solution coffee break :)
Encode solution Encode info in KB
Encode problem instance as data Encode problem instance as data
Apply program to data Ask queries
Wow, logic programming is really amazing.
But, reality is reality. The following process is:
Ordinary programming Logic programming
Apply program to data Ask queries
Start the program Start the program
Program finished Running
Global travel Running
# Most Constrained Variables And Least Constraining Value
There are two important ordering methods for solving CSP problems using backtracking search:
• Minimum Remaining Values (MRV)
• Least Constraining Value (LCV)
The former is for ordering variables. It tells us that it’s better to consider the variables with minimum remaining values, so it’s also known as the most constrained variables method. The latter is for ordering values. After we decided to consider one variable, there are many values to consider. It tells us that it’s better to consider the least constraining value firstly.
Most of the people believe that MRV leads the searching fail-fast, and LCV leads the searching succeed-fast. Choose the fail-fast variable but choose the succeed-fast value. Choose the most constrained variable but choose the least constraining value. It seems that the two ordering methods are not consistent, right?
Actually, this question confuses me a lot. After struggling hard, I believe the truth is in my mind now. Both methods want the searching end-fast. MRV leads the searching not only fail-fast but also succeed-fast. LCV leads the searching succeed-fast, but not fail-fast. When considering to choose variables, choosing the most constrained variable makes the searching tree thin, so it leads the searching end-fast. When considering to choose values, there are two cases. Suppose that we have $n$ values to consider, that means we have $n$ sub searching trees currently. If all the $n$ sub searching trees fail in the end, then it’s no matter how we order the values. If one of the sub searching trees succeeds, it’s better to consider this succeeding sub searching tree. We believe the least constraining value has the most probability to succeed, so we consider the least constraining value firstly, and so this leads succeed-fast.
# Pacman Died Finally
After fighting for an hour, our little Ghost killed the eval Pacman. :P
Actually, it’s a task in UCB CS188 course Pacman project. In this task, we write a method controlling the pacman to earn more scores. The method in the above screencast is a depth-3 expectimax search with a tricky evaluation function. It’s hard for me to get more scores.
# My Favorite Alpha Beta Pruning Bug
The following $\alpha \beta$ pruning algorithm looks clever, however, it has a tricky bug. Can you find where the bug is?
def alpha_beta(state, alpha, beta):
if state.finished():
return state.calcScore()
for next_state in state.next():
alpha = max(alpha, -alpha_beta(next_state, -beta, -alpha))
if alpha >= beta:
break
return alpha
Tips: This code is able to find the correct solution, but may be more time-consuming.
# The Way To Good Heuristics
In UCB CS188 course Pacman project, we should give some heuristics to search efficiently. Heuristics are easy to find. For example, the two trivial heuristics are always zero and the true minimum cost to the goal. The former is easy to compute but without boosting the searching efficiency. The latter is very hard to compute (actually it’s our original searching problem), but it gives the best searching efficiency. Thus, a good heuristic is a trade-off between the heuristic computing time and the approximate accuracy.
Here I give my way to find a good heuristic:
• Relax the problem to ensure admissive and make it efficient to compute.
• Check the consistency.
In the relaxing step, we can relax the problem to make it efficient to solve, for example, by the greedy algorithm, by dynamic programming, etc. The relaxing is to ignore some restrictions in the original problem. But be careful, we should relax the problem slightly, to approximate accurately. The relaxing can ensure admissive.
In the checking step, we should make sure our heuristic is consistent. The consistency is defined as heuristic “arc” cost <= actual arc cost for each arc, formally:
For the grid search problem, every arc cost is $1$, so we should check whether $h(u)-h(v)\leq 1$ for every grid point $u$ and its neighbour $v$. If we use Manhattan distance as the heuristic, it’s consistent, since the Manhattan distance cannot decrease more than $1$ in one move. It’s no hurt to the consistency if the heuristic increases in any move. | 2019-02-22 03:23:27 | {"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": 0.51886385679245, "perplexity": 2124.299865783393}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-09/segments/1550247512461.73/warc/CC-MAIN-20190222013546-20190222035546-00074.warc.gz"} |
http://photography.seldonhunt.com/the-lost-nfsrmfg/ba669c-land-for-sale-bates-county%2C-mo | Questions : Represent the following inequalities in the interval notation: (i) x ⥠â1 and x < 4 (ii) x ⤠5 and x ⥠â3 (iii) x < â1 or x < 3 (iv) -2x > 0 or 3x - 4 < 11. For example, consider the set of numbers that are all greater than 5. Write the following in interval notation x < 8. Showing if the beginning and end number are included is important; There are three main ways to show intervals: Inequalities, The Number Line and Interval Notation. Interval notation combines inequality, or set notation, with its graph, and allows us to accurately express an interval with easy to understand symbols. Interval Notation and Set Builder Notation Calculator: This calculator determines the interval notation and set builder notation for a given numerical statement. Solved Graph The Function And Give Its Domain And Range U. DOWNLOAD IMAGE. A closed interval is one that includes its endpoints: for example, the set { x | â 3 ⤠x ⤠1 } . To write this interval in interval notation, we use closed brackets [ ]: The various types of numerical statements are noted below. The figure below shows a graph of f (x) = 1 2 x â 6 on the given interval: Example 4. In advanced mathematics, interval notation is the preferred method of representing domain and range and is cleaner and easier to use and interpret. â8 2â1 2 2 6 3 8 4 4 â3 -2â1 0 1 4 3 4 âIS â4 â10 2 o o 3 2 o 2 Name the domain and range of each relation using interval notation⦠Intervals can be expressed with appropriate interval notation in Math. {eq}\left \{ x\mid x \leq 5 \right \} {/eq} Inequatilities. Solution for Graph the set and write the interval notation : a) (-00, -1]U (-3, 12] b) [0, 8) n (3, 12] But it's important to make sure that you're describing a specific interval correctly. Really clear math lessons (pre-algebra, algebra, precalculus), cool math games, online graphing calculators, geometry art, fractals, polyhedra, parents and teachers areas too. [-1, 4) Now, if the -1 was an open circle and the 4 had a solid dot, then it would be: (-1, 4]. Graph 1. Inequalities solution can be expressed by a graph or using interval notation. Interval Notation Domain Kampa Luckincsolutions Org. Sketch the graph of the function f (x) = 1 2 x â 6 on the interval [-4, 12). DOWNLOAD IMAGE. [-10, 10. First notice that the interval does not include the number 3, but does include the number 5. Graph 2. We use interval notation to represent subsets of real numbers. What is the domain of a function? Interval notation is a common way to express the solution set to an inequality, and itâs important because itâs how you express solution sets in calculus. Hence the domain, in interval notation, is written as At the left end of each interval, use [ with each end value to be included in the set (solid dot) or ( for each excluded end value (open dot). 9. Given a line graph, describe the set of values using interval notation. Intervals and interval notation. Interval Notation DRAFT. Write interval notation using a beginning bracket or parenthesis and an ending bracket or parenthesis. Interval Notation Interval notation is a way of writing subsets of the real number line . You can use interval notation to express where a set of solutions begins and where it ends. C) Determine the vertical asymptote. y = log_5(x + 3). Example: Representing Inequalities in Interval notations - Examples. Interval Notation 1 - Cool Math has free online cool math lessons, cool math games and fun math activities. Closed Interval. 11) 15) 12) 16) 13) 17) 14) 18) 8 Day 2 â Solving Multi-Step Linear Inequalities Warm â Up . In an equation, the 2 expressions are deemed equal which is shown by the symbol =. Usually, this is used to describe a certain span or group of spans of numbers along a axis, such as an x-axis. â)â means the end point of 10 is NOT included in the interval. How To Find Domain And Range On A Graph Using Interval Notation DOWNLOAD IMAGE. For example, write the set of numbers between 3 and 20 as (3, 20). Note: The âOpen-Intervalâ form may resemble an âordered pairâ used in graphing. Letâs see another chart that will help you translate between the verbal description, inequality notation, and interval notation. Notice that your solution set is just $(2, \infty)$ though, since $3$ is already in the set $(2, \infty)$. Then, the open interval (a,b) represents the set of all real numbers between a and b, except a and b. Solution to example 1 the graph starts at x 4 and ends x 6. Most pre-calculus books and some pre-calculus teachers now require all sets to [â¦] Interval Notation Made Easy W 14 Step By Step Examples. (c) The open intervals on which f is concave upward. The interval graphs include all proper interval graphs, graphs defined in the same way from a set of unit intervals.. a < x < b is the inequality description. ... which is less than or equal to 5. Open Interval. Identify the intervals to be included in the set by determining where the heavy line overlays the real line. Write the inequality shown by each graph and express in interval notation. Written in interval notation, it would be as follows. Graph 3. -4x+8 > -12 pass the 8 on the appropriate area via subtracting 8 from the two aspects -4x > -20 divide the two area via -4 x > 5 X could desire to be extra suitable than 5 2. (a) The open intervals on which f is increasing. Open Interval: An interval that does not include itâs end values or points is termed as an open interval. (-4,00) The solution set in set notation is {x:( (Use integers or fractions for any numbers in the expression.) However, this notation can be used to describe any group of numbers. Note In addition to the keyboard shortcuts listed in this topic, some symbols can be typed using the keyboard shortcuts for your operating system; for example, you can press ALT + 0247 on Windows to type ÷. It is usually represented with the help of parenthesis (round brackets). Find the domain of the graph of the function shown below and write it in both interval and inequality notations. Examples with detailed solutions example 1. For all x between -4 and 6, there points on the graph. If you really want to use interval notation, you could also denote this as $[3,3]$. The context of the question will tell you whether (1, 4) means âall numbers between 1 and 4â or âa point that is over 1, up 4 from the originâ on a graph. Next, write the beginning number of the set followed by a comma and the ending number of the set. Interval notation. Solution for Use interval notation to express solution sets and graph solution set on a number line.Solve linear inequality. Solution to Example 1 The graph starts at x = - 4 and ends x = 6. For more review on set notation and interval notation, visit this tutorial on set-builder and interval notation. Free math problem solver answers your algebra, geometry, trigonometry, calculus, and statistics homework questions with step-by-step explanations, just like a math tutor. Closed Interval: Suppose that a and b are real numbers such that a < b. B) Write the domain and range in interval notation. 8. (Enter your answer using interval notation.) Domain of a Graph; Examples with Detailed Solutions Example 1. Why should we care? (b) The open intervals on which f is decreasing. answer choices . Brackets indicate that the set includes the beginning and/or ending number. Interval notation is a method of writing down a set of numbers. What is interval notation. Worked example: domain and range from graph. An interval in Math is, is the group of all real numbers in between two specified values, each of which are the end points of the interval. SWBAT write the domain and range of a graph using interval notation by comparing it to set builder notation. The chart below will show you all of the possible ways of utilizing interval notation. Write the inequality in interval notation and graph it. State the domain of this graph using interval notation: Algebra: Oct 1, 2012: graph solution and interval notation: Algebra: Sep 26, 2009: Interval Notation Of An Irregular Graph: Pre-Calculus: Aug 27, 2009: Graphing Inequalities, Interval Notation: Algebra: Mar 12, 2009 Preview this quiz on Quizizz. We use the symbol â to indicate "infinity" or the idea that an interval does not have an endpoint. Interval Notation in Math. Since â is not a number, it should not be used with a square bracket. An Interval is all the numbers between two given numbers. What is the range of a function? Play this game to review Algebra I. Solution. We can represent not including a number with an open circle and including a number with a closed circle. Express the interval in set notation and graph the interval on a number line. Interval graphs are chordal graphs and perfect graphs.They can be recognized in linear time, and an optimal graph coloring or maximum clique in these graphs can be found in linear time. Big Idea To have a deeper understanding of interval notation from inequalities and how the square bracket includes the number and the parentheses does not include it. 5[3(2 - 3x) - 2(5 - x)] - 6[5(x -⦠Use the given graph of f over the interval (0,7) to find the following. Equations and inequalities are mathematical sentences formed by relating 2 expressions to each other. 9th grade. 5) 8) 6) 9) 7) 10) 7 Solve each inequality and graph the solutions. Set and Interval Notation The following set and interval notation can be entered in Show My Work boxes. (Enter your answer using interval notation.) If you give me an x anywhere in between negative 2 and 5, I can look at this graph to see where the function is defined. Inequalities, Interval Notation, & Number line Inequalities. Express the following in Interval Notation -8 ⤠x < 8. Graph the inequality: \(3 | 2022-10-04 08:02:59 | {"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": 0.8306417465209961, "perplexity": 549.7610119102053}, "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-2022-40/segments/1664030337480.10/warc/CC-MAIN-20221004054641-20221004084641-00042.warc.gz"} |
https://rtemis.netlify.app/resampling.html | # 13 Resampling
library(rtemis)
.:rtemis 0.8.0: Welcome, egenn
[x86_64-apple-darwin17.0 (64-bit): Defaulting to 4/4 available cores]
Documentation & vignettes: https://rtemis.netlify.com
Resampling refers to a collection of techniques for selecting cases from a sample. It is central to many machine learning algorithms and pipelines. The two core uses of resampling are:
• Model selection (a.k.a. tuning) - Find a combination of hyperparameters that works well
• Model assessment - Assessing how well an a performs
By convention, we use the terms training and validation sets when refering to model selection, and training and testing sets when refering to model assessment. The terminology is unfortunately not intuitive and has led to confusion. Some people reverse the terms, but we use the terms training, validation, and testing as they are used in the Elements of Statistical Learning (p. 222, Second edition, 12th printing)
## 13.1 Model Selection and Assessment
1. Model Selection aka Hyperparameter tuning
Resamples of the training set are drawn. For each resample, a combination of hyperparameters is used to train a model. The mean validation-set error across resamples is calculated. The combination of hyperparameters with the minimum loss on average across validation-set resamples is selected to train the full training sample.
2. Model assessment
Resamples of the full sample is split into multiple training - testing sets. A model is trained on each training set and its performance assessed on the corresponding test set. Model performance is averaged across all test sets.
Nested resampling or nested crossvalidation is the procedure where 1. and 2. are nested so that hyperparameter tuning (resampling of the training set) is performed within each of multiple training resamples and performance is tested in each corresponding test set. [elevate] performs automatic nested resampling and is one of the core supervised learning functions in rtemis.
## 13.2 The resample function
The resample function is responsible for all resampling in rtemis.
x <- rnorm(500)
res <- resample(x)
[[ Resampling Parameters ]]
n.resamples: 10
resampler: strat.sub
stratify.var: y
train.p: 0.75
strat.n.bins: 4
[2020-06-23 08:22:22 resample] Created 10 stratified subsamples
class(res)
[1] "resample" "list"
It outputs a list which is an S3 object of class resample, with print and plot methods.
res
.:rtemis resample object
N: 10
type: strat.sub
train.p: 0.75
strat.n.bins: 4
plot(res)
The teal-colored lines represent the training cases selected for each resample, the white are testing cases (held out).
resample supports 5 types of resampling:
1. k-fold crossvalidation (Stratified)
You split the cases into k sets (folds). Each set is used once as the validation or testing set. This means each cases is left out exactly once and there is no overlap between different validation/test sets. In rtemis, the folds are also stratified by default on the outcome unless otherwise chosen. Stratification tries to maintain the full sample’s distribution in both training and left-out sets. This is crucial for non-normally distributed continuous outcomes or imbalanced datasets. 10 is a common value for k, called 10-fold. Note that the size of the training and left-out sets depends on the sample size.
res.10fold <- resample(x, 10, "kfold")
[[ Resampling Parameters ]]
n.resamples: 10
resampler: kfold
stratify.var: y
strat.n.bins: 4
[2020-06-23 08:39:48 resample] Created 10 independent folds
1. Stratified subsampling
Draw n.resamples stratified samples from the data given a certain probability (train.p) that each case belongs to the training set. Since you are randomly sampling from the full sample each time, there will be overlap in the test set cases, but you control the training-to-testing ratio and number of resamples independently, unlike in k-fold resampling.
res.25ss <- resample(x, 25, "strat.sub")
[[ Resampling Parameters ]]
n.resamples: 25
resampler: strat.sub
stratify.var: y
train.p: 0.75
strat.n.bins: 4
[2020-06-23 08:39:48 resample] Created 25 stratified subsamples
1. Bootstrap
The bootstrap: random sampling with replacement. Since cases are replicated, you should use bootstrap as the outer resampler if you will also have inner resampling for tuning, since the same case may end up in both training and validation sets.
res.100boot <- resample(x, 100, "bootstrap")
[[ Resampling Parameters ]]
n.resamples: 100
resampler: bootstrap
[2020-06-23 08:39:49 resample] Created 100 bootstrap resamples
1. Stratified Bootstrap
This is stratified subsampling with random replication of cases to match the length of the original sample. Same as the bootstrap, do not use if you will be further resampling each resample.
res.100sboot <- resample(x, 100, "strat.boot")
[[ Resampling Parameters ]]
n.resamples: 100
resampler: strat.boot
stratify.var: y
train.p: 0.75
strat.n.bins: 4
target.length: 500
[2020-06-23 08:39:49 resample] Created 100 stratified bootstraps
1. Leave-One-Out-Crossvalidation (LOOCV)
This is k-fold crossvalidation where $$k = N$$, where $$N$$ is number of data points/cases in the whole sample. It has been included for experimentation and completenes, but it is not recommended either for model selection or assessment over the other resampling methods.
res.loocv <- resample(x, resampler = "loocv")
[[ Resampling Parameters ]]
n.resamples: 500
resampler: loocv
[2020-06-23 08:39:49 resample] Created 500 independent folds (LOOCV)
## 13.3 Example: Stratified vs random sampling in a binomial distribution
Imagine y is the outcome of interest where events occur with a probability of .1 - a common scenario in many fields.
set.seed(2020)
x <- rbinom(100, 1, .1)
mplot3.x(x)
freq <- table(x)
prob <- freq[2] / sum(freq)
res.nonstrat <- lapply(seq(10), function(i) sample(seq(x), .75*length(x)))
res.strat <- resample(x)
[[ Resampling Parameters ]]
n.resamples: 10
resampler: strat.sub
stratify.var: y
train.p: 0.75
strat.n.bins: 4
[2020-06-23 08:39:50 strat.sub] Using max n bins possible = 2
[2020-06-23 08:39:50 resample] Created 10 stratified subsamples
prob.nonstrat <- sapply(seq(10), function(i) {
freq <- table(x[res.nonstrat[[i]]])
freq[2]/sum(freq)
})
prob.strat <- sapply(seq(10), function(i) {
freq <- table(x[res.strat[[i]]])
freq[2]/sum(freq)
})
prob.nonstrat
1 1 1 1 1 1 1
0.09333333 0.08000000 0.08000000 0.06666667 0.06666667 0.10666667 0.10666667
1 1 1
0.10666667 0.09333333 0.08000000
sd(prob.nonstrat)
[1] 0.0156505
prob.strat
1 1 1 1 1 1 1
0.08108108 0.08108108 0.08108108 0.08108108 0.08108108 0.08108108 0.08108108
1 1 1
0.08108108 0.08108108 0.08108108
sd(prob.strat)
[1] 0
As expected, the random sampling resulted in different event probability in each resample, whereas stratified subsampling maintained a constant probability across resamples. | 2020-12-02 21:12:09 | {"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": 0.39675167202949524, "perplexity": 5901.91581543828}, "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-2020-50/segments/1606141716970.77/warc/CC-MAIN-20201202205758-20201202235758-00326.warc.gz"} |
https://www.freemathhelp.com/forum/threads/finding-area-within-a-circle-from-a-point-not-at-the-centre.117842/ | # Finding area within a circle from a point NOT at the centre
#### Coach Sanders
##### New member
Hey, I've been working on this question for years and have been unable to find anything regarding area within a circle from a point on a chord.
It involves cutting a pizza into 5 by starting with a segment equal to one fifth the area of the pizza (this is estimated since I haven't solved this problem). Then, from the midpoint of that chord, cut the remaining four fifths in half, and those 2 in half again (from that same point). The result needs to be 5 equal pieces.
The question then is: what is the height of the segment? and the measure of the angles dividing the other 4 pieces?
#### Dr.Peterson
##### Elite Member
Do you mean something like this?
I've worked out approximate values for CD and the angles shown; they can't be calculated exactly, but require numerical methods (I used Desmos).
Now, what help do you want from us? The main formula I used was for the area of a circular segment (I used equation 14). I also used formulas for area of a sector and of a triangle.
#### Coach Sanders
##### New member
Thank you,
This is exactly what I mean, so I'm wondering how you come up with those measurements. If they can't be measured exactly, why is that? The greatest thing about math, is that there should always be an answer; a method. If a computer can generate an image like that, where all 5 pieces have the same area, shouldn't there be a way to get there?
#### Dr.Peterson
##### Elite Member
Thank you,
This is exactly what I mean, so I'm wondering how you come up with those measurements. If they can't be measured exactly, why is that? The greatest thing about math, is that there should always be an answer; a method. If a computer can generate an image like that, where all 5 pieces have the same area, shouldn't there be a way to get there?
You can get there, but not exactly, and not by a formula. Computers are great at approximating things!
I wrote a couple equations, as I described, and then used a computer program to find approximate solutions. It is a dirty little secret that algebra can't solve every equation you can write; in fact, the only equations we usually show students are those that we know how to solve. The equations for the height of the first segment, and for the angles in the others, are transcendental equations, which require numerical approximation methods.
I could also have constructed the picture (in GeoGebra) by just placing points and moving them around until I got the right areas; I chose instead to get numerical values first. I can't show you the equations I used at the moment, because I'm not at home where I did them.
#### Coach Sanders
##### New member
Thank you very much Dr. Peterson. It actually makes me feel better that, despite many years of trying to solve what seemed to be a simple math problem, this is a problem "Math" has not tried to figure out. Aside from a desire to get 5 equal pieces of pizza (which I will generally eat by myself anyway), it must not have any other useful applications.
I will sleep better now.
#### Dr.Peterson
##### Elite Member
Just to be complete, here are the two equations I used, where I'm taking the radius as 1, d is the distance from the center to my point D, C is my angle at point C, and D is my angle at D:
$$\displaystyle \cos^{-1}(d) - d\sqrt{1 - d^2} = \pi/5$$
$$\displaystyle C + d\sin(C) = 2\pi/5$$
$$\displaystyle D = \tan^{-1}\frac{\sin(C)}{d + \cos(C)}$$
The approximate solution is d = 0.492, C = 0.878 radians = 50.3 degrees, D = 34.23 degrees. | 2019-09-15 22:51:55 | {"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": 0.7280417084693909, "perplexity": 440.395104263657}, "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-2019-39/segments/1568514572436.52/warc/CC-MAIN-20190915215643-20190916001643-00436.warc.gz"} |
http://mathematica.stackexchange.com/questions/linked/2639 | 79 views
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If a list contains duplicate elements, for example list = {a, 1, 5, 3, 5, x^2, x^2}, how can the duplicate elements be removed? The result would be ... | 2015-07-04 07:02:14 | {"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": 0.43156346678733826, "perplexity": 849.0449125850353}, "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/1435375096579.52/warc/CC-MAIN-20150627031816-00138-ip-10-179-60-89.ec2.internal.warc.gz"} |
http://dlmf.nist.gov/1.13 | # §1.13 Differential Equations
## §1.13(i) Existence of Solutions
A domain in the complex plane is simply-connected if it has no “holes”; more precisely, if its complement in the extended plane is connected.
The equation
where , a simply-connected domain, and , are analytic in , has an infinite number of analytic solutions in . A solution becomes unique, for example, when and are prescribed at a point in .
### ¶ Fundamental Pair
Two solutions and are called a fundamental pair if any other solution is expressible as
1.13.2
where and are constants. A fundamental pair can be obtained, for example, by taking any and requiring that
1.13.3
### ¶ Wronskian
The Wronskian of and is defined by
1.13.4
Then
where is independent of . If , then the Wronskian is constant.
The following three statements are equivalent: and comprise a fundamental pair in ; does not vanish in ; and are linearly independent, that is, the only constants and such that
are .
## §1.13(ii) Equations with a Parameter
Assume that in the equation
and belong to domains and respectively, the coefficients and are continuous functions of both variables, and for each fixed (fixed ) the two functions are analytic in (in ). Suppose also that at (a fixed) , and are analytic functions of . Then at each , , and are analytic functions of .
## §1.13(iii) Inhomogeneous Equations
The inhomogeneous (or nonhomogeneous) equation
with , , and analytic in has infinitely many analytic solutions in . If is any one solution, and , are a fundamental pair of solutions of the corresponding homogeneous equation (1.13.1), then every solution of (1.13.8) can be expressed as
1.13.9
where and are constants.
### ¶ Variation of Parameters
With the notation of (1.13.8) and (1.13.9)
1.13.10
## §1.13(iv) Change of Variables
### ¶ Transformation of the Point at Infinity
The substitution in (1.13.1) gives
where
1.13.12
### ¶ Elimination of First Derivative by Change of Dependent Variable
The substitution
in (1.13.1) gives
where
1.13.15
### ¶ Liouville Transformation
Let satisfy (1.13.14), be any thrice-differentiable function of , and
Then
Here dots denote differentiations with respect to , and is the Schwarzian derivative:
1.13.20
## §1.13(v) Products of Solutions
The product of any two solutions of (1.13.1) satisfies
If and are respectively solutions of
then is a solution of
For extensions of these results to linear homogeneous differential equations of arbitrary order see Spigler (1984).
## §1.13(vi) Singularities
For classification of singularities of (1.13.1) and expansions of solutions in the neighborhoods of singularities, see §2.7.
## §1.13(vii) Closed-Form Solutions
For an extensive collection of solutions of differential equations of the first, second, and higher orders see Kamke (1977). | 2013-05-19 17:53:38 | {"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": 0.9433201551437378, "perplexity": 1737.4382031411376}, "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-2013-20/segments/1368697917013/warc/CC-MAIN-20130516095157-00042-ip-10-60-113-184.ec2.internal.warc.gz"} |
http://forums.dc3.com/showthread.php?10431-Rotator-amp-focuser-problem-!!!-(same-device)&s=2be99b4201e37c39404c19a5d4e47e6c | ## Rotator & focuser problem !!! (same device)
Dear Bob and all
I use a rotator which is also focuser .
It is called "ROTOFOC" and is made by Italian company called Reginato.
http://www.reginato.it/accessory.html
Until now in tests i had no problem using only the rotator (set up and used correctly)
Tonight i tried to use also autofocus with focusmax.
The focuser and rotator use the same ascom driver and working unde the same software .
My problem is that when rotator is on (connected) when acp try to connect focuser i get error (see attached)
Any idea how can i solve it?
ROTATOR.jpg | 2020-09-29 14:04:48 | {"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": 0.8024446368217468, "perplexity": 6564.500034745925}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-40/segments/1600401643509.96/warc/CC-MAIN-20200929123413-20200929153413-00329.warc.gz"} |
https://www.groundai.com/project/approximability-of-covering-cells-with-line-segments/ | Approximability of Covering Cells with Line SegmentsPaz Carmi is supported by Grant 2016116 from the United States-Israel Binational Science Foundation. Anil Maheshwari is supported in part by Natural Sciences and Engineering Research Council of Canada (NSERC). Saeed Mehrabi is supported by a Carleton-Fields postdoctoral fellowship.
# Approximability of Covering Cells with Line Segments††thanks: Paz Carmi is supported by Grant 2016116 from the United States-Israel Binational Science Foundation. Anil Maheshwari is supported in part by Natural Sciences and Engineering Research Council of Canada (NSERC). Saeed Mehrabi is supported by a Carleton-Fields postdoctoral fellowship.
Paz Carmi Department of Computer Science, Ben-Gurion University of the Negev, Beer-Sheva, Israel. carmip@cs.bgu.ac.il Anil Maheshwari School of Computer Science, Carleton University, Ottawa, Canada. anil@scs.carleton.ca, saeed.mehrabi@carleton.ca Saeed Mehrabi School of Computer Science, Carleton University, Ottawa, Canada. anil@scs.carleton.ca, saeed.mehrabi@carleton.ca Luís Fernando Schultz Xavier da Silveira School of Computer Science and Electrical Engineering, University of Ottawa, Ottawa, Canada. schultz@ime.usp.br
###### Abstract
In COCOA 2015, Korman et al. studied the following geometric covering problem: given a set of line segments in the plane, find a minimum number of line segments such that every cell in the arrangement of the line segments is covered. Here, a line segment covers a cell if is incident to . The problem was shown to be -hard, even if the line segments in are axis-parallel, and it remains -hard when the goal is cover the “rectangular” cells (i.e., cells that are defined by exactly four axis-parallel line segments).
In this paper, we consider the approximability of the problem. We first give a for the problem when the line segments in are in any orientation, but we can only select the covering line segments from one orientation. Then, we show that when the goal is to cover the rectangular cells using line segments from both horizontal and vertical line segments, then the problem is -hard. We also consider the parameterized complexity of the problem and prove that the problem is when parameterized by the size of an optimal solution. Our algorithm works when the line segments in have two orientations and the goal is to cover all cells, complementing that of Korman et al. [9] in which the goal is to cover the “rectangular” cells.
## 1 Introduction
Set Cover is a well-studied problem in computer science. The input to the problem is a ground set of elements and a set of subsets of ; that is, such that for all . The objective is to find a minimum-cardinality subset of whose union is . Set Cover is known to be -hard [5] and even hard to approximate [8].
In this paper, we consider a geometric variant of the set cover problem that was first studied by Korman et al. [9]. A set of line segments in the plane is said to be non-overlapping if any two line segments from the set intersect in at most one point. Given a set of non-overlapping line segments in the plane, a cell in the arrangement of is a maximally connected region that is not intersected by any line segment in [9]. Then, the objective of the Line Segment Covering () problem is to select a minimum number of line segments such that every cell in the arrangement of the line segments is covered. Here, a cell is covered by a line segment if it is incident to the line segment (i.e., the line segment is in the set of line segments defining the boundary of the cell). We assume that at most two line segments may share a fixed point in the plane.
### Related work.
Korman et al. [9] proved that when the line segments are only horizontal and vertical, the problem is -hard and it remains -hard when the goal is to cover the “rectangular” cells. By a closer look at their hardness proof, one can see that the problem is -hard even if we are only allowed to select the line segments from one orientation (they only select vertical line segments when constructing a solution from a given truth assignment for the corresponding 3SAT problem). Moreover, the authors gave an -time algorithm for covering the rectangular cells when parameterized by , the size of an optimal solution. However, the algorithm does not work when the goal is to cover all cells of the arrangement. The authors leave open studying the approximability of the problem.
The problem is closely related to a guarding problem studied by Bose et al. [3]. Given a set of lines in the plane, they studied the problems of guarding cells of the arrangement by selecting a minimum number of lines, or guarding the lines by selecting a minimum number of cells. Here, “guarding” has the same meaning as “covering” in the problem. However, their results do not extend to the problem, because (as also noted by Korman et al. [9]) they use some properties of lines that are not true for the case of line segments.
### Our results.
In this paper, we prove the following results.
• We give a for the problem when the line segments in can have any arbitrarily orientations, but we are allowed to select the covering line segments from only one orientation. Given the -hardness of the problem [9], this settles the complexity of this variant of the problem.
• When we allow selecting the covering line segments from more than one direction, we show that the problem is -hard when the line segments in have two orientations and the goal is to cover the rectangular cells.
• We give an algorithm for the problem when the line segments in have only two orientations and the goal is to cover all cells of the arrangement. This complements the algorithm of Korman et al. [9] as we do not restrict the covering only to rectangular cells.
### Organization.
In Section 2, we give some definitions and revisit some necessary background. We show our in Section 3 and the -hardness result in Section 4. Finally, the algorithm is given in Section 5 and we conclude the paper in Section 6.
## 2 Preliminaries
In the following, we revisit some techniques and background that are used throughout this paper.
### Local search.
Our for the problem is based on the local search technique, which was introduced independently by Mustafa and Ray [11], and Chan and Har-Peled [4]. Consider an optimization problem in which the objective is to compute a feasible subset of a ground set whose cardinality is minimum over all such feasible subsets of . Moreover, it is assumed that computing some initial feasible solution and determining whether a subset is a feasible solution can be done in polynomial time. The local search algorithm for a minimization problem is as follows. Fix some fixed parameter , and let be some initial feasible solution for the problem. In each iteration, if there are and such that , and is a feasible solution, then set and re-iterate. The algorithm returns and terminates when no such local improvement is possible.
Clearly, the local search algorithm runs in polynomial time. Let and be the solutions returned by the algorithm and an optimal solution, respectively. The following result establishes the connection between local search technique and obtaining a .
###### Theorem 2.1 ([4, 11]).
Consider the solutions and for a minimization problem, and suppose that there exists a planar bipartite graph that satisfies the local exchange property: for any subset , is a feasible solution, where denotes the set of neighbours of in . Then, the local search algorithm yields a for the problem.
The local search was used by Mustafa and Ray [11] to obtain a for geometric hitting set problem and by Chan and Har-Peled [4] to obtain a for geometric independent set problem. Since then, the technique has been used to get a for several other geometric problems, such as geometric dominating set [2] and unique covering [1].
### Fixed-parameter tractability.
The theory of parameterized complexity was developed by Downey and Fellows [6]. Let be a finite alphabet. Then, a parameterized problem is a language in which the second component is called the parameter of the problem. A parameterized problem is said to be fixed-parameter tractable or , if the question “” can be decided in time , where is an arbitrary function. We call an algorithm with such running time , an algorithm.
For the rest of this paper, we denote a set of line segments in the plane by (i.e., ) and the resulting arrangement by .
## 3 PTAS
In this section, we show that the problem admits a when the line segments in are in any orientations, but we can select line segments from only one orientation to cover the cells. To this end, we run the local search algorithm with parameter for some , where is a constant. Let be the solution returned by the algorithm and let be an optimal solution. We can assume that . This is because if , then we can consider the sets and , and analyze the algorithm with and . Here, we mark the faces covered by a line segment in as “covered” so as we do not need to cover then in the new variant of the problem. This guarantees that the approximation factor of the original instance is upper bounded by that of the new instance with these two new sets and .
We now construct a planar bipartite graph that satisfies the local exchange property, hence proving that the problem admits a by Theorem 2.1. To this end, we first construct an auxiliary planar graph and then show how to obtain from by edge contraction. For each cell , let and be two line segments that cover ; we select a point and and connect them by a curve that lies in the interior of (except its endpoints and ). Notice that since both and are feasible solutions, we know that contains at least one line segment that covers and also contains at least one line segment that covers , for all . We add and to and to . We complete the definition of by connecting every pair of consecutive points in , for all , by an edge that is exactly the portion of that lies between the pair of points. See Figure 1(a) for an example. Clearly, is planar because the first set of edges are drawn in the interior of cells and each cell contains at most one edge. Moreover, the second set of edges are aligned with the line segments in . Since the line segments in are non-overlapping and all have the same orientation, the second set of edges are also non-crossing. To obtain the graph , for each segment , we contract the edges of that are contained in such that we get a single point corresponding to ; see Figure 1(b). So, . Graph is planar since remains planar after this edge contraction. Moreover, is a bipartite graph as the edges of with both endpoints belonging to a line segment in or both endpoints belonging to a line segment in are collapsed into a single point (i.e., ).
###### Lemma 3.1.
Graph is planar and bipartite.
We next show that satisfies the exchange property.
###### Lemma 3.2.
Graph satisfies the local exchange property.
###### Proof.
It is sufficient to show that for every cell , there are vertices and such that both segments corresponding to these vertices cover and . Take any cell and let be the set of all line segments that cover . Notice that and because and are each a feasible solution. Then, by definition, there must be a and for which . This completes the proof of the lemma. ∎
Putting everything together, we have the main result of this section.
###### Theorem 3.1.
There exists a for the line segment covering problem when the line segments in can have any orientation and we are allowed to select the covering line segments from only one orientation.
## 4 APX-Hardness
In this section, we show that the problem is -hard when the line segments in have only two orientations and the goal is to cover the rectangular cells. To this end, we give an -reduction from the Minimum Vertex Cover () problem on graphs with maximum-degree three to this variant of the problem. Our reduction is inspired by the construction of Mehrabi [10]. As a reminder, we first give a formal definition of -reduction [12], which is one of the gap-preserving reductions. Let and be two optimization problems with the cost functions and , respectively. We say that -reduces to if there are two polynomial-time computable functions and such that the followings hold.
1. For any instance of , is an instance of .
2. If is a solution to , then is a solution to .
3. There exists a constant such that
OPTΠ′(f(x))≤αOPTΠ(x),
where denotes the cost of an optimal solution for problem on its instance .
4. There exists a constant such that for every solution for ,
|OPTΠ(x)−cΠ(g(y))|≤β|OPTΠ′(f(x))−cΠ(y)|,
where denotes the absolute value of .
###### Lemma 4.1.
The minimum vertex cover problem on graphs with maximum-degree three is -reducible to the problem, where is a set of horizontal and vertical line segments and the goal is to cover the rectangular cells of .
###### Proof.
Let be an instance of on graphs of maximum-degree three; let be the graph corresponding to and let be the size of the smallest vertex cover in . First, let be an arbitrary ordering of the vertices of , where . In the following, we give a polynomial-time computable function that takes as input and outputs an instance of the problem.
We first describe the vertex gadgets. For each vertex , , construct a horizontal line segment and a vertical line segment , and connect them as shown in Figure 2. We call the (blue) horizontal line segment used in the connection of and the horizontal connector of . Moreover, there are four (small, dashed) line segments used in the connection of and that we call the small connectors of . Notice that these five “connectors” along with and form exactly two rectangular cells. For each edge , where , we add two small line segments, one horizontal and one vertical, at the intersection point of and such that they intersect each other as well as each intersects one of and , hence forming a rectangular cell; see the two (red, dashed) line segments at the intersection of and in Figure 2 for an example. We call such a pair edge line segments and denote them by . Finally, for every rectangular cell whose four sides are all defined by the line segments corresponding to a 4-subset of (i.e., the cell is not covered by a horizontal connector or edge line segments), we insert a vertical line segment into the cell so as to make it non-rectangular; see the vertical (red) line segment in Figure 2. This ensures that every rectangular cell is incident either to a horizontal connector or to edge line segments for some and . This gives the instance of the problem. Notice that is a polynomial-time computable function. In the following, we denote an optimal solution for the instance of a problem by . We now prove that all the four conditions of -reduction hold.
First, let be a vertex cover of of size . Denote by the set of horizontal line segments induced by and define analogously. Moreover, let be the set of horizontal connectors whose corresponding vertex is not in . We show that is a feasible solution for covering all the rectangular cells of . Let be a rectangular cell. Then, must be incident either to a horizontal connector or to edge line segments for some and . First, if is incident to a horizontal connector , then either or and by the construction of and so is covered either way. Next, if is incident to edge line segments for some and , where w.l.o.g. , then either or because we know that or . So, is again covered in this case. Therefore, is a feasible solution.
Second, let be any feasible solution for . Notice that we can construct a feasible solution for such that and consists of only and for some , or a horizontal connector. This is because (i) any rectangular cell covered by a small connector is also covered by a horizontal connector, and (ii) any cell covered by a pair of edge line segments (for some and ) is also covered by and . For (ii), if exactly one of the line segments in is in , then we replace it with exactly one of or . Otherwise, if both line segments of are in , then we replace both of them with and . So, and is a feasible solution for . Now, let . To show that is a vertex cover for , consider any edge , where . Then, we know that there exists a rectangular cell at the intersection of and that must be covered by . Since none of the two edge line segments of are in , we conclude that at least one of and is in , which means that or . Hence, is a vertex cover.
Third, observe that and also . Given that has degree three, and so .
We now prove the last condition of -reduction. First, define ; that is, the paths of a vertex , where both its horizontal and vertical line segments appear in . Also, define to be the remaining line segments corresponding to either or for some ; i.e., those of , where only one of its line segments appears in . Take any vertex . To cover the two rectangular cells incident to the horizontal connector of , we must have or ; this is true for all . Then, . Moreover, since is a vertex cover of . Therefore, . By this and our earlier inequality , we have . Now, suppose that for some . Then,
|F|−|s∗(f(I))|=c ⇒|F|−(n+k)=c ⇒|F′|−(n+k)≤c ⇒|One[F′]|+|Both[F′]|/2+n−(n+k)≤c ⇒|One[F′]|+|Both[F′]|/2−k≤c ⇒|M|−|s∗(I)|≤c.
That is, . This concludes our -reduction from on graphs of maximum-degree three to with and . ∎
###### Theorem 4.1.
The line segment covering problem is -hard when the line segments in are either horizontal or vertical and the goal is to cover the rectangular cells of .
## 5 FPT
In this section, we show that the problem is fixed-parameter tractable (parametrized by the size of an optimal solution) when the line segments in are either horizontal or vertical, and the goal is to cover all the cells in . This complements the result of Korman et al. [9], where the goal is to cover the rectangular cells. Throughout this section, let be the size of an optimal solution.
Our follows the framework of Korman et al. [9]. That is, we formulate the problem as a hitting set problem and argue that we only need to hit an number of sets; hence, obtaining a kernel of size for the problem. The of Korman et al. [9] is based on the fact that any three orthogonal line segments can cover at most two “rectangular” cells (i.e., at most two rectangular cells can be incident to all the three line segments). As an analogous result, we prove in Lemma 5.1 that the number of such cells can be at most six when the goal is to cover all cells, including non-rectangular ones. We will then apply this result to obtain the desired kernel.
###### Lemma 5.1.
Let be a set of axis-parallel line segments in the plane. Then, for any three line segments , there are at most six cells in that can be covered by all three line segments and .
###### Proof.
Take any three line segments and in and let be the set of all cells in that are covered by all three line segments and . We need to show that . To this end, we construct a planar graph corresponding to and the cells in and will then argue that this graph must contain a subdivision of if . We next give the details. Let be a cell in . Consider a point in the interior of as well as a distinct point in , for all (notice that is on the boundary of ). These points together form the set of vertices of ; that is, . Now, for each , consider an ordering of the points on , , and connect every two consecutive points by an edge, which is exactly the portion of that lies between the two points. Moreover, for each cell , we connect to by a curve that lies strictly in the interior of (except at its endpoints) for all . Then, the edge set of consists of the set of all edges connecting the consecutive points as we as the curves , for all and . Clearly, is a planar graph. In the following, we consider several cases depending on whether the line segments and intersect each other; observe that there can be at most two intersection points between them.
### Case 1.
There is no intersection point; that is, the line segments and are pairwise disjoint. In this case, we show that in fact . To this end, suppose for a contradiction that . Take any three cells and consider the subgraph of induced by . Now, consider the graph constructed from as follows. For each , , we place a new vertex close to and connect it to the three vertices for all such that the resulting graph remains planar. One can easily verify that this is doable since the three line segments are disjoint and so there are a few cases for where to place depending on which side of the three cells lie; see Figure 3. Observe that the resulting graph is a planar drawing of a subdivision of , which is not possible. So, .
### Case 2.
There is exactly one intersection point; assume w.o.l.g. that is horizontal, and are vertical and intersects . Here, we show that . Again, suppose for a contradiction that . Then, considering the graph , there must be at least three vertices in that lie w.l.o.g. to the right of . Take any three such vertices and denote the corresponding cells by . We can now construct the graph analogous to the one in Case 1 with these three cells and so obtain a planar drawing of a subdivision of , which is a contradiction.
### Case 3.
There are two intersection points. Here, we show that and we use a similar argument to those in the previous cases. Denote the endpoints of by and , and let and be the intersection points of with and ; assume w.l.o.g. that is the left endpoint of and that lies to the left of . If , then at least one of the line segments and must contain three vertices of ; assume w.l.o.g. that it is . Then, take any three such vertices on and consider the three cells and corresponding to these vertices. We can now construct the graph analogous to the one in Case 1 and so obtain a planar drawing of a subdivision of , which is a contradiction. As such, .
By the three cases described above, we conclude that . ∎
We note that the upper bound in Lemma 5.1 is tight as Figure 4 shows an example with three line segments that cover six cell. We now apply Lemma 5.1 to obtain our . We first formulate the problem as a hitting set problem as follows. The ground set is and for each cell in , there exists a set that contains the line segments that cover the cell. Let be the resulting set of subsets of . Then, the problem is equivalent to selecting a minimum number of elements from such that each set in is hit by at least one selected element.
We first reduce the set to a set as follows. For every pair of line segments , if they appear in more than sets , then we remove all such sets form and add the set to . Let be the resulting set.
###### Lemma 5.2.
A set with is a minimum-size cover of if and only if it is a minimum-size cover of .
###### Proof.
We prove the lemma by an argument similar to the one by Korman et al. [9]. The lemma clearly follows if . So, assume that and are two non-empty sets. Let with be a minimum-size cover for . First, is also a cover for because for every set there exists a pair of line segment and such that both and are in and we have . We now prove that is also a cover of minimum size for .
Suppose for a contradiction that there exists a cover for such that . Then, cannot be a cover because is a cover of minimum size for . Since covers , there must exist such that neither nor is in . But, we introduced the set into because there were more than sets containing both and . If neither nor is in , then every other line segment can cover at most six of such subsets by Lemma 5.1. Therefore, — a contradiction. By a similar argument, we can show that a minimum-size cover of is also a minimum-size cover for . This completes the proof of the lemma. ∎
Next, we reduce to a new set as follows. For each line segment , we count how many sets in contain . If appears in more than , then we remove all those sets and add the set to . Let denote the resulting set.
###### Lemma 5.3.
A set with is a minimum-size cover for if and only if it is a minimum-size cover for .
###### Proof.
The lemma follows if . So, assume that and are two non-empty sets. Let with be a minimum-size cover for . For any set , there exists a singleton set in whose member is in . This means that is also a cover for . We next show that is also a minimum-size cover for .
Suppose for a contradiction that there exists a cover for such that . Therefore, is not a cover of . Since covers , there must exist a set in that is not covered by . Notice that this set must be of size 1 from the construction of ; let be such a set, where . The reason we have the set in is that because there were more than sets in containing . If is not in , then all such sets of must be covered by other line segments. But, from the construction of , every pair of line segments can appear in at most sets. So, must be greater than , which is a contradiction. A similar argument can be used to show that a minimum-size cover for is also is minimum-size cover for . This completes the proof of the lemma. ∎
Consider the set . By Lemma 5.3, no line segment of appears in more than sets in . Therefore, if , then the problem does not have a cover of size at most . Since the construction of can be done in polynomial time, we have the following result.
###### Lemma 5.4.
For the problem on a set of axis-parallel line segments, in polynomial time, we can either obtain a kernel of size or conclude that the problem does not have a cover of size at most , where is the size of an optimal cover.
Since having a kernel of size implies that the problem is [7], we have the main result of this section.
###### Theorem 5.1.
The line segment covering problem on a set of axis-parallel line segments is with respect to the size of an optimal cover.
## 6 Conclusion
In this paper, we considered the problem of covering the cells in the arrangement of a set of line segments in the plane. We proved that the problem admits a when the covering line segments can be selected from only one orientation. We then showed that if we allow selecting the covering line segments from more than one orientation, then the problem is -hard when we are interested in covering the rectangular cells. Finally, we gave an algorithm for the problem when the line segments have only two orientations, but the goal is to cover all the cells. Our -hardness rules out the possibility of a for “covering rectangular faces” variant of the problem, but is there a 2-approximation algorithm for the problem? For the more general variant, where the line segments are in any orientation, covering line segments can be selected from any orientation and the goal is to cover all the cells, can we obtain a -approximation algorithm for some small constant ?
## References
• [1] Pradeesha Ashok, Sudeshna Kolay, Neeldhara Misra, and Saket Saurabh. Unique covering problems with geometric sets. In Computing and Combinatorics - 21st International Conference, COCOON 2015, Beijing, China, August 4-6, 2015, Proceedings, pages 548–558, 2015.
• [2] Sayan Bandyapadhyay, Anil Maheshwari, Saeed Mehrabi, and Subhash Suri. Approximating dominating set on intersection graphs of rectangles and L-frames. In MFCS 2018, to appear, 2018. arXiv version available at https://arxiv.org/abs/1803.06216.
• [3] Prosenjit Bose, Jean Cardinal, Sébastien Collette, Ferran Hurtado, Matias Korman, Stefan Langerman, and Perouz Taslakian. Coloring and guarding arrangements. Discrete Mathematics & Theoretical Computer Science, 15(3):139–154, 2013.
• [4] Timothy M. Chan and Sariel Har-Peled. Approximation algorithms for maximum independent set of pseudo-disks. Discrete & Computational Geometry, 48(2):373–392, 2012.
• [5] Thomas H. Cormen, Charles E. Leiserson, Ronald L. Rivest, and Clifford Stein. Introduction to Algorithms, Second Edition. The MIT Press and McGraw-Hill Book Company, 2001.
• [6] Rodney G. Downey and Michael R. Fellows. Parameterized Complexity. Springer, 1999.
• [7] Rodney G. Downey and Michael R. Fellows. Fundamentals of Parameterized Complexity. Texts in Computer Science. Springer, 2013.
• [8] Uriel Feige. A threshold of ln for approximating set cover. Journal of ACM, 45(4):634–652, 1998.
• [9] Matias Korman, Sheung-Hung Poon, and Marcel Roeloffzen. Line segment covering of cells in arrangements. Inf. Process. Lett., 129:25–30, 2018.
• [10] Saeed Mehrabi. Approximating domination on intersection graphs of paths on a grid. In 15th International Workshop on Approximation and Online Algorithms (WAOA 2017), Vienna, Austria, Revised Selected Papers, pages 76–89, 2017.
• [11] Nabil H. Mustafa and Saurabh Ray. Improved results on geometric hitting set problems. Discrete & Computational Geometry, 44(4):883–895, 2010.
• [12] Christos H. Papadimitriou and Mihalis Yannakakis. Optimization, approximation, and complexity classes. J. Comput. Syst. Sci., 43(3):425–440, 1991.
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Test description | 2020-10-24 09:11:03 | {"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": 0.8649820685386658, "perplexity": 332.7996751634399}, "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-2020-45/segments/1603107882103.34/warc/CC-MAIN-20201024080855-20201024110855-00344.warc.gz"} |
https://www.clutchprep.com/chemistry/practice-problems/79673/predict-the-products-of-each-of-the-following-double-displacement-reactions-in-w | # Problem: Predict the products of each of the following double-displacement reactions in words. Then change each equation into a balanced formula equation. All reactions occur in aqueous solution. Ammonium phosphate reacts with barium nitrate.
###### Problem Details
Predict the products of each of the following double-displacement reactions in words. Then change each equation into a balanced formula equation. All reactions occur in aqueous solution.
Ammonium phosphate reacts with barium nitrate. | 2020-06-01 05:54:13 | {"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": 0.9112479090690613, "perplexity": 3594.239944793298}, "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/1590347414057.54/warc/CC-MAIN-20200601040052-20200601070052-00481.warc.gz"} |
https://socratic.org/questions/how-do-you-find-the-vertical-horizontal-and-slant-asymptotes-of-y-8x-48-x-2-13x- | # How do you find the vertical, horizontal and slant asymptotes of: y=(8x-48)/(x^2-13x+42)?
