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https://software.broadinstitute.org/wdl/documentation/article?id=6751 | # (howto) Generate a JSON file describing inputsTutorials | Created 2016-01-09 | Last updated 2017-07-29
The simplest way to specify values for the input variables (such as file names and parameters) to the commands in your WDL scripts is to hard-code them, i.e. write them in the script itself. However, doing so forces you to make a new copy and edit the inputs every time you want to run your script on a new batch of data -- which undermines the advantages of setting up a pipeline script in the first place.
A much better way to proceed is to specify all the values for the input variables that you want to be able to customize from run to run in a JSON file (a structured text format a bit like XML but better -- certainly more readable). Then all you need to do is create a new file of inputs for each new batch of data that you want to run through your pipeline. The execution engine will use that JSON file to fill in the values of inputs to commands in your script where appropriate.
Still, on the face of it you might think that putting together the JSON file of inputs (specifically, structuring it correctly and not forgetting any command's inputs) would be a tedious and/or daunting task, especially in a command-line-only world with no point-and-click GUI.
But fear not! Help is at hand. WDL comes with a utility function (in the wdltool package) that will parse your WDL script and automatically generate a template JSON file containing the appropriate input file and parameter definitions. All you need to do then is to populate a copy of the file with the actual values that you want to use in a given run of the pipeline. When you are ready to run your script on your chosen execution engine, you'll simply provide the inputs file along with the script.
### Generating the template JSON
To generate the template of inputs for your WDL script, simply call the wdltool inputs function on your script:
java -jar wdltool.jar inputs myWorkflow.wdl > myWorkflow_inputs.json
This will create a file called myWorkflow_inputs.json that lists all the inputs to all the tasks in your script following the pattern below:
{
"<workflow name>.<task name>.<variable name>": "<variable type>"
}
This saves you from having to compile a list of all the tasks and their variables manually! Pretty nifty, right?
If you omit the > myWorkflow_inputs.json part of the command, the template content will be output to the terminal instead of being written to file.
### Customizing the inputs file for a particular run
Every time you want to run the script on some new data or with some different parameters, you simply open this file (or better, a copy) in a text editor and replace the part on the right of the colon with the value that you want.
In case you're wondering, the "<variable type>" bit in the original template is just there to remind you what type of variable the task will expect to see. In the same spirit, we do recommend giving your tasks and variables names that will be meaningful when you're going through your inputs file filling in filenames and parameter values. Otherwise you'll find yourself having to refer back to the pipeline script itself often, as it's not possible to add comments in a JSON file.
#### Example
Let's say the myWorkflow.wdl script describes a workflow called myWorkflowName. This workflow includes a task called stepA that takes two inputs, a File called input_file and a String called sample_name. The input template generated by the command above would look like this:
{
"myWorkflowName.stepA.input_file": "File"
"myWorkflowName.stepA.sample_name": "String"
}
So to run this script on a file called input.bam where the sample name of interest is NA12878, you would change it to:
{
"myWorkflowName.stepA.input_file": "~/path/to/input.bam"
"myWorkflowName.stepA.sample_name": "NA12878"
} | 2018-09-26 03:14:35 | {"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": 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": 0.33127516508102417, "perplexity": 1220.1247900461908}, "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-2018-39/segments/1537267163146.90/warc/CC-MAIN-20180926022052-20180926042452-00402.warc.gz"} |
https://astronomy.stackexchange.com/questions/12556/why-does-spectral-class-and-u-v-colour-correlate-slightly-differently-for-main-s | # Why does spectral class and U-V colour correlate slightly differently for main sequence, giant and supergiant stars?
In this diagram the $x$-axis is spectral type for stars and the $y$-axis is B-V colour.
Taking Series 1 as the Main Sequence, Series 2 as Giants and Series 3 as Super-giants. Why the spectral class for all these stars differ from the other ones?
If we were to say differences, for instance, how would one express that information?
• What is measured on the vertical axis? – James K Nov 19 '15 at 17:52
• Hi there.The vertical axis has the star colour B - V. – J. Astro Nov 20 '15 at 0:46
• This question doesn't strike me as clear, but I think what's being asked is "Why do stars of the same spectral type have different colours at different brightnesses?" In his case, for example, a K5 dwarf has B-V=0.8ish, vs a K5 giant with B-V=1.1ish. (Based on the plot above. No idea where it comes from.) – Warrick Nov 20 '15 at 7:56
• This is clearly not the Hertzsprung Russell diagram en.wikipedia.org/wiki/Hertzsprung%E2%80%93Russell_diagram – Eubie Drew Nov 20 '15 at 23:49
• Okay, the vertical axis is the B-V colour index, as explained here en.wikipedia.org/wiki/Color_index – Eubie Drew Nov 20 '15 at 23:54 | 2021-03-07 08:52:09 | {"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": 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": 0.6496835350990295, "perplexity": 1675.2514881371592}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-10/segments/1614178376206.84/warc/CC-MAIN-20210307074942-20210307104942-00523.warc.gz"} |
https://www.mathdoubts.com/problems/square-root-10-power-2-plus-half-log-16/ | # Find the value of $\sqrt{10^{ 2+\frac{1}{2} \log_{10} 16}}$
$10$ is a base number of an exponential term and its exponent is the sum of number $2$ and half of the common logarithm of $16$. The entire exponential term is placed under a square root. The value of this radical has to find in this problem.
$\sqrt{\large 10^{\displaystyle \normalsize 2+\frac{1}{2} \log_{10} 16}}$
###### Step: 1
The exponent contains a logarithm term and the logarithmic term is multiplied by rational number $\dfrac{1}{2}$. It can be cancelled if the number of the logarithmic term is expressed in exponential notation.
Therefore, write the number $16$ in exponential notation on the basis of $4$.
$= \,\,$ $\sqrt{\large 10^{\displaystyle \normalsize 2+\frac{1}{2} \log_{10} 4^2}}$
Use power rule of logarithms to shift the power of the exponential term of the logarithm.
$= \,\,$ $\sqrt{\large 10^{\displaystyle \normalsize 2+\frac{2}{2} \log_{10} 4}}$
$= \,\,$ $\sqrt{\large 10^{\displaystyle \normalsize 2+\require{cancel} \frac{\cancel{2}}{\cancel{2}} \log_{10} 4}}$
$= \,\,$ $\sqrt{\large 10^{\displaystyle \normalsize 2+ \log_{10} 4}}$
###### Step: 2
The radicand can be split as two multiplying factors in exponential notation by the product rule of exponents.
$= \,\,$ $\sqrt{\large 10^{\displaystyle \normalsize 2} \times \large 10^{\displaystyle \normalsize \log_{10} 4}}$
###### Step: 3
The value of second exponential term is $4$ as per the fundamental logarithm identity.
$= \,\,$ $\sqrt{\large 10^{\displaystyle \normalsize 2} \times 4}$
$= \,\,$ $\sqrt{\large 10^{\displaystyle \normalsize 2} \times 2^2}$
$= \,\,$ $\sqrt{\large 10^{\displaystyle \normalsize 2}} \times \sqrt{\large 2^2}$
$= \,\,$ $10 \times 2$
$= \,\,$ $20$
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You can get the latest updates from us by following to our official page of Math Doubts in one of your favourite social media sites. | 2022-06-30 13:31:39 | {"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": 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": 0.6986483335494995, "perplexity": 979.5716420794297}, "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/1656103821173.44/warc/CC-MAIN-20220630122857-20220630152857-00041.warc.gz"} |
https://blog.savageevan.com/posts/2012-11-02-persistent-location-tracking-looking-for-a-few-good-data-points/ | # Persistent Location Tracking: Looking For A Few Good Data Points
In this post, I revisit the question of whether Google Latitude meets my
persistent location tracking needs. In my previous post, I compared
Google Latitude to InstaMapper and concluded that the latter is too
battery-intensive. By looking at maps and base-level insights from the data,
I suggest that Google Latitude optimizes for battery life at the expense of
data quality.
## Exhibit A: Some Maps #
I started gathering data on Oct. 3, 2012:
Since then, Valkyrie Savage and I have travelled to Boston and Chicago.
Our stopover in Phoenix is clearly visible at this scale. You can barely make
out our day trip to Mount Monadnock, NH over near Boston. Here's a
closer look at that trip:
Ouch. The data is noisy in some areas, sparse in others. It's fairly clear that
we took Hwy 2 over, but some of the GPS readings are miles off. Let's zoom in
on that hike:
Only five data points actually lie within the park/mountain boundaries. That's
five data points for a four-hour hike. Our Boston data is somewhat more
accurate:
Still, the red line cuts through city blocks with reckless abandon. Either
we're flying, or we're packing some incredibly efficient demolition equipment.
Here's the map for one of my more itinerant Bay Area days:
I cycled to a doctor's appointment, visited
BiD to hear Mary Czerwinski speak about emotion tracking, worked
from home for a bit, went into San Francisco to meet up with
Lev Popov, and finally dragged myself home again.
The BART ride into San Francisco is understandably sparse: most of it is
separated from cell towers and GPS satellites by rock and/or water.
Most of my travel is on foot, by bike, or via public transit. Not content
with the Mount Monadnock hike data, I tried another quick drive up into
Tilden:
Google Latitude captured just four points during the 20-minute drive.
## Exhibit B: Some Analysis #
You can see the code for this analysis
here
and here.
After trudging through several lackluster map views, I'm left with a
nagging impression:
This data isn't that useful.
This impression deserves further analysis, so I grab the KML to answer some
of my questions. First off: how often is Google Latitude checking my location?
About every two minutes. GPS is a huge battery drain;
increasing the time between updates can help by allowing the GPS radio to
enter an idle state. How are those location readings scheduled?
Google Latitude really likes spacing its readings out by a whole number of
minutes.
How accurate is the data? The KML doesn't provide accuracy estimates
for its locations. Fortunately, the Google Latitude API does, so
I retrieve my data using this script and look at the accuracy readings:
Actually, the readings have fairly high accuracy. Only 7% of readings have a
reported error radius greater than 100m.
The maps above suggest that location readings are less accurate while
travelling at high speed. Is that true? The API provides speed estimates
for some readings, but this data is kind of sparse:
\$ python speed.py < history.api
found 7429 speed values among 20898 readings
I try a different method: the Haversine distance formula, which
gives me the distance between two points on the Earth's surface:
def haversineDistance(A, B): """ Distance (in meters) between two Locations. Uses the Haversine formula. See http://www.movable-type.co.uk/scripts/latlong.html for corresponding JavaScript implementation. """ # Earth's radius in meters R = 6371009 dLat = math.radians(B.lat - A.lat) dLon = math.radians(B.lng - A.lng) lat1 = math.radians(A.lat) lat2 = math.radians(B.lat) sLat = math.sin(dLat / 2.0) sLon = math.sin(dLon / 2.0) a = sLat * sLat + sLon * sLon * math.cos(lat1) * math.cos(lat2) c = 2.0 * math.atan2(math.sqrt(a), math.sqrt(1.0 - a)) return R * c
I use this distance formula to get a plot of accuracy versus travelling speed:
No clear correlation here; there are low-quality readings at both low and high
speeds. There are several possible explanations:
• Confirmation bias: I mistakenly extrapolated a small handful of
low-quality readings taken at high speeds to a general pattern.
perhaps the error radius doesn't mean what I think it does.
• Different location sources: Location accuracy is
relatively well-defined for GPS, but I'm not sure what happens when
cell towers or WiFi access points are incorporated into location fixes.
• Longer sampling interval: Maybe Google Latitude assumes that precise
location tracking is less important when driving.
To test this last hypothesis, I also plot sampling interval versus speed:
Nothing conclusive there.
## Conclusion #
The problem appears to be sampling frequency. To reduce battery usage, Google
Latitude polls about once every two minutes.
While it has some mechanism for
polling more often in periods of high activity, it's unclear how that works.
Reliance on fixes from cell towers and WiFi may be reducing location quality
in more remote areas. Testing this hypothesis is difficult: how do you
quantify remote? One possibility is to compute nearest-neighbor distance
against a database of cities. Another confounding factor is the
reliability of those accuracy values. Improving upon that would likely
involve manual labelling.
## Why Do This? #
Accuracy is not binary.
In Quantified Self applications, we use personal data to drive changes in our
lives.
We put a lot of trust in the accuracy and relevance of that data, and
we extend that trust to the tools and services that collect it.
We trust Fitbit to track our fitness.
We trust Zeo to improve our sleep.
We trust Lumosity to train our perception and attentiveness.
In giving so much trust to these tools, we sometimes forget that data are not
infallible.
Physics guarantees that there is no such thing as perfect data. All
data contain error.
As a system consisting of geosynchronous satellites that
travel at relativistically significant speeds and beam data
through our multilayered atmosphere to tiny chip radios sandwiched between
layers of dense circuitry, GPS is understandably error-prone.
When your chosen tools and services add noise on top of that, it's reasonable | 2021-03-07 04:55:12 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.21542005240917206, "perplexity": 8188.380914994426}, "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-2021-10/segments/1614178376144.64/warc/CC-MAIN-20210307044328-20210307074328-00059.warc.gz"} |
http://www.openproblemgarden.org/op/goldbach_conjecture | # Goldbach conjecture
Importance: Outstanding ✭✭✭✭
Author(s): Goldbach, Christian
Subject: Number Theory » Additive Number Theory
Posted by: Benschop on: November 20th, 2007
\begin{conjecture} Every even integer greater than 2 is the sum of two primes. \end{conjecture}
This famous conjecture is one of the oldest unsolved problems in mathematics. It arose originally out of a correspondence (1742) between Goldbach and Euler. See \Def[Wikipedia's Goldbach's conjecture]{goldbach's conjecture} for more.
### Weak conjecture now solved
Note that Harald Helfgott has proved the ternary version of Goldbach's conjecture, that every odd number greater than 7 is the sum of three odd primes. See http://arxiv.org/abs/1205.5252 (minor arc estimates) and http://arxiv.org/abs/1305.2897 (major arc estimates).
### Goldbach conjecture
Those interested in a suggestion for a proof via semigroup theory and carry extension, see: "Additive structure of Z(.) mod [\prod first k primes], with carry extension to prime pair sums". - http://home.iae.nl/users/benschop/ngb0203.pdf (10 pgs, submitted for publication)
### Goldbach conjecture
In addition to the above link to the paper with Goldbach-Conj. proof, the abstract and links to an intro-article are at http://home.iae.nl/users/benschop/ng-abstr.htm | 2018-01-17 07:26:27 | {"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": 0.8106099963188171, "perplexity": 3882.5249203093417}, "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-2018-05/segments/1516084886830.8/warc/CC-MAIN-20180117063030-20180117083030-00606.warc.gz"} |
http://www.koreascience.or.kr/article/JAKO201823955285150.page | # REMARKS ON ISOMORPHISMS OF TRANSFORMATION SEMIGROUPS RESTRICTED BY AN EQUIVALENCE RELATION
• Namnak, Chaiwat (Department of Mathematics Faculty of Science Naresuan University) ;
• Sawatraksa, Nares (Department of Mathematics Faculty of Science Naresuan University)
• Accepted : 2017.12.28
• Published : 2018.07.31
• 254 6
#### Abstract
Let T(X) be the full transformation semigroup on a set X and ${\sigma}$ be an equivalence relation on X. Denote $$E(X,{\sigma})=\{{\alpha}{\in}T(X):{\forall}x,\;y{\in}X,\;(x,y){\in}{\sigma}\;\text{implies}\;x{\alpha}=y{\alpha}\}.$$. Then $E(X,{\sigma})$ is a subsemigroup of T(X). In this paper, we characterize two semigroups of type $E(X,{\sigma})$ when they are isomorphic.
#### Keywords
transformation semigroup;isomorphism theorem;equivalence
#### Acknowledgement
Supported by : Naresuan University
#### References
1. P. Jitjankarn and T. Rungratgasame, A note on isomorphism theorems for semigroups of order-preserving transformations with restricted range, Int. J. Math. Math. Sci. 2015 (2015), Art. ID 187026, 6 pp.
2. Y. Kemprasit, W. Mora, and T. Rungratgasame, Isomorphism theorems for semigroups of order-preserving partial transformations, Int. J. Algebra 4 (2010), no. 17-20, 799-808.
3. S. Mendes-Goncalves and R. P. Sullivan, Semigroups of transformations restricted by an equivalence, Cent. Eur. J. Math. 8 (2010), no. 6, 1120-1131. https://doi.org/10.2478/s11533-010-0066-8
4. H. Pei, Regularity and Green's relations for semigroups of transformations that preserve an equivalence, Comm. Algebra 33 (2005), no. 1, 109-118. https://doi.org/10.1081/AGB-200040921
5. T. Saito, K. Aoki, and K. Kajitori, Remarks on isomorphisms of regressive transformation semigroups, Semigroup Forum 53 (1996), no. 1, 129-134. https://doi.org/10.1007/BF02574127
6. J. Sanwong and W. Sommanee, Regularity and Green's relations on a semigroup of trans- formations with restricted range, Int. J. Math. Math. Sci. 2008 (2008), Art. ID 794013, 11 pp.
7. A. Umar, Semigroups of order-decreasing transformations: the isomorphism theorem, Semigroup Forum 53 (1996), no. 2, 220-224. https://doi.org/10.1007/BF02574137 | 2019-12-06 22:43:05 | {"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": 0.8745846152305603, "perplexity": 2896.9102203788552}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-51/segments/1575540491491.18/warc/CC-MAIN-20191206222837-20191207010837-00138.warc.gz"} |
http://mca.ignougroup.com/2017/04/solved-complexity-of-hamiltonian-path.html | World's most popular travel blog for travel bloggers.
# [Solved]: Complexity of Hamiltonian path and clique problem
, ,
Problem Detail:
I came across this question. If we want to check if a graph contains both Hamiltonian path and clique. Would this problem be NPC.
I knew that clique contains a Hamiltonian path and both problems are NPC, but I am uncertain if something would be different if we check it in same time.
In the clique problem we are required to determine if there exists a clique of a certain size (given as input), so the observation that every clique contains a Hamiltonian path won't help much (a graph $G$ with $n$ vertices may contain cliques of size $<n$, but not have a Hamiltonian path).
$Clique=\left\{\langle G,k\rangle | \hspace{1mm} G \text{ contains a clique of size }\ge k\right\}$
$HPath=\left\{G | \hspace{1mm} G \text{ contains a Hamiltonian path}\right\}$
Since the instances to the problems are not of the same type, it isn't really interesting to talk about their intersection (which is empty), so what you probably have in mind is the set:
$L=\left\{\langle G,k\rangle | \hspace{1mm} G \text{ contains a clique of size }\ge k \land G \text{ contains a Hamiltonian path}\right\}$.
$L$ can be shown to be NP-complete using a simple reduction from $HPath$ (think about how to change the instance to the HPath problem, to an instance of $L$, without ruining the properties of the graph). | 2020-04-05 06:29:27 | {"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": 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": 0.5789064168930054, "perplexity": 308.93736401974473}, "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-2020-16/segments/1585370529375.49/warc/CC-MAIN-20200405053120-20200405083120-00497.warc.gz"} |
http://archives.gentoo.org/gentoo-dev/msg_14cbb5ee854d8f581888dc03e2bc9501.xml | Note: Due to technical difficulties, the Archives are currently not up to date. GMANE provides an alternative service for most mailing lists.
c.f. bug 424647
List Archive: gentoo-dev
On Fri, 23 Mar 2012 11:58:47 -0400 Mike Gilbert wrote: > > oasis_src_compile() { > > oasis_src_compile_no_doc > > if has doc \${IUSE} && use doc; then > > ocaml setup.ml -doc || die > > fi > > } > > This should probably call use_if_iuse from eutils.eclass, which > handles IUSE="[+-]doc". Actually, neither way works. The spec says: Global variables must only contain invariant values (see~\ref{sec:metadata-invariance}). If a global variable's value is invariant, it may have the value that would be generated at any given point in the build sequence. So you can't rely upon IUSE having the "merged" value in an eclass. -- Ciaran McCreesh | 2013-05-19 13:44:34 | {"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": 0.42311891913414, "perplexity": 8314.144507048732}, "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/1368697552127/warc/CC-MAIN-20130516094552-00064-ip-10-60-113-184.ec2.internal.warc.gz"} |
http://www.koreascience.or.kr/article/ArticleFullRecord.jsp?cn=E1BMAX_2012_v49n4_775 | A NOTE ON OSTROWSKI TYPE INEQUALITIES RELATED TO SOME s-CONVEX FUNCTIONS IN THE SECOND SENSE
Title & Authors
A NOTE ON OSTROWSKI TYPE INEQUALITIES RELATED TO SOME s-CONVEX FUNCTIONS IN THE SECOND SENSE
Liu, Zheng;
Abstract
Some errors in literatures are pointed out and corrected. A generalization of Ostrowski type inequalities for functions whose derivatives in absolute value are s-convex in the second sense is established. Special cases are discussed.
Keywords
Ostrowski type inequality;convex function;s-convex function;H$\small{\ddot{o}}$lder inequality;averaged midpoint-trapezoid inequality;
Language
English
Cited by
1.
INEQUALITIES FOR THE RIEMANN-STIELTJES INTEGRAL OF PRODUCT INTEGRATORS WITH APPLICATIONS,;
대한수학회지, 2014. vol.51. 4, pp.791-815
1.
Some perturbed Ostrowski type inequalities for functions of bounded variation, Asian-European Journal of Mathematics, 2015, 08, 04, 1550069
2.
Some Ostrowski Type Inequalities for Harmonicallys, m-Convex Functions in Second Sense, International Journal of Analysis, 2015, 2015, 1
3.
INEQUALITIES FOR THE RIEMANN-STIELTJES INTEGRAL OF PRODUCT INTEGRATORS WITH APPLICATIONS, Journal of the Korean Mathematical Society, 2014, 51, 4, 791
4.
Remarks on some inequalities for s-convex functions and applications, Journal of Inequalities and Applications, 2015, 2015, 1
5.
On new general integral inequalities for s-convex functions, Applied Mathematics and Computation, 2014, 246, 306
References
1.
M. Alomari, M. Darus, S. S. Dragomir, and P. Cerone, Ostrowski type inequalities for functions whose derivatives are s-convex in the second sense, Appl. Math. Lett. 23 (2010), no. 1, 1071-1076.
2.
S. S. Dragomir and R. P. Agarwal, Two inequalities for differentiable mappings and applications to special means of real numbers and to trapezoidal formula, Appl. Math. Lett. 11 (1998), no. 5, 91-95.
3.
S. S. Dragomir, P. Cerone, and J. Roumeliotis, A new generalization of Ostrowski integral inequality for mappings whose derivatives are bounded and applications in numerical integration and for special means, Appl. Math. Lett. 13 (2000), no. 1, 19-25.
4.
H. Hudzik and L. Maligranda, Some remarks on s-convex functions, Aequationes Math. 48 (1994), no. 1, 100-111.
5.
U. S. Kirmaci, M. K. Bakula, M. E. Ozdemir, and J. Pecaric, Hadamard-type inequalities for s-convex functions, Appl. Math. Comput. 193 (2007), no. 1, 26-35.
6.
C. E. M. Pearce and J. Pecaric, Inequalities for differentiable mappings with applications to special means and quadrature formulae, Appl. Math. Lett. 13 (2000), no. 2, 51-55. | 2018-08-19 10:04:31 | {"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": 1, "/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": 0.6203540563583374, "perplexity": 1724.2495666696884}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-34/segments/1534221215075.58/warc/CC-MAIN-20180819090604-20180819110604-00104.warc.gz"} |
https://stats.stackexchange.com/questions/497910/why-is-the-delta-method-defined-the-way-it-is/497912 | # Why is the delta method defined the way it is?
The delta method begins with the assumption of $$\sqrt{n} \left[X_n - \theta\right] \stackrel{D}{\to} \mathcal{N}(0, \sigma^2)$$. Why is this? Wouldn't it make more sense to start in the more familiar arrangement of $$X_n \stackrel{D}{\to} \mathcal{N}\left(\theta, \frac{\sigma^2}{n} \right)$$ ? Or even better, replace $$\theta$$ with $$\mu$$. Now it's a normal distribution in terms of mean and standard error.
I have the same question for the conclusion. Wouldn't the conclusion of $$g(X_n) \stackrel{D}{\to} \mathcal{N}\left(g(\theta),\frac{\sigma^2[g'(\theta)]^2}{n}\right)$$ make more sense to more readers for the same reason?
Is my math wrong? Is there some history of the delta method that dictates the original form? Thanks!
There are a few misconceptions to clear up first.
First, to clear up the biggest misconception. The final expression you gave has the limiting distribution also varying with $$n$$; this cannot be! Your limiting distribution must be one that is independent of $$n$$. Of course, you can say that $$g(x_n)$$ is approximately distributed as the normal distribution you provide. Here, no asymptotic statements are being made, so we don't have the undesirable situation where our limit varies with $$n$$ as well. In fact, one could say when we're talking about approximate distributions, we want our distribution on the RHS to depend on $$n$$, but, that's besides the point.
Secondly, the (univariate) delta method begins with that assumption you give because we want to know the limiting distribution of
$$\frac{\sqrt{n}(g(X_n) - g(\theta))}{\sigma^2}$$.
If we assume that the limiting distribution of $$\sqrt{n}(X_n - \theta)$$ is $$\mathcal{N}(0, \sigma^2)$$, then the limiting distribution of the desired expression is normal, just with different parameters. These parameters are given via a first-order Taylor expansion of $$g(X_n)$$ about $$\theta$$, which is why we have the added assumption that $$g'$$ is continuous at $$\theta$$.
Notice, though that $$\theta$$ is any parameter, and $$X_n$$ is a sequence of estimators for that parameter. $$\theta$$ need not be any mean, say.
• Yes, $\mu = E(X)$ was my assumption. Nov 25, 2020 at 0:30
• So, in short, we don't write it that way, because we don't want to create a normal distribution (approximate, asymptotic, or otherwise) that is dependent on sample size? Nov 25, 2020 at 0:33
• Almost! If the normal distribution is the limit of a sequence of random variables converging in distribution, it needs to be fixed with respect to the variable that is "going to infinity". You can certainly have an approximate distribution that depends on sample size, though. Nov 25, 2020 at 1:40
• Think in terms of sequences of real numbers. I can have a sequence of real numbers $a_n$ that converges to a fixed real number $a$, but the notion of a sequence of real numbers $a_n$ "converging to" a sequence of real numbers $b_n$ is nonsense! You can certainly say that $a_n$ "for large n" is approximately $b_n$, though. Analogous statements are made in the previous comment. Nov 25, 2020 at 1:45 | 2022-07-06 09:33:17 | {"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": 20, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8669065237045288, "perplexity": 169.31368706432443}, "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-2022-27/segments/1656104669950.91/warc/CC-MAIN-20220706090857-20220706120857-00669.warc.gz"} |
http://www.mscs.dal.ca/~selinger/quipper/doc/Algorithms-TF-Definitions.html | The Quipper System
Algorithms.TF.Definitions
Description
This module provides global definitions for the Triangle Finding Algorithm.
Synopsis
# Qram abstraction
data Qram Source #
A data structure to hold a Qram implementation. This provides operations for fetching and storing quantum data from a quantum array, addressed by a quantum integer. One implementation is given by algorithms a8_FetchT, a9_StoreT and a10_FetchStoreT.
Constructors
Qram Fieldsqram_fetch :: forall qa. QData qa => QDInt -> IntMap qa -> qa -> Circ (QDInt, IntMap qa, qa) qram_store :: forall qa. QData qa => QDInt -> IntMap qa -> qa -> Circ (QDInt, IntMap qa, qa) qram_swap :: forall qa. QData qa => QDInt -> IntMap qa -> qa -> Circ (QDInt, IntMap qa, qa)
# Types for the Triangle Finding Algorithm
type CNode = [Bit] Source #
A node of the graph (classical circuit type).
type QNode = [Qubit] Source #
A node of the graph (quantum circuit type).
type QWTFP_spec = (Int, Int, QNode -> QNode -> Qubit -> Circ Qubit, Qram) Source #
The type of problem specifications for the Triangle Finding Problem. A problem specification consists of:
• an integer n which determines the number N=2n of nodes of the graph,
• an integer r which determines the size R=2r of tuples in the Hamming graph,
• a function edge_oracle which inputs two graph nodes and a qubit and flips the qubit if the nodes are connected by an edge and
• additional options, for selecting, e.g., which qRAM implementation should be used.