Jul 1, 2016
vertical asymptote x = 7
horizontal asymptote y = 0
#### Explanation:
The first step here is to factorise and simplify y.
$\frac{8 x - 48}{{x}^{2} - 13 x + 42} = \frac{8 \cancel{\left(x - 6\right)}}{\left(x - 7\right) \cancel{\left(x - 6\right)}} = \frac{8}{x - 7}$
The denominator of this rational function cannot be zero as this would lead to division by zero which is undefined.By setting the denominator equal to zero and solving for x we can find the value that x cannot be and if the numerator is also non-zero for this value of x then it must be a vertical asymptote.
solve : x - 7 = 0 → x = 7 is the asymptote
Horizontal asymptotes occur as
${\lim}_{x \to \pm \infty} , y \to c \text{ (a constant)}$
divide terms on numerator/denominator by x
$\frac{\frac{8}{x}}{\frac{x}{x} - \frac{7}{x}} = \frac{\frac{8}{x}}{1 - \frac{7}{x}}$
as $x \to \pm \infty . y \to \frac{0}{1 - 0}$
$\Rightarrow y = 0 \text{ is the asymptote}$
Slant asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here (numerator-degree 0 ,denominator-degree 1 )Hence there are no slant asymptotes.
graph{8/(x-7) [-20, 20, -10, 10]} | 2019-06-18 16:42:38 | {"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": 0.9444429278373718, "perplexity": 685.3341169324831}, "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/1560627998808.17/warc/CC-MAIN-20190618163443-20190618185443-00052.warc.gz"} |
https://jmbh.github.io/Predictability-in-network-models/ | Network models have become a popular way to abstract complex systems and gain insights into relational patterns among observed variables in many areas of science. The majority of these applications focuses on analyzing the structure of the network. However, if the network is not directly observed (Alice and Bob are friends) but estimated from data (there is a relation between smoking and cancer), we can analyze - in addition to the network structure - the predictability of the nodes in the network. That is, we would like to know: how well can a given node in the network predicted by all remaining nodes in the network?
Predictability is interesting for several reasons:
1. It gives us an idea of how practically relevant edges are: if node A is connected to many other nodes but these only explain, let’s say, only 1% of its variance, how interesting are the edges connected to A?
2. We get an indication of how to design an intervention in order to achieve a change in a certain set of nodes and we can estimate how efficient the intervention will be
3. It tells us to which extent different parts of the network are self-determined or determined by other factors that are not included in the network
In this blogpost, we use the R-package mgm to estimate a network model and compute node wise predictability measures for a dataset on Post Traumatic Stress Disorder (PTSD) symptoms of Chinese earthquake victims. We visualize the network model and predictability using the qgraph package and discuss how the combination of network model and node wise predictability can be used to design effective interventions on the symptom network.
The datasets contains complete responses to 17 PTSD symptoms of 344 individuals. The answer categories for the intensity of symptoms ranges from 1 ‘not at all’ to 5 ‘extremely’. The exact wording of all symptoms is in the paper of McNally and colleagues.
## Estimate Network Model
We estimate a Mixed Graphical Model (MGM), where we treat all variables as continuous-Gaussian variables. Hence we set the type of all variables to type = 'g' and the number of categories for each variable to 1, which is the default for continuous variables level = 1:
For more info on how to estimate Mixed Graphical Models using the mgm package see this previous post or the mgm paper.
## Compute Predictability of Nodes
After estimating the network model we are ready to compute the predictability for each node. Node wise predictability (or error) can be easily computed, because the graph is estimated by taking each node in turn and regressing all other nodes on it. As a measure for predictability we pick the propotion of explained variance, as it is straight forward to interpret: 0 means the node at hand is not explained at all by other nodes in the nentwork, 1 means perfect prediction. We centered all variables before estimation in order to remove any influence of the intercepts. For a detailed description of how to compute predictions and to choose predictability measures, have a look at this paper. In case there are additional variable types (e.g. categorical) in the network, we can choose an appropriate measure for these variables (e.g. % correct classification, for details see ?predict.mgm).
We calculated the percentage of explained variance ($R^2$) for each of the nodes in the network. Next, we visualize the estimated network and discuss its structure in relation to explained variance.
## Visualize Network & Predictability
We provide the estimated weighted adjacency matrix and the node wise predictability measures as arguments to qgraph() to obtain a network visualization including the predictability measure $R^2$:
The mgm-package also allows to compute predictability for higher-order (or moderated) MGMs and for (mixewd) Vector Autoregressive (VAR) models. For details see this paper. For an early paper looking into the predictability of symptoms of different psychological disorders, see this paper. | 2021-08-01 07:52:43 | {"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": 0.8125913739204407, "perplexity": 759.1264119979708}, "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/1627046154163.9/warc/CC-MAIN-20210801061513-20210801091513-00439.warc.gz"} |
https://proxieslive.com/tag/modeling/ | ## Best practice for modeling data that is both general (default) and entity-specific
I have tried searching for good guidance on this already, but without much luck. Still, apologies in advance if this is duplicated elsewhere.
# The Problem
In a nutshell, we have external contractors that work on cases for our clients. We already have tables with contractor and client information in our SQL Server database. Going forward we’d like to store billing info in there too. Billing rates can differ for each client and contractor, but usually each client has a general “default” pay rate that applies to most contractors.
# Option A
The initial proposal was to create a new table with the following basic design:
clientContractorPay
• clientID – foreign key to client table
• contractorID – foreign key to contractor table
• basePay – pay rate for this client-contractor combination
• ... – several more (10+ and likely to grow) columns with supplemental pay rate details
• A unique index to help optimize lookup and also prevent multiple rows for a given client-contractor combination.
Contractor-specific pay rates would naturally be linked to the relevant contractor (and client). General (default) pay for a client would be stored in a row where contractorID is NULL. This is to avoid having to duplicate the same default pay for all contractors that don’t have specific exceptions.
# Option B
However, one of our senior devs has strong reservations about Option A. Their main argument is that using NULL in the contractorID column to mean “this is the default pay rate row” is unintuitive and/or confusing. In other words, it’s bad to assign meaning to NULL values.
Their counter proposal was to duplicate these new pay rate columns in the client table. The data stored there would indicate the default pay for each client, while contractor-specific exceptions would still live in the new table above.
# What To Do?
It seems clear both proposals would work just fine, but I have my own reservations about the second. Mainly it seems wrong to store the same type of data (client-contractor pay rate details) in multiple places, not to mention more complex logic to read/write this data. I also don’t like duplicating these new columns in both tables, since it would force us to add any future pay rate columns to both tables.
However, I can see my colleague’s point about potentially misusing NULL in this case. At the very least, it’s not immediately obvious that rows with a NULL contractorID contain default pay rates.
It’s been far too long since my database programming courses, so I’m not sure what the current best practice for this type of entity relationship is? I’m open to whatever is best long term, and would appreciate any expert guidance, especially with links to additional resources.
## Modeling a three-way association with optional relation
I have three tables (Parties, Categories and Products) which representing the following relationships:
• A product is classified by zero-one-to-many categories
• A category classifies zero-one-or-many products
Then, I have the party relationships:
• A product is classified by one-to-one party
• A party classifies one-to-many products
In other words, a product doesn’t have to be assigned a category.
## Design proposal
I have based my design on the proposal found here, but it’s not entirely applicable since want to enforce party_id for both Products and Categories:
How to model a three-way association that involves Product, Category and Label?
## Question
Is the usage of the three-way association table correct in my proposal to avoid the risk of having the application layer assigning a product to a category without enforcing the party_id?
## Threat modeling for visitor access control
I am trying to understand threat modeling but it seems too elasti from restrictive requirements to general requirements.
Now i am trying to understand it with some realistic examples. The first example which comes to my mind is physical access control of an office premise in which visitors have preapproved restrictive access and employees have unrestricted access. Each employee and legitimate visitor is given an id card to prevent this. Any official laptop should not go outside office without permission. Each laptop has an rfid tag to prevent this.
Can somebody help me understand threat model in this example? Or can somebody point me to where similar analysis has been done?
## Modeling cutaway prep scenes in Fate
Many crime stories feature a particular kind of foreshadowing and plot twist: Early in the story, you see a character doing something significant, but the scene cuts away before you find out what it is. You don’t find out what happened until the climax, when the character reveals the crazy plan that they set up after the cutaway. There’s often a brief flashback to the missing part of the establishing scene. It’s a staple of heist stories – Leverage does it almost every episode.
How can I model these cutaway prep scenes in Fate?
Are there any drawbacks or better alternatives of this non relational model ? I note that it remains easy to understand but the concern is with the code interacting with.
First, as I was introduced to the no-SQL world, in many occasions I confronted a one-to-many relationship between entities. Today, I have a really relatively cascading example that might grow in the future.
Based on functionalities assumptions, I came up with a simple Snowflake model. Specifically a cascading one-to-many relationships with some describing data.
[User] 1 — * [Session] 1 — * [Execution] 1 — * [Report]
The data model as it seems at first is easy to deal with, but I finally found that acting on data using Mongoose (a NodeJS library) can become complex and less performant, especially in a web application context (request and response cycle). The first way of thinking is to simple refer to parents by children in a normalization fashion. Another way to implement this data model is using document embedding approach: https://docs.mongodb.com/manual/tutorial/model-embedded-one-to-many-relationships-between-documents/ which is easier to interact with if you just model all in one entity; However this comes at the expense of performance; Because whenever you load a user, you load all sessions, executions and reports with it.
I found a compromise between a normalized model and the one using embedded documents; Modeled here:
The compromise consist of embedding a minimal variant of the child entity like Executions of type ExecutionsMini in Sessions. While maintaining the child entity Executions separate.
The concern grows because between Users and Loggings, there might be other entities added, in a one-to-many kind or not, and this could complex more the solution (not the data model).
## Modeling and managing attack surface around individual finance [on hold]
I want to protect myself from fraud and identity theft.
While there are on the Internet plenty of arbitrary collections of precautionary tips, I want to make rational, fully informed choices to manage the risk that I suffer from financial crime. (I’m not an unusually valuable target for crime; I just want to make responsible choices.)
Essentially, I want to know how my choices will affect the attack surface around my individual finances and crime-relevant information. Having a good model of this attack surface would allow me to answer, for example, these questions:
• How do I evaluate a bank or credit union for its information security practices?
• How do I choose among email service providers and email information security practices?
• What practices around financial transactions minimize this attack surface?
I’m not looking for answers to these questions in particular, but rather how to model the attack surface they are asking about.
So, my question is:
When a security expert wants to model a complex attack surface across multiple institutions and information systems, how does he or she go about doing it? What steps does he or she go through? Can a technically capable but non-expert follow these steps?
## Modeling a set of probabilistic concurrent processes
I’m looking into discrete-time Markov chains (DTMCs) for use in analyzing a probabilistic consensus protocol. One basic thing I haven’t been able to figure out is how to model a set of independent processes: consider $$N$$ processes. These processes will concurrently execute a series of identical instructions labeled $$0, 1, 2, 3,$$ etc. and all are starting in instruction $$0$$. When probability is not involved, modeling this is simple: it’s a state machine which branches nondeterministically off the start state to $$N$$ different states, where in each of those $$N$$ states a different process was the first to execute instruction $$0$$. What do we do when probability is involved? Do we do the same thing with $$N$$ states branching from the start state, where the probability of transitioning to each state is $$\frac{1}{N}$$? As in, it’s uniformly random which process was the first one to execute instruction $$0$$?
Is this like taking the product of the state machines of each process?
I’m using a DTMC here, would I gain anything by moving to a CTMC if I don’t care about anything beyond the global order of execution?
Bonus question: assigning probabilities to whichever action (process executing an instruction) is taken first seems like a generalization of the non-probabilistic notion of fairness; if it is, what is the formal definition of this generalized notion of probabilistic fairness?
## The Challenges that Practitioners face with on Software Modeling
We have recently released a survey on understanding the challenges that practitioners face with in their software modeling activities. The survey takes approximately 2-5 minutes to complete.
We would be so grateful if you could separate a few minutes of you to participate in our research.
## User experience and Agile Modeling
I have a question regarding architect modeling and how an UX designer can use them to deliver better documentation to the developers/software architects. Has anyone used a component diagram to visualize the physical components in a system?
Thank you.
## Cassandra data modeling with mutable attribute rows
We set up da timeseries database. Every measurement has also attributes,that may change later, after storage (eg. deviceName may changed) .
As cassandra recomends an query first approach, I thought about pack that attributes into the table, but how can I handle updates?
• Database update may result in inconsistency and heavy code.
• Joins are not possible in cassandra
• Merging the mutable attributes after the database reads in application layer seems possible, but is it the best way? | 2020-06-06 13:28:59 | {"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": 9, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.24487051367759705, "perplexity": 1727.0281193764165}, "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/1590348513321.91/warc/CC-MAIN-20200606124655-20200606154655-00277.warc.gz"} |
https://groupprops.subwiki.org/w/index.php?title=Importance_of_characteristicity&mobileaction=toggle_view_mobile | # Importance of characteristicity
## Introduction
Characteristicity is one of the most important subgroup properties, probably second only to normality in the frequency with which it occurs in group theory.
One of the many basic questions in group theory is the deep similarity as well as contrast in the definitions of normal and characteristic subgroup: a priori, the notion of characteristic subgroup (as being a subgroup invariant under all automorphisms) seems more natural than the notion of normal subgroup (as being a subgroup invariant under only the inner automorphisms). However, it is rather surprising that characteristicity is not as common or important as it may seem -- it is normality that plays the more crucial role (refer importance of normality).
Nonetheless, characteristicity is still fairly important, as we shall explore in this article.
## Subgroup-defining functions yield characteristic subgroups
### What is a subgroup-defining function?
Further information: subgroup-defining function
A subgroup-defining function is a function that takes in a group and outputs a unique subgroup of that group. For instance:
### All subgroup-defining functions yield characteristic subgroups
Subgroup-defining functions satisfy the condition of being invariant under isomorphism: any isomorphism of groups, preserves the function. Hence, in particular, the subgroup must be invariant under all automorphisms, and hence, must be a characteristic subgroup.
A natural reverse question is: does every characteristic subgroup arise as the image of some subgroup-defining function? The answer is yes if we are allowed unlimited power in describing the subgroup-defining function. However, if we are restricted to certain kind of languages (for instance, the language of first-order logic, the language of second-order logic, the language of set theory) then we cannot hope to describe every characteristic subgroup of a group.
## Importance as the left transiter of normality
One major defect with normality is that it is not transitive. That is, if $H \triangleleft K \triangleleft G$ are subgroups, it is not necessary that $H \triangleleft G$. This lack of trnaisitivity of normality often comes in the way of proving results inductively, as well as in trying to use normality at one place to force normality at another.
Characteristic subgroups provide a neat way out. This essentially stems from the fact that every characteristic subgroup of a normal subgroup is normal. In particular, if $H \le G$ is normal, then the image of $H$ under any subgroup-defining function, is also normal in $G$. Thus, the center of $H$, the commutator subgroup of $H$ are all normal in $G$. | 2019-10-18 01:39:22 | {"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": 8, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8823835849761963, "perplexity": 408.84906449808983}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-43/segments/1570986677412.35/warc/CC-MAIN-20191018005539-20191018033039-00416.warc.gz"} |
https://www.physicsforums.com/threads/limit-of-0-0-type.347673/ | # Limit of 0/0 type
1. Oct 21, 2009
### faen
1. The problem statement, all variables and given/known data
lim x->2 (t^3 - 8)/(t^4 - 16)
2. Relevant equations
3. The attempt at a solution
Well, i just cant find the common factor in the numerator and denomenator. I can split (t^4 - 16) to (t^2 - 2)^4 but i cant find any other factors in the numerator.
Thx for any help :)
2. Oct 21, 2009
### emyt
multiply by a conjugate or try (t^2+4)(t^2-4)
3. Oct 21, 2009
### Fightfish
There's a rather obvious common factor there. Try factorising 8 and 16.
Alternatively you could use l'Hôpital's rule
4. Oct 21, 2009
### faen
Yeah i tried (t^2+4)(t^2-4), but i cant find the same factor in the numerator. The t^3 term complicates the matter cause what can i multiply with itself to get t^3 without getting a more complicated factor.. In other words, im unable to express; (t^3 - 8) in any other way.
I tried conjugate but didnt work for me, and we didnt learn L'Hopitals yet so not allowed to ues it.
5. Oct 21, 2009
### Staff: Mentor
t4 - 16 $\neq$(t2 - 2)4
Note that the polynomial on the left side is of degree 4, while the one on the right is of degree 8. That should have been a clue that something is wrong.
You should be thinking "difference of squares" and "difference of cubes" for your factoring.
6. Oct 21, 2009
### Staff: Mentor
t2 - 4 can be factored. Also, the difference of cubes can be factored. a3 - b3 = (a - b)(a2 + ab + b2).
7. Oct 21, 2009
### faen
Ah, yeah, i was thinking that it would equal to (t^2 - 4)^2 but the minus sign would be different among the factors.
Anyway i still cant figure it out, can u help me a bit more :p?
8. Oct 21, 2009
### faen
now i think i got it.. thanks a lot! :D
9. Oct 21, 2009
### faen
Ok i found that (t-2)(t^2 +2t +4) = t^3 - 8, and now the two t-2 factors in the numerator and denomenator cancels. however im still stuck with (t^2 - 4) factor which tends to 0 while t tends to 2. So im stuck again.
If someone could just solve the: lim x->2 (t^3 - 8)/(t^4 - 16) itd be of great help.
Last edited: Oct 21, 2009
10. Oct 21, 2009
### Staff: Mentor
Then your factoring of t4 - 16 is incorrect. Show me how you factored this.
11. Oct 21, 2009
### faen
you are right, ok i finally solved it. Thanks :) | 2018-02-22 13:17:17 | {"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": 0.6532663106918335, "perplexity": 1768.3340946984338}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-09/segments/1518891814105.6/warc/CC-MAIN-20180222120939-20180222140939-00205.warc.gz"} |
https://support.bioconductor.org/p/99110/ | Question: Is the CPM filtering method from NOISeq unsupervised?
0
2.2 years ago by
Mau20
Mau20 wrote:
Hi,
Previous questions (for example A: Why does the filtering of lowly expressed genes for analysis with edgeR must be or edgeR cpm filter with >1 factor) and at least this paper have stressed the importance of using unsupervised filtering methods for lowly expressed genes in a DE analysis. To my understanding, this means that such filters must have no knowledge of which condition is applied to each sample.
I recently came across the Bioconductor package NOISeq. I am interested in its functions to explore data pre-DE analysis. Section 4.2 "Low-count filtering" of its manual explains the filtering methods available in the package. Method 1 says:
"CPM (method 1): The user chooses a value for the parameter counts per million (CPM) in a sample under which a feature is considered to have low counts. The cutoff for a condition with s samples is CPM × s. Features with sum of expression values below the condition cutoff in all conditions are removed."
From this, I understand that filtering is done considering the samples in each condition. I was wondering if this filtering strategy is unsupervised, or statistically sound?
Thank you
modified 2.2 years ago by James W. MacDonald51k • written 2.2 years ago by Mau20
Answer: Is the CPM filtering method from NOISeq unsupervised?
0
2.2 years ago by
United States
James W. MacDonald51k wrote:
It's fine, so long as you don't use the part about filtering on coefficient of variation within condition, which obviously uses information about your groups to filter genes. The idea is that you don't use information about your groups to filter, because that might bias you towards selecting genes that fulfill the criteria you are going to use to test for differential expression.
A simple heuristic is to avoid any filtering method that requires you to say what group each sample is in, as well as any method that uses measures of variance for each gene. The former may bias you towards genes that have a higher likelihood of being significant (e.g., you are 'snooping'), and the latter may bias you towards genes that have a higher variance (and most of the RNA-Seq methods out there expect to get an unbiased measure of variance).
Thanks for your input. I agree with your general rule, but isn't the criteria I asked about breaking it? It seems to me that it filters based on expression levels within each condition (group)
OK, yes. I missed the part about CPM per condition. I usually just do what Ryan Thompson suggested at one point, which is
plot(density(rowSums(cpm(<your counts go here>, log = TRUE))))
There is usually a bimodal distribution, and I make the assumption that the two peaks represent unexpressed and expressed genes, respectively. If you then choose a rowSum that splits the two, then you are +/- keeping the expressed genes, and getting rid of the unexpressed genes. | 2019-10-15 23:58:50 | {"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": 0.5945026278495789, "perplexity": 1506.7751683892645}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-43/segments/1570986660829.5/warc/CC-MAIN-20191015231925-20191016015425-00489.warc.gz"} |
https://stats.stackexchange.com/questions/66081/difficulty-in-understanding-a-vector-quantization-algorithm | # Difficulty in understanding a vector quantization algorithm
Neural gas for vector quantizationpaper explains a technique for symbolizing or quantizing data. Algorithm presents the algorithm in Section 4. An application in EEG data symbolization is presented Application. In the Application it is shown that an n- dimensional data like EEG recordings is vectorized to 1D. But Neural gas is basically an unsupervised clustering algorithm where data can be assigned to more than one cluster.
The definition of Vector Quantization (VQ) is the process of quantizing n-dimensional input vectors to a limited set of n-dimensional output vectors referred to as code-vectors. The set of possible code-vectors is called the codebook. The codebook is usually generated by clustering a given set of training vectors (called training set), the codebook is then used to quantize input vectors.
What I do not understand is that how a multi dimensional data is converted to single dimension as in the Application paper and the module where the symbols are assigned. In the Algorithm, I did not understand what is "reference vector $w_{c_i}$, the set $A$ and Step 3 where it says to order all elements of $A$ according to distance. What is the distance measure here?
Can somebody please explain the algorithm in simple terms how the symbolization is done? Thank you
Neural Gas is just a soft version of k-means clustering. in k-means , you just move the closest reference vector towards your training pattern, in neural gas, you are also moving the other reference vectors ( weighted according to their "closeness" , ie 1 -nearest, ...,k kth-nearest refence vector).
the point is that this soft version can [potentially/and more slowly] find better clusters because it avoids getting stuck in local minima... so after training you are just using the nearest reference vector to symbolise the data
reference vector w_{c_i} is then one of the reference vectors (code-vectors), in order of closeness ( just Euclidean distance) to the current training vector.
so "where it says to order all elements of A according to distance. What is the distance measure here?" -. A is just your codebook (of refence vectors). for each training pattern you rank the refence vectors according to how close they are to that training pattern, and apply the training rule. the distance measure is just euclidean distance.
symbolisation is done by just returning the index into your codebook of vectors... it is not turning it into a 1 dimensional space . its just like reading off the position in a dictionary - the position does not encode any "useful" information (ie about the meaning off the word)
• Thank you for your reply.(A)I thought the code vectors are the number of symbols used to represent the data say symbols from 1-10 range.Reference vector is still unclear.(B)If symbolization does not provide any useful information then how is it used in the Application paper for activity recognition.There is another symbolization method saxproject.org (symbolic aggregate) which has been used in dimensional reduction,anomaly detection etc.All these are basically clustering algo.Could you please let me know of the applications of using neural gas or any other clustering algorithm. – SKM Aug 1 '13 at 15:45
• code vectors=reference vectors is the multi dimensional "prototype"...eg you have a salesman travelling around america, and your vector is the geographical position the salesman is each night ( x,y coordinates say). then your codebook vectors may well end up being the geographic position of the major cities. your symbol is then the city name. then each element in your codebook is a city, and the code vector is the x,y coordinates of the city, and the symbolisation is the city name. so the "1-d" sequence is New York, Boston, LA, Chicago... – seanv507 Aug 1 '13 at 19:25
• symbolisation: a) clearly the name "New York" doesn't tell you anything about geographic position,distance between cities etc (or anything else). however, the sequence of names is still providing useful information... – seanv507 Aug 1 '13 at 19:31
• Thank you for your clarification,shall be grateful for few more(A)size of code vectors = size of input data?(B)How does the algorithm decide upon the number of symbols to represent the data?So,let a sequence be 12145, then both the 1's would refer to the same information in the code vector?(C)I read that neural gas assigns index of the nearest winning code vector, but in this case there will be ambiguity.If vectorization does not help in dimension reduction and it just gives information about temporal ordering, then how come it is used in data mining when the actual info abt "data" is lost? – SKM Aug 1 '13 at 23:38
• each code vector corresponds to a centre around which the data cluster, a prototypical point[eg New Yok City Centre coordinates], so yes is exactly the same dimension as the input patterns. B) you have to decide the number of codebook vectors before hand, then you run neural gas [ ie trial and error!] . You are replacing vectors (eg x,y coordinates) by a symbol, eg New York, Boston,... so the symbols 1,2,1,4,5 = New York, Boston, New York, LA, Chicago... no ambiguity... and in terms of categorising sequences/machine learning, the symbol description may be sufficient – seanv507 Aug 2 '13 at 8:43 | 2019-07-23 00:46:14 | {"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": 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": 0.4566277265548706, "perplexity": 1347.1800720979577}, "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/1563195528635.94/warc/CC-MAIN-20190723002417-20190723024417-00010.warc.gz"} |
http://blog.codebender.cc/2014/03/07/lesson-1-inputs-and-outputs/ | [tutorial] Lesson 1: Inputs and Outputs
Introduction
A microcontroller (uC) is a computing device. Its main job is to perform computations… it takes some data from its memory, manipulates them, and then stores them back to memory. But having data in a uC’s RAM doesn’t do much good. In order to make use of that information, we have to be able to take it out of the uC and realize it in some way. To do that, a uC has outputs that can transfer out bits of information to devices that can display it, make sound or light, cause something to move, etc.
In many occasions though, we don’t have our data ready in advance. Some times, we want to transfer data in the uC as they become available. Other times, we want to let the user interact with the uC, or we want to sense different aspects of the environment around us. For those reasons, a uC also has inputs that let it take in information.
We are given access to the inputs and outputs (IO) of a uC through the pins of its IC package. The circuit behind each IO pin is built such that the user can configure that pin as either input or output.
One pin, multiple uses
A uC has many capabilities that require many pins to accommodate them all. Often that number of pins is more than what a specific IC package can carry. To solve that problem, some pins are being multiplexed. This means that different circuits are guided to the same pin. Then, depending on the operation we are carrying out, one of those circuits is connected to the pin at any time. For example, in the ATMEGA32U4 we see that pin 21 can be used either as an IO pin (D1), or as a serial output line (TX).
On an Arduino board, the pins of the uC are broken out and connected to female pin headers so they are more accessible. They are organized into groups and are given new names with which they are going to be referenced in the code.
An analog signal describes a physical variable that varies continuously with another variable (often time). Examples of analog signals are the intensity of light of an LED as its current increases, the current going through a resistor as we vary the voltage, or the temperature in a room as the time passes.
An analog signal can take on infinite values. Unfortunately a uC cannot represent an infinite number of values. So what we do when we want to sample an analog signal is perform an Analog to Digital Conversion (ADC). This basically means that we pick specific values in a region and then we use a series of bits to represent those values only. Any intermediate values are truncated to the closest representable value.
Let’s say we have an analog sensor that gives an output in the region of $0-5 V$ and an A-to-D converter with a $10bit$ resolution. This means that we can represent $2^{10}$ or $1024$ values. The procedure goes as follows: We divide the $0-5 V$ region into $2^{10}$ sub-regions of $4.88 mV$ width, and assign to each one a bit combination. So we get: $[0, 4.88mV) \rightarrow 0000000000,$ $[4.88mV, 9.76mV) \rightarrow 0000000001,$ $[9.76mV, 14.64mV) \rightarrow 0000000010…$ and we keep adding ones $(1)$ up to $[4.995V, 5V] \rightarrow 1111111111$.
Now, let’s say we sample the sensor’s output and the value we get is $2.932 V$. This value falls into the $[2.930V, 2.935V)$ sub-region. So the bit combination we get back from the A-to-D converter is $1001011000$, which is the number $600$ in decimal. (If interested in how to convert a binary number to a decimal, read [here](http://www.wikihow.com/Convert-from-Binary-to-Decimal).)
The analog pins on an Arduino board are named Ax, where x is a number. We use these names to reference the analog pins in the code. We can read a value from an analog input pin with the analogRead function. This function requires a pin number as an argument, and returns an int value (the value read from the A-to-D converter).
Analog Outputs
An analog output is produced by a process called Digital to Analog Conversion (DAC). A D-to-A converter takes in a binary number and outputs an analog signal (a constant value, e.g. $0100011111 \rightarrow 1.4V$). Unfortunately, unlike reading analog inputs (ADC), writing analog outputs (DAC) isn’t something that all uC support. If that is the case with your own uC then you will need to buy a DAC module.
On the other hand, a common feature on many uC is the Pulse Width Modulation (PWM) capability. A PWM(odulated) signal is a square wave (shown above), a.k.a. pulse train. It is a periodic signal, so it has a frequency $f$, measured in either $kHz$ or $MHz$, and a period $T=1/f$, measured in either $ms$ or $us$. A pulse in a PWM signal has an on time and an off time. The on time $t_1$ is the time that the pulse stays HIGH. The off time $t_2=T-t_1$ is the time that the pulse stays LOW. The last but not least parameter that describes a PWM signal is the duty cycle which defines the percentage of time that one pulse stays on, $\frac{100t_1}{T} \%$.
More on PWM Signals
The interesting thing about the duty cycle is that varying it from $0$ to $100\%$, it changes the average value of the PWM signal.
A PWM signal is not an analog signal, but we can make one out of it. We can use a special circuit to smooth the PWM signal out and make its average value appear (but this is a subject for a whole other tutorial). Also, this is sort of where the analogWrite function got its name from.
The PWM pins on an Arduino board are designated with a ~ symbol. In the code, we use the analogWrite function to output a PWM signal. Its arguments are the pin number to which to write, and the duty cycle defined here as a number between 0 and 255.
Digital Inputs
A digital signal, unlike an analog signal, cannot take on infinite values. The digital signals of interest to us take only on two values referred to as 0 and 1, or LOW and HIGH, respectively. Examples of digital signals are the on-off state of an oven, the movement-no movement state of a PIR sensor, or the pressed-not pressed state of a push button.
We declare an IO pin as input or output through the use of the pinMode function. This function takes two arguments. A pin number, and one of the following constants to declare the pin’s mode: INPUT, OUTPUT, or INPUT_PULLUP.
Pull-up/down Resistors
A very simple form of input component is a switch. A switch can be either open or closed. It has two leads, one of which we connect to a point of reference (GND or VCC), and the other to an input pin on our uC. When the switch is closed, the input pin is connected to the point of reference and we read an appropriate value. But when the switch is open, the input pin is left floating and is picking up electrical noise from the environment. As a result reading from the input pin when the switch is open gets us back a random value. To solve this problem, we make use of a resistor that we connect to another point of reference and the input pin. Now, when the switch is open, the input pin connects to a point of reference and we read a known value. A resistor, when used in this manner, is called either pull-up or pull-down, because it pulls the input pin either up to VCC or down to GND.
Arduino boards include internal pull-up resistors that we can make use of, if necessary. In that case, we can actually connect a switch alone on an input pin. We just have to remember to enable the pin’s pull-up resistor in the code.
We can read a digital input with the digitalRead function. It takes a pin number as argument, and returns HIGH or LOW.
Digital Outputs
Finally, we get to digital outputs. Not much is left to say. In the code, we declare an IO pin as output with the pinMode function, and we write to a digital output with the digitalWrite function. The digitalWrite function takes two arguments. One is a pin number and the other is the state to write to the pin, HIGH or LOW.
Conclusion
Inputs or outputs, analog or digital, signals of any form… now you know enough to interface them all. Now you have the power. I know these are a lot to take in, but hang in there. The possibilities that open up just with those things alone are countless. Next time, we’ll see already some applications! 😉
Images of specific components are courtesy of arduino.cc, sparkfun.com, and adafruit.com. Others are in the public domain.
Images are CC BY-NC-SA 3.0.
Stay tuned:
* we promise that we won´t spam you, never. | 2019-08-20 09:39:54 | {"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": 28, "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": 0.45034605264663696, "perplexity": 786.4988486019612}, "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/1566027315321.52/warc/CC-MAIN-20190820092326-20190820114326-00513.warc.gz"} |
http://mathhelpforum.com/number-theory/211607-inequality-primes-print.html | # Inequality with primes
Let $a$ be a positive integer and $q_i,i=1,2..,n$; $n\ge2$ distinct primes
If $\prod_{i=1}^{n}q_i = 3^a-2$
then $\prod_{i=1}^{n}(q_i-1) > \frac{2}{3}3^{a}$ | 2016-07-30 04:34:44 | {"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": 0.7630282044410706, "perplexity": 1611.0843270992953}, "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-2016-30/segments/1469257832939.48/warc/CC-MAIN-20160723071032-00161-ip-10-185-27-174.ec2.internal.warc.gz"} |
https://socratic.org/questions/how-do-you-simplify-x-3-8-x-2-x-2-2x-4-div-x-2-4-x-6 | How do you simplify (x^3+8)*(x-2)/(x^2-2x+4)div(x^2-4)/(x-6)?
Sep 27, 2016
$x - 6$
Explanation:
$\textcolor{red}{\left({x}^{3} + 8\right)} \cdot \frac{x - 2}{\textcolor{b l u e}{{x}^{2} - 2 x + 4}} \div \frac{\textcolor{g r e e n}{{x}^{2} - 4}}{x - 6}$
In algebraic fractions you want to factorize as much as possible.
$\frac{\textcolor{red}{\left(x + 2\right) \left({x}^{2} - 2 x + 4\right)}}{1} \times \frac{x - 2}{\textcolor{b l u e}{{x}^{2} - 2 x + 4}} \times \frac{\textcolor{g r e e n}{x - 6}}{\left(x + 2\right) \left(x - 2\right)}$
Now that everything is expressed as factors you may cancel:
$\frac{\cancel{x + 2} \cancel{{x}^{2} - 2 x + 4}}{1} \times \frac{\cancel{x - 2}}{\cancel{{x}^{2} - 2 x + 4}} \times \frac{x - 6}{\cancel{x + 2} \cancel{x - 2}}$
$= \left(x - 6\right)$ | 2019-05-23 01:00:57 | {"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": 0.8786067366600037, "perplexity": 2442.106299647496}, "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-2019-22/segments/1558232256997.79/warc/CC-MAIN-20190523003453-20190523025453-00308.warc.gz"} |
https://socratic.org/questions/how-can-i-calculate-the-percent-composition-of-c4h4o | # How can I calculate the percent composition of C4H4O?
##### 1 Answer
Jun 4, 2014
To find the percent composition of C₄H₄O, you divide the total mass of each atom by the molecular mass and multiply by 100 %.
% by mass = $\text{mass of component"/"total mass}$ × 100 %
Mass of 4 C atoms = 4 C atoms × $\left(12.01 \text{u")/(1"C atom}\right)$ = 48.04 u.
Mass of 4 H atoms = 4 H atoms × $\left(1.008 \text{u")/(1"H atom}\right)$ = 4.032 u
Mass of 1 O atom = 1 O atom × $\left(16.00 \text{u")/(1"O atom}\right)$ = 16.00 u
Mass of 1 C₄H₄O molecule = (48.04 + 4.032 + 16.00) u = 68.07 u
% of C = $\text{mass of C"/"total mass}$ × 100 % = $\left(48.04 \text{u")/(68.07"u}\right)$ × 100 % = 70.57 %
% of H = $\text{mass of H"/"total mass}$ × 100 % = $\left(4.032 \text{u")/(68.07"u}\right)$ × 100 % = 5.923 %
% of O = $\text{mass of O"/"total mass}$ × 100 % = $\left(16.00 \text{u")/(68.07"u}\right)$ × 100 % = 23.50 % | 2019-10-21 09:47:20 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 10, "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": 0.7649385333061218, "perplexity": 1441.953433506798}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-43/segments/1570987769323.92/warc/CC-MAIN-20191021093533-20191021121033-00043.warc.gz"} |
https://novelsthatencourage.com.au/its-celebration-time-a-quiz-giveaway-sale/?replytocom=1366 | # It’s Celebration Time–A Quiz, Giveaway + Sale!
It’s been a WHOLE YEAR since I released my first novel, Jayne’s Endeavour!!!
I can hardly believe it’s been that long already. It’s seems like not so long ago when I was eagerly opening my first boxful of Jayne’s Endeavour, signing copies, sending them out, and nervously anticipating readers’ reactions. 😀
And I decided that today was worth celebrating! God has definitely blessed me richly throughout this whole adventure! So I pulled together a little celebration for my readers (who I very much appreciate!). Because who doesn’t like a chance to celebrate and have fun together!?
## It’s time for a Book Anniversary Party!!
And (in case any of you were wondering), here’s what’s on today’s party list:
• A quiz for Jayne’s Endeavour fans to test their knowledge (and memories!)
• A great sale on both the ebook and paperback versions of Jayne’s Endeavour (I think it’s the biggest discount I’ve had so far)
• And last but not least–a JOY themed giveaway (you’ll have to keep scrolling to find out the details!)
Since quizzes are fun (and I haven’t done one on my blog before), let’s start with that first! 😉 And make sure you remember how many points you get, because you can use your score to get extra entries in the giveaway later on!
Welcome to the Jayne's Endeavour Anniversary Quiz!!!
Click on "next" to see the first question!
If you haven’t read Jayne’s Endeavour yet, here’s your chance to grab either the paperback or the ebook version at a bargain price (it won’t appear like it’s on sale on Amazon–but it is!). For the next few days, US readers can buy a copy for only $10.95 (normally$14.95) or an ebook for just \$3.49. 😀 And Australian readers can purchase their copy straight from my store.
Sale for American Readers: paperback | ebook
Sale for Australian Readers: paperback | ebook
## And now . . . it’s time for the giveaway!!
Since Jayne’s Endeavour is the first book of The JOY Series, I thought it was appropriate to do a joy-themed giveaway for my readers as a thank you.
I’m really, really grateful to everyone who has taken the time to read my novel, left reviews on Amazon, Goodreads, my website etc, and I’m also really grateful to the many people who have been so encouraging–both before and after the release of Jayne’s Endeavour. Your enthusiasm and kind words have meant a lot and been such a blessing to me, so thank you!
Today’s giveaway is open to international entries, but only entrants in America, Australia, or Canada will receive the physical prize. 🙂 The physical prize is a “Joy of the Lord” wooden plaque (11″ by 8″) and a “Joy of the Lord” violet pen. A winner from a different country than the ones I mentioned above will receive an ebook of my choice or a voucher. 😀
a Rafflecopter giveaway
Thank you for joining in the celebration!! I hope you’ve had a fun! Don’t forget to grab a copy of Jayne’s Endeavour before the sale ends. And be sure to let me know in the comments what score you got in the quiz!
Oh! And before you go, don’t forget to help yourself to a piece of strawberry sponge cake . . . I’ve heard it’s delicious 😛
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1. 21 points! Not bad for having only read the book once a few months ago, if I do say so myself Love the quiz, thanks for putting it together!
Still looking forward to Book 2….
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Good job!! 21 points is quite impressive 😉 I’m so glad you enjoyed taking the quiz. I had fun coming up with the questions!
And I’m looking forward to sharing it with you!!
2. Congratulations! What a great milestone. Thanks for the fun quiz and the piece of sponge. yum. I am very much looking forward to reading book 2. Keep writing!
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Thank you, Mum!! I’m glad you enjoyed the quiz (and the delicious sponge!! 😉 ). Thanks for the special pizza last night to help celebrate!! And thank you for always encouraging me ❤️
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I’m glad to hear you had fun, Katja!! I’m enjoying hearing how people did on the quiz! 😀 And re-reading books can be fun. You notice more things the second time around!
3. That’s so fun Lauren! Congratulations I took the quiz and got 14 points… which shocked me because I’ve never read the book!
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4. Congratulations, Lauren! So exciting that it’s been a year!! I can’t wait to read the next book in the series, and like Katja said – I should re-read Jayne’s Endeavour!
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Thank you so much, Chelsea! 😀 It’s been a lot of fun be reminded of the release!! Lots of happy memories. Happy re-reading (I can’t wait to release book 2!)!
5. I canNOT believe it’s already been a year!!! I am SO looking forward to books 2 and 3. I only got 13 points on the quiz, so I’m thinkin’ it’s time for a re-read =)
And that sponge cake in your post looks SO DELICIOUS!!! Wowza! Is that something you made, or did you just find a picture of it? Because if you made that, I’d surely like the recipe! 😀 =)
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I know!!! Time certainly seems to fly by! Thank you for being so eager for the next books–it really spurs me on! It’s so fun seeing how many points different people are getting. My family were sitting around last night taking the quiz and competing to get the highest scores. My dad and one of my sisters tied for 23 points. 😉
Oh! I know!! Every time I see the picture of the sponge cake, I think how yummy it looks!! But I’m so sorry that I don’t have the recipe–otherwise I would certainly send it to you. 😀
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6. Yay!!!!!!! This book was such a joy to read. Congratulations Lauren!!!! I’m so very happy for you!!!
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7. A year! Wow! Congratulations, Lauren! My oldest niece keeps asking me when Book 2 is coming out. 🙂
I was able to get 16 right on the quiz which surprised me since I haven’t read the book in a year and can’t remember some things. 🙂 Time to join Katja and Chelsea in a reread. 😉
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Thank you so much, Rebekah! That’s so nice to hear that your niece is so keen for book 2! 😀 Haha! I’m not sure how good my memory would be on this sort of quiz after a year either 😉 So well done on getting 16 points!! 😀 😀
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Thank you for stopping by, Anna Grace! 😀 Oh, that sounds exciting (and it’s on sale now for a great price)! 😉 I hope it’s a blessing to you whenever you get to read it!
8. Wow! Can’t believe it’s been a year!!
I just did the quiz and was shocked to get 20 points, considering I hadn’t read it for almost a year! Thanks for putting it together. It was fun!
Looking forward to book 2. All the best with that!
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I know!! It’s amazing that it’s been a year already. 😀 Great job on the quiz!! 20 points is quite impressive! Aside from family members, you have one of the highest scores so far 😉
Thank you–can’t wait to release it so you can read it! 😀 | 2022-08-17 20:40:29 | {"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": 0.2522577941417694, "perplexity": 3725.2315350342574}, "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-2022-33/segments/1659882573104.24/warc/CC-MAIN-20220817183340-20220817213340-00465.warc.gz"} |
https://www.transtutors.com/questions/vogel-inc-an-s-corporation-for-five-years-distributes-a-tract-of-land-held-as-an-inv-2576195.htm | # Vogel, Inc., an S corporation for five years, distributes a tract of land held as an investment t...
Vogel, Inc., an S corporation for five years, distributes a tract of land held as an investment to Jamari, its majority shareholder. The land was purchased for $45,000 ten years ago and is currently worth$120,000. As a result of the distribution, what is Vogel's recognized capital gain? How much is reported as a distribution to shareholders? What is the net effect of the distribution on Vogel's AAA? Assume instead that the land had been purchased for $120,000 and was currently worth$45,000. How much would Vogel recognize as a loss? What would be the net effect on Vogel's AAA? What would be Jamari's basis in the land? | 2018-06-20 10:55:34 | {"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": 0.20710903406143188, "perplexity": 3855.815851332267}, "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/1529267863518.39/warc/CC-MAIN-20180620104904-20180620124904-00359.warc.gz"} |
https://biz.libretexts.org/Under_Construction/Map%3A_Business_Strategy/24%3A_Monetary_Policy_Transmission_Mechanisms/24.2%3A_How_Important_Is_Monetary_Policy%3F | # 24.2: How Important Is Monetary Policy?
learning objectives
1. What types of evidence can strengthen researchers’ conviction that a reduced-form model has the direction of causation right, say, from M to Y? How?
2. What evidence is there that money matters?
Early Keynesians believed that monetary policy did not matter at all because they could not find any evidence that interest rates affected planned business investment. Milton Friedman and Anna Schwartz, another monetarist, countered with a huge tome called A Monetary History of the United States, 1867–1960 which purported to show that the Keynesians had it all wrong, especially their kooky claim that monetary policy during the Great Depression had been easy (low real interest rates and MS growth). Nominal rates on risky securities had in fact soared in 1930–1933, the depths of the depression. Because the price level was falling, real interest rates, via the Fisher Equation, were much higher than nominal rates. If you borrowed $100, you’d have to repay only$102 in a year, but those 102 smackers could buy a heck of a lot more goods and services a year hence. So real rates were more on the order of 8 to 10 percent, which is pretty darn high. The link between interest rates and investment, the monetarists showed, was between investment and real interest rates, not nominal interest rates.
As noted above, the early monetarists relied on MV = PY, a reduced-form model. To strengthen their conviction that causation indeed ran from M to Y instead of Y to M or some unknown variables A…Z to M and Y, the monetarists relied on three types of empirical evidence: timing, statistical, and historical. Timing evidence tries to show that increases in M happen before increases in Y, and not vice versa, relying on the commonplace assumption that causes occur before their effects. Friedman and Schwartz showed that money growth slowed before recessions, but the timing was highly variable. Sometimes slowing money growth occurred sixteen months before output turned south; other times, only a few months passed. That is great stuff, but it is hardly foolproof because, as Steve Miller points out, time keeps on slipping, slipping, slipping, into the future.www.lyricsfreak.com/s/steve+miller/fly+like+an+eagle_20130994.html Maybe a decline in output caused the decline in the money supply by slowing demand for loans (and hence deposits) or by inducing banks to decrease lending (and hence deposits). Changes in M and Y, in other words, could be causing each other in a sort of virtuous or pernicious cycle or chicken-egg problem. Or again maybe there is a mysterious variable Z running the whole show behind the scenes.
Statistical evidence is subject to the same criticisms plus the old adage that there are three types of untruths (besides Stephen Colbert’s truthiness,en.wikipedia.org/wiki/Truthiness of course): lies, damn lies, and statistics. By changing starting and ending dates, conflating the difference between statistical significance and economic significance,www.deirdremccloskey.com/articles/stats/preface_ziliak.php manipulating the dates of structural breaks, and introducing who knows how many other subtle little fibs, researchers can make mountains out of molehills, and vice versa. It’s kinda funny that when monetarists used statistical tests, the quantity theory won and money mattered, but when the early Keynesians conducted the tests, the quantity theory looked, if not insane, at least inane.
But Friedman and Schwartz had an empirical ace up their sleeves: historical evidence from periods in which declines in the money supply appear to be exogenous, by which economists mean “caused by something outside the model,” thus eliminating doubts about omitted variables and reverse causation. White-lab-coat scientists (you know, physicists, chemists, and so forth—“real” scientists) know that variables change exogenously because they are the ones making the changes. They can do this systematically in dozens, hundreds, even thousands of test tubes, Petri dishes, atomic acceleration experiments, and what not, carefully controlling for each variable (making sure that everything is ceteris paribus), then measuring and comparing the results. As social scientists, economists cannot run such experiments. They can and do turn to history, however, for so-called natural experiments. That’s what the monetarists did, and what they found was that exogenous declines in MS led to recessions (lower Y*) every time. Economic and financial history wins! (Disclaimer: The author of this textbook is a financial historian.) While they did not abandon the view that C, G, I, NX, and T also affect output, Keynesians now accept money’s role in helping to determine Y. (A new group, the real-business-cycle theorists associated with the Minneapolis Fed, has recently challenged the notion that money matters, but those folks haven’t made it into the land of undergraduate textbooks quite yet, except in passing.)
key takeaways
• Timing, statistical, and historical evidence strengthen researchers’ belief in causation.
• Timing evidence attempts to show that changes in M occur before changes in Y.
• Statistical evidence attempts to show that one model’s predictions are closer to reality than another’s.
• The problem with stats, though, is that those running the tests appear to rig them (consciously or not), so the stats often tell us more about the researcher than they do about reality.
• Historical evidence, particularly so-called natural experiments in which variables change exogenously and hence are analogous to controlled scientific experiments, provide the best sort of evidence on the direction of causation.
• The monetarists showed that there is a strong correlation between changes in the MS and changes in Y and also proffered timing, statistical, and historical evidence of a causal link.
• Historical evidence is the most convincing because it shows that the MS sometimes changed exogenously, that is, for reasons clearly unrelated to Y or other plausible causal variables, and that when it did, Y changed with the expected sign (+ if MS increased, − if it decreased). | 2019-07-22 07:33:01 | {"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": 0.376209557056427, "perplexity": 3445.6887736542317}, "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/1563195527828.69/warc/CC-MAIN-20190722072309-20190722094309-00522.warc.gz"} |
https://deepai.org/publication/intersecting-edge-distinguishing-colorings-of-hypergraphs | # Intersecting edge distinguishing colorings of hypergraphs
An edge labeling of a graph distinguishes neighbors by sets (multisets, resp.), if for any two adjacent vertices u and v the sets (multisets, resp.) of labels appearing on edges incident to u and v are different. In an analogous way we define total labelings distinguishing neighbors by sets or multisets: for each vertex, we consider labels on incident edges and the label of the vertex itself. In this paper we show that these problems, and also other problems of similar flavor, admit an elegant and natural generalization as a hypergraph coloring problem. An ieds-coloring (iedm-coloring, resp.) of a hypergraph is a vertex coloring, in which the sets (multisets, resp.) of colors, that appear on every pair of intersecting edges are different. We show upper bounds on the size of lists, which guarantee the existence of an ieds- or iedm-coloring, respecting these lists. The proof is essentially a randomized algorithm, whose expected time complexity is polynomial. As corollaries, we derive new results concerning the list variants of graph labeling problems, distinguishing neighbors by sets or multisets. We also show that our method is robust and can be easily extended for different, related problems. We also investigate a close connection between edge labelings of bipartite graphs, distinguishing neighbors by sets, and the so-called property B of hypergraphs. We discuss computational aspects of the problem and present some classes of bipartite graphs, which admit such a labeling using two labels.
## Authors
• 11 publications
• 29 publications
• ### Optimal Adjacent Vertex-Distinguishing Edge-Colorings of Circulant Graphs
A k-proper edge-coloring of a graph G is called adjacent vertex-distingu...
04/27/2020 ∙ by Sylvain Gravier, et al. ∙ 0
• ### Quasi-polynomial Algorithms for List-coloring of Nearly Intersecting Hypergraphs
A hypergraph H on n vertices and m edges is said to be nearly-intersect...
04/04/2019 ∙ by Khaled Elbassioni, et al. ∙ 0
• ### Choosability in bounded sequential list coloring
The list coloring problem is a variation of the classical vertex colorin...
12/31/2018 ∙ by Simone Gama, et al. ∙ 0
• ### Classification of distributed binary labeling problems
We present a complete classification of the deterministic distributed ti...
11/29/2019 ∙ by Alkida Balliu, et al. ∙ 0
• ### Edge-sum distinguishing labeling
In this paper we study edge-sum distinguishing labeling, a type of label...
04/15/2018 ∙ by Jan Bok, et al. ∙ 0
• ### Distinguishing numbers and distinguishing indices of oriented graphs
A distinguishing r-vertex-labelling (resp. r-edge-labelling) of an undir...
10/28/2019 ∙ by Kahina Meslem, et al. ∙ 0
• ### A note on the neighbour-distinguishing index of digraphs
In this note, we introduce and study a new version of neighbour-distingu...
09/23/2019 ∙ by Eric Sopena, et al. ∙ 0
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## 1 Introduction
Among the variants of graph coloring, there is a prominent family of problems, where the coloring of vertices of is not given explicitly, but derived from some other function. Usually, this function is some labeling of edges or vertices of , and the color of the vertices is based on the labels assigned to incident edges or adjacent vertices (or both in case of total labeling).
Perhaps the most famous problem of this kind was proposed by Karoński, Łuczak, and Thomason [39]. They were labeling the edges of a graph with integers , so that for every two adjacent vertices , the sums of labels assigned to edges incident to each of and are different. We call such a labeling neighbor sum distinguishing. Observe a neighbor sum distinguishing labeling exists if and only if has no isolated edge, such graphs will be called nice. Karoński, Łuczak, and Thomason showed that each nice graph has a neighbor sum distinguishing labeling with 183 labels, if we allow real numbers as labels. They also showed that if the minimum degree of is large, 30 (real) labels suffice, and conjectured that every nice graph has a neighbor sum distinguishing labeling with labels . This problem, known as the 1-2-3 conjecture, raised significant interest in the graph theory community. Addario-Berry, Dalal, McDiarmid, Reed, and Thomason [2] showed that integer labels are sufficient to find a neighbor sum distinguishing labeling of any nice graph. The upper bound on the largest label was subsequently improved: to 16 by Addario-Berry, Dalal, and Reed [3], then to 13 by Wang and Yu [61], and then to 6 by Kalkowski, Karoński, and Pfender [37]. Currently best bound for general graphs is 5 and was shown by Kalkowski, Karoński, and Pfender [38]. Bartnicki, Grytczuk, and Niwczyk [8] proposed a stronger conjecture, that for any assignment of 3-element lists to edges of , one can find a neighbor sum distinguishing labeling, such that every edge gets a label from its list. A constant bound on the size of lists that guarantee the existence of such a labeling is known for some special classes of graphs, e.g. complete graphs, complete bipartite graphs, or nice trees [8].
Dudek and Wajc [20] considered the computational problem of deciding whether a given nice graph has a neighbor sum distinguishing edge labeling with labels . They proved that the problem is NP-complete for and for . This was later extended by Dehghan, Sadeghi, and Ahadi [19] to all pairs , even if the input graph is cubic.
Let us have a closer look at the already mentioned result by Karoński, Łuczak, and Thomason [39], that a constant number of real labels is sufficient to distinguish adjacent vertices by sums of labels on incident edges. We will say that an edge labeling distinguishes neighbors by multisets, if for every pair of adjacent vertices, the multisets of labels appearing on the edges incident to and are different, see Figure 1 a). Notice that different sums always imply different multisets, but it is possible to have different multisets that give the same sum. The authors of [39] proved that every graph has an edge labeling, which distinguishes neighbors by multisets, and uses a constant number of labels and then, by choosing labels that satisfy certain independence properties, one can obtain a labeling that distinguishes sums. Edge labelings distinguishing neighbors by multisets were further studied by Addario-Berry, Aldred, Dalal, and Reed [1], who showed every nice graph has such a labeling using four labels and, if the minimum degree of is large enough, three labels suffice. Observe that if is regular, then an edge labeling using two labels is neighbor sum distinguishing if and only if it distinguishes neighbors by multisets. Thus, by the already mentioned result by Dehghan, Sadeghi, and Ahadi [19], it is NP-complete to decide whether a given graph has an edge labeling distinguishing neighbors by multisets.
Instead of considering sums or multisets of labels appearing on edges incident to adjacent vertices, one can also consider sets. We say that an edge labeling distinguishes neighbors by sets if for any two adjacent vertices and , the sets of labels on edges incident to and are different, see Figure 1 b). Observe that different sets of colors always imply different multisets, but not the other way around. On the other hand, sets are sums are incomparable: one may have different multisets that give the same sum and different sets, or the same set and different sums. By the generalized neighbor-distinguishing index of a graph , denoted by , we mean the minimum number of labels in an edge labeling distinguishing neighbors by sets. This parameter was introduced by Győri, Horňák, Palmer, and Woźniak [30], who proved that for every nice graph we have . This bound was later refined by Horňak and Sotak [34] and by Győri and Palmer [29], who proved that if , then . Horňák and Woźniak considered a list variant of the problem, where each edge is equipped with a list of possible labels and we ask for the existence of a labeling distinguishing neighbors by multisets and respecting these lists. They proved tight bounds on the size of lists, that guarantee the existence of such a labeling in paths and cycles, and showed that for trees lists of size three are sufficient. They also showed that lists of size three may be necessary, even for trees with .
Inspired by the 1-2-3 conjecture, Woźniak and Przybyło [49] suggested a closely related problem considering total labelings of , i.e., labelings of edges and vertices. They considered the problem of finding a total labeling using minimum number of labels, which distinguishes adjacent vertices by sums of labels appearing on incident edges and the vertex itself. Such a labeling is called neighbor sum distinguishing total labeling. Woźniak and Przybyło conjectured that every graph has a neighbor sum distinguishing total labeling, using labels only, this problem is known as the 1-2 conjecture. Kalkowski [36] showed that each graph has a neighbor sum distinguishing total labeling with labels , in which the label 3 does not appear on any vertex. Wong and Zhu [63], and Przybyło and Woźniak [50] conjectured that the 1-2 conjecture holds even in the list variant. Recently, Wong and Zhu [64] showed a list version of the theorem by Kalkowski: a list neighbor sum distinguishing total labeling exists if each vertex has a list of size 2 and every edge has a list of size 3. Observe that in an analogous way one may define total labelings distinguishing neighbors by multisets and distinguishing neighbors by sets, see Figure 2.
Finally, let us mention a similar problem, considered by Seamone and Stevens [54]. Let be a graph and suppose that its edges are linearly ordered. A labeling of edges distinguishes neighbors by sequences, if the sequences of colors (implied by the global ordering of edges), appearing on edges incident to adjacent vertices, are different.
Seamone and Stevens showed that if the ordering of edges can be chosen, then for any nice graph lists of size 2 are sufficient to find a list edge labeling, distinguishing neighbors by sequences. If the ordering of edges is fixed, the lists of size 3 suffice, provided that the minimum degree is large enough, compared to the maximum degree. In particular, lists of size 3 are sufficient for a -regular graph with .
Another variants of the mentioned problems have also been studied. For example, one can ask for an edge-labeling, which distinguishes neighbors by sums/multisetes/sums, and is also required to be proper: for distinguishing neighbors by sums of labels, see e.g. Przybyło [47, 48], Bonamy and Przybyło [14], Hocquard and Przybyło [33]; for distinguishing neighbors by sets of labels, see Zhang, Liu, and Wang [65], Balister, Győri, Lehel, and Schelp [6], Edwards, Horňák, and Woźniak [22], Bonamy, Bousquet, and Hocquard [13] or Hatami [31]; for list edge labelings distinguishing neighbors by multisets see a recent exciting result by Kwaśny and Przybyło [40].
There is also some work on edge labelings, that distinguish neighbors by products of labels (see Skowronek-Kaziów [55, 56, 57]). For a more detailed overview on the related problems, we refer the reader to the recent book by Zhang [65], and a survey by Seamone [53].
### 1.1 Our contribution
Let us introduce the main character of this paper. For a hypergraph , we say that a vertex coloring is intersecting edge distinguishing by multisets (or, in short, is an iedm-coloring), if the multisets of colors appearing in intersecting edges are different. Similarly, a vertex coloring is intersecting edge distinguishing by sets (or, in short, is an ieds-coloring), if the sets of colors appearing in intersecting edges are different.
It is perhaps interesting to mention the special case of graphs, i.e., 2-uniform hypergraphs. It is straightforward to verify that in this case iedm- and ieds-colorings are equivalent. Moreover, a vertex coloring distinguishes intersecting edges by sets (or, equivalently, multisets) if and only if no two vertices with a common neighbor receive the same color. Such a concept is already known in graph theory and usually referred to as an -labeling. The motivation to study this kind of a coloring came from the hidden terminal problem in telecommunication [10, 42]. Optimal (i.e., using the minimum number of colors) -labelings are known for simple classes of graphs, like paths, cycles, grids (see Makansi [42] and Jin, Yeh [35]), hypercubes (see Wan [60]), and complete binary trees (see Bertossi and Bonuccelli [10]). Bodlaender, Kloks, Tan, van Leeuwen [11] showed some bounds on the number of colors required to find an -labeling of for some special classes of graphs: bounded-treewidth graphs, permutation graphs, outerplanar graphs, split graphs, and bipartite graphs. On the complexity side, it is known that the decision problem whether an input graph has an -labeling with 3 colors is NP-complete for planar graphs [10] and split graphs [11]. The parameterized complexity of this problem was considered by Fiala, Golovach, and Kratochvíl [27], who showed that the problem is -hard, when parameterized by treewidth, but FPT, when parameterized by the vertex cover number (we refer the reader to the book by Cygan et al. [16] for more information about parameterized complexity).
In Section 2.2 we argue that colorings of hypergraphs, that distinguishing intersecting edges by sets or multisets, are natural common generalization of edge and total labelings distinguishing neighbors by sets and multisets. We are interested in the list variants of both problems. As the main result, in Theorem 1 we show upper bounds on the size of lists that guarantee the existence of a list ieds-coloring or a list iedm-coloring of a given hypergraph.
The main part of the paper, i.e., Section 3, is devoted to the proof of Theorem 1. The proof uses the so-called entropy compression method, which is a variant of the Lovász Local Lemma [25]. The Local Lemma is essentially non-constructive, but several algorithmic versions have also been developed (see Alon [4], Molloy, Reed [43], and Moser, Tardos [45]). Entropy compression originates in the algorithmic version of the Local Lemma by Moser and Tardos, and was first used by Grytczuk, Kozik, and Micek [28] to study the list version of the problem of Thue. Then the method was successfully applied in many other contexts (see e.g. Esperet, Parreau [26] or Dujmović, Joret, Kozik, and Wood [21]). The main idea of our proof is similar to the one of Bosek, Czerwiński, Grytczuk, and Rzążewski [15], but there are two significant differences. First, the authors of [15] considered the so-called harmonious colorings, where all pairs of edges need to be distinguished. Second, they were considering colorings in which no edge contained two vertices in the same color. If we drop this restriction, distinguishing edges by sets is significantly more difficult and makes the argument more complicated.
The proof of Theorem 1 is essentially a randomized algorithm which finds a list ieds- or iedm-coloring of a given hypergraph. In Section 3.4 we consider its computational complexity and show that the expected number of steps in the execution of the algorithm is polynomial.
In Section 4 we discuss the applicability of Theorem 1 and the method used in the proof. As corollaries from Theorem 1, in Section 4.1 we obtain several bounds on the size of lists, which guarantee the existence of a list edge/total labeling, distinguishing neighbors of a regular graph by sets or multisets. In particular, in Corollaries 8 and 9 we show that if is sufficiently large, then lists of size are sufficient to find an edge or total labeling distinguishing neighbors by sets. In case of distinguishing by multisets, in Corollaries 10 and 11 we prove that lists of size suffice, if is large enough. We also consider the case of the so-called configurations. In Section 4.2, we show that our method can be easily extended and adapted to other problem of similar kind. In particular, we show how to construct an algorithm, which finds a list coloring of vertices of a hypergraph, distinguishing intersecting edges by sequences. We omit most of the details of the proof, as it is essentially the same as the proof of Theorem 1. We focus on highlighting the crucial issues that have to be considered, when adapting our approach to a new problem. As a side result, we improve the result of Seamone and Stevens [54] for regular graphs. In particular, we show that given a -regular graph with , with a fixed ordering of edges, then lists of size 2 are sufficient to choose a list edge labeling, in which every two adjacent vertices have distinct sequences of colors appearing on incident edges.
In Section 5, we investigate an interesting relation between the generalized neighbor-distinguishing index of bipartite graphs and two well-known problems, i.e., a variant of the satisfiability problem called Not-All-Equal Sat, and the so-called property B of hypergraphs. The paper is concluded in Section 6 with several open problems and suggestions for future work.
## 2 Preliminaries
For an integer , by we denote the set . By we denote the number of total preorders of . In other words, is the number of orderings of with possible ties. Observe that
fn=n∑i=0i!{ni}, (1)
where denotes the Stirling number of the second kind, i.e., the number of partitions of into non-empty subsets. The value is sometimes called an ordered Bell number or a Fubini number (see the corresponding OEIS entry [58]). For any function , and any subset , by we denote the multiset of images of elements of . This is in contrast with the usual notation , which is the image of , i.e., the set of images of elements of . For two disjoint sets and a function , such that , we say that a function is color-preserving if for every it holds that .
Consider a hypergraph . For every vertex , let be the set of edges containing . The degree of a vertex is defined as . By we denote the maximum degree, i.e., . A hypergraph is -uniform, for , if for every . Clearly, graphs are 2-uniform hypergraphs. For a -uniform hypergraph , define
I(H):={|P∖Q|:P,Q∈E}∩[k−1].
### 2.1 Graph labeling problems
Let be a simple, undirected graph, i.e., a 2-uniform hypergraph and consider an (unrestricted) edge labeling of . We say that distinguishes neighbors by sets (respectively, by multisets) if (respectively, ) for every pair of adjacent vertices and . Clearly, in both cases, such a labeling exists if and only if and only if the graph is nice, i.e., it does not have an isolated edge. Recall that by we denote the minimum number of labels used in an edge labeling of , distinguishing neighbors by sets.
Similarly, let be a total labeling of , i.e., a labeling of its edges and vertices. We say it distinguishes neighbors by sets (respectively, by multisets) if (respectively, ) for every pair of adjacent vertices and . Note that such a labeling exists for every graph , as it is enough to use a different label for every vertex, and one extra label for all edges.
In list variants of all four problems, edges (or edges and vertices in total labelings) are equipped with lists of possible labels, and we ask for a labeling, where the label of every edge (or edge and vertex in total labelings) belongs to the appropriate list.
It is interesting to note a very close connection between total and edge labelings distinguishing neighbors by sets. Consider a nice graph and let denote the minimum number of labels in a total labeling distinguishing neighbors by sets. First, observe that , because we can extend any edge labeling, which is neighbor distinguishing by sets, to a total labeling, by assigning to each vertex a label that appears on some edge incident to . On the other hand, consider a total labeling of , which distinguishes neighbors by sets and uses labels. By the argument analogous to the one used by Horňák and Soták [34], we observe that
{c−1(M)∪c−1([m]∖M):M⊆[m]},
is a proper vertex coloring of , which implies that and thus . Recall that Győri and Palmer [29] showed that if , then , which implies that .
Finally, if is a bipartite graph then , because we can extend its proper vertex coloring using colors to a total labeling, by assigning label 1 to every edge of . The inverse of this statement is also true: if has a total labeling , distinguishing neighbors by sets, then is bipartite. This is because the sets and form a bipartition of .
Observe that this reasoning does not show equivalence of the list variants of the problems, so it still makes sense to consider them separately.
Finally, note that if we are interested in edge/total labelings of a disconnected graph , then we can label each connected component of independently, as the distinguishing constraints are local. Thus we will focus on connected graphs.
### 2.2 Distinguishing neighbors via colorings of hypergraphs
Let be a hypergraph, we do not allow multiple edges. We say that a coloring distinguishes intersecting edges by sets (or, in short, is an ieds-coloring) if for every pair of distinct intersecting edges it holds that
φ(P)≠φ(Q). (2)
Analogously, distinguishes intersecting edges by multisets (or is an iedm-coloring), if for every pair of distinct intersecting edges it holds that
φ⟨P⟩≠φ⟨Q⟩. (3)
See Figure 3 for an example. Note that (2) implies (3), so every ieds-coloring is also an iedm-coloring. Moreover, if , then (3) is satisfied for any coloring .
By a list ieds-coloring (respectively, a list iedm-coloring) we mean an ieds-coloring (respectively, an iedm-coloring), in which the color of each vertex is chosen from a list which is assigned to the vertex . The lists come with a graph and are assumed to be a part of the instance.
Now, for the graph , let be its dual hypergraph, i.e., the hypergraph whose vertex set is the set of edges of , and edges of correspond to vertices of in the following way: . Observe that in such a hypergraph each vertex belongs to exactly two edges and also, if is -regular, then is -uniform and . Let be an edge labeling of a graph . We observe that distinguishes neighbors by sets (respectively, by multisets) if and only if it is an ieds-coloring (respectively, iedm-coloring) of the dual hypergraph of , see Figure 4.
Analogously to the previous case, we define the total hypergraph of , i.e., the hypergraph , whose vertices are both vertices and edges of . Every edge of is the set of all edges incident to a vertex in and the vertex itself, i.e., . Note that and, if is -regular, then is -uniform and . Now, if is a total labeling of , it distinguishes neighbors by sets (respectively, by multisets) if and only if it is an ieds-coloring (respectively, an iedm-coloring) of .
## 3 Upper bound
In this section we prove the main results of the paper, i.e., upper bounds on the size of lists, which guarantee the existence of an ieds-coloring or an iedm-coloring of a given -uniform hypergraph.
###### Theorem 1.
Let and let be a -uniform hypergraph with and , whose every vertex is equipped with a list of at least colors. Define and for every . The following hold:
For the rest of the section, let be a fixed -uniform hypergraph on vertices, such that and for some . Note that if , then there are no intersecting edges and thus the problem is trivial. We assume that sets and are linearly ordered, clearly these orderings induce a linear ordering on any subset of or . For every , denote by a position of element in the set , determined by this ordering. Suppose that every vertex of is assigned with a list of at least colors.
For a subset of , a function is called a partial ieds-coloring if for every pair of distinct, intersecting edges such that , it holds that
φ(P∩U)≠φ(Q∩U) (4)
and for every vertex . Analogously, is called a partial iedm-coloring if for every such pair of edges we have that
φ⟨P∖Q⟩≠φ⟨Q∖P⟩ (5)
and for every . A partial ieds-coloring (iedm-coloring, respectively) is complete if , note that for the condition (4) is exactly (2) and the condition (5) is equivalent to (3). Moreover, notice that conditions (4) or (5) are necessary, if we want to extend a partial ieds-coloring or a partial iedm-coloring to a complete one, see Figure 5.
Both statements in Theorem 1 can be shown in a very similar way. We will discuss the proof in a detail and point out the differences between the cases of sets and multisets. By a (partial) ied-coloring we will mean a (partial) ieds-coloring or a (partial) iedm-coloring, depending whether we want to show the statement (a) or (b).
We will construct an appropriate coloring of iteratively, ensuring that after each step the current partial coloring is a partial ied-coloring. In general, the algorithm works as follows. We fix a large and a sequence of numbers from . In each step of the algorithm we color an uncolored vertex, using the next number from , and check if the obtained partial coloring is a partial ied-coloring. If so, we proceed to the next iteration. If not, it means that the condition (4) or (5) (depending on the case of sets or multisets) is violated for some pair of edges . In this situation we say that a conflict appears on these two edges. We erase colors of some already colored vertices, to ensure that the current coloring is a partial ied-coloring. Moreover, we use an additional table to register all information about occurring conflicts. The algorithm terminates when all vertices of are colored or all numbers from the sequence are used (i.e., after iterations). In the first case the algorithm returns a complete ied-coloring. In the second one, it returns a pair , where is the table of conflicts and is the partial ied-coloring obtained after the last iteration.
For the contradiction, assume that does not have any ied-coloring, respecting the lists . This means that for every possible sequence the algorithm does not return a complete list ied-coloring of , but some pair . We will show that there is a bijection between all possible sequences and all possible pairs . Then, we will show that if is sufficiently large, then the number of all pairs is strictly smaller than , which is a number of possible sequences . This leads to a contradiction, so there is at least one sequence , for which the algorithm successfully returns a complete list ied-coloring of .
### 3.1 The algorithm
Let be a large integer and let be a sequence of integers from . At the beginning of the procedure all vertices of are uncolored, call this (empty) partial coloring . Also, if a list for some has more than elements, we truncate it, so that all lists are of size exactly . For , by we denote the partial ied-coloring after the -th iteration.
Recall that the vertices of are linearly ordered. For each , the -th iteration of the algorithm consists of the following steps.
Step 1.
Find the smallest uncolored vertex in , call it .
Step 2.
Assign the -th color from the list to and denote by the obtained partial coloring.
Step 3.
If is a partial ied-coloring, write in and set . If is a complete ied-coloring, return it and terminate, otherwise, proceed to the next iteration.
Step 4.
If is not a partial ied-coloring, then a conflict appeared on a pair of intersecting edges , and belongs to at least one of them, say . This is because was a partial ied-coloring and a conflict was caused by coloring . If there is more than one conflict, we can choose any of them. Let , so , and define the set
The final step differs in case of distinguishing by sets and distinguishing by multisets.
Step 5 (variant (a): sets).
There are two types of possible conflicts that may occur, we consider them separately.
1. First, consider the case that . This means that all vertices from were colored in previous iterations, and . Let and . In write the quadruple , where is a color-preserving function from to (later we will specify how to choose , for now it is enough to know that it is color-preserving). After that, uncolor all vertices from , denoting the obtained coloring by , and proceed to the next iteration.
2. Now, consider the case that , clearly all vertices of are already colored. Observe that belongs to exactly one of the sets and ), because was a partial ieds-coloring. Without loss of generality assume that , otherwise switch the names of and .
Let be the smallest vertex of , it exists, because does not have multiple edges. Let and . In write the quadruple , where is a color-preserving function from to . After that, uncolor all vertices from , denoting the obtained coloring by , and proceed to the next iteration.
Step 5 (variant (b): multisets).
Note that now , because was a partial iedm-coloring. Recall that . Let and . In write the triple , where is a color-preserving function from to . After that, uncolor all vertices from , denoting the obtained coloring by , and proceed to the next iteration.
Observe that after uncoloring (and possibly some additional vertices) there are no more conflicts, so is a partial ied-coloring for every . Moreover, note that the execution of the algorithm does not depend on the content of . The information stored there will only be used in the proof.
It is clear that the algorithm either returns an ied-coloring, or runs for iterations and terminates without success. In the latter situation, we say that the pair is produced from the sequence , note that this pair is uniquely determined by . Let (for input) denote the set of all possible sequences , and (for output) denote the set of all pairs that can be possibly produced from a sequence . Assume that the algorithm never returns a complete ied-coloring, which means that for every it produces some pair . This implies that .
### 3.2 Equicardinality of sets I and O
We will show that if a pair belongs to , then there is exactly one sequence that produces . By we denote the set of uncolored vertices after the -th iteration of the algorithm and by we denote the set of vertices that are colored after the -th iteration of the algorithm. Moreover, we define and . Notice that for every the sets and form a partition of , i.e., and .
First, let us prove that using the entries of only, we can reconstruct sets for all .
###### Lemma 2.
For every the set is uniquely determined by and .
###### Proof.
Let be the smallest vertex in the set . In the -th iteration we assigned a color to . Depending on and, obviously, the variant of coloring we consider, we have the following possibilities:
Case 1: (both variants (a) and (b): sets and multisets).
In this case, no conflicts arose after coloring in the -th iteration and no vertex was uncolored. Therefore
Case 2: (variant (a): sets).
This means that there was a conflict on some intersecting edges and , such that . We know that is the -th element of the set . From the size of the domain of , we can determine the value of . Knowing , , and , we can uniquely determine the set , and thus also the edge , which is the -th element of . After the conflict occurred, we uncolored all the vertices from , so
Case 3: (variant (a): sets).
This means that there was a conflict on some edges and , such that . We know that is the -th element of , and is the -th element of . The vertex is the smallest vertex in . After the conflict occurred, we uncolored all vertices from , so
Case 4: (variant (b): multisets).
This case is very similar to the second one. Again we have a conflict on edges and , such that . Using , we can determine , and is given by the size of the domain of . Having , , and , we can compute , and then, using , we can find . After the conflict occurred, we uncolored all the vertices from , so
Using 2, we can reconstruct sets . Now we show that using the pair that was returned by the algorithm and sets for , we can reconstruct all elements of .
###### Lemma 3.
For every , the partial ied-coloring and the number are uniquely determined by , , and the sets .
###### Proof.
For every we have . Let be the smallest vertex in , this is the vertex that was colored in the -th iteration. We know that agrees with on the set . Again we consider the cases.
Case 1: (both variants (a) and (b): sets and multisets).
This means that coloring in the -th iteration did not cause any conflict. Thus for every we have . From we get , and is the position of the color in the list .
Case 2: (variant (a): sets).
Recall that this means that there was a conflict on some edges and , where . We determine and as in 2. To get , we need to recover colors of the vertices from , and also the number . Observe that all colors in appear on vertices of , and since is color-preserving, we can easily reconstruct the coloring on . In the same way we reconstruct the color that was given to in the -th iteration, is its position in .
Case 3: (variant (a): sets).
This means there was a conflict on some edges and , where . We find , , and as in 2. As in the case above, we use to reconstruct colors of to get , and also the color that was assigned to to get .
Case 4: (variant (b): multisets).
Again, we find and as in 2, and then use to reconstruct and .∎
Finally, we can use 2 and 3 to get the following corollary.
###### Corollary 4.
The sets and have the same cardinality.
###### Proof.
From 2 it follows that knowing , we can determine the sets for every . Using them, the table , and the partial ied-coloring , we can reconstruct all partial ied-colorings and the sequence , as shown in 3. This implies that every possible pair is produced by a unique sequence , which means that . Since we know that also holds, the proof is complete. ∎
### 3.3 Cardinality of O
Now we want to estimate the number of possible pairs
that may be produced by the algorithm. Let us start with estimating the number of possible entries in the table , other than just a sign. We use the notation from the previous section. Consider an iteration , where a conflict appeared on two intersecting edges and , such that , and , and let be the coloring obtained by assigning the color to .
We will consider the cases of sets and multisets separately. For each , by (, respectively), we denote the set of all possible entries other than +, that may appear in in case of ieds-coloring (iedm-coloring, respectively). Clearly if , then there are no edges and for which holds, so there are no possible conflicts. Therefore in such a case we have . Now consider .
#### Variant (a): sets.
Then each entry is a quadruple or . Since , clearly we have .
#### Case 1: T(j)=(1,xP,xQ,γ).
Note that in this case , recall , where . Equivalently, contains all edges, which do not contain , and have exactly common vertices with . There are at most edges intersecting , which implies that there are at most elements of , so
xQ∈[⌊(Δ−1)(k−1)k−i⌋].
Finally, we need to estimate the number of possible color-preserving functions , recall and we want to store the information about for every . Let and , clearly . Also, define and , so we have . Observe that and . Note that and may be considered fixed, as they only depend on , , and . Fix a linear ordering of elements of , which is implied by the linear ordering of vertices of : each color is represented by the smallest vertex in this color.
We observe that if the condition (4) does not hold, every color from must appear on some vertex of . On the other hand, colors from may appear on vertices in , but do not have to. Let and . Now is a surjective function from to and can be chosen in at most
g1∑g′1=0(g1g′1)surj(i,g′1+g2)≤k−i∑g′1=0(k−ig′1)surj(i,g′1+g2)
ways, where denotes the number of surjective functions from an -element set to a -element set. We observe that
surj(i,g′1+g2)=(g′1+g2)!⋅{ig′1+g2}≤fi,
where the last inequality follows from (1). Summing up, we obtain that the number of possible functions is at most
k−i∑g′1=0(k−ig′1)surj(i,g′1+g2)≤k−i∑g′1=0(k−ig′1)fi=2k−ifi. (6)
#### Case 2: T(j)=(2,xP,xQ,γ).
If , then , so and thus clearly . The bound on the number of functions is obtained in a way analogous to the previous case. Again, we are interested in bounding the number of surjective functions from to , where is the number of colors from that appear in , and . We observe that this number is bounded by (6).
Summing up, we conclude that
|Si|≤ (Δ⋅⌊(Δ−1)(k−1)k−i⌋⋅2k−ifi)Case 1.+(Δ⋅(Δ−1)⋅2k−ifi)Case 2. ≤ 2Δ⋅(Δ−1)(k−1)k−i⋅2k−ifi=Δ(Δ−1)(k−1)k−i⋅2k−i+1fi.
#### Variant (b): multisets.
This variant is significantly simpler. The only possible entry in is a triple . Just as in the Case 1 in the previous variant, we have
xP∈[Δ] and xQ∈[⌊(Δ−1)(k−1)k−i⌋].
By condition (5), can be assumed to be a bijection, so it can be chosen in ways. So, summing up, we obtain the following bound:
|Mi|≤Δ(Δ−1)(k−1)k−i⋅i!.
The rest of the proof is exactly the same in both variants: distingushing by sets and multisets. For every , by let us denote , if we are interested in finding an ieds-coloring, or , if we are interested in finding an iedm-coloring. Define .
Now, let us bound the number of all possible tables that could be produced by the algorithm, denote it by . By denote a number of symbols in . For every let be the number of appearances of the elements of in ; if , then clearly . Notice that . Denote by
(Np,t1,t2,…,tk−1)=N!p!t1!t2!…tk−1!
the number of partitions of an -element set into the subsets of cardinalities .
###### Lemma 5.
The number of all possible tables is bounded from above by
N∑s=N−n+1∑t1+⋯+(k−1)tk−1=s (Np,t1,…,tk−1)at11⋅…⋅atk−1k−1. (7)
###### Proof.
For every , the total number of appearances of elements of in is . It implies that for fixed , the number of ways to fill the table is at most
(Np,t1,…,tk−1)at11⋅…⋅atk−1k−1.
Observe that if an element of appears in , it means that in the -th iteration we uncolored vertices that were colored in previous iterations. Thus to each occurrence of an element of in we can assign iterations when we wrote into , and each symbol is assigned at most once. Let be the number of iterations when a conflict occurred, or when we colored a vertex that was uncolored later. There are iterations when a conflict occured, and iterations when we colored a vertex which was uncolored later, so
s=(t1+t2+…+tk−1)+(t2+2t3+…+(k−2)tk−1)=t1+2t2+…+(k−1)tk−1.
Clearly is at most , which is the total number of iterations. On the other hand, after steps the algorithm returns a partial ied-coloring, where at most vertices are colored. Therefore . ∎
We will need the following technical lemma shown in [15, Lemma 2.7].
###### Lemma 6 (Bosek, Czerwiński, Grytczuk, Rz. [15]).
If are non-negative integers such that and , then
(Np,t1,t2,…,tk−1)≤(Ns)(st1,2t2,…,(k−1)tk−1)qt11⋅q2t22⋅…⋅q(k−1)tk−1k−1,
where and
qi=ii−1i√i−1,
for
Now we are ready to prove the final lemma.
###### Lemma 7.
Let
R:=⌈2+k−1∑i=1qii√ai⌉,
where and for Then there exists such that for every the number of elements of the set is strictly smaller than .
###### Proof.
From Lemma 5 it follows that the number of all possible tables that may be produced is bounded by (7). Using 6, we have that
#T≤N∑s=N−n+1∑t1+⋯+(k−1)tk−1=s (Np,t1,…,tk−1)at11⋅…⋅atk−1k−1≤N∑s=N−n+1(Ns)∑t1+⋯+(k−1)tk−1=s(st1,2t2,…,(k−1)tk−1)k−1∏i=1(qii√ai)iti. (8)
Because , we can use the Multinomial Theorem to write
∑t1+…+(k−1)tk−1=s(st1,2t2,…,(k−1)tk−1)k−1∏i=1(qii√ai)iti=(k−1∑i=1qii√ai) | 2021-09-18 20:34:56 | {"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": 0.9221087694168091, "perplexity": 548.6015999671413}, "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/1631780056572.96/warc/CC-MAIN-20210918184640-20210918214640-00150.warc.gz"} |
https://snaildove.github.io/2018/06/02/Building+a+Recurrent+Neural+Network+-+Step+by+Step+-+v3/ | ## Note
This is one of my personal programming assignments after studying the course nlp sequence models at the 1st week and the copyright belongs to deeplearning.ai.
# Building your Recurrent Neural Network - Step by Step
Welcome to Course 5’s first assignment! In this assignment, you will implement your first Recurrent Neural Network in numpy.
Recurrent Neural Networks (RNN) are very effective for Natural Language Processing and other sequence tasks because they have “memory”. They can read inputs $x^{\langle t \rangle}$ (such as words) one at a time, and remember some information/context through the hidden layer activations that get passed from one time-step to the next. This allows a uni-directional RNN to take information from the past to process later inputs. A bidirection RNN can take context from both the past and the future.
Notation:
• Superscript $[l]$ denotes an object associated with the $l^{th}$ layer.
• Example: $a^{[4]}$ is the $4^{th}$ layer activation. $W^{[5]}$ and $b^{[5]}$ are the $5^{th}$ layer parameters.
• Superscript $(i)$ denotes an object associated with the $i^{th}$ example.
• Example: $x^{(i)}$ is the $i^{th}$ training example input.
• Superscript $\langle t \rangle$ denotes an object at the $t^{th}$ time-step.
• Example: $x^{\langle t \rangle}$ is the input x at the $t^{th}$ time-step. $x^{(i)\langle t \rangle}$ is the input at the $t^{th}$ timestep of example $i$.
• Lowerscript $i$ denotes the $i^{th}$ entry of a vector.
• Example: $a^{[l]}_i$ denotes the $i^{th}$ entry of the activations in layer $l$.
We assume that you are already familiar with numpy and/or have completed the previous courses of the specialization. Let’s get started!
Let’s first import all the packages that you will need during this assignment.
## 1 - Forward propagation for the basic Recurrent Neural Network
Later this week, you will generate music using an RNN. The basic RNN that you will implement has the structure below. In this example, $T_x = T_y$.
Figure 1: Basic RNN model
Here’s how you can implement an RNN:
Steps:
1. Implement the calculations needed for one time-step of the RNN.
2. Implement a loop over $T_x$ time-steps in order to process all the inputs, one at a time.
Let’s go!
## 1.1 - RNN cell
A Recurrent neural network can be seen as the repetition of a single cell. You are first going to implement the computations for a single time-step. The following figure describes the operations for a single time-step of an RNN cell.
Figure 2: Basic RNN cell. Takes as input $x^{\langle t \rangle}$ (current input) and $a^{\langle t - 1\rangle}$ (previous hidden state containing information from the past), and outputs $a^{\langle t \rangle}$ which is given to the next RNN cell and also used to predict $y^{\langle t \rangle}$
Exercise: Implement the RNN-cell described in Figure (2).
Instructions:
1. Compute the hidden state with tanh activation: $a^{\langle t \rangle} = \tanh(W_{aa} a^{\langle t-1 \rangle} + W_{ax} x^{\langle t \rangle} + b_a)$.
2. Using your new hidden state $a^{\langle t \rangle}$, compute the prediction $\hat{y}^{\langle t \rangle} = softmax(W_{ya} a^{\langle t \rangle} + b_y)$. We provided you a function: softmax.
3. Store $(a^{\langle t \rangle}, a^{\langle t-1 \rangle}, x^{\langle t \rangle}, parameters)$ in cache
4. Return $a^{\langle t \rangle}$ , $y^{\langle t \rangle}$ and cache
We will vectorize over $m$ examples. Thus, $x^{\langle t \rangle}$ will have dimension $(n_x,m)$, and $a^{\langle t \rangle}$ will have dimension $(n_a,m)$.
a_next[4] = [ 0.59584544 0.18141802 0.61311866 0.99808218 0.85016201 0.99980978
-0.18887155 0.99815551 0.6531151 0.82872037]
a_next.shape = (5, 10)
yt_pred[1] = [ 0.9888161 0.01682021 0.21140899 0.36817467 0.98988387 0.88945212
0.36920224 0.9966312 0.9982559 0.17746526]
yt_pred.shape = (2, 10)
Expected Output:
a_next[4]: [ 0.59584544 0.18141802 0.61311866 0.99808218 0.85016201 0.99980978 -0.18887155 0.99815551 0.6531151 0.82872037] a_next.shape: (5, 10) yt[1]: [ 0.9888161 0.01682021 0.21140899 0.36817467 0.98988387 0.88945212 0.36920224 0.9966312 0.9982559 0.17746526] yt.shape: (2, 10)
## 1.2 - RNN forward pass
You can see an RNN as the repetition of the cell you’ve just built. If your input sequence of data is carried over 10 time steps, then you will copy the RNN cell 10 times. Each cell takes as input the hidden state from the previous cell ($a^{\langle t-1 \rangle}$) and the current time-step’s input data ($x^{\langle t \rangle}$). It outputs a hidden state ($a^{\langle t \rangle}$) and a prediction ($y^{\langle t \rangle}$) for this time-step.
Figure 3: Basic RNN. The input sequence $x = (x^{\langle 1 \rangle}, x^{\langle 2 \rangle}, …, x^{\langle T_x \rangle})$ is carried over $T_x$ time steps. The network outputs $y = (y^{\langle 1 \rangle}, y^{\langle 2 \rangle}, …, y^{\langle T_x \rangle})$.
Exercise: Code the forward propagation of the RNN described in Figure (3).
Instructions:
1. Create a vector of zeros ($a$) that will store all the hidden states computed by the RNN.
2. Initialize the “next” hidden state as $a_0$ (initial hidden state).
3. Start looping over each time step, your incremental index is $t$ :
• Update the “next” hidden state and the cache by running rnn_cell_forward
• Store the “next” hidden state in $a$ ($t^{th}$ position)
• Store the prediction in y
• Add the cache to the list of caches
4. Return $a$, $y$ and caches
a[4][1] = [-0.99999375 0.77911235 -0.99861469 -0.99833267]
a.shape = (5, 10, 4)
y_pred[1][3] = [ 0.79560373 0.86224861 0.11118257 0.81515947]
y_pred.shape = (2, 10, 4)
caches[1][1][3] = [-1.1425182 -0.34934272 -0.20889423 0.58662319]
len(caches) = 2
Expected Output:
a[4][1]: [-0.99999375 0.77911235 -0.99861469 -0.99833267] a.shape: (5, 10, 4) y[1][3]: [ 0.79560373 0.86224861 0.11118257 0.81515947] y.shape: (2, 10, 4) cache[1][1][3]: [-1.1425182 -0.34934272 -0.20889423 0.58662319] len(cache): 2
Congratulations! You’ve successfully built the forward propagation of a recurrent neural network from scratch. This will work well enough for some applications, but it suffers from vanishing gradient problems. So it works best when each output $y^{\langle t \rangle}$ can be estimated using mainly “local” context (meaning information from inputs $x^{\langle t’ \rangle}$ where $t’$ is not too far from $t$).
In the next part, you will build a more complex LSTM model, which is better at addressing vanishing gradients. The LSTM will be better able to remember a piece of information and keep it saved for many timesteps.
## 2 - Long Short-Term Memory (LSTM) network
This following figure shows the operations of an LSTM-cell.
Figure 4: LSTM-cell. This tracks and updates a “cell state” or memory variable $c^{\langle t \rangle}$ at every time-step, which can be different from $a^{\langle t \rangle}$.
Similar to the RNN example above, you will start by implementing the LSTM cell for a single time-step. Then you can iteratively call it from inside a for-loop to have it process an input with $T_x$ time-steps.
#### - Forget gate
For the sake of this illustration, lets assume we are reading words in a piece of text, and want use an LSTM to keep track of grammatical structures, such as whether the subject is singular or plural. If the subject changes from a singular word to a plural word, we need to find a way to get rid of our previously stored memory value of the singular/plural state. In an LSTM, the forget gate lets us do this:
$$\Gamma_f^{\langle t \rangle} = \sigma(W_f[a^{\langle t-1 \rangle}, x^{\langle t \rangle}] + b_f)\tag{1}$$
Here, $W_f$ are weights that govern the forget gate’s behavior. We concatenate $[a^{\langle t-1 \rangle}, x^{\langle t \rangle}]$ and multiply by $W_f$. The equation above results in a vector $\Gamma_f^{\langle t \rangle}$ with values between 0 and 1. This forget gate vector will be multiplied element-wise by the previous cell state $c^{\langle t-1 \rangle}$. So if one of the values of $\Gamma_f^{\langle t \rangle}$ is 0 (or close to 0) then it means that the LSTM should remove that piece of information (e.g. the singular subject) in the corresponding component of $c^{\langle t-1 \rangle}$. If one of the values is 1, then it will keep the information.
#### - Update gate
Once we forget that the subject being discussed is singular, we need to find a way to update it to reflect that the new subject is now plural. Here is the formulat for the update gate:
$$\Gamma_u^{\langle t \rangle} = \sigma(W_u[a^{\langle t-1 \rangle}, x^{\{t\}}] + b_u)\tag{2}$$
Similar to the forget gate, here $\Gamma_u^{\langle t \rangle}$ is again a vector of values between 0 and 1. This will be multiplied element-wise with $\tilde{c}^{\langle t \rangle}$, in order to compute $c^{\langle t \rangle}$.
#### - Updating the cell
To update the new subject we need to create a new vector of numbers that we can add to our previous cell state. The equation we use is:
$$\tilde{c}^{\langle t \rangle} = \tanh(W_c[a^{\langle t-1 \rangle}, x^{\langle t \rangle}] + b_c)\tag{3}$$
Finally, the new cell state is:
$$c^{\langle t \rangle} = \Gamma_f^{\langle t \rangle}* c^{\langle t-1 \rangle} + \Gamma_u^{\langle t \rangle} *\tilde{c}^{\langle t \rangle} \tag{4}$$
#### - Output gate
To decide which outputs we will use, we will use the following two formulas:
$$\Gamma_o^{\langle t \rangle}= \sigma(W_o[a^{\langle t-1 \rangle}, x^{\langle t \rangle}] + b_o)\tag{5}$$
$$a^{\langle t \rangle} = \Gamma_o^{\langle t \rangle}* \tanh(c^{\langle t \rangle})\tag{6}$$
Where in equation 5 you decide what to output using a sigmoid function and in equation 6 you multiply that by the $\tanh$ of the previous state.
### 2.1 - LSTM cell
Exercise: Implement the LSTM cell described in the Figure (3).
Instructions:
1. Concatenate $a^{\langle t-1 \rangle}$ and $x^{\langle t \rangle}$ in a single matrix: $concat = \begin{bmatrix} a^{\langle t-1 \rangle} \\ x^{\langle t \rangle} \end{bmatrix}$
2. Compute all the formulas 1-6. You can use sigmoid() (provided) and np.tanh().
3. Compute the prediction $y^{\langle t \rangle}$. You can use softmax() (provided).
a_next[4] = [-0.66408471 0.0036921 0.02088357 0.22834167 -0.85575339 0.00138482
0.76566531 0.34631421 -0.00215674 0.43827275]
a_next.shape = (5, 10)
c_next[2] = [ 0.63267805 1.00570849 0.35504474 0.20690913 -1.64566718 0.11832942
0.76449811 -0.0981561 -0.74348425 -0.26810932]
c_next.shape = (5, 10)
yt[1] = [ 0.79913913 0.15986619 0.22412122 0.15606108 0.97057211 0.31146381
0.00943007 0.12666353 0.39380172 0.07828381]
yt.shape = (2, 10)
cache[1][3] = [-0.16263996 1.03729328 0.72938082 -0.54101719 0.02752074 -0.30821874
0.07651101 -1.03752894 1.41219977 -0.37647422]
len(cache) = 10
Expected Output :
a_next[4]: [-0.66408471 0.0036921 0.02088357 0.22834167 -0.85575339 0.00138482 0.76566531 0.34631421 -0.00215674 0.43827275] a_next.shape: (5, 10) c_next[2]: [ 0.63267805 1.00570849 0.35504474 0.20690913 -1.64566718 0.11832942 0.76449811 -0.0981561 -0.74348425 -0.26810932] c_next.shape: (5, 10) yt[1]: [ 0.79913913 0.15986619 0.22412122 0.15606108 0.97057211 0.31146381 0.00943007 0.12666353 0.39380172 0.07828381] yt.shape: (2, 10) cache[1][3]: [-0.16263996 1.03729328 0.72938082 -0.54101719 0.02752074 -0.30821874 0.07651101 -1.03752894 1.41219977 -0.37647422] len(cache): 10
### 2.2 - Forward pass for LSTM
Now that you have implemented one step of an LSTM, you can now iterate this over this using a for-loop to process a sequence of $T_x$ inputs.
Figure 4: LSTM over multiple time-steps.
Exercise: Implement lstm_forward() to run an LSTM over $T_x$ time-steps.
Note: $c^{\langle 0 \rangle}$ is initialized with zeros.
a[4][3][6] = 0.172117767533
a.shape = (5, 10, 7)
y[1][4][3] = 0.95087346185
y.shape = (2, 10, 7)
caches[1][1[1]] = [ 0.82797464 0.23009474 0.76201118 -0.22232814 -0.20075807 0.18656139
0.41005165]
c[1][2][1] -0.855544916718
len(caches) = 2
Expected Output:
a[4][3][6] = 0.172117767533 a.shape = (5, 10, 7) y[1][4][3] = 0.95087346185 y.shape = (2, 10, 7) caches[1][1][1] = [ 0.82797464 0.23009474 0.76201118 -0.22232814 -0.20075807 0.18656139 0.41005165] c[1][2][1] = -0.855544916718 len(caches) = 2
Congratulations! You have now implemented the forward passes for the basic RNN and the LSTM. When using a deep learning framework, implementing the forward pass is sufficient to build systems that achieve great performance.
The rest of this notebook is optional, and will not be graded.
## 3 - Backpropagation in recurrent neural networks (OPTIONAL / UNGRADED)
In modern deep learning frameworks, you only have to implement the forward pass, and the framework takes care of the backward pass, so most deep learning engineers do not need to bother with the details of the backward pass. If however you are an expert in calculus and want to see the details of backprop in RNNs, you can work through this optional portion of the notebook.
When in an earlier course you implemented a simple (fully connected) neural network, you used backpropagation to compute the derivatives with respect to the cost to update the parameters. Similarly, in recurrent neural networks you can to calculate the derivatives with respect to the cost in order to update the parameters. The backprop equations are quite complicated and we did not derive them in lecture. However, we will briefly present them below.
### 3.1 - Basic RNN backward pass
We will start by computing the backward pass for the basic RNN-cell.
Figure 5: RNN-cell’s backward pass. Just like in a fully-connected neural network, the derivative of the cost function $J$ backpropagates through the RNN by following the chain-rule from calculas. The chain-rule is also used to calculate $(\frac{\partial J}{\partial W_{ax}},\frac{\partial J}{\partial W_{aa}},\frac{\partial J}{\partial b})$ to update the parameters $(W_{ax}, W_{aa}, b_a)$.
#### Deriving the one step backward functions:
To compute the rnn_cell_backward you need to compute the following equations. It is a good exercise to derive them by hand.
The derivative of $\tanh$ is $1-\tanh(x)^2$. You can find the complete proof here. Note that: $\text{sech}(x)^2 = 1 - \tanh(x)^2$
Similarly for $\frac{ \partial a^{\langle t \rangle} } {\partial W_{ax}}, \frac{ \partial a^{\langle t \rangle} } {\partial W_{aa}}, \frac{ \partial a^{\langle t \rangle} } {\partial b}$, the derivative of $\tanh(u)$ is $(1-\tanh(u)^2)du$.
The final two equations also follow same rule and are derived using the $\tanh$ derivative. Note that the arrangement is done in a way to get the same dimensions to match.
Expected Output:
#### Backward pass through the RNN
Computing the gradients of the cost with respect to $a^{\langle t \rangle}$ at every time-step $t$ is useful because it is what helps the gradient backpropagate to the previous RNN-cell. To do so, you need to iterate through all the time steps starting at the end, and at each step, you increment the overall $db_a$, $dW_{aa}$, $dW_{ax}$ and you store $dx$.
Instructions:
Implement the rnn_backward function. Initialize the return variables with zeros first and then loop through all the time steps while calling the rnn_cell_backward at each time timestep, update the other variables accordingly.
gradients["dx"][1][2] = [-2.07101689 -0.59255627 0.02466855 0.01483317]
Expected Output:
## 3.2 - LSTM backward pass
### 3.2.1 One Step backward
The LSTM backward pass is slighltly more complicated than the forward one. We have provided you with all the equations for the LSTM backward pass below. (If you enjoy calculus exercises feel free to try deriving these from scratch yourself.)
### 3.2.2 gate derivatives
$$d \Gamma_o^{\langle t \rangle} = da_{next}*\tanh(c_{next}) * \Gamma_o^{\langle t \rangle}*(1-\Gamma_o^{\langle t \rangle})\tag{7}$$ $$d\tilde c^{\langle t \rangle} = dc_{next}*\Gamma_u^{\langle t \rangle}+ \Gamma_o^{\langle t \rangle} (1-\tanh(c_{next})^2) * i_t * da_{next} * \tilde c^{\langle t \rangle} * (1-\tanh(\tilde c)^2) \tag{8}$$ $$d\Gamma_u^{\langle t \rangle} = dc_{next}*\tilde c^{\langle t \rangle} + \Gamma_o^{\langle t \rangle} (1-\tanh(c_{next})^2) * \tilde c^{\langle t \rangle} * da_{next}*\Gamma_u^{\langle t \rangle}*(1-\Gamma_u^{\langle t \rangle})\tag{9}$$ $$d\Gamma_f^{\langle t \rangle} = dc_{next}*\tilde c_{prev} + \Gamma_o^{\langle t \rangle} (1-\tanh(c_{next})^2) * c_{prev} * da_{next}*\Gamma_f^{\langle t \rangle}*(1-\Gamma_f^{\langle t \rangle})\tag{10}$$
### 3.2.3 parameter derivatives
$$dW_f = d\Gamma_f^{\langle t \rangle} * \begin{pmatrix} a_{prev} \\ x_t\end{pmatrix}^T \tag{11}$$ $$dW_u = d\Gamma_u^{\langle t \rangle} * \begin{pmatrix} a_{prev} \\ x_t\end{pmatrix}^T \tag{12}$$ $$dW_c = d\tilde c^{\langle t \rangle} * \begin{pmatrix} a_{prev} \\ x_t\end{pmatrix}^T \tag{13}$$ $$dW_o = d\Gamma_o^{\langle t \rangle} * \begin{pmatrix} a_{prev} \\ x_t\end{pmatrix}^T \tag{14}$$ To calculate $db_f, db_u, db_c, db_o$ you just need to sum across the horizontal (axis= 1) axis on $d\Gamma_f^{\langle t \rangle}, d\Gamma_u^{\langle t \rangle}, d\tilde c^{\langle t \rangle}, d\Gamma_o^{\langle t \rangle}$ respectively. Note that you should have the keep_dims = True option. Finally, you will compute the derivative with respect to the previous hidden state, previous memory state, and input. $$da_{prev} = W_f^T*d\Gamma_f^{\langle t \rangle} + W_u^T * d\Gamma_u^{\langle t \rangle}+ W_c^T * d\tilde c^{\langle t \rangle} + W_o^T * d\Gamma_o^{\langle t \rangle} \tag{15}$$ Here, the weights for equations 13 are the first n_a, (i.e. $W_f = W_f[:n_a,:]$ etc...) $$dc_{prev} = dc_{next}\Gamma_f^{\langle t \rangle} + \Gamma_o^{\langle t \rangle} * (1- \tanh(c_{next})^2)*\Gamma_f^{\langle t \rangle}*da_{next} \tag{16}$$ $$dx^{\langle t \rangle} = W_f^T*d\Gamma_f^{\langle t \rangle} + W_u^T * d\Gamma_u^{\langle t \rangle}+ W_c^T * d\tilde c_t + W_o^T * d\Gamma_o^{\langle t \rangle}\tag{17}$$ where the weights for equation 15 are from n_a to the end, (i.e. $W_f = W_f[n_a:,:]$ etc...) **Exercise:** Implement lstm_cell_backward by implementing equations $7-17$ below. Good luck! :)
Expected Output:
### 3.3 Backward pass through the LSTM RNN
This part is very similar to the rnn_backward function you implemented above. You will first create variables of the same dimension as your return variables. You will then iterate over all the time steps starting from the end and call the one step function you implemented for LSTM at each iteration. You will then update the parameters by summing them individually. Finally return a dictionary with the new gradients.