# TF integers
## Types
We define a QData family of integer datatypes (QIntTF, CIntTF, IntTF). These are similar to (QDInt, CInt, IntM), except that the integers are considered to be mod 2m-1 instead of 2m.
In general, functions on these types should be able to handle both 00…00 and 11…11, and should treat them equally, essentially regarding IntTF, CIntTF, and the computational basis of QIntTF as formal quotients. Some operations are not perfect. One should keep in mind, for example, that specifying a control on a QIntTF of the form q .==. 0 will compare the bitwise representation to 0, and not the logical quotient.
data XIntTF x Source #
All three types QIntTF, CIntTF, and IntTF are special cases of a more general type XIntTF x, parameterized by a type x of bits. It is an abstract type, and details of its implementation is not exposed to user-level code.
Constructors
XIntTF (XInt x)
Instances
# Methods(==) :: IntTF -> IntTF -> Bool #(/=) :: IntTF -> IntTF -> Bool # # MethodsshowsPrec :: Int -> IntTF -> ShowS #show :: IntTF -> String #showList :: [IntTF] -> ShowS # Show x => Show (XIntTF x) # MethodsshowsPrec :: Int -> XIntTF x -> ShowS #show :: XIntTF x -> String #showList :: [XIntTF x] -> ShowS # QCLeaf x => QCData (XIntTF x) # Methodsqcdata_mapM :: Monad m => XIntTF x -> (q -> m q') -> (c -> m c') -> QCType q c (XIntTF x) -> m (QCType q' c' (XIntTF x)) Source #qcdata_zip :: XIntTF x -> q -> c -> q' -> c' -> QCType q c (XIntTF x) -> QCType q' c' (XIntTF x) -> ErrMsg -> QCType (q, q') (c, c') (XIntTF x) Source #qcdata_promote :: BType (XIntTF x) -> XIntTF x -> ErrMsg -> BType (XIntTF x) Source # QCLeaf x => Labelable (XIntTF x) String # Methodslabel_rec :: XIntTF x -> String -> LabelMonad () Source # type QTypeB IntTF # type QTypeB IntTF = QIntTF type QCType x y (XIntTF z) # type QCType x y (XIntTF z) = XIntTF (QCType x y z)
The type of fixed-length m-qubit quantum integers, regarded modulo 2m-1.
The type of fixed-length m-bit classical integers, regarded modulo 2m-1.
The type of fixed-length m-bit integer parameters, regarded modulo 2m-1. A value of type IntTF may have indeterminate length, similarly to IntM.
## Operations for IntTF
Convert an IntTF of length m to an Integer in the range {0, …, 2m-2}. If the IntTF has indeterminate length, return the original Integer.
Convert an Integer to an IntTF of indeterminate length.
Convert an Integer to an IntTF of length m.
Return the length of an IntTF, or Nothing if indeterminate.
Set the length of an IntTF to m ≥ 0. This operation is only legal if the input (a) has indeterminate length or (b) has determinate length already equal to m. In particular, it cannot be used to change the length from anything other than from indeterminate to determinate.
If both arguments already have determinate lengths, and they do not coincide, throw an error. The String argument is used as an error message in that case.
Try to set the length of an IntTF to that of another XIntTF value (which could be a QIntTF, a CIntTF, or another IntTF). This will fail with an error if both numbers already have determinate lengths that don't coincide. In this case, the string argument is used as an error message. The promotion is done modulo 2m-1.
Convert an IntTF to human readable form. We show the bit value, i.e., 0 and 2m-1 are shown as different values.
## Operations for QIntTF
Convert a QIntTF to a list of qubits. The conversion is little-headian, i.e., the head of the list holds the least significant digit.
Convert a list of qubits to a QIntTF. The conversion is little-headian, i.e., the head of the list holds the least significant digit.
Return a piece of shape data to represent an m-qubit QIntTF. Please note that the data can only be used as shape; it will be undefined at the leaves.
## Auxiliary functions
The low-level isomorphism from XInt x to XIntTF x. Note that "isomorphism" is between the underlying raw types, and does not respect the arithmetic operations.
The low-level isomorphism from XIntTF x to XInt x. Note that "isomorphism" is between the underlying raw types, and does not respect the arithmetic operations.
Like xint_of_xinttf, but first try to promote the length of the IntTF to that of the given XIntTF.
# Miscellaneous circuit-building functions
phaseFlipIf :: ControlSource ctrl => ctrl -> Circ () Source #
Controlled phase flip of -1.
phaseFlipUnless :: ControlSource ctrl => ctrl -> Circ () Source #
Variant of phaseFlip that performs a phase flip unless all controls are in the given state.
qor :: Qubit -> [(Qubit, Bool)] -> Circ Qubit Source #
qor q c: Applies "not" to q, if any of the control qubits in c is in specified state.
# Arithmetic functions
Increment a standard QDInt (i.e. big-endian, mod 2).
Decrement a standard QDInt (i.e. big-endian, mod 2).
increment_big :: [Qubit] -> Circ [Qubit] Source #
Increment a bit-string, considered as a big-endian integer mod 2.
decrement_big :: [Qubit] -> Circ [Qubit] Source #
Decrement a bit-string, considered as a big-endian integer mod 2.
increment_little :: [Qubit] -> Circ [Qubit] Source #
Increment a bit-string, considered as a little-endian integer mod 2.
decrement_little :: [Qubit] -> Circ [Qubit] Source #
Decrement a bit-string, considered as a little-endian integer mod 2.
choose :: Integral a => a -> a -> a Source #
The standard “combinations” function “n choose k”.
# IntMaps as QData
addKeys :: IntMap a -> IntMap (Key, a) Source #
Replace an IntMap f with the IntMap mapping each key k to (k,f(k)). An auxiliary function for defining mapWithKeyM, etc.
mapWithKeyM :: Monad m => (Key -> a -> m b) -> IntMap a -> m (IntMap b) Source #
Analogous to mapM, but allows the function to use the key. Particularly useful for mapping in parallel over two (or more) IntMaps assumed to have the same domain.
mapWithKeyM_ :: Monad m => (Key -> a -> m b) -> IntMap a -> m () Source #
Analogous to mapM_, but allows the function to use the key.
intMap_replicate :: Int -> a -> IntMap a Source #
Analogous to replicate on lists.
(!) :: IntMap a -> Key -> a infixl 9 Source #
Convenient syntax for accessing elements of an IntMap. Left associative, and binds very strongly, like '(!!)'.
# Orphan instances
QCData a => QCData (IntMap a) # Methodsqcdata_mapM :: Monad m => IntMap a -> (q -> m q') -> (c -> m c') -> QCType q c (IntMap a) -> m (QCType q' c' (IntMap a)) Source #qcdata_zip :: IntMap a -> q -> c -> q' -> c' -> QCType q c (IntMap a) -> QCType q' c' (IntMap a) -> ErrMsg -> QCType (q, q') (c, c') (IntMap a) Source #qcdata_promote :: BType (IntMap a) -> IntMap a -> ErrMsg -> BType (IntMap a) Source # # Methodslabel_rec :: IntMap a -> String -> LabelMonad () Source # Labelable a s => Labelable (IntMap a) (IntMap s) # Methodslabel_rec :: IntMap a -> IntMap s -> LabelMonad () Source # | 2017-11-18 19:34:43 | {"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": 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": 0.3915156424045563, "perplexity": 13997.199408595869}, "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/1510934805023.14/warc/CC-MAIN-20171118190229-20171118210229-00221.warc.gz"} |
https://www.semanticscholar.org/paper/Uniformization-of-semistable-bundles-on-elliptic-Li-Nadler/e28b21828cb6de3813d734491f16be655b3fa0e8 | # Uniformization of semistable bundles on elliptic curves
@article{Li2015UniformizationOS,
title={Uniformization of semistable bundles on elliptic curves},
journal={arXiv: Representation Theory},
year={2015}
}
• Published 29 October 2015
• Mathematics
• arXiv: Representation Theory
7 Citations
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## References
SHOWING 1-10 OF 51 REFERENCES
• Mathematics
• 1997
Let $G$ be a simple and simply connected complex Lie group. We discuss the moduli space of holomorphic semistable principal $G$ bundles over an elliptic curve $E$. In particular we give a new proof
• Mathematics
• 2005
This paper is a sequel to math.AG/0505148, where the Hall algebra U^+_E of the category of coherent sheaves on an elliptic curve E defined over a finite field was explicitly described, and shown to
Let g be a complex reductive Lie algebra with Cartan algebra h. Hotta and Kashiwara defined a holonomic D-module M, on g x h, called Harish-Chandra module. We relate gr(M), an associated graded
• Mathematics
• 2009
We apply the ideas of derived algebraic geometry and topological field theory to the representation theory of reductive groups. Our focus is the Hecke category of Borel-equivariant D-modules on the
The article is a contribution to the local theory of geometric Langlands duality. The main result is a categorification of the isomorphism between the (extended) affine Hecke algebra associated to a
• Mathematics
Compositio Mathematica
• 2010
Abstract We exhibit a strong link between the Hall algebra HX of an elliptic curve X defined over a finite field 𝔽l (or, more precisely, its spherical subalgebra U+X) and Cherednik’s double affine
• Mathematics
• 2009
In this paper we compute the convolution algebra in the equivariant K-theory of the Hilbert scheme of A^2. We show that it is isomorphic to the elliptic Hall algebra, and hence to the spherical DAHA
Introduction THE primary purpose of this paper is the study of algebraic vector bundles over an elliptic curve (defined over an algebraically closed field k). The interest of the elliptic curve lies
• Mathematics
• 2000
This paper continues the study of holomorphic semistable principal G-bundles over an elliptic curve. In this paper, the moduli space of all such bundles is constructed by considering deformations of | 2023-03-26 21:15:59 | {"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": 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": 0.7618893384933472, "perplexity": 736.3207237848042}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-14/segments/1679296946535.82/warc/CC-MAIN-20230326204136-20230326234136-00230.warc.gz"} |
http://mathhelpforum.com/advanced-algebra/28344-solved-abstract-algebra-integers-modulo-n.html | # Math Help - [SOLVED] Abstract Algebra: Integers Modulo n
1. ## [SOLVED] Abstract Algebra: Integers Modulo n
Güten Tag!
Here's another one. This is one of those questions where the answer seems so obvios to me that I don't know how they expect me to do it.
Problem:
Prove that $[a] \odot ([b] \oplus [c]) = ([a] \odot [b]) \oplus ([a] \odot [c])$ for all $[a],[b],[c] \in \mathbb{Z}_n$
Things that may come in handy:
Nothing really. I assume that anyone who would be able to help me on this will already know what all those symbols I used mean. We really don't need to know anything but their defintions I think.
What I've Tried:
Stared at the problem for many minutes wondering..."Uh, isn't this obvios? $\odot$ and $\oplus$ are associative and commutative on $\mathbb{Z}_n$. Does the result not immediately follow from that? Maybe they are asking me to prove: $[a] \odot [b + c] = [ab] \oplus [ac]$, in which case, the solution is again obvios and trivial! (and when i say that, you know it's true. I'm only a Nascent Mathematician, after all)"
Can anyone help me to rise above my inclinations and be rigorous with this problem?
Thanks guys and gals (of course, for future reference, unless otherwise stated, gals refers to JaneBennet -- are there anymore female mathematicians otu there?)
2. Originally Posted by Jhevon
Prove that $[a] \odot ([b] \oplus [c]) = ([a] \odot [b]) \oplus ([a] \odot [c])$ for all $[a],[b],[c] \in \mathbb{Z}_n$
I assume by $\odot$ you mean multiplication by congruences classes. And $\oplus$ you mean addition of congruences classes. Then use the fact that $[a] + [b] = [a+b]$ is well-defined operations. Thus, $[a]([b]+[c]) = [a]([b+c]) = [ab+ac] = [ab]+[ac]=[a][b]+[a][c]$.
3. Originally Posted by ThePerfectHacker
I assume by $\odot$ you mean multiplication by congruences classes. And $\oplus$ you mean addition of congruences classes. Then use the fact that $[a] + [b] = [a+b]$ is well-defined operations. Thus, $[a]([b]+[c]) = [a]([b+c]) = [ab+ac] = [ab]+[ac]=[a][b]+[a][c]$.
That is exactly what i was thinking of doing. but it seemed too easy. here is how i would write it out (because my professor is the type that wants you to give a reason for anything you do--which is not necessarily bad, i guess).
we wish to show that $[a] \odot ([b] \oplus [c]) = ([a] \odot [b]) \oplus ([a] \odot [c])
$
Note that: $[a] \odot ([b] \oplus [c]) = [a] \odot [b + c]$ ..........................by the definition of $\oplus$
............................................ $= [a(b + c)]$ ..............................by the definition of $\odot$
............................................ $= [a \cdot b + a \cdot c]$ ........................ by the the distributive property of real numbers
............................................ $= [a \cdot b] \oplus [a \cdot c]$ .......................by the definition of $\oplus$
............................................ $= ([a] \odot [b]) \oplus ([a] \odot [c])$ ........by the definition of $\odot$
as was to be shown | 2014-10-25 21:34:09 | {"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": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 26, "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": 0.8962822556495667, "perplexity": 687.8991373659246}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-42/segments/1414119650516.39/warc/CC-MAIN-20141024030050-00182-ip-10-16-133-185.ec2.internal.warc.gz"} |
http://taoofmac.com/space/apps/Textpander | # Textpander
Textpander is an utility that allows you to build pre-defined snippets of text and graphics (a step beyond the usual sort of abbreviation managers): | 2016-12-08 11:50:37 | {"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": 0.9884798526763916, "perplexity": 2543.260991863843}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-50/segments/1480698542588.29/warc/CC-MAIN-20161202170902-00256-ip-10-31-129-80.ec2.internal.warc.gz"} |
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### Contents
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Metric bolts. metric bolts are sized according to diameter, distance or spacing (in millimeters) between threads and length. For example a size of 12×1.75×30 translates to a bolt that is 12mm in diameter, has a distance of 1.75 between the threads and is 30mm long.. Metric bolt sizes can also be listed as M6x1x25.
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https://www.physicsforums.com/threads/how-to-integrate-int-x-2-4-1-2-dx-is-it-a-substitution.79670/ | # How to integrate int (x^2 + 4)^(-1/2) dx Is it a substitution?
1. Jun 20, 2005
### gordda
I need help antidiffing this equation:
(x^2+4)^(-1/2)
i have tried subbing u=x^2+4 i have tried subbing u= (x^2+4)^(-1/2).
i have tried making x the subject. even tried to use partial fraction, with no avail, because i could not figure out how to use partially factorize it. | 2017-10-18 06:15:01 | {"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": 0.8033755421638489, "perplexity": 2854.735587245976}, "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-2017-43/segments/1508187822747.22/warc/CC-MAIN-20171018051631-20171018071631-00478.warc.gz"} |
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# The probability of a man hitting a bulls eye in one fire is
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07 Aug 2012, 06:49
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The probability of a man hitting a bulls eye in one fire is 1/4. What is the least number of times that the man should fire in order to take the probability of hitting the bulls eye at least once to more than 50%.
A. 5
B. 7
C. 3
D. 12
E. 6
[Reveal] Spoiler: OA
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Last edited by SOURH7WK on 07 Aug 2012, 07:10, edited 2 times in total.
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07 Aug 2012, 09:13
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SOURH7WK wrote:
The probability of a man hitting a bulls eye in one fire is 1/4. What is the least number of times that the man should fire in order to take the probability of hitting the bulls eye at least once to more than 50%.
A. 5
B. 7
C. 3
D. 12
E. 6
It is a Typical Probability Question
IN Probability we have a rule ( which is logic actually) that
P(>1) = 1-P(0)
That is Probability of hitting the bulls eye at least once out of N trials is 1- probability of missing it all the times. & P of missing it is 3/4
That means$$1-(\frac{3}{4})^n$$ > $$\frac{1}{2}$$
Solve it , n > 3 satisfies.
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SOURH7WK wrote:
The probability of a man hitting a bulls eye in one fire is 1/4. What is the least number of times that the man should fire in order to take the probability of hitting the bulls eye at least once to more than 50%.
A. 5
B. 7
C. 3
D. 12
E. 6
OA will follow after few discussions
The probability that the man hits the bulls eye at least once in 2 tries is 1-(3/4)^2=7/16, so less than 1/2 (the probability of hitting the bulls eye at least once is 1 minus the probability of the opposite event, so the probability that the men misses in all 2 tries).
The probability that the man hits the bulls eye at least once in 3 tries is 1-(3/4)^3=37/64, so more than 1/2.
Answer: C.
P.S. You must indicate OA, rule #7: rules-for-posting-please-read-this-before-posting-133935.html
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Re: The probability of a man hitting a bulls eye in one fire is [#permalink]
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08 Aug 2012, 10:20
very helpful Bunuel/mandyr, need to revise the "at least" rule and this helps, thank you both
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Re: The probability of a man hitting a bulls eye in one fire is [#permalink]
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14 Jul 2013, 00:10
Expert's post
Bumping for review and further discussion.
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# The probability of a man hitting a bulls eye in one fire is
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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®. | 2016-04-30 08:14:18 | {"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": 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": 0.5772006511688232, "perplexity": 5275.129613135117}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-18/segments/1461860111620.85/warc/CC-MAIN-20160428161511-00110-ip-10-239-7-51.ec2.internal.warc.gz"} |
https://www.physicsforums.com/threads/an-elementary-observation-regarding-definability.974001/ | # I An Elementary Observation Regarding Definability
#### SSequence
Few months ago there was a discussion in the topic(Complex numbers in QM) regarding the notion of definable real numbers. The discussion was in the first 3 or 4 pages of that topic.
Anyway, I thought of a reasonably interesting observation about it. Since the main theme of that topic seems quite different, I have posted this as a separate thread. I think I have gotten some idea of post#80. Below is one way I thought about it:
Consider, in the language of set-theory, we consider all well-formed-formulas that have one free variable. Since the formulas are easily counted we could use the notation $\phi_n(x)$ (where $n \in \mathbb{N}$) to denote $n$-th formula. $x$ is supposed to be the free variable in the formula. Now if we consider the set of reals $\mathbb{R}$ (say, as subset of naturals), then we say that a given formula $\phi_i(x)$ defines the real number $r$ iff $r$ is the only real number satisfying that formula.
Now suppose that we have a well-ordering of reals available to us and say its order-type is $\beta \geq \omega_1$. Now, using that well-ordering, for every formula $\phi_i(x)$ we can test whether it is satisfied by only one real or not. So, for example, we first test whether $\phi_0(x)$ is satisfied by a unique real or not ........ and then $\phi_1(x)$, $\phi_2(x)$, $\phi_3(x)$, $\phi_4(x)$ etc. This way we can pick all those reals which are definable. Then we diagonalise through these definable real numbers giving us a new real number $R$ which is different from every definable real.
===================
It is quite informal, but it seems plausible to me that if there is a formula (in language of set-theory) for the well-ordering of $\mathbb{R}$ then it can be shown that there must also be a formula which is only satisfied by the real number $R$ above (and no other real number). But that would mean that $R$ is definable (which it isn't supposed to be since it was formed from diagonalisation of definable reals). Hence, it seems to me, that we also have to assume that there can't be any formula (in set-theory language) describing the well-order of $\mathbb{R}$?
===================
Does any of this sound remotely reasonable? There is one other thing, which I don't quite get. In this question it is mentioned (in the accepted answer):
there is a model of ZFC in which every real number and indeed every set-theoretic object is definable.
Does it go against the first paragraph written in post#80 or am I misinterpreting something:
As far as I know, nobody tries to do mathematics using only definable objects, because the usual mathematical axioms don't hold when restricted to definable objects. However, the set of reals is certainly definable.
Sorry I didn't want to link the above question (since it seems far too advanced for this particular discussion, where I am just trying to get a sense of things). Also, quoting a specific sentence is kind of nitpicking, but nevertheless, the two statements felt at odds to me, so it seemed reasonable to inquire.
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#### A. Neumaier
The diagonalizatio you refer to happens on the objectlevel, wile the construction of the reals happens on the metalevel so the model of the reals is metacountable but intrinsically uncountable. There are not enough intrinsic maps to make the metbijection needed.
As to your other question, if a set is definable it does not mean that all its elements are individually definable.
#### SSequence
Below, denoting the language of (first-order) set theory as $L_S$.
As to your other question, if a set is definable it does not mean that all its elements are individually definable.
I assume that this is the kind of thing that's happening with $\mathbb{R}$. In general, knowing a diagonalisation through reals definable in $L_S$ doesn't mean that we can write down a formula for the diagonalised real [uniquely satisfied by that real alone in the set $\mathbb{R}$] in $L_S$. Right? At least that's what I gathered from discussion/posts in previous thread.
======
Also, to re-iterate my question in the first half of OP: My question was that if we assume that there is a formula for a well-order of $\mathbb{R}$ in $L_S$, it seems plausible that we can conclude that there is also a (unique) formula for a real $R$ which is not definable (in $L_S$)? I am not familiar enough with logical constructions/machinery to know for sure.
Hence, by argument in prev. paragraph, can we conclude that there is no formula for a well-order of $\mathbb{R}$ in $L_S$? And furthermore is the line of reasoning in previous paragraph correct?
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#### A. Neumaier
Hence, by argument in prev. paragraph, can we conclude that there is no formula for a well-order of $\mathbb{R}$ in $L_S$? And furthermore is the line of reasoning in previous paragraph correct?
In no ZFC model of the reals, there is a formula for well-ordering.
#### SSequence
I see. So it seems reasonable to assume that when people talk about formula for well-ordering of $\mathbb{R}$, they are talking about a language that is different (presumably richer?) from $L_S$ in some way.
#### A. Neumaier
I see. So it seems reasonable to assume that when people talk about formula for well-ordering of $\mathbb{R}$, they are talking about a language that is different (presumably richer?) from $L_S$ in some way.
In general, existence is nonconstructive, unlike formulas.
"An Elementary Observation Regarding Definability"
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• Solo and co-op problem solving | 2019-07-19 22:59:07 | {"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": 0.9096308350563049, "perplexity": 379.35581887139466}, "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/1563195526386.37/warc/CC-MAIN-20190719223744-20190720005744-00402.warc.gz"} |
http://www.gradesaver.com/the-open-window/q-and-a/bernie--why-do-you-bound--313564 | # Bernie , Why do you bound ?
In the line "Bernie , Why do you bound ?" , Is Bernie a singer or Mrs. Sappleton's real name ?
##### Answers 1
Does this have anything to do with The Open Window? | 2017-06-23 09:02:02 | {"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": 0.9062768816947937, "perplexity": 5782.159230429107}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "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-2017-26/segments/1498128320040.36/warc/CC-MAIN-20170623082050-20170623102050-00602.warc.gz"} |
https://en.wikipedia.org/wiki/Shamir%27s_Secret_Sharing | # Shamir's secret sharing
(Redirected from Shamir's Secret Sharing)
Shamir's secret sharing (SSS) is an efficient secret sharing algorithm for distributing private information (the "secret") in such a way that a single member of a group cannot derive any information about the secret without a quorum of the group acting in agreement. To achieve this, the secret is converted into parts (the "shares") from which the secret can be reassembled only when a sufficient number of shares are combined. SSS has the property of information-theoretic security, meaning an adversary without enough shares cannot reconstruct the secret even with infinite time and computing capacity.
Shamir's secret sharing finds use in some applications to share the access keys to a master secret.
## High-level explanation
SSS is used to secure a secret in a distributed form, most often to secure encryption keys. The secret is split into multiple shares, which individually do not give any information about the secret.
To reconstruct a secret secured by SSS, a number of shares is needed, called the threshold. No information about the secret can be gained from any number of shares below than the threshold (a property called perfect secrecy). In this sense, SSS is a generalisation of the one-time pad (which can be viewed as SSS with a two-share threshold and two shares in total).
### Application example
A company needs to secure their vault. If a single person knows the code to the vault, the code might be lost or unavailable when the vault needs to be opened. If there are several people who know the code, they may not trust each other to always act honestly.
SSS can be used in this situation to generate shares of the vault's code which are distributed to authorized individuals in the company. The minimum threshold and number of shares given to each individual can be selected such that the vault is accessible only by (groups of) authorized individuals. If fewer shares than the threshold are presented, the vault cannot be opened.
By accident or as an act of opposition, some individuals might present incorrect information for their shares. If the total of correct shares fails to meet the minimum threshold, the vault remains locked.
### Use cases
Shamir's secret sharing is, for example, used to
## Properties and weaknesses
SSS has useful properties, but also weaknesses[4] that mean that it is unsuited to some uses.
Useful properties include:
1. Secure: The scheme has information-theoretic security.
2. Minimal: The size of each piece does not exceed the size of the original data.
3. Extensible: For any given threshold, shares can be dynamically added or deleted without affecting existing shares
4. Dynamic: Security can be enhanced without changing the secret, but by changing the polynomial occasionally (keeping the same free term) and constructing a new share for each of the participants.
5. Flexible: In organizations where hierarchy is important, each participant can be issued different numbers of shares according to their importance inside the organization. For instance, with a threshold of 3, the president could unlock the safe alone if given three shares, while three secretaries with one share each must combine their shares to unlock the safe.
Weaknesses include:
1. No verifiable secret sharing: During the share reassembly process, SSS does not provide a way to verify the correctness of each share being used. Verifiable secret sharing aims to verify that shareholders are honest and not submitting fake shares.
2. Single point of failure: The secret must exist in one place when it is split into shares, and again in one place when it is reassembled. These are attack points, and other schemes including multisignature eliminate at least one of these single points of failure.
## History
Adi Shamir first formulated the scheme in 1979.[5]
## Mathematical principle
The scheme exploits the Lagrange interpolation theorem, specifically that ${\displaystyle k}$ points on the polynomial uniquely determines a polynomial of degree less than or equal to ${\displaystyle k-1}$. For instance, 2 points are sufficient to define a line, 3 points are sufficient to define a parabola, 4 points to define a cubic curve and so forth.
## Mathematical formulation
Shamir's secret sharing is an ideal and perfect ${\displaystyle \left(k,n\right)}$-threshold scheme based on polynomial interpolation over finite fields. In such a scheme, the aim is to divide a secret ${\displaystyle S}$ (for example, the combination to a safe) into ${\displaystyle n}$ pieces of data ${\displaystyle S_{1},\ldots ,S_{n}}$ (known as shares) in such a way that:
1. Knowledge of any ${\displaystyle k}$ or more shares ${\displaystyle S_{i}}$ makes ${\displaystyle S}$ computable. That is, the entire secret ${\displaystyle S}$ can be reconstructed from any combination of ${\displaystyle k}$ shares.
2. Knowledge of any ${\displaystyle k-1}$ or fewer shares ${\displaystyle S_{i}}$ leaves ${\displaystyle S}$ completely undetermined, in the sense that the possible values for ${\displaystyle S}$ remain as likely with knowledge of up to ${\displaystyle k-1}$ shares as with knowledge of ${\displaystyle 0}$ shares. The secret ${\displaystyle S}$ cannot be reconstructed with fewer than ${\displaystyle k}$ shares.
If ${\displaystyle n=k}$, then all of the shares are needed to reconstruct the secret ${\displaystyle S}$.
One can draw an infinite number of polynomials of degree 2 through 2 points. 3 points are required to uniquely determine a polynomial of degree 2. This image is for illustration purposes only — Shamir's scheme uses polynomials over a finite field, which are not easy to represent in a 2-dimensional plane.
Assume that the secret ${\displaystyle S}$ can be represented as an element ${\displaystyle a_{0}}$ of a finite field ${\displaystyle GF(q)}$ (where ${\displaystyle q}$ is greater than the number of shares being generated). Randomly choose ${\displaystyle k-1}$ elements, ${\displaystyle a_{1},\cdots ,a_{k-1}}$, from ${\displaystyle GF(q)}$ and construct the polynomial ${\displaystyle f\left(x\right)=a_{0}+a_{1}x+a_{2}x^{2}+a_{3}x^{3}+\cdots +a_{k-1}x^{k-1}}$. Compute any ${\displaystyle n}$ points out on the curve, for instance set ${\displaystyle i=1,\ldots ,n}$ to find points ${\displaystyle \left(i,f\left(i\right)\right)}$. Every participant is given a point (a non-zero input to the polynomial, and the corresponding output).[6] Given any subset of ${\displaystyle k}$ of these pairs, ${\displaystyle a_{0}}$ can be obtained using interpolation, with one possible formula for doing so being ${\displaystyle a_{0}=f(0)=\sum _{j=0}^{k-1}y_{j}\prod _{\begin{smallmatrix}m\,=\,0\\m\,\neq \,j\end{smallmatrix}}^{k-1}{\frac {x_{m}}{x_{m}-x_{j}}}}$, where the list of points on the polynomial is given as ${\displaystyle k}$ pairs of the form ${\displaystyle (x_{i},y_{i})}$. Note that ${\displaystyle f(0)}$ is equal to the first coefficient of polynomial ${\displaystyle f(x)}$.
## Example calculation
The following example illustrates the basic idea. Note, however, that calculations in the example are done using integer arithmetic rather than using finite field arithmetic to make the idea easier to understand. Therefore the example below does not provide perfect secrecy and is not a proper example of Shamir's scheme. The next example will explain the problem.
### Preparation
Suppose that the secret to be shared is 1234 ${\displaystyle (S=1234)\,\!}$.
In this example, the secret will be split into 6 shares ${\displaystyle (n=6)\,\!}$, where any subset of 3 shares ${\displaystyle (k=3)\,\!}$ is sufficient to reconstruct the secret. ${\displaystyle k-1=2}$ numbers are taken at random. Let them be 166 and 94.
This yields coefficients ${\displaystyle (a_{0}=1234;a_{1}=166;a_{2}=94),\,\!}$ where ${\displaystyle a_{0}}$ is the secret
The polynomial to produce secret shares (points) is therefore:
${\displaystyle f(x)=1234+166x+94x^{2}\,\!}$
Six points ${\displaystyle D_{x-1}=(x,f(x))}$ from the polynomial are constructed as:
${\displaystyle D_{0}=(1,1494);D_{1}=(2,1942);D_{2}=(3,2578);D_{3}=(4,3402);D_{4}=(5,4414);D_{5}=(6,5614)\,\!}$
Each participant in the scheme receives a different point (a pair of ${\displaystyle x\,\!}$ and ${\displaystyle f(x)\,\!}$). Because ${\displaystyle D_{x-1}}$ is used instead of ${\displaystyle D_{x}}$ the points start from ${\displaystyle (1,f(1))}$ and not ${\displaystyle (0,f(0))}$. This is necessary because ${\displaystyle f(0)}$ is the secret.
### Reconstruction
In order to reconstruct the secret, any 3 points are sufficient
Consider using the 3 points${\displaystyle \left(x_{0},y_{0}\right)=\left(2,1942\right);\left(x_{1},y_{1}\right)=\left(4,3402\right);\left(x_{2},y_{2}\right)=\left(5,4414\right)\,\!}$.
Computing the Lagrange basis polynomials:
${\displaystyle \ell _{0}(x)={\frac {x-x_{1}}{x_{0}-x_{1}}}\cdot {\frac {x-x_{2}}{x_{0}-x_{2}}}={\frac {x-4}{2-4}}\cdot {\frac {x-5}{2-5}}={\frac {1}{6}}x^{2}-{\frac {3}{2}}x+{\frac {10}{3}}}$
${\displaystyle \ell _{1}(x)={\frac {x-x_{0}}{x_{1}-x_{0}}}\cdot {\frac {x-x_{2}}{x_{1}-x_{2}}}={\frac {x-2}{4-2}}\cdot {\frac {x-5}{4-5}}=-{\frac {1}{2}}x^{2}+{\frac {7}{2}}x-5}$
${\displaystyle \ell _{2}(x)={\frac {x-x_{0}}{x_{2}-x_{0}}}\cdot {\frac {x-x_{1}}{x_{2}-x_{1}}}={\frac {x-2}{5-2}}\cdot {\frac {x-4}{5-4}}={\frac {1}{3}}x^{2}-2x+{\frac {8}{3}}}$
Using the formula for polynomial interpolation, ${\displaystyle f(x)}$ is:
{\displaystyle {\begin{aligned}f(x)&=\sum _{j=0}^{2}y_{j}\cdot \ell _{j}(x)\\[6pt]&=y_{0}\ell _{0}(x)+y_{1}\ell _{1}(x)+y_{2}\ell _{2}(x)\\[6pt]&=1942\left({\frac {1}{6}}x^{2}-{\frac {3}{2}}x+{\frac {10}{3}}\right)+3402\left(-{\frac {1}{2}}x^{2}+{\frac {7}{2}}x-5\right)+4414\left({\frac {1}{3}}x^{2}-2x+{\frac {8}{3}}\right)\\[6pt]&=1234+166x+94x^{2}\end{aligned}}}
Recalling that the secret is the free coefficient, which means that ${\displaystyle S=1234\,\!}$, and the secret has been recovered.
### Computationally efficient approach
Using polynomial interpolation to find a coefficient in a source polynomial ${\displaystyle S=f(0)}$ using Lagrange polynomials is not efficient, since unused constants are calculated.
Considering this, an optimized formula to use Lagrange polynomials to find ${\displaystyle f(0)}$ is defined as follows:
${\displaystyle f(0)=\sum _{j=0}^{k-1}y_{j}\prod _{\begin{smallmatrix}m\,=\,0\\m\,\neq \,j\end{smallmatrix}}^{k-1}{\frac {x_{m}}{x_{m}-x_{j}}}}$
### Problem of using integer arithmetic
Although the simplified version of the method demonstrated above, which uses integer arithmetic rather than finite field arithmetic, works, there is a security problem: Eve gains information about ${\displaystyle S}$ with every ${\displaystyle D_{i}}$ that she finds.
Suppose that she finds the 2 points ${\displaystyle D_{0}=(1,1494)}$ and ${\displaystyle D_{1}=(2,1942)}$. She still does not have ${\displaystyle k=3}$ points, so in theory she should not have gained any more information about ${\displaystyle S}$. But she could combine the information from the 2 points with the public information: ${\displaystyle n=6,k=3,f(x)=a_{0}+a_{1}x+\cdots +a_{k-1}x^{k-1},a_{0}=S,a_{i}\in \mathbb {N} }$. Doing so, Eve could perform the following algebra:
1. Fill the formula for ${\displaystyle f(x)}$ with ${\displaystyle S}$ and the value of ${\displaystyle k:f(x)=S+a_{1}x+\cdots +a_{3-1}x^{3-1}\Rightarrow {}f(x)=S+a_{1}x+a_{2}x^{2}}$
2. Fill (1) with the values of ${\displaystyle D_{0}}$'s ${\displaystyle x}$ and ${\displaystyle f(x):1494=S+a_{1}1+a_{2}1^{2}\Rightarrow {}1494=S+a_{1}+a_{2}}$
3. Fill (1) with the values of ${\displaystyle D_{1}}$'s ${\displaystyle x}$ and ${\displaystyle f(x):1942=S+a_{1}2+a_{2}2^{2}\Rightarrow {}1942=S+2a_{1}+4a_{2}}$
4. Subtract (3)-(2): ${\displaystyle (1942-1494)=(S-S)+(2a_{1}-a_{1})+(4a_{2}-a_{2})\Rightarrow {}448=a_{1}+3a_{2}}$ and rewrite this as ${\displaystyle a_{1}=448-3a_{2}}$. Eve knows that ${\displaystyle a_{2}\in \mathbb {N} }$ so she starts replacing ${\displaystyle a_{2}}$ in (4) with 0, 1, 2, 3, ... to find all possible values for ${\displaystyle a_{1}}$:
${\displaystyle a_{2}=0\rightarrow {}a_{1}=448-3\times 0=448}$
${\displaystyle a_{2}=1\rightarrow {}a_{1}=448-3\times 1=445}$
${\displaystyle a_{2}=2\rightarrow {}a_{1}=448-3\times 2=442}$
${\displaystyle \,\,\,\,\,\,\,\,\,\vdots }$
${\displaystyle a_{2}=148\rightarrow {}a_{1}=448-3\times 148=4}$
${\displaystyle a_{2}=149\rightarrow {}a_{1}=448-3\times 149=1}$
5. After checking ${\displaystyle a_{2}=149}$, she stops because would get negative values for ${\displaystyle a_{1}}$ with larger values of ${\displaystyle a_{2}}$ (which is impossible because ${\displaystyle a_{1}\in \mathbb {N} }$). Eve can now conclude ${\displaystyle a_{2}\in [0,1,\dots ,148,149]}$
6. Now, Eve can replace ${\displaystyle a_{1}}$ by (4) in (2): ${\displaystyle 1494=S+(448-3a_{2})+a_{2}\Rightarrow {}S=1046+2a_{2}}$. Now, replacing ${\displaystyle a_{2}}$ in (6) by the values found in (5), she gets ${\displaystyle S\in [1046+2\times 0,1046+2\times 1,\dots ,1046+2\times 148,1046+2\times 149]}$ which leads her to the information: ${\displaystyle S\in [1046,1048,\dots ,1342,1344].}$
Eve now only has 150 numbers to guess from instead of an infinite quantity of natural numbers.
### Solution using finite field arithmetic
This is a polynomial curve over a finite field — the order of the polynomial has seemingly little to do with the shape of the graph.
The above attack exploits constraints on the values that the polynomial may take by virtue of how it was constructed: the polynomial must have coefficients that are natural numbers, and the polynomial must take a natural number as value when evaluated at each of the coordinates used in the scheme. This reduces its possible values at unknown points, including the resultant secret, given fewer than ${\displaystyle k}$ shares.
This problem can be remedied by using finite field arithmetic. A field of size ${\displaystyle p\in \mathbb {P} :p>S,p>n}$ is used as an example. The figure shows a polynomial curve over a finite field. In contrast to a smooth curve it appears disorganised and disjointed.
In practice this is only a small change. The order of the field (i.e. the number of values that it has) must be chosen to be is greater than the number of participants and the number of values of the secret ${\displaystyle a_{0}=S}$). All calculations involving the polynomial must also be calculated over the field (mod p in our example) instead of over the integers. Both the choice of the field and the mapping of the secret to a value in this field are considered to be publicly known. In our example, the value ${\displaystyle p}$ is the order of the field.
For this example, choose ${\displaystyle p=1613}$, so the polynomial becomes ${\displaystyle f(x)=1234+166x+94x^{2}{\bmod {1613}}}$ which gives the points: ${\displaystyle (1,1494);(2,329);(3,965);(4,176);(5,1188);(6,775)}$
This time Eve doesn't gain any information when she finds a ${\displaystyle D_{x}}$ (until she has ${\displaystyle k}$ points).
Suppose again that Eve finds ${\displaystyle D_{0}=\left(1,1494\right)}$ and ${\displaystyle D_{1}=\left(2,329\right)}$, and the public information is: ${\displaystyle n=6,k=3,p=1613,f(x)=a_{0}+a_{1}x+\dots +a_{k-1}x^{k-1}\mod {p},a_{0}=S,a_{i}\in \mathbb {N} }$. Attempting the previous attack, Eve can:
1. Fill the ${\displaystyle f(x)}$-formula with ${\displaystyle S}$ and the value of ${\displaystyle k}$ and ${\displaystyle p}$: ${\displaystyle f(x)=S+a_{1}x+\dots +a_{3-1}x^{3-1}\mod 1613\Rightarrow {}f(x)=S+a_{1}x+a_{2}x^{2}-1613m_{x}:m_{x}\in \mathbb {N} }$
2. Fill (1) with the values of ${\displaystyle D_{0}}$'s ${\displaystyle x}$ and ${\displaystyle f(x):1494=S+a_{1}1+a_{2}1^{2}-1613m_{1}\Rightarrow {}1494=S+a_{1}+a_{2}-1613m_{1}}$
3. Fill (1) with the values of ${\displaystyle D_{1}}$'s ${\displaystyle x}$ and ${\displaystyle f(x):1942=S+a_{1}2+a_{2}2^{2}-1613m_{2}\Rightarrow {}1942=S+2a_{1}+4a_{2}-1613m_{2}}$
4. Subtracts (3)-(2): ${\displaystyle (1942-1494)=(S-S)+(2a_{1}-a_{1})+(4a_{2}-a_{2})+(1613m_{1}-1613m_{2})\Rightarrow {}448=a_{1}+3a_{2}+1613(m_{1}-m_{2})}$ and rewrites this as ${\displaystyle a_{1}=448-3a_{2}-1613(m_{1}-m_{2})}$
5. Using ${\displaystyle a_{2}\in \mathbb {N} }$ so she starts replacing ${\displaystyle a_{2}}$ in (4) with 0, 1, 2, 3, ... to find all possible values for ${\displaystyle a_{1}}$:
${\displaystyle a_{2}=0\rightarrow {}a_{1}=448-3\times 0-1613(m_{1}-m_{2})=448-1613(m_{1}-m_{2})}$
${\displaystyle a_{2}=1\rightarrow {}a_{1}=448-3\times 1-1613(m_{1}-m_{2})=445-1613(m_{1}-m_{2})}$
${\displaystyle a_{2}=2\rightarrow {}a_{1}=448-3\times 2-1613(m_{1}-m_{2})=442-1613(m_{1}-m_{2})}$
${\displaystyle \,\,\,\,\,\,\,\,\,\vdots }$
This time she is not able to stop because ${\displaystyle (m_{1}-m_{2})}$ could be any integer modulo ${\displaystyle p}$ (even negative if ${\displaystyle m_{2}>m_{1}}$) so there are ${\displaystyle p}$ possible values for ${\displaystyle a_{1}}$. She knows that ${\displaystyle [448,445,442,\ldots ]}$ always decreases by 3, so if ${\displaystyle 1613}$ was divisible by ${\displaystyle 3}$ she could conclude ${\displaystyle a_{1}\in [1,4,7,\ldots ]}$. However, ${\displaystyle p}$ is prime she can not conclude this. Thus, using a finite field avoids this possible attack.
## Python code
"""
The following Python implementation of Shamir's secret sharing is
released into the Public Domain under the terms of CC0 and OWFa:
https://creativecommons.org/publicdomain/zero/1.0/
http://www.openwebfoundation.org/legal/the-owf-1-0-agreements/owfa-1-0
See the bottom few lines for usage. Tested on Python 2 and 3.
"""
from __future__ import division
from __future__ import print_function
import random
import functools
# 12th Mersenne Prime
_PRIME = 2 ** 127 - 1
# The 13th Mersenne Prime is 2**521 - 1
_RINT = functools.partial(random.SystemRandom().randint, 0)
def _eval_at(poly, x, prime):
"""Evaluates polynomial (coefficient tuple) at x, used to generate a
shamir pool in make_random_shares below.
"""
accum = 0
for coeff in reversed(poly):
accum *= x
accum += coeff
accum %= prime
return accum
def make_random_shares(secret, minimum, shares, prime=_PRIME):
"""
Generates a random shamir pool for a given secret, returns share points.
"""
if minimum > shares:
raise ValueError("Pool secret would be irrecoverable.")
poly = [secret] + [_RINT(prime - 1) for i in range(minimum - 1)]
points = [(i, _eval_at(poly, i, prime))
for i in range(1, shares + 1)]
return points
def _extended_gcd(a, b):
"""
Division in integers modulus p means finding the inverse of the
denominator modulo p and then multiplying the numerator by this
inverse (Note: inverse of A is B such that A*B % p == 1). This can
be computed via the extended Euclidean algorithm
http://en.wikipedia.org/wiki/Modular_multiplicative_inverse#Computation
"""
x = 0
last_x = 1
y = 1
last_y = 0
while b != 0:
quot = a // b
a, b = b, a % b
x, last_x = last_x - quot * x, x
y, last_y = last_y - quot * y, y
return last_x, last_y
def _divmod(num, den, p):
"""Compute num / den modulo prime p
To explain this, the result will be such that:
den * _divmod(num, den, p) % p == num
"""
inv, _ = _extended_gcd(den, p)
return num * inv
def _lagrange_interpolate(x, x_s, y_s, p):
"""
Find the y-value for the given x, given n (x, y) points;
k points will define a polynomial of up to kth order.
"""
k = len(x_s)
assert k == len(set(x_s)), "points must be distinct"
def PI(vals): # upper-case PI -- product of inputs
accum = 1
for v in vals:
accum *= v
return accum
nums = [] # avoid inexact division
dens = []
for i in range(k):
others = list(x_s)
cur = others.pop(i)
nums.append(PI(x - o for o in others))
dens.append(PI(cur - o for o in others))
den = PI(dens)
num = sum([_divmod(nums[i] * den * y_s[i] % p, dens[i], p)
for i in range(k)])
return (_divmod(num, den, p) + p) % p
def recover_secret(shares, prime=_PRIME):
"""
Recover the secret from share points
(points (x,y) on the polynomial).
"""
if len(shares) < 3:
raise ValueError("need at least three shares")
x_s, y_s = zip(*shares)
return _lagrange_interpolate(0, x_s, y_s, prime)
def main():
"""Main function"""
secret = 1234
shares = make_random_shares(secret, minimum=3, shares=6)
print('Secret: ',
secret)
print('Shares:')
if shares:
for share in shares:
print(' ', share)
print('Secret recovered from minimum subset of shares: ',
recover_secret(shares[:3]))
print('Secret recovered from a different minimum subset of shares: ',
recover_secret(shares[-3:]))
if __name__ == '__main__':
main() | 2023-02-04 21:14:00 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 144, "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": 0.6857064962387085, "perplexity": 1091.179158158181}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-06/segments/1674764500154.33/warc/CC-MAIN-20230204205328-20230204235328-00465.warc.gz"} |
https://courses.wikinana.org/math121b/04-01 | # Lecture Notes
math121b:04-01
## 2020-04-01, Wednesday
Recall the first time we dipped in Chapter 13, partial differential equations, we discussed about the 'method of separation of variables'. We covered Boas 13.1 and 13.2. Today, we plan to talk about 13.3 and 13.4, the heat equation, Schroedinger equation and the wave equation.
### Heat Equation
$$\d_t u = \Delta u$$ This equation discribe the diffusion of heat.
#### Example: heat diffusion on a circle
Consider 1-dimensional example, on a circle. Then function to be solved is $u(t, \theta)$ with $\theta \in [0, 2\pi]$. Suppose we are given the intial conditino $u_0(\theta)$ at $t=0$, then we try to solve for the general solution.
The basic idea is to find the eigenvalue of the operator $\Delta$ acting on function on $S^1$. The eigenvalues are $$\lambda = 0 \z{ and } -n^2, n=1,2,\cdots$$ The eigen-function (up to scale) for $\lambda=0$ is $1$, and the eigenfunctino for $\lambda = -n^2$ is $\sin(n\theta)$ and $\cos(n\theta)$.
Hence, we have the following solution to the heat equation (ignoring the initial condition for now) $$1, \z{ and } e^{-n^2 t} \sin(n \theta), e^{-n^2 t} \cos(n \theta)$$ Thus, if we decompose the initial condition $u_0$ as $$u_0(\theta) = c + \sum_{n = 1}^\infty a_n \cos(n \theta) + b_n \sin(n \theta)$$ then we have $$u(t, \theta) = c + \sum_{n = 1}^\infty a_n e^{-n^2 t} \cos(n \theta) + b_n e^{-n^2 t} \sin(n \theta).$$
#### Example: Heat diffusion on a square
Let $u = u(t,x,y)$, $x \in [0,1], y \in [0,1]$. $$\d_t u = \d_x^2 u + \d_y^2 u$$ and suppose the temperature on the boundary of the interval is held at a constant temperature $T$, and we have initial condition $u_0(x,y)$ given, compatible with the boundary condition. Let's try to find the evolution.
Let's first remove the boundary condition. We may write $$u(t,x,y) = T + v(t,x,y)$$ then $v$ satisfies the same equation as $u$, and with initial condition given $v(0,x,y) = u_0(x,y) - T.$ We may now write down the basis of general solution to the equation (ignoring the initial condition) $$v_{n,m}(t,x,y) = e^{- \pi^2 (n^2 + m^2) t} (\sin (n \pi x) \sin(n \pi y)), \quad n, m \geq 1$$ Then the general solution can be written as $$v(x,y) = \sum_{n,m} a_{n,m} v_{n,m}(t,x,y)$$ To fix the coefficients $v_{n,m}$, we use the initial conditions $$v(0,x,y) = \sum_{n,m} a_{n,m} v_{n,m}(0,x,y)$$ so multiply both sides by $v_{n,m}(0,x,y)$ and integrate, only one term on the RHS contribute, and we get $$a_{n,m} = \frac{\int_{[0,1]^2} v(0,x,y) v_{n,m}(0,x,y) dx dy}{\int_{[0,1]^2} v^2_{n,m}(0,x,y) dx dy}.$$
Remark: if the boundary temperature is not constant $T$ (but still time-independent), we may still find a special solution first, a function $U(x,y)$ that satisfies the boundary condition, and $\Delta U(x,y) = 0$. Such function $U(x,y)$ exists and is unique, it is called the harmonic extension of the boundary value to the interior. Then we can still get rid of the boundary condition by setting $$u(t,x,y) = U(x,y) + v(t,x,y)$$ where $v(t,x,y)$ now has boundary condition $0$, and initial condition $v(0,x,y) = u(0,x,y) - U(x,y)$.
### Schroedinger Equation (without potential)
$$i \d_t u = - \Delta u$$
We may reuse the analysis for the heat equation, except replacing $t$ in heat equation to $it$. Thus, exponential decay now become oscillation.
### Wave equation
$$\d_t^2 u = \Delta u$$ Suppose $u$ lives on a domain $D$ with boundary value zero, or $u$ lives on a space without boundary, e.g $S^1$ or a torus. We may then consider eigenvalue of $\Delta$, $\lambda_1 \leq \lambda_2 \leq \cdots,$ with $\lambda_n \geq 0$, (repeated with multiplicity), with eigenfunction $u_1(x), u_2(x), \cdots,$ then we may write the general solution $$u(t,x) = \sum_{n=1}^\infty (a_n \cos(\sqrt{\lambda_n} t) + b_n \sin(\sqrt{\lambda_n} t) ) u_n(x)$$ (if $\lambda_n=0$, then we may set $b_n=0$.) To fix the coefficient, we use initial condition $u(0, x)$ and $\dot u(0, x)$: $$\int_D u(0,x) u_n(x) dx = a_n \int u_n^2 dx$$ $$\int_D \dot u(0,x) u_n(x) dx = \sqrt{\lambda_n} b_n \int u_n^2 dx$$
Example: 1-dim string vibration on an interval. | 2021-09-16 22:01:39 | {"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": 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": 0.969208300113678, "perplexity": 201.92947761115542}, "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/1631780053759.24/warc/CC-MAIN-20210916204111-20210916234111-00514.warc.gz"} |
https://locandafilomarino.it/1469/vibrating-screen-principle-amplitude/2020-03-20.html | AS a leading global manufacturer of crushing and milling equipment, we offer advanced, rational solutions for any size-reduction requirements. We also supply individual crushers and mills as well as spare parts of them.
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# vibrating screen principle amplitude
• #### Vibrating Screen Working Principle
IntroductionNamingMechanismApplicationsPurposeCauseUseSpecificationsSafetyOperationAssessmentCriticismsAnalysisExampleAdvantagesConstructionTypesDesignThe simplest Vibrating Screen Working Principle can be explained using the single deck screen and put it onto an inclined frame. The frame is mounted on springs. The vibration is generated from an unbalanced flywheel. A very erratic motion is developed when this wheel is rotated. You will find these simple screens in smaller operations and rock quarries where sizing isnt as critical. As the performance of this type of screen isnt good enough to meet the requirements of most mining operations two variations of this scre
• #### Vibration frequency, screen inclination of the vibrating
Apr 27, 2019· Usually, the amplitude of the linear vibrating screen is A=3.5~6mm. 4. The increase of vibration frequency ω vibration frequency can increase the number of times the material beats on the screen surface, which increases the probability of material permeability.
• #### Vibrating Screen Types & Working Principle [How To Choose
Working Principle : The two motor axes of the linear vibrating screen have an angle of inclination in the vertical direction relative to the screen panel. Under the combined force of the exciting force and the self gravity of the material, the material is thrown on
• #### Checking the amplitude of a vibrating screen Quarry
Dec 06, 2018· The amplitude of vibration does not depend on the rotation of the vibrating mechanisms. Amplitude combined with rotation imparts a certain amount of acceleration to the equipment, and that acceleration is responsible for the conveyance of the material.
• #### how to check vibration screens amplitude, gold mining
amplitude for vibrating screen klimaatwebsite.be Circle-throw vibrating screens operate on an inclined surface, When vibration amplitude is within the range of 3 to 35mm, the equipment, Once the area of the unit is known, the screen's length and width must be calculated so that a ratio, More details » Get Price.
• #### 5 Vibrating Screen Common Problems And How To Solve? M&C
Poor screening quality. There are many factors affecting the screening effect, including the nature
• #### Vibrating Screen for All Kinds of Sands and Stones
Vibrating Screen Working Principle: Many customers may do not understand screen machine’s working principle. Here, Aimix will give you a detailed description. The actual screening process of a round vibrating screen is: a large number and different sizes of mixed materials will be sent to the sieve surface, and only a part of the particles will touch the screen surface.
• #### High-frequency vibrating screens Wikipedia
OverviewCharacteristics assessmentRange of applicationsAdvantages and limitationsAvailable designsMain process characteristicsPossible heuristics to be used during design of the processPost-treatment systems
The purpose of the vibrating screen is that particles are introduced to the gaps in the screens repeatedly. The frequency of the screen must be high enough so that it prevents the particles from blocking the apertures and the maximum height of the particle trajectory should occur when the screen surface is at its lowest point. Based on the principle, there is an optimum frequency and amplitude of vibration Transmission refers to the fraction of desired particle that passes through the apertures in the screen. At low frequency, screening efficiency is high but blinding is severe. Blinding will decrease as freq
• New content will be added above the current area of focus upon selection
The purpose of the vibrating screen is that particles are introduced to the gaps in the screens repeatedly. The frequency of the screen must be high enough so that it prevents the particles from blocking the apertures and the maximum height of the particle trajectory should occur when the screen surface is at its lowest point. Based on the principle, there is an optimum frequency and amplitude of vibration Transmission refers to the fraction of desired particle that passes through the apertures in the screen. At low frequency, screening efficiency is high but blinding is severe. Blinding will decrease as frequency increases but the particles will have difficulty going through the apertures. When designing a high frequency vibrating screen, an optimum point of frequency and amplitude must be chosen, depending on the specific applications. The separation efficiency is simply a measure of the amount of material removed by the screen compared to the theoretical amount that should have been removed. Screen efficiency can be obtained using different equation, which depends on whether the desired product is the oversize or undersize fraction from the screen. The screen efficiency based on the oversize (Eo) is given by: $${\displaystyle E_{o}={\frac {Q_{ms}(o)\;[1-M_{u}(o)]}{Q_{ms}(f)\;[1-M_{u}(f)]}}}$$The screen efficiency based on the undersize (Eu) is then given by: $${\displaystyle E_{u}={\frac {Q_{ms}(u)\;M_{u}(u)}{Q_{ms}(f)\;M_{u}(f)}}}$$where Qms(o) is the mass flow rate of solid in the screen overflow, Qms(f) is the mass flow rate of solid feed, Qms(u) is the mass flow rate of solid in the screen underflow, Mu(o) is the mass fraction of undersize in the overflow, Mu(f) is the mass fraction of undersize in the feed, Mu(u) is the mass fractio
Wikipedia · Text under CC-BY-SA license
• #### Vibrating Screen Crushing & Mining Screen JXSC Mine
Vibrating screen working principle: The vibratory screening machine is to utilize reciprocating vibration of the vibration generator produced. The processing of the screen separates the different size material by a single- or triple-deck screen. That is, according to the size of particles to separate.
• #### Construction, Working and Maintenance of Electric
A vibrating screen is vibrated in order to throw particles off the screening surface so that they can again be presented to the screen, and to convey the particles along the screen. Vibration also lifts the large particles out of the holes (If the large particles are not lifted out, it will plug the screen).
• #### Vibrating Screen Types & Working Principle [How To Choose
Working Principle: After the rotary screen is started, the eccentric blocks of different phases at the upper and lower ends of the vibrating motor generate a composite inertia force, which forces the vibrating body of the rotary screen machine to rotate again and again, and the screen frame continuously moves to and fro under the action of the
• #### vibrating screen amplitude measurement
Based on the principle, there is an optimum frequency and amplitude of vibration Transmission refers to the fraction of desired particle that passes through the apertures in the screen. At low frequency, screening efficiency is high but blinding is severe.