Instructions: Implement the lstm_backward function. Create a for loop starting from $T_x$ and going backward. For each step call lstm_cell_backward and update the your old gradients by adding the new gradients to them. Note that dxt is not updated but is stored.
gradients["dx"][1][2] = [-0.00173313 0.08287442 -0.30545663 -0.43281115]
Expected Output: | 2020-01-22 01:45:26 | {"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": 0.6077609658241272, "perplexity": 959.5135454064052}, "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-2020-05/segments/1579250606269.37/warc/CC-MAIN-20200122012204-20200122041204-00528.warc.gz"} |
https://jeopardylabs.com/play/math-in-focus-ch-8-review | Add Decimals 1 Add Decimals 2 Subtract Decimals 1 Subtract Decimals 2 Word Problems!
3.11
2.31 + 0.8 =
1.26
0.4 + 0.86 =
46.22
49.62 - 3.4 =
$1.70$8.10 - $6.40 = 100 62.4 + 3.5 + 6.04 = 71.94 Jasmine started at 62.4 on a number line. She jumped forward 3.5 and then 6.04. Where did she land? 200 9.71 7.95 + 1.76 = 200 6.01 3.94 + 2.07 200 16.59 Subtract 12.84 from 29.43 200 2.08 6.4 - 4.32 = 200 4.23 + 5.83 = 10.06 meters A length of pipe is 4.23 meters long. Another length of pipe is 1.6 meters longer. What is the total length of the two pipes? 300 13.64 Add 6.24 and 7.4 300 27.53 What is 0.43 and 27.1 altogether? 300 54.47 84.3 - 29.83 300 13.05 Subtract 27.16 from 34.21 300$36.50 - $12.97 =$23.53
Ryan spent $36.50. He spent$12.97 more than Kevin. How much did Kevin spend?
$15.12$6.48 + $8.64 = 400 99.88 Add 30.41, 61.38, and 8.09 400 40.81 Subtract 2.19 from 43 400 5.63 What is the difference between 48.7 and 43.07? 400$18.99 + $15.57 =$34.56
A bag cost $3.42 less than sunglasses. The sunglasses cost$18.99. How much would the bag and sunglasses cost altogether?
500
$161.08 Add$41.97, $36.49, and$82.62
500
60.3 + 31.06 = 91.36 mph
A car travels at 60.3 mph. A second car passes by going 31.06 mph faster than the first car. What was the speed of the second car?
500
$178.06 What is the difference between$138.75 and \$316.81?
500
14.37 - 2.37 - 4.06 = 7.94
Meg starts at 14.37 on a number line. She jumps back 2.37 and then another 4.06. Where does she land?
500
1.3 + 1.52 + 1.26 = 4.08 meters
Jeremy is 0.04 meters taller than Catherine. Mark is 0.22 meters taller than Jeremy. Jeremy is 1.3 meters. What is their total height?
Math in Focus Ch. 8 Review
Press F11 for full-screen mode | 2017-10-22 08:03:50 | {"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": 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": 0.4237147271633148, "perplexity": 6213.806819648751}, "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-2017-43/segments/1508187825154.68/warc/CC-MAIN-20171022075310-20171022095310-00686.warc.gz"} |
https://ccssmathanswers.com/into-math-grade-4-module-15-lesson-4-answer-key/ | # Into Math Grade 4 Module 15 Lesson 4 Answer Key Rename Mixed Numbers to Subtract
We included HMH Into Math Grade 4 Answer Key PDF Module 15 Lesson 4 Rename Mixed Numbers to Subtract to make students experts in learning maths.
## HMH Into Math Grade 4 Module 15 Lesson 4 Answer Key Rename Mixed Numbers to Subtract
I Can rename mixed numbers to subtract fractions with like denominators to solve real-world problems.
At a petting zoo, a baby goat is on a special diet. She receives 3$$\frac{1}{4}$$ cups of food each day. The baby goat eats 1$$\frac{3}{4}$$ cups of the food in the morning. How much is left for her evening feeding?
Represent the problem with a fraction model.
1$$\frac{2}{4}$$ = 1$$\frac{1}{2}$$
Explanation:
3$$\frac{1}{4}$$ – 1$$\frac{3}{4}$$
= $$\frac{13}{4}$$ – $$\frac{7}{4}$$
= $$\frac{13 – 7}{4}$$
= $$\frac{6}{4}$$
= 1$$\frac{2}{4}$$
Turn and Talk A classmate says that you just need to subtract the whole number from the whole number and the fraction from the fraction to solve this problem. How would you respond?
$$\frac{2}{4}$$
Explanation:
3$$\frac{1}{4}$$ – 1$$\frac{3}{4}$$
subtract the whole number from the whole number
3 – 1 = 2
the fraction from the fraction to solve this problem.
= $$\frac{1}{4}$$ – $$\frac{3}{4}$$
= $$\frac{1 – 3}{4}$$
= $$\frac{2}{4}$$
Build Understanding
Question 1.
Bessie the cow drinks 4$$\frac{1}{8}$$ gallons of water each day. She drinks 1$$\frac{7}{8}$$ gallons before noon. How much water will Bessie still need to drink before the end of the day?
A. Model the situation with a visual representation.
2 $$\frac{2}{8}$$
B. What equation can you use to model the problem?
Use d to represent the amount Bessie still needs to drink.
2 $$\frac{2}{8}$$
Explanation:
d= 4$$\frac{1}{8}$$ – 1$$\frac{7}{8}$$
d= $$\frac{33}{8}$$ – $$\frac{15}{8}$$
d= $$\frac{33 – 15}{8}$$
d = $$\frac{18}{8}$$
d = 2 $$\frac{2}{8}$$
C. Explain how you could model the problem with an equation using only fractions greater than one. Show your equation.
Subtract the whole number from the whole number
4 – 1 = 3
Explanation:
d= 4$$\frac{1}{8}$$ – 1$$\frac{7}{8}$$
d= $$\frac{33}{8}$$ – $$\frac{15}{8}$$
d= $$\frac{33 – 15}{8}$$
d = $$\frac{18}{8}$$
d = 2 $$\frac{2}{8}$$
D. What mixed number equals the number of gallons of water that are left for Bessie to drink?
2 $$\frac{2}{8}$$
Explanation:
4$$\frac{1}{8}$$ – 1$$\frac{7}{8}$$
= $$\frac{33}{8}$$ – $$\frac{15}{8}$$
= $$\frac{33 – 15}{8}$$
= $$\frac{18}{8}$$
= 2 $$\frac{2}{8}$$
Turn and Talk What equation could you have written to model the problem using another operation? How do the two equations relate to each other?
2 $$\frac{2}{8}$$
Explanation:
4$$\frac{1}{8}$$ – 1$$\frac{7}{8}$$
= $$\frac{33}{8}$$ – $$\frac{15}{8}$$
= $$\frac{33 – 15}{8}$$
= $$\frac{18}{8}$$
= 2 $$\frac{2}{8}$$
Step It Out
Question 2.
Two kittens get 4$$\frac{2}{4}$$ ounces of food. The bigger kitten gets 2$$\frac{3}{4}$$ ounces. How much food does the smaller kitten get?
You sometimes need to rename mixed numbers to subtract. One way ¡s to rename a mixed number so the fractional part is greater than 1.
A. Rename the mixed number that is the amount of food given to the pair so its fractional part is greater than 1. Show your work.
1$$\frac{3}{4}$$
Explanation:
4$$\frac{2}{4}$$ – 2$$\frac{3}{4}$$
= $$\frac{18}{4}$$ – $$\frac{11}{4}$$
= $$\frac{18 – 11}{4}$$
= $$\frac{7}{4}$$
= 1$$\frac{3}{4}$$
B. Model the problem with an equation using the renamed mixed number and solve the problem.
1$$\frac{3}{4}$$
Explanation:
4$$\frac{2}{4}$$ – 2$$\frac{3}{4}$$
= $$\frac{18}{4}$$ – $$\frac{11}{4}$$
= $$\frac{7}{4}$$
= 1$$\frac{3}{4}$$
Another way is to rename both mixed numbers as fractions greater than 1, and then subtract.
C. Write an equation modeling the problem and its solution using only fractions greater than 1.
$$\frac{7}{4}$$
Explanation:
4$$\frac{2}{4}$$ – 2$$\frac{3}{4}$$
4 – 2 = 2
4$$\frac{2}{4}$$ – 2$$\frac{3}{4}$$
= $$\frac{18}{4}$$ – $$\frac{11}{4}$$
= $$\frac{7}{4}$$
Turn and Talk Which method do you prefer if you have to rename a mixed number to subtract: renaming one mixed number or renaming both mixed numbers? Why?
One way ¡s to rename a mixed number so the fractional part is greater than 1.
To rename a mixed number if the fraction we are subtracting is larger than the first fraction. We have two mixed numbers.
Explanation:
For example, we find common denominators first, convert fractions into decimals and then back into fractions, or simplify fractions first and then subtract.
Check Understanding
Question 1.
Max has 6$$\frac{3}{6}$$ pounds of seed. He used some and now has 4$$\frac{5}{6}$$ pounds left. How many pounds of seed did Max use? Write an equation to model the problem.
6$$\frac{3}{6}$$ – 4$$\frac{5}{6}$$ = 1$$\frac{4}{6}$$
Explanation:
Max has 6$$\frac{3}{6}$$ pounds of seed.
He used some and now has 4$$\frac{5}{6}$$ pounds left.
Total pounds of seed Max use = 6$$\frac{3}{6}$$ – 4$$\frac{5}{6}$$
= $$\frac{39}{6}$$ – $$\frac{29}{6}$$
= $$\frac{10}{6}$$ = 1$$\frac{4}{6}$$
Question 2.
Pablo spends 3$$\frac{1}{6}$$ hours painting, and Miguel spends 1$$\frac{5}{6}$$ hours painting. How many fewer hours does Miguel spend painting than Pablo spends? Show your work.
$$\frac{8}{6}$$ hours
Explanation:
Pablo spends 3$$\frac{1}{6}$$ hours painting,
Miguel spends 1$$\frac{5}{6}$$ hours painting.
Total fewer hours Miguel spend painting than Pablo spends,
= 3$$\frac{1}{6}$$ – 1$$\frac{5}{6}$$
= $$\frac{19}{6}$$ – $$\frac{11}{6}$$
= $$\frac{8}{6}$$
Question 3.
Open-Ended Write three mixed numbers that are between 8 and 10 that would require renaming in order to subtract 7$$\frac{6}{8}$$ from them.
$$\frac{6}{8}$$
Explanation:
x = 8 – 7$$\frac{6}{8}$$
= 8 – $$\frac{62}{8}$$
= $$\frac{64 – 62}{8}$$ = $$\frac{2}{8}$$= $$\frac{6}{8}$$
Question 4.
Reason Susie has 5$$\frac{2}{8}$$ cups of flour. She uses 2$$\frac{7}{8}$$ cups of flour to make pasta. To make bread, she needs 2$$\frac{4}{8}$$ cups of flour. Does she have enough flour to make both? Explain.
No, she don’t have enough flour to make.
Explanation:
Susie has 5$$\frac{2}{8}$$ cups of flour.
She uses 2$$\frac{7}{8}$$ cups of flour to make pasta.
= 5$$\frac{2}{8}$$ – 2$$\frac{7}{8}$$
= $$\frac{42}{8}$$ – $$\frac{23}{8}$$
= $$\frac{42-23}{8}$$ = $$\frac{19}{8}$$
= 2$$\frac{3}{8}$$
Find the difference. If possible, write your answer as a mixed number.
Question 5.
1$$\frac{6}{8}$$ – $$\frac{7}{8}$$
$$\frac{7}{8}$$
Explanation:
1$$\frac{6}{8}$$ – $$\frac{7}{8}$$
= $$\frac{14}{8}$$ – $$\frac{7}{8}$$
= $$\frac{14-7}{8}$$ – $$\frac{7}{8}$$
Question 6.
3$$\frac{2}{8}$$ – 2$$\frac{7}{8}$$
$$\frac{3}{8}$$
Explanation:
3$$\frac{2}{8}$$ – 2$$\frac{7}{8}$$
= $$\frac{26}{8}$$ – $$\frac{23}{8}$$
= $$\frac{26-23}{8}$$ – $$\frac{3}{8}$$
Question 7.
4$$\frac{6}{12}$$ – 1$$\frac{11}{12}$$
2$$\frac{7}{12}$$
Explanation:
4$$\frac{6}{12}$$ – 1$$\frac{11}{12}$$
= $$\frac{54}{12}$$ – $$\frac{23}{12}$$
= $$\frac{54-23}{12}$$ = $$\frac{31}{12}$$
= 2$$\frac{7}{12}$$
Question 8.
9$$\frac{1}{4}$$ – $$\frac{3}{4}$$
8$$\frac{2}{4}$$
Explanation:
9$$\frac{1}{4}$$ – $$\frac{3}{4}$$
= $$\frac{37}{4}$$ – $$\frac{3}{4}$$
= $$\frac{37-3}{4}$$ = $$\frac{34}{4}$$
= 8$$\frac{2}{4}$$
Question 9.
6$$\frac{2}{5}$$ – 4$$\frac{4}{5}$$
1$$\frac{3}{5}$$
Explanation:
6$$\frac{2}{5}$$ – 4$$\frac{4}{5}$$
= $$\frac{32}{5}$$ – $$\frac{24}{5}$$
= $$\frac{32-24}{5}$$ = $$\frac{8}{5}$$
= 1$$\frac{3}{5}$$
Question 10.
8$$\frac{2}{6}$$ – 4$$\frac{5}{6}$$
3$$\frac{3}{6}$$
Explanation:
8$$\frac{2}{6}$$ – 4$$\frac{5}{6}$$
= $$\frac{50}{6}$$ – $$\frac{29}{6}$$
= $$\frac{50-29}{6}$$ = $$\frac{21}{6}$$
= 3$$\frac{3}{6}$$
I’m in a Learning Mindset!
How does a fraction model help me when subtracting mixed numbers that need renaming? | 2022-07-01 22:52:35 | {"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": 0.45720380544662476, "perplexity": 3244.486011676984}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-27/segments/1656103947269.55/warc/CC-MAIN-20220701220150-20220702010150-00031.warc.gz"} |
https://www.wyzant.com/resources/answers/topics/discrete-probability-distribution | 9 Answered Questions for the topic Discrete Probability Distribution
Discrete Probability Distribution Statistics Probability Probability Distribution
08/01/18
The cumulative distribution of the random variable X is given by G(x) = 0 for x<0 3x2-2x3 for 0≤x<1 1 for x ≥1 a)Find the... more
Discrete Probability Distribution Statistics Probability Joint Probability
01/21/17
#### Need Help finding constant C in joint pmf for two discrete variables
Determine the constant c so that f (x, y) satisfies the conditions of being a joint pmf for two discrete random variables X and Y: f (x, y) = c(1/4)^x * (1/3)^y, where x = 1, 2, . . . , y = 1,... more
Discrete Probability Distribution Discrete Math Mathematics Discrete Mathematics
01/02/17
#### Home Work Assignmet
1. How many different positive integers can be made from the digits {2, 4, 6, 8} if repetitions are allowed? 2. What is the telescoping form of f(x) = x4 + 7x3 - x2 + 2x + 1 ? 3. Let r =... more
Discrete Probability Distribution Statistics Question Distribution Probability Distribution
12/15/16
#### Discrete probability distributions
If you select a red diamond picture card (i,e.) a card that doesn't have a number on it) you win $100. It costs$10 to play this game. What is your expected earnings? if you played this game 10,000... more
Discrete Probability Distribution Division Discrete Mathematics Properties Of Divisibility
10/08/15
#### Properties of Divisibility
Let a, b, and c be integers with a≠0. Prove that if a|b then ab|c.
#### what proportion of a normal distribution is within one standard deviation of the mean
how do i figure out what proportion of a normal distribution is within one standard deviation of the mean
Discrete Probability Distribution Probability
12/01/14
#### Discrete Probability - Need Help Please!
Five slips of paper containing the numbers 5, 7, 9, 13, 21 are placed in a hat. If the experiment consists of drawing one number, and if the experiment is repeated a large number of times, what is... more
Discrete Probability Distribution Probability
12/01/14
#### Discrete Probability - Need Help Please!
Suppose an experiment has seven possible outcomes for x: 0, 1, 2, 3, 4, 5, 6. The probability that each of these outcomes occurs is x/21. What is the expected value of x for the experiment?
Discrete Probability Distribution
10/28/14
#### A certain lock has five tumblers, and each tumbler can assume six positions. How many different possibilities are there?
I am doing most of my questions for my homework on probability and I Do not know how solve any of them
## Still looking for help? Get the right answer, fast.
Get a free answer to a quick problem.
Most questions answered within 4 hours.
#### OR
Choose an expert and meet online. No packages or subscriptions, pay only for the time you need. | 2021-07-25 05:49:45 | {"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": 0.7840152382850647, "perplexity": 1156.9312235159828}, "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/1627046151638.93/warc/CC-MAIN-20210725045638-20210725075638-00383.warc.gz"} |
https://ccjou.wordpress.com/page/2/ | ## 每週問題 July 18, 2016
Prove that each of the following statements is true.
(a) If $A=[a_{ij}]$ is skew symmetric, then $a_{ii}=0$ for each $i$.
(b) If $A=[a_{ij}]$ is skew Hermitian, then each $a_{ii}$ is a pure imaginary number.
(c) If $A$ is real and symmetric, then $B=\mathrm{i}A$ is skew Hermitian, where $\mathrm{i}=\sqrt{-1}$.
| | 發表留言
## 每週問題 July 11, 2016
Suppose that $A$ is the coefficient matrix for a homogeneous system of four equations in six unknowns and suppose that $A$ has at least one nonzero row.
(a) Determine the fewest number of free variables that are possible.
(b) Determine the maximum number of free variables that are possible.
| 標記 , , | 發表留言
## 每週問題 July 4, 2016
Suppose that $\begin{bmatrix} A|\mathbf{b}\end{bmatrix}$ is reduced to a matrix $\begin{bmatrix} E|\mathbf{c} \end{bmatrix}$.
(a) Is $\begin{bmatrix} E|\mathbf{c} \end{bmatrix}$ in row echelon form if $E$ is?
(b) If $\begin{bmatrix} E|\mathbf{c} \end{bmatrix}$ is in row echelon form, must $E$ be in row echelon form?
| 標記 | 2 則迴響
## 每週問題 June 27, 2016
Let $A$ be an $n\times n$ matrix. If $A\mathbf{x}=\mathbf{0}$ has nonzero solutions, is it possible that $A^T\mathbf{x}=\mathbf{b}$ has a unique solution for some vector $\mathbf{b}$?
| 標記 | 發表留言
## 每週問題 June 20, 2016
Let $\{\mathbf{q}_1,\mathbf{q}_2,\mathbf{q}_3\}$ be an orthonormal set in $\mathbb{R}^3$ and $\mathbf{q}_3$ be the cross product of $\mathbf{q}_1$ and $\mathbf{q}_2$, i.e., $\mathbf{q}_3=\mathbf{q}_1\times\mathbf{q}_2$. A linear transformation $T:\mathbb{R}^3\to\mathbb{R}^3$ is defined by
$T(\mathbf{x})=\mathbf{x}\times \mathbf{q}_1+(\mathbf{q}_2^T\mathbf{x})\mathbf{q}_1$.
Determine the rank of $T$.
## 每週問題 June 13, 2016
Let $\boldsymbol{\beta}=\{\mathbf{x}_1,\ldots,\mathbf{x}_k\}$ and $\boldsymbol{\gamma}=\{\mathbf{y}_1,\ldots,\mathbf{y}_k\}$ be bases for a subspace $\mathcal{V}$ in $\mathbb{R}^n$. Let $X=\begin{bmatrix} \mathbf{x}_1&\cdots&\mathbf{x}_k \end{bmatrix}$ and $Y=\begin{bmatrix} \mathbf{y}_1&\cdots&\mathbf{y}_k \end{bmatrix}$. Show that the change of coordinates matrix from $\boldsymbol{\beta}$ to $\boldsymbol{\gamma}$ is
$P=(Y^TY)^{-1}Y^TX$.
## 每週問題 June 6, 2016
Let $A$ be an $m\times n$ complex matrix. If $A\mathbf{x}=\mathbf{b}$ is consistent for some $\mathbf{b}$, prove that there exists a unique solution $\mathbf{x}$ in the column space of $A^\ast$.
## 每週問題 May 30, 2016
Let $A$ be an $m\times n$ matrix and $S$ be the solution set for a consistent system of linear equations $A\mathbf{x}=\mathbf{b}$ for some $\mathbf{b}\neq\mathbf{0}$.
(a) If $S_{\max}$ is a maximal independent subset of $S$ and $\mathbf{x}_p$ is any particular solution, show that
$\hbox{span}(S_{\max})=\hbox{span}\{\mathbf{x}_p\}+N(A)$,
where $N(A)$ denotes the nullspace of $A$.
(b) If $\hbox{rank}A=r$, show that $A\mathbf{x}=\mathbf{b}$ has at most $n-r+1$ independent solutions.
## 每週問題 May 23, 2016
Let $A$ and $B$ be $m\times n$ matrices. If $\hbox{rank}A=r$ and $\hbox{rank}B=k\le r$, show that
$r-k\le \hbox{rank}(A+B)\le r+k$.
In words, a perturbation of rank $k$ can change the rank by at most $k$.
## Cayley-Hamilton 定理的一個錯誤「證明」
\begin{aligned} p(\lambda)&=\det(A-\lambda I)=\begin{vmatrix} a-\lambda&b\\ c&d-\lambda \end{vmatrix}\\ &=\lambda^2-(a+d)\lambda+ad-bc,\end{aligned}
Cayley-Hamilton 定理宣稱
$p(A)=A^2-(a+d)A+(ad-bc)I=0$
Cayley-Hamilton 定理有很多種證法 (見“Cayley-Hamilton 定理”)，但其中幾乎挑不出一個簡單的證明。底下這個看似快捷實乃錯誤的「證明」曾經不斷地被初學者重複發現：
$p(A)=\det(A-AI)=\det(A-A)=\det 0=0$ | 2016-09-26 12:13:14 | {"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": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 84, "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": 0.9371249675750732, "perplexity": 310.60441663074107}, "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-2016-40/segments/1474738660760.0/warc/CC-MAIN-20160924173740-00085-ip-10-143-35-109.ec2.internal.warc.gz"} |
https://socratic.org/questions/how-do-you-solve-5x-4-3-3 | # How do you solve |5x-4| +3 = 3?
Apr 29, 2017
See the solution process below:
#### Explanation:
First, subtract $\textcolor{red}{3}$ from each side of the equation to isolate the absolute value function while keeping the equation balanced:
$\left\mid 5 x - 4 \right\mid + 3 - \textcolor{red}{3} = 3 - \textcolor{red}{3}$
$\left\mid 5 x - 4 \right\mid + 0 = 0$
$\left\mid 5 x - 4 \right\mid = 0$
Normally an absolute value equality would produce two answers. However, because the absolute value function is equal to $0$ there is only one solution.
We can equate the term within the absolute value to $0$ and solve for $x$:
$5 x - 4 = 0$
$5 x - 4 + \textcolor{red}{4} = 0 + \textcolor{red}{4}$
$5 x - 0 = 4$
$5 x = 4$
$\frac{5 x}{\textcolor{red}{5}} = \frac{4}{\textcolor{red}{5}}$
$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{5}}} x}{\cancel{\textcolor{red}{5}}} = \frac{4}{5}$
$x = \frac{4}{5}$ | 2019-11-12 05:17:55 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 14, "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": 0.8956325054168701, "perplexity": 364.0985065014567}, "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-2019-47/segments/1573496664752.70/warc/CC-MAIN-20191112051214-20191112075214-00404.warc.gz"} |
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Journal of Global Optimization [SJR: 0.992] [H-I: 60] [4 followers] Follow Hybrid journal (It can contain Open Access articles) ISSN (Print) 1573-2916 - ISSN (Online) 0925-5001 Published by Springer-Verlag [2352 journals]
• An efficient strategy for the activation of MIP relaxations in a multicore
global MINLP solver
• Authors: Kai Zhou; Mustafa R. Kılınç; Xi Chen; Nikolaos V. Sahinidis
Pages: 497 - 516
Abstract: Abstract Solving mixed-integer nonlinear programming (MINLP) problems to optimality is a NP-hard problem, for which many deterministic global optimization algorithms and solvers have been recently developed. MINLPs can be relaxed in various ways, including via mixed-integer linear programming (MIP), nonlinear programming, and linear programming. There is a tradeoff between the quality of the bounds and CPU time requirements of these relaxations. Unfortunately, these tradeoffs are problem-dependent and cannot be predicted beforehand. This paper proposes a new dynamic strategy for activating and deactivating MIP relaxations in various stages of a branch-and-bound algorithm. The primary contribution of the proposed strategy is that it does not use meta-parameters, thus avoiding parameter tuning. Additionally, this paper proposes a strategy that capitalizes on the availability of parallel MIP solver technology to exploit multicore computing hardware while solving MINLPs. Computational tests for various benchmark libraries reveal that our MIP activation strategy works efficiently in single-core and multicore environments.
PubDate: 2018-03-01
DOI: 10.1007/s10898-017-0559-0
Issue No: Vol. 70, No. 3 (2018)
• A proximal bundle method for constrained nonsmooth nonconvex optimization
with inexact information
• Authors: Jian Lv; Li-Ping Pang; Fan-Yun Meng
Pages: 517 - 549
Abstract: Abstract We propose an inexact proximal bundle method for constrained nonsmooth nonconvex optimization problems whose objective and constraint functions are known through oracles which provide inexact information. The errors in function and subgradient evaluations might be unknown, but are merely bounded. To handle the nonconvexity, we first use the redistributed idea, and consider even more difficulties by introducing inexactness in the available information. We further examine the modified improvement function for a series of difficulties caused by the constrained functions. The numerical results show the good performance of our inexact method for a large class of nonconvex optimization problems. The approach is also assessed on semi-infinite programming problems, and some encouraging numerical experiences are provided.
PubDate: 2018-03-01
DOI: 10.1007/s10898-017-0565-2
Issue No: Vol. 70, No. 3 (2018)
• Completely positive and completely positive semidefinite tensor
relaxations for polynomial optimization
• Authors: Xiaolong Kuang; Luis F. Zuluaga
Pages: 551 - 577
Abstract: Abstract Completely positive (CP) tensors, which correspond to a generalization of CP matrices, allow to reformulate or approximate a general polynomial optimization problem (POP) with a conic optimization problem over the cone of CP tensors. Similarly, completely positive semidefinite (CPSD) tensors, which correspond to a generalization of positive semidefinite (PSD) matrices, can be used to approximate general POPs with a conic optimization problem over the cone of CPSD tensors. In this paper, we study CP and CPSD tensor relaxations for general POPs and compare them with the bounds obtained via a Lagrangian relaxation of the POPs. This shows that existing results in this direction for quadratic POPs extend to general POPs. Also, we provide some tractable approximation strategies for CP and CPSD tensor relaxations. These approximation strategies show that, with a similar computational effort, bounds obtained from them for general POPs can be tighter than bounds for these problems obtained by reformulating the POP as a quadratic POP, which subsequently can be approximated using CP and PSD matrices. To illustrate our results, we numerically compare the bounds obtained from these relaxation approaches on small scale fourth-order degree POPs.
PubDate: 2018-03-01
DOI: 10.1007/s10898-017-0558-1
Issue No: Vol. 70, No. 3 (2018)
• An accelerated extended cutting plane approach with piecewise linear
approximations for signomial geometric programming
• Authors: Yiduo Zhan; Qipeng P. Zheng; Chung-Li Tseng; Eduardo L. Pasiliao
Pages: 579 - 599
Abstract: Abstract This paper presents a global optimization approach for solving signomial geometric programming (SGP) problems. We employ an accelerated extended cutting plane (ECP) approach integrated with piecewise linear (PWL) approximations to solve the global optimization of SGP problems. In this approach, we separate the feasible regions determined by the constraints into convex and nonconvex ones in the logarithmic domain. In the nonconvex feasible regions, the corresponding constraint functions are converted into mixed integer linear constraints using PWL approximations, while the other constraints with convex feasible regions are handled by the ECP method. We also use pre-processed initial cuts and batched cuts to accelerate the proposed algorithm. Numerical results show that the proposed approach can solve the global optimization of SGP problems efficiently and effectively.
PubDate: 2018-03-01
DOI: 10.1007/s10898-017-0563-4
Issue No: Vol. 70, No. 3 (2018)
• A trajectory-based method for mixed integer nonlinear programming problems
• Authors: Terry-Leigh Oliphant; M. Montaz Ali
Pages: 601 - 623
Abstract: Abstract A local trajectory-based method for solving mixed integer nonlinear programming problems is proposed. The method is based on the trajectory-based method for continuous optimization problems. The method has three phases, each of which performs continuous minimizations via the solution of systems of differential equations. A number of novel contributions, such as an adaptive step size strategy for numerical integration and a strategy for updating the penalty parameter, are introduced. We have shown that the optimal value obtained by the proposed method is at least as good as the minimizer predicted by a recent definition of a mixed integer local minimizer. Computational results are presented, showing the effectiveness of the method.
PubDate: 2018-03-01
DOI: 10.1007/s10898-017-0570-5
Issue No: Vol. 70, No. 3 (2018)
• A Frank–Wolfe based branch-and-bound algorithm for mean-risk
optimization
• Authors: Christoph Buchheim; Marianna De Santis; Francesco Rinaldi; Long Trieu
Pages: 625 - 644
Abstract: Abstract We present an exact algorithm for mean-risk optimization subject to a budget constraint, where decision variables may be continuous or integer. The risk is measured by the covariance matrix and weighted by an arbitrary monotone function, which allows to model risk-aversion in a very individual way. We address this class of convex mixed-integer minimization problems by designing a branch-and-bound algorithm, where at each node, the continuous relaxation is solved by a non-monotone Frank–Wolfe type algorithm with away-steps. Experimental results on portfolio optimization problems show that our approach can outperform the MISOCP solver of CPLEX 12.6 for instances where a linear risk-weighting function is considered.
PubDate: 2018-03-01
DOI: 10.1007/s10898-017-0571-4
Issue No: Vol. 70, No. 3 (2018)
• Order-based error for managing ensembles of surrogates in mesh adaptive
direct search
• Authors: Charles Audet; Michael Kokkolaras; Sébastien Le Digabel; Bastien Talgorn
Pages: 645 - 675
Abstract: Abstract We investigate surrogate-assisted strategies for global derivative-free optimization using the mesh adaptive direct search (MADS) blackbox optimization algorithm. In particular, we build an ensemble of surrogate models to be used within the search step of MADS to perform global exploration, and examine different methods for selecting the best model for a given problem at hand. To do so, we introduce an order-based error tailored to surrogate-based search. We report computational experiments for ten analytical benchmark problems and three engineering design applications. Results demonstrate that different metrics may result in different model choices and that the use of order-based metrics improves performance.
PubDate: 2018-03-01
DOI: 10.1007/s10898-017-0574-1
Issue No: Vol. 70, No. 3 (2018)
• Partial inverse maximum spanning tree in which weight can only be
decreased under $$l_p$$ l p -norm
• Authors: Xianyue Li; Zhao Zhang; Ding-Zhu Du
Pages: 677 - 685
Abstract: Abstract The maximum or minimum spanning tree problem is a classical combinatorial optimization problem. In this paper, we consider the partial inverse maximum spanning tree problem in which the weight function can only be decreased. Given a graph, an acyclic edge set, and an edge weight function, the goal of this problem is to decrease weights as little as possible such that there exists with respect to function containing the given edge set. If the given edge set has at least two edges, we show that this problem is APX-Hard. If the given edge set contains only one edge, we present a polynomial time algorithm.
PubDate: 2018-03-01
DOI: 10.1007/s10898-017-0554-5
Issue No: Vol. 70, No. 3 (2018)
• Inertial projection and contraction algorithms for variational
inequalities
• Authors: Q. L. Dong; Y. J. Cho; L. L. Zhong; Th. M. Rassias
Pages: 687 - 704
Abstract: Abstract In this article, we introduce an inertial projection and contraction algorithm by combining inertial type algorithms with the projection and contraction algorithm for solving a variational inequality in a Hilbert space H. In addition, we propose a modified version of our algorithm to find a common element of the set of solutions of a variational inequality and the set of fixed points of a nonexpansive mapping in H. We establish weak convergence theorems for both proposed algorithms. Finally, we give the numerical experiments to show the efficiency and advantage of the inertial projection and contraction algorithm.
PubDate: 2018-03-01
DOI: 10.1007/s10898-017-0506-0
Issue No: Vol. 70, No. 3 (2018)
constrained quadratic programming via surrogate constraint
• Authors: Xiaojin Zheng; Yutong Pan; Xueting Cui
Abstract: Abstract We investigate in this paper nonconvex binary quadratically constrained quadratic programming (QCQP) which arises in various real-life fields. We propose a novel approach of getting quadratic convex reformulation (QCR) for this class of optimization problem. Our approach employs quadratic surrogate functions and convexifies all the quadratic inequality constraints to construct QCR. The price of this approach is the introduction of an extra quadratic inequality. The “best” QCR among the proposed family, in terms that the bound of the corresponding continuous relaxation is best, can be found via solving a semidefinite programming problem. Furthermore, we prove that the bound obtained by continuous relaxation of our best QCR is as tight as Lagrangian bound of binary QCQP. Computational experiment is also conducted to illustrate the solution efficiency improvement of our best QCR when applied in off-the-shell software.
PubDate: 2018-02-26
DOI: 10.1007/s10898-017-0591-0
• On branching-point selection for trilinear monomials in spatial
branch-and-bound: the hull relaxation
• Authors: Emily Speakman; Jon Lee
Abstract: Abstract In Speakman and Lee (Math Oper Res 42(4):1230–1253, 2017), we analytically developed the idea of using volume as a measure for comparing relaxations in the context of spatial branch-and-bound. Specifically, for trilinear monomials, we analytically compared the three possible “double-McCormick relaxations” with the tight convex-hull relaxation. Here, again using volume as a measure, for the convex-hull relaxation of trilinear monomials, we establish simple rules for determining the optimal branching variable and optimal branching point. Additionally, we compare our results with current software practice.
PubDate: 2018-02-23
DOI: 10.1007/s10898-018-0620-7
• Convex envelopes of bivariate functions through the solution of KKT
systems
• Authors: Marco Locatelli
Abstract: Abstract In this paper we exploit a slight variant of a result previously proved in Locatelli and Schoen (Math Program 144:65–91, 2014) to define a procedure which delivers the convex envelope of some bivariate functions over polytopes. The procedure is based on the solution of a KKT system and simplifies the derivation of the convex envelope with respect to previously proposed techniques. The procedure is applied to derive the convex envelope of the bilinear function xy over any polytope, and the convex envelope of functions $$x^n y^m$$ over boxes.
PubDate: 2018-02-23
DOI: 10.1007/s10898-018-0626-1
• Minimal curvature-constrained networks
• Authors: D. Kirszenblat; K. G. Sirinanda; M. Brazil; P. A. Grossman; J. H. Rubinstein; D. A. Thomas
Abstract: Abstract This paper introduces an exact algorithm for the construction of a shortest curvature-constrained network interconnecting a given set of directed points in the plane and a gradient descent method for doing so in 3D space. Such a network will be referred to as a minimum Dubins tree, since its edges are Dubins paths (or slight variants thereof). The problem of constructing a minimum Dubins tree appears in the context of underground mining optimisation, where the objective is to construct a least-cost network of tunnels navigable by trucks with a minimum turning radius. The Dubins tree problem is similar to the Steiner tree problem, except the terminals are directed and there is a curvature constraint. We propose the minimum curvature-constrained Steiner point algorithm for determining the optimal location of the Steiner point in a 3-terminal network. We show that when two terminals are fixed and the third varied in the planar version of the problem, the Steiner point traces out a limaçon.
PubDate: 2018-02-21
DOI: 10.1007/s10898-018-0625-2
• Global optimization algorithm for capacitated multi-facility continuous
location-allocation problems
• Authors: Cristiana L. Lara; Francisco Trespalacios; Ignacio E. Grossmann
Abstract: Abstract In this paper we propose a nonlinear Generalized Disjunctive Programming model to optimize the 2-dimensional continuous location and allocation of the potential facilities based on their maximum capacity and the given coordinates of the suppliers and customers. The model belongs to the class of Capacitated Multi-facility Weber Problem. We propose a bilevel decomposition algorithm that iteratively solves a discretized MILP version of the model, and its nonconvex NLP for a fixed selection of discrete variables. Based on the bounding properties of the subproblems, $$\epsilon$$ -convergence is proved for this algorithm. We apply the proposed method to random instances varying from 2 suppliers and 2 customers to 40 suppliers and 40 customers, from one type of facility to 3 different types, and from 2 to 32 potential facilities. The results show that the algorithm is more effective at finding global optimal solutions than general purpose global optimization solvers tested.
PubDate: 2018-02-21
DOI: 10.1007/s10898-018-0621-6
• Integrality gap minimization heuristics for binary mixed integer nonlinear
programming
• Authors: Wendel Melo; Marcia Fampa; Fernanda Raupp
Abstract: Abstract We present two feasibility heuristics for binary mixed integer nonlinear programming. Called integrality gap minimization algorithm (IGMA)—versions 1 and 2, our heuristics are based on the solution of integrality gap minimization problems with a space partitioning scheme defined over the integer variables of the problem addressed. Computational results on a set of benchmark instances show that the proposed approaches present satisfactory results.
PubDate: 2018-02-20
DOI: 10.1007/s10898-018-0623-4
• Bi-objective decision making in global optimization based on statistical
models
• Authors: Antanas Žilinskas; James Calvin
Abstract: Abstract A global optimization problem is considered where the objective functions are assumed “black box” and “expensive”. An algorithm is theoretically substantiated using a statistical model of objective functions and the theory of rational decision making under uncertainty. The search process is defined as a sequence of bi-objective selections of sites for the computation of the objective function values. It is shown that two well known (the maximum average improvement, and the maximum improvement probability) algorithms are special cases of the proposed general approach.
PubDate: 2018-02-19
DOI: 10.1007/s10898-018-0622-5
• MultiGLODS: global and local multiobjective optimization using direct
search
• Authors: A. L. Custódio; J. F. A. Madeira
Abstract: Abstract The optimization of multimodal functions is a challenging task, in particular when derivatives are not available for use. Recently, in a directional direct search framework, a clever multistart strategy was proposed for global derivative-free optimization of single objective functions. The goal of the current work is to generalize this approach to the computation of global Pareto fronts for multiobjective multimodal derivative-free optimization problems. The proposed algorithm alternates between initializing new searches, using a multistart strategy, and exploring promising subregions, resorting to directional direct search. Components of the objective function are not aggregated and new points are accepted using the concept of Pareto dominance. The initialized searches are not all conducted until the end, merging when they start to be close to each other. The convergence of the method is analyzed under the common assumptions of directional direct search. Numerical experiments show its ability to generate approximations to the different Pareto fronts of a given problem.
PubDate: 2018-02-19
DOI: 10.1007/s10898-018-0618-1
• Efficient multicriterial optimization based on intensive reuse of search
information
• Authors: Victor Gergel; Evgeny Kozinov
Abstract: Abstract This paper proposes an efficient method for solving complex multicriterial optimization problems, for which the optimality criteria may be multiextremal and the calculations of the criteria values may be time-consuming. The approach involves reducing multicriterial problems to global optimization ones through minimax convolution of partial criteria, reducing dimensionality by using Peano curves and implementing efficient information-statistical methods for global optimization. To efficiently find the set of Pareto-optimal solutions, it is proposed to reuse all the search information obtained in the course of optimization. The results of computational experiments indicate that the proposed approach greatly reduces the computational complexity of solving multicriterial optimization problems.
PubDate: 2018-02-19
DOI: 10.1007/s10898-018-0624-3
• Surrogate-based feasibility analysis for black-box stochastic simulations
with heteroscedastic noise
• Authors: Zilong Wang; Marianthi Ierapetritou
Abstract: Abstract Feasibility analysis has been developed to evaluate and quantify the capability that a process can remain feasible under uncertainty of model inputs and parameters. It can be conducted during the design stage when the objective is to get a robust design which can tolerate a certain amount of variations in the process conditions. Also, it can be used after a design is fixed when the objective is to characterize its feasible region. In this work, we have extended the usage of feasibility analysis to the cases in which inherent stochasticity is existent in the model outputs. With a surrogate-based adaptive sampling framework, we have developed and compared three algorithms that are promising to make accurate predictions on the feasible regions with a limited sampling budget. Both the advantages and limitations are discussed based on the results from five benchmark problems. Finally, we apply such methods to a pharmaceutical manufacturing process and demonstrate its potential application in characterizing the design space of the process.
PubDate: 2018-02-16
DOI: 10.1007/s10898-018-0615-4
• Corrections to: Differentiable McCormick relaxations
• Authors: Kamil A. Khan; Matthew Wilhelm; Matthew D. Stuber; Huiyi Cao; Harry A. J. Watson; Paul I. Barton
Abstract: Abstract These errata correct various errors in the closed-form relaxations provided by Khan, Watson, and Barton in the article “Differentiable McCormick Relaxations” (J Glob Optim, 67:687–729, 2017). Without these corrections, the provided closed-form relaxations may fail to be convex or concave and may fail to be valid relaxations.
PubDate: 2018-01-25
DOI: 10.1007/s10898-017-0601-2
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APIs | 2018-04-23 07:43:59 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 2, "mathjax_asciimath": 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": 0.5955237150192261, "perplexity": 1550.2239926555442}, "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/1524125945855.61/warc/CC-MAIN-20180423070455-20180423090455-00127.warc.gz"} |
http://inwestbudwl.pl/02xyvu3k/correlation-and-regression-formula-cdc2d4 | # correlation and regression formula
A plot of the data may reveal outlying points well away from the main body of the data, which could unduly influence the calculation of the correlation coefficient. In our correlation formula, both are used with one purpose - get the number of columns to offset from the starting range. Example $$\PageIndex{6}$$ doing a correlation and regression analysis using r. Example $$\PageIndex{1}$$ contains randomly selected high temperatures at various cities on a single day and the elevation of the city. The rest of the labs can be found here. 2. A paediatric registrar has measured the pulmonary anatomical dead space (in ml) and height (in cm) of 15 children. 11.3 If the values of x from the data in 11.1 represent mean distance of the area from the hospital and values of y represent attendance rates, what is the equation for the regression of y on x? Correlation and Regression are the two most commonly used techniques for investigating the relationship between two quantitative variables.. These represent what is called the “dependent variable”. The formula for the best-fitting line (or regression line) is y = mx + b, where m is the slope of the line and b is the y-intercept.This equation itself is the same one used to find a line in algebra; but remember, in statistics the points don’t lie perfectly on a line — the line is a model around which the data lie if a strong linear pattern exists. N = Number of values or elements Note this does not mean that the x or y variables have to be Normally distributed. ΣY2 = Sum of Square of Second Scores, x and y are the variables. The first of these is its distance above the baseline; the second is its slope. The second, regression, Correlation As mentioned above correlation look at global movement shared between two variables, for example when one variable increases and the other increases as well, then these two variables are said to be … Correlation, and regression analysis for curve fitting. Regression uses correlation and estimates a predictive function to relate a dependent variable to an independent one, or a set of independent variables. We already have to hand all of the terms in this expression. You will find Formulas List of Correlation and Regression right from basic to advanced level. The independent variable is not random. The first of these, correlation, examines this relationship in a symmetric manner. That the prediction errors are approximately Normally distributed. In R we can build and test the significance of linear models… ΣXm = Mean of First (X) Data Set The intercept is often close to zero, but it would be wrong to conclude that this is a reliable estimate of the blood pressure in newly born male infants! In this context “regression” (the term is a historical anomaly) simply means that the average value of y is a “function” of x, that is, it changes with x. The Formula for Spearman Rank Correlation $$r_R = 1 – \frac{6\Sigma_i {d_i}^2}{n(n^2 – 1)}$$ where n is the number of data points of the two variables and d i is the difference in the ranks of the i th element of each random variable considered. Correlation describes the strength of an association between two variables, and is completely symmetrical, the correlation between A and B is the same as the correlation between B and A. Also referred to as least squares regression and ordinary least squares (OLS). The correlation coefficient of 0.846 indicates a strong positive correlation between size of pulmonary anatomical dead space and height of child. That the scatter of points about the line is approximately constant – we would not wish the variability of the dependent variable to be growing as the independent variable increases. The residual (error) values follow the normal distribution. 4. The number of pairs of observations was 15. | 2021-08-06 04:37:53 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 2, "mathjax_asciimath": 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": 0.6703816056251526, "perplexity": 330.3288063575119}, "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/1627046152112.54/warc/CC-MAIN-20210806020121-20210806050121-00403.warc.gz"} |
http://stats.stackexchange.com/questions/51450/confidence-intervals-using-standard-deviation-mistake-or-misunderstanding | # Confidence intervals using standard deviation - mistake or misunderstanding?
Hedges et al. 1999, Ecology 80: 1150-1156 reintroduces the old concept of using the natural logarithm of response ratio for ecologists as preferred statistics over p-values in comparative experiments and meta-analyses.
Response ratio, $R = \bar{X}_{E =treatment}/\bar{X}_{C=control}$; $L = ln(R) = ln(\bar{X}_{E}) - ln(\bar{X}_{C})$.
They indicate that taking a logarithm of the response ratio helps to tone down the violations against assumptions of normal distribution and homoscedasticity, which are evident in most experimental data sets.
Authors write: "If $\bar{X}_{E}$ and $\bar{X}_{C}$ are normally distributed and $\bar{X}_{C}$ is unlikely to be negative, then L is approximately normally distributed with mean approximately equal to the true log response ratio and variance, v, approximately equal to"
$\frac{(SD_{E})^2}{n_{E}\bar{X}_{E}^2}$ + $\frac{(SD_{C})^2}{n_{C}\bar{X}_{C}^2}$
They continue: "An approximate 100(1-$\alpha$)% confidence interval for the individual log response ratio parameter $\lambda$ is given by"
$L - z_{\alpha/2}\sqrt{v}\leq \lambda \leq L + z_{\alpha/2}\sqrt{v}$
"where $z_{\alpha/2}$ is the 100(1-$\alpha/2$)% point of the standard normal distribution, and the corresponding confidence interval for the (unlogged) response ratio $\rho$ is obtained by taking the antilogs of the confidence limits for the log response ratio."
In here, as I understand it, the authors indicate that standard deviation instead of standard error should be used to calculate the confidence intervals. Is this my misunderstanding or a mistake in the article?
-
## 1 Answer
I think the equation is correct. v is computed from the two standard deviations but also from the sample sizes. The square root of v is the standard deviation of L. But the standard deviation of a computed parameter is a standard error, just like the standard deviation of a sample mean is the standard error of the mean (which is very different than the standard deviation of the data or the distribution itself).
When you look at a set of values, the standard deviation of those values is very different than the standard error of the mean of those values. But when you look at a estimated (via calculations) parameter, the "standard deviation" of that parameter and the "standard error" of that parameter are really the same. Or more precisely, the standard error is an estimate of the unknown standard deviation.
-
Well, that's a oversimplification. Standard deviation, or the square root of the variance, is a population parameter. Here, the distribution of $L$ is unknown, as it contains both unknown mean and unknown variance. Standard error is the estimate of this unknown standard deviation of the sample statistic. – StasK Mar 6 '13 at 14:36
@StasK: Thanks. I updated the answer accordingly. – Harvey Motulsky Mar 6 '13 at 15:27
Thank you for clarification Harvey. Does this mean that v is a good estimate of standard error if used together with the second equation, but if the equation is changes, for instance, if I replace z-distribution with t-distribution it may not be such a good estimate any longer? – Mikko Mar 7 '13 at 6:49
@largh: I'm sure about SD vs. SEM as explained in my answer. Thanks for pointing out the issue of z vs. t. I'm not sure about that, but think that probably t is the right distribution to use. – Harvey Motulsky Mar 8 '13 at 0:30 | 2015-04-26 08:12:53 | {"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": 0.9154596924781799, "perplexity": 446.27367491257843}, "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-2015-18/segments/1429246654114.44/warc/CC-MAIN-20150417045734-00190-ip-10-235-10-82.ec2.internal.warc.gz"} |
https://www.warriortrading.com/dtc-03/ | Welcome to your ACTIVE - DTC: #3 Quiz
Name Email
1. How many technical characteristics must a stock have to be worth trading for Momentum or Gap and Go Strategies?
2. Trader A says that by trading stocks with high relative volume we ensure that thousands of other day traders are also trading these same stocks. Trader B says that High Relative Volume is not the result of a News Catalyst, but always the result of a technical breakout. Who is correct?
3. Trader A says that trading low volume stocks is asking for trouble. Trader B says that low volume stocks are more likely to be manipulated by a small number of traders or HFT traders. Who is correct?
4. A follow through day is considered:
5. What is a Stock Buyback?
6. Which best describes an activist investor?
7. In 2010, over ___ percent of trading in the market was HFT trading. Trades executed by computer programs and designed to profit from the market.
8. Large Cap stocks are preferred by large investors because they can trade with large amounts of capital. That makes large cap stocks often dominated by High Frequency Trading Algorithms. These Algo's can, make stock price action unpredictable and at times, flat out erratic. What type of stocks are comparatively more predictable, and as a result, more popular among retail traders with smaller accounts?
9. We look for stocks that will likely be trading on high relative volume. What causes high relative volume?
10. In the past when we've seen stocks go from $2-50 within a period of hours or days, what do these stocks all have in common? 11. Do parabolic stocks (stocks that go from$2-50 for example), require a news catalyst to make that type of move?