• #### Principle of vibrating screen Schalker Freunde Rotenburg
Efficient Circular Vibrating Screen Fote Machinery. Working principle of circular vibrating screen The circular vibrating screen uses the ordinary motor to drive the eccentric block of the vibrator to produce highspeed rotation The running eccentric block generates a large centrifugal force, which stimulates the screen box to generate a periodic circular motion of a certain amplitude along the
• #### Vibrating Screen Working Principle For Coal Vibration
Vibrating screen working principle through the motor force and adjusting the amplitude with the eccentric shaft vibrating screen moves quickly being a circle working principle of vibrating screen Fruitful group vibrating screen is mainly used in metallurgy mining coal building materials electric power stone and other industries.
• #### 「amplitude of a grizzly vibro screen」
Vibrating Screen Amplitude Test Van Iseghem Armand. In fact, the amplitude of every vibro screen is different, thats why we cant make the amplitude Details Vibration Screen Vibration Machine Laboratory Test . 2019 9 11 · JXSC provide sieve shaker equipment since 1985，types of sieve shaker includes horizontal vibrating screen, linear vibrating screen, circular vibrating screen separator
• #### Types Of Vibrating Screens Introduction And Difference
Inclined screen. Vibrating inclined screen is the most popular types of vibrating screens. The screen fix to the tilted frame at an angle of 15 to 30 degrees. The inclined screen is usually designed in a multi-layer structure and can be classified into 2-5 grades. And the screen can be adjusted to provide overall performance and efficiency.
• #### Working principle of vibrating screen zotvanbier.be
YKN Vibrating ScreenVibrating Screen Working Principle. Working Principle of YKN Vibrating Screen This series of vibrating screen with eccentric block as the excitation source to drive the sieve box for circular vibration its working principle is the sieve box through the elastic support on the base or base the eccentric block is arranged on the main shaft the main shaft through a bearing and
• #### Vibrating screen,linear vibrating screen,dewatering
Double frequency screen adopts the principle of combined vibration frequence. In order to realize efficient screening of material which are easily to block deck mesh, PU deck or composite mesh grid deck are used to increase the contact area between material and screen mesh, also to increase air permeability, resulting in efficient screening.
• #### Screening Theory and Practice Triple/S Dynamics
B. The Shaking Screen: 475 rpm, 1” stroke, zero pitch, 6 deg. slope. C. The Inclined Vibrating Screen: 1200 rpm, 1/4” vertical circle dia. D. The Horizontal Vibrating Screen: 840 rpm,1/2” stroke at 45°. Each has a .063” dia. wire screen with 1/8” clear opening, moving under a particle travelling
• #### Vibratory Screeners Cleveland Vibrator
Linear Vibration. Sure, using one motor would cost less. But, 99% of vibratory equipment applications perform best with linear vibration, in line with gravitational forces, that are easily achieved with two synchronized vibrator motors, using the Dual Motor Principle.
• #### Vibrating Screen Crushing & Mining Screen JXSC Mine
Vibrating screen working principle: The vibratory screening machine is to utilize reciprocating vibration of the vibration generator produced. The processing of the screen separates the different size material by a single- or triple-deck screen.
• #### Electromagnetic high frequency vibrating screen-China
The amplitude of the screen surface can reach 1-3mm, and the vibration intensity is 3-4 times as strong as that of general screening equipment, so the screen mesh can never be blocked. The screen surface has an excellent self-cleaning ability, high efficiency and a large capacity.
• #### Circle-throw vibrating machine Wikipedia
Vibration isolation is a control principle employed to mitigate transmission. On circle-throw vibrating screens, passive vibration isolation in the form of mechanical springs and suspension are employed at the base of the unit, which provides stability and control of motor vibration.
• #### gyratory vibrating screen, gyratory vibrating screen
gyratory vibrating screen, buy gyratory vibrating screen products from global gyratory vibrating screen manufacturer and gyratory vibrating screen supplier at vibratingscreen.cc 4.Working principle- Gyratory reciprocating motion ensures material constantly contact with screen surface reach higher screening efficiency. Amplitude (mm
• #### circular vibrating screen working principles,
Working Principle Of Vibrating Screen Pdf. Circular vibrating screen working principle.1, the working principle circular vibrating screen, also known as single-axis shaker, sieve boxes at various points along the trajectory is a circle or a similar.Get price and support online vibrating sieve working principle fairytime.Working principle of vibrating screen slideshare.
• #### Vibrating Screen Working Principle For Coal Vibration
Vibrating screen working principle through the motor force and adjusting the amplitude with the eccentric shaft vibrating screen moves quickly being a circle working principle of vibrating screen Fruitful group vibrating screen is mainly used in metallurgy mining coal building materials electric power stone and other industries.
• #### Vibrating Screen For Sale FTM Machinery
The vibrating screen, also named stone screening machine, is a new type of high efficiency vibrating screen with a multilayer structure. The circular vibrating screen adopts the cylinder type eccentric shaft exciter or the eccentric block to adjust the amplitude, which realizes the long dripping line of the screen and various screening standards.
• #### Working principle of vibrating screen zotvanbier.be
YKN Vibrating ScreenVibrating Screen Working Principle. Working Principle of YKN Vibrating Screen This series of vibrating screen with eccentric block as the excitation source to drive the sieve box for circular vibration its working principle is the sieve box through the elastic support on the base or base the eccentric block is arranged on the main shaft the main shaft through a bearing and
• #### vibrating screen amplitude measurement
Based on the principle, there is an optimum frequency and amplitude of vibration Transmission refers to the fraction of desired particle that passes through the apertures in the screen. At low frequency, screening efficiency is high but blinding is severe.
• #### Principle of vibrating screen Schalker Freunde Rotenburg
Working principle of circular vibrating screen The circular vibrating screen uses the ordinary motor to drive the eccentric block of the vibrator to produce highspeed rotation The running eccentric block generates a large centrifugal force, which stimulates the screen box to generate a periodic circular motion of a certain amplitude along the
• #### Vibrating screens circular vibrating screen Korea Crusher
Vibrating Screens operating principle. Vibrating screens are a kind of sieving equipment of international advanced level, developed by Samyoung Crusher plant Manufacturer on the basis of carrying on the advantages of traditional screens and absorbing the outstanding technology from abroad.. Vibrating screen is widely used for grading and screening materials in the minerals, quarry,
• #### gyratory vibrating screen, gyratory vibrating screen
gyratory vibrating screen, buy gyratory vibrating screen products from global gyratory vibrating screen manufacturer and gyratory vibrating screen supplier at vibratingscreen.cc 4.Working principle- Gyratory reciprocating motion ensures material constantly contact with screen surface reach higher screening efficiency. Amplitude (mm
• #### Ultrasonic Vibrating Sieve Unitfine Machinery Co.,Ltd
The ultrasonic vibrating sieve successfully breaks this static bonding between the powder particles; thus preventing the clogging of the screen mesh. Does not dust The ultrasonic vibrating sieve operates at high frequency and low amplitude. Since it is at low amplitude, there are no
• #### Dewatering Screen Machine Manufacturer Gaofu
Dewatering screen, also known as high-frequency dewatering screen, changing the water surface tension of pulp through the exciting force, pulp water through the screen to become the undersize material, and fine material is blocked by the screen to form the filter layer by the vibration of the forward movement of discharge.
• #### Dynamic Characteristics of a Vibrating Flip-Flow Screen
Due to the limitation of the structure of the VFFS, the highest vibration frequency point that can be measured is only 97.37 rad/s, at which frequency the amplitude of the main screen frame is 1.89 mm, the amplitude of the floating screen frame is 14.89 mm, and the relative amplitude is 15.90 mm.
• #### 「amplitude of a grizzly vibro screen」
Vibrating Screen Amplitude Test Van Iseghem Armand. In fact, the amplitude of every vibro screen is different, thats why we cant make the amplitude Details Vibration Screen Vibration Machine Laboratory Test . 2019 9 11 · JXSC provide sieve shaker equipment since 1985，types of sieve shaker includes horizontal vibrating screen, linear vibrating screen, circular vibrating screen separator
• #### Vibrating Screen Amplitude Test-magnetic Separator
5 Vibrating Screen Common Problems And How To Solve. 4 technical indicators do not meet the requirements the technical indicators of the operation of the sieve include the rotational speed vibration force amplitude and frequency of the sieve etc common fault types are the screen can not start or the amplitude is too small the speed is not enough the vibration force is weak the fourpoint
• #### Linear Vibrating Screen,Screening Machines,Henan Pingyuan
Introduction Linear vibrating screen is designed with dual-vibration motor drive,two synchronous motors are reversely placed so that the exciter generate reverse excitation force, the exciting force generated by eccentric block cancel each other out on the parallel direction of motor axis,and stack together with the perpendicular direction of motor axis, so its trajectory is linear.Linear | 2021-01-19 18:58:10 | {"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": 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": 0.31401580572128296, "perplexity": 4275.126061635079}, "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-2021-04/segments/1610703519600.31/warc/CC-MAIN-20210119170058-20210119200058-00684.warc.gz"} |
http://math.stackexchange.com/questions/195564/sending-vector-space-to-dual-is-a-functor | # Sending vector space to dual is a functor
In the category of finite dimensional vector spaces over a field and linear maps between them, the map that sends each space to its dual and linear map to its transpose is a functor, right?
But this doesn't make sense to me. Call the map $F$, and let $f: V \to W$ and $g: W \to Z$ be linear maps. So,
$$F(g \circ f) = (g \circ f)^{*} :Z^* \to V^*$$
needs to be $F(g) \circ F(f)$ but this is not defined, $F(f) \circ F(g)$ is defined however. Can someone tell me what I'm missing here?
-
## 1 Answer
The functor $F$ is a contravariant functor, that is why it reverses composition. Check out the definitions at
http://en.wikipedia.org/wiki/Functor
-
Thanks for pointing this out. – Jonah777 Sep 14 '12 at 22:41 | 2015-01-31 19:54:46 | {"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": 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": 0.9133999347686768, "perplexity": 126.72791905212523}, "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-2015-06/segments/1422122030742.53/warc/CC-MAIN-20150124175350-00073-ip-10-180-212-252.ec2.internal.warc.gz"} |
https://quantumcomputing.stackexchange.com/questions/12598/what-is-the-shortest-sequence-of-decomposition-a-given-single-qubit-unitary-gate | # What is the shortest sequence of decomposition a given single-qubit unitary gate
Given a single-qubit unitary matrix, can we find the shortest sequence of Clifford + T gates that correspond to that unitary?
According to Fast and efficient exact synthesis of single qubit unitaries generated by Clifford and T gates , which is Solovay-Kitaev decomposition, I learned single-qubit decomposition may need $$O(\log^{3.97} (1/\delta))$$ clifford+T gates with the accuracy $$\delta$$.
And later many optimization is worked on it. For example: Synthesis of unitaries with Clifford+T circuits
So I want to know if there exists a shortest sequence of Clifford + T gates that correspond to decompose any single-qubit unitary into Clifford+T? If it existed, what is commonly used in current compiler?
In other words, this is the best that can be done. It yields the shortest possible sequence. This best is $$O(\log\frac{1}{\delta})$$ Hadamard + T. | 2021-04-23 12:08:00 | {"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": 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": 3, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.858323872089386, "perplexity": 730.6204269768235}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-17/segments/1618039617701.99/warc/CC-MAIN-20210423101141-20210423131141-00570.warc.gz"} |
http://tex.stackexchange.com/questions/86043/in-biblatex-make-title-sentence-case-but-not-journal-name/86051 | # In biblatex, make title sentence case but not journal name
To force reference titles to a sentence case I use the following line in my .bbx style file:
\DeclareFieldFormat[article,inbook,incollection,inproceedings,patent,thesis,unpublished]{titlecase}{\MakeSentenceCase*{#1}}
But this also causes the journal name to be written in sentence case, which is unwanted.
Why is that and how can I circumvent that?
A minimal example:
\documentclass{article}
\usepackage{filecontents}
\begin{filecontents}{\jobname.bib}
@article{ref1,
author = {Doe, J. and Dane, D., and Dewy, R.},
year = {2000},
title = {This and That},
journal = {Journal of Deep Understanding of Things},
}
@article{ref2,
author = {Doe, J. and Dewy, D., and Dane, R.},
year = {2000},
title = {The Other},
journal = {Journal of deep understanding of things},
}
\end{filecontents}
\usepackage[style=authoryear-comp,natbib=true,
maxcitenames = 2,
mincitenames = 1,
firstinits = true,
labelyear=true,
uniquename=false,
uniquelist=false,
terseinits = false,
backend=biber]{biblatex}
\DeclareFieldFormat[article,inbook,incollection,inproceedings,patent,thesis,unpublished]{titlecase}{\MakeSentenceCase*{#1}}
\begin{document}
Some text and a ref \citep{ref1}.
Then another ref with same first author and year \citep{ref2}
\printbibliography
\end{document}
gives:
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Closely related: tex.stackexchange.com/q/22980 – Audrey Dec 8 '12 at 3:05
The original definition of the bibmacro used to print the journal information is
\newbibmacro*{journal}{%
\iffieldundef{journaltitle}
{}
{\printtext[journaltitle]{%
\printfield[titlecase]{journaltitle}%
\setunit{\subtitlepunct}%
\printfield[titlecase]{journalsubtitle}}}}
Thus the instruction \DeclareFieldFormat[article]{titlecase}{\MakeSentenceCase*{#1}} impact the journal title as well. The solution is to modify the definition of the journal bib macro
\newbibmacro*{journal}{%
\iffieldundef{journaltitle}
{}
{\printtext[journaltitle]{%
\printfield[myplain]{journaltitle}%
\setunit{\subtitlepunct}%
\printfield[myplain]{journalsubtitle}}}}
where we can define a myplain field format that just produce an unformatted value.
\DeclareFieldFormat{myplain}{#1}
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A directive identical to myplain comes predefined in biblatex.def - it's called noformat. – Audrey Dec 8 '12 at 16:34
If you are trying to get the IEEE style you may want to use the following settings (details):
\usepackage[
bibstyle=ieee,% IEEE citation style
citestyle=numeric-comp,% citing multiple papers will produce format similar to [2,4-8,12] instead of [2,4,5,6,7,8,12] (optional)
sorting=none,
backend=biber,
maxnames=100,% show up to 100 authors per author in the bibliography (optional)
isbn=false,url=false,doi=false% remove extra info (optional)
] {biblatex}
It will produce the intended results but it might modify other things too.
- | 2016-07-26 04:39:27 | {"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": 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": 0.6590620279312134, "perplexity": 7272.665486120508}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-30/segments/1469257824624.99/warc/CC-MAIN-20160723071024-00011-ip-10-185-27-174.ec2.internal.warc.gz"} |
http://yvyi.laroccashop.it/factoring-trinomials-a-not-1-worksheet.html | # Factoring Trinomials A Not 1 Worksheet
Key Terms Use the vocabulary terms listed below to complete each statement in Exercises 1–4. Book of Gina Wilson Factoring Trinomials Worksheet Answers, as an amazing reference becomes what you need to get. That will being reported, all of us provide a number of very simple however useful content. Number of terms (monomials connected by + or – signs) term count terminology name 1 1-term monomial (product of a coefficient and variables) 2 2-term binomial. Factoring trinomials worksheet factoring trinomials when a is not 1 worksheet worksheets for all printable. x2 2x 35 3. Here is a graphic preview for all of the Monomials and Polynomials Worksheets. The coloring portion makes a symmetric design that helps students self check their answers and makes it easy for teachers to grade. The resulting factors will be (x + r) and (x + s). There are many sections in later chapters where the first step will be to factor a polynomial. 5) x 2 − 7x − 8 6) k 2 + 5k − 5. 25 problems on factoring quadratic expressions with leading coefficients of 1 and factors of 9 or less. In general, multiplying binomials gives four terms, one corresponding to each letter of the FOIL acronym. Quadratic trinomials are in the form of a {x^2} x2 + b x + c, and the a, b, and c all stands for a number. Factoring trinomials of the form ax 2 + bx + c: (i) 2m 2 + 11m + 12 (ii) 3m 2 + 8m + 4. Do you remember how to check this?. Unit 7 Polynomials And Factoring Homework 1 Classifying Poly - Displaying top 8 worksheets found for this concept. Since in the Student Worksheet about 90% of the articles of the entire book are questions, both numerous selection and answer questions which are not available. Best 14 Best Images of Kuta Software Factoring Trinomials Worksheet Answer Key Kuta Software. 1) b2 + 8b + 7 2) n2 − 11 n + 10 3) m2 + m − 90 4) n2 + 4n − 12 5) n2 − 10 n + 9 6) b2 + 16 b + 64 7) m2 + 2m − 24 8) x2 − 4x + 24 9) k2 − 13 k + 40 10) a2 + 11 a + 18 11) n2 − n − 56 12) n2 − 5n + 6-1-. Various skills included in designing a worksheet, forms of worksheets, and sample worksheets are explained in detail. teacherspayteachers. Some of the worksheets displayed are Factoring trinomials a 1 date period, Factoring trinomials a 1 date period, Factoring polynomials, Factoring practice, Foil and factoring trinomials period, Factoring polynomials gcf and quadratic expressions, Factoring polynomials, 5 math 51 factoring trimonials. One problem is not factorable, one is a perfect square, and seven problems require students to factor out a common integer prior to factoring the trinomial. 22x2y3 + 14xy 2. [PDF] Factoring Quadratic Trinomials Worksheet Answers Trinomials Factoring. L Worksheet by Kuta Software LLC Kuta Software - Infinite Algebra 1 Name_____ Factoring Trinomials (a > 1) Date_____ Period____. The square root of the last term 5y. x 3 + 2x 2 − 5x − 6. Clear fractions (by multiplying through by the common denominator) 2. There are many sections in later chapters where the first step will be to factor a polynomial. Factoring Trinomials (a = 1) Date Period. Worksheet: Practice in Factoring Quadratics Name _____ Solve the following equations by factoring, where possible. In this factoring trinomials instructional activity, students factor twenty-six trinomial expressions. Some of the worksheets for this concept are Classifying polynomials, Unit 7 polynomials, Factoring polynomials, Polynomials notes 1, Polynomials classifying polynomials, Adding and subtracting polynomials date period, Dividing polynomials, Basic polynomial operations date period. The following diagrams show how to factor trinomials where the leading coefficient is 1 (a = 1). Factoring When Leading Coefficient Is Not 1 - Displaying top 8 worksheets found for this concept. r2 – 8r + 16 YES 2. 2 of your textbook) Use the Factor Theorem to show that for every real number c, (x c) is not a factor of x4 +x2 +1. Should you seek help on multiplying and dividing rational or maybe inverse, Algebra1help. Polynomials Worksheet #1 Author: Des Last modified by: jbrown Created Date: 11/18/2016 6:37:00 PM Company: Microsoft Other titles: Polynomials Worksheet #1. Write the first and last term in the first and last box respectively. This is just one of the solutions for you to be successful. Polynomials with one variable Polynomials with multiple variables. Factoring is a process of splitting the algebraic expressions into factors that can be multiplied. Page 3 Lesson 1: Factoring using the Greatest Common Factor Factor each expression by factoring out the GCF. We hope these Factoring Trinomials a 1 Worksheet images gallery can be a guide for you, give you more references and of course bring you what you need. This activity is self checking because they only have a certain set of binomia. Factor the common factor out of each expression. Factoring Polynomials Any natural number that is greater than 1 can be factored into a product of prime numbers. Scroll down the page for more examples and solutions of factoring trinomials. Remember, it always factors into 2 2 b x 5) Use the principle of square roots 6) Solve the remaining equation 7) Check your answer in the original equation. Before factoring a polynomial equation, it is better to rewrite it in descending order. Factoring Trinomials Case 1: lead coefficient is one. students will solve problems related to polynomial functions answer multiple choice questions topics include factoring polynomials puzzle from just mathematics on teachersnotebook com 2 pages complex number imaginary maze review worksheet teacherspayteachers great practice trinomials with a coefficient for my algebra students! this is 7 bundle of popular activities! love the fun 1000 images. 2x(3x2 5) 6x3 10x. Our team involving resourceful internet writers have exceptional abilities inside verbal and also written communication, which usually convert for you to the sort of written content you simply will not discover wherever else. We allow you this proper as capably as simple pretension to get those all. Some of the worksheets displayed are factoring quadratic expressions factoring trinomials a 1 date period factoring practice cp algebra 2 unit 2 1 factoring and solving quadratics factoring polynomials gcf and quadratic expressions factoring and solving quadratic equations work factoring polynomials. Factoring worksheets from K5 Learning. com includes practical strategies on factoring trinomial worksheet, subtracting and absolute and other algebra topics. long division or factoring to nd the others. Factor completely. Instructions to tutor: Read instructions under “Activity” and follow all steps for each problem exactly as given. Check your work by multiplying your factors to get the original polynomial. Regardless of what your company planning objectives, cash flow remains the most essential resource in the organization, and managing money is the one small. This section is about factoring trinomials (three-term polynomial). 2(2x)(5y)= 20xy 20xy = 10xy 4x2-16xy+8y2 There is a. Gallery of 50 Factoring Trinomials A 1 Worksheet. Recall that factoring is the process of taking a sum (things added) and rewriting it as a product (things multiplied). Step #1: Write out the trinomial in descending Ex. 3 3-term trinomial. Factoring trinomials is actually the result of ‘working backwards’ from ‘FOIL’. Type of Resume and sample, factoring polynomials color by number worksheet answers. Read online Worksheet: Factoring Trinomials (a=1) book pdf free download link book now. com makes available useful answers on algebra open sentences worksheets, college algebra and matrix and other algebra topics. In this factoring trinomials worksheet, students factor twenty-six trinomial expressions. In fact, this is not even a trinomial because there are 2 terms. 28 Factoring Polynomials Practice Worksheet with Answers- Rather than inserting the exact same text, modifying font styles or correcting margins every time you begin a new document, opening a personalized template will let you get directly to work on the content instead of wasting time tweaking the styles. Based on the signs of the trinomial, we can determine whether we want factors that add up or subtract to c (see method 1). If we reverse the problem, ) ( , we say we have factored into. ©3 52n011A2j DKHuntwae XSkoBfbtRwMacrHeV OLlLCCX. y t VMHaQdneY Pw3iNtuh0 4Ijn9fIignzi4tIel JAClUgceqbGrpar 018. Worksheets are very critical for every student to practice his/ her concepts. If so, factor out the GCF. -1-CLASS EXAMPLES: Factor each completely. 7 Lesson 5: Factoring Binomials that are the Difference of Two Perfect. We deal with GCF, Difference of Squares, and the two types of Trinomials. Example 6: 2. 4) 27 + 8x. Students, teachers and parents can download all CBSE educational material and extremely well prepared worksheets from this website. Kuta Software - Infinite Algebra 1 Name_____ Factoring Trinomials (a = 1) Date_____ Period____ Factor each completely. Connection to the internet is not require. 97x + 43883. 1) 3x2 + 14x + 152) 3x2 + 14x + 15. M i sANlnln MrJi Ygch ptDsC Mrmels le Grxv desdi. Put ax2in the upper left box and c in the lower right box. Download link is at the bottom of the post. ©i bKkubtda7 zS2oCfftwwCa3rBeb yLrLxCb. x2 2x 35 3. All quizzes. Polynomial Worksheet Exercise 1 State whether the following algebraic expressions are polynomials or not. Practice the worksheet on factoring trinomials to know how to factorize the quadratic expression of the form ax 2 + bx + c. Some of the worksheets displayed are factoring quadratic expressions factoring trinomials a 1 date period factoring practice cp algebra 2 unit 2 1 factoring and solving quadratics factoring polynomials gcf and quadratic expressions factoring and solving quadratic equations work factoring polynomials. Step 1: Make sure that the trinomial is written in the correct order; the trinomial must be written in descending order from highest power to lowest power. In the multiplication problem ( ) , are the factors and is the product. So the solutions must be x=-2 and x=-3. Here is a graphic preview for all of the Monomials and Polynomials Worksheets. Then you should click on the trace button to trace the trinomial. In this factoring trinomials worksheet, students factor twenty-six trinomial expressions. Factoring Trinomials of the Form x 2 + bx + c and x 2 - bx + c. 4) Factor the Perfect Square Trinomial on the left side of the equation and simplify the right side. 125x3 64 15. After having gone through the stuff given above, we hope that the students would have understood, "Basic Concepts in Polynomials Worksheet" Apart from the stuff given in this section, if you need any other stuff in math, please use our google custom search here. Keep in mind, you may only in a position to delete the worksheet and you also aren't capable to restore it any longer, there is not any kind of Undo function for your removal of worksheet, so that you can wish to be cautious upon what you intend to. 2 Factoring Trinomials of the Type x2 + bx + c Learning Objectives A Factor trinomials of the type x2 + bx + c by examining the constant term c. Factoring a sum/difference of cubes Factoring by grouping Factoring quadratic form Factoring using all techniques Factors and Zeros The Remainder Theorem Irrational and Imaginary Root Theorems Descartes' Rule of Signs More on factors, zeros, and dividing The Rational Root Theorem Polynomial equations Basic shape of graphs of polynomials. j A tARl9lg EreiIg jhKtRsy Lr de Ps de vrMv3e ed 6. 14r2s - 21rst 4. The worksheet really should be short, crisp, simple child friendly. GCF: If the polynomial has a greatest common factor other than 1, then factor out the greatest common factor. Trinomials are algebraic expressions that has three terms in it. Example 1: To factor trinomial $2x^2+x-3$, type 2x^2 + x - 3 Example 2: To factor trinomial $6a^2-13ab-5b^2$ ,go into "multiple variable" mode and then type 6a^2 - 13ab - 5b^2. Part 1 of the lesson consists of “circle puzzles” as a preparation for the actual factoring lesson in Part 2. V Worksheet by Kuta Software LLC Integrated Algebra I ID: 1 Name_____ Date_____ Period____ ©y n2y0 W101 M wKGuLtsa s tS5ofet Dw 9aerIe y 3L7L PCg. Now we must. FACTORING TRINOMIALS WHOSE LEADING COEFFICIENT IS NOT 1 In this section we will learn methods for factoring trinomials of the form ax bx c a or2 ++ ≠;01: i. Geometry Worksheets. Solving Quadratic Equations by Factoring. Showing top 8 worksheets in the category - Factoring When Leading Coefficient Is Not 1. All quizzes. Factor the given trinomials. Algebra 1 - Monomials and Polynomials Worksheets Factoring Quadratics Worksheets. 5 Exercises Factoring Worksheet Trinomials FactoringTrinomials Factorthefollowingcompletely. Quiz not found! BACK TO EDMODO. One problem is not factorable, one is a perfect square, and seven problems require students to factor out a common integer prior to factoring the. Elementary Algebra Skill Factoring 2-Variable Trinomial Squares with Leading Coefficient Not 1 Factor each completely. Whenever a quadratic has constants 3, 2, −1, then for any argument, the factoring will be (3 times the argument − 1) (argument + 1). (Exercise 67 from Section 4. ©L Y2A0B1I2 v nK0udt Ia3 9SRoRfKtSw ka QrFex 5L 3L qCF. $$5x ^{\red 3} + 6x^2 + 9$$. 