12. Would we consider OPK a stock worth looking at for a day trade?
13. Would we consider AVGR for a day trade?
14. When a stock is on our gap scanner, and we check the news and see it's a buyout or a merger, would we still consider it for a day trade?
15. Why do companies conduct reverse stock splits?
16. What are important qualities of a stock we're considering trading?
17. We like to trade stocks that are obvious because that means other traders will also be watching the same stock and it will be more likely to trade in a predictable manor. What indicates the stock may be "obvious"?
18. As a small cap momentum trader, there will be periods where the market is extremely hot and we'll see 50-100% movers several times a week. Then there will be periods where the market is very slow and we'll see maybe only one or two 20-30% moves each week. What is the best way to handle the ebb and flow of these cycles?
19. During a hot market cycle, we can get spoiled because buying stocks high, or chasing them, may pay off. How can we balance the desire to maximize on hot streaks while at the same time preventing the creation of bad habits that will then hurt us during cold markets?
20. True or false: By trading only the strongest stocks we can minimize our risk of loss. Trade the best, leave the rest. | 2018-06-21 06:22:00 | {"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": 0.19235698878765106, "perplexity": 2735.1184894364465}, "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-2018-26/segments/1529267864039.24/warc/CC-MAIN-20180621055646-20180621075646-00572.warc.gz"} |
https://codereview.stackexchange.com/questions/1219/print-an-integer-and-its-digits-reversed | # Print an integer and its digits reversed
This Common Lisp program is an exercise to print an integer and its digits reversed to the screen:
(defun read-number () (format t "Enter a number: ~%") (read))
(defun reverse-string (the-string)
(if (eq (length the-string) 0)
""
(concatenate 'string (reverse-string (subseq the-string 1)) (subseq the-string 0 1))))
(defun reverse-digits (the-number) (reverse-string (write-to-string the-number)))
(format t "N->: ~a~%<-N: ~a~%" the-number (reverse-digits the-number)))
• Wow, lots of Lisp exercises. Is this for school, or did you just pick it up for the hell of it? :) – Inaimathi Mar 10 '11 at 2:24
• Just picked it up for the heck of it :) I've always heard Lisp is a good language to learn. – jaresty Mar 10 '11 at 6:27
This problem is more about numbers than strings, so I felt the need to post a non-string-based solution. I've got my original Scheme version, and a Common Lisp adaptation of same.
Scheme version:
(define (reverse-digits n)
(let loop ((n n) (r 0))
(if (zero? n) r
(loop (quotient n 10) (+ (* r 10) (remainder n 10))))))
Common Lisp translation of the Scheme version:
(defun reverse-digits (n)
(labels ((next (n v)
(if (zerop n) v
(multiple-value-bind (q r)
(truncate n 10)
(next q (+ (* v 10) r))))))
(next n 0)))
• I meanwhile wrote my own version of quotient, idiv. remainder and module seem to be synonyms. – user unknown Mar 10 '11 at 4:16
• @user unknown: remainder and modulo have different behaviour if the dividend is negative. (remainder -5 3) => -2, whereas (modulo -5 3) => 1. – Chris Jester-Young Mar 10 '11 at 4:24
• Yeah, I didn't test negative values. For such, I would use a wrapper method. (if (< 0 x) (* -1 (redigit (* -1 x) 0))) – user unknown Mar 10 '11 at 7:12
• +1 - Just keep in mind that TCO isn't guaranteed in CL the way it is in Scheme. – Inaimathi Mar 10 '11 at 15:18
• @Inaimathi: Of course. But how many digits is your number going to have? ;-) – Chris Jester-Young Mar 10 '11 at 15:19
You're talking about digits and integer, but to me, as an non-lisper, it looks as if you operate on Strings.
• I would take the number, modulo 10, print that digit.
• If the rest is > 0 recursively call the function with the (number - digit) / 10.
In most languages with integer-arithmetic you can omit the subtraction, since 37/10 => 3 :==: (37-7)/10 => 3
In scala it would look like this:
def redigit (n: Int, sofar: Int = 0) : Int = {
if (n == 0) sofar else
redigit (n / 10, sofar * 10 + n % 10)}
redigit (123)
321
It uses default arguments. First tests with DrScheme didn't succeed. Here is what I came up with:
;; redigit : number number -> number
(define (redigit in sofar)
(cond [(< in 1) sofar]
[else (redigit (/ in 10) (+ (* 10 sofar) (modulo in 10)))])
)
(redigit 134 0)
But the division is precise and doesn't cut off the digits behind the floting point. I get this error:
modulo: expects type as 1st argument, given: 67/5; other arguments were: 10
I looked for a function toInt, asInt, floor, round and so on for the term (/ in 10) but I didn't find something useful. Maybe you know it yourself?
• You need to use quotient and remainder. See my answer. :-) – Chris Jester-Young Mar 10 '11 at 4:08 | 2020-02-17 23:29:19 | {"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": 0.21642878651618958, "perplexity": 8596.766831961952}, "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-2020-10/segments/1581875143373.18/warc/CC-MAIN-20200217205657-20200217235657-00409.warc.gz"} |
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<3 | 2022-12-06 15:26:58 | {"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": 0.8350664377212524, "perplexity": 1592.973578133695}, "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-2022-49/segments/1669446711108.34/warc/CC-MAIN-20221206124909-20221206154909-00739.warc.gz"} |
https://livingthing.danmackinlay.name/hilbert_space.html | # Inner product spaces
### The most highly developed theory of squaring things
The most well-worn tool in the functional analysis kit. Let’s walk through the classic setup, as a refresher to my dusty brains. Many details will be skipped, much violence will be done.
## Normed spaces
We work on normed spaces which are vector spaces endowed with a notion of distance.
We will be considering particular types of normed spaces, which have the additional structure of inner products, and norms induced by their inner products.
But first!
We take a normed space $$\cc{F}$$, which will be a space of functions $$\cc{T}\to\cc{V}$$. When I say functions I mean ODE solutions, signals, vectors and even the trivial case of scalars, which are constant functions, but since there are functions of functions in the mix, we need to keep constant in mind what we are working with here. I call the$$\cc{T}$$ the index set, which we may as well take $$\mathbb{C}^N,$$ and $$\cc{V}$$ the state set, which can be $$\mathbb{C}^M$$ why not? The other case I care about often is $$\cc{T}=\mathbb{Z}$$ but… later.
### Operators
Let’s say we have two normed spaces, $$\cc{F}$$ and $$\cc{G}.$$ An operator $$\Phi$$ on $$\cc{F}$$ is a map $$\cc{F}\to\cc{G}.$$ We think of it, for the current purpose, as a function-valued function of functions. We call the operator continuous at $$x$$ if for any $$\epsilon>0$$ there exists $$\delta>0$$ such that $$\|x-y\|_{\cc{F}}<\delta\Rightarrow\|\Phi x-\Phi y\|_{\cc{G}}<\epsilon.$$ We are specially interested in a particular type of operator, the functional.
### Functionals
A functional $$\phi$$ on $$\cc{F}$$, is a map $$\cc{F}\to\mathbb{C}$$ (or $$\to\mathbb{R}$$, but let’s come back to that later.). A linear functional $$\phi$$ is one satisfying $$\phi(af+bg)=a\phi(x)+b\phi(g)$$ for each $$f,g\in\cc{F}.$$ A functional is bounded iff there exists a finite $$A$$ such that $$|\phi(f)|\leq A\|f\|$$ for each $$f\in\cc{F}.$$ Note that this does not exclude functionals that are not themselves bounded. It’s easy to show that for linear functionals, boundedness and continuity are the same. (So, what is an example of non-continuous linear functional? Long story.)
Ok, that’s enough of that.
Inner products.
## Inner product space
If $$\cc{F}$$ is also an inner product space over $$\bb{F}$$ then it is associated with an inner product $$\inner{\cdot}{\cdot}:\cc{F} \times \cc{F} \rightarrow \bb{F},$$ which satisfies the following properties for any $$f,g,h\in \cc{F}$$ and $$\alpha \in \bb{F}$$:
1. $$\inner{f}{f}\geq 0$$ with equality iff $$f=0$$
2. $$\inner{f}{g}=\overline{\inner{g}{f}}$$
3. $$\inner{af}{g}=a\inner{f}{g}$$
4. $$\inner{f}{g+h}=\inner{f}{g}+\inner{f}{h}$$
The inner product induces an associated norm $$\norm{f}:=\sqrt{\inner{f}{f}},$$ so the inner product space is also a normed space, but one where the norm has special properties. The most special property is the Cauchy-Schwarz inequality: $|\inner{f}{g}|\leq \norm{f}\norm{g}$ which we can show in the “classic” proof. We assume that $$\inner{g}{ g}>0$$ or there is nothing to do. Next we note that for all $$\alpha,$$ $$\inner{f-\alpha g}{f-\alpha g}$$ is real and positive. Thence, \begin{aligned} 0 &\leq\inner{f-\alpha g}{f-\alpha g}\\ &=\inner{f}{f} + \inner{-\alpha g}{f} + \inner{f}{-\alpha g} + \inner{-\alpha g}{-\alpha g}\\ &=\inner{f}{f} + \inner{f}{-\alpha g}^* + \inner{f}{-\alpha g} + |\alpha|^2\inner{g}{ g}\\ &=\inner{f}{f} + 2\cc{R}(\inner{f}{-\alpha g}) + |\alpha|^2\inner{g}{ g}\\ &=\inner{f}{f} - 2\alpha^*\cc{R}(\inner{f}{g}) + |\alpha|^2\inner{g}{ g}\\ &=\inner{f}{f} - 2\left(\frac{\inner{f}{g}}{\inner{g}{g}}\right)^*\cc{R}(\inner{f}{g}) + \left|\frac{\inner{f}{g}}{\inner{g}{g}}\right|^2\inner{g}{ g} &\text{choosing $$\alpha=\inner{f}{g}/\inner{g}{g}$$ } \\ &=\inner{f}{f} - 2\frac{\inner{f}{g}^*}{\inner{g}{g}}\cc{R}(\inner{f}{g}) + \frac{|\inner{f}{g}|^2}{\inner{g}{g}} \\ 0&\leq\inner{f}{f} \inner{g}{g} - 2\inner{f}{g}^*\cc{R}(\inner{f}{g}) + |\inner{f}{g}|^2 \\ |\inner{f}{g}|^2 &=\inner{f}{f} \inner{g}{g} - 2\inner{f}{g}^*\cc{R}(\inner{f}{g}) \\ &=\inner{f}{f} \inner{g}{g} - 2\inner{f}{g}\inner{f}{g}&\text{ since all terms are real } \\ &\leq\inner{f}{f} \inner{g}{g} \\ \end{aligned}
If $$\cc{F}$$ is complete (i.e. it contains the limit of all convergent sequences within it) we call it a Hilbert space.
The choice of norm has given us lots of magical properties.
Let’s consider the functionals generated by the inner product; take any $$h\in\cc{F}.$$ Then $\phi:f\mapsto\inner{f}{h}$ is clearly a linear functional and it’s simple to show (by Cauchy-Schwarz) that it’s bounded and hence continuous.
The famous Riesz representation theorem, workhorse of statistical learning, is a converse. | 2019-08-24 07:46:52 | {"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": 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": 0.9927845001220703, "perplexity": 573.3998950844663}, "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-2019-35/segments/1566027319915.98/warc/CC-MAIN-20190824063359-20190824085359-00115.warc.gz"} |
http://talkstats.com/threads/little-problem-that-im-struggling-to-solve.67463/ | # Little problem that I'm struggling to solve...
#### jackjack583
##### New Member
The proportion of smokers in a certain population is 25%. What is the probability
that smokers' proportion in a sample of 40 people randomly selected from this
population will be below (strictly less than) 20%.
Surely you would use the binomial table N = 40 P = 0.25 X = 8 (since below 20% = less than 8 people from the sample of 40)
I tried using the binomial table to find P(X<8) then use the binomial equation P(X=8).
So to complete the question I did P(X<8) - P(X=8) = 0.1819.. and I thought 18.19% is the correct answer but I seem to be mistaken? Any help?
The correct answer is 0.2296, why?
Appreciate any help thanks!
#### katxt
##### Active Member
Book answers are not always right. | 2022-05-16 11:52:17 | {"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": 0.8036876320838928, "perplexity": 1064.3958819906481}, "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/1652662510117.12/warc/CC-MAIN-20220516104933-20220516134933-00341.warc.gz"} |
https://answers.yahoo.com/question/index?qid=20091220000010KK09813 | # 抽象代數-conjugate&normal subgroup
1.
Suppose that a is conjugate to b in a group G.
Show that C(a) is also conjugate to C(b) in G
2.
Show that if H and K are normal subgroups of a group G such that
H聯集K={e} , then hk=kh for all h屬於H and k屬於K
Update:
To 淡淡:
Update 2:
!!!
Rating
1. a conjugate to b in G <=> there exists g in G such that g a g^-1 = b
for all A in c(a) , i.e. aA=Aa, try to show gAg^-1 commute with b.
gAg^-1 b = g A (g^-1 b g) g^-1 = g A (a ) g^-1 = g a A g^-1
= g a g^-1 g A g^-1 =b g A g^-1. so we done.
我是物理系的學生，想嘗試修數學系純數的課
常常課本都覺得很有道理
但是一做習題就沒輒...
(你沒說我不知道原來我的問題都很基本....) | 2020-12-01 21:28:22 | {"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": 0.9287163615226746, "perplexity": 4750.461665566436}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-50/segments/1606141681524.75/warc/CC-MAIN-20201201200611-20201201230611-00074.warc.gz"} |
https://www.hsph.harvard.edu/donna-spiegelman/software/meta_subtype_trend/ | # %subtype_trend
The %subtype_trend macro tests whether the exposure-subtype association has a trend across the ordinal cancer subtypes. The user runs separate Cox (for cohort studies) or conditional logistic models (for nested case-control studies) for each subtype, and then tests the heterogeneity hypothesis using the outputs from the separate models, or the user takes the estimates (and standard errors) from the literature and test the heterogeneity hypothesis. In the subtype-specific analysis, the confounders-disease associations are allowed to be different among the subtypes. | 2022-10-06 04:35:11 | {"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": 0.8564193248748779, "perplexity": 6011.622472060428}, "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-2022-40/segments/1664030337723.23/warc/CC-MAIN-20221006025949-20221006055949-00519.warc.gz"} |
https://homework.cpm.org/category/MN/textbook/cc1mn/chapter/2%20Unit%202/lesson/CC1:%202.2.3/problem/2-49 | ### Home > CC1MN > Chapter 2 Unit 2 > Lesson CC1: 2.2.3 > Problem2-49
2-49.
complete the histogram below to represent families of various sizes. Remember that data falling on a tick mark ($3, 6, 9, …$) goes in the bin to the right of that mark. What range do most families fall between? Do you see any families that are much larger or much smaller than other families?
Student # of Family Members LaTrese $4$ James $8$ Phu $7$ Byron $3$ Evan $2$ Diamond $11$ Jackie $5$ Antonio $5$ Shinna $6$ Ryan $8$
Refer to problem 2-13 that you completed in class for help with making a histogram. Try creating your own graph before checking the answer.
Begin by counting how many students have families with less than $3$ people, with $3$ to $5$ people, with $6$ to $8$ people, and with $9$ to $11$ people (since none have more than $12$).
Families with less than $3$ people: $1$
Families with $3$ to $5$ people: $4$
Families with $6$ to $8$ people: $4$
Families with $9$ to $11$ people: $1$
In what range do the most number of family members fall? Remember to answer all of the questions. | 2020-05-24 23:30:38 | {"extraction_info": {"found_math": true, "script_math_tex": 30, "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": 0.23431441187858582, "perplexity": 1679.7766607285423}, "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/1590347385193.5/warc/CC-MAIN-20200524210325-20200525000325-00006.warc.gz"} |
http://blog.mdda.net/ai/2019/04/16/presentation-at-tensorflow | Sam Witteveen and I started the TensorFlow and Deep Learning Singapore group on MeetUp in February 2017, and the twenty-fourth MeetUp, aka TensorFlow and Deep Learning : Images and CNNs, was again hosted by Google Singapore.
My talk was titled “First steps in Deep Learning with TensorFlow 2.0 : CNNs”, and was intended to introduce the elements of a CNN model before explaining the intricacies of back-propagtion (i.e. in some ways it was a different approach from the typical “build from the ground up” NN introductory pitch).
Timothy Liu talked about how to use pretrained models (including from tf.keras.applications), and Sam explored how Progressively growing training works for ResNet-50 (from scratch).
The slides for my talk (as a 400Kb PDF) are here :
If there are any questions about the presentation please ask below, or contact me using the details given on the slides themselves.
PS: And if you liked the content, please ‘star’ my Deep Learning Workshop repo :: | 2019-09-16 03:01:38 | {"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": 0.23161660134792328, "perplexity": 3830.972641542214}, "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-2019-39/segments/1568514572471.35/warc/CC-MAIN-20190916015552-20190916041552-00512.warc.gz"} |
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In this module, students extend their study of functions to include function notation and the concepts of domain and range. Practice e. 3 Pensions 9. , initial value, growth vs decay, rate of increase or decrease) within the context of the situation for (4 out of 5) functions. 5: Interpret the parameters in a linear or exponential function in terms of a context. Students will explore and interpret the characteristics of functions, using graphs, tables, and simple algebraic techniques. An example of an exponential function with one independent variable is: y = a x Does the table represent a linear or an exponential function? X - 1, 2, 3, 4. Standard: A1. The differences are startling. includes all the same ideas as used for the linear and quadratic models. SWL - Science. , initial value, growth vs decay, rate of increase or decrease) within the context of the situation for (4 out of 5) functions. If 0 b 1 the function represents exponential decay. Videos, worksheets, 5-a-day and much more. This video. These functions model things that shrink over time, such as the. Worksheet For Analytical Calibration Curve Worksheet For Analytical Calibration Curve Worksheet For Analytical Calibration Curve Linear And Quadratic Systems — Harder Example Video Identify Where A Function Is Linear Increasing Or Decreasing Interpreting Graphs Of Linear And Nonlinear Functions Linear Equations In Science Math In Science Worksheet For Analytical Calibration Curve. Use our printable 9th grade worksheets in your classroom as part of your lesson plan or hand them out as homework. Thanks for contributing an answer to Mathematics Stack Exchange! Please be sure to answer the question. Exponential functions always have some positive number other than 1 as the base. v Worksheet by Kuta Software LLC Kuta Software - Infinite Algebra 1 Name_____ Exponential Functions Date_____ Period____. But most functions are of other types, and their graphs often involve different, or at least additional, issues in order to complete the graphs clearly and correctly. See more ideas about Exponential functions, Exponential, Algebra 1. Use the menu links on the left-hand side to find resources for a particular topic. - distinguish exponential functions from linear and quadratic functions by making. The graphs of exponential functions are used to analyze and interpret data. Exploration of Functions in Everyday Situations. Some of the worksheets for this concept are Interpreting function graphs algebra, 7 functionswork, Interpreting functions, , Lessonunit plan name key features of graphs swbat, Mathematics ii unit 5 step and piecewise functions part 1, A guide to algebraic functions, Name reading and. Middle School Balanced And Unbalanced Forces Worksheet Answers. This means that as the input increases by 1, the output value will be the product of the base and the previous output, regardless of the value of a. Published by Google Sheets – Report Abuse – Updated automatically every 5 minutesGoogle Sheets – Report Abuse – Updated automatically every 5 minutes. 04 - I can interpret and explain how changing values of the numerical terms in exponential functions impacts the graphs, table, contextual situation, recursive rule or function rule. 5 #1a, 2, 6abdf, 8b, 10be) Transformations of Parent Functions interactive website Combining Transformations worksheet (1. [email protected] Exponential growth is a specific way that a quantity may increase over time. This quiz/worksheet combo will test your ability to solve problems involving exponential and logarithmic functions. About This Quiz & Worksheet. Exponential decay functions also cross the y-axis at (0, 1), but they go up to the left forever, and crawl along the x-axis to the right. The inverses of exponential functions are logarithmic functions. ; We can use a formula to find the derivative of , and the relationship allows us to extend our differentiation formulas to include logarithms with arbitrary bases. THe topic is exponential growth in the form of compounding of money (again). We call the base 2 the constant ratio. Exponential Function 3. Writing Exponential Functions from Sets/Tables d. A Very Big Branch Worksheet Answers. In addition to linear, quadratic, rational, and radical functions, there are exponential functions. Math courses include algebra, geometry, algebra 2, precalculus, and calculus. Evaluate an exponential function at a given value Shopping Center Planning: Looking at Exponential and Linear Models Find the growth factor of an exponential function from a table Create an exponential model for a data set algebraically Use regression to find a linear model Interpret the meaning of the slope of a linear function. Displaying all worksheets related to - Exponential Functions And Graphs. Composition Of Functions Word Problems Worksheet With Answers. 03 - I can translate a graph of an exponential or linear function and rewrite the function rule to reveal the translation. You can check your answer in Sage. 2: Construct exponential functions given a graph, a description of a relationship, or a table. An exponential function is defined by the formula f(x) = a x, where the input variable x occurs as an exponent. Create, interpret and/or analyze tables, charts, and graphs involving data. We are working hard on a new platform for setting, building and monitoring homework. Exponential growth is a specific way that a quantity may increase over time. Extend the properties of exponents to rational exponents. 6 Calculate and interpret the average rate of change of a function (presented symbolically or as a table) over a specified interval. A simple way to know differentiate between the two is to look at the output values when plugging in a number for an unknown variable. f(-3) Evaluate the following for + = 21, 1 3, 1 3 62, 3 xx fx x xx: 4. Some of the worksheets displayed are Exponential functions date period, Work logarithmic function, 11 exponential and logarithmic functions work, Concept 17 write exponential equations, Exponential functions work 1, Exponential functions, Review exponential and logorithmic functions date, 4 1 exponential. Just as in any exponential expression, b is called the base and x is called the exponent. 11 Exponential and Logarithmic Functions Worksheet Concepts: • Rules of Exponents • Exponential Functions – Power Functions vs. We call the base 2 the constant ratio. 2 Use function notation, evaluate functions for inputs in their domains, and interpret statements that use function notation in terms of a context. Learn vocabulary, terms, and more with flashcards, games, and other study tools. In this section, you will find resources that will support your teaching of our AS Maths specification. If you don't see any interesting for you, use our search form on bottom ↓. When b is less than 1, you have an exponential decay function. Use the properties of exponents to interpret expressions for exponential functions. A Very Big Branch Worksheet Answers. For example, f(x)=3xis an exponential function, and g(x)=(4 17. Graphing exponential functions | Exponential and logarithmic functions Interpret an Exponential Equation Modeling Depreciation - Duration: 3:03. Free Algebra 1 worksheets created with Infinite Algebra 1. Exponential Functions. 2 Solving Equations 1. 6 Stretches of Functions 3. (b) Interpret the constants, coefficients, and bases in the context of the problem. Interpret the structure of expressions. The y-intercept of the graph of y = abx is a. The range is the set of all positive numbers if a 0 and all negative numbers if. y = a / (1 + b e-kx), k > 0. For example, the function A = s2 giving the area of a square as a function of its side length is not linear because its graph contains the points (1,1), (2,4) and (3,9), which are not on a straight line. For example, f (x) =2x^3 or f (x) = (x+1)/ (x-1) for x is not equal to 1. Ordering Rational Numbers. DFM is a huge bank of free educational resources for teaching mathematics, with full sets of slides, worksheets, games and assessments that span Year 7 to Further Maths and enrichment resources with a Maths Challenge/Olympiad focus. Linear function word problem U. (The function values can be made as close as desired by taking sufficiently close values of the domain. A Guide to Advanced Algebraic Functions The section, functions, is an incredibly important part of the CAPS curriculum. These Probability Worksheets will produce problems with simple numbers, sums, differences, multiples, divisors, and factors using a pair of dice. The general equation for exponential decay is,. Notes: Discovering Parent Functions/KEY Notes: Graphing Transformations of Functions/KEY Notes: Continuity, Piecewise Functions/KEY Piecewise Worksheet/KEY OPTIONAL Functions Quiz Review WS/KEY Notes: Function Properties: Symmetry, Even, Odd, Neither/KEY Notes: Function Operations/KEY Notes: Function Composition/KEY Notes: Inverse Functions/KEY. The task is an introduction to the graphing of exponential functions. (Limit to linear and exponential functions. Graphing Exponential Functions. In this logarithmic function worksheet, learners solve and graph logarithmic and exponential functions. This unit is about exponential functions, so this is found in Chapter 4. is a quadratic function, and its graph is a parabola. This lesson unit is intended to help you assess how well students are able to interpret exponential and linear functions and in particular, to identify and help students who have the following difficulties: Translating between descriptive, algebraic, tabular, and graphical representation of the functions. The inverses of exponential functions are logarithmic functions. Interpreting Linear and Exponential Functions Core Guide Secondary Math I Understand the concept of a linear or exponential function and use function notation. Interpret expressions for function in terms of the situation they model. Clusters and Outliers - Interpretation. Some of the worksheets for this concept are Graphing exponential, Exponential functions date period, Graphing exponential functions, Concept 17 write exponential equations, Lesson 8 3 of 5 for problem solving and data analysis, A guide to advanced algebraic. Covers the following skills: use linear functions, linear equations, and systems of linear equations to represent, analyze, and solve a variety of problems. AS Maths e-library. ) )𝑓( =3 −123+5 SQUARE ROOT FUNCTION CUBIC FUNCTION The graph of which equation is the graph of f ( x ) = 0 x 0 reflected in the x -axis, translated 8 units left, vertically. 01)^(12t), y = (1. An exponential function usually takes the form of y=a^x. Name Class Date 9-3 Exponential Growth and Decay Going Deeper Essential question: How do you write, graph, and interpret exponential growth and decay functions? When you graph a function f (x) in a coordinate plane, the x-axis represents the independent variable and the y-axis represents the dependent variable. Interpreting Graphs of Functions Interpreting Graphs of Quadratic Equations ( GMAT / GRE / CAT / Bank PO / SSC CGL) Interpreting Graphs of Quadratic Equations is quite easy. If the problems appear in the form of a graph or a table, the following instructions will help you navigate through them. Scaffolded Math and Science. Booklet Day 5: # 1 - 3 Booklet Day 6: # 1 - 3 (whatever you can. Bernstein-gamma functions and exponential functionals of Lévy Processes Article (PDF Available) in ELECTRONIC JOURNAL OF PROBABILITY 23 · July 2018 with 767 Reads How we measure 'reads'. Worksheet by Kuta Software LLC Algebra 2 Practice- Converting from Logarithm to Exponential Name_____ ID: 1 ©G r2K0i1U5U kKHust^aR eS_ovfntCwaafrfev zLJLgCr. Exponential Functions - Notes. ) )𝑓( =3 −123+5 SQUARE ROOT FUNCTION CUBIC FUNCTION The graph of which equation is the graph of f ( x ) = 0 x 0 reflected in the x -axis, translated 8 units left, vertically. Comparing relative magnitudes of functions and their rates of change (for example, contrasting exponential growth, polynomial growth, and logarithmic growth). Showing top 8 worksheets in the category - Exponential Functions. 3 Identify the effect on the graph of replacing f(x) by f(x) + k, k f(x), f(kx), and. Algebra 1 Unit 4: Exponential Functions EOC Review Resources Must Do: Complete the Unit 4 Sample Questions and check your answers. Evaluate a linear function U. Lesson: Exponential Growth and Decay Holt McDougal Larson Algebra 1 • 1st Edition • Glencoe • Algebra 2 • we will learn how to set up and solve exponential growth and decay equations and how to interpret their solutions. a) PlayerFind an exponential model for this data. tionship between two quantities when given in graphical or tabular form. Algebra I Name: _____ Function Notation Worksheet Hour: _____ Date: _____ 1. The table addressing various values of the parameter b in the equation y = ab^x is a scaffolded support to help student inquire into how to interpret the structure of exponential functions. Continuity as a property of functions. Sketch the graph of a function from a verbal description showing key features. 10 Explain that the graph of an equation in two variables is the set of all its solutions plotted in the coordinate plane. Interpret expressions for functions in terms of the situation they model MCC9‐12. (Elsewhere on Purplemath are lessons for graphing exponential and logarithmic functions. The variable b represents the growth or decay factor. Standard: A1. The range is the set of all positive numbers if a # 0 and all negative. exponential functions. An exponential function is a nonlinear function that has the form of. • Apply the slope formula. Graphing Exponential Functions. Some bacteria. Outlier : This is a data point that is completely different from the rest of the data in the set. 13_1__day_1__exponential_and_logarithmic_forms_homework_worksheet__key_. y = a b x, w h e r e a ≠ 0, b > 0. 3 — Interpret the equation y = mx + b as defining a linear function, whose graph is a straight line; give examples of functions that are not linear. Course Material Related to This Topic: Read lecture notes, section 1 on pages 1–2. Graph 2: Average Toe. This worksheet has students identify the vertex (turning point), axis of symmetry, roots (solutions) of quadratic functions. Exponential Functions In this chapter, a will always be a positive number. Amplitude and period for sine and cosine functions worksheet. If f is a function, x is the input (an element of the domain), and f(x) is the output (an element of the range). 2 Function Notation THURSDAY LESSON 1. Point-Slope Form of Linear Equations Slope-Intercept Form of Linear Equations Standard Form of Linear Equations Exponential Growth Graphs of Exponential Functions Function Rules based on Graphs Forms of Linear Equations Exponential Decay Geometric Sequences and Exponential Functions Slope of a Line Using Two Points Write a Function in Slope. Interpreting Graphs and Tables Students learn how to read information from a graph and also the important components that most graphs should have. 5 Logarithmic. Lesson Planet. FUNCTIONS (F) 107 INTERPRETING FUNCTIONS (F. pdf doc ; Exponential Functions - Recognizing exponential functions and their. De nition 1. The Human Digestive System Worksheet Answers Key Fill In The Blank. A sequence of numbers in which the ratio between any two. Interpreting Graphs, Function Notation, and the Vertical Line Test. Exponential Growth and Decay Word Problems & Functions - Algebra & Precalculus - Duration: 12:49. Point-Slope Form of Linear Equations Slope-Intercept Form of Linear Equations Standard Form of Linear Equations Exponential Growth Graphs of Exponential Functions Function Rules based on Graphs Forms of Linear Equations Exponential Decay Geometric Sequences and Exponential Functions Slope of a Line Using Two Points Write a Function in Slope. CHAPTER 7 286 CHAPTER TABLE OF CONTENTS 7-1 Laws of Exponents 7-2 Zero and Negative Exponents 7-3 Fractional Exponents 7-4 Exponential Functions and Their Graphs 7-5 Solving Equations Involving Exponents 7-6 Solving Exponential Equations 7-7 Applications of Exponential Functions Chapter Summary Vocabulary Review Exercises Cumulative Review EXPONENTIAL FUNCTIONS. Area; area under a curve, area between two curves; Volume; volumes by slicing. 1 Write a function that describes a relationship between two quantities. Another type of function, called the logistic function, occurs often in describing certain kinds of growth. This could be a desk, the dinner table, or even a portable clipboard and an armchair. If you don't see any interesting for you, use our search form on bottom ↓. 4pm – Exponential and Logarithmic Functions. b) Understand situations using these combinations (F. Some of the worksheets displayed are Graphing exponential, Exponential functions date period, Graphing exponential functions, Concept 17 write exponential equations, Lesson 8 3 of 5 for problem solving and data analysis, A guide to advanced algebraic functions. Pay-per-use plan C. Choose and apply measures of central tendency (mean, median, and mode) and variability (range and visual displays of distribution). Can be used for Science / C. 5 Interpret the parameters in a linear or exponential function in terms of a context. Composition Of Functions Word Problems Worksheet With Answers. UNIT 6 - EXPONENTIAL FUNCTIONS Linear vs. 13_1__day_1__exponential_and_logarithmic_forms_homework_worksheet__key_. Graphing Exponential Functions Worksheet Answers and 32 New Exponential Growth and Decay Word Problems. Sectionalism Worksheet Answers. ) Build new functions from existing functions MCC9-12. For any positive number a>0, there is a function f : R ! (0,1)called an exponential function that is defined as f(x)=ax. Interpreting and Using Exponential Functions A scientist models the number of grams of a radioactive substance with the expression 15(1/2)^(d/5), where d is the number of days since the sample was measured. 7th Grade Math Worksheets and Answer key, Study Guides. Writing Exponential Functions from Sets/Tables d. If b > 1 the function represents exponential growth. 2 Graph and describe the basic shape of the graphs and analyze the general form of the equations for the following families of functions: linear, quadratic, exponential, piece-wise, and absolute value (use technology when appropriate. 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This lesson unit is intended to help you assess how well students are able to interpret exponential and linear functions and in particular, to identify and help students who have the following difficulties: Translating between descriptive, algebraic, tabular, and graphical representation of the functions. The material was further updated by Zeph Grunschlag. Recognize arithmetic and geometric seq uences as examples of linear and exponential functions (F. WORKSHEETS: AI: Regents-Modeling Linear Functions AI: 9: TST PDF DOC TNS: Regents-Modeling Exponential Functions 1 AI/IA: 10/1: TST PDF DOC TNS: AII: Regents-Modeling Exponential Functions 2 AII: 4: TST PDF DOC TNS: LESSON PLANS. Identify the percent rate of change. Graphing Exponential And Logarithmic Functions For Teachers 10th - 12th. It occurs when the instantaneous rate of change (that is, the derivative) of a quantity with respect to time is proportional to the quantity itself. f(x) = a x. Our 9th grade math worksheets cover topics from pre-algebra, algebra 1, and more!. The general equation for exponential decay is,. Interpreting Functions. WORKSHEETS: AI/AII: Regents-Modeling Linear Functions AI/IA/A: 3/1/4: TST PDF DOC TNS: Regents-Modeling Exponential. Videos, worksheets, 5-a-day and much more. Functions of this kind are called "linear" because their graphs are straight lines: output. This means that as the input increases by 1, the output value will be the product of the base and the previous output, regardless of the value of a. July 24, 2018 Exponential functions can model the rate of change of many situations, including population growth, radioactive decay, bacterial growth, compound interest, and much more. Interpreting information - verify that you can read information regarding functions and square roots and interpret it correctly Knowledge application - use your knowledge to answer questions about. 11 Exponential and Logarithmic Functions Worksheet Concepts: • Rules of Exponents • Exponential Functions – Power Functions vs. 2 1 ) growing or decaying?. We derive the derivatives of inverse exponential functions using implicit differentiation. A Very Big Branch Worksheet Answers. • Sketch and identify key characteristics of the graphs of linear functions. Exponential Functions. Function representing problem: f(x)= 10x + 20. Exponential Functions And Graphs. For example the expression 1. Displaying all worksheets related to - Exponential Graphs. Showing top 8 worksheets in the category - Exponential Graphs. 11_exponential_regression_mini_lesson. Content Strand: Interpreting Functions. If you don't see any interesting for you, use our search form on bottom ↓. Exponential functions are function where the variable x is in the exponent. Composition Of Functions Word Problems Worksheet With Answers. What pattern do you notice in the sequence? This is an example of a geometric sequence. We are working hard on a new platform for setting, building and monitoring homework. Grade 11 maths Here is a list of all of the maths skills students learn in grade 11! These skills are organised into categories, and you can move your mouse over any skill name to preview the skill. Berkeley’s calculus course. Graphs Of Exponential Functions And Logarithic. The graph below shows the exponential functions corresponding to these two geometric sequences. Graphing, Substitution, Elimination, and Systems of Inequalities. of a cooling body by adding a constant function to a decaying exponential, and relate these functions to the model. For example, identify percent rate of change in functions such as y = (1. Watch a video or use a hint. Exponential functions are commonly written with a base of \(e \approx 2. 97)^t, y = (1. Math courses include algebra, geometry, algebra 2, precalculus, and calculus. Linear and nonlinear functions page 2 answer key. • Interpret exponential functions that arise in applications in terms of the context. Learn the properties of odd functions and how to. Lesson 6-1: Interpreting Graphs and Functions Examples Absolute Minimum Absolute Maximum Absolute Minimum Reading Graphs (Extrema) Types of Data Learning Targets Extrema - all maximum and minimum values y-Intercept: the point at which the graph of the function intersects the. education2020. Graphing exponential functions is used frequently, we often hear of situations that have exponential growth or exponential decay. Sketch the graph of a function from a verbal description showing key features. See more ideas about Exponential functions, Exponential, Algebra 1. First function End Behavior: At VA:. 8 Exponential Equations. Exponential Function An. The task is an introduction to the graphing of exponential functions. Differential Equations; Other Bases and Exponential Growth; Applications of Integration. • Construct exponential functions given a graph, a description of a relationship, or a table. The Corbettmaths Practice Questions on Exponential Graphs. Exponential decay describes a function that decreases by a factor every time increases by. SMART notebook lesson. Interpret the Parameters in a Linear Function. Worksheet containing pra. In this function, a represents the starting value such as the starting population or the starting dosage level. Interpreting Functions - Displaying top 8 worksheets found for this concept. , initial value, growth vs decay, rate of increase or decrease) within the context of the situation for (4 out of 5) functions. 17 Know that the logarithm and exponential functions are inverses and use this information to solve real-world problems. Use the properties of exponents to interpret expressions for exponential functions. Exponential Function An. Exponential decay functions also cross the y-axis at (0, 1), but they go up to the left forever, and crawl along the x-axis to the right. pdf: File Size: 102 kb: File Type: pdf. 8 Write a function defined by an expression in different but equivalent forms to reveal and explain different properties of the function. It can be used as a worksheet function (WS) in Excel. The function is continuous and one-to-one. Try our Free Online Math Solver! Online Math Solver. Doppler Effect. Exercise 2It has been observed that a particular plant's growth is directly proportional to time. (Framing Text): Interpret expressions for functions in terms of the situation they model. Some of the worksheets displayed are Graphing exponential, Exponential functions date period, 11 exponential and logarithmic functions work, Graphing exponential functions, Concept 17 write exponential equations, 4 1 exponential functions and their graphs, Graphing exponential functions work, Work logarithmic. The rate of exponential growth or decay is the ratio between two consecutive output values in an exponential function. CHAPTER 7 286 CHAPTER TABLE OF CONTENTS 7-1 Laws of Exponents 7-2 Zero and Negative Exponents 7-3 Fractional Exponents 7-4 Exponential Functions and Their Graphs 7-5 Solving Equations Involving Exponents 7-6 Solving Exponential Equations 7-7 Applications of Exponential Functions Chapter Summary Vocabulary Review Exercises Cumulative Review EXPONENTIAL FUNCTIONS. Mathispower4u 1,805 views. 01) 12t , y = (1. Include equations arising from linear and quadratic functions, and simple rational and exponential functions. When conducting any statistical analysis it is important to evaluate how well the model fits the data and that the data meet the assumptions of the model. Properties of Exponential Graphs of the Form f x b x: (p. Thus we define an exponential function to be any function of the form. is a quadratic function, and its graph is a parabola. It is beyond the scope of this Handbook to discuss more than a few of these. ) Build a function that models a relationship between two quantities MGSE9‐12. o y wAMldl k urMihg jhYt Xse FrqensPeur tvze hd 9. Linear and nonlinear functions page 2 answer key Linear and nonlinear functions page 2 answer key. Interpreting Relationships Presented in Scatterplots, Graphs, Tables, and Equations. We derive the derivatives of inverse exponential functions using implicit differentiation. v Worksheet by Kuta Software LLC Kuta Software - Infinite Algebra 1 Name_____ Exponential Functions Date_____ Period____. For example, f(x)=3xis an exponential function, and g(x)=(4 17. 5) READY, SET, GO Homework: Linear and Exponential Functions 2. Ball players have been signing ever-larger contracts. Section 7: Exponential Functions Section 7 - Topic 1 Geometric Sequences Consider the sequence 3,6,12,24, …. The table addressing various values of the parameter b in the equation y = ab^x is a scaffolded support to help student inquire into how to interpret the structure of exponential functions. Covers the following skills: Understand patterns, relations, and functions. Graphically, in the function f(x) = ab x. Using the universal formula for writing an explicit formula for a geometric sequence, we come up with this function. 2 Function Notation THURSDAY LESSON 1. 1) 2) Find the slope of the line through each pair of poi nts. Graphing Exponential Functions The graph of a function y = abx is a vertical stretch or shrink by a factor of ∣ a ∣ of the graph of the parent function y = bx. The variable b represents the growth or decay factor. Interpreting Exponential Graphs Worksheet. • Know the constant percentage rate of growth or decay of an exponential function. Mathematics Vision Project | MVP - Mathematics Vision Project. Exponential Growth and Decay Word Problems & Functions - Algebra & Precalculus - Duration: 12:49. The Human Digestive System Worksheet Answers Key Fill In The Blank. Today we worked through examples 1, 2 and 3. Interpreting Graphs and Tables Students learn how to read information from a graph and also the important components that most graphs should have. Applications Of Log And Exponential Equations Worksheet Answers. Engaging and rigorous. _Write an exponential function using function notation to represent the situation. Learn the properties of odd functions and how to. Algebra 1 Unit 4: Exponential Functions EOC Review Resources Must Do: Complete the Unit 4 Sample Questions and check your answers. Write a rule for a function table U. Include equations arising from linear and quadratic functions, and simple rational and exponential functions. Linear and nonlinear functions page 2 answer key. Videos, worksheets, 5-a-day and much more. If 0 b 1 the function represents exponential decay. Real Life Graphs Intro sleep. The rate of exponential growth or decay is the ratio between two consecutive output values in an exponential function. ) )𝑓( =3 −123+5 SQUARE ROOT FUNCTION CUBIC FUNCTION The graph of which equation is the graph of f ( x ) = 0 x 0 reflected in the x -axis, translated 8 units left, vertically. In this worksheet, we will practice setting up and solving exponential growth and decay equations and interpreting their solutions. Earthquake Log Worksheet , Answers ü 3103. 4-6) Standard III. Graphing Exponential Functions Worksheet Answers and 32 New Exponential Growth and Decay Word Problems. e is the base rate of growth shared by all continually growing processes. 6 ~ Linear Function Definition and Identification I can identify linear and nonlinear functions represented different ways (verbal description, table, graph, equation). Exponential. Engaging and rigorous. Interpreting Relationships Presented in Scatterplots, Graphs, Tables, and Equations. K 2/6 Desmos Activity 2/7 Sine/Cosine Day 1 2/8 Sine/Cosine Day 2 2/9 Guess My Equation 2/10 Application Day 1 2/13 Period 3-4 Presentations Period 11-12 Presentations. Probability With a Deck of Cards Worksheet These Probability Worksheets will produce problems about a standard 52 card deck without the Jokers. WORKSHEETS: AI/AII: Regents-Modeling Linear Functions AI/IA/A: 3/1/4: TST PDF DOC TNS: Regents-Modeling Exponential. com for a Google drive link to the lesson plan. This video. Lesson Planet. Writing Exponential Functions from Graphs b. This book is suitable for algebra 2 or precalculus students. Exponential functions are commonly written with a base of \(e \approx 2. 4) 01 b: is an exponential function. From section 1. For example, the function A = s² giving the area of a square as a function of its side length is not linear because its graph contains the points (1,1), (2,4) and (3,9), which are not on a. Showing top 8 worksheets in the category - Exponential Graphs. This session reveals that exponential functions are expressed in constant ratios between successive outputs and that quadratic functions have constant second differences. ) )𝑓( =3 −123+5 SQUARE ROOT FUNCTION CUBIC FUNCTION The graph of which equation is the graph of f ( x ) = 0 x 0 reflected in the x -axis, translated 8 units left, vertically. Standard: A1. Sieling for login info) An explanation of how to write an exponential equation from a table 3. Sectionalism Worksheet Answers. First function End Behavior: At VA:. •The domain of a function is the set of all inputs, or x-values of a function. HW Solutions Day 8: 9: Applications of Exponential Functions - Finding and interpreting data - Lab Activity. Graphically, the graph is y = f(x). 2 Graph and describe the basic shape of the graphs and analyze the general form of the equations for the following families of functions: linear, quadratic, exponential, piece-wise, and absolute value (use technology when appropriate. Generally statisticians (which I am not but I. Some examples of exponential functions are f(x) = 2 x , f(x) = 5 x – 2 , or f(x) = 9 2x + 1. Algebra 1 Unit 4: Exponential Functions EOC Review Resources Must Do: Complete the Unit 4 Sample Questions and check your answers. Berkeley’s calculus course. Probability distribution. Find the slope from an. 1 - Interpreting Graphs. Learners should be taught how quadratic equations, factorising and transformations form part of this section. The equation can be written in the form f x =a (1+r)x or abx where b = 1+r Where a is the initial or starting value of. Sectionalism Worksheet Answers. This behavior is different from the behavior of polynomials or rational functions, which behave similarly for large inputs regardless of whether the input is large positive or large negative. Solve using elimination an example, factors of 100, translate algebraic expressions solver, numerical skills pre algebra, rules of linear. Use properties of exponents to interpret expressions for exponential functions. Related Topics: Common Core (Functions) Common Core for Mathematics. Key Components Key Components 2. Mid-Year Exam Review Materials and Practice Exams. 3A (Exponential Functions) Exploring Exponential Functions Worksheet…. BAol Oln GrIi fg2h UtxsL arye Usefr Av6e 7d1. Berkeley's calculus course. Exponential Graphs. exponential). 10/31 Harry Potter Exponential Filled out Worksheet 11/1 Modeling with Exponential Functions Day 1 Filled out Worksheet 11/2 Modeling with Exponential Functions Day 2 Filled out Worksheet 11/3 Schedule Classes 11/6 Go Over Formative 6 Key 11/7 Solved "I Have Who Has" (you can print off and do it again!) 11/8 Summative 6 on Exponential Functions. Example 1: Determine which functions are exponential functions. By (date), when given an exponential function modeling a real world scenario and an exponential growth or decay function with key parts labeled, (name) will identify the key features of the function (e. Chapter Outline 6. Graphing exponential functions | Exponential and logarithmic functions Interpret an Exponential Equation Modeling Depreciation - Duration: 3:03. 472 Exponential and Logarithmic Functions Using this de nition of eand a little Calculus, we can take Equation6. •Recognize,evaluate, and graph exponential functions with base e. Some of the worksheets for this concept are Graphing exponential, Exponential functions date period, Graphing exponential functions, Concept 17 write exponential equations, Lesson 8 3 of 5 for problem solving and data analysis, A guide to advanced algebraic functions, Interpreting function graphs algebra, Math 1a calculus work. Name Class Date 9-3 Exponential Growth and Decay Going Deeper Essential question: How do you write, graph, and interpret exponential growth and decay functions? When you graph a function f (x) in a coordinate plane, the x-axis represents the independent variable and the y-axis represents the dependent variable. Graphing Exponential Functions The graph of a function y = abx is a vertical stretch or shrink by a factor of ∣ a ∣ of the graph of the parent function y = bx. View Notes - Exponential Functions and Their Graphs hmwk ans from MATHEMATIC Honors alg at Central Bucks High School South. Published by Google Sheets – Report Abuse – Updated automatically every 5 minutesGoogle Sheets – Report Abuse – Updated automatically every 5 minutes. In this unit students learn to recognize exponential functions defined by !"=\$%&, where %>0,. This quiz/worksheet combo will test your ability to solve problems involving exponential and logarithmic functions. Interpreting Graphs and Tables Students learn how to read information from a graph and also the important components that most graphs should have. E Graph exponential and logarithmic functions, showing intercepts and end behavior… HSF. 3 — Interpret the equation y = mx + b as defining a linear function, whose graph is a straight line; give examples of functions that are not linear. Video Is the exponential function. The y-intercept of the graph of y = abx is a. | 2020-08-03 20:59:29 | {"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": 0.3556925356388092, "perplexity": 1212.0394351250304}, "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/1596439735833.83/warc/CC-MAIN-20200803195435-20200803225435-00331.warc.gz"} |
http://www.biomedsearch.com/nih/Evaluation-catheter-positioning-neurally-adjusted/19652950.html | Document Detail
From MEDLINE®/PubMed®, a database of the U.S. National Library of Medicine
Full Text Journal Information Journal ID (nlm-ta): Intensive Care Med ISSN: 0342-4642 ISSN: 1432-1238 Publisher: Springer-Verlag, Berlin/Heidelberg Article Information Download PDF © The Author(s) 2009 Received Day: 26 Month: 4 Year: 2009 Accepted Day: 9 Month: 7 Year: 2009 Electronic publication date: Day: 4 Month: 8 Year: 2009 Print publication date: Month: 10 Year: 2009 Volume: 35 Issue: 10 First Page: 1809 Last Page: 1814 ID: 2749172 PubMed Id: 19652950 Publisher Id: 1587 DOI: 10.1007/s00134-009-1587-0 issue-copyright-statement: © Copyright jointly hold by Springer and ESICM 2009
Evaluation of the catheter positioning for neurally adjusted ventilatory assist Jürgen BarwingAff1 Markus AmboldAff1 Nadine LindenAff1 Michael QuintelAff1 Onnen MoererAff1 Address: +49-551-399561 omoerer@gwdg.de Department of Anaesthesiology, Emergency and Intensive Care Medicine, Georg-August University of Göttingen, Robert-Koch-Str. 40, 37075 Göttingen, Germany
Introduction
Neurally adjusted ventilatory assist (NAVA) is a new mode of assisted mechanical ventilation [15]. During NAVA the ventilator is triggered by the electrical activity of the diaphragm (EAdi), assessed by a special gastric tube (EAdi-catheter). Patient–ventilator synchronization is increased by neural control [6]. Pressure support is applied in proportion to the amplitude of the EAdi, which represents direct control by the patients own respiratory center [15, 7].
During NAVA reliable positioning of the EAdi-catheter is mandatory in order to trace a representative EAdi signal from the diaphragm. One method to predict the correct position of a gastric feeding tube is based on the measure from the nose to the ear lobe to the xiphoid process of the sternum (NEX distance) [8]. Using the NEX distance the gastric feeding tube was placed trans nasally more than 10 cm into the stomach in 26% of 99 adult cadavers and five normal adult volunteers. For EAdi-catheter positioning the manufacturer suggests a modified NEX distance which aims to predict the ideal position of the electrode array. There are no studies that evaluated this approach. Therefore we aimed to determine whether or not it is sufficient to place the catheter purely by using the modified NEX distance.
Methods
This observational study was performed after ethics committee approval at the anaesthesiological ICU, University of Goettingen, Germany. NAVA was initiated with the intention to improve patient respirator synchrony and to facilitate respirator weaning.
Patients were treated with a ventilator capable of NAVA (Servo-i, Maquet Critical Care, Solna, Sweden). Pressure support ventilation was used prior to EAdi-catheter positioning.
Patients were in supine position with the upper part of the body elevated in a 30° angle and gastric content was drained via the nasogastric tube. Afterwards the standard tubing was replaced by the EAdi-catheter (16 french diameter, 125 cm long; Maquet Critical Care, Solna, Sweden), inserted nasally to a maximum distance of 80 cm.