338 - 342. Factoring and multiplying polynomials - both processes are inverses of each other. Note that the coefficient of the x^2 term is 1. The first step is to factor out any common factor, if any. Key Terms Use the vocabulary terms listed below to complete each statement in Exercises 1–4. This is an introduction to the trial factors method. 28 Factoring Polynomials Practice Worksheet with Answers- Rather than inserting the exact same text, modifying font styles or correcting margins every time you begin a new document, opening a personalized template will let you get directly to work on the content instead of wasting time tweaking the styles. 4x2-10xy+25y2 The square root of the first term is 2x. Number of terms (monomials connected by + or – signs) term count terminology name 1 1-term monomial (product of a coefficient and variables) 2 2-term binomial. Keep in mind, you may only in a position to delete the worksheet and you also aren't capable to restore it any longer, there is not any kind of Undo function for your removal of worksheet, so that you can wish to be cautious upon what you intend to. This is typically an Algebra 1 skill but we place it in Algebra 2. One problem is not factorable, one is a perfect square, and seven problems require students to factor out a common integer prior to factoring the. 3 Lesson 2: Solving Literal Equations by Factoring pg. Then you should click on the trace button to trace the trinomial. Some of the worksheets displayed are Factoring trinomials a 1 date period, Factoring trinomials a 1 date period, Factoring polynomials, Factoring practice, Foil and factoring trinomials period, Factoring polynomials gcf and quadratic expressions, Factoring polynomials, 5 math 51 factoring trimonials. FACTORING TRINOMIALS OBJECTIVES. Find two numbers that multiply to give c and add to give b. We could use the Quadratic Formula to find the factors. chapter 19. (Thus a no longer has to be equal to 1 as in the previous section but we still assume that the discriminant is a perfect square. No, because neither 1 nor −1 will make that polynomial equal to 0. 7) a 2 + 14a + 48 8) n 2 − 15n + 50. Factoring When Leading Coefficient Is Not 1. com and study equations by factoring, rational and countless additional math subject areas. Factoring Trinomials Worksheet with Answer Key Worksheet: Factoring Trinomials (a=1) Write each trinomial in factored form (as the product of two binomials). teacherspayteachers. ) Factoring by Grouping: used when a. The procedure to use the factoring trinomials calculator is as follows: Step 1: Enter the trinomial function in the input field Step 2: Now click the button "Factor" to get the result Step 3: Finally, the factors of a trinomial will be displayed in the new window. Example 2 Factor the trinomial 9x 2. You may select which type of polynomials to factor, and whether you want some non-factorable expressions or not. Name:___Date:___Block:___3. 9 - 4x 2 = 3 2 - (2x) 2 = (3 - 2x)(3 + 2x) As a practice, multiply (3 - 2x)(3 + 2x) to obtain the given expression. 30a4b2 + 45 ab3 3. If you’re in school, your teacher will often have you complete as much as possible up to the point where it cannot be factored. However, if both binomials have the same variables to the same powers, then it is true. 6w3x2y - 9w2x3 - 3x4y2. Beside that, we also come with more related things as follows algebra 2 chapter 1 worksheet, adding polynomials worksheet printable and factoring trinomials worksheet. x 3 + 2x 2 − 5x − 6. Note that the coefficient of the x^2 term is 1. ) Trial and Error: used when a and c are prime numbers. Z Worksheet by Kuta Software LLC Factoring Name_____ Trinomials where a is not 1 Factor each completely. 4x2 +16x 3. : not a polynomial since there is a variable on the radical. 3) p 2 − 4 p − 45 4) b 2 − 8b + 12. This page will try to solve a quadratic equation by factoring it first. As understood, success does not recommend that you have. M Worksheet by Kuta Software LLC Kuta Software - Infinite Algebra 1 Name_____ Factoring Trinomials (a = 1) Date_____ Period____. Factoring polynomials is the reverse procedure of multiplication of factors of polynomials. Factoring and multiplying polynomials - both processes are inverses of each other. ax 2 + bx + c = 0. Some of the worksheets displayed are Factoring trinomials a 1 date period, Factoring trinomial squares with leading coefficient, Factoring trinomials a 1 date period, Wksht 27, Factoring trinomial squares with leading coefficient of 1, Factoring quadratic. Factoring Perfect Square Trinomials - Displaying top 8 worksheets found for this concept. This is most helpful in solving Solve and Expand problems or anytime in Factor problems to ensure tiles are properly removed. 7) 9k2 - 25 8) 4n2 - 1 9) 25 - n4 10) 16m6 - 9 SAMPLE: Factor each completely. Unit 7 Polynomials And Factoring Homework 1 Classifying Poly - Displaying top 8 worksheets found for this concept. 7 c kMgabdceJ mwNiI tHhR tIMnkfMiYnbiUtse K GAfl Eg jeObMrLaZ z1 P. Factoring Trinomials (a = 1) Task Cards Students will practice factoring trinomials (where a = 1) by working through these 24 task cards. b2 – 5b + 36 YES YES NO NO Yes or No? 7. Check to see if the factored polynomial can be factored further. x2 + 2x + 1 5. Our main purpose is that these Factoring Trinomials Worksheets with Answers pictures gallery can be a guidance for you, bring you more samples and also present you what you need. There are many sections in later chapters where the first step will be to factor a polynomial. Learn with flashcards, games, and more — for free. Binomials: If the polynomial has two terms (it is a binomial), then see if it is the difference of two squares: a 2. So we'll need to add together 2+3+1. **When there are 3 terms, we are factoring. Worksheets. You can remember these two factored forms by remembering that the sign in the binomial is always the same as the sign in the original expression, the first sign in the trinomial is the opposite of the sign in the original expression, and the second sign in the trinomial is always a plus sign. Some trinomials include a greatest common factor. Factoring Trinomial Squares with Leading Coefficient of 1. To begin this lesson, it is important for you to understand the process of multiplying binomials using the FOIL method. You may select which type of polynomials to factor, and whether you want some non-factorable expressions or not. Steps for Factoring where a = 1. This algebra video tutorial shows you how to factor trinomials in the form ax2+bx+c when a, the leading coefficient, is not 1. Factoring When Leading Coefficient Is Not 1. Nonmonic quadratic trinomials are expressions where the leading coeffecient (a) is not equal to 1. $$5x ^{\red 3} + 6x^2 + 9$$. Both of these polynomials have similar factored patterns: A sum of cubes: A difference of cubes: Example 1. Displaying top 8 worksheets found for - Factoring When Leading Coefficient Is Not 1. Worksheet Factoring Trinomials A 1 Grass Fedjp Study Site , Algebraic Identities Of Polynomials A Plus Topper , Factoring Polynomials Practice Review Worksheet Test Prep , 35 best images about Math Integers, Negative Numbers on , Circuit Training. Furthermore, the equation x 3 − 1 = 0, which factors as (x − 1)(x 2 + x + 1 = 0), has only one real root since the quadratic x 2 + x + 1 = 0 has no solutions. Worksheet by Kuta Software LLC Algebra 1 - Mr. 4x2-10xy+25y2 The square root of the first term is 2x. Factoring Trinomials Worksheet with Answer Key Worksheet: Factoring Trinomials (a=1) Write each trinomial in factored form (as the product of two binomials). ) Trial and Error: used when a and c are prime numbers. leading coefficient negative positive prime 1. 5x 40x factoring quadratic expressions x 7x 12 x 10x 25. Factoring Trinomials 1 (+) Factor each completely. You can think of this as unFOILing the trinomial. Factoring Sum/Differences of Cubes Worksheets Mixture of Both Types Elementary Algebra Skill. One problem is not factorable, one is a perfect square, and seven problems require students to factor out a common integer prior to factoring the trinomial. 3x2 – 20x +12 = 0 3. Perfect Square Trinomial 1. Factoring Trinomials where a = 1 Trinomials =(binomial) (binomial) Hint:You want the trinomial to be in descending order with the leading coefficient positive. t ] fAulWlu TrKisgohdtGs[ arxeWsfe\rxvme_dv. Factor completely. Factoring and multiplying polynomials - both processes are inverses of each other. Best 14 Best Images of Kuta Software Factoring Trinomials Worksheet Answer Key Kuta Software. Some of the worksheets displayed are Factoring trinomials a 1 date period, Factoring trinomials a 1 date period, Factoring polynomials, Factoring practice, Foil and factoring trinomials period, Factoring polynomials gcf and quadratic expressions, Factoring polynomials, 5 math 51 factoring trimonials. Title: Factoring Trinomials Using the Grouping Method. H Worksheet by Kuta Software LLC Answers to Worksheet: Factoring Trinomials (a=1) 1) (. (All are factorable. Factor_Trinomials_A_is_Not_1. Trinomial : 2x5. Factoring Trinomials A 1 Worksheet | akademiexcel. For example 20 = (2)(2)(5) and 30 = (2)(3)(5). • Multiply a*c. Scroll down the page for more examples and solutions on how to factor trinomials. Factoring Polynomials by Grouping: Slopes of Perpendicular Lines: Linear Equations: Roots - Radicals 1: Graph of a Line: Sum of the Roots of a Quadratic: Writing Linear Equations Using Slope and Point: Factoring Trinomials with Leading Coefficient 1: Writing Linear Equations Using Slope and Point: Simplifying Expressions with Negative Exponents. : not a polynomial since there is a variable on the radical. Factoring Trinomials (a = 1) Date Period - Kuta Software LLC ©1 t2t0 w1v2 Y PKOuct 4aN IS po 9fbt ywGaZr 2eh 3L DLNCR. Factoring Trinomials (Leading Coefficient Not Equal To 1) Use this method to factor a trinomial in the form of when “a” does not equal 1. 5 Lesson 3: Finding Factors, Sums, and Differences pg. Find two numbers that multiply to give c and add to give b. trinomial (three terms) and figure out what two binomials could have led to this when multiplied. While the goal of the lesson is for students to be able to factor quadratic trinomials without devices such as circle. a2 + 2ab + b2. Factoring a sum/difference of cubes Factoring by grouping Factoring quadratic form Factoring using all techniques Factors and Zeros The Remainder Theorem Irrational and Imaginary Root Theorems Descartes' Rule of Signs More on factors, zeros, and dividing The Rational Root Theorem Polynomial equations Basic shape of graphs of polynomials. One problem is not factorable, one is a perfect square, and seven problems require students to factor out a common integer prior to factoring the trinomial. Download Free Factoring Quadratic Trinomials Worksheet AnswersAs this Factoring Trinomials Worksheets With Answers, it ends happening innate one of the favored books Factoring Trinomials Worksheets With Answers collections that we have. Factors should be exact. leading coefficient negative positive prime 1. Unit 7 Polynomials And Factoring Homework 1 Classifying Poly - Displaying top 8 worksheets found for this concept. What we are being asked to do is find two binomials which multiply to give the original expression. The trinomials on the left have the same constants 1, −3, −10 but different arguments. Factoring Polynomials Example Factor the polynomial x2 – 6x + 10. 4 (& up) 4-term (& up) none. We start by letting A=2 and C=1, since. Factoring When Leading Coefficient Is Not 1. The following diagrams show how to factor trinomials where the leading coefficient is 1 (a = 1). Factor the common factor out of each expression. Before factoring a polynomial equation, it is better to rewrite it in descending order. Some of the worksheets for this concept are Classifying polynomials, Unit 7 polynomials, Factoring polynomials, Polynomials notes 1, Polynomials classifying polynomials, Adding and subtracting polynomials date period, Dividing polynomials, Basic polynomial operations date period. They are made so that the cards form a loop of question, answer, new question, answer, new question, etc. Just before preaching about Factoring Trinomials A 1 Worksheet, you need to know that Education and learning is usually the key to a greater next week, plus finding out does not only quit as soon as the university bell rings. WORKSHEET 1 M FACTORING POLYNOMIALS''Factoring Trinomials Practice MathBitsNotebook April 18th, 2018 - MathBitsNotebook Algebra 1 CCSS Lessons and Factoring Trinomials Practice The cartoon people may or may not be helpful Check each answer when'. The first step is to factor out any common factor, if any. 4 (& up) 4-term (& up) none. In order to factor trinomials, you'll have to work to find two numbers that will multiply to equal the "c" from the quadratic form above, and also add up to equal "b". 9 - 4x 2 = 3 2 - (2x) 2 = (3 - 2x)(3 + 2x) As a practice, multiply (3 - 2x)(3 + 2x) to obtain the given expression. The worksheet really should be short, crisp, simple child friendly. Worksheet Factoring Trinomials A 1 Grass Fedjp Study Site , Algebraic Identities Of Polynomials A Plus Topper , Factoring Polynomials Practice Review Worksheet Test Prep , 35 best images about Math Integers, Negative Numbers on , Circuit Training. com supplies great facts on Trinomial Factoring Calculator, subtracting fractions and rational numbers and other math subject areas. Factor each completely. Some of the worksheets for this concept are Factoring trinomials a 1 date period, Factoring trinomial squares with leading coefficient, Factoring trinomials a 1 date period, Wksht 27, Factoring trinomial squares with leading coefficient of 1, Factoring quadratic. Factor trinomials of the form x^2 + bx + c, where the constant term (c) is positive. In this factoring trinomials worksheet, students factor twenty-six trinomial expressions. where x is the variable and a, b & c are constants. G K uAvIrla Sr1iWg2hltysp TrSéGsGe5ur5vye5dI. x2 +4x 12 5. We could use the Quadratic Formula to find the factors. net provides vital answers on factoring trinomials worksheet a is not 1, course syllabus for intermediate algebra and addition and other algebra topics. b2 – 5b + 36 YES YES NO NO Yes or No? 7. com You need to understand how to project cash flow. 8x2 15x+2 12. Book of Gina Wilson Factoring Trinomials Worksheet Answers, as an amazing reference becomes what you need to get. Author: Sharareh Masooman. 2y2 11y 12 3. Factoring trinomials of the form ax2 bx worksheet answers best c. Factor Completely. This lesson will concentrate on the second step of factoring:Factoring Trinomials. Improve your math knowledge with free questions in "Factor polynomials" and thousands of other math skills. WORKSHEET 1 M FACTORING POLYNOMIALS''Factoring Trinomials Practice MathBitsNotebook April 18th, 2018 - MathBitsNotebook Algebra 1 CCSS Lessons and Factoring Trinomials Practice The cartoon people may or may not be helpful Check each answer when'. Any time you have to have advice on factoring as well as percents, Algbera. If the leading coefficient is 1, as it is here, the process is simple. p v dMnaMdfev lw TiSt1h t HIbnZf difngikt le O sAOl1g fe Gb8r6a e Q1Y. Factor x 6 – y 6. 5, – 7 – 2 So x2 – 2x – 35 = (x + 5)(x – 7). 11) 50v4 - 32v2 12) 48n6 - 27n2 Factor each completely. Elementary Algebra Skill Factoring Trinomial Squares with Leading Coefficient Different from 1 Factor each completely. Factoring Trinomials Worksheet Doodle ing Math by from factoring trinomials a 1 worksheet , image source: www. This is not merely an esoteric exercise. Name:&&_____&&Date:&&_____&&Block:&&_____ Worksheet)Factoring)Trinomials Factor)the)followingcompletely. Type of Resume and sample, factoring polynomials color by number worksheet answers. Factoring Trinomials A 1 Worksheet Answers Example Factoring from factoring trinomials worksheet with answer key , source:bonanycats. Some of the worksheets displayed are Factoring trinomials a 1 date period, Factoring trinomials a 1 date period, Factoring polynomials, Factoring practice, Foil and. Related posts: Ocean Zones. Factoring Trinomials (a = 1) Task Cards Students will practice factoring trinomials (where a = 1) by working through these 24 task cards. ©i bKkubtda7 zS2oCfftwwCa3rBeb yLrLxCb. I’ll refer to the general equation below while describing the process of factoring polynomials: When given an equation and asked to factor it, you first need to choose A and C so that. Factoring Trinomials 1 (+) Factor each completely. Step 1: Make sure that the trinomial is written in the correct order; the trinomial must be written in descending order from highest power to lowest power. d2 + 50d + 225 4. Factoring When Leading Coefficient Is Not 1. Factoring Quadratic Expressions Color Worksheet #2. 8 Worksheet: Factoring Perfect Square Trinomials Factor each completely. long division or factoring to nd the others. In order to factor trinomials, you'll have to work to find two numbers that will multiply to equal the "c" from the quadratic form above, and also add up to equal "b". Each group got one pre-cut sheet of cards. Sometimes one or more roots of a polynomial are known, perhaps having been found using the rational root theorem. Download Worksheet: Factoring Trinomials (a=1) book pdf free download link or read online here in PDF. The general form of a quadratic equation is. Some of the worksheets displayed are Factoring trinomials a 1 date period, Factoring trinomials a 1 date period, Factoring polynomials, Factoring practice, Foil and factoring trinomials period, Factoring polynomials gcf and quadratic expressions, Factoring polynomials, 5 math 51 factoring trimonials. Some of the worksheets displayed are Factoring trinomials a 1 date period, Factoring trinomials a 1 date period, Factoring practice, Wksht 27, Foil and factoring trinomials period, Factoring polynomials gcf and quadratic expressions, Factoring polynomials, Factoring quadratic expressions. Factor x 3 + 125. 2 Factoring Trinomials of the Type x2 + bx + c Learning Objectives A Factor trinomials of the type x2 + bx + c by examining the constant term c. y t VMHaQdneY Pw3iNtuh0 4Ijn9fIignzi4tIel JAClUgceqbGrpar 018. So, if you can’t factor the polynomial then you won’t be able to even start the problem let alone finish it. What's for is this book?. When I did factoring with the Algebra 2 kids, we did stations (much to their dismay, they're not big fans of walking around the classrooms). Displaying top 8 worksheets found for - Factoring When Leading Coefficient Is Not 1. H Worksheet by Kuta Software LLC Answers to Worksheet: Factoring Trinomials (a=1) 1) (. You may select which type of polynomials to factor whether you want some non factorable expressions or not and the coefficient of the first term. 3 3-term trinomial. 7 Lesson 5: Factoring Binomials that are the Difference of Two Perfect. Factoring Trinomials in the form x 2 + bx + c. We now know that the product of two binomials of the form (x + d) and. Prime factors trees are also introduced. 2x(3x2 5) 6x3 10x. 2 we talked about factors of whole numbers. Factoring - Trinomials where a 1 Objective: Factor trinomials using the ac method when the coefficient of x2 is not one. Some of the worksheets for this concept are Classifying polynomials, Unit 7 polynomials, Factoring polynomials, Polynomials notes 1, Polynomials classifying polynomials, Adding and subtracting polynomials date period, Dividing polynomials, Basic polynomial operations date period. ©Q Q2i0N1e3g EKju gtSa K JS5o5fLt Pwta PrEe o f LLf Cl. M Worksheet by Kuta Software LLC Kuta Software - Infinite Algebra 1 Name_____ Factoring Trinomials (a = 1) Date_____ Period____. Do not try to fill in the box with a number, because we do not know what the numbers would be, just work with the sign. Should you seek help on multiplying and dividing rational or maybe inverse, Algebra1help. Showing top 8 worksheets in the category - Trinomials Factoring. This algebra video tutorial shows you how to factor trinomials in the form ax2+bx+c when a, the leading coefficient, is not 1. One problem is not factorable, one is a perfect square, and seven problems require students to factor out a common integer prior to factoring the. MAT 0024 Ch 13 Factoring Review Worksheet Instructor: C. Download Free Factoring Quadratic Trinomials Worksheet AnswersAs this Factoring Trinomials Worksheets With Answers, it ends happening innate one of the favored books Factoring Trinomials Worksheets With Answers collections that we have. U Worksheet by Kuta Software LLC Kuta Software - Infinite Algebra 1 Name_____ Factoring Trinomials (a = 1) Date_____ Period____ Factor each completely. 1) 7 x2 + 10xy + 3y2 2) 3x2 − 10xy − 48y2 3) 2x2 − 11xy + 9y2 4) 5x2 − 27xy + 28y2. Factoring Trinomials in the form x 2 + bx + c. For problems 1 – 4 factor out the greatest common factor from each polynomial. worksheet answers along with it is not directly done, you could admit even more something like this life, as regards the world. Factoring Perfect Square Trinomials Worksheet 1. 4 (& up) 4-term (& up) none. 30a4b2 + 45 ab3 3. long division or factoring to nd the others. M Worksheet by Kuta Software LLC. Now, can you do the factoring? Factoringtrinomial a2 + 2ab + b2which is a perfect square gives the following famous formula: (a + b) 2 = a2 + 2ab + b2. 9) p 2 + 4 p + 4 10) x 2 − 6x − 27. This method works well for factoring small integers, but is inefficient for larger integers. Math Worksheets Examples, solutions, videos, worksheets, and activities to help Algebra and Grade 9 students learn about factoring standard trinomials for a > 1. A zero pair is a pair of tiles that sums to 0, such as a yellow 1 tile and a red. Factoring Trinomials when a = 1 Worksheet by Mr Greenlaw from factoring trinomials worksheet answer key , image source: www. In this worksheet we will factor polynomials. Kuta Software - Infinite Algebra 1 Name_____ Factoring Trinomials (a = 1) Date_____ Period____ Factor each completely. Factor x 3 + 125. Math Worksheets Factoring Polynomials Worksheet mathematics at home printable games for kindergarten multiplying fractions by fractions printable math worksheets year 2 3 digit addition and subtraction games The most important thing about these math worksheets is that they are used for tutoring and not for the main course studies. Algebra 2 Factoring numbers Polynomials Factoring monomials. Usually, there’s just one or two questions about polynomials sa exam. ))Lookfor)aGCFfirst. The procedure to use the factoring trinomials calculator is as follows: Step 1: Enter the trinomial function in the input field Step 2: Now click the button "Factor" to get the result Step 3: Finally, the factors of a trinomial will be displayed in the new window. Factoring trinomials (aka quadratic functions) is one of the bedrocks of Algebra. So, if you can’t factor the polynomial then you won’t be able to even start the problem let alone finish it. Learn with flashcards, games, and more — for free. To factor these trinomials, we must again find roots x 1 and x 2 such that ax 2 +bx+c = a(x - x 1)(x - x 2). First, we pull out the GCF, if possible. com includes practical strategies on factoring trinomial worksheet, subtracting and absolute and other algebra topics. By solving the problems, students get the answer to the joke: "why was the cat so small?" Read more. For example 20 = (2)(2)(5) and 30 = (2)(3)(5). Step by step I understood all the topics, and then I was able to solve the hardest of the exercises without the program. chapter 19. Math Worksheets Examples, solutions, videos, worksheets, and activities to help Algebra and Grade 9 students learn about factoring standard trinomials for a > 1. WORKSHEET 1 M FACTORING POLYNOMIALS''Factoring Trinomials Practice MathBitsNotebook April 18th, 2018 - MathBitsNotebook Algebra 1 CCSS Lessons and Factoring Trinomials Practice The cartoon people may or may not be helpful Check each answer when'. We start by letting A=2 and C=1, since. Factoring Complex Trinomials Worksheet? Below is a number of search phrases that visitors used recently to get to our algebra help pages. The first step is to factor out any common factor, if any. Factoring When Leading Coefficient Is Not 1. Factoring Trinomials 3 – Negative B and C – Factoring Trinomials 4 – Negative B, Positive C – Factoring Trinomials 5 – Factor Out – Factoring Trinomials 6 – Difference of Two Squares – Factoring Trinomials 7 – Binomial Squares – Factoring Trinomials 8 – A > 1 and C = 1 – Factoring Trinomials 9 – Factor Out and Factor –. Luna Republic of the Philippines | DEPARTMENT OF EDUCATION Region III | Division of City Schools | West District Tamarind St. Check your work by multiplying your factors to get the original polynomial. One problem is not factorable, one is a perfect square, and seven problems require students to factor out a common integer prior to factoring the trinomial. FACTORING TRINOMIALS WHOSE LEADING COEFFICIENT IS NOT 1 In this section we will learn methods for factoring trinomials of the form ax bx c a or2 ++ ≠;01: i. ©Q Q2i0N1e3g EKju gtSa K JS5o5fLt Pwta PrEe o f LLf Cl. Worksheet by Kuta Software LLC Algebra 1 - Mr. Some of the worksheets displayed are factoring quadratic expressions factoring trinomials a 1 date period factoring practice cp algebra 2 unit 2 1 factoring and solving quadratics factoring polynomials gcf and quadratic expressions factoring and solving quadratic equations work factoring polynomials solve each equation with the quadratic. ©1 t2t0 w1v2 Y PKOuct 4aN IS po 9fbt ywGaZr 2eh 3L DLNCR. Factor 8 x 3 – 27. Usually, there’s just one or two questions about polynomials sa exam. Showing top 8 worksheets in the category - Factoring When Leading Coefficient Is Not 1. Name_____ (leading coefficient is one) Factor each trinomial. Factoring Polynomials by Grouping: Slopes of Perpendicular Lines: Linear Equations: Roots - Radicals 1: Graph of a Line: Sum of the Roots of a Quadratic: Writing Linear Equations Using Slope and Point: Factoring Trinomials with Leading Coefficient 1: Writing Linear Equations Using Slope and Point: Simplifying Expressions with Negative Exponents. Factoring by GCF Lesson 8-2 Learning Target: I can factor polynomials by using the greatest common factor. This is why you remain in the best website to look the incredible book to have. Mathematics 8 Worksheet Algebra 1st Quarter Mr. When you multiply two binomials together, you use the FOIL method, multiplying the F irst, then the O uter, then the I nner, and finally the L ast terms of the two binomials into a trinomial. Some of the worksheets for this concept are Classifying polynomials, Unit 7 polynomials, Factoring polynomials, Polynomials notes 1, Polynomials classifying polynomials, Adding and subtracting polynomials date period, Dividing polynomials, Basic polynomial operations date period. Math Worksheets Factoring Polynomials Worksheet mathematics at home printable games for kindergarten multiplying fractions by fractions printable math worksheets year 2 3 digit addition and subtraction games The most important thing about these math worksheets is that they are used for tutoring and not for the main course studies. G I xMAaxd Xev jw ki ft Jh H XIVn1fUisnki Ot0eN 2AZlhgre ib FrKaP l1 5. These worksheets ask students to find all the factors of different sized numbers. Note that the coefficient of the x^2 term is 1. To use this method all that we do is look at all the terms and determine if there is a factor that is in common to all the terms. Come to Sofsource. Factoring Perfect Square Trinomials - Displaying top 8 worksheets found for this concept. You may select which type of polynomials to factor, and whether you want some non-factorable expressions or not. Scroll down the page for more examples and solutions of factoring trinomials. Worksheets. 4x2 +16x 3. If we reverse the problem, ) ( , we say we have factored into. Steps: Example: Multiply leading coefficient and constant. Gallery of 50 Factoring Trinomials Worksheet Answer Key. If there are 4 or more terms, try factoring by grouping. ©1 t2t0 w1v2 Y PKOuct 4aN IS po 9fbt ywGaZr 2eh 3L DLNCR. Some of the worksheets for this concept are Classifying polynomials, Unit 7 polynomials, Factoring polynomials, Polynomials notes 1, Polynomials classifying polynomials, Adding and subtracting polynomials date period, Dividing polynomials, Basic polynomial operations date period. Greatest Common Factor (GCF) Find the GCF of the numbers. Remove common factors if possible. com INFO POLYNOMIALS WORKSHEET GRADE 9 FREE ZIP DOWNLOAD PRINT. In this worksheet we will factor polynomials. If a = 1, look for factors of c that will add up to b. Chapter 13 POLYNOMIALS: FACTORING 13. Factor Completely. Class: Math 100. First you must factor out a common monomial factor, then factor the the remaining trinomial. 15 Review More Practice Factoring with Pizzazz worksheets pg. This worksheet will give students practice with this skill. In other words, our answer must be to properly multiply out to the trinomial given in this question. What is Meant by Factoring Trinomials? In Mathematics, trinomial is a. Practice the worksheet on factoring trinomials to know how to factorize the quadratic expression of the form ax 2 + bx + c. Factoring When Leading Coefficient Is Not 1. Both ways require “trial and error” to find the right binomial factors. For example, x 2 + 3x + 2 = (x + 1)(x + 2). The general form of a quadratic equation is. com is certainly the right place to have a look at!. All quizzes. (All are factorable. 3x 2x 5x 7. First, notice that x 6 – y 6 is both a difference of squares and a difference of cubes. long division or factoring to nd the others. Factoring trinomials (specifically quadratics here) with a leading coefficient not equal to 1. Worksheet: Factoring Trinomials with Leading Coefficient > 1 Factoring Trinomials with Leading Coefficient > 1 Worksheet: pin. leading coefficient negative positive prime 1. Factoring Trinomials (a = 1) Date Period - Kuta Software LLC ©1 t2t0 w1v2 Y PKOuct 4aN IS po 9fbt ywGaZr 2eh 3L DLNCR. Factor 2 x 3 + 128 y 3. ©1 t2t0 w1v2 Y PKOuct 4aN IS po 9fbt ywGaZr 2eh 3L DLNCR. Look at the number of terms: 2 Terms: Look for the Difference of 2 Squares ; 3 Terms: Factor the Trinomial ; 4 Terms: Factor by Grouping ; 3. The cartoon people may, or may not, be helpful!! Check each answer when finished. If you're behind a web filter, please make sure that the domains *. This is an introduction to the trial factors method. Some of the worksheets for this concept are Factoring trinomials a 1 date period, Factoring trinomial squares with leading coefficient, Factoring trinomials a 1 date period, Wksht 27, Factoring trinomial squares with leading coefficient of 1, Factoring. Download Worksheet: Factoring Trinomials (a=1) book pdf free download link or read online here in PDF. Factor x 6 – y 6. Software for math teachers that creates exactly the worksheets you need in a matter of minutes. (Factoring Trinomials Of The Form , and ) Given a quadratic expression with the terms , we are often asked to factor. *Factor out the GCF of 2 y We are not finished, we can still factor the trinomial. The square root of the last term 5y. Book of Gina Wilson Factoring Trinomials Worksheet Answers, as an amazing reference becomes what you need to get. Name_____ (leading coefficient is one) Factor each trinomial. -1-CLASS EXAMPLES: Factor each completely. Factoring Trinomials Worksheet Doodle ing Math by from factoring trinomials a 1 worksheet , image source: www. Some of the worksheets for this concept are Classifying polynomials, Unit 7 polynomials, Factoring polynomials, Polynomials notes 1, Polynomials classifying polynomials, Adding and subtracting polynomials date period, Dividing polynomials, Basic polynomial operations date period. v Y gAhlcll XrBiug GhWtdsd Frle Zsve pr7v Qexd C. 50x2 372 9. 25 scaffolded questions that start relatively easy and end with some real challenges. Do not try to fill in the box with a number, because we do not know what the numbers would be, just work with the sign. Steps for Factoring where a = 1. Greatest common factor gcf find the gcf of the numbers. Geometry Worksheets. WORKSHEET 1 M FACTORING POLYNOMIALS''Factoring Trinomials Practice MathBitsNotebook April 18th, 2018 - MathBitsNotebook Algebra 1 CCSS Lessons and Factoring Trinomials Practice The cartoon people may or may not be helpful Check each answer when'. Not all polynomials are factorable. Remember to always look at the problem to make sure there is nothing else you can do. Some of the worksheets for this concept are Factoring trinomials a 1 date period, Factoring trinomial squares with leading coefficient, Factoring trinomials a 1 date period, Wksht 27, Factoring trinomial squares with leading coefficient of 1, Factoring quadratic. We know, in order to factorize the expression ax 2 + bx + c, we have to find two numbers p and q, such that p + q = b and p × q = ac. Part 1 of the lesson consists of “circle puzzles” as a preparation for the actual factoring lesson in Part 2. Scroll down the page for more examples and solutions on how to factor trinomials. p v dMnaMdfev lw TiSt1h t HIbnZf difngikt le O sAOl1g fe Gb8r6a e Q1Y. Factoring Trinomials Worksheet free download; Identifying Common Factors of Factoring trinomials worksheet. Factoring Quadratic Expressions Color Worksheet #2. Just before referring to Factoring Trinomials Worksheet Algebra 2, you need to be aware that Knowledge is definitely our own crucial for an even better down the road, as well as finding out does not only avoid as soon as the institution bell rings. Try: Case 2: Lead coefficient not equal to one, Trial method. That staying stated, many of us give you a number of straightforward however helpful reports and web templates built made for almost any educative purpose. A quiz and full answer keys are also provided. Demonstrate how to factor using algebra tiles and the attached Teacher Resource for Factoring Polynomials. ©Q Q2i0N1e3g EKju gtSa K JS5o5fLt Pwta PrEe o f LLf Cl. Factoring trinomials (aka quadratic functions) is one of the bedrocks of Algebra. net is the ideal destination to head to!. We will see how to solve non-factorable quadratics using the square root method or the quadratic formula. In fact, this is not even a trinomial because there are 2 terms. Keywords/Tags: Factor, factoring trinomials, grouping method, ac method, splitting middle term. The two versions on one card do not match up. 30a4b2 + 45 ab3 3. y t VMHaQdneY Pw3iNtuh0 4Ijn9fIignzi4tIel JAClUgceqbGrpar 018. Keep in mind, you may only in a position to delete the worksheet and you also aren't capable to restore it any longer, there is not any kind of Undo function for your removal of worksheet, so that you can wish to be cautious upon what you intend to. So the solutions must be x=-2 and x=-3. $$5x ^{\red 3} + 6x^2 + 9$$. net provides vital answers on factoring trinomials worksheet a is not 1, course syllabus for intermediate algebra and addition and other algebra topics. Add to give b. Use these two numbers to factor the trinomial. Some of the worksheets for this concept are Work factoring perfect square trinomials date period, Factoring perfect square trinomials, Perfect squares and factoring examples, Factoring practice, Work factoring perfect square trinomials date period, Algebra 1, Cp algebra 2 unit 2 1 factoring and solving. v Y gAhlcll XrBiug GhWtdsd Frle Zsve pr7v Qexd C. Upon completing this section you should be able to: Mentally multiply two binomials. 1) b2 + 8b + 7 (b + 7)(b + 1). In the stations, the kids did this exact process with a premade 4 square foldable (I created a word doc with a 2x2 table prelabeled with the different methods and spots for steps/examples/notes). We start by letting A=2 and C=1, since. Your most common factoring task, aside from greatest common factoring, is changing a quadratic trinomial into the product of two linear binomials. Showing top 8 worksheets in the category - Trinomials Factoring. Trinomials with a leading coefficient of one Example 1. To factor a trinomial with a leading coefficient which is not 1, use trial and error to find the binomial factors. 2x(3x2 5) 6x3 10x. If you’re in school, your teacher will often have you complete as much as possible up to the point where it cannot be factored. Factoring Polynomials Any natural number that is greater than 1 can be factored into a product of prime numbers. The following diagrams show how to factor trinomials where the leading coefficient is 1 (a = 1). 1) 7 m2 + 6m − 1 2) 3k2 − 10k + 7 3) 5x2 − 36x − 81 4) 2x2 − 9x − 81. When factoring in general this will also be the first thing that we should try as it will often simplify the problem. 3 Factoring Trinomials with Leading Coefficients of 1 Factoring a trinomial of form $$x^2+bx+c\text{,}$$ where $$b$$ and $$c$$ are integers, is essentially the reversal of a FOIL process. If there are 4 or more terms, try factoring by grouping. Factoring Trinomials Case 1: lead coefficient is one. ©r G2b0X1 U3S 0K euEtQa P KSRorf WtCwmaKrxeL PLQLRC9. • Place the factors as coefficients to the x-variable in the remaining boxes. com supplies great facts on Trinomial Factoring Calculator, subtracting fractions and rational numbers and other math subject areas. The following diagram shows how to factor trinomials with no guessing. ©3 52n011A2j DKHuntwae XSkoBfbtRwMacrHeV OLlLCCX. leading coefficient negative positive prime 1. After having gone through the stuff given above, we hope that the students would have understood, "Basic Concepts in Polynomials Worksheet" Apart from the stuff given in this section, if you need any other stuff in math, please use our google custom search here. Check to see if the factored polynomial can be factored further. Worksheets. 4) 27 + 8x. GINA WILSON FACTORING TRINOMIALS WORKSHEET ANSWERS PDF DOWNLOAD: GINA WILSON FACTORING TRINOMIALS WORKSHEET ANSWERS PDF New updated! The latest book from a very famous author finally comes out. Download Free Factoring Quadratic Trinomials Worksheet AnswersAs this Factoring Trinomials Worksheets With Answers, it ends happening innate one of the favored books Factoring Trinomials Worksheets With Answers collections that we have. L-2-Worksheet by Kuta Software program LLC Kuta Software program - Infinite Algebra 1 Title Factoring Trinomials (á gt; 1) Date Time period Aspect each completely. B N UMOaFd qeu Iw DiDtMhh YIen UfOisndi 6t Ve0 uA xl Hgae Gber 8aU Z2Q. Math Worksheets Examples, solutions, videos, worksheets, and activities to help Algebra and Grade 9 students learn about factoring standard trinomials for a > 1. When factoring trinomials we used the ac method to split the middle term and then factor by grouping. They are not flash cards. Pay particular attention to any factor that is greater than a first degree. com supplies great facts on Trinomial Factoring Calculator, subtracting fractions and rational numbers and other math subject areas. Factoring Trinomials 1 (+) Factor each completely. 2x 28x 7 2 4. Needed: 2 x 2 tiles 3 rectangular x tiles 1 + tile. G K uAvIrla Sr1iWg2hltysp TrSéGsGe5ur5vye5dI. The procedure to use the factoring trinomials calculator is as follows: Step 1: Enter the trinomial function in the input field Step 2: Now click the button “Factor” to get the result Step 3: Finally, the factors of a trinomial will be displayed in the new window. Demonstrate how to factor using algebra tiles and the attached Teacher Resource for Factoring Polynomials. Check my multiplying. Factoring quadratic trinomials. Find a quiz. 1) 7 m2 + 6m − 1 2) 3k2 − 10k + 7 3) 5x2 − 36x − 81 4) 2x2 − 9x − 81. 3 x6 + 2 x3 − 1 = (3 x3 − 1) (x3 + 1). So the solutions must be x=-2 and x=-3. teacherspayteachers. students will solve problems related to polynomial functions answer multiple choice questions topics include factoring polynomials puzzle from just mathematics on teachersnotebook com 2 pages complex number imaginary maze review worksheet teacherspayteachers great practice trinomials with a coefficient for my algebra students! this is 7 bundle of popular activities! love the fun 1000 images. Remove Zero Pairs allows only zero pairs to be removed from the workspace. FACTORING TRINOMIALS WHOSE LEADING COEFFICIENT IS NOT 1 In this section we will learn methods for factoring trinomials of the form ax bx c a or2 ++ ≠;01: i. Use these zeros to help you factor this function. April 4: Trinomial Factoring Practice Worksheet April 3: Factoring Trinomials of the Form ax 2 + bx + c where a is not equal to 1 pp. : not a polynomial since there is a variable on the radical. All quizzes. Factoring Trinomials Worksheet With Answer Key in a learning medium may be used to check pupils skills and understanding by addressing questions. By solving the problems, students get the answer to the joke: "why was the cat so small?" Read more. 5th Grade Math 6th Grade Math Pre-Algebra Algebra 1 Geometry Algebra 2 College Students learn that a trinomial in the form ax^2 + bx + c, such as x^2 + 7x + 10, can be factored as the product of two binomials, in this case (x + 5)(x + 2). If the coefficient of the x 2 term is 1, then x 2 + bx + c = (x + n)(x + m), where n and m i. Harder Trinomials - Undoing FOIL 1 - Cool Math has free online cool math lessons, cool math games and fun math activities. When factoring trinomials we used the ac method to split the middle term and then factor by grouping. Scroll down the page for more examples and solutions of factoring trinomials. When factoring in general this will also be the first thing that we should try as it will often simplify the problem. , Malabanias, Angeles City School Year 2018-2019 NAME: _____ SCORE: _____ GRADE & SECTION: _____ DATE: _____ 1st Quarter Worksheet #9. Monomial : 5, 6x, x and etc. 13) 100m2 - 4 14) 4a2 - 16. In the event that you need to have guidance on solving linear equations or formula, Mathpoint. When I did factoring with the Algebra 2 kids, we did stations (much to their dismay, they're not big fans of walking around the classrooms). Algebra worksheets to improve student performance through fun riddles, activities and games. 13) x 2 − 8x − 9 14) p 2. 2 Factoring Trinomials of the Type x2 + bx + c Learning Objectives A Factor trinomials of the type x2 + bx + c by examining the constant term c. worksheet answers along with it is not directly done, you could admit even more something like this life, as regards the world. Q y hMLaGdUeK mwqi ltsh4 iI knXf1i on 5i8t Oee EA lcg xeeb wrDa1 J1 N. You must choose the format of your resume depending on your work and personal background. Check by Multiplying. 11) r 2 − r + 4 12) x 2 + x − 72. FACTORING TRINOMIALS WHOSE LEADING COEFFICIENT IS NOT 1 In this section we will learn methods for factoring trinomials of the form ax bx c a or2 ++ ≠;01: i. 25 scaffolded questions that start relatively easy and end with some real challenges. To use this method all that we do is look at all the terms and determine if there is a factor that is in common to all the terms. So the solutions must be x=-2 and x=-3. Factor 2x 2 + 3x + 1 Algebra tiles are great for factoring quadratic trinomials where the A value is not 1. Connection to the internet is not require. $$x^{\red 3} + 2x + 1$$ this is not a quadratic trinomial because there is an exponent that is $$\red { \text{ greater than 2} }$$ $$2x + 4$$ this is not a quadratic trinomial because there is not exponent of 2. 3x2 – 20x +12 = 0 3. If so, factor out the GCF. 358 - 364 April 2: Factoring Trinomials of the Form ax 2 + bx + c where a = 1 pp. While the goal of the lesson is for students to be able to factor quadratic trinomials without devices such as circle.
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http://maps.thefullwiki.org/Cross_product | # Cross product: Map
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In mathematics, the cross product is a binary operation on two vector in a three-dimensional Euclidean space that results in another vector which is perpendicular to the plane containing the two input vectors. The algebra defined by the cross product is neither commutative nor associative. It contrasts with the dot product which produces a scalar result. In many engineering and physics problems, it is desirable to be able to construct a perpendicular vector from two existing vectors, and the cross product provides a means for doing so. The cross product is also useful as a measure of "perpendicularness"—the magnitude of the cross product of two vectors is equal to the product of their magnitudes if they are perpendicular and scales down to zero when they are parallel. The cross product is also known as the vector product, or Gibbs vector product.
The cross product is only defined in three or seven dimensions. Like the dot product, it depends on the metric of Euclidean space. Unlike the dot product, it also depends on the choice of orientation or "handedness". Certain features of the cross product can be generalized to other situations. For arbitrary choices of orientation, the cross product must be regarded not as a vector, but as a pseudovector. For arbitrary choices of metric, and in arbitrary dimensions, the cross product can be generalized by the exterior product of vectors, defining a two-form instead of a vector.
## Definition
The cross product of two vectors a and b is denoted by In physics, sometimes the notation ab is used, though this is avoided in mathematics to avoid confusion with the exterior product.
In a three-dimensional Euclidean space, with a right-handed coordinate system, a × b is defined as a vector c that is perpendicular to both a and b, with a direction given by the right-hand rule and a magnitude equal to the area of the parallelogram that the vectors span.
The cross product is defined by the formula
\mathbf{a} \times \mathbf{b} = a b \sin \theta \ \mathbf{\hat{n}}
where θ is the measure of the smaller angle between a and b (0° ≤ θ ≤ 180°), a and b are the magnitudes of vectors a and b, and \scriptstyle\mathbf{\hat{n}} is a unit vector perpendicular to the plane containing a and b in the direction given by the right-hand rule as illustrated. If the vectors a and b are collinear (i.e., the angle θ between them is either 0° or 180°), by the above formula, the cross product of a and b is the zero vector 0.
The direction of the vector \scriptstyle\mathbf{\hat{n}} is given by the right-hand rule, where one simply points the forefinger of the right hand in the direction of a and the middle finger in the direction of b. Then, the vector \scriptstyle\mathbf{\hat{n}} is coming out of the thumb (see the picture on the right). Using this rule implies that the cross-product is anti-commutative, i.e., b × a = -(a × b). By pointing the forefinger toward b first, and then pointing the middle finger toward a, the thumb will be forced in the opposite direction, reversing the sign of the product vector.
Using the cross product requires the handedness of the coordinate system to be taken into account (as explicit in the definition above). If a left-handed coordinate system is used, the direction of the vector \scriptstyle\mathbf{\hat{n}} is given by the left-hand rule and points in the opposite direction.
This, however, creates a problem because transforming from one arbitrary reference system to another (e.g., a mirror image transformation from a right-handed to a left-handed coordinate system), should not change the direction of \scriptstyle\mathbf{\hat{n}}. The problem is clarified by realizing that the cross-product of two vectors is not a (true) vector, but rather a pseudovector. See cross product and handedness for more detail.
## Computing the cross product
### Coordinate notation
The unit vectors i, j, and k from the given orthogonal coordinate system satisfy the following equalities:
i × j = k j × k = i k × i = j
Together with the skew-symmetry and bilinearity of the cross product, these three identities are sufficient to determine the cross product of any two vectors. In particular, the following identities are also seen to hold
j × i = −k k × j = −i i × k = −j
i × i = j × j = k × k = 0.
With these rules, the coordinates of the cross product of two vectors can be computed easily, without the need to determine any angles: Let
a = a1i + a2j + a3k = (a1, a2, a3)
and
b = b1i + b2j + b3k = (b1, b2, b3).
The cross product can be calculated by distributive cross-multiplication:
a × b = (a1i + a2j + a3k) × (b1i + b2j + b3k)
a × b = a1i × (b1i + b2j + b3k) + a2j × (b1i + b2j + b3k) + a3k × (b1i + b2j + b3k)
a × b = (a1i × b1i) + (a1i × b2j) + (a1i × b3k) + (a2j × b1i) + (a2j × b2j) + (a2j × b3k) + (a3k × b1i) + (a3k × b2j) + (a3k × b3k).
Since scalar multiplication is commutative with cross multiplication, the right hand side can be regrouped as
a × b = a1b1(i × i) + a1b2(i × j) + a1b3(i × k) + a2b1(j × i) + a2b2(j × j) + a2b3(j × k) + a3b1(k × i) + a3b2(k × j) + a3b3(k × k).
This equation is the sum of nine simple cross products. After all the multiplication is carried out using the basic cross product relationships between i, j, and k defined above,
a × b = a1b1(0) + a1b2(k) + a1b3(−j) + a2b1(−k) + a2b2(0) + a2b3(i) + a3b1(j) + a3b2(−i) + a3b3(0).
This equation can be factored to form
a × b = (a2b3a3b2) i + (a3b1a1b3) j + (a1b2a2b1) k = (a2b3a3b2, a3b1a1b3, a1b2a2b1).
### Matrix notation
The definition of the cross product can also be represented by the determinant of a matrix:
\mathbf{a}\times\mathbf{b}=\det \begin{bmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\a_1 & a_2 & a_3 \\b_1 & b_2 & b_3 \\\end{bmatrix}.
This determinant can be computed using Sarrus' rule. Consider the table\begin{matrix}\mathbf{i} & \mathbf{j} & \mathbf{k} & \mathbf{i} & \mathbf{j} & \mathbf{k} \\a_1 & a_2 & a_3 & a_1 & a_2 & a_3 \\b_1 & b_2 & b_3 & b_1 & b_2 & b_3\end{matrix} From the first three elements on the first row draw three diagonals sloping downward to the right (for example, the first diagonal would contain i, a2, and b3), and from the last three elements on the first row draw three diagonals sloping downward to the left (for example, the first diagonal would contain i, a3, and b2). Then multiply the elements on each of these six diagonals, and negate the last three products. The cross product would be defined by the sum of these products:\mathbf{i}a_2b_3 + \mathbf{j}a_3b_1 + \mathbf{k}a_1b_2 - \mathbf{i}a_3b_2 - \mathbf{j}a_1b_3 - \mathbf{k}a_2b_1.
## Properties
### Geometric meaning
Figure 1: The area of a parallelogram as a cross product
Figure 2: The volume of a parallelepiped using dot and cross-products; dashed lines show the projections of c onto a × b and of a onto b × c, a first step in finding dot-products.
The magnitude of the cross product can be interpreted as the positive area of the parallelogram having a and b as sides (see Figure 1):
A = | \mathbf{a} \times \mathbf{b}| = | \mathbf{a} | | \mathbf{b}| \sin \theta. \,\!
Indeed, one can also compute the volume V of a parallelepiped having a, b and c as sides by using a combination of a cross product and a dot product, called scalar triple product (see Figure 2):
V = |\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c})|.
Figure 2 demonstrates that this volume can be found in two ways, showing geometrically that the identity holds that a "dot" and a "cross" can be interchanged without changing the result. That is:
V =\mathbf{a \times b \cdot c} = \mathbf{a \cdot b \times c} \ .
Because the magnitude of the cross product goes by the sine of the angle between its arguments, the cross product can be thought of as a measure of "perpendicularness" in the same way that the dot product is a measure of "parallelness". Given two unit vectors, their cross product has a magnitude of 1 if the two are perpendicular and a magnitude of zero if the two are parallel.
### Algebraic properties
The cross product is anticommutative,
a × b = −b × a,
a × (b + c) = (a × b) + (a × c),
and compatible with scalar multiplication so that
(r a) × b = a × (r b) = r (a × b).
It is not associative, but satisfies the Jacobi identity:
a × (b × c) + b × (c × a) + c × (a × b) = 0.
It does not obey the cancellation law:
If a × b = a × c and a0 then:
(a × b) − (a × c) = 0 and, by the distributive law above:
a × (bc) = 0
Now, if a is parallel to (bc), then even if a0 it is possible that (bc) ≠ 0 and therefore that bc.
However, if both a · b = a · c and a × b = a × c, then it can be concluded that b = c. Indeed,
a · (b - c) = 0, and
a × (b - c) = 0
so that b - c is both parallel and perpendicular to the non-zero vector a. This is only possible if b - c = 0.
The distributivity, linearity and Jacobi identity show that R3 together with vector addition and cross product forms a Lie algebra. In fact, the Lie algebra is that of the real orthogonal group in 3 dimensions, SO.
Further, two non-zero vectors a and b are parallel if and only if a × b = 0.
It follows from the geometrical definition above that the cross product is invariant under rotations about the axis defined by a×b.
There is also this property relating cross products and the triple product:
(a × b) × (a × c) = (a · (b × c)) a.
The cross product obeys this identity under matrix transformations:
(M\mathbf{a}) \times (M\mathbf{b}) = (\det M) M^{-T}(\mathbf{a} \times \mathbf{b}) \,\!
where \scriptstyle M is a 3 by 3 matrix and \scriptstyle M^{-T} is the transpose of the inverse
The cross product of two vectors in 3-D always lies in the null space of the matrix with the vectors as rows. In other words
\mathbf{a} \times \mathbf{b} \in NS\left(\begin{bmatrix}\mathbf{a} \\ \mathbf{b}\end{bmatrix}\right)
### Triple product expansion
The triple product expansion, also known as Lagrange's formula, is a formula relating the cross product of three vectors (called the vector triple product) with the dot product:
a × (b × c) = b(a · c) − c(a · b).
The mnemonic "BAC minus CAB" is used to remember the order of the vectors in the right hand member. This formula is used in physics to simplify vector calculations. A special case, regarding gradients and useful in vector calculus, is given below.
\begin{align}
\nabla \times (\nabla \times \mathbf{f})& {}= \nabla (\nabla \cdot \mathbf{f} ) - (\nabla \cdot \nabla) \mathbf{f} \\& {}= \mbox{grad }(\mbox{div } \mathbf{f} ) - \Delta \mathbf{f}.\end{align} This is a special case of the more general Laplace-de Rham operator \scriptstyle\Delta = d \delta + \delta d.
The following identity also relates the cross product and the dot product:
|\mathbf{a} \times \mathbf{b}|^2 + |\mathbf{a} \cdot \mathbf{b}|^2 = |\mathbf{a}|^2 |\mathbf{b}|^2.
This is a special case of the multiplicativity \scriptstyle |\mathbf{vw}| = |\mathbf{v}| |\mathbf{w}| of the norm in the quaternion algebra, and a restriction to \scriptstyle\mathbb{R}^3 of Lagrange's identity.
## Alternative ways to compute the cross product
### Quaternions
The cross product can also be described in terms of quaternions, and this is why the letters i, j, k are a convention for the standard basis on \scriptstyle\mathbf{R}^3: it is thought of as the imaginary quaternions.
For instance, the above given cross product relations among i, j, and k agree with the multiplicative relations among the quaternions i, j, and k. In general, if a vector [a1, a2, a3] is represented as the quaternion a1i + a2j + a3k, the cross product of two vectors can be obtained by taking their product as quaternions and deleting the real part of the result. The real part will be the negative of the dot product of the two vectors.
Alternatively and more straightforwardly, using the above identification of the 'purely imaginary' quaternions with \scriptstyle\mathbf{R}^3, the cross product may be thought of as half of the commutator of two quaternions.
### Conversion to matrix multiplication
A cross product between two vectors (which can only be defined in three-dimensional space) can be rewritten in terms of pure matrix multiplication as the product of a skew-symmetric matrix and a vector, as follows:
\mathbf{a} \times \mathbf{b} = [\mathbf{a}]_{\times} \mathbf{b} = \begin{bmatrix}\,0&\!-a_3&\,\,a_2\\ \,\,a_3&0&\!-a_1\\-a_2&\,\,a_1&\,0\end{bmatrix}\begin{bmatrix}b_1\\b_2\\b_3\end{bmatrix}
\mathbf{a} \times \mathbf{b} = [\mathbf{b}]^\top_{\times} \mathbf{a} = \begin{bmatrix}\,0&\,\,b_3&\!-b_2\\ -b_3&0&\,\,b_1\\\,\,b_2&\!-b_1&\,0\end{bmatrix}\begin{bmatrix}a_1\\a_2\\a_3\end{bmatrix}
where
[\mathbf{a}]_{\times} \stackrel{\rm def}{=} \begin{bmatrix}\,\,0&\!-a_3&\,\,\,a_2\\\,\,\,a_3&0&\!-a_1\\\!-a_2&\,\,a_1&\,\,0\end{bmatrix}.
Also, if \scriptstyle\mathbf{a} is itself a cross product:
\mathbf{a} = \mathbf{c} \times \mathbf{d}
then
[\mathbf{a}]_{\times} = (\mathbf{c}\mathbf{d}^\top)^\top - \mathbf{c}\mathbf{d}^\top.
This notation provides another way of generalizing cross product to the higher dimensions by substituting pseudovectors (such as angular velocity or magnetic field) with such skew-symmetric matrices. It is clear that such physical quantities will have n(n-1)/2 independent components in n dimensions, which coincides with number of dimensions for three-dimensional space, and this is why vectors can be used (and most often are used) to represent such quantities.
This notation is also often much easier to work with, for example, in epipolar geometry.
From the general properties of the cross product follows immediately that
[\mathbf{a}]_{\times} \, \mathbf{a} = \mathbf{0} and \mathbf{a}^\top \, [\mathbf{a}]_{\times} = \mathbf{0}
and from fact that \scriptstyle[\mathbf{a}]_{\times} is skew-symmetric it follows that
\mathbf{b}^\top \, [\mathbf{a}]_{\times} \, \mathbf{b} = 0.
The above-mentioned triple product expansion (bac-cab rule) can be easily proven using this notation.
The above definition of \scriptstyle[\mathbf{a}]_{\times} means that there is a one-to-one mapping between the set of 3×3 skew-symmetric matrices, also known as the Lie algebra of SO, and the operation of taking the cross product with some vector \scriptstyle\mathbf{a} .
### Index notation
The cross product can alternatively be defined in terms of the Levi-Civita symbol, \scriptstyle\varepsilon_{ijk}\mathbf{a \times b} = \mathbf{c}\Leftrightarrow\ c_i = \sum_{j=1}^3 \sum_{k=1}^3 \varepsilon_{ijk} a_j b_kwhere the indices \scriptstyle i,j,k correspond, as in the previous section, to orthogonal vector components. This characterization of the cross product is often expressed more compactly using the Einstein summation convention as\mathbf{a \times b} = \mathbf{c}\Leftrightarrow\ c_i = \varepsilon_{ijk} a_j b_kin which repeated indices are summed from 1 to 3. Note that this representation is another form of the skew-symmetric representation of the cross product:
\varepsilon_{ijk} a_j = [\mathbf{a}]_\times.
In classical mechanics: representing the cross-product with the Levi-Civita symbol can cause mechanical-symmetries to be obvious when physical-systems are isotropic in space. (Quick example: consider a particle in a Hooke's Law potential in three-space, free to oscillate in three dimensions; none of these dimensions are "special" in any sense, so symmetries lie in the cross-product-represented angular-momentum which are made clear by the abovementioned Levi-Civita representation).
### Mnemonic
The word xyzzy can be used to remember the definition of the cross product.
If
\mathbf{a} = \mathbf{b} \times \mathbf{c}
where:
\mathbf{a} = \begin{bmatrix}a_x\\a_y\\a_z\end{bmatrix},\mathbf{b} = \begin{bmatrix}b_x\\b_y\\b_z\end{bmatrix},\mathbf{c} = \begin{bmatrix}c_x\\c_y\\c_z\end{bmatrix}
then:
a_x = b_y c_z - b_z c_y \,
a_y = b_z c_x - b_x c_z \,
a_z = b_x c_y - b_y c_x. \,
The second and third equations can be obtained from the first by simply vertically rotating the subscripts, xyzx. The problem, of course, is how to remember the first equation, and two options are available for this purpose: either to remember the relevant two diagonals of Sarrus's scheme (those containing i), or to remember the xyzzy sequence.
Since the first diagonal in Sarrus's scheme is just the main diagonal of the above-mentioned \scriptstyle 3 \times 3 matrix, the first three letters of the word xyzzy can be very easily remembered.