For EAdi measurement this tube is mounted with an electrode array of nine electrode rings on the distal part of the catheter at intervals of 16 mm, starting 120 mm from the tip.
Catheter positioning was monitored by a special tool implemented in the ventilator. It displays an EAdi curve and four raw leads not filtered for ECG activity. The electrical activity used for generating the EAdi signal is highlighted. The position of the electrodes in relation to the heart and diaphragm can be estimated by evaluating the different leads for presence/absence of p-wave and QRS complex. During the placement procedure the catheter was pulled out in steps of 1 cm. EAdi signal and electrical activity from raw leads were recorded at each step using special software (NAVA tracker, Maquet Critical Care, Solna, Sweden) for offline analysis until ECG-signals disappeared. The “optimal” catheter position (OPT) was identified by checking the recording for the following three criteria: (1) stable EAdi signal, (2) electrical activity highlighted in central leads of the catheter positioning tool, and (3) absence of p-wave in distal lead. Within the catheter positions that fulfilled the above mentioned criteria we chose the one with the maximum EAdi value as OPT.
Afterwards we calculated the insertion distance of the EAdi-catheter (NEXmod) by the formula proposed by the manufacturer (Maquet catheter packaging).
[Formula ID: Equa]
[\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$${\text{NEX}}_{\bmod } = \;\left( {{\text{NEX}}\; \times \;0. 9} \right) + 1 8\left( {{\text{for}}\;{\text{the}}\; 1 6\;{\text{F}} .\;{\text{EAdi}}\hbox{-}{\text{catheter}}} \right)$$\end{document}]
In the formula the measured NEX distance is multiplied by a correction factor (NEX × 0.9 for nasal catheter insertion) in order to predict the distance to the crural diaphragm. Depending on the catheter size used, a constant is added to compensate for the varying electrode array localisation (“+18” cm for the 16 F, 125 cm catheter).
For NEX measurement the patient was placed supine and the outer canthus of the patient’s eye was vertically aligned with the tragus of the ear to decrease errors resulting from head tilt.
Data on NEXmod and OPT position were compared by Wilcoxon matched pairs.
Results
Patient characteristics of 26 enrolled patients are presented in Table 1.
EAdi-catheter placement was possible in all patients and enteral nutrition via the EAdi-catheter was continued uneventfully during NAVA.
p-Waves were absent in seven patients (six due to atrial fibrillation). In one patient (25 + 1), treated with an extracorporal cardiac assist device, there was no detectable EAdi signal although the patient was able to trigger pressure support ventilation (PSV) by his auxiliary respiratory muscles. Since cervical magnetic stimulation of the phrenic nerves did not result in any diaphragmatic response, bilateral injury was suspected and the patient excluded from further data analysis.
In patient 19 the EAdi-catheter was positioned endotracheally first. The suspected malposition was detected at once by complete lack of stable EAdi and ECG-signals and corrected immediately after verification by chest X-ray.
In patient 24 (8 years post gastrectomy) the catheter positioning tool did not highlight the electrical activity of the diaphragm comprehensibly although it was present.
At NEXmod the EAdi signal was suitable for running NAVA in 18 patients (72%). The remaining seven patients (28%) had an inadequate signal quality, thus NAVA was impossible (Fig. 1). Six of these patients showed the alarm “check catheter position”. NAVA was possible at OPT position in all patients. The NEXmod position was identical with the OPT position in four patients (16%). In general we found the OPT position caudal of the NEXmod (Fig. 2). The median difference was 2 cm and the difference ranged from 3 cm too cranial to a position 12 cm too caudal (P < 0.01). Patient 7 (12 cm difference) had a short neck and low lying ears which led to the suspected diagnosis of a Laurence–Moon–Biedl–Bardet–Syndrom [9, 10] and should be regarded as an outlier. At NEXmod position the catheter positioning tool highlighted the central leads in ten cases (42%). p-Wave was absent in the distal lead in nine cases (47% of patients with p-waves) (Fig. 1). The criterion “highlighted electrical activity in central leads” narrowed the OPT position down to an area of 2–7 cm. In consideration of “absence of p-wave”, the OPT position could additionally be narrowed down to an area of 2–5 cm.
Discussion
During NAVA, correct placement of the EAdi-catheter is mandatory to deduce a reliable EAdi signal for respirator control. The modified NEX distance successfully predicted the EAdi-catheter insertion distance in 18 of 25 patients and was identical with the OPT position in four patients. In combination with the catheter positioning tool, catheter placement was possible in all 25 patients.
Former clinical investigations regarding EAdi-catheter positioning refer to the work of Beck [11], where the optimal catheter position was defined by an electromyographic signal from the central electrodes highest in center frequency and reduced in root mean square. Brander [3] additionally judged the correct position by ECG interpretation.
In this study, different information provided by the catheter positioning tool were considered. The anatomical proximity of the right atrium and diaphragm allows an ECG guided positioning of the EAdi-catheter. The progressively decreasing amplitude of the p-wave from the cranial to the caudal electrodes and its absence in the distal lead indicates a position caudal of the right atrium [3]. The catheter positioning tool highlights the leads used for generating the EAdi signal. A catheter position at which the signal for respirator control is taken from the center of the electrode array is advantageous, because displacement of the diaphragm during in- and expiration will most likely not exceed beyond the electrode array. The alarm “check catheter position” appears in this case and proved to be a reliable indicator of a bad catheter position.
We found a median difference of 2 cm with the OPT situated distal of the NEXmod position. Although NEXmod and OPT positions were identical in only four patients, running the NAVA mode was possible in 18 patients at NEXmod position. The diaphragm forms a muscular “tunnel” of 2–3 cm covering the oesophagus [12]. This explains why we found stable signals over a certain range and could use NAVA at most NEXmod positions. However, in seven patients with a NEXmod to OPT difference greater than 3 cm, running NAVA was impossible at NEXmod position due to impaired EAdi signal detection. Thus the positioning tool did not only ensure the correct catheter position but also allowed running NAVA in those patients were NEXmod failed to predict a good position.
The position of the diaphragm depends on the application of PEEP, body position and intra-abdominal pressure (IAP), respectively [1315]. The patients included in our study were placed supine with the upper part of the body elevated in a 30° angle for catheter positioning and had a median PEEP level of 8 cmH2O. Pressure support ventilation with zero PEEP compared to a PEEP level of 10 cm H2O results in a 2 cm shift caudal regarding the middle part of the diaphragm [14]. Although we didn’t measure intra-abdominal pressure there was no clinical suspicion of such. The body position and application of PEEP may have contributed to the fact, that OPT position was found distal of the NEXmod position.
In one patient missing EAdi led to the diagnosis of bilateral phrenic nerve injury. The patient with suspected Laurence–Moon–Biedl–Bardet–Syndrom represented an obvious problem with NEXmod calculation that was predictable due to the askew anatomy. In one patient (post gastric surgery) the catheter positioning tool did not provide reliable data, nevertheless we were able to treat the patient with NAVA. Thus gastric surgery might be considered as a relative contraindication for EAdi-catheter placement.
Conclusion
Positioning the EAdi-catheter using NEXmod gives a good approximation of the catheter position in most patients. The additional tools for catheter positioning are needed to ensure an optimal position, especially in those patients were NEXmod fails. Further studies based on a larger set of patients are required to enhance the accuracy of the NEX formula taking into account PEEP, body position and IAP, respectively.
Open Access
References
Figures
Tables
[TableWrap ID: Tab1] Table 1
Basic patient characteristics including gender, age, height, body mass index (BMI), Reason for ICU admission (Reason adm.), severity off illness on admission evaluated by SAPS II, positive endexpiratory pressure (PEEP) during EAdi-catheter placement and heart rhythm
Patient Gender Age (years) Height (cm) BMI Reason adm. SAPS II PEEP (cmH2O) Heart rhythm NEXmod (cm) OPT (cm) Diff. (cm)
1 M 56 182 37.7 CHD 29 8 SR 70.2 67 −3.2
2 M 41 190 23.5 TBI 51 8 SR 66.6 71 4.4
3 F 76 165 29.4 CPR 65 8 SR 56.7 65 8.3
4 F 79 150 29.3 AS 52 11 SR 60.3 60 −0.3
5 M 33 190 27.7 TBI 48 7 SR 66.6 66 −0.6
6 F 75 160 30.1 ARDS 79 10 Afib 58.5 61 2.5
7 F 72 160 25.4 Trauma 66 6 SR 54 66 12
8 M 73 174 26.4 TBI 59 8 SR 65.7 66 0.3
9 M 41 190 24.9 AI 54 8 Afib 69.3 71 1.7
10 M 66 174 29.4 Trauma 32 8 SR 66.6 68 1.4
11 M 63 170 26.0 TBI 51 10 SR 61.2 63 1.8
12 M 78 176 24.2 CHD 38 8 Afib 61.8 66 4.2
13 F 64 162 24.8 ARF 59 10 SR 58.2 57 −1.2
14 M 86 175 26.1 Trauma 63 8 SR 64.8 69 4.2
15 F 87 163 24.1 CHF 45 8 SR 59 63 4
16 M 54 189 23.8 TBI 20 7 SR 65 67 2
17 F 64 165 22.0 SAH 58 8 SR 62.1 65 2.9
18 M 68 170 29.4 CHD 19 8 SR 63.9 66 2.1
19 M 82 178 27.5 CHD 24 10 Afib 71.1 69 −2.1
20 M 78 167 19.7 AS 35 6 Afib 61.2 61 −0.2
21 M 35 180 23.1 Sepsis 47 12 SR 68.4 69 0.6
22 F 64 169 26.6 Sepsis 93 10 Afib 63.9 65 1.1
23 F 81 158 22.0 CHD 75 8 SR 61.2 66 4.8
24 M 77 165 29.4 ICH 53 10 SR 63 67 4
25 M 40 185 27.8 Trauma 64 12 SR 69 69 0
25 + 1 M 40 179 32.6 CHF 72 BVAD
Median (Q25/Q75) 68 (55/78) 170 (165/180) 26 (24.1/29.4) 52 (40/64) 8 (8/10) 63.9 (61.2/66.6) 66 (65/68) 1.8 (0/4)
Mean ± SD 65.3 ± 16.0 172.3 ± 11.0 26.4 ± 3.6 51.2 ± 18.0 8.7 ± 1.6 63.5 ± 4.3 65.7 ± 3.3 2.2 ± 3.1
The median exact catheter position estimated by NEXmod was 64 cm compared to 66 cm detected by optimal catheter positioning (OPT) based on a stable EAdi signal, electrical activity highlighted in the central leads of the catheter positioning tool and the absence of p-waves in the distal lead
M male, F female, CHD coronary heart disease, TBI traumatic brain injury, CPR post cardiopulmonary resucitation, AS aortic valve stenosis, ARDS adult respiratory distress syndrome, Trauma severe trauma other than TBI, AI aortic valve insufficiency, ARF acute renal failure, CHF congestive heart failure, SAH subarachnoid hemorrhage, ICH intracerebral hemorrhage, Afib atrial fibrillation, BVAD biventricular cardiac assist device, SR sinus rhythm
Article Categories:Physiological and Technical Notes Keywords: Keywords Neurally adjusted ventilatory assist (NAVA), Neural control, Electrical activation of the diaphragm (EAdi).
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https://math.stackexchange.com/questions/659448/the-fox-and-the-duck-puzzle?noredirect=1 | # The Fox And The Duck Puzzle
A duck, pursued by a fox, escapes to the center of a perfectly circular pond. The fox cannot swim, and the duck cannot take flight from the water. The fox is four times faster than the duck. Assuming the fox and duck pursue optimum strategies, is it possible for the duck to reach the edge of the pond and fly away without being eaten? If so, how?
• The purpose of boldface and italics is to make certain words stand out, to make them special over the other words and catch the eye of the reader on a cursory glance. If everything is boldface and italicized then all the words are standing out, they are all special, in the same way. Therefore none stand out, and none is special. Therefore using boldface and/or italics through the entire post is... illogical. – Asaf Karagila Feb 1 '14 at 12:42
• Must be a duplicate, though I can't find it fast. Check dcg.ethz.ch/members/roger/puzzles – Macavity Feb 1 '14 at 12:43
• Perhaps, this. – David Mitra Feb 1 '14 at 12:45
## Definitions
Let $R$ be the radius of the pond. Let the velocities be $v$ for the duck, and $4v$ for the fox (see diagram).
## Phase 1 - The Headstart
As long as the duck stays with a circle of radius $\frac{R}{4}$, he can ensure that he keeps the fox as far away as possible (on a diametral line to himself) by turning in a spiral, where his maximum outward velocity is given by:
$\dot{r} = v\sqrt{1 - \dfrac{16r^2}{R^2}}$
## Phase 2 - The Escape
Assume now that the duck has reached the point $D$ (as shown) at a radius $r$ from the center (with the fox at point $F$), and wants to begin phase 2. His fastest route to shore takes him to point $S$ and covers a distance of $R-r$, while the fox must cover arc length $R\pi$ to reach $S$. Hence, for the two times:
$t_D = \dfrac{R-r}{v}$ for the duck, and $t_F = \dfrac{R\pi}{4v}$ for the fox.
If the duck is to make safety we need
$\dfrac{R-r}{v} < \dfrac{R\pi}{4v}$ or $r > (1 - \dfrac{\pi}{4}) R \approx 0.2146 R$. Since this is within the spiral zone $(r < \dfrac{R}{4})$,
the duck will be able to safely reach the shore. | 2020-01-19 02:40:10 | {"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": 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": 0.6917223930358887, "perplexity": 581.4514694851499}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-05/segments/1579250594101.10/warc/CC-MAIN-20200119010920-20200119034920-00306.warc.gz"} |
https://tosakaucw.github.io/tags/%E5%80%8D%E5%A2%9E/ | # 「Codeforces 739B」Alyona and a tree - 倍增 + 差分
1 \le n \le 2 \times 10^5
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https://socratic.org/questions/58f6390611ef6b365a190268 | # Question #90268
Apr 19, 2017
Because the positive and negative charges cancel out, and you are left with only neutral charges.
#### Explanation:
An atom has three different subatomic particles:
• Proton - positive charge
• Neutron - neutral charge
• Electron - negative charge
The protons and neutrons live in the nucleus of an atom (its center), and contain most of the atom's mass. The electrons 'orbit' the nucleus in shells, or energy levels.
In a neutral atom, the number of protons = the number of electrons.
Extra:
You can't change the number of protons in the atom of and element without changing that element. If you change the number of neutrons, you have something called an isotope. When you change the number of electrons you have an ion. This positive/negative relationship is seen in the ion charges of atoms.
For example, ${\text{Na}}^{+}$ is a neutral $\text{Na}$ atom with one electron taken away. Normally, a neutral $\text{Na}$ atom has $11$ protons and $11$ electrons.
When it loses one electron, it becomes ${\text{Na}}^{+}$. Why? Well, you started with $11$ protons and $11$ electrons. Take one electron away, and you have $11$ protons, which have a positive charge, and $10$ electrons, which have a negative charge.
Since there is one more positive charge than negative charge (i.e. the sum of positive $11$ and negative $10$), the charge of ${\text{Na}}^{+}$ is positive 1, or $+$. | 2019-08-21 17:54:54 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 14, "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": 0.6128818392753601, "perplexity": 625.9875023370571}, "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-2019-35/segments/1566027316150.53/warc/CC-MAIN-20190821174152-20190821200152-00522.warc.gz"} |
https://msp.org/gt/2022/26-5/p01.xhtml | Recent Issues
The Journal About the Journal Editorial Board Editorial Interests Editorial Procedure Subscriptions Submission Guidelines Submission Page Policies for Authors Ethics Statement ISSN (electronic): 1364-0380 ISSN (print): 1465-3060 Author Index To Appear Other MSP Journals
Topological dualities in the Ising model
Daniel S Freed and Constantin Teleman
Geometry & Topology 26 (2022) 1907–1984
Abstract
We relate two classical dualities in low-dimensional quantum field theory: Kramers–Wannier duality of the Ising and related lattice models in $2$ dimensions, with electromagnetic duality for finite gauge theories in $3$ dimensions. The relation is mediated by the notion of boundary field theory: Ising models are boundary theories for pure gauge theory in one dimension higher. Thus the Ising order/disorder operators are endpoints of Wilson/’tHooft defects of gauge theory. Symmetry breaking on low-energy states reflects the multiplicity of topological boundary states. In the process we describe lattice theories as (extended) topological field theories with boundaries and domain walls. This allows us to generalize the duality to nonabelian groups; to finite, semisimple Hopf algebras; and, in a different direction, to finite homotopy theories in arbitrary dimension.
Keywords
nonabelian Ising model, topological field theory, Turaev–Viro theories, duality
Mathematical Subject Classification 2010
Primary: 57R56, 81T25, 82B20 | 2023-02-06 16:06:26 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 2, "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": 0.6359880566596985, "perplexity": 2797.394839144691}, "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-2023-06/segments/1674764500356.92/warc/CC-MAIN-20230206145603-20230206175603-00337.warc.gz"} |
https://www.authorea.com/users/5445/articles/189999-a-lambert-like-transformation-with-a-logarithmic-kernel/_show_article | # Abstract
This work strongly relates to Lambert series, but uses an alternative kernel in the transformation. Many results are shown from experimental techniques and tabulated below.
# Main
From the product representation of $$e^z$$ we can write $x = \sum_{k=1}^\infty \frac{-\mu(k)}{k} \log(1-x^k)$ or perhaps more nicely $x = \sum_{k=1}^\infty \frac{\mu(k)}{k} \log\left(\frac{1}{1-x^k}\right)$ I then wonder what other functions can be written in terms of this, for example $e^x -1 = \sum_{k=1}^\infty \frac{a_k}{k!}\log(1-x^k)$ where the $$a_k$$ starting from $$a_1$$ are $$-1,0,1,5,23,59,719,839$$ which appear to be A253901 minus $$1$$, an then we have $e^x \approx 1+\sum_{k=1}^\infty \frac{1}{k!}\left(\prod_{d|k}(d-1)!^{\mu\left(\frac{k}{d}\right)}-1\right)\log(1-x^k)$ but this seems to be wrong for the $$12^{th}$$ power and some others, so this is not the full story! We seem to have $x^2 = \sum_{k=1}^\infty \frac{-\mu(k)}{k} \log(1-x^{2k})$ in fact it seems $x^n = \sum_{k=1}^\infty \frac{-\mu(k)}{k} \log(1-x^{nk})$ which is fairly obvious from taking the power of the argument, however we could also have squared the entire sum, so we know that $\sum_{k=1}^\infty \frac{-\mu(k)}{k} \log(1-x^{2k}) = \left(\sum_{k=1}^\infty \frac{-\mu(k)}{k} \log(1-x^{k})\right)^2$ and more generally $\sum_{k=1}^\infty \frac{-\mu(k)}{k} \log(1-x^{nk}) = \left(\sum_{k=1}^\infty \frac{-\mu(k)}{k} \log(1-x^{k})\right)^n$ interestingly we can write $\log(1-x^m)= \sum_{n=1}^\infty \sum_{k=1}^\infty \frac{\mu(k)}{nk}\log(1-x^{mnk})$ and we must have the expansion $e^x =1 -\sum_{n=1}^\infty \sum_{k=1}^\infty \frac{\mu(k)}{n!k}\log(1-x^{nk})$ we can get from this the generating function of the Bell Numbers $e^{e^x-1}= \prod_{n=1}^\infty \prod_{k=1}^\infty (1-x^{nk})^{\frac{-\mu(k)}{kn!}}$ of course this is all for $$x<1$$. The product seems to converge for constants for example $$e^{e^{1/2}-1}$$. Other interesting results $\frac{x}{(1-x)} = \sum_{n=1}^\infty \sum_{k=1}^\infty \frac{-\mu(k)}{k}\log(1-x^{nk})$ $\frac{x}{(1-x)^2} = \sum_{n=1}^\infty \sum_{k=1}^\infty -n\frac{\mu(k)}{k}\log(1-x^{nk})$ $O\left[\frac{1}{\zeta(s-1)}\right] = \sum_{n=1}^\infty \sum_{k=1}^\infty \frac{n\mu(n)\mu(k)}{k}\log\left(\frac{1}{1-x^{nk}}\right)$ $\sum_{n=1}^\infty A101035(n)\frac{x^n}{n} = \sum_{n=1}^\infty \sum_{k=1}^\infty \mu(n)\mu(k)\log\left(\frac{1}{1-x^{nk}}\right)$ \begin{aligned} -\log(1-x) = \sum_{n=1}^\infty \sum_{k=1}^\infty \mu(n)\log\left(\frac{1}{1-x^{nk}}\right)\\ -\log(1-x) = \sum_{n=1}^\infty \sum_{k=1}^\infty \mu(k)\log\left(\frac{1}{1-x^{nk}}\right)\\ -\log(1-x) = \sum_{n=1}^\infty \sum_{k=1}^\infty \frac{\mu(n)}{nk}\log\left(\frac{1}{1-x^{nk}}\right)\\ -\log(1-x) = \sum_{n=1}^\infty \sum_{k=1}^\infty \frac{\mu(k)}{nk}\log\left(\frac{1}{1-x^{nk}}\right)\\\end{aligned} $\sum_{n=1}^\infty \sum_{d|n} d\tau(d)x^n = \sum_{n=1}^\infty \sum_{k=1}^\infty \log\left(\frac{1}{1-x^{nk}}\right)\\$
where $$O$$ hypothetically maps a Dirichlet generating function to an ordinary generating function. The last one is a statement that $\prod_{n=1}^\infty e^{\sum_{d|n} d \tau(d) x^n}= \prod_{n=1}^\infty \prod_{k=1}^\infty \frac{1}{1-x^{n k}}$ this has a few similarities to the Euler product of the zeta function $\zeta(s) = \prod_{\text{primes}} \frac{1}{1-p^{-s}}$ if we let $$x=1/p$$, then we get $\prod_{n=1}^\infty e^{\sum_{d|n} d \tau(d) p^{-n}}= \prod_{n=1}^\infty \prod_{k=1}^\infty \frac{1}{1-p^{-n k}}$ $\prod_{n=1}^\infty \prod_{d|n}e^{d \tau(d) p^{-n}}= \prod_{n=1}^\infty \prod_{k=1}^\infty \frac{1}{1-p^{-n k}}$ this then is a relation $\prod_{n=1}^\infty \prod_{d|n}e^{d \tau(d) p^{-n}}= \prod_{n=1}^\infty \frac{1}{(\frac{1}{p^n};\frac{1}{p^n})_\infty}$ interestingly we could write $\prod_{n=1}^\infty \prod_{d|n}e^{d \tau(d) p^{-n}}= \left(\prod_{k=1}^\infty \frac{1}{1-p^{-k}}\right)\prod_{n=2}^\infty \prod_{k=1}^\infty \frac{1}{1-p^{-n k}}$ then take the product over all primes $\prod_{\text{primes}}\frac{\prod_{n=1}^\infty \prod_{d|n}e^{d \tau(d) p^{-n}}}{\prod_{n=2}^\infty \prod_{k=1}^\infty \frac{1}{1-p^{-n k}}}= \prod_{\text{primes}}\left(\prod_{k=1}^\infty \frac{1}{1-p^{-k}}\right)$ $\prod_{\text{primes}}\frac{\prod_{n=1}^\infty \prod_{d|n}e^{d \tau(d) p^{-n}}}{\prod_{n=2}^\infty \prod_{k=1}^\infty \frac{1}{1-p^{-n k}}}= \prod_{k=1}^\infty\zeta(k)$ of course this is probably nonsense
# Other Interesting Sums
we can change $$\mu(n)$$ to be something else for example we seem to have $\sum_{n=1}^\infty\sum_{k=1}^\infty \frac{\varphi(k)}{k}\log\left(\frac{1}{1-x^{kn}}\right) = \sum_{n=1}^\infty d(n)x^n$ $\sum_{n=1}^\infty\sum_{k=1}^\infty \frac{\varphi(k)}{nk}\log\left(\frac{1}{1-x^{kn}}\right) = \sum_{n=1}^\infty \frac{\sigma_1(n)}{n}x^n$ $\sum_{n=1}^\infty\sum_{k=1}^\infty \frac{\varphi(k)}{1}\log\left(\frac{1}{1-x^{kn}}\right) = \sum_{n=1}^\infty \frac{\sigma_2(n)}{n}x^n$ for Euler totient function $$\varphi(n)$$ and divisors $$d(n)$$. $\sum_{n=1}^\infty\sum_{k=1}^\infty k\log\left(\frac{1}{1-x^{kn}}\right) = \sum_{n=1}^\infty \sum_{d|n}\frac{d \sigma_1(d)}{n}x^n$ $\sum_{n=1}^\infty\sum_{k=1}^\infty \sigma_0(k)\log\left(\frac{1}{1-x^{kn}}\right) = \sum_{n=1}^\infty \sum_{d|n}\frac{d \sigma_1(\frac{n}{d})\tau(d)}{n}x^n$ $\sum_{n=1}^\infty\sum_{k=1}^\infty k^2\log\left(\frac{1}{1-x^{kn}}\right) = \sum_{n=1}^\infty \sum_{d|n}\frac{d^3 \sigma_1(\frac{n}{d})}{n}x^n$ $\sum_{n=1}^\infty\sum_{k=1}^\infty \frac{1}{k}\log\left(\frac{1}{1-x^{kn}}\right) = \sum_{n=1}^\infty \sum_{d|n}\frac{\sigma_1(d)}{n}x^n$ other results are,$$k/n$$ transforms to $$A007433$$,$$(k-1)/k$$ transforms to $$A069914/n$$.
We seem to have $\sum_{n=1}^\infty\sum_{k=1}^\infty \frac{1}{n}\left(\sum_{d|n}\mu(d)\mu\left(\frac{n}{d}\right)\right)\log\left(\frac{1}{1-x^{kn}}\right) = \sum_{n=1}^\infty \frac{\varphi(n)}{n}x^n$ $\sum_{n=1}^\infty\sum_{k=1}^\infty \left(\sum_{d|n}\mu(d)\mu\left(\frac{n}{d}\right)\right)\log\left(\frac{1}{1-x^{kn}}\right) = \sum_{n=1}^\infty \frac{1}{n}\left(\sum_{d|n}d \mu(d)\right)x^n$ $\sum_{n=1}^\infty\sum_{k=1}^\infty \lambda(n)\log\left(\frac{1}{1-x^{kn}}\right) = \sum_{n=1}^\infty \frac{\text{sum of square divisors}(n)}{n}x^n$
# Other
$\sum_{n=1}^\infty\sum_{k=1}^\infty \lambda_3(n)\log\left(\frac{1}{1-x^{kn}}\right) = -\log(1-x)+\sum_{n=2}^\infty x^{n^3} + \sum_{k=2}^\infty \sum_{n=2}^\infty \frac{x^{k^3 n}}{n}$
where $$\lambda_3(n)$$ is the equivalent of the Liouville function for squares but for cubes. | 2017-11-25 03:51:26 | {"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": 0.9487918615341187, "perplexity": 223.32785408484273}, "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-2017-47/segments/1510934809392.94/warc/CC-MAIN-20171125032456-20171125052456-00778.warc.gz"} |
http://www.phy.ntnu.edu.tw/ntnujava/msg.php?id=8261 | two anti-parallel vectors $\vec{A}$ and $\vec{B}$ means $\vec{B}=-\vec{A}$
The sum of two vectors $\vec{A}$ and $\vec{B}$ is $\vec{A}+\vec{B}= \vec{A}+(-\vec{A})=\vec{0}$ | 2018-07-20 20:37:18 | {"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": 6, "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": 0.9240513443946838, "perplexity": 89.74153735850794}, "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-2018-30/segments/1531676591831.57/warc/CC-MAIN-20180720193850-20180720213850-00363.warc.gz"} |
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# Outlining his strategy for nursing the troubled conglomerate
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Outlining his strategy for nursing the troubled conglomerate [#permalink] 03 Jan 2013, 13:56
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Hi,
I am a non-native English speaker. I was going through the practice questions on OG 13th edition and I found myself confused when came across the below:
OG 13th Edition SC Q68 -
Outlining his strategy for nursing the troubled conglomerate back to health, the chief executive's plans were announced on Wednesday fpr cutting the company'shuge debt bv selling nearly $12 billion in assets over the next 18 months. A executive's plans were announced on Wednesday for cutting the company's huge debt byselling nearly$12 billion in assets overthe next 18 months
B executive's plans, which are to cut the company's huge debt by selling nearly $12 billion in assets over the next 18 months, were announced on Wednesday C executive's plans for cutting the company's huge debt by selling nearly$12 billion in assets over the next 18 months were announced on Wednesday
D executive announced plans Wednesday to cut the company's huge debt by selling nearly $12 billion in assets over the next 18 months E executive announced plans Wednesday that are to cut the company's huge debt by selling nearly$12 billion in assets over the next 18 months
OA: executive announced plans Wednesday to cut the company's huge debt by selling nearly $12billion in assets over the next 18 months. I got the answer wrong because I thought there should be 'on' preceding Wednesday. Could someone explain this to me please? Thanks a lot for your time [Reveal] Spoiler: OA Kaplan Promo Code Knewton GMAT Discount Codes GMAT Pill GMAT Discount Codes Moderator Joined: 01 Sep 2010 Posts: 2740 Followers: 598 Kudos [?]: 4631 [0], given: 755 Re: Outlining his strategy for nursing the troubled conglomerate [#permalink] 03 Jan 2013, 14:57 Expert's post This question is a mess. and difficult But what you have to do as first thing is to ask your self what the CEO did ?? announced something ----> so the plans do not announced itself also the first threee choices are in passive voice and are incorrect Quote: A executive's plans were announced on Wednesday for cutting the company's huge debt byselling nearly$12 billion in assets overthe next 18 months
B executive's plans, which are to cut the company's huge debt by selling nearly $12 billion in assets over the next 18 months, were announced on Wednesday C executive's plans for cutting the company's huge debt by selling nearly$12 billion in assets over the next 18 months were announced on Wednesday
So now we remain with D and E
As such, E has the problem the words: $$that are to cut the company's huge$$ seems that the plans cut the huge debt by itself
So D is the best choice
Thanks
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Re: Outlining his strategy for nursing the troubled conglomerate [#permalink] 03 Jan 2013, 15:21
Hi Carcass, thanks for the quick response. I understood how to eliminate the choices from OG's explanation section but my question was more fundamental than that - ie is it correct to say "CEO announced plans Wednesday" rather than "CEO announced plans on Wednesday". Am I missing a basic concept of English here?
p.s - apologies for not posting the answer choices with my question.
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Re: Outlining his strategy for nursing the troubled conglomerate [#permalink] 03 Jan 2013, 15:52
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nave81 wrote:
Hi Carcass, thanks for the quick response. I understood how to eliminate the choices from OG's explanation section but my question was more fundamental than that - ie is it correct to say "CEO announced plans Wednesday" rather than "CEO announced plans on Wednesday". Am I missing a basic concept of English here?
p.s - apologies for not posting the answer choices with my question.
as far as I know is a question of style
The preposition of time ON is used with sunday monday and so on......BUT (just yesterday this thing was said by E-Gmat instructor here on the board) is a question of PURE American english style to omit it
I hope is clear now to you
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Re: Outlining his strategy for nursing the troubled conglomerate [#permalink] 03 Jan 2013, 15:52
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# Outlining his strategy for nursing the troubled conglomerate
Powered by phpBB © phpBB Group and phpBB SEO Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®. | 2016-02-10 09:59:48 | {"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": 0.22629354894161224, "perplexity": 9759.51473367947}, "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/1454701159031.19/warc/CC-MAIN-20160205193919-00135-ip-10-236-182-209.ec2.internal.warc.gz"} |
https://space.stackexchange.com/questions/18449/what-limits-thrust-to-weight-ratio-of-ion-thusters-beside-power-density-of-ene | # What limits thrust-to-weight ratio of ion thusters? (beside power density of energy source)
If we consider nuclear-rocket technology not feasible (whether due to technological or political limitations), the second best options for future transportation in solar system are perhaps solar-electric ships... there are many concepts to use ion-thrusters in combination with photovoltaics in large scale - both historical and more recent.
Thin-film solar cells can ultimately provide power source with density something like 6W/g. In future it can be in principle developed may be up to 100-400 W/g (10-40% conversion efficiency in 1 micron foil at Earth orbit).
Therefore it makes sense to ask what are the inherent limits (maximum thrust-to-weight ratio) of the ion thruster, assuming the problem of sufficiently light power source is solved.
Now what about Ion thrusters? What makes them heavy? I guess current experimental realizations do not try to push thrust-to-weight up hard enough, simply because they don't have enough power to feed them. But in combination with these thin-film solar cell, the bottleneck starts to be the weight of thrusters.
Intuitively I assume classical ion thrusters with grid electrodes can be made lighter than Hall thrusters. While the grids can be made of thin metallic foils or wires, the Hall thrusters need quite heavy electromagnets. But I don't really know the engineering challenges which have to be solved in order to make them lighter.
What I really like with this respect is this concept where the thrusters are distributed over the solar array
• Do you want to account for the solar panels mass ? They consume lots of power. Oct 2, 2016 at 11:47
• I'm guessing that a good answer will mention the need for copper and possibly iron for confinement magnets, and electronics for DC as well as RF power supplies.
– uhoh
Nov 2, 2016 at 5:59
• Amtzi - I explainded why I think mass of solar panels could be reduced so much that bottleneck start to be elsewhere Nov 2, 2016 at 7:53
• It's also worth noting that solar powered ion thrusters loose efficiency as they move further from the sun. I remember reading somewhere that for this reason, currently ion probes are out for anything beyond the asteroid belt. Nov 10, 2018 at 0:14
• I suspect part of the problem is the space charge limit, which makes it hard to fit very much charged particle beam in a small space. Jan 5, 2020 at 23:01
The engines aren't particularly heavyweight, but we're handling lots of power in a very tiny volume. Lots of power means cooling. Gas accelerated to these energies becomes an extremely corrosive plasma, best held at bay by magnetic fields because otherwise the engine will be burning through itself. So - heavy electromagnets to guide the propellant. The electrical subsystems handle pretty high power at very high voltages, so again, not really lightweight. But overall, the engines aren't particularly much heavier than chemical engines of comparable power. The 'weight' part of thrust-to-weight is so-so, nothing really out of ordinary.
But now let's look at the thrust.
$$E_k = {1 \over 2} m v^2 \ \ (1)$$
$$p = mv \ \ (2)$$
These are the equations for kinetic energy and momentum.
Ion engines are all about maximizing performance; specific impulse. $$I_{sp}={ v_e\over g_0}$$.
To achieve maximum performance, you want to maximize the exhaust gas velocity. You only have a certain amount of energy to handle, your solar panels or other energy source as input. Take equation (1) To get as much performance - as much exhaust velocity of the gas, given constant accessible energy, you must reduce mass - specifically, take less propellant, apply same electric energy squeezing it into lower amount of propellant, achieve higher acceleration of the propellant, higher exhaust velocity. And still you're getting only $$v = \sqrt{2E_k \over m}$$ - a square root growth of velocity with both increase of energy/power or reduction of mass; diminishing returns although still worth it. In other words, an engine of twice the performance of another will either require 4x as much power, or 1/4 the fuel flow, the amount of propellant used per unit of time - and the engine, its power sources, the structure, doesn't get any lighter in the process of improving the performance.
And now let's look at how that impacts thrust. The rocket motion is based off conservation of momentum. There are many fancy equations that describe it in terms of differential time, change of mass over time etc, but it all boils down to the simplest approach, momentum from equation (2) is conserved: $$v_{rocket} m_{rocket} = v_{exhaust} m_{exhaust}$$.
Now what did we just do to get the most performance out of our engine? We reduced $$m_{exhaust}$$ linearly, to increase $$v_{exhaust}$$ in a square root proportion. The better performance the lower the right-hand term of the above equation. $$m_{rocket}$$ didn't improve, our engine is just as heavy as less efficient. Therefore $$v_{rocket}$$ suffers. Over a unit of time, our rocket gains less velocity, so it accelerated less - we lost thrust.
And this, the fact that with given a certain accessible engine power, increasing performance by a factor of $$\sqrt{n}$$ you reduce exhaust mass flow by a factor of $$n$$, leads inevitably to loss of thrust (and no weight savings), therefore the better the engine performance, for given power, the worse TWR is to be expected.
Ion thrusters need a power source. And power sources can be massive.
This was a major objection to Franklin Chang Diaz' claim that VASIMR could get to Mars in 39 days. He assumed an alpha of .5 kg/KWe. Which isn't doable with present day state of the art. So what would a power source look like that cranks out a kilowatt electricity per half kilogram? I tried to illustrate it. A screen capture from my The Need For A Better Alpha
Dominique is a 60 kilogram girl. If she had that kind of alpha she could do the work of a Ford Focus' engine along with the gasoline and oxygen.
I believe thin film photovoltaic arrays have the potential for a good alpha. If we can get solar arrays that deliver 250 watts per kilogram I believe 1 mm/sec^2 acceleration is doable.
• In the original question I sketched some power source (thin foil phtovoltaics) which can achieve such power density. Other possibilities may be high temperature nuclear reactors with magnetohydrodynamic or thermoelectric generators. It is not of-the-shelf thechnology, but it does not require any theoretical breakthrough to develop. As I explain in original question, the possibility to construct such power source is the very motivation why to thing about possible boundaries to minimize mass of ion engine. Jan 5, 2020 at 14:28
• I guess the text to the orginal question was not clear enough, so I modified it now. Jan 5, 2020 at 14:34
There are 2 things that limit basically all electric rockets. One is the power supply. An electric rocket is only as good as it’s power supply. Whether it is solar or nuclear, in many designs, the weight of the power generation is so great that it usurps the advantage of needing little propellant. As you pointed out, the increasing performance and light weight of solar cells is encouraging.
The other big limitation is the density of the exhaust stream. Because these are extremely hot, they are naturally of low mass and large volume. So the problem becomes how to compress the exhaust stream so you can have more thrust.
I don’t know that the weight of the engine itself is as much of a problem. As you pointed out a design with magnetic coils (like the VASMIR) is heavier, but the coils enable it to squeeze the exhaust stream to make it more dense, so maybe it cancels out the weight burden.
• increasing density of exhaust stream is one possibility, other is to increase area of the exhaust "nozzle". What I tried to suggest - if electrostatic ion thruster needs basically just 2 electrode grids ... those grids can be very lightweight (like alluminium foil) ... why not make ion thuster like 10g/m^2 ? ... but in many current electrostatic ion thuster desings there is lot of other heavy stuff besides the grids ... is it really necessary? Or there is just not enough motivation for aggressive weight reduction. Oct 3, 2016 at 21:44
• Using a very lightweight electrode grid is not possible due to grid erosion.
– Uwe
Oct 24, 2016 at 9:05
• I always thought the biggest limitation comes from erosion of the electrode by high energy ions.
– Aron
Nov 2, 2016 at 3:51
• Can you support the following sentences: 1) "Because these are extremely hot, they are naturally of low mass and large volume." and 2): "So the problem becomes how to compress the exhaust stream so you can have more thrust." Or at least add links to clearer explanations?
– uhoh
Nov 2, 2016 at 5:12
• I suspect that even if you handwaved the mass of the thruster itself down to zero, the limits of chemical, solar or RTG power supply mass would leave you with very poor performance, and I would be surprised if even nuclear fission couldn't deliver 1:1 TWR (i.e. < 10 m/s^2). Someone other than me should run the math. Nov 2, 2016 at 5:20
I agree with the other comments regarding thin film photovoltaics being a good source of power. I built an ion thruster that is patented for lifting its power supply against Earth's gravity. A normal xenon ion thruster or "ion lifter," can't lift its power supply because the thrust to weight or mass ratio is too low. In a xenon thruster for instance, electrical energy is used to knock electrons loose from the xenon atoms which requires a fair amount of wattage and that creates power losses especially when considering the added losses in the electrical system and heat generated in the exhaust. Those engines also have erosion issues and normally must run for long periods of time to have much effect. "Ion lifters," were basically toys that required a large external power supply and rely on ambient O2 molecules primarily as the propellant.
The Ion Propelled Vehicle or Self-Contained Ion Powered "Aircraft," just adds electrons to a small percentage of either ambient O2 molecules or to O2 or SF6 supplied by optional onboard propellant tanks. O2 has a strong affinity for gaining extra electrons and so does not require the same ionization energy, it just absorbs electrons produced by the power supply. While the voltage of the power supply is much higher than in a xenon thruster, the current and wattage is far lower, so it was possible to reduce the mass of the power supply significantly. The erosion on the collector surfaces is minimized because it works by a different principal with very little wattage required. | 2022-07-03 23:23:59 | {"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": 11, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5413315296173096, "perplexity": 1187.5103299312073}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-27/segments/1656104277498.71/warc/CC-MAIN-20220703225409-20220704015409-00114.warc.gz"} |
http://meta.stackexchange.com/questions/133653/tooltips-for-user-suspension-annotation-view | # Tooltips for user suspension/annotation view
As you can see, user one has a grey box with a one in it and user two has a red box with a two in it.
For some reason they are in different font sizes and alignments as well.
But...uh...I really have no idea what this means. I assume the red box means "bad" but with no tool tip I don't know what these counts refer to. Could we get a little title tool tip explaining "1 somethings" on hover? I assumed the count in Timed Suspension meant how many previous suspensions they have, but a user with a previous suspension still shows with 0 there (identical to one who's been suspended for the first time)
-
In "Latest User Annotations" the number is the number of annotations and the background is red if the user is currently suspended. </educated guess> – Yannis May 25 '12 at 15:27
@YannisRizos they're both suspended. I was hoping for a tooltip though, since an explanation only helps people that read the Q – Ben Brocka May 25 '12 at 15:49
Then it could be a bug, because in the "Latest User Annotations" on ProgSE only the currently suspend ones show with a red background. – Yannis May 25 '12 at 15:52
@YannisRizos doesn't match up to only current suspensions on either site I mod – Ben Brocka May 25 '12 at 20:09 | 2015-01-27 14:25:55 | {"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": 0.4429737329483032, "perplexity": 2884.184532458612}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-06/segments/1422121981339.16/warc/CC-MAIN-20150124175301-00237-ip-10-180-212-252.ec2.internal.warc.gz"} |
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#### SAS~Skylla
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##### logic gates and switches
« on: May 26, 2018, 08:16:14 AM »
Hi Gents,
what until now has been missing badly in the BoX ME were some kind of logic components. I have created a few switch and gate groups that should help to get around this.
All these groups are based on a query principle. That means, incoming links set the status of a switch or gate and this status gets saved within the group. The outgoing signal only gets produced on request.
This is due to the fact that incoming signals will only very seldom come in at the exact same time. Also this way you have to chance to "remember" in your mission if a certain signal was already sent or not.
Below you can find a description of the components and how to use them. I hope they will be as useful for your missioneering as they are for my
1. Boolean switch
2. Ten way switch
3. Switch concatenation
4. Logic gates
4.1 General structure
4.2 Gate concatenation
1. Boolean switch:
The basic idea behind it is pretty simple: When you are building a mission, you might want to base some decisions about what is going to happen next on wheter a certain signal was already sent in the past. For this there is the boolean switch that can either result in a true or a false output.
The switch looks like this:
As you can see, the switch has three inputs and two outputs:
• IN: set true: Will set the switch to state true, but will produce no output signal
• IN: set false: Will set the switch to state false, but will produce no output signal
• IN: query: Triggering this will query the state of the switch and will either lead to a false or a true output signal
• OUT: true: Indicates that the switch state is true
• OUT: false: Indicates that the switch state is false
The switch is based on the fact that in the BoX ME a Timer Trigger can have two states: it can either be activated or deactivated. When the switch is queried and the testfalse timer is active, it will deactivate the true output and trigger a false output. If it got deactivated, the waittrue timer will produce a true output after 10 ms. Cool, huh?
Please note that the switch will lead to a false output by default.
2. Ten way switch:
Sometimes just two states might not be sufficient to trace what's going on in a mission. That's where the next group, the ten way switch might come in handy (sorry, but I had to leave out the MCU names because it would get messy otherwise):
The switch has eleven inputs and eleven outputs. The topmost row of timer MCUs are the inputs, the lowest row are the outputs.
The leftmost timer queries the switch state and will lead to one of the outputs to fire, depending on the switch state. The other inputs will set the switch state to state 1 ... 10, beginning from the second timer MCU. Please note that those will only set the switch state and will not trigger an output singal.
The lowest MCU row are the outputs. If the switch got queried, they will show the switch state from 0 (leftmost) to 10 (rightmost).
As you may have noticed it is impossible to set the switch state to 0. In fact this state shows that the switch has not yet been used. It will only be active if none of the states 1 to 10 has been set during the mission yet.
The idea behind the switch is again based on timer (de)activation: If a status is active, it's check state timer (middle row) will be activated, all other check state timers will be deactivated and therefore cannot produce an output signal.
3. Switch concatenation:
If you should ever come around an occasion where even ten switch states are not enough, you can also concatenate two (or more ...) ten way switches. It works like this:
We sacrifice the state 10 of switch 1. It's output gets target linked to the query input of switch 2. All set state inputs of switch two get an additional target link pointing to the set state 10 input of switch 1.
Now if you choose a state of switch two and query the overall state (with the query input of switch 1!), switch 1 will lead to a state ten output. This one will query the state of switch 2 which will then lead to the desired state output.
Please note that in- and outputs of state 10 / switch 1 must not be used for any further signal processing!
4. Logic gates:
Now that we have the possibility to save a signal or a signal state, we will also like to process them further. If you make a dynamic mission you will soon come to the point where you will want to base your decisions on whether certain signals have been sent and others haven't, or whether some switch is in one position or not in another ..
That's where logic gates come into play. Depending on two input signals (each is either false or true) they produce one output signal which can be false or true.
I have created six logic gate groups: AND, OR, NOT-AND (NAND), NOT-OR (NOR), exclusive OR (XOR) and exclusive NOT-OR (XNOR). You can find their truth tables below:
4.1 General structure:
The signal processing in the logic gates is a little more complex and differs from gate to gate, therefore I will just discuss their structure in general here:
The basis of each logic gate are two boolean switches as introduced earlier.
Therefore each logic gate has five inputs and two outputs:
• IN: State 1 TRUE: Will set the state of input signal 1 to true, but will produce no output signal
• IN: State 1 FALSE: Will set the state of input signal 1 to false, but will produce no output signal
• IN: State 2 TRUE: Will set the state of input signal 2 to true, but will produce no output signal
• IN: State 2 FALSE: Will set the state of input signal 2 to false, but will produce no output signal
• IN: query: Triggering this will start the query process and will either lead to a false or a true output signal
• OUT: TRUE: Gets triggered if the result is true
• OUT: FALSE: Gets triggered if the result is false
4.2 Gate concatenation:
Just like the switches the gates can also be concatenated - may it be to just allow more input signals for "one" gate or in order to combine different gates with each other. Below you can see how it is done:
The outputs of gate 1 are connected to the corresponding signal inputs of gate 2. Furthermore they are connected to a 5 ms timer which will start the query process in gate 2. In this example we therefore get an AND gate that allows three input signals.
Credits
• Idea and implementation: SAS~Skylla
• Many thanks to SAS~Storebror for beta testing!
+---------------------------------------------------------------------------------------------| Logic gates and switches for BoX ME, v1.0+---------------------------------------------------------------------------------------------Contents:- boolean switch group- ten way switch group- AND gate group- OR gate group- NAND gate group- NOR gate group- XOR gate group- XNOR gate group+---------------------------------------------------------------------------------------------| Installation:+---------------------------------------------------------------------------------------------Import the groups via ME.+---------------------------------------------------------------------------------------------| How To Use It:+---------------------------------------------------------------------------------------------1. Boolean switch: The switch has three inputs and two outputs: - IN: set true: Will set the switch to state true, but will produce no output signal. - IN: set false: Will set the switch to state false, but will produce no output signal. - IN: query: Triggering this will query the state of the switch and will either lead to a false or a true output signal. - OUT: true: Indicates that the switch state is true. - OUT: false: Indicates that the switch state is false.2. Ten way switch: The switch has eleven inputs and eleven outputs: - IN: query: Triggering this will query the state of the switch and will lead to an output signal. - IN: set state 1 .. 10: will set the switch state, but will produce no output signal. - OUT: state 0: This output will only get triggered if no state has been set on the switch yet. - OUT: state 1 .. 10: Indicates the switch state. You can concatenate two (or more) then way switches in order to get more possible states. In order to do this, target link the state 10 output of switch 1 to the query input of switch 2. All set state inputs of switch 2 get an additional target link pointing to the set state 10 input of switch 1. Now if you choose a state of switch two and query the overall state (with the query input of switch 1!), switch 1 will lead to a state ten output. This one will query the state of switch 2 which will then lead to the desired state output. Please note that in- and outputs of state 10 / switch 1 must not be used for any further signal processing!3. Logic gates: In general each logic gate has five inputs and two outputs: - IN: State 1 TRUE: Will set the state of input signal 1 to true, but will produce no output signal. - IN: State 1 FALSE: Will set the state of input signal 1 to false, but will produce no output signal. - IN: State 2 TRUE: Will set the state of input signal 2 to true, but will produce no output signal. - IN: State 2 FALSE: Will set the state of input signal 2 to false, but will produce no output signal. - IN: query: Triggering this will start the query process and will either lead to a false or a true output signal. - OUT: TRUE: Gets triggered if the result is true. - OUT: FALSE: Gets triggered if the result is false. You can find the truth tables for the switches below. "0" stands for false, "1" for true. AND OR NAND NOR XOR XNOR A | B | AB A | B | AB A | B | AB A | B | AB A | B | AB A | B | AB 0 | 0 | 0 0 | 0 | 0 0 | 0 | 1 0 | 0 | 1 0 | 0 | 0 0 | 0 | 1 0 | 1 | 0 0 | 1 | 1 0 | 1 | 1 0 | 1 | 0 0 | 1 | 1 0 | 1 | 0 1 | 0 | 0 1 | 0 | 1 1 | 0 | 1 1 | 0 | 0 1 | 0 | 1 1 | 0 | 0 1 | 1 | 1 1 | 1 | 1 1 | 1 | 0 1 | 1 | 0 1 | 1 | 0 1 | 1 | 1 You can also concatenate logic gates. For this, the outputs of gate 1 are connected to the corresponding signal inputs of gate 2 (true to true, false to false). Furthermore they are connected to a 5 ms timer which will start the query process in gate 2. +---------------------------------------------------------------------------------------------| Versions:+---------------------------------------------------------------------------------------------- v1.0: initial public release+---------------------------------------------------------------------------------------------| Credits:+---------------------------------------------------------------------------------------------- Idea and implementation: SAS~Skylla- Many thanks to SAS~Storebror for beta testing!+---------------------------------------------------------------------------------------------| Bug reports:+---------------------------------------------------------------------------------------------The only spot I will look out for bug reports are this Mod's release topics on www.sas1946.com and on www.forum.il2sturmovik.comI will NOT care for bug reports that do not contain both of the following:- a DETAILED description of the problem and how it can be reproduced (minimal working example)- the acknowledgment that you have read this readme. This means you should literally write "I have read and understood the readme" somewhere in your post.I repeat, bug reports missing any of these points will get ignored.To those who may think of writing a help request via PM to me: Don't you dare! Use the forum so others can benefit from the answer as well ;) >>>>> I hope you'll have a blast with this! See you in the skies! <<<<<______________________________________________________________________________________________26.05.2018, SAS~Skylla @ www.sas1946.com
Best Regards,
skylla
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#### SAS~Skylla
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##### Re: logic gates and switches
« Reply #1 on: May 26, 2018, 08:16:26 AM »
reserved
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#### SAS~Storebror
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##### Re: logic gates and switches
« Reply #2 on: May 26, 2018, 11:14:20 AM »
These are extremely useful Groups and they're very well crafted.