## Applications
### Computational geometry
The cross product can be used to calculate the normal for a triangle or polygon, an operation frequently performed in computer graphics.
In computational geometry of the plane, the cross product is used to determine the sign of the acute angle defined by three points \scriptstyle p_1=(x_1,y_1), \scriptstyle p_2=(x_2,y_2) and \scriptstyle p_3=(x_3,y_3). It corresponds to the direction of the cross product of the two coplanar vector defined by the pairs of points \scriptstyle p_1, p_2 and \scriptstyle p_1, p_3, i.e., by the sign of the expression \scriptstyle P = (x_2-x_1)(y_3-y_1)-(y_2-y_1)(x_3-x_1). In the "right-handed" coordinate system, if the result is 0, the points are collinear; if it is positive, the three points constitute a negative angle of rotation around \scriptstyle p_2 from \scriptstyle p_1 to \scriptstyle p_3, otherwise a positive angle. From another point of view, the sign of \scriptstyle P tells whether \scriptstyle p_3 lies to the left or to the right of line \scriptstyle p_1, p_2.
### Mechanics
Moment of a force \scriptstyle\mathbf{F_B} applied at point B around point A is given as:
: \mathbf{M_A} = \mathbf{r_{AB}} \times \mathbf{F_B} \,
### Other
The cross product occurs in the formula for the vector operator curl.It is also used to describe the Lorentz force experienced by a moving electrical charge in a magnetic field. The definitions of torque and angular momentum also involve the cross product.
The trick of rewriting a cross product in terms of a matrix multiplication appears frequently in epipolar and multi-view geometry, in particular when deriving matching constraints.
## Cross product as an exterior product
The cross product in relation to the exterior product.
In red are the unit normal vector, and the "parallel" unit bivector.
The cross product can be viewed in terms of the exterior product. This view allows for a natural geometric interpretation of the cross product. In exterior calculus the exterior product (or wedge product) of two vectors is a bivector. A bivector is an oriented plane element, in much the same way that a vector is an oriented line element. Given two vectors a and b, one can view the bivector ab as the oriented parallelogram spanned by a and b. The cross product is then obtained by taking the Hodge dual of the bivector ab, identifying 2-vectors with vectors:
a \times b = * (a \wedge b) \,.
This can be thought of as the oriented multi-dimensional element "perpendicular" to the bivector. Only in three dimensions is the result an oriented line element – a vector – whereas, for example, in 4 dimensions the Hodge dual of a bivector is two-dimensional – another oriented plane element. So, in three dimensions only is the cross product of a and b the vector dual to the bivector ab: it is perpendicular to the bivector, with orientation dependent on the coordinate system's handedness, and has the same magnitude relative to the unit normal vector as ab has relative to the unit bivector; precisely the properties described above.
## Cross product and handedness
When measurable quantities involve cross products, the handedness of the coordinate systems used cannot be arbitrary. However, when physics laws are written as equations, it should be possible to make an arbitrary choice of the coordinate system (including handedness). To avoid problems, one should be careful to never write down an equation where the two sides do not behave equally under all transformations that need to be considered. For example, if one side of the equation is a cross product of two vectors, one must take into account that when the handedness of the coordinate system is not fixed a priori, the result is not a (true) vector but a pseudovector. Therefore, for consistency, the other side must also be a pseudovector.
More generally, the result of a cross product may be either a vector or a pseudovector, depending on the type of its operands (vectors or pseudovectors). Namely, vectors and pseudovectors are interrelated in the following ways under application of the cross product:
* vector × vector = pseudovector
* vector × pseudovector = vector
* pseudovector × pseudovector = pseudovector
Because the cross product may also be a (true) vector, it may not change direction with a mirror image transformation. This happens, according to the above relationships, if one of the operands is a (true) vector and the other one is a pseudovector (e.g., the cross product of two vectors). For instance, a vector triple product involving three (true) vectors is a (true) vector.
A handedness-free approach is possible using exterior algebra.
## Generalizations
There are several ways to generalize the cross product to the higher dimensions.
### Lie algebra
The cross product can be seen as one of the simplest Lie products,and is thus generalized by Lie algebras, which are axiomatized as binary products satisfying the axioms of multilinearity, skew-symmetry, and the Jacobi identity. Many Lie algebras exist, and their study is a major field of mathematics, called Lie theory.
For example, the Heisenberg algebra gives another Lie algebra structure on \scriptstyle\mathbf{R}^3. In the basis \scriptstyle\{x,y,z\}, the product is \scriptstyle [x,y]=z, [x,z]=[y,z]=0.
### Using octonions
A cross product for 7-dimensional vectors can be obtained in the same way by using the octonions instead of the quaternions. The nonexistence of such cross products of two vectors in other dimensions is related to the result that the only normed division algebras are the ones with dimension 1, 2, 4, and 8.
### Wedge product
In general dimension, there is no direct analogue of the binary cross product. There is however the wedge product, whichhas similar properties, except that the wedge product of two vectors is now a 2-vector instead of an ordinary vector. As mentioned above, the cross product can be interpreted as the wedge product in three dimensions after using Hodge duality to identify 2-vectors with vectors.
The wedge product and dot product can be combined to form the Clifford product.
### Multilinear algebra
In the context of multilinear algebra, the cross product can be seen as the (1,2)-tensor (a mixed tensor) obtained from the 3-dimensional volume form,By a volume form one means a function that takes in n vectors and gives out a scalar, the volume of the parallelotope defined by the vectors: \scriptstyle V\times \cdots \times V \to \mathbf{R}. This is an n-ary multilinear skew-symmetric form. In the presence of a basis, such as on \scriptstyle\mathbf{R}^n, this is given by the determinant, but in an abstract vector space, this is added structure. In terms of G-structures, a volume form is an \scriptstyle SL-structure. a (0,3)-tensor, by raising an index.
In detail, the 3-dimensional volume form defines a product \scriptstyle V \times V \times V \to \mathbf{R}, by taking the determinant of the matrix given by these 3 vectors.By duality, this is equivalent to a function \scriptstyle V \times V \to V^*, (fixing any two inputs gives a function \scriptstyle V \to \mathbf{R} by evaluating on the third input) and in the presence of an inner product (such as the dot product; more generally, a non-degenerate bilinear form), we have an isomorphism \scriptstyle V \to V^*, and thus this yields a map \scriptstyle V \times V \to V, which is the cross product: a (0,3)-tensor (3 vector inputs, scalar output) has been transformed into a (1,2)-tensor (2 vector inputs, 1 vector output) by "raising an index".
Translating the above algebra into geometry, the function "volume of the parallelepiped defined by \scriptstyle (a,b,-)" (where the first two vectors are fixed and the last is an input), which defines a function \scriptstyle V \to \mathbf{R}, can be represented uniquely as the dot product with a vector: this vector is the cross product \scriptstyle a \times b. From this perspective, the cross product is defined by the scalar triple product, \scriptstyle\mathrm{Vol}(a,b,c) = (a\times b)\cdot c.
In the same way, in higher dimensions one may define generalized cross products by raising indices of the n-dimensional volume form, which is a \scriptstyle (0,n)-tensor.The most direct generalizations of the cross product are to define either:
• a \scriptstyle (1,n-1)-tensor, which takes as input \scriptstyle n-1 vectors, and gives as output 1 vector – an \scriptstyle (n-1)-ary vector-valued product, or
• a \scriptstyle (n-2,2)-tensor, which takes as input 2 vectors and gives as output skew-symmetric tensor of rank n−2 – a binary product with rank n−2 tensor values. One can also define \scriptstyle (k,n-k)-tensors for other k.
These products are all multilinear and skew-symmetric, and can be defined in terms of the determinant and parity.
The \scriptstyle (n-1)-ary product can be described as follows: given \scriptstyle n-1 vectors \scriptstyle v_1,\dots,v_{n-1} in \scriptstyle\mathbf{R}^n, define their generalized cross product \scriptstyle v_n = v_1 \times \cdots \times v_{n-1} as:
• perpendicular to the hyperplane defined by the \scriptstyle v_i,
• magnitude is the volume of the parallelotope defined by the \scriptstyle v_i, which can be computed as the Gram determinant of the \scriptstyle v_i,
• oriented so that \scriptstyle v_1,\dots,v_n is positively oriented.
This is the unique multilinear, alternating product which evaluates to \scriptstyle e_1 \times \dots \times e_{n-1} = e_n, \scriptstyle e_2 \times \dots \times e_n = e_1, and so forth for cyclic permutations of indices.
In coordinates, one can give a formula for this n-ary analogue of the cross product in Rn+1 by:
\bigwedge(\mathbf{v}_1,\dots,\mathbf{v}_n)=\begin{vmatrix}v_1{}^1 &\cdots &v_1{}^{n+1}\\\vdots &\ddots &\vdots\\v_n{}^1 & \cdots &v_n{}^{n+1}\\\mathbf{e}_1 &\cdots &\mathbf{e}_{n+1}\end{vmatrix}.
This formula is identical in structure to the determinant formula for the normal cross product in R3 except that the row of basis vectors is the last row in the determinant rather than the first. The reason for this is to ensure that the ordered vectors (v1,...,vn,Λ(v1,...,vn)) have a positive orientation with respect to (e1,...,en+1). If n is even, this modification leaves the value unchanged, so this convention agrees with the normal definition of the binary product. In the case that n is odd, however, the distinction must be kept. This n-ary form enjoys many of the same properties as the vector cross product: it is alternating and linear in its arguments, it is perpendicular to each argument, and its magnitude gives the hypervolume of the region bounded by the arguments. And just like the vector cross product, it can be defined in a coordinate independent way as the Hodge dual of the wedge product of the arguments.
## History
In 1773, Joseph Louis Lagrange introduced the component form of both the dot and cross products in order to study the tetrahedron in three dimensions. In 1843 the Irish mathematical physicist Sir William Rowan Hamilton introduced the quaternion product, and with it the terms "vector" and "scalar". Given two quaternions [0, u] and [0, v], where u and v are vectors in R3, their quaternion product can be summarized as [−u·v, u×v]. James Clerk Maxwell used Hamilton's quaternion tools to develop his famous electromagnetism equations, and for this and other reasons quaternions for a time were an essential part of physics education.
However, Oliver Heaviside in England and Josiah Willard Gibbs in Connecticut felt that quaternion methods were too cumbersome, often requiring the scalar or vector part of a result to be extracted. Thus, about forty years after the quaternion product, the dot product and cross product were introduced—to heated opposition. Pivotal to (eventual) acceptance was the efficiency of the new approach, allowing Heaviside to reduce the equations of electromagnetism from Maxwell's original 20 to the four commonly seen today.
Largely independent of this development, and largely unappreciated at the time, Hermann Grassmann created a geometric algebra not tied to dimension two or three, with the exterior product playing a central role. William Kingdon Clifford combined the algebras of Hamilton and Grassmann to produce Clifford algebra, where in the case of three-dimensional vectors the bivector produced from two vectors dualizes to a vector, thus reproducing the cross product.
The cross notation, which began with Gibbs, inspired the name "cross product". Originally appearing in privately published notes for his students in 1881 as Elements of Vector Analysis, Gibbs's notation—and the name—later reached a wider audience through Vector Analysis , a textbook by a former student. Edwin Bidwell Wilson rearranged material from Gibbs's lectures, together with material from publications by Heaviside, Föpps, and Hamilton. He divided vector analysis into three parts:
Two main kinds of vector multiplications were defined, and they were called as follows:
• The direct, scalar, or dot product of two vectors
• The skew, vector, or cross product of two vectors
Several kinds of triple products and products of more than three vectors were also examined. The above mentioned triple product expansion was also included. | 2019-11-21 13:12:43 | {"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": 0.9754615426063538, "perplexity": 610.6796818510895}, "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/1573496670821.55/warc/CC-MAIN-20191121125509-20191121153509-00529.warc.gz"} |
https://yourpastquestions.com/product/young-children-in-family-therapy-a-look-at-frequency-of-inclusion-therapist-training-and-therapy-content/ | # Young Children in Family Therapy: A Look at Frequency of Inclusion, Therapist Training, and Therapy Content
0
## Description
Young Children in Family Therapy: A Look at Frequency of Inclusion, Therapist Training, and Therapy Content, Is A Well-Researched Topic, It Is To Be Used As A Guide Or Framework For Your Research.
## INTRODUCTION
In the mid-1950’s a change began to occur when the family, rather than the individual client, became a potential unit of treatment in psychotherapy. With this change the attention of the therapist focused on the dysfunctional system in which the “identified patient” is seen as but a symptom, and the -family, or system, is viewed as the client (Corsini & Wedding, 1989). Consequently, family therapists have a
difficult job in that they are expected to understand and incorporate in their treatment all members of the system.
Nathan Ackerman (1970) stated that, ” … psychotherapists as a group are
candid in confessing their preferred patient ages”. Many feel uncomfortable working with children and feel that they require too much energy while others relate easily to children and feel uncomfortable working with adults. In individual therapy this may
pose no problem yet, in family therapy the problem cannot be avoided (Ackerman, 1970). Thus, the difficulty exists in fully involving all members of the family simultaneously.
ACKNOWLEDGEMENTS ……………………………. 111
LIST OF TABLES …………………………………. v1
Chapter
I. INTRODUCTION . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
Reasons Cited for Excluding Young Children . . . . . . . . . . . . . . . . . 2
Reasons Cited for Including Young Children . . . . . . . . . . . . . . . . . 4
When to Include Young Children . . . . . . . . . . . . . . . . . . . . . . . . . 5
Appropriate Topics for Young Children . . . . . . . . . . . . . . . . . . . . . 6
Developmental Distinctions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8
The Use of Play . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9
The Present Study …………… : ……………… 12
II. METHOD …………………………………. 16
Subjects …………………………………. 16
Materials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17
Procedure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19
III. RESULTS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22
Demographics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22
In Your Practice . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24
In Your Practice – Open-Ended Responses ………………. 27
Attitudes/Beliefs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31
Attitudes/Beliefs – Open-Ended Responses . . . . . . . . . . . . . . . . . . . 39
Training …………………………………. 42
IV. DISCUSSION ……………………………….. 45
A Brief Review . . . . . . . . . . . . . . . . ; . . . . . . . . . . . . . . . . . . 45
Findings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 7
Limitations of the Study ………………………… 55
Future Suggestions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57
Implications ……………………………….. 58
Appendix
A. YOUNG CHILDREN IN FAMILY THERAPY QUESTIONNAIRE …. 60
B. COVER LETTER TO FIRST MAILING ………………… 68
C. FOLLOW-UP POSTCARD – SECOND MAILING …………… 70
D. COVER LETTER TO THIRD MAILING ………………… 72
REFERENCE LIST ………………………………… 74
VITA ……………………………………….. 78 | 2021-05-07 01:49:29 | {"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": 0.9485962390899658, "perplexity": 71.86796892825753}, "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/1620243988774.18/warc/CC-MAIN-20210506235514-20210507025514-00332.warc.gz"} |
https://gsebsolutions.com/gseb-solutions-class-6-maths-chapter-14-ex-14-6/ | # GSEB Solutions Class 6 Maths Chapter 14 Practical Geometry Ex 14.6
Gujarat Board GSEB Textbook Solutions Class 6 Maths Chapter 14 Practical Geometry Ex 14.6 Textbook Questions and Answers.
## Gujarat Board Textbook Solutions Class 6 Maths Chapter 14 Practical Geometry Ex 14.6
Question 1.
Draw ∠POQ of measure 75° and find its line of symmetry.
Solution:
Steps of construction:
Step I: Construct BOA = 90°, where ∠AOQ = 60° and ∠BOQ = 30°.
Step II: Draw $$\overrightarrow{\mathrm{OP}}$$, the angle bisector of ∠BOQ, such that ∠QOP = $$\frac { 1 }{ 2 }$$ ∠BOQ.
∠QOP = $$\frac { 1 }{ 2 }$$ (30°) = 15°
Step III: Since, 60° + 15° = 75°
So, ∠AOQ + ∠QOP = ∠POA Thus, ∠POA = 75°.
Question 2.
Draw an angle of measure 147° and construct its bisector.
Solution:
Steps of construction:
Step I: Draw a ray $$\overrightarrow{\mathrm{AB}}$$.
Step II: Using a protractor construct ∠BAC = 147°.
Step III: With centre ‘A’ and a convenient radius, draw an arc which intersects the arms $$\overrightarrow{\mathrm{AB}}$$ and $$\overrightarrow{\mathrm{AC}}$$ at P and Q respectively.
Step IV: With centre P and radius more than half of PQ, draw an arc.
Step V: With centre Q and keeping the same radius, draw another arc that intersects the previous arc at R.
Step VI: Join A and R. Produce AR.
Thus, $$\overrightarrow{\mathrm{AR}}$$ is the required angle bisector of ∠BAC.
Question 3.
Draw a right angle and construct its bisector.
Solution:
Steps of construction:
Step I: Draw a line l and mark a point ‘O’ on it.
Step II: With centre ‘O’ and a convenient radius, draw an arc which intersects l at A and B.
Step III: With centres at A and B and radius more than half of AB, draw two arcs which intersect each other at C.
Step IV: Join OC, such that ∠AOC is a right angle.
Step V: A and D as centres and radius more than half of AD, draw two arcs which cut each other at E.
Step VI: Join OE and produce it.
Thus, $$\overrightarrow{\mathrm{OE}}$$ is the required angle bisector of right angle ∠AOC, i.e. $$\overrightarrow{\mathrm{OE}}$$ is the bisector of the right angle.
Question 4.
Draw an angle of measure 153° and divide it into four equal parts.
Solution:
Steps of construction:
Step I: Draw a ray $$\overrightarrow{\mathrm{AB}}$$.
Step II: Using a protractor, construct ∠BAC = 153°.
Step III: Draw $$\overrightarrow{\mathrm{AD}}$$, bisector of ∠BAC.
Step IV: Again, draw $$\overrightarrow{\mathrm{AE}}$$, bisector of ∠DAC.
Step V: Also, draw $$\overrightarrow{\mathrm{AF}}$$, bisector of ∠BAD.
Thus, $$\overrightarrow{\mathrm{AE}}$$, $$\overrightarrow{\mathrm{AD}}$$ and $$\overrightarrow{\mathrm{AF}}$$, divide the given ∠BAC into four equal parts.
Question 5.
Construct with ruler and compasses, angles of following measures:
(a) 60°
Solution:
In figure ∠AOB = 60°
(b) 30°
Solution:
∵ $$\frac { 1 }{ 2 }$$ (60°) = 30°
∴ In the figure, ∠AOC = 30°
(c) 90°
Solution:
In the figure, ∠POQ = 90°
(d) 120°
Solution:
Since 2 × (60°) = 120°
∴ In the figure, ∠AOC = 120°
(e) 45°
Solution:
∵ $$\frac { 1 }{ 2 }$$ (90°) = 45°
∴ In the figure, ∠POR = 45°
(f) 135°
Solution:
Since, 90° + 45° = 135°
i.e. 90° + $$\frac { 1 }{ 2 }$$(90°) = 135°
.’. In the figure, ∠AOD = 135°.
Question 6.
Draw an angle of measure 45° and bisect it.
Solution:
Steps of construction:
Step I: Draw a ray OA.
Step II: Construct ∠AOQ = 45°.
Step III: With centre C and radius more than CD, draw an arc.
Step IV: With the same radius and centre at D, draw another arc which intersects the arc at B.
Step V: Join OB and produce it.
Thus, $$\overrightarrow{\mathrm{OB}}$$, bisects ∠AOQ into two equal parts.
i.e. ∠AOB = $$\frac { 1 }{ 2 }$$(45°) =22$$\frac{1^{\circ}}{2}$$
Question 7.
Draw an angle of measure 135° and bisect it.
Solution:
Steps of construction:
Step I: Draw a line AB and mark a point O on it.
Step II: Construct ∠AOC = 135°.
Step III: Draw $$\overrightarrow{\mathrm{OD}}$$, bisector ∠AOC.
Thus, ∠AOD = $$\frac { 1 }{ 2 }$$ (135°) = 67$$\frac{1^{\circ}}{2}$$
Question 8.
Draw an angle of 70°. Make a copy of it using only a straight edge and compasses.
Solution:
Steps of construction:
Step I: Draw a line 1 and mark a point O on it.
Step II: Using a protractor construct ∠AOB = 70°.
Step III: With centre O and a suitable radius, draw an arc which intersects $$\overrightarrow{\mathrm{OA}}$$ and $$\overrightarrow{\mathrm{OB}}$$ at E and F respectively.
Step IV: Draw a ray $$\overrightarrow{\mathrm{PQ}}$$.
Step V: Keeping the same radius and centre P, draw an arc intersecting $$\overrightarrow{\mathrm{PQ}}$$ at R.
Step VI: With centre R and radius equal to EF, draw an arc intersecting the previous arc at S.
Step VII: Join PS and produce it.
Thus, ∠QPS is a copy of ∠AOB = 70°.
Question 9.
Draw an angle of 40°. Copy its supplementary angle.
Solution:
Steps of construction:
Step I: By using protractor draw ∠AOB = 40°.
∠COF is the supplementary angle.
Step II: With centre O and a convenient radius, draw an arc which intersects $$\overrightarrow{\mathrm{OC}}$$ and $$\overrightarrow{\mathrm{OB}}$$ at E and F respectively.
Step III: Draw a ray $$\overrightarrow{\mathrm{QR}}$$
Step IV: With centre Q and same radius, draw an arc intersecting $$\overrightarrow{\mathrm{PQ}}$$ at L.
Step V: With centre L and radius equal to EF draw an arc which intersects the previous arc at S.
Step VI: Join QS and produce it.
Thus, ∠PQS is the copy of the supplementary angle ∠COB | 2022-10-06 20:35:08 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.5874204039573669, "perplexity": 5676.37435080829}, "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/1664030337855.83/warc/CC-MAIN-20221006191305-20221006221305-00745.warc.gz"} |
http://stats.stackexchange.com/questions/33918/is-there-an-optimal-bandwidth-for-a-kernel-density-estimator-of-derivatives | # Is there an optimal bandwidth for a kernel density estimator of derivatives?
I need to estimate the density function based on a set of observations using the kernel density estimator. Based on the same set of observations, I also need to estimate the first and second derivatives of the density using the derivatives of the kernel density estimator. The bandwidth will certainly have a great effect of the final result.
First, I know there are a couple of R functions that give the KDE bandwidth. I am not sure which one is more preferred. Can anyone recommend one out of these R functions for KDE bandwidth?
Secondly, for the derivative of KDE, should I choose the same bandwidth?
-
For a density the choice of bandwidth is always somewhat subjective. It is a question of what is too narrow and therefore causes variation in the curve that is essentially following the noise vs too wide where the curve is too smooth and misses some real features in the curve. But you estimate the density to find out the shape. So how smooth the estimate should be is not easy to know. For derivatives I would think it depends on what feature of the derivative you want to know about. – Michael Chernick Aug 8 '12 at 15:15
The optimal bandwidth for derivative estimation will be different from the bandwidth for density estimation. In general, every feature of a density has its own optimal bandwidth selector.
If your objective is to minimize mean integrated squared error (which is the usual criterion) there is nothing subjective about it. It is a matter of deriving the value that minimizes the criterion. The equations are given in Section 2.10 of Hansen (2009).
The tricky part is that the optimal bandwidth is a function of the density itself, so this solution is not directly useful. There are a number of methods around to try to deal with that problem. These usually approximate some functionals of the density using normal approximations. (Note, there is no assumption that the density itself is normal. The assumption is that some functionals of the density can be obtained assuming normality.)
Where the approximations are imposed determines how good the bandwidth selector is. The crudest approach is called the "normal reference rule" which imposes the approximation at a high level. The end of Section 2.10 in Hansen (2009) gives the formula using this approach. This approach is implemented in the hns() function from the ks package on CRAN. That's probably the best you will get if you don't want to write your own code. So you can estimate the derivative of a density as follows (using ks):
library(ks)
h <- hns(x,deriv.order=1)
den <- kdde(x, h=h, deriv.order=1)
A better approach, usually known as a "direct plug in" selector, imposes the approximation at a lower level. For straight density estimation, this is the Sheather-Jones method, implemented in R using density(x,bw="SJ"). However, I don't think there is a similar facility available in any R package for derivative estimation.
Rather than use straight kernel estimation, you may be better off with a local polynomial estimator. This can be done using the locpoly() function from the ks package in R. Again, there is no optimal bandwidth selection implemented, but the bias will be smaller than for kernel estimators. e.g.,
den2 <- locpoly(x, bandwidth=?, drv=1) # Need to guess a sensible bandwidth
-
Thanks a million, Rob. I will likely use SJ bandwidth for density estimation. – user13154 Aug 9 '12 at 0:31 For derivative estimation, if I use h <- hns(x); den <- kdde(x, h=h, deriv.order=1), is h the optimal bandwidth used for the estimation. I am asking this because h <- hns(x) is chosen without specifying the derivative order. Thanks. – user13154 Aug 9 '12 at 0:31 I checked the formula given at the end of Section 2.10 in Hansen (2009). It looks like the bandwidth depends on the order the derivative, say rth derivative. h <- hns(x) does not seem to depend on the r. – user13154 Aug 9 '12 at 1:02 I just found out that hns function has a deriv.order argument where I can specify the order the derivatives. Thanks again very very much, Rob. – user13154 Aug 9 '12 at 1:12 Sorry. I left that out. Now fixed. – Rob Hyndman Aug 9 '12 at 2:23 | 2013-05-21 09:25:12 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.8813781142234802, "perplexity": 506.1408174155419}, "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-2013-20/segments/1368699856050/warc/CC-MAIN-20130516102416-00067-ip-10-60-113-184.ec2.internal.warc.gz"} |
http://physics.stackexchange.com/questions/82014/purposes-of-qec-stabilizers | Purposes of QEC stabilizers
I am going through the idea of stabilizer formalism.
Defined what is a Pauli group $P_n$ and its properties, we describe a stabilizer set $S$ as: $$S\subset P_n$$
The stabilizer set establishes valid codewords for a state if the equation $$s\left|\psi\right\rangle=\left|\psi\right\rangle,\;\;\;\forall s \in S \;\;\;\;\; (1)$$ is satisfied. That means $\left|\psi\right\rangle$ is a +1 eigenstate of $s$.
Each valid codeword belongs to $V$, that is a set of qubits stabilized by $S$. Therefore, if $(1)$ is satisfied, then $\left|\psi\right\rangle \in V$.
Let's consider the Steane code of 7 qubits. The followings are the stabilizer codes for such encode: $$K^1 = IIIXXXX$$ $$K^2 = XIXIXIX$$ $$K^3 = IXXIIXX$$ $$K^4 = IIIZZZZ$$ $$K^5 = ZIZIZIZ$$ $$K^6 = IZZIIZZ$$
These reduce the $2^7$ Hilbert space into a two-dimensional subspace. These stabilizers generate valid codewords for the Steane code: $$\left|0\right\rangle_L \equiv \frac{1}{\sqrt{8}}(\left|0000000\right\rangle + \left|1010101\right\rangle + \left|0110011\right\rangle + \left|1100110\right\rangle + \left|0001111\right\rangle + \left|1011010\right\rangle + \left|0111100\right\rangle + \left|1101001\right\rangle)$$ $$\left|1\right\rangle_L \equiv \frac{1}{\sqrt{8}}(\left|1111111\right\rangle + \left|0101010\right\rangle + \left|1001100\right\rangle + \left|0011001\right\rangle + \left|1110000\right\rangle + \left|0100101\right\rangle + \left|1000011\right\rangle + \left|0010110\right\rangle)$$
Here my doubt come; Each stabilizer is used as "filters of the input", so if an input, on which are applied one or more of those stabilizers, does not satisfy the equation $(1)$ ($\left|\psi\right\rangle$ -1 eigenvalue of $s$?), then we can say that an error occurred. Through syndrome measurement we can identify where the error occurred and correct it.
Another issue: verifying $(1)$ means, for example, $\;K^1 \left|1010101\right\rangle = \left|1011010\right\rangle$. Since both $\left|1010101\right\rangle$ and $\;\left|1011010\right\rangle$ represent $\left|0\right\rangle_L$, we say that $(1)$ is satisfied?
Finally: $\;K^4 \left|1010101\right\rangle = ?$
Thank you.