Thanks a lot for sharing your work with us
Mike
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##### Re: logic gates and switches
« Reply #3 on: July 07, 2018, 12:53:51 AM »
One question (I'm on the run, that's why I didn't test myself):
Say I want to set/reset the logic gate with output.
The logical way would be to trigger both the "set" input (true or false) and the "query" input from an MCU or whatever else at the same time.
Will this generate a race condition (i.e. will I have to introduce a further timer before triggering the "query" input) or will it work well?
Mike
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##### Re: logic gates and switches
« Reply #4 on: July 07, 2018, 11:47:55 AM »
Alright, now I've had time to test.
Turns out that indeed there is a race condition, so in order to set/query at the same time, add a 100ms timer in front of the query input or change the query input properties (which is a timer itself) to 100ms delay.
Mike
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##### Re: logic gates and switches
« Reply #5 on: July 08, 2018, 02:00:17 AM »
sorry, was not on the PC yesterday ..
you are right, on the logic gates this will not work without an extra timer trigger in front of the query input. This is shown (maybe a little hidden) in the gate concatenation section.
When I tested the concenated gates I found a delay of 5 ms in front of the query input to be sufficient.
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Page created in 0.032 seconds with 25 queries. | 2019-05-22 04:10:21 | {"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": 0.5875424146652222, "perplexity": 4460.435352117711}, "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-2019-22/segments/1558232256724.28/warc/CC-MAIN-20190522022933-20190522044933-00180.warc.gz"} |
https://search.r-project.org/CRAN/refmans/catSurv/html/estimateThetas.html | estimateThetas {catSurv} R Documentation
## Estimates of Ability Parameters for a Dataset of Response Profiles
### Description
Estimates the expected value of the ability parameter \theta, conditioned on the observed answers, prior, and the item parameters for complete response profiles for a dataset of respondents.
### Usage
## S4 method for signature 'Cat'
estimateThetas(catObj, responses)
### Arguments
catObj An object of class Cat responses A dataframe of complete response profiles
### Details
Estimation approach is specified in estimation slot of Cat object.
The expected a posteriori approach is used when estimation slot is "EAP". This method involves integration. See Note for more information.
The modal a posteriori approach is used when estimation slot is "MAP". This method is only available using the normal prior distribution.
The maximum likelihood approach is used when estimation slot is "MLE". When the likelihood is undefined, the MAP or EAP method will be used, determined by what is specified in the estimationDefault slot in Cat object.
The weighted maximum likelihood approach is used when estimation slot is "WLE". Estimating \theta requires root finding with the “Brent” method in the GNU Scientific Library (GSL) with initial search interval of [-5,5].
### Value
The function estimateThetas returns a vector containing respondents' estimated ability parameters.
### Note
This function is to allow users to access the internal functions of the package. During item selection, all calculations are done in compiled C++ code.
This function uses adaptive quadrature methods from the GNU Scientific Library (GSL) to approximate single-dimensional integrals with high accuracy. The bounds of integration are determined by the lowerBound and upperBound slots of the Cat object.
### Author(s)
Haley Acevedo, Ryden Butler, Josh W. Cutler, Matt Malis, Jacob M. Montgomery, Tom Wilkinson, Erin Rossiter, Min Hee Seo, Alex Weil
Cat-class, estimateTheta
### Examples
## Loading ltm Cat object
data(ltm_cat)
## Set different estimation procedures and estimate ability parameter
data(npi)
setEstimation(ltm_cat) <- "EAP"
estimateThetas(ltm_cat, responses = npi[1:25, ])
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http://www.mathnet.ru/php/archive.phtml?wshow=paper&jrnid=mzm&paperid=705&option_lang=eng | RUS ENG JOURNALS PEOPLE ORGANISATIONS CONFERENCES SEMINARS VIDEO LIBRARY PERSONAL OFFICE
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Mat. Zametki: Year: Volume: Issue: Page: Find
Mat. Zametki, 2001, Volume 69, Issue 6, Pages 912–918 (Mi mz705)
On Groups with Finite Involution and Locally Finite 2-Isolated Subgroup of Even Period
A. I. Sozutov
Abstract: A proper subgroup $H$ of a group $G$ is said to be strongly isolated if it contains the centralizer of any nonidentity element of $H$ and 2-isolated if the conditions $C_G(g)\cap H\ne1$ and $2\in\pi(C_G(g))$ imply that $C_G(g)\le H$. An involution $i$ in a group $G$ is said to be finite if $|ii^g|<\infty$ ($\forall g\in G$). In the paper we study a group $G$ with finite involution $i$ and with a 2-isolated locally finite subgroup $H$ containing an involution. It is proved that at least one of the following assertions holds:
• 1) all 2-elements of the group $G$ belong to $H$;
• 2) $(G,H)$ is a Frobenius pair, $H$ coincides with the centralizer of the only involution in $H$, and all involutions in $G$ are conjugate;
• 3) $G=F\leftthreetimes C_G(i)$ is a locally finite Frobenius group with Abelian kernel $F$;
• 4) $H=V\leftthreetimes D$ is a Frobenius group with locally cyclic noninvariant factor $D$ and a strongly isolated kernel $V$, $U=O_2(V)$ is a Sylow 2-subgroup of the group $G$, and $G$ is a $Z$-group of permutations of the set $\Omega=\{U^g\mid g\in G\}$.
DOI: https://doi.org/10.4213/mzm705
Full text: PDF file (190 kB)
References: PDF file HTML file
English version:
Mathematical Notes, 2001, 69:6, 833–838
Bibliographic databases:
UDC: 512.544
Citation: A. I. Sozutov, “On Groups with Finite Involution and Locally Finite 2-Isolated Subgroup of Even Period”, Mat. Zametki, 69:6 (2001), 912–918; Math. Notes, 69:6 (2001), 833–838
Citation in format AMSBIB
\Bibitem{Soz01} \by A.~I.~Sozutov \paper On Groups with Finite Involution and Locally Finite 2-Isolated Subgroup of Even Period \jour Mat. Zametki \yr 2001 \vol 69 \issue 6 \pages 912--918 \mathnet{http://mi.mathnet.ru/mz705} \crossref{https://doi.org/10.4213/mzm705} \mathscinet{http://www.ams.org/mathscinet-getitem?mr=1861573} \zmath{https://zbmath.org/?q=an:1029.20017} \elib{http://elibrary.ru/item.asp?id=5022586} \transl \jour Math. Notes \yr 2001 \vol 69 \issue 6 \pages 833--838 \crossref{https://doi.org/10.1023/A:1010290717481} \isi{http://gateway.isiknowledge.com/gateway/Gateway.cgi?GWVersion=2&SrcApp=PARTNER_APP&SrcAuth=LinksAMR&DestLinkType=FullRecord&DestApp=ALL_WOS&KeyUT=000169913100024}
• http://mi.mathnet.ru/eng/mz705
• https://doi.org/10.4213/mzm705
• http://mi.mathnet.ru/eng/mz/v69/i6/p912
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Citing articles on Google Scholar: Russian citations, English citations
Related articles on Google Scholar: Russian articles, English articles
This publication is cited in the following articles:
1. V. I. Senashov, A. I. Sozutov, V. P. Shunkov, “Investigation of groups with finiteness conditions in Krasnoyarsk”, Russian Math. Surveys, 60:5 (2005), 805–848
2. Jabara E., “A Note on Groups Covered by Conjugates of a Proper Subgroup”, J. Algebra, 370 (2012), 171–175
• Number of views: This page: 113 Full text: 34 References: 9 First page: 1 | 2019-03-24 12:26:50 | {"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": 0.2636823356151581, "perplexity": 3547.704936187403}, "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/1552912203438.69/warc/CC-MAIN-20190324103739-20190324125739-00274.warc.gz"} |
https://topanswers.xyz/tex?q=587 | Charles (imported from SE)
After my reading of a nice topic <https://tex.stackexchange.com/questions/59742/progress-bar-for-latex-beamer> on progress bar, I want to know if it is possible to do a progress bar which take into account all the whole slides and also the number of slides in the current subsection.
To be more precise, I would like a very simple progress bar as in <https://tex.stackexchange.com/questions/413760/change-the-color-or-the-progress-bar-indicator-in-the-beamer-metropolis-theme>, but when we can distinguish the different subsection. The aim of this is to allow people to know where we are in the subsection and in the entire presentation.
To be more precise with what I imagine, the principle would be the same as the one with the circle progress bar, you can tell how many slides remains in the current subsection and in the entire presentation. So I would like to do the same but with a progress bar, maybe by adding some vertical bars indicating the limits between the differents sections (subsections).
samcarter (imported from SE)
One could draw a simple progress bar with subsection marks in tikz:
\documentclass{beamer}
\usepackage{totcount}
\newtotcounter{mysub}
\newcounter{foo}
\usepackage{refcount}
\usepackage{tikz}
\setbeamertemplate{footline}{%
\begin{tikzpicture}
\draw[ultra thick] (0,0) -- (\thepage/\insertdocumentendpage*\paperwidth,0);
\foreach \x in {1,...,\totvalue{mysub}}{%
\setcounterpageref{foo}{mysub:\x}
\draw[thick] (\thefoo/\insertdocumentendpage*\paperwidth,0) -- (\thefoo/\insertdocumentendpage*\paperwidth,0.5);
}
\end{tikzpicture}
}
\begin{document}
\section{section name}
\subsection{section name}
\begin{frame}
\end{frame}
\begin{frame}
\end{frame}
\subsection{section name}
\begin{frame}
\end{frame}
\subsection{section name}
\begin{frame}
\end{frame}
\section{section name}
\subsection{section name}
\begin{frame}
\end{frame}
\begin{frame}
\end{frame}
\end{document}
[![enter image description here][1]][1]
[1]: https://i.stack.imgur.com/qd6Ba.gif
Enter question or answer id or url (and optionally further answer ids/urls from the same question) from
Separate each id/url with a space. No need to list your own answers; they will be imported automatically. | 2022-06-25 14:57:01 | {"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": 0.7538095116615295, "perplexity": 1174.4816505707277}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-27/segments/1656103035636.10/warc/CC-MAIN-20220625125944-20220625155944-00604.warc.gz"} |
https://edmundlth.github.io/posts/introduction-to-singular-models/ | Introduction to Singular Learning Theory
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# Introduction
Let’s first set the context. Imagine we are given a data generating process $q(x)$ where we can ask for $N \in \N$ samples Throughout, we assume that the process generates i.i.d. samples. In particular, $$X_i \sim q$$ for all $$i$$, with $$q$$ unchanging as we ask for more samples. However, we note that this is a simplification: one would imagine that an (artificial) intelligent "student" would judiciously ask for more informative examples from a "teacher" process. , $D_N = { X_1, \dots, X_N }$. Our goal is to learn a distribution $p(x)$ from which we can make inferences about the data generating process itself. In other words, the task of a statistical learning machine is to discover structures and properties $q(x)$ from training examples $D_N$.
Some examples:
• Deterministic data: If $q$ generates the result of “$1 + 1$”, we can set $p(x_1) = 1$ where $x_1 = 2$ is the first “measurement” or request we made to the data generating process. Here the learning process recover everything we wish to know about $q$ just from the first data point, i.e. $p = q$. As such, there is no reason to deviate from this learning process.
• Deterministic with measurement error If $q$ generates the results of ballot count by humans, the above learning process would still be reasonable, but we should perhaps account for human error. We could, for instance, ask for lots of recount set $p(\text{most frequently occuring count}) = 1$. Or perhaps a deterministic result doesn’t sit well with us when we know that error can occur, we can set $p(x) =$ proportion of recount that turns out to be $x$.
• Experiments in empirical science If $q$ generates experimental measurements of physical quantities $(x, y)$ that is governed by some law of nature $y = f_\alpha(x)$ that depends on some parameter $\alpha$ and experimental measurements is marred by (normally distributed) random error, then we have $Y_i - f_\alpha(X_i) \sim N(0, \sigma)$. The value of $\alpha$ can be estimated given a learnt model $p$.
• Generalised Linear Models
• AI agents
• ## etc 1
In general, we instantiate a large space of hypothesis,
$\Delta = \left\{p = p(x|w) \, \, : \,w \in W\right\}$
parametrised by $w \in W \subset \R^d$ equipped with a prior $\varphi(w)$ and cast the learning process as an optimisation procedure that finds the best hypothesis that explains the observed samples. One way to define “best” is to select $p$ that minimises the Kullback-Leibler divergence between $q$ and $p$, i.e.choose $p(x) = p(x \mid \hat{w})$ such that $\hat{w}$ minimises
\begin{aligned} K(w) = \E_q\left[\log \frac{q(x)}{p(x| w)}\right] = \int_X q(x) \log \frac{q(x)}{p(x| w)}dx \end{aligned}
We will investigate the properties of learning machine of this form. Properties of a learning machine that we might care about:
• Error rate. Generalisation. Generalisation gap: $$B_g, G_g, B_t, G_t$$.
• Data efficiency. Compute efficiency. Behaviour in overparametrised regime. Scaling laws. Double descent.
• Training behaviour. Stochastic noise.
# Real Log Canonical Threshold
For a given statistical model $(p(x \mid w), q(x), \varphi(w))$, the following are equivalent definitions for its real log canonical threshold (RLCT), $\lambda$ and its order $\theta$.
1. Largest pole of zeta function of $$K$$
Define the zeta function of $K$ as$$\zeta$$ analytically continues to a meromorphic function with poles on the negative real axis.:
\begin{aligned} & \zeta: \C \to \C &\zeta(z) = \int_W K(w)^z \varphi(w)dw. \end{aligned}
The RLCT $\lambda$ is the largest pole of $\zeta$ and $\theta$ the order of the pole at $\lambda$.
2. Convergence rate of Laplace integral of $K$
$(\lambda, \theta)$ governs the asymptotic behaviour as $n \to \infty$ of the Laplace integralwhich is the deterministic version of the (normalised) evidence $$Z^0_n = \int_W \exp\left(-nK_N(w)\right)\varphi(w) dw$$. Note that the limiting variable $$n$$ is different from the number of training samples $$N$$. This is one place where inverse temperature $$\beta$$ can come in: set $$n = \beta k$$.:
\begin{aligned} \int_W \exp\left(-nK(w)\right)\varphi(w) dw \stackrel{n \to \infty} \sim Cn^{-\lambda}\left(\log n\right)^{\theta -1} \end{aligned}
for some positive real constant $C$.
3. Convergence rate of free energy
Taking the negative logarithm of the previous asymptotic expression givesthe stochastic version translate as $$F^0_n = \lambda \log n - (\theta -1) \log \log n +$$ stochastic process of constant order.
\begin{aligned} \log \int_W \exp\left(-nK(w)\right)\varphi(w) dw \stackrel{n \to \infty} \sim \lambda \log n - \left(\theta -1\right) \log \log n + O(1). \end{aligned}
4. Asymptotic expansion of density of states near $W_0$
The density of state
\begin{aligned} v(t) = \int_W \delta\left(t - K(w)\right) \varphi(w) dw \end{aligned}
has asymptotic expansion as $t \to 0$
\begin{aligned} v(t) \sim C t^{\lambda -1} (- \log(t))^{\theta -1} \end{aligned}
for some positive real constant $C$.
5. Volume codimension $W_0$
\begin{aligned} \lambda = \lim_{t \to 0^+} \log_a \frac{V(at)}{V(t)} \end{aligned}
where $1 \neq a > 0$ and
$V: \R_{\geq 0} \to \R_{\geq 0}$
is the volume measure of neighbourhoods of $W_0$
\begin{aligned} V(t) = \int_{K(w) < t} \varphi(w) dw. \end{aligned}
6. From resolution of singularity
Hironaka’s resolution of singularity for the real analytic function $K(w)$ gives us a proper birational map2 $g: U \to W$ such that in the neighbourhood of $w_0 \in W_0$, the zero set of $K$
\begin{aligned} K(g(u) - w_0) &= u^{2k} = u_1^{2k_1}u_2^{2k_2} \dots u_d^{2k_d}\\ g'(u) &= b(u)u^h = b(u)u_1^{h_1}u_2^{h_2} \dots u_d^{h_d} \end{aligned}
for some $u, k \in \N^d$ and analytic $b(u) \neq 0$. We then have
\begin{aligned} \lambda = \inf_{w \in W_0} \min_{1 \leq j \leq d}\frac{h_j + 1}{2k_j} \end{aligned}
and $\theta$ is given by the number of times the above minimum is achieved.3
7. RLCT of ideals of analytic functions
TODO: there are various square roots involved in this that I don’t really understand
# RLCT for Regular Models
The RLCT of a regular realisable model is given by
$\lambda = \frac{d}{2} \quad \theta = 1.$
We shall use the Laplace integral characterisation of RLCT. We want to show that
$Z^0_n = \int_W \exp\left(-nK(w)\right) \varphi(w) d{w} \sim C n^{-\frac{d}{2}}$
for some positive constant $$C$$ as $$n \to \infty$$. Since the model is realisable and identifiable, it has unique minimum at $$w^* \in \mathrm{supp}(\varphi)$$. Taylor expansion of $$K$$ centered around $$w^*$$ up to order 2 gives
\begin{aligned} K(w) &= K(w^*) + \nabla K(w^*) \cdot (w - w^*) + \frac{1}{2} (w - w^*)^T \nabla^2K(w^*)(w - w^*) + O(\left|\, w - w^* \,\right|^3) \end{aligned}
where $\nabla^2K(w^)$ is the Hessian of $K$ at $w^$. That $w^$ realises the true model and is a local minimum gives us $K(w^) = 0$ and $\nabla K(w^*) =0$, reducing the above to
\begin{aligned} K(w) &= \frac{1}{2} (w - w^*)^T \nabla^2K(w^*)(w - w^*) + O(\left|\, w - w^* \,\right|^3). \end{aligned}
Substituting the above into the integral, we get, in the limit as $$n \to \infty$$
\begin{aligned} Z^0_n &\sim \int_W \exp\left(-\frac{n}{2} (w - w^*)^T \nabla^2K(w^*)(w - w^*) \right) \varphi(w) d{w} \end{aligned}
which we recognise as a Gaussian integral with precision matrix $$n \nabla^2K(w^*)$$ which is positive definite by assumption. Therefore, we conclude that
\begin{aligned} Z^0_n \sim \varphi(w^*)\sqrt{\frac{(2\pi)^d}{\det\left(n \nabla^2K(w^*)\right)}} = \varphi(w^*)\sqrt{\frac{(2\pi)^d}{\det\left(\nabla^2K(w^*)\right)}} n^{-\frac{d}{2}}. \end{aligned}
We shall use the characterisation that for any positive $$a \neq 1$$, $$\lambda = \lim_{t \to 0^+} \log_a \frac{V(at)}{V(t)}$$ where $$V$$ is the volume function
\begin{aligned} V(t) = \int_{K(w) \leq t} \varphi(w) dw. \end{aligned}
By regularity assumption, we have that $$w^*$$ is a non-degenerate critical point of $$K$$ and hence by Morse lemma, there is a local chart $$x(w) = (x_i(w))_{i = 1, \dots, d}$$ in a small enough neighbourhood of $$w^*$$ such that $$K(w) = \cancelto{0}{K(w^*)} + \sum_i x_i(w)^2$$. Therefore, for small enough $$t > 0$$,
\begin{aligned} V(t) = \int_{\sum_i x_i^2 \leq t} \varphi(x) dx \end{aligned}
which is proportional to the volume of a $$d$$-dimensional ball with radius $$\sqrt{t}$$, i.e. $$V(t) \propto t^{d}$$.[^16] Finally,
\begin{aligned} \lambda = \lim_{t \to 0^+} \log_a \frac{(at)^{d/2}}{t^{d/2}} = \frac{d}{2}. \end{aligned}
# Footnotes
1. TODO: more examples. Contrast different inference tasks.
2. obtained via recursive blow up.
3. This deep result shows that $$(\lambda, \theta) \in \Q \times \Z$$. | 2023-02-03 11:03:33 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9939938187599182, "perplexity": 1114.93325288173}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.3, "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-2023-06/segments/1674764500044.66/warc/CC-MAIN-20230203091020-20230203121020-00771.warc.gz"} |
https://www.degruyter.com/document/doi/10.1515/fhep-2019-0021/html | Jeffrey Sullivan, Tiffany M. Shih, Emma van Eijndhoven, Yash J. Jalundhwala, Darius N. Lakdawalla, Cindy Zadikoff, Jennifer Benner, Thomas S. Marshall and Kavita R. Sail
# The Social Value of Improvement in Activities of Daily Living among the Advanced Parkinson’s Disease Population
Open Access
De Gruyter | Published online: November 25, 2020
# Abstract
## Objectives
Quantify the value of functional status (FS) improvements consistent in magnitude with improvements due to levodopa-carbidopa intestinal gel (LCIG) treatment, among the advanced Parkinson’s disease (APD) population.
## Methods
The Health Economic Medical Innovation Simulation (THEMIS), a microsimulation that estimates future health conditions and medical spending, was used to quantify the health and cost burden of disability among the APD population, and the value of quality-adjusted life-years gained from FS improvement due to LCIG treatment compared to standard of care (SoC). A US-representative Parkinson’s disease (PD)-comparable cohort was constructed in THEMIS based on observed PD patient characteristics in a nationally representative dataset. APD was defined from the literature and clinical expert input. The PD and APD cohorts were followed from 2010 over their remaining lifetimes. All individuals were ages 65 and over at the start of the simulation. To estimate the value of FS improvement due to LCIG treatment, decreases in activities of daily living (ADL) limitations caused by LCIG treatment were calculated using data from a randomized, controlled, double-blind, double-dummy clinical trial and applied to the APD population in THEMIS.
## Results
Total burden of disability associated with APD was $17.7 billion (B). From clinical trial data, LCIG treatment versus SoC lowers the odds of difficulties in walking, dressing, and bathing by 76%, 42% and 39%, respectively. Among the APD population, these reductions generated$2.6B in value to patients and cost savings to payers. The added value was 15% of the burden of disability associated with APD and offsets 15% of the cost of LCIG treatment.
## Conclusions
FS improvements, consistent with improvements due to LCIG treatment, in the APD population created health benefits and reduced healthcare costs in the US.
## 1 Introduction
### Figure 2:
Change in outcomes from reducing disability due to APD.
B billion, QALY quality-adjusted life year.
Figure 3 provides the effects of LCIG treatment on walking, dressing, and bathing, as estimated from the clinical trial data (Olanow et al. 2014). The effects of LCIG treatment compared to SoC lowered the odds of difficulty in walking, dressing, and bathing by 76%, 43% and 38%, respectively. Table 2 provides the health and cost outcomes resulting from LCIG treatment under the baseline and sensitivity scenarios. When applied simultaneously, reductions in difficulty with walking, dressing, and bathing due to LCIG treatment in the APD cohort generated 15,946 patient life-years, 10,631 QALYs (equivalent to 8.1% of the total life-year burden and 11.8% of the total QALY burden associated with disability due to by APD), and $0.5B in savings in medical expenditures by all payers over the lifetime of all APD patients, relative to SoC. The value of health improvement and cost savings from improving all three ADLs due to LCIG treatment totaled$2.6B in value across all payers and patients relative to SoC. This value was equivalent to 15% of the total disability burden of APD and offset 15% of LCIG treatment costs among the APD population. Medicaid spending decreased by $0.2B, partially due to 3692 fewer patient-years spent by LCIG treated APD patients in nursing homes and 7450 fewer person-years of Medicaid enrollment. ### Figure 3: Odds ratio of experiencing FS limitations for individuals treated with LCIG versus individuals under SoC, based on 12-week clinical trial results. FS functional status, LCIG levodopa-carbidopa intestinal gel, SoC standard of care The clinical trial UPDRS Part II secondary endpoint results were disaggregated and then used to estimate the proportion of subjects with difficulty in each ADL, by treatment arm (Olanow et al. 2014). Specifically, for each ADL, the odds ratio of difficulty for individuals receiving LCIG treatment versus individuals receiving SoC was calculated using the 12-week trial results as follows: Odds ratio = ( Proportion of subjects with limitation , LCIG treatment Proportion of subjects without limitation , LCIG treatment ) ( Proportion of subjects with limitation , SoC Proportion of subjects without limitation , SoC ) . Table 2: The effect of improved activity from LCIG treatment for the APD population relative to SoC. APD classification ADLs Baselined All three Baselined Walking Baselined Dressing Baselined Bathing Sensitivitye All three Health outcomes Patient life-years (not quality-adjusted) 15,954 2,041 12,619 1,448 41,502 QALYs (discounted to 2015) 10,631 4,203 5,030 1,167 26,160 Monetized QALYs (2015$ millions, $200 K/QALY) 2,126 841 1,006 233 5,232 Key expenditure components Nursing home patient-years −3,692 −2,735 −891 −47 −9,431 Medicaid enrollment person-years −7,450 −5,543 −1,395 −770 −19,732 Medicaid expenditures (2015$ millions) −208 −133 −65 −21 −534
Total medical expendituresa (2015$millions) −519 −584 83b −2 −1,274 Total social valuec(2015$ millions) 2,645 1,425 923 236 6,506
ADL activities of daily living, APD advanced Parkinson’s disease, K thousand, QALY quality-adjusted life year.aTotal medical expenditures measured all medical payments made for a patient by all payers, and are not limited to Medicaid expenditures. This measure was obtained directly from survey respondents and was cross-checked using Medicare claims; LCIG treatment cost was not included.bTotal medical expenditures may increase due to increased longevity from improved FS.cChange in total social value is the sum of changes in total medical expenditures and monetized QALYs.dBaseline classification of APD is repeated falls and use of an ambulatory assistance device as measured in the HRS within the PD-comparable cohort.eSensitivity classification of APD first identifies APD as PD with repeated falls and ambulatory assistance device use as observed in the MCBS, then imputes APD status into the HRS based on the characteristics of the APD population in the MCBS.
## 4 Discussion
This study estimated the value of FS improvements that are consistent in magnitude with improvements due to LCIG treatment, over the lifetime of the APD population, using a microsimulation model that predicts long-term economic and health outcomes based on longitudinal transitions in the current population. Results from a randomized, controlled, double-blind, double-dummy clinical trial were leveraged to quantify the effect of LCIG treatment on FS (Olanow et al. 2014). Because the model was parameterized using nationally representative data, the results of the study are generalizable to the broader APD population in the US.
Prior work has shown that the costs of PD are substantial, particularly in the advanced stages. Johnson et al. (2013b) found that in the year following institutionalization or the adoption of an ambulatory assistance device, privately insured PD patients had 6–7 times higher healthcare costs than their non-PD counterparts (Johnson et al. (2013b)). Similarly, Kaltenboeck et al. (2012) showed that Medicare beneficiaries with early and advanced PD experienced higher direct medical costs and mortality than their non-PD counterparts (Kaltenboeck et al. 2012). The current study builds upon this prior work, but also provides new findings by (1) isolating the burden of disability in APD, (2) incorporating impacts on QoL in addition to medical expenditures, and mortality to estimate overall value, and (3) providing value estimates aggregated over a lifetime horizon. The results demonstrated that the disability component of APD alone impose a sizeable economic burden to society, equivalent to $17.7B. This study also found that LCIG treatment can reduce this burden by improving FS. The combined reductions in difficulty in walking, dressing, and bathing due to LCIG treatment generated a total lifetime value of$2.6B for the current APD population in the baseline estimates. Overall, improvement in these three activities reduced the total APD disability burden by 15%. The impact on monetized QALYs due to improvement in FS from LCIG treatment constituted 80% of the overall social value. Reduced total medical expenditures comprised the remaining 20% of the value from improved FS due to LCIG treatment. To characterize contributors to total medical expenditures, impacts of LCIG treatment on nursing home patient-years and Medicaid enrollment and expenditures were estimated over the patients’ lifetimes. The probability of nursing home and Medicaid enrollment both increase for those with three or more ADL limitations. The FS improvement from LCIG treatment therefore decreased total time spent in nursing homes and Medicaid expenditures.
Furthermore, the individual components of value from LCIG treatment were identified. In particular, in the clinical trial results, LCIG treatment had the greatest impact on walking difficulty. As a result, the effect of LCIG treatment on reducing walking difficulty led to the greatest increase in social value among the three individual ADLs analyzed in this study. Improvement in walking and dressing led to similar gains in QALYs, while bathing led to smaller QALY gains. A comparison of the impact on patient life-years versus the impact on QALYs reveals that improvement in walking primarily resulted in improvement in QoL, while improvement in dressing was primarily associated with increases in patient life-years. Since dressing extended life in a relatively poor health states, it also resulted in higher medical spending. As a result of these factors, the overall value from improved dressing was lower than the value from improved walking. The analysis of individual ADLs thereby revealed that walking improvement were the key driver of the value of LCIG treatment.
There are some limitations to this study. First, PD status could not be directly observed in the HRS data. As a result, a PD cohort was imputed based on characteristics (demographics and ADLs) of the observed PD population in the MCBS. This imputation may not perfectly identify PD patients, thus creating error in our estimates. A sensitivity analysis was performed defining APD directly within the observed PD population in the MCBS data, and APD status was imputed into the HRS data to be utilized in THEMIS. Relative to the baseline results, the absolute burden of disability due to APD and value of improved FS due to LCIG treatment both increased in the sensitivity analysis. However, the proportion of the APD burden alleviated by LCIG treatment decreased slightly, from 15 to 13%.
Second, the FS questionnaire in the clinical trial differed from that used in the HRS data. As a result, a restricted set of ADLs was included in the analysis, which may omit important FS determinants of value. For example, prior research has found that difficulty eating may also be influential in outcomes for the elderly population (Fong, Mitchell, and Koh 2015). Furthermore, we were only able to capture binary changes between any difficulty versus no difficulty in ADLs in THEMIS, so that actual LCIG treatment may lead to additional benefits from reductions in the severity of difficulty.
Third, a challenge was to properly define APD in order to estimate social value for this population. While there is no single definition of APD agreed upon by clinicians, for this study the definition of having both repeated falls and being unable to walk unassisted was used, based on clinical expert input and the results of a Delphi panel of movement disorder specialists, which was conducted to identify primary characteristics of APD patients (Antonini et al. 2015). However, estimates of the value of improved FS within the APD population may change under an alternative definition of APD.
Fourth, ADL limitations affect multiple health outcomes beyond the endpoints in this study. ADL limitations were included in the transition models of different health outcomes when the literature showed them to be a predictor of the health outcome. However, this correlation between ADL limitations and health outcomes might be due to unobserved factors. As the value of LCIG treatment is measured through its effect on ADL limitations, these other associations might lead to an overestimation of the social value of LCIG treatment.
Lastly, the current version of THEMIS does not yet have the capability to compute confidence intervals around simulation estimates. There are many sources of uncertainty in the model, since many of the transition models are based on regression estimates. This means we are unable translate the variation around a particular parameter directly into variation of the output.
## 5 Conclusion
This study found that disability due to APD imposes a substantial economic burden on society through its effects on lifespan, QoL, and medical expenditures among the APD population. Reduced limitations in FS that are of a magnitude consistent with LCIG treatment, improved health outcomes and reduced healthcare costs for APD patients in the US over their lifetimes. Future research should consider the comprehensive effect of APD treatments in reducing NHA risk and social value, incorporating its effects on a broader set of ADLs and IADLs, and should perform further sensitivity analyses around the definition of APD.
Funding source: AbbVie
# Acknowledgments
We would like to thank Suepattra May-Slater, employee of PRECISIONheor, and Warren Stevens and Shalak Gunjal, former employees of PRECISIONheor, for their support of this study. Financial support for their services was provided by AbbVie.
Declaration of funding: This study and manuscript were funded by AbbVie. The design, study conduct, and financial support for the study were provided by AbbVie. AbbVie participated in the study design, research, interpretation of data, writing, reviewing, and approving the manuscript. No honoraria or payments were made for authorship.
Declaration of financial/other relationships: Kavita Sail and Yash J. Jalundhwala are employees of AbbVie and may own AbbVie stock or stock options. Thomas Marshall was an employee of AbbVie at the time of the study and may own AbbVie stock or stock options. Jeffrey Sullivan and Emma van Eijndhoven are employees of PRECISIONheor. Jennifer Benner and Tiffany Shih were employees of PRECISIONheor at the time of the study. PRECISIONheor is part of Precision for Medicine Group, which receives compensation from various biopharmaceutical companies, including AbbVie. Darius N. Lakdawalla holds equity in Precision Medicine Group. Cindy Zadikoff currently is an employee of AbbVie and may own AbbVie stock or stock options. When the study was being conducted, Cindy Zadikoff was affiliated with Feinberg School of Medicine, Northwestern University, Chicago, Illinois, United States and has previously received honoraria for consulting and lecturing from AbbVie. In addition, she has received consulting honoraria from various biopharmaceutical companies.
Appendix
Appendix Table 1:
Advanced Parkinson’s disease model in the MCBS used to impute into the HRS (sensitivity identification strategy).
Coefficient SE
Race
Black 0.282 0.230
Hispanic −0.585** 0.264
Education
Less than high school −0.085 0.150
Some college and above 0.097 0.155
Male 0.008 0.133
Functional status
Two-year lag of difficulty with walking 0.489*** 0.151
Two-year lag of difficulty with dressing 0.136 0.183
Two-year lag of difficulty getting out of bed and a chair 0.195 0.167
Two-year lag of difficulty using the telephone 0.116 0.164
Two-year lag of difficulty handling money 0.286* 0.164
Two-year lag of difficulty using the toilet 0.248 0.210
Two-year lag of currently smoking −0.115 0.248
Two-year lag of being widowed −0.027 0.147
BMI
Two-year lag of underweight: bmi <18.5 kg/m2 0.419 0.536
Two-year lag of overweight: bmi 25–30 kg/m2 −0.210 0.140
Two-year lag of obese: bmi ≥30kg/m2 −0.038 0.171
Age
Two-year lag of ages 65 and-younger −0.008 0.017
Two-year lag of ages 65–74 0.036 0.027
Two-year lag of ages 65 and older 0.013 0.014
Constant −0.710 1.015
BMI body mass index, SE standard error.
### Appendix Figure 1:
Study design flow chart
APD advanced Parkinson’s disease, FS functional status, HRS Health and Retirement Study, MCBS Medicare Current Beneficiary Survey, PD Parkinson’s disease, QALYs quality-adjusted life-years, RCT randomized controlled trial; THEMIS The Health Economics Medical Innovation Simulation
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https://scioly.org/forums/viewtopic.php?f=208&t=7697&start=105 | ## Astronomy C
astro124
Member
Posts: 44
Joined: November 23rd, 2012, 1:44 pm
Division: C
State: AZ
Location: Phoenix, AZ
### Re: Astronomy C
Hey everyone! It's been awhile since I've last been on here but I'm glad to see a lot of familiar names still around.
Anyways, I'm having a little bit of trouble with the Union County 2016 practice test (number 25a)
How would we go about solving the tangential velocity. For part b, I figured that we would use Newton's equation for gravitational force. Would the orbital radius be given based on our answer for tangential velocity in part A?
Thanks!
2012 Season: Reach for the Stars-3rd (State) / Keep the Heat-19th (State)
2013 Season: Astronomy-2nd (State) / Disease Detectives-15th (State)
2014 Season: Astronomy-1st (State) / Experimental Design-20th (State)/ Anatomy and Physiology-16th (State)
2015 Season: Astronomy
Thunder
andrewwski
Posts: 952
Joined: January 12th, 2007, 7:36 pm
State: -
### Re: Astronomy C
That question cannot actually be answered as written.
You are given the masses of the two bodies and the mean separation between them. The mean separation is equal to the semimajor axis of an ellipse. In order to determine the tangential velocity, though, you need to know the position along the ellipse (or the distance "r" between the masses at the time in question).
IF and only if the orbit is circular, then the separation is constant and equal to "r", in which case all velocity is tangential and you can use the Vis-Viva equation, which would reduce to:
$v=\sqrt{\frac{GM}{r}}$
But the answer in the key (to part a) makes no sense in this case. Intuitively it doesn't make any sense - it says 0.119 m/s. This isn't even near the right order of magnitude! Consider that the earth orbits the sun at ~30 km/s, and in this problem the star is twice as massive and the separation is half!
Likewise, if (and only if) you were to assume the orbit is circular, then you can solve part (b) with the given information. But even then, the answer in the key is wrong (it's off by 6 orders of magnitude, should be $5.637 x 10^{23} N$).
But since it was not stated to assume that the orbit is circular, there is not enough information to answer the question.
Skink
Exalted Member
Posts: 947
Joined: February 8th, 2009, 12:23 pm
Division: C
State: IL
### Re: Astronomy C
I'm back; this of this as part one of two, as I still have a list of National test images to ask about. This is higher priority, though. If anyone can help, my team and I would really appreciate it!
National Site NY Test #27
We've had difficulty locating the phrase "inclination derived" and are unclear how inclination, itself, relates to the rest, particularly what we're looking for.
National Site MIT Test #24(e)
There's clearly some relation we don't know about in order to find this ratio.
Thanks.
syo_astro
Exalted Member
Posts: 591
Joined: December 3rd, 2011, 9:45 pm
State: NY
Contact:
### Re: Astronomy C
I'm back; this of this as part one of two, as I still have a list of National test images to ask about. This is higher priority, though. If anyone can help, my team and I would really appreciate it!
National Site NY Test #27
We've had difficulty locating the phrase "inclination derived" and are unclear how inclination, itself, relates to the rest, particularly what we're looking for.
National Site MIT Test #24(e)
There's clearly some relation we don't know about in order to find this ratio.
Thanks.
Hey, I wrote that first test, and sorry about the bad wording in retrospect. Most of the questions were meant to be straight-forward. Typically, though, I try to make the questions have some sense that they are coming from somewhere (eg. the inclination was derived or modeled somehow working with data and theory) rather than just saying the value is blah (and in retrospect again I failed at that considering how I just gave other values in those problems, but again this wasn't meant to be the most impossible questions >.>). The other thing I phrased admittedly badly was the ratio of velocities. This one I meant (but didn't express well) to write that the ratio of velocities were determined assuming no inclination. Accounting for inclination modifies it slightly because then we are actually viewing a component of the star's radial velocity in reality, so you have to multiply it by a factor of sin(i).
But the basis of this question relates well to your other problem in 24e on the MIT test. To summarize, it is based around conserving momentum (m1 * v1 = m2 * v2). It takes an extra step, but it's nothing too bad. Let's do it twice!
m_a * v_a = m_planet * v_planet_a; m_b * v_b = m_planet * v_planet_b
Dividing one equation by the other gets us that
(m_a / m_b) * (v_a / v_b) = (v_planet_a / v_planet_b)
Still confused how to get the velocities within the question? I think it should be clear at least you need the masses (given) and the velocities of the planets (I assume these aren't the same if you were to calculate them). It's not a relation that gives the answer outright, but it certainly involves fundamentals! Any questions about this?
B: Crave the Wave, Environmental Chemistry, Robo-Cross, Meteorology, Physical Science Lab, Solar System, DyPlan (E and V), Shock Value
C: Microbe Mission, DyPlan (Earth's Fresh Waters), Fermi Questions, GeoMaps, Gravity Vehicle, Scrambler, Rocks, Astronomy
Skink
Exalted Member
Posts: 947
Joined: February 8th, 2009, 12:23 pm
Division: C
State: IL
### Re: Astronomy C
Thank you. I solved your problem now and followed your algebra for the second. I'm not sure how to calculate the velocities of the planet for each star (the only things I need to get the requested ratio), but I'll see if the team can figure it out.
And, as a side note: that effort doesn't go unnoticed. We all strive to write questions and scenarios that have purpose (or, at least, more color) to them versus saying "27. Mass is this. Volume is that. Calculate density.". Riveting problem there.
astro124
Member
Posts: 44
Joined: November 23rd, 2012, 1:44 pm
Division: C
State: AZ
Location: Phoenix, AZ
### Re: Astronomy C
I just got back from my state's Astro test and was probably the weirdest test I've taken in Science Olympiad.
Instead of an actual test we were given a computer simulation to find habitable planets. The program, which was sponsored by ASU and NASA Astrobiology, was the final for the college's Astronomy 106 class. When we started, there was a screen of over 600 'stars'. You click on one and you're given some data like parallax, apparent magnitude, peak wavelength, etc. From there you're supposed to calculate distance (easy), luminosity, and Temperature. Now came the tricky part. You click on the second screen and start imputing data for the planet. At first I thought, the gave you some background data, but nope. Nothing. After 20 minutes I was completely lost. The proctor told me that you have to use some of your 'funds' (we started with 50,000 USD) to purchase analysis on planets.
Anyways, it didn't go very well. Points were awarded for finding certain planets correctly and supposedly, there was one habitable planet among all ~600 stars.
Has anyone else had a test like this? I'm still trying to find it online but with no luck.
2012 Season: Reach for the Stars-3rd (State) / Keep the Heat-19th (State)
2013 Season: Astronomy-2nd (State) / Disease Detectives-15th (State)
2014 Season: Astronomy-1st (State) / Experimental Design-20th (State)/ Anatomy and Physiology-16th (State)
2015 Season: Astronomy
Thunder
syo_astro
Exalted Member
Posts: 591
Joined: December 3rd, 2011, 9:45 pm
State: NY
Contact:
### Re: Astronomy C
Well, weird tests like that make me feel slightly better about my tests, but I do wish people tried to cover the whole rules...and no, I haven't quite heard of a test like that astro124. I guess if I find anything I'll say.
Also glad to hear that Skink, if they can't figure out how to get the velocities, Tad posted a bit back (on March 1st) about it. As a reminder, just remember that speed is distance divided by time, planets orbit in circles (in simplified cases, so what's the distance around a circle? Parts a and b relate to this), and time is defined by orbital period (look to part c).
B: Crave the Wave, Environmental Chemistry, Robo-Cross, Meteorology, Physical Science Lab, Solar System, DyPlan (E and V), Shock Value
C: Microbe Mission, DyPlan (Earth's Fresh Waters), Fermi Questions, GeoMaps, Gravity Vehicle, Scrambler, Rocks, Astronomy
AlphaTauri
Staff Emeritus
Posts: 829
Joined: September 11th, 2009, 1:41 pm
State: PA
Location: 04h 35m 55.239s, +16° 30′ 33.49″
Contact:
### Re: Astronomy C
Just uploaded the test I wrote for MI Region 8 to the test exchange.
Based on the score distribution, it was probably a little too hard for a typical Regionals, but it's still good practice :)
Hershey Science Olympiad 2009 - 2014
Volunteer for Michigan SO 2015 - 2018
]\/[ Go Blue!
asdfqwerzzz2
Member
Posts: 35
Joined: June 12th, 2013, 7:57 pm
Division: C
State: -
### Re: Astronomy C
Just uploaded the test I wrote for MI Region 8 to the test exchange.
Based on the score distribution, it was probably a little too hard for a typical Regionals, but it's still good practice
Just for reference when I take it, what were the top raw scores?
AlphaTauri
Staff Emeritus
Posts: 829
Joined: September 11th, 2009, 1:41 pm
State: PA
Location: 04h 35m 55.239s, +16° 30′ 33.49″
Contact:
### Re: Astronomy C
Top was just under 40%. I was aiming for ~70%, but uh, that didn't happen...
Hershey Science Olympiad 2009 - 2014
Volunteer for Michigan SO 2015 - 2018
]\/[ Go Blue! | 2019-10-14 16:32:17 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 2, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6758780479431152, "perplexity": 2654.4342035230775}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-43/segments/1570986653876.31/warc/CC-MAIN-20191014150930-20191014174430-00464.warc.gz"} |
https://cis300.cs.ksu.edu/io/advanced-text-file/ | Though the File.ReadAllText and File.WriteAllText methods provide simple mechanisms for reading and writing text files, they are not always the best choices. For one reason, files can be very large — too large to fit into memory, or possibly even larger than the maximum length of a string in C# (2,147,483,647 characters). Even when it is possible to store the entire contents of a file as a string, it may not be desirable, as the high memory usage may degrade the overall performance of the system.
For the purpose of handling a sequence of input or output data in more flexible ways, the .NET Framework provides streams. These streams are classes that provide uniform access to a wide variety of sequences of input or output data, such as files, network connections, other processes, or even blocks of memory. The StreamReader and StreamWriter classes (in the System.IO namespace) provide read and write, respectively, access to text streams, including text files.
Some of the more useful public members of the StreamReader class are:
• A constructor that takes a string giving a file name as its only parameter and constructs a StreamReader to read from that file.
• A Read method that takes no parameters. It reads the next character from the stream and returns it as an int. If it cannot read a character because it is already at the end of the stream, it returns -1 (it returns an int because -1 is outside the range of char values).
• A ReadLine method that takes no parameters. It reads the next line from the stream and returns it as a string. If it cannot read a line because it is already at the end of the stream, it returns null.
• An EndOfStream property that gets a bool indicating whether the end of the stream has been reached.
With these members, we can read a text file either a character at a time or a line at a time until we reach the end of the file. The StreamWriter class has similar public members:
• A constructor that takes a string giving a file name as its only parameter and constructs a StreamWriter to write to this file. If the file already exists, it is replaced by what is written by the StreamWriter; otherwise, a new file is created.
• A Write method that takes a char as its only parameter. It writes this char to the end of the stream.
• Another Write method that takes a string as its only parameter. It writes this string to the end of the stream.
• A WriteLine method that takes no parameters. It writes a line terminator to the end of the stream (i.e., it ends the current line of text).
• Another WriteLine method that takes a char as its only parameter. It writes this char to the end of the stream, then terminates the current line of text.
• Yet another WriteLine method that takes a string as its only parameter. It writes this string to the end of the stream, then terminates the current line of text.
Thus, with a StreamWriter, we can build a text file a character at a time, a line at a time, or an arbitrary string at a time. In fact, a number of other Write and WriteLine methods exist, providing the ability to write various other types, such as int or double. In each case, the given value is first converted to a string, then written to the stream.
Streams are different from other classes, such as strings or arrays, in that they are unmanaged resources. When a managed resource, such as a string or an array, is no longer being used by the program, the garbage collector will reclaim the space that it occupies so that it can be allocated to new objects that may need to be constructed. However, after a stream is constructed, it remains under the control of the program until the program explicitly releases it. This has several practical ramifications. For example, the underlying file remains locked, restricting how other programs may use it. In fact, if an output stream is not properly closed by the program, some of the data written to it may not actually reach the underlying file. This is because output streams are typically buffered for efficiency — when bytes are written to the stream, they are first accumulated in an internal array, then written as a single block when the array is full. When the program is finished writing, it needs to make sure that this array is flushed to the underlying file.
Both the StreamReader and StreamWriter classes have Dispose methods to release them properly; however, because I/O typically requires exception handling, it can be tricky to ensure that this method is always called when the I/O is finished. Specifically, the try-catch may be located in a method that does not have access to the stream. In such a case, the catch-block cannot call the stream’s Dispose method.
To handle this difficulty, C# provides a using statement. A using statement is different from a using directive, such as
using System.IO;
A using statement occurs within a method definition, not at the top of a code file. Its recommended form is as follows:
using ( /* declaration and initialization of disposable variable(s) */ )
{
/* Code that uses the disposable variables(s) */
}
Thus, if we want to read and process a text file whose name is given by the string variable fileName, we could use the following code structure:
using (StreamReader input = new StreamReader(fileName))
{
/* Code that reads and process the file accessed by the
}
This declares the variable input to be of type StreamReader and initializes it to a new StreamReader to read the given file. This variable is only visible with the braces; furthermore, it is read-only — its value cannot be changed to refer to a different StreamReader. The using statement then ensures that whenever control exits the code within the braces, input’s Dispose method is called.
More than one variable of the same type may be declared and initialized within the parentheses of a using statement; for example:
using (StreamReader input1 = new StreamReader(fileName1),
{
/* Code that reads from input1 and input2 */
}
The type of variable(s) declared must be a subtype of IDisposable. This ensures that the variables each have a Dispose method.
As a complete example of the use of a StreamReader and a StreamWriter, together with a using statement for each, suppose we want to write a method that takes as its parameters two strings giving the name of an input file and the name of an output file. The method is to reproduce the input file as the output file, but with each line prefixed by a line number and a tab. We will start numbering lines with 1. The following method accomplishes this:
/// <summary>
/// Copies the file at inFileName to outFileName with each line
/// prefixed by its line number followed by a tab.
/// </summary>
/// <param name="inFileName">The path name of the input file.</param>
/// <param name="outFileName">The path name of the output file.</param>
private void AddLineNumbers(string inFileName, string outFileName)
{
{
using (StreamWriter output = new StreamWriter(outFileName))
{
int count = 0;
while (!input.EndOfStream)
{
We can call the above method within a try-block to handle any exceptions that may be thrown during the I/O. The catch-block will not have access to either input or output, but it doesn’t need it. If an exception is thrown during the I/O, the two using statements will ensure that the Dispose methods of both the StreamReader and the StreamWriter are called. | 2022-10-07 12:31:56 | {"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": 0.31382712721824646, "perplexity": 1148.892484955662}, "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-2022-40/segments/1664030338073.68/warc/CC-MAIN-20221007112411-20221007142411-00565.warc.gz"} |
http://www.jiskha.com/display.cgi?id=1359157877 | # Science
posted by on .
The rms(root-mean-square) value of a periodic signal s(t) is defined to be
rms [s]=(1/T Ss^2t dt)^1/2
where T is defined to be the signal's period: the smallest positive number such that s(t)=s(t+T).
What is the rms value of the sinusoid s(t)=Asin(2πf0t)? (Again, write A as A and f0 as f0.)
• Science - ,
The rms value of a sine function of amplitude A is A/sqrt2 = 0.70711 A.
That can be derived from the definition you were given.
### Related Questions
More Related Questions
Post a New Question | 2017-05-23 18:52:18 | {"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": 0.9597961902618408, "perplexity": 3071.97498693111}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-22/segments/1495463607649.26/warc/CC-MAIN-20170523183134-20170523203134-00483.warc.gz"} |
https://www.zbmath.org/?q=an%3A0903.68215 | # zbMATH — the first resource for mathematics
Automated model selection for simulation based on relevance reasoning. (English) Zbl 0903.68215
Summary: Constructing an appropriate model is a crucial step in performing the reasoning required to successfully answer a query about the behavior of a physical situation. In the compositional modeling approach of Falkenhainer and Forbus, a system is provided with a library of composable pieces of knowledge about the physical world called model fragments. The model construction problem involves selecting appropriate model fragments to describe the situation. Model construction can be considered either for static analysis of a single state or for simulation of dynamic behavior over a sequence of states. The latter is significantly more difficult than the former since one must select model fragments without knowing exactly what will happen in the future states. The model construction problem in general can advantageously be formulated as a problem of reasoning about relevance of knowledge that is available to the system using a general framework for reasoning about relevance described by Levy (1993) and Levy and Sagiv (1993). In this paper, we present a model formulation procedure based on that framework for selecting model fragments efficiently for the case of simulation. For such an algorithm to be useful, the generated model must be adequate for answering the given query and, at the same time, as simple as possible. We define formally the concepts of adequacy and simplicity and show that the algorithm in fact generates an adequate and simplest model.
##### MSC:
68U20 Simulation (MSC2010)
Full Text:
##### References:
[1] Addanki, S.; Cremonini, R.; Penberthy, J., Reasoning about assumptions in graphs of models, (), 443-446 · Zbl 0719.68066 [2] Bobrow, D.; Falkenhainer, B.; Farquhar, A.; Fikes, R.; Forbus, K.; Gruber, T.; Iwasaki, Y.; Kuipers, B., A compositional modeling language, (), 12-21 [3] also: AAAI Tech. Rept. WS-96-01. [4] Bobrow, D.; Falkenhainer, B.; Farquhar, A.; Fikes, R.; Forbus, K.; Gruber, T.; Iwasaki, Y.; Kuipers, B., Cml: a compositional modeling language, () [5] Carnap, R., () [6] Crawford, J.; Farquhar, A.; Kuipers, B., A compiler from physical models into qualitative differential equations, (), 365-372 [7] Denn, M.M., () [8] Falkenhainer, B.; Forbus, K.D., Compositional modeling: finding the right model for the job, Artificial intelligence, 51, 95-143, (1991) [9] Forbus, K.D., Qualitative process theory, Artificial intelligence, 24, 178-219, (1984) [10] Gärdenfors, P., On the logic of relevance, Synthese, 37, 351-367, (1978) · Zbl 0395.03005 [11] Genesereth, M.R.; Fikes, R.E., Knowledge interchange format, version 3.0 reference manual, () [12] Iwasaki, Y.; Low, C.M., Model generation and simulation of device behavior with continuous and discrete change, Intelligent systems engineering, 1, 2, 115-145, (1993) [13] Iwasaki, Y.; Simon, H.A., Causality in device behavior, Artificial intelligence, 29, 3-32, (1986) [14] Iwasaki, Y.; Farquhar, A.; Fikes, R.; Rice, J., A web-based compositional modeling system for sharing of physical knowledge, () [15] Iwasaki, Y.; Low, C.M., Device modeling environment: an integrated model-formulation and simulation environment for continuous and discrete phenomena, (), 141-146 [16] Keynes, J.M., () [17] Levy, A.Y., Irrelevance reasoning in knowledge base systems, () [18] Levy, A.Y., Creating abstractions using relevance reasoning, (), 588-594 [19] Levy, A.Y.; Fikes, R.E.; Sagiv, S., Speeding up inferences using relevance reasoning: a formalism and algorithms, Artificial intelligence, 97, 83-136, (1997) · Zbl 0904.68163 [20] Levy, A.Y.; Rajaraman, A.; Ordille, J.J., Query answering algorithms for information agents, (), 40-47 [21] Levy, A.Y.; Sagiv, Y., Constraints and redundancy in Datalog, (), 67-80 [22] Levy, A.Y.; Sagiv, Y., Exploiting irrelevance reasoning to guide problem solving, (), 138-144 [23] Macaulay, D., () [24] Nayak, P.P., Automated model selection, () [25] Nayak, P.P., Causal approximations, Artificial intelligence, 70, 277-334, (1994) · Zbl 0938.68845 [26] Nayak, P.P.; Joskowicz, L., Efficient compositional modeling for generating causal explanations, Artificial intelligence, 83, 193-227, (1996) [27] Rickel, J.; Porter, B., Automated modeling for answering prediction questions: exploiting interaction paths, (), 82-95 [28] Rickel, J.; Porter, B., Automated modeling for answering prediction questions: selecting the time scale and system boundary, (), 1191-1198 [29] Rickel, J.; Porter, B., Automated modeling of complex systems for answering prediction questions, Artificial intelligence, 93, 201-260, (1997) · Zbl 1017.93500 [30] Simon, H.A.; Rescher, N., Cause and counterfactual, Philos. sci., 33, 323-340, (1966) [31] Subramanian, D.; Genesereth, M.R., The relevance of irrelevance, (), 416-422 [32] Subramanian, D., A theory of justified reformulations, () · Zbl 0792.68169 [33] Weld, D.S., Approximation reformulation, (), 407-412 [34] Williams, B.C., Capturing how things work: constructing critical abstractions of local interactions, (), 163-174 [35] Williams, B.C., Interaction-based invention: designing novel devices from first principles, (), 349-356
This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching. | 2021-08-02 08:38:59 | {"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": 0.8390750288963318, "perplexity": 14753.335322653269}, "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-2021-31/segments/1627046154310.16/warc/CC-MAIN-20210802075003-20210802105003-00159.warc.gz"} |
https://www.ustermetrics.com/publication-type/4/ | # 4
## tseries: Time Series Analysis and Computational Finance
tseries: Time Series Analysis and Computational Finance
## Efficient Estimation of Volatility Using High Frequency Data
Efficient Estimation of Volatility Using High Frequency Data
## Cointegration and Exchange Market Efficiency: An Analysis of High Frequency Data
Cointegration and Exchange Market Efficiency: An Analysis of High Frequency Data
## On the ergodicity and stationarity of the ARMA (1,1) recurrent neural network process
On the ergodicity and stationarity of the ARMA (1,1) recurrent neural network process
## Stationary and Integrated Autoregressive Neural Network Processes
Stationary and Integrated Autoregressive Neural Network Processes
## IFNN Manual: Integrated Framework for Neural Network and Conventional Modelling
IFNN Manual: Integrated Framework for Neural Network and Conventional Modelling
## SIM user's manual. A flexible toolbox for spatial interaction modelling
IFNN Manual: Integrated Framework for Neural Network and Conventional Modelling | 2022-07-02 20:18:45 | {"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": 0.8487268686294556, "perplexity": 7008.030011930495}, "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-2022-27/segments/1656104204514.62/warc/CC-MAIN-20220702192528-20220702222528-00474.warc.gz"} |
https://forums.powershell.org/t/invoke-command-and-copy-item/5016 | # Invoke-Command and copy-item
I am trying to copy files on a remote machine to another location on the same machine, effectively a backup. I build the source and destination paths from various bits of data, and use Invoke-Command to complete the action from a remote machine. From the command line, everything works.
But from within a scrip:
Invoke-Command -ComputerName Server6 { Copy-Item -Path C:\program files\software\BIN* -Destination E:\sw_install\1\BIN_6.2\ -Recurse -Force }
Container cannot be copied onto existing leaf item.
+ CategoryInfo : InvalidArgument: (C:\program file…BIN\Plugins:String) [Copy-Item], PSArgumentException
+ FullyQualifiedErrorId : CopyContainerItemToLeafError,Microsoft.PowerShell.Commands.CopyItemCommand
The files are all copied, but the sub folders and their contents are not.
What do I need to do?
Just get rid of “*” in
`Copy-Item -Path C:\program files\software\BIN\* -Destination E:\sw_install\1\BIN_6.2\ -Recurse -Force `
It really was that easy! Many Thanks! | 2021-05-16 13:08:52 | {"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": 0.8027418851852417, "perplexity": 12073.357683794295}, "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-2021-21/segments/1620243991269.57/warc/CC-MAIN-20210516105746-20210516135746-00020.warc.gz"} |
https://simple.m.wiktionary.org/wiki/iridium | # iridium
The Simple English Wikipedia has an article on:
osmium${\displaystyle \longleftarrow }$ 77 ${\displaystyle \longrightarrow }$platinum
## Noun
Pluralnone
1. (uncountable) Iridium is a metallic (meaning made of metal) element with an atomic number of 77 and symbol Ir. | 2021-10-22 20:10:06 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 2, "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": 0.5495487451553345, "perplexity": 10062.150112729842}, "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/1634323585518.54/warc/CC-MAIN-20211022181017-20211022211017-00159.warc.gz"} |
https://www.themathdoctors.org/pattern-and-sequence-puzzles/ | # Pattern and Sequence Puzzles
One of the harder types of question to answer effectively is a puzzle, which as I define it means that there is no routine way to solve it, so any hint would likely give away the answer. But sometimes these are only “puzzles” to us, because we don’t know the context that would have told us what kind of answer to expect. I have collected here several different sorts of sequence or pattern puzzles, showing various ways we can approach them. (These are often done before learning algebra, so that although variables may be used in the answers, no algebra is needed to solve them.)
## Finding a formula for a sequence
Here is a typical question from 2001:
Terms and Rules
My step daughter is in sixth grade and she has been doing a pattern journal where she has two columns of numbers: the first column is the n, and the second colum is the term, and she has to find the rule (e.g. n^2), etc. Is there some place I can find out the basics for this concept?
The one example offered may or may not be typical of the particular kind of “rule” she is to look for, so I started with general suggestions and a request for examples:
I don't know that there is much to say in general about this kind of problem. A lot depends on the level of difficulty; most likely she has been given a set of problems that all have a similar kind of pattern, so that a method can be developed for solving them. The problems can differ in the presentation (whether consecutive terms are given, for example), and in the complexity of the rule (a simple multiplication, a more complicated calculation from n, a recursive rule - based on the previous term - or even a weird trick rule like "the number of letters in the English word for the number n"). For that reason, it would be very helpful if you could send us a couple sample questions so we could help more specifically.
I started by offering a particularly simple example:
I don't see any good examples in our archives at the elementary level - probably just because we've never felt that any one such answer was useful to help others. Let's try a few samples. Here's an easy one:
n | term
---+------
1 | 3
2 | 6
3 | 9
4 | 12
Here you may just be able to see that the terms are a column in a multiplication table (or, more simply, that all the terms are multiples of 3); or you might look at the differences between successive terms (a very useful method at a higher level, called "finite differences") and see that the terms are "skip-counting" by 3's, a clue that the rule involves multiplication by 3. However you see it, the rule (using "*" for the multiplication sign) is
3 * n
At that level, there is no method needed for solving; mere recognition is often enough.
Here's a slightly more complicated one:
n | term
---+------
1 | 6
2 | 11
3 | 16
4 | 21
Here the differences are all 5, suggesting multiplication by 5. You might want to add a new column to the table so you can compare the given term with 5n:
n | 5n | term
---+----+------
1 | 5 | 6
2 | 10 | 11
3 | 15 | 16
4 | 20 | 21
Now you can see that each term is one more than 5n, so the rule is
5n + 1
I particularly like this method, which I think of as building up the formula step by step. The “skip-counting” tells us the formula should be related to multiplication by 5, so we can just compare the given sequence to 5n, and observe that we have to add 1. Any linear sequence can be found this way.
Now they might get still more complicated:
n | term
---+------
1 | 2
2 | 5
3 | 10
4 | 17
Here the differences aren't all the same (3, 5, 7), so something besides multiplication and addition must be going on. I happen to know that when the differences are themselves increasing regularly (by 2 each time in this case) that there is a square involved; try adding a column for the square of n and see if you can figure it out.
This just gives a small taste of what these puzzles might be like. They can get much harder. Actually, in many harder cases I think the problems can be very unfair, because in reality (when you don't know ahead of time what kind of rule to expect) the rule could be absolutely anything, such as "the nth number on a page of random numbers I found in a book"! It's really just guesswork, and sometimes it's really hard to say which of several possible answers is what the author of the problem might have had in mind. But at your daughter's level you don't have to worry about that; it may be mostly just a matter of recognizing familiar patterns such as skip counting or squares. It's only where I sit, seeing problems completely out of context, and having to make a wild guess as to what sort of rule to look for, that these problems are always challenging.
## Finding a formula for a process
William responded to my request for examples:
Thank you, Dr. Peterson. That is exactly what I was talking about and you gave me great examples. One question I do remember was connected pentagons 1 = 5, sides -2 = 8 - 3 = 11 etc. where n is the number of pentagons and the term is the number of sides exposed, and then we needed to find the rule for 100 pentagons. Your answer definitely helped.
This is really an entirely different type: not just a list of numbers, but numbers generated by a defined process. This is no longer just a puzzle, since we have not just the first few terms, but a way to make all of them. There is definitely one correct answer.
He’s describing (a little cryptically) a table:
n | term
--+-----
1 | 5
2 | 8
3 | 11
Presumably the problem was to count the “exposed sides” of a sequence of pentagons like these:
Even without a formula, by continuing to add more pentagons in a row we could get as many terms as we want. I responded:
My answer is most helpful if you are just given a list of terms, and have to guess the rule. Your example is actually easier in some respects, and harder in others. It's common to just make a table of terms based on the geometry of the problem, and then try to guess a rule from that. But how do you know that the rule is really the right one, when it's not just an arbitrary list of terms, but one generated by a real situation? You really need to find some reason for connecting the rule to the objects you are counting. For that reason, I recommend trying to find a rule in the counting process itself.
So you can use the guessing approach if you want, and I suspect many texts and teachers expect that, but it's much more satisfying to skip the tables and really know you have the right answer!
So the rule, from the process I described, is $$a_n = 5 + 3(n – 1)$$, which can be simplified to $$a_n = 3n + 2$$.
Using my table method, we compare to 3n because terms increase by 3, and get
n | 3n | term = 3n + 2
--+----+-----
1 | 3 | 5
2 | 6 | 8
3 | 9 | 11
This is correct; but it does not in itself guarantee that it will work for all n.
## Tricky sequence puzzles
Here is another question (2002) requesting general guidelines for sequence puzzles, but turning out to need an entirely different type of answer:
Finding the Next Number in a Series
I haven't been able to find much information about an approach or method in determining the "next" number is a given series of numbers, e.g.,
9, 5, 45, 8, 6, 48, 6, 7... What is the next number?
I can usually figure it out but if there is a formal way that makes it easier I would love to know about it!
In the previous question, we were looking at fairly routine types of sequences. This one looked different. I started with some general comments about non-routine puzzles:
Mathematically speaking, problems like this are impossible. Literally!
That's because there is no restriction on what might come next in a sequence; ANY list of numbers, chosen for no reason at all, forms a sequence. So the next number can be anything.
A question like this is really not a math question, but a psychology question with a bit of math involved. You are not looking for THE sequence that starts this way, but for the one the asker is MOST LIKELY to have chosen - the most likely one that has a particularly simple RULE. And there is no mathematical definition for that.
I meant that literally: the next number could be anything at all, and we could find a rule to fit the sequence:
If you just wanted _a_ sequence that starts this way, but can be defined by _some_ mathematical rule, there is a technique that lets you find an answer without guessing. This is called "the method of finite differences," and you can find it by searching our site (using the search form at the bottom of most pages) for the phrase. It assumes (as is always possible) that the sequence you want is defined by a polynomial, and finds it. Sometimes this is what the problem is really asking for.
As we explain in detail elsewhere, any list of n numbers can be generated by a polynomial with degree n-1 or less. But the coefficients will often be ugly, and it is very unlikely that it is the intended answer for a puzzle of this type. That, however, is not a mathematical conclusion, but a psychological one!
But often, especially when many terms are given, there is a much simpler rule that is not of polynomial form. Then you are being asked to use your creativity to find a nice rule. Sometimes starting with finite differences gives you a good clue, even if you don't end up with a polynomial; just seeing a pattern in the differences can reveal something about the sequence. Other times it is helpful to factor the numbers, or to look at successive ratios. Here you are doing a more or less orderly search, in order to find something that may not turn out to be orderly.
Some puzzles like this are really just tricks. The "rule" may be that the numbers are in alphabetical order, or that each number somehow "describes" the one before, or even that they are successive digits of pi. In such cases, you have to ignore all thoughts of rules and orderly solutions, and just let your mind wander. This is sometimes called "lateral thinking," and it's entirely incompatible with "formal methods"!
I’ll be looking at some of these ideas in a later post. In this case, after suggesting things that might work, I got around to trying the specific sequence we’d been given, and got a surprise:
I first assumed the specific sequence you gave was just a random list of numbers, rather than a real problem, so I shouldn't bother looking for a pattern. But glancing at it, I see that it is not random:
9, 5, 45, 8, 6, 48, 6, 7, ...
I see some multiplications here:
9 * 5 = 45, 8 * 6 = 48, 6 * 7 = __
I can't recall what chain of reasoning my mind went through to see that, but it may have helped that my kids asked me to go through a set of multiplication flash cards an hour or two ago. And focusing on the few larger numbers, thinking about how large numbers might pop up (multiplication makes bigger changes than addition), probably led me in the right direction. I don't recall seeing anything quite like this presented as a sequence problem before, but seeing factors, one of my usual techniques, was the key.
As I had suggested, I just let my mind wander. And once I saw it, I realized I couldn’t just give a hint, unless it was as vague as, “Think about multiplication”. But I also realized something else:
In this case, you were apparently just asked to find the NEXT number, so we're done as soon as you fill in my blank. It may well be that there is no pattern beyond that; the choice of 9, 5, 8, 6, 6, 7 may be random. That's a good reminder that we have to read the problem carefully and not try to solve more than we were asked. We weren't told that there was any pattern beyond the next number!
## A general procedure
Here’s another question that elicited a collection of ideas from Doctor Wilko, in 2004:
Thinking about Number Pattern Problems
How do I figure out the next 2 numbers in the pattern 1, 8, 27, 64, ____, ____?
Using addition, the pattern doesn't add up. If I use multiplication, I'm stumped. I get 1x1; 4x2; 9x3; 16x4;, but I don't understand what the next number would be to use for ___x5 and then ____ x6.
Teresa was familiar with some particular kind of pattern; this didn’t fit. What do you do when you run out of ideas? Doctor Wilko started with something like what Teresa had probably done:
Sequence or pattern problems can be anywhere from very challenging to fairly easy. This is a matter of perspective and depends on how many problems like this you've been exposed to. The more you do and see, the less challenging these types of problems become.
I'll tell you how I approach a problem like this.
The first thing is that I know there is a pattern, so that is my goal,to find the pattern.
I might see if there is a pattern using addition.
1, 8, 27, 64, __, __,
\/ \/ \/ \/ \/
+7 +19 +37 ? ?
Hmmm, I don't see any pattern using addition.
He is not necessarily looking for the same number being added, but for some sort of pattern in the differences. This is a classic first step.
Next, I might try to use multiplication.
The question is: Is there a number that I can multiply each term by to get the next term? To get from 1 to 8, there is only one number that I can multiply 1 by to get 8 - it's 8. But I also notice that I can't multiply 8 by any other number to get exactly 27 (no remainder).
1, 8, 27, 64, __, __,
\/ \/ \/ \/ \/
*8 *?
So far, I don't see a pattern with using addition from term to term and I don't see a number that I can multiply each term by to get the
next.
I would quickly do a similar check for subtraction and division, but I don't find anything that works there, either.
Very often the ratio of one term to the next is not nice, so we can give up this idea quickly.
Now, I'm going to look at the position of each number, as sometimes this will reveal a pattern. 1 is in the 1st position, 8 is in the 2nd position, ...(see below)
1, 8, 27, 64, __, __,
| | | | | |
1 2 3 4 5 6
Does 1 have anything to do with 1?
Does 2 have anything to do with 8?
Does 3 have anything to do with 27?
Does 4 have anything to do with 64?
This might go somewhere!
1 is a divisor of 1.
2 is a divisor of 8.
3 is a divisor of 27.
4 is a divisor of 64.
So maybe multiplication will work, but not in the way I tried above! I'm going to factor each term in the sequence to look at the divisors.
1: 1 = 1
8: (4 * 2) = (2 * 2 * 2)
27: (9 * 3) = (3 * 3 * 3)
64: (16 * 4) = (4 * 4 * 4)
The 4s could be factored further, but I'm starting to see a pattern emerge, so I'll keep it as (4 * 4 * 4).
If this pattern continues, can you find the 5th and 6th term of the sequence? (I rewrote the 1's to match the pattern.)
1 = (1 * 1 * 1)
8 = (2 * 2 * 2)
27 = (3 * 3 * 3)
64 = (4 * 4 * 4)
? = (? * ? * ?)
? = (? * ? * ?)
Some people will immediately recognize the numbers in this sequence as cubes; others will need to go through the factoring and notice that factors come in threes. Your level of experience determines how hard this will be.
Notice, though, that Teresa had done almost exactly what Doctor Wilko did. If she had noticed that the multipliers 1, 4, 9, and 16 she found are squares — in fact, the square of n — she might have had the answer.
Doctor Wilko concluded:
Does this help? The basic idea in number pattern problems is to keep trying possible patterns. As I said at the top, the more of these you do the easier it gets to see the patterns. As with most things, experience and practice pays off!
Once you have seen enough patterns, you will start recognizing those you have seen before, and have a longer list of things to try in the future.
### 3 thoughts on “Pattern and Sequence Puzzles”
This site uses Akismet to reduce spam. Learn how your comment data is processed. | 2021-05-18 11:32:55 | {"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": 0.6001845002174377, "perplexity": 360.3579203648965}, "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-2021-21/segments/1620243989819.92/warc/CC-MAIN-20210518094809-20210518124809-00316.warc.gz"} |
https://www.gradesaver.com/textbooks/math/algebra/college-algebra-10th-edition/chapter-8-review-exercises-page-635/57 | ## College Algebra (10th Edition)
Let $x$ be the number of small boxes, $y$ be the number of medium boxes, $z$ be the amount of large boxes. Then our equations: (for oatmeal raisin): $x+2y+2z=15$ (for chocolate chip): $x+y+2z=10$ (for shortbread): $y+3z=11$ Deducing the second equation from the first one we get: $y=5$, then substituting this into the third, $z=2$, then $x=1$. So we buy $1$ small, $5$ medium and $2$ big boxes. | 2019-11-16 00:58:07 | {"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": 0.9223690032958984, "perplexity": 832.3292902925847}, "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-2019-47/segments/1573496668716.69/warc/CC-MAIN-20191116005339-20191116033339-00016.warc.gz"} |
http://www.r-bloggers.com/page/6/?s=regression | # 1036 search results for "regression"
## Moving Average Representation of VAR
March 10, 2013
By
A vector autoregression (VAR) process can be represented in a couple of ways. The usual form is as follows: The above (AR process) is what we often see and use in practice. However, I recently see more and … Continue reading
## Veterinary Epidemiologic Research: Linear Regression Part 2 – Checking assumptions
March 6, 2013
By
We continue on the linear regression chapter the book Veterinary Epidemiologic Research. Using same data as last post and running example 14.12: Now we can create some plots to assess the major assumptions of linear regression. First, let’s have a look at homoscedasticity, or constant variance of residuals. You can run a statistical test, the
## Visualizing neural networks from the nnet package
March 4, 2013
By
Neural networks have received a lot of attention for their abilities to ‘learn’ relationships among variables. They represent an innovative technique for model fitting that doesn’t rely on conventional assumptions necessary for standard models and they can also quite effectively handle multivariate response data. A neural network model is very similar to a non-linear regression
## Tools for making a paper
March 1, 2013
By
Since it seems to be the fashion, here’s a post about how I make my academic papers. Actually, who am I trying to kid? This is also about how I make slides, letters, memos and “Back in 10 minutes” signs to pin on the door. Nevertheless it’s for making academic papers that I’m going to
## How to make a scientific result disappear
February 27, 2013
By
$How to make a scientific result disappear$
Nathan Danneman (a co-author and one of my graduate students from Emory) recently sent me a New Yorker article from 2010 about the “decline effect,” the tendency for initially promising scientific results to get smaller upon replication. Wikipedia can summarize the phenomenon as well as I can: In his article, Lehrer gives several examples where
## R Bootcamp Materials!
February 24, 2013
By
Learn about ColoRs in R!Analyze model results with custom functions.Good and Bad GraphicsTo train new employees at the Wisconsin Department of Public Instruction, I have developed a 2-3 day series of training modules on how to get work done in R. These...
## the BUGS Book [guest post]
February 24, 2013
By
(My colleague Jean-Louis Fouley, now at I3M, Montpellier, kindly agreed to write a review on the BUGS book for CHANCE. Here is the review, en avant-première! Watch out, it is fairly long and exhaustive! References will be available in the published version. The additions of book covers with BUGS in the title and of the corresponding
## A slightly different introduction to R, part IV
February 21, 2013
By
Now, after reading in data, making plots and organising commands with scripts and Sweave, we’re ready to do some numerical data analysis. If you’re following this introduction, you’ve probably been waiting for this moment, but I really think it’s a good idea to start with graphics and scripting before statistical calculations. We’ll use the silly
## Working with R2MLwiN Part 2
February 19, 2013
By
## Specifying the model
This is the second part of a series of notes demonstrating use of the R package, R2MLwiN, an R command interface to the multilevel modelling software package, MLwiN (see the MLwiN site for getting access to MLwiN). The first set of notes showed how to get started with R2MLwiN. In these notes,... | 2013-05-19 15:31:14 | {"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": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 1, "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": 0.5341957807540894, "perplexity": 2507.7372283053787}, "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-2013-20/segments/1368697745221/warc/CC-MAIN-20130516094905-00045-ip-10-60-113-184.ec2.internal.warc.gz"} |
https://www.jiskha.com/questions/1203644/the-term-formula-mass-is-a-general-term-that-can-be-applied-to-both-ionic-compounds-and | # chemistry cp
The term formula mass is a general term that can be applied to both ionic compounds and molecular compounds because
(a) ionic compounds exist as individual molecules
(b) molecular formulas are always empirical formulas
(c) not all formulas represent individual molecules
(d) molecular formulas are always Lewis formulas
1. 👍
2. 👎
3. 👁
4. ℹ️
5. 🚩
1. I don't think so; in fact, answer a is the reason some profs like to use formula mass instead of molar mass. That is, ionic compounds do NOT exist as individual molecules.
1. 👍
2. 👎
3. ℹ️
4. 🚩
2. so it it letter c?
1. 👍
2. 👎
3. ℹ️
4. 🚩
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Find the first three non-zero terms and the general term for the power series representation of f(x) = 1/(x^2 + 3). I came up with 1/3 - x^2/9 + x^4/27, but I don't know the general term. | 2022-05-17 21:38:30 | {"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": 0.8399429321289062, "perplexity": 963.7514402694178}, "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/1652662520817.27/warc/CC-MAIN-20220517194243-20220517224243-00088.warc.gz"} |
http://math.stackexchange.com/questions/335929/problem-about-subsets-of-1-2-dots-n | # Problem about subsets of $\{1, 2,\dots,n\}$
Let $A=\{1, 2,\dots,n\}$ What is the maximum possible number of subsets of $A$ with the property that any two of them have exactly one element in common ?
I strongly suspect the answer is $n$, but can't prove it.
-
I think answer should be n but i m not sure – kalpeshmpopat Mar 20 '13 at 15:33
Well, {1}, {1,2}, {1,3}....{1,n} form such a collection of n subsets. The subsets don't have to have the same cardinality. – Cosmonut Mar 20 '13 at 15:35
I would like to point out that most of the answers are missing the case $\{1, 2\}, \{2, 3\}, \{3, 1\}$, so there is an error in their logic. – Calvin Lin Mar 20 '13 at 16:14
@Inceptio No, but as Thomas mentioned, you can for $n = 13$. I.e. the projective plane of order 3. – Calvin Lin Mar 20 '13 at 16:24
@Thomas I've undeleted it. – Thomas Andrews Mar 20 '13 at 21:10
This is a well known type of problem in combinatorics. (Try googling "exact intersections".) The (slight) generalisation of your claim, in which we require $|A\cap B|=\ell>0$ whenever $A\neq B$, is apparently due to Fisher.
Let $\mathcal{A}$ be a family of subsets of $\{1,2,\dots,n\}$ such that for every two distinct $A,B\in\mathcal{A}$ we have $|A\cap B|=\ell$. We claim $|\mathcal{A}|\leq n$. Certainly we're done if some $A\in\mathcal{A}$ has size $\ell$ (this is where we use $\ell>0$), so assume otherwise.
For each $A\in\mathcal{A}$ consider the "indicator vector" $1_A\in\mathbf{R}^n$ given by $1_A(x)=1$ if $x\in A$ and $0$ otherwise. I claim that $\{1_A : A\in\mathcal{A}\}$ is a linearly independent set, so $|\mathcal{A}|\leq n$.
Suppose $\lambda_A\in\mathbf{R}$ are some coefficients such that
$$\sum_{A\in\mathcal{A}} \lambda_A 1_A = 0.$$
Taking the scalar product with $1_B$, noting $1_A\cdot 1_B = \ell$ when $A\neq B$, we have
$$\lambda_B |B| + \sum_{A\neq B} \lambda_A \ell = 0.$$
Rearranging slightly,
$$\lambda_B (|B| - \ell) = - \ell\sum_{A\in\mathcal{A}} \lambda_A.$$
Conclusion: either $\sum\lambda_A = 0$, in which case every $\lambda_B=0$ (since $|B|>\ell$ for all $B\in\mathcal{A}$), or $\sum\lambda_A\neq 0$, in which case all $\lambda_B$ are nonzero and opposite in sign to $\sum\lambda_A$, impossible.
-
Yes, I mean the {scalar,dot,inner,...}-product. – Sean Eberhard Mar 20 '13 at 21:13
Wait, $\lambda_B$ are not all equal, but rather $\lambda_B=\frac{1}{|B|-\ell}C$? If all the $|B|$ are equal, clearly it is true that the $\lambda_B$ are equal, but we can't conclude that here, can we? – Thomas Andrews Mar 20 '13 at 21:15
Ah, thanks, correcting now. – Sean Eberhard Mar 20 '13 at 21:16
What we really know is that they are all the same sign, since $|B|>\ell$, which is enough. – Thomas Andrews Mar 20 '13 at 21:17
nice proof .... – wece Mar 20 '13 at 21:19
Hint: For a set of length $i > 2$, no other set can have any subset of that set $i \ge 2$ as a subset.
- | 2016-05-26 09:11:29 | {"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": 0.9567028284072876, "perplexity": 292.1459082569333}, "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/1464049275764.90/warc/CC-MAIN-20160524002115-00043-ip-10-185-217-139.ec2.internal.warc.gz"} |
http://www.physicsforums.com/showthread.php?p=3403993 | # Fortran90 reading in a formatted input file
by Rowdy5000
Tags: file, formatted, fortran90, input, reading
P: 73
What you can try is switching your read statements leading up to your DO/ENDDO loop with write statements and seeing what the file looks like. It would look like
open(10,file='test.txt')
write(10,*)a
write(10,*)b
write(10,*)c,d
write(10,*)e,f,g,h
write(10,*)i
write(10,*)j
write(10,*)ntimes
close(10)
With arbitrary values assigned to a-j and ntimes.
Just an idea.
P: 5 Thanks for that; it did show me how basic of an input I could use. I am having some trouble with this in debugging though. I get a ".exe triggered a breakpoint" error at the line of the first read statement. I'll do some more headscratching and then return with more details if I can't get it. Thanks again.
Mentor
P: 20,409
## Fortran90 reading in a formatted input file
"Triggering a breakpoint" isn't an error. It seems that you set a breakpoint at that statement, which causes the debugger to stop.
HW Helper
P: 6,900
Quote by Mark44 "Triggering a breakpoint" isn't an error.
It could be an error, if the compiler generates breakpoint instructions (INT 3) or uses an invalid address to access memory in a failure case.
It's possible that the open statement failed. Is there anyway to check the status of the open statement?
P: 5 rcgldr: I think you might be close to home w/ suspecting the open statement. In Visual Studio 2005 I get a little green arrow pointing at the first read statement. When I hover over that for detail, I get the following message: "This code has called into another function. When that function is finished, this is the next statement that will be executed." And when I go to disassembly, I've got a little yellow arrow pointing at the line just after a line that reads: "00431340 int 3" I don't really know yet how to check the status of the open statement. Thanks to all for the support.
Mentor
P: 20,409
Quote by Rowdy5000 rcgldr: I think you might be close to home w/ suspecting the open statement. In Visual Studio 2005 I get a little green arrow pointing at the first read statement. When I hover over that for detail, I get the following message: "This code has called into another function. When that function is finished, this is the next statement that will be executed." And when I go to disassembly, I've got a little yellow arrow pointing at the line just after a line that reads: "00431340 int 3"
This is the breakpoint instruction that rcgldr mentioned.
Have you set a breakpoint? If so, it will show up as a red circle in the left margin in Visual Studio. You can delete all breakpoints by clicking Delete All Breakpoints in the the Debug menu in VS. This menu item doesn't appear if there are no breakpoints set.
Quote by Rowdy5000 I don't really know yet how to check the status of the open statement. Thanks to all for the support.
P: 5 I think the compiler is automatically setting a breakpoint or something...I got a disassembly display and a call stack that mean nothing to me and I think I've hit a wall with self-diagnosis. If anyone can suggest a new lead or what I could post here to clue us in, I'll gladly oblige.
Mentor P: 20,409 AFAIK, the compiler doesn't automatically set breakpoints. Whatever, your open statement doesn't look right to me, especially that form = "..." part. Here's a link to a page that has several examples of open statements. Try modifying your open statement and see if that makes a difference. http://www.livephysics.com/computati...-handling.html For additional examples, you can do what I did, which was enter "fortran open" in the browser.
P: 3,007
Quote by Mark44 AFAIK, the compiler doesn't automatically set breakpoints. Whatever, your open statement doesn't look right to me, especially that form = "..." part. Here's a link to a page that has several examples of open statements. Try modifying your open statement and see if that makes a difference. http://www.livephysics.com/computati...-handling.html For additional examples, you can do what I did, which was enter "fortran open" in the browser.
In addition to the link Mark44 provided, I have also found this site to be infinitely helpful.
Mentor
P: 20,409
Here's another link to a page with more than you want to know about the OPEN statement - http://www.hpc.unimelb.edu.au/doc/f90lrm/lrm0545.htm.
This page describes the FORM parameter, which I didn't remember using (it's been a long while since I did any Fortran programming...).
I don't know if you noticed it, but there's an extra space in the code you showed (in 'format ted'). If that's actually in your code, it's probably what's causing the compiler to complain.
Quote by Rowdy5000 open(10,file='input',status='unknown',form='format ted')
P: 5 Thanks to everyone for the useful links and support with this. I actually resolved this finally by toggling status in the open statement to 'old'. It didn't really make sense to me based upon what I'd read about the status character, but I tried this and it worked.
Related Discussions Engineering, Comp Sci, & Technology Homework 31 Programming & Computer Science 5 Programming & Computer Science 2 Programming & Computer Science 0 Programming & Computer Science 2 | 2014-03-07 16:07:53 | {"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": 0.2745136618614197, "perplexity": 1281.2736424464183}, "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/1393999645498/warc/CC-MAIN-20140305060725-00019-ip-10-183-142-35.ec2.internal.warc.gz"} |
https://www.jobilize.com/online/course/3-8-summary-of-key-concepts-basic-operations-with-real-numbers-by-open?qcr=www.quizover.com | # 3.8 Summary of key concepts
Page 1 / 1
This module is from Elementary Algebra by Denny Burzynski and Wade Ellis, Jr. The basic operations with real numbers are presented in this chapter. The concept of absolute value is discussed both geometrically and symbolically. The geometric presentation offers a visual understanding of the meaning of |x|. The symbolic presentation includes a literal explanation of how to use the definition. Negative exponents are developed, using reciprocals and the rules of exponents the student has already learned. Scientific notation is also included, using unique and real-life examples.This module contains a summary of the key concepts in the chapter "Basic Operations with Real Numbers".
## Positive and negative numbers ( [link] )
A number is denoted as positive if it is directly preceded by a " $+$ " sign or no sign at all. A number is denoted as negative if it is directly preceded by a " $-$ " sign.
Opposites are numbers that are the same distance from zero on the number line but have opposite signs.
## Double-negative property ( [link] )
$-\left(-a\right)=a$
## Absolute value (geometric) ( [link] )
The absolute value of a number $a$ , denoted $|a|$ , is the distance from $a$ to 0 on the number line.
## Absolute value (algebraic) ( [link] )
$|a|=\left\{\begin{array}{cc}a& \text{if}\text{\hspace{0.17em}}a\ge 0\\ -a& \text{if}\text{\hspace{0.17em}}a<0\end{array}$
like signs , add the absolute values of the numbers and associate the common sign with the sum.
unlike signs , subtract the smaller absolute value from the larger absolute value and associate the sign of the larger absolute value with the difference.
$0+\text{any}\text{\hspace{0.17em}}\text{number}=\text{that}\text{\hspace{0.17em}}\text{particular}\text{\hspace{0.17em}}\text{number}$ , that is, $0+a=a$ for any real number $a$ .
Since adding 0 to a real number leaves that number unchanged, 0 is called the additive identity.
## Definition of subtraction ( [link] )
$a-b=a+\left(-b\right)$
## Subtraction of signed numbers ( [link] )
To perform the subtraction $a-b$ , add the opposite of $b$ to $a$ , that is, change the sign of $b$ and add.
## Multiplication and division of signed numbers ( [link] )
$\begin{array}{lll}\left(+\right)\text{\hspace{0.17em}}\left(+\right)=+\hfill & \frac{\left(+\right)}{\left(+\right)}=+\hfill & \frac{\left(+\right)}{\left(-\right)}=-\hfill \\ \left(-\right)\text{\hspace{0.17em}}\left(-\right)=+\hfill & \hfill & \hfill \\ \left(+\right)\text{\hspace{0.17em}}\left(-\right)=-\hfill & \hfill & \hfill \\ \left(-\right)\text{\hspace{0.17em}}\left(+\right)=-\hfill & \frac{\left(-\right)}{\left(-\right)}=+\hfill & \frac{\left(-\right)}{\left(+\right)}=-\hfill \end{array}$
Two numbers are reciprocals of each other if their product is 1. The numbers 4 and $\frac{1}{4}$ are reciprocals since $\left(4\right)\text{\hspace{0.17em}}\left(\frac{1}{4}\right)=1$ .
## Negative exponents ( [link] )
If $n$ is any natural number and $x$ is any nonzero real number, then ${x}^{-n}=\frac{1}{{x}^{n}}$ .
## Writing a number in scientific notation ( [link] )
To write a number in scientific notation:
1. Move the decimal point so that there is one nonzero digit to its left.
2. Multiply the result by a power of 10 using an exponent whose absolute value is the number of places the decimal point was moved. Make the exponent positive if the decimal point was moved to the left and negative if the decimal point was moved to the right.
## Converting from scientific notation: Positive exponent ( [link] )
To convert a number written in scientific notation to a number in standard form when there is a positive exponent as the power of 10, move the decimal point to the right the number of places prescribed by the exponent on the 10.
## Negative exponent ( [link] )
To convert a number written in scientific notation to a number in standard form when there is a negative exponent as the power of 10, move the decimal point to the left the number of places prescribed by the exponent on the 10.
where we get a research paper on Nano chemistry....?
what are the products of Nano chemistry?
There are lots of products of nano chemistry... Like nano coatings.....carbon fiber.. And lots of others..
learn
Even nanotechnology is pretty much all about chemistry... Its the chemistry on quantum or atomic level
learn
da
no nanotechnology is also a part of physics and maths it requires angle formulas and some pressure regarding concepts
Bhagvanji
Preparation and Applications of Nanomaterial for Drug Delivery
revolt
da
Application of nanotechnology in medicine
what is variations in raman spectra for nanomaterials
I only see partial conversation and what's the question here!
what about nanotechnology for water purification
please someone correct me if I'm wrong but I think one can use nanoparticles, specially silver nanoparticles for water treatment.
Damian
yes that's correct
Professor
I think
Professor
Nasa has use it in the 60's, copper as water purification in the moon travel.
Alexandre
nanocopper obvius
Alexandre
what is the stm
is there industrial application of fullrenes. What is the method to prepare fullrene on large scale.?
Rafiq
industrial application...? mmm I think on the medical side as drug carrier, but you should go deeper on your research, I may be wrong
Damian
How we are making nano material?
what is a peer
What is meant by 'nano scale'?
What is STMs full form?
LITNING
scanning tunneling microscope
Sahil
how nano science is used for hydrophobicity
Santosh
Do u think that Graphene and Fullrene fiber can be used to make Air Plane body structure the lightest and strongest. Rafiq
Rafiq
what is differents between GO and RGO?
Mahi
what is simplest way to understand the applications of nano robots used to detect the cancer affected cell of human body.? How this robot is carried to required site of body cell.? what will be the carrier material and how can be detected that correct delivery of drug is done Rafiq
Rafiq
if virus is killing to make ARTIFICIAL DNA OF GRAPHENE FOR KILLED THE VIRUS .THIS IS OUR ASSUMPTION
Anam
analytical skills graphene is prepared to kill any type viruses .
Anam
Any one who tell me about Preparation and application of Nanomaterial for drug Delivery
Hafiz
what is Nano technology ?
write examples of Nano molecule?
Bob
The nanotechnology is as new science, to scale nanometric
brayan
nanotechnology is the study, desing, synthesis, manipulation and application of materials and functional systems through control of matter at nanoscale
Damian
Is there any normative that regulates the use of silver nanoparticles?
what king of growth are you checking .?
Renato
What fields keep nano created devices from performing or assimulating ? Magnetic fields ? Are do they assimilate ?
why we need to study biomolecules, molecular biology in nanotechnology?
?
Kyle
yes I'm doing my masters in nanotechnology, we are being studying all these domains as well..
why?
what school?
Kyle
biomolecules are e building blocks of every organics and inorganic materials.
Joe
how did you get the value of 2000N.What calculations are needed to arrive at it
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https://www.jobilize.com/online/course/7-4-example-module-for-use-of-cnxml-mathml-tags-by-openstax?qcr=www.quizover.com | # 7.4 Example module for use of cnxml/mathml tags
Page 1 / 2
A paragraph is the place for text. You can also include vocabulary terms .
Do not use <emphasis> for vocabulary terms.
There are two ways to present definitions - using the glossary to define the term as above or using the definition tag within the text.
problem
an intricate unsettled question; a source of perplexity, distress, or vexation; difficulty in understanding oraccepting
## Section name
A document can have sections, however they are not required.
## Subsection name
Sections can have subsections. You can include quotes in paragraphs.
The seasons alter: hoary-headed frosts Fall in the fresh lap of the crimsonrose, And on old Hiems thin and icy crown An odorous chaplet of sweet summer buds Is, as in mockery, set. The spring, thesummer, The childing autumn, angry winter, change Their wonted liveries, and the mazed world, By their increase, nowknows not which is which. William Shakespeare; A Midsummer Night's Dream
You can also denote words from another language, such as biological genus and species, E.coli , or the southern German greeting, GrGott .
Paragraphs can contain many other tags such as lists and figures. shows how a figure will display in our system.
There are four different options for list. One type of list is a named-item list.
• ## First name
description of first item
• ## Second name
description of second item
• ## Third name
description of third item
• ## Fourth name
description of fourth item
## Optional name of example
Here is where you would put an example that relates to what the previous paragraphs were discussing. In anexample, you can include any tags that are allowed in any other paragraph including tables (see ).
Temperatures in 5 cities on 11/16/2002
Mean 54.000 12.22
Median 54.000 12.22
Variance 330.00 18.166
SD 101.852 10.092
Houston 54 12.22
Chicago 37 2.78
Minneapolis 31 -0.56
Miami 78 25.56
Phoenix 70 21.11
In the connexions system, it is important to use frame="all" with the table tag and colsep="1" rowsep="1" with the tbody tag to properly display the table in the print system.
If you would like a block of material to display exactly as you type it, you can use <code type='block'> ; as you would when discussing computer programming. >> syms t >> laplace(exp(t)) ans = 1/(s-1) >> laplace(t*exp(-t)) ans = 1/(s+1>^2
You can also include math in your document. Math can be displayed in three ways in our system: inline, block, ornumbered equations. Here are examples of an inline variable, $x$ , and an inline equation, $0< x< \pi$ . Math that you want set apart from the text can be numbered when in an equation tag or simply set apart. $\lim_{n\to }n\to$ x n x 0
## Optional equation name
$P(f)=\int_{()} \,d t$ p t 2 f t t 0 2 f t 1 2 f 2 f 1
See our discussion of Content MathML for a more basic discussion of math. See also our MathML Extensions page for more csymbol options.
## Pythagorean theorem
For a right triangle with legs $a$ and $b$ and hypotenuse $c$ , $a^{2}+b^{2}=c^{2}$
The proof would go here.
What is a composite number?
A composite number $n$ is a positive integer, $n> 1$ , which is not prime.
what is the stm
is there industrial application of fullrenes. What is the method to prepare fullrene on large scale.?
Rafiq
industrial application...? mmm I think on the medical side as drug carrier, but you should go deeper on your research, I may be wrong
Damian
How we are making nano material?
what is a peer
What is meant by 'nano scale'?
What is STMs full form?
LITNING
scanning tunneling microscope
Sahil
how nano science is used for hydrophobicity
Santosh
Do u think that Graphene and Fullrene fiber can be used to make Air Plane body structure the lightest and strongest. Rafiq
Rafiq
what is differents between GO and RGO?
Mahi
what is simplest way to understand the applications of nano robots used to detect the cancer affected cell of human body.? How this robot is carried to required site of body cell.? what will be the carrier material and how can be detected that correct delivery of drug is done Rafiq
Rafiq
what is Nano technology ?
write examples of Nano molecule?
Bob
The nanotechnology is as new science, to scale nanometric
brayan
nanotechnology is the study, desing, synthesis, manipulation and application of materials and functional systems through control of matter at nanoscale
Damian
Is there any normative that regulates the use of silver nanoparticles?
what king of growth are you checking .?
Renato
What fields keep nano created devices from performing or assimulating ? Magnetic fields ? Are do they assimilate ?
why we need to study biomolecules, molecular biology in nanotechnology?
?
Kyle
yes I'm doing my masters in nanotechnology, we are being studying all these domains as well..
why?
what school?
Kyle
biomolecules are e building blocks of every organics and inorganic materials.
Joe
anyone know any internet site where one can find nanotechnology papers?
research.net
kanaga
sciencedirect big data base
Ernesto
Introduction about quantum dots in nanotechnology
what does nano mean?
nano basically means 10^(-9). nanometer is a unit to measure length.
Bharti
do you think it's worthwhile in the long term to study the effects and possibilities of nanotechnology on viral treatment?
absolutely yes
Daniel
how to know photocatalytic properties of tio2 nanoparticles...what to do now
it is a goid question and i want to know the answer as well
Maciej
Abigail
for teaching engĺish at school how nano technology help us
Anassong
How can I make nanorobot?
Lily
Do somebody tell me a best nano engineering book for beginners?
there is no specific books for beginners but there is book called principle of nanotechnology
NANO
how can I make nanorobot?
Lily
what is fullerene does it is used to make bukky balls
are you nano engineer ?
s.
fullerene is a bucky ball aka Carbon 60 molecule. It was name by the architect Fuller. He design the geodesic dome. it resembles a soccer ball.
Tarell
what is the actual application of fullerenes nowadays?
Damian
That is a great question Damian. best way to answer that question is to Google it. there are hundreds of applications for buck minister fullerenes, from medical to aerospace. you can also find plenty of research papers that will give you great detail on the potential applications of fullerenes.
Tarell
how did you get the value of 2000N.What calculations are needed to arrive at it
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http://www.ck12.org/book/CK-12-Algebra-II-with-Trigonometry-Concepts/section/10.0/ | <meta http-equiv="refresh" content="1; url=/nojavascript/"> Conic Sections | CK-12 Foundation
# Chapter 10: Conic Sections
Created by: CK-12
0 0 0
## Introduction
Conic sections are four shapes; parabolas, circles, ellipses, and hyperbolas, created from the intersection of a plane with a cone or two cones.
In this chapter, we will study these four conic sections and place them in the $x-y$ plane. For each shape, we will analyze the parts, find the equation and graph. Lastly, we will introduce the general conic section equation and solve systems with conics and lines.
## Summary
This chapter covers parabolas, circles, ellipses, and hyperbolas. You will learn how to graph and analyze these conic sections. In the last concept, we will solve systems with conics and lines.
Oct 08, 2013 | 2014-10-02 09:23:22 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 1, "texerror": 0, "math_score": 0.2864263355731964, "perplexity": 1870.4819347537389}, "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/1412037663739.33/warc/CC-MAIN-20140930004103-00239-ip-10-234-18-248.ec2.internal.warc.gz"} |
http://mathforum.org/mathimages/index.php/Classification_Theorem_for_Compact_Surfaces | # Classification Theorem for Compact Surfaces
This is a Helper Page
This is a preliminary page that needs development.
Taken by Htasoff 17:00, 18 July 2011 (UTC). Rough diagram of how the connected sum of a Real Projective Plane and a Torus is homeomorphic to that of a Real Projective Plane and a Klein bottle. Since a Klein bottle is homeomorpic to the connected sum of two Projective Planes (not demonstrated), the connected sum of a Real Projective Plane and a Torus is homeomorphic to that of three Real Projective Planes.
Thus, as stated in the pictures (# means: the connected sum):
$RPP~\#~Torus = RPP~\#~Klein~bottle$
And (not described in the diagram):
$RPP~\#~RPP = Klein~bottle$
Thus:
$RPP~\#~Torus = RPP~\#~RPP~\#~RPP$
It is crucial to note, however, that you cannot simply subtract a RPP from both sides, because:
$Torus \neq Klein~bottle$
Thus, the connected sum of any number of tori and real projective planes can be reduced solely to a sum of real projective planes. This proof turns the and/ or to exclusively to or when added to the proof that any closed surface is homeomorphic to a sphere, and/ or to a connected sum of tori, and/ or to a connected sum of projective planes (also not yet proven on this page).
This diagram is based off of the book in this footnote[1].
I hope this will lead to an explanation of the claims made in the Why It's Interesting sections of the Real Projective Plane and Torus pages.
## Instructions for the Future
This page need to be written. I believe this to be interesting, and worth devoting a page to. Unfortunately, due to time constraints, I only had time to create the page. Here is what's needed:
• A quality visual proof like the one in the pictures or the book I've referenced.
• An explanation of how the location of the twist in the Mobius strip modeling the RPP renders the Klein bottle and torus effectively the same.
• I recommend using the 'equations' I wrote in this explanation.
• A proof of the broader statement: that any closed surface is homeomorphic to a sphere, and/ or to a connected sum of tori, and/ or to a connected sum of projective planes.
• Further development as you see fit.
## References
1. Massey, William. (1991). A Basic Course in Algebraic Topology (Graduate Texts in Mathematics). New York: Springer-Verlag. | 2015-09-01 12:26:18 | {"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": 4, "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": 0.7764578461647034, "perplexity": 423.7577505505048}, "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/1440645176794.50/warc/CC-MAIN-20150827031256-00017-ip-10-171-96-226.ec2.internal.warc.gz"} |