Added Last trouble:
The state of the system is represented by:
$$\left|\psi\right\rangle_F={1\over 2}(\left|\psi\right\rangle_I+U\left|\psi\right\rangle_I)\left|0\right\rangle + {1\over 2}(\left|\psi\right\rangle_I-U\left|\psi\right\rangle_I)\left|1\right\rangle$$
We apply $K^1,K^2,K^3$ to the input and we measure the ancilla qubits to verify the integrity of the input (if $\left|\psi\right\rangle_I$ is +1 eigenstate of $K^1,K^2,K^3$). If the equation $(1)$ is not satisfied, then the corrupted qubit is corrected with a $Z$ gate addressed by syndrome measurement of ancilla qubits. This is how does the system work?
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1 Answer
1) If there is an error $E_j$, the new states $E_j|0\rangle_L$ and $E_j|1\rangle_L$ are eigenvectors, with eigenvalue $-1$, of all the stabilizers $s_j$ belonging to some set subset $S_j$ of $S$. (the elements of $S_j$ anticommute with $E_j$).This subset $S_j$ identifies uniquely the error $E_j$.
2) $|0\rangle_L$ and $|1\rangle_L$ are eigenvectors, with eigenvalue $1$, of all the stabilizers $s$ belonging to $S$ (this is not true for the "components" of $|0\rangle_L$ and $|1\rangle_L$ like, for instance, $|1010101\rangle$).For a stabilizer $s$, you just calculate $s|0\rangle_L$ and $s|1\rangle_L$, and you check that the result is $|0\rangle_L$ or $|1\rangle_L$.
For instance :
$K^1\left|0\right\rangle_L = (IIIXXXX) \\\frac{1}{\sqrt{8}}(\left|0000000\right\rangle + \left|1010101\right\rangle + \left|0110011\right\rangle + \left|1100110\right\rangle + \left|0001111\right\rangle + \left|1011010\right\rangle + \left|0111100\right\rangle + \left|1101001\right\rangle)= \\ \frac{1}{\sqrt{8}}(\left|0001111\right\rangle + \left|1011010\right\rangle + \left|0111100\right\rangle + \left|1101001\right\rangle + \left|0000000\right\rangle + \left|1010101\right\rangle + \left|0110011\right\rangle + \left|1100110\right\rangle)\\ =\left|0\right\rangle_L$
3) $K^4 \left|1010101\right\rangle = IIIZZZZ |1010101\rangle$. With $Z |0\rangle = |0\rangle$, and $Z |1\rangle = -|1\rangle$, you get :
$K^4 \left|1010101\right\rangle = |1010101\rangle$
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it is clear. Thank you. I edited my question with an example, perhaps you can tell me if I am wrong or not – Antonio Verlotta Oct 25 '13 at 13:00
@AntonioVerlotta : Please, make a new question for this. – Trimok Oct 25 '13 at 17:38 | 2015-07-07 02:49:55 | {"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": 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": 0.9330002069473267, "perplexity": 217.12931000747838}, "config": {"markdown_headings": false, "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-2015-27/segments/1435375098987.83/warc/CC-MAIN-20150627031818-00297-ip-10-179-60-89.ec2.internal.warc.gz"} |
https://support.bioconductor.org/p/90143/ | Question: R package "metaboliteData" for package "pcaMethods"
1
2.4 years ago by
lukasz.galecki10 wrote:
Dear all,
I'm working for a project at the university of Cologne for which I want to use the package "pcaMethods". It contains a dataset named "metaboliteData".
Does anyone know what the columns exactly describe in this matrix?
Best regards,
Lukasz Galecki
modified 2.4 years ago by Valerie Obenchain6.7k • written 2.4 years ago by lukasz.galecki10
Answer: R package "metaboliteData" for package "pcaMethods"
1
2.4 years ago by
United States
Valerie Obenchain6.7k wrote:
The man page has a description:
?metaboliteData
Valerie | 2019-04-23 14:14:39 | {"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": 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": 0.23949678242206573, "perplexity": 11548.662544996441}, "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-2019-18/segments/1555578605510.53/warc/CC-MAIN-20190423134850-20190423160850-00496.warc.gz"} |
https://forum.allaboutcircuits.com/threads/looking-for-someone-to-build-a-circuit-board.104223/ | # LOOKING FOR SOMEONE TO BUILD A CIRCUIT BOARD.
#### RVW
Joined Dec 3, 2014
8
I am looking for someone who can help with a small project. I would like a small circuit board that will activate a switch with a whistle or two claps. I know they sell kits that will do this, but I want to choose the sound that activates the switch. I think this could be accomplished with fairly simple circuits. I'm not looking for advice on how to do it, I am looking for someone to build it for me. If you are interested please let me know, and we can talk. Thanks
#### ISB123
Joined May 21, 2014
1,238
#### RVW
Joined Dec 3, 2014
8
It would depend on how much but yes I would be willing to pay for time and materials. As long as the cost isn't too high.
#### ISB123
Joined May 21, 2014
1,238
For what do you need the switch?What kind of load are you planning to switch?
#### RVW
Joined Dec 3, 2014
8
For what do you need the switch?What kind of load are you planning to switch?
It would be for a small electric dc lock would be nice if I could run the circuitry and the lock from a 9 volt battery, so probably a small load.
#### ISB123
Joined May 21, 2014
1,238
So you only want two kinds of sounds to activate the switch(whistle and 2x claps which should be selectable) and not jumbo jet passing by?Just to be clear.
#### jpanhalt
Joined Jan 18, 2008
8,507
I have never seen one of these types of threads lead anywhere productive.
What do you consider too much for a complete, custom device, including PCB? $100,$500, $1000,$2000... ?
What avenues have you investigated for a commercially available product? What do you want the command to be? For example, a specific sequence of claps or a spoken word?
John
#### RVW
Joined Dec 3, 2014
8
So you only want two kinds of sounds to activate the switch(whistle and 2x claps which should be selectable) and not jumbo jet passing by?Just to be clear.
Yes that's what I am looking for, a select switch for either the whistle or the two clap activation.
#### ISB123
Joined May 21, 2014
1,238
#### RVW
Joined Dec 3, 2014
8
What would constitute a cost being too high? Keep in mind that engineering/development time is not cheap. It's unlikely that you would be looking at less than a couple thousand dollars and that's if you don't want something that is nicely packaged.
What are your reliability requirements? If you whistle and nothing happens, how bad is that? Is it okay if you occasionally have to whistle multiple times? Is it okay if it occasionally actives due to some sound that it thought was an acceptable whistle?
This is just for a project, so reliability isn't an issue, but what I spend on it is, not looking to have something engineered, just put together.
#### WBahn
Joined Mar 31, 2012
25,115
Well, the kit to build a clap activated switch is about $15.00 and similar cost for a whistle activated switch. I kind of wanted to combine the two. I would build it myself, but I am disabled. I was hoping materials and labor would be small, because of this seemed to be fairly simple, at least I thought. The problem is NRE -- non-recurring engineering -- costs. If you are making 100,000 kits and it costs you$10,000 to design the product, then the design cost is only $0.10/unit and can almost be forgotten compared to the cost of the actual parts (the Bill of Materials) and the labor to assemble it. But in a one-off design, the NRE has to be covered by the single part that is actually made. It's the same for some of the materials -- the cost of a PCB has to include the NRE to lay out the board and for the fab house to prepare the masks. Just the fab costs for a basic board that includes a protective solder mask is probably going to be pushing$100 for the smallest boards.
#### WBahn
Joined Mar 31, 2012
25,115
This is just for a project, so reliability isn't an issue, but what I spend on it is, not looking to have something engineered, just put together.
So are you going to provide the design for someone else to put together? If not, then someone is going to have to design it -- that's known as engineering. Doesn't mean that it's done by an engineer -- it can be done by a teenage kid that's just kinda interested in doing it for fun, but you can't just put something together unless someone has designed it first. Even if it is something simple enough that someone with the right background can throw it together on the fly based on their experience, they are still designing it in addition to putting it together.
#### mcgyvr
Joined Oct 15, 2009
5,394
clap on...... clap off....... clap on..clap off.. the clapper
#### ISB123
Joined May 21, 2014
1,238
Someone could build the thing for like 100$for fun but you have another problem which is shipping cost that means if someone isn't from your state/country shipping could easily increase the cost by 30%. Thread Starter #### RVW Joined Dec 3, 2014 8 So are you going to provide the design for someone else to put together? If not, then someone is going to have to design it -- that's known as engineering. Doesn't mean that it's done by an engineer -- it can be done by a teenage kid that's just kinda interested in doing it for fun, but you can't just put something together unless someone has designed it first. Even if it is something simple enough that someone with the right background can throw it together on the fly based on their experience, they are still designing it in addition to putting it together. I agree, but I think someone has already designed this, at least a clap and whistle activated switch, maybe not on one board, but it shouldn't be too hard to combine, even if its two separate boards in one enclosure. The kits and directions already exist. #### Papabravo Joined Feb 24, 2006 12,774 I have never seen one of these types of threads lead anywhere productive. What do you consider too much for a complete, custom device, including PCB?$100, $500,$1000, \$2000... ?
What avenues have you investigated for a commercially available product? What do you want the command to be? For example, a specific sequence of claps or a spoken word?
John
I too have tried numerous times to respond to these threads. They never go anywhere because people refuse to work for what it is worth to the people who propose the projects. What we do is viewed as nearly worthless by the vast majority of the general public. I'm in agreement with the others who say stop wasting our time with requests to work for free. Pay us or fahgeddaboudit.
#### ISB123
Joined May 21, 2014
1,238
You could use SPDT switch to choose which board gets the power but that would require some playing with soldering iron and rewiring.
RVW
#### GopherT
Joined Nov 23, 2012
8,012
I
I agree, but I think someone has already designed this, at least a clap and whistle activated switch, maybe not on one board, but it shouldn't be too hard to combine, even if its two separate boards in one enclosure. The kits and directions already exist.
i think you are confused to assume that a clap-on-clap off would not respond to two whistles. And the whistle activated device would not respond to any screech or chalkboard scratch.
My advice, buy some type of solder kit assembly project and a clapper. That way, you have something to assemble and a clap-activated switch.
Kits available at jameco or tayda and many other places. Clapper at walgreens. | 2020-02-27 06:05:44 | {"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": 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": 0.2575719356536865, "perplexity": 1610.5757500176032}, "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/1581875146647.82/warc/CC-MAIN-20200227033058-20200227063058-00416.warc.gz"} |
https://www.lesswrong.com/users/avery-2 | Sorted by New
# Wiki Contributions
1. Meteors. This one isn’t exactly right since the equilibrium point either happens or gets skipped over entirely but nonetheless...imagine a meteoroid flying through the solar system towards Earth. If it crosses into the atmosphere, it becomes a meteor. And then depending on the composition of the meteor and its size, it may burn up in the atmosphere or it may make it to the surface (at that point, it’s a meteorite).
In this case, you can think of the “zone of attraction” as the entire journey to the atmosphere. Once it hits the atmosphere, it quickly is decelerated by air resistance. And if the meteor burns up, then its velocity “stabilizes” at 0. So “burned up in the atmosphere” is sort of like an equilibrium point. If it doesn’t burn up (e.g. because it’s larger), then it plummets to the surface. In this case, the other side of the equilibrium point is not really repulsive. Rather, it’s just a continuation of the pre-equilibrium point. And then when it hits the Earth the entire system is changed.
2. Nuclear Weapon Acquisition. You could argue the initial development of nuclear weapons triggered an “attractive force” for major world powers to stock up on warheads (e.g. an arms race). Eventually, the development and acquisition slowed down and even regressed (as in nuclear disarmament). But supplies were not totally eliminated. You could say an equilibrium point has been reached. The reasoning continues for the “repulsive force”: the equilibrium could be broken if any country were to deploy nuclear attacks on another. It would (probably?) lead to retaliatory strikes and the acquisition of warheads may accelerate drastically. (related: apparently the semistability of nuclear deterrence has a name! stability–instability paradox)
3. My Steam library on a weekend night. On weekend nights when I have nothing going on, I am attracted to my steam library. I often arrive at an equilibrium where I will sift through my games but don’t actually play anything. Then, I either decide against playing games, at which point I usually back away from the equilibrium point for the night. Or some “external force” nudges me into playing, in which case +4 hours of my night are gone.
Unsurprisingly, friends-wanting-to-play-something-as-a-group is almost always enough to convince me to go past the point of no return. Similarly, I am also likely to break equilibrium if I am playing through a recently acquired game.
Again, I don’t think my first example is a true semistable equilibrium (neither the second example?) but maybe there’s something there. All in all, this was great. I just came across the Framing Practicum Sequences and love them! Thanks for the post AllAmericanBreakfast!
...now off to sift through some games :p | 2023-03-24 02:50:07 | {"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": 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": 2, "x-ck12": 0, "texerror": 0, "math_score": 0.5102062821388245, "perplexity": 1356.0699075135963}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-14/segments/1679296945242.64/warc/CC-MAIN-20230324020038-20230324050038-00358.warc.gz"} |
https://quant.stackexchange.com/questions/30345/volar-higher-order-parametrizations | # Volar Higher Order Parametrizations
I came across this presentation from volar.io. The authors show fitting examples for a flexible volatility smile parametrization in 5 to 8 parameters which is also able to fit the locally concave market implied volatility smiles around special events.
Does anybody know the details of their parametrization and can you provide a reference? In particular, is it a simple extension of their C3 parametrization where the Cn curve is given by
$$\sigma^2(z) = \sigma_0^2 \left( 1 + \sum_{i = 1}^{n - 1} \frac{1}{n!} \xi_i z^n \right)$$
with
$$z = \frac{\ln(K / F)}{\sigma_0 \sqrt{T}}.$$
I suppose this is not the case and there is more to it. Some reasons:
• Their examples look very stable on the wings which I would not expect from higher order polynomials. While they do not show too much extrapolation, their C5 and C6 curves on slides 31 and 32 look fairly well behaved in the wings (where they loose some quality of fit though).
• It might be possible that they define a lower and upper cutoff beyond which they use a different tail function (e.g. linear in variance) and impose smoothness in these points. However on slide 10, they explicitly write that they don't like "hacks" in the wings.
• Another "goal" states on slide 10 is for no-arbitrage constraints to be easy to incorporate. In the above setup, absence of butterfly arbitrage at all strikes used for the calibration creates non-linear constraints for an otherwise nice linear problem.
• Interesting question, I was not aware of such developments. I wonder if no (static) arbitrage conditions can easily be accounted for under such a parametrisation. – Quantuple Sep 28 '16 at 13:44
• @Quantuple: That will exactly be my follow up question once we know how the parametrization looks like. It is usually not difficult to compute the density by just differentiating - either by hand (Mathematica) or using AD. The question is if we obtain sth. tractable that can be easily incorporated as a constraint in the fitting. Also - can it be ensured for all strikes or just a selected set? – LocalVolatility Sep 28 '16 at 13:53
• Completely agree with you. – Quantuple Sep 28 '16 at 13:57
• @LocalVolatility There seems more to it, they claim this can be done in real-time so Mathematica or AD is not a likely solution. – pyCthon Sep 30 '16 at 21:19
• It's just a Taylor expansion, given its fitted using data in the wings, it's still going to be stable there. It's only extrapolation where it starts to become bad. – will Oct 29 '16 at 18:12 | 2019-05-25 15:14:21 | {"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": 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": 2, "x-ck12": 0, "texerror": 0, "math_score": 0.6878277063369751, "perplexity": 762.6155717812584}, "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-2019-22/segments/1558232258120.87/warc/CC-MAIN-20190525144906-20190525170906-00531.warc.gz"} |
https://istopdeath.com/simplify-1-1-cosx1-1cosx/ | # Simplify 1/(1-cos(x))+1/(1+cos(x))
To write as a fraction with a common denominator, multiply by .
To write as a fraction with a common denominator, multiply by .
Write each expression with a common denominator of , by multiplying each by an appropriate factor of .
Combine.
Combine.
Reorder the factors of .
Combine the numerators over the common denominator.
Simplify the numerator.
Multiply by .
Multiply by . | 2023-01-30 23:57:05 | {"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": 0.9128178358078003, "perplexity": 1377.9394569013436}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-06/segments/1674764499831.97/warc/CC-MAIN-20230130232547-20230131022547-00771.warc.gz"} |
https://collegephysicsanswers.com/openstax-solutions/what-average-power-output-heart-defibrillator-dissipates-400-j-energy-100-ms-b | Change the chapter
Question
(a) What is the average power output of a heart defibrillator that dissipates 400 J of energy in 10.0 ms? (b) Considering the high-power output, why doesn’t the defibrillator produce serious burns?
1. $4.00\times 10^{4}\textrm{ W}$
2. Power dissipated as heat in the skin is proportional to the resistance of the skin. Skin resistance is complicated - see Wikipedia article. The defibrillator voltage and frequency are such that skin resistance is reduced, which prevents the skin from dissipating heat and burning. Also, gel is put between the paddles and skin which prevents the air gap that would otherwise heat up and burn the skin.
Solution Video | 2023-02-02 18:44:51 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.5505256056785583, "perplexity": 1526.1721616258164}, "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-2023-06/segments/1674764500035.14/warc/CC-MAIN-20230202165041-20230202195041-00717.warc.gz"} |
https://developer.synthetix.io/tokens/ | # Tokens¶
Here are the different kinds of tokens used in the Synthetix ecosystem.
## Collateral¶
### SNX¶
SNX is the primary form of collateral backing the synthetic assets available in the Synthetix protocol. Its stakers are entitled to fees generated by Synth trades on Synthetix.Exchange.
### Ether¶
A three-month trial allowing Ether (ETH) to be staked as collateral in the Synthetix system has been completed and further trials will be run to test demand and implementation. The initial trial allowed ETH holders to borrow sETH against their staked ETH at a 150% Collateralisation Ratio with an APR of 5%.
## Synthetic assets (Synths)¶
Synths are synthetic assets, minted against the value of cryptoassets staked as collateral. The values of Synths are provided by external price feeds fed on-chain.
### Forex¶
Forex Synths track the price of forex currencies via price feeds supplied by Chainlink's oracle network of distributed node operators.
### Commodity¶
Commodity Synths track the price of commodities via price feeds supplied by Chainlink's oracle network of distributed node operators.
### Crypto¶
Crypto Synths track the price of cryptocurrencies via price feeds currently supplied by a Synthetix oracle. Crypto Synths will soon be transitioned to use Chainlink's oracle network.
### Index¶
Index Synths track the price of indices via price feeds supplied by Chainlink's oracle network of distributed node operators. These indices can either be available in off-chain finance, such as sNIKKEI, or they can be custom designed by the Synthetix community, such as sDEFI.
### Inverse (iSynths)¶
iSynths inversely track the price of assets via price feeds supplied by either Synthetix's oracle or Chainlink's oracle network of distributed node operators. They allow traders to effectively take a short position, and are currently available for Crypto Synths and Index Synths.
Each iSynth has three important points: an entry point, an upper limit, and a lower limit. Its entry point is the price at which it is entered into the system.
#### An example of how iSynths work¶
Let's say iBTC was added into the system when sBTC was at $5000. That would mean iBTC's entry point would be $5000, and its value would inversely track that of sBTC, so if the price of sBTC drops to $4900 then iBTC would be at $5100, and vice versa.
Let's also say that iBTC's upper limit is $7500 and lower limit is $2500 (these are programmed when it's added to the system). These represents the points at which this iSynth gets frozen. If sBTC reaches $2500, then iBTC reaches its upper limit of $7500 and is frozen at that value. Any further fluctuations in the value of sBTC will not be reflected in the value of iBTC. The Synthetix core contributors will then endeavour to redeploy the frozen iSynth as soon as possible. This involves 'purging' everyone who holds the frozen iSynth into sUSD, after which a new iBTC is then deployed with a new entry point and new limits.
#### iSynth leverage¶
The primary reason iSynths need upper and lower limits is because the further an iSynth gets from its entry point, the more effective leverage there is for each movement due to design.
#### An example of how iSynth leverage works¶
As BTC moves away from the entry point iBTC starts to behave more like a leveraged token. If iBTC was at $3750, halfway to its lower limit of $2500, then each $1 of iBTC purchased generates $1.67 of price movement when BTC moves $1, since 6250 / 3750 = 1.6667. With iBTC at $2500, it would generate $3 of price movement (if it wasn't frozen), since 7500 / 2500 = 3. To use a different and more extreme example without limits, if iBTC's entry point was at $5000, and sBTC got up to $9999, leaving iBTC at $1, then you get around 10,000x leverage at that point, because for every 1:10,000 movement in sBTC you get a 1:1 movement in iBTC. | 2020-12-04 15:12:38 | {"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": 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": 0.34037160873413086, "perplexity": 8480.560038787664}, "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/1606141737946.86/warc/CC-MAIN-20201204131750-20201204161750-00452.warc.gz"} |
https://www.hackmath.net/en/math-problem/407 | # Cube corners
From the cube of edge 5 cm, cut off all vertices so that each cutting plane intersects the edges 1 cm from the nearest vertice.
How many edges will have this body?
n = 36
### Step-by-step explanation:
$n=3\cdot 8+3\cdot 4=36$
Did you find an error or inaccuracy? Feel free to write us. Thank you!
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## Related math problems and questions:
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• Cutting the prism
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• Magnified cube
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• Cutting circles
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• Cube 1-2-3
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The cube has an edge of 25 cm. We cut it into small cubes of 5 cm long side. How many of these little ones left when we build a new cube of 20 cm in length?
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The regular triangular prism, whose edges are identical, has a surface of 2514 cm ^ 2 (square). Find the volume of this body in cm3 (l).
• Hole
In the center of the cube with edge 14 cm we will drill cylinder shape hole. Volume of the hole must be 27% of the cube. What drill diameter should be chosen? | 2021-10-20 19:15:31 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 1, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5615328550338745, "perplexity": 918.6911570150899}, "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-2021-43/segments/1634323585348.66/warc/CC-MAIN-20211020183354-20211020213354-00700.warc.gz"} |
https://www.physicsforums.com/threads/direct-product-of-cyclic-groups.780437/ | # Direct Product of Cyclic Groups
1. Nov 6, 2014
### Bashyboy
Hello everyone,
I was wondering if the following claim is true:
Let $G_1$ and $G_2$ be finite cyclic groups with generators $g_1$ and $g_2$, respectively. The group formed by the direct product $G_1 \times G_2$ is cyclic and its generator is $(g_1,g_2)$.
I am not certain that it is true. If I make the following stipulation
Let $G_1$ and $G_2$ be finite cyclic groups with generators $g_1$ and $g_2$, respectively, and the group formed by the direct product $G_1 \times G_2$ is cyclic, then it has the generator $(g_1,g_2)$.
this might be true. However, I would like to hear from you before I try to go prove something that is false.
2. Nov 6, 2014
### WWGD
The product is cyclic iff the orders are relatively-prime. In the direct product, you have:
$|G_1 \times G_2|=|G_1| \times |G_2|$ . Then, if you can find an element in the product with that
order, you are done.
3. Nov 6, 2014
### Bashyboy
More importantly, I am interested in knowing if the generator of $G_1 \times G_2$ is based off the generators of the individual groups $G_1$ and $G_2$.
Or is this not true in general?
4. Nov 6, 2014
### WWGD
Notice that the generator of a cyclic group is not necessarily unique.
RE your question on relation between individual generators and generators of the product: yes, it is. Let $g_1, g_2$ be generators for $G_1, G_2$ respectively. Then there are positive integers $m,n$ with
$g_1^n=e_{G_1} , g_2^m = e_{G_2}$. What is then the order of $(g_1, g_2)$ ?
5. Nov 7, 2014
### Bashyboy
Let's see if I understand this correctly: Let $|G_1| = x$ and $|G_2| = y$, and let both be both cyclic with the generators alluded in the above posts. If this is so, then $\langle g_1 \rangle = G_1 \implies |\langle g \rangle | = x$, with a similar thing being true of $G_2$. Furthermore, $|G_1 \times G_2 | = xy$. There is a theorem that I proved which shows that the direct product of two subgroups produces a subgroup of the direct product of the larger group(s). Therefore, $\langle g_1 \rangle \times \langle g_2 \rangle$ is a subgroup, and its order is $| \langle g_1 \rangle \times \langle g_2 \rangle | = xy$. Thus, the element $(g_1,g_2)$ has an order of $xy$, and it must generate $G_1 \times G_2$.
Does this sound right?
6. Nov 9, 2014
### TheMathSorcerer
The order need not be $xy$. Take $Z_2$ which is cyclic with generator $1$. Then in the group $Z_2 \times Z_2$, the element $(1,1)$ has order two since $(1, 1) \neq (0,0)$, and $(1, 1) + (1, 1) = (0, 0)$, yet $Z_2 \times Z_2$ has order $4$, so $(1,1)$ does not generate $Z_2 \times Z_2$. Note $Z_2 \times Z_2$ is isomorphic to the Klein 4-group which is not cyclic. I think perhaps a look at the previous post is a good idea as it really says it all. What is the order of $(g_1, g_2)$? It is worth looking at. LaTeX fixed, thanks Greg:)
Last edited: Nov 10, 2014
7. Nov 10, 2014 | 2018-06-20 15:25:29 | {"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": 0.8998096585273743, "perplexity": 183.82890387996855}, "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/1529267863650.42/warc/CC-MAIN-20180620143814-20180620163814-00188.warc.gz"} |
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# In the figure above, if angle ABC = 110 degrees, and angle AED = 100 d
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In the figure above, if angle ABC = 110 degrees, and angle AED = 100 d [#permalink]
### Show Tags
Updated on: 23 Jun 2011, 20:27
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In the figure above, if angle ABC = 110 degrees, and angle AED = 100 degrees, then angle ADE is how many degrees smaller than angle BCD?
(1) Angle DAB = 45 degrees
(2) Angle ADC + angle BCD = 205 degrees
Hello. I am doing the PR1012, and I came across this question. I got it wrong to begin with, but I also do not understand their brief explanation. Please help me. Just for the record I am in the college class of 2015, but I need to take the GMAT before the end of my sophomore year for a combined degree program (shooting for 700; I know it is really hard!). Eventually I may want an MBA, though. I am studying now to ease into it. I am pretty smart for the most part, but I do badly on tests. I got a 1380 SAT (670M, 710CR) with a little studying. Anyway, here is the question.
After one day of studying I already hate the data questions.
In the figure above, if angle ABC = 110 degrees, and angle AED = 100 degrees, then angle ADE is how many degrees smaller than angle BCD?
1. Angle DAB = 45 degrees
2. Angle ADC + angle BCD = 205 degrees
OPEN DISCUSSION OF THIS QUESTION IS HERE: this-ds-i-just-cannot-figure-out-i-am-getting-e-99063.html
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Originally posted by jw on 23 Jun 2011, 14:31.
Last edited by jw on 23 Jun 2011, 20:27, edited 1 time in total.
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Joined: 24 Mar 2011
Posts: 341
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Re: In the figure above, if angle ABC = 110 degrees, and angle AED = 100 d [#permalink]
### Show Tags
23 Jun 2011, 15:10
st-2 is fairly simple to get BCD - ADE, so st is sufficient.
as per st -1, BCD = 360-110-100-45-DAE
BCD = 105-DAE
combining both eqns, BCD = 105-80+ADE
==> BCD - ADE = 25
sufficient.
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Re: In the figure above, if angle ABC = 110 degrees, and angle AED = 100 d [#permalink]
### Show Tags
23 Jun 2011, 22:47
My apologies for mis-posting.
Please move this to the data sufficiency forum.
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Re: In the figure above, if angle ABC = 110 degrees, and angle AED = 100 d [#permalink]
### Show Tags
24 Jun 2011, 13:28
St 2 is very simple, as discussed in the above post. But I solved the first stmt like below, I hope it'll help you!
Let <BCD = y; <ADE = x; therefore <CDA = 180 - x (since, <CDA and <ADE are supplimentary)
Now sum of angles of quadrilateral ABCD is 360;
<DAB + <ABC + <BCD + <CDA = 360 => 45 + 110 + Y + 180 - X = 360
You can get the difference b/t y and X which is the answer for the question. Therefore the answer is D
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Re: In the figure above, if angle ABC = 110 degrees, and angle AED = 100 d [#permalink]
### Show Tags
24 Jun 2011, 14:39
I still do not see how statement two is so simple.
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Posts: 1772
Re: In the figure above, if angle ABC = 110 degrees, and angle AED = 100 d [#permalink]
### Show Tags
24 Jun 2011, 14:50
jw wrote:
I still do not see how statement two is so simple.
Sum of angles in a straight line is 180.
2.
Sufficient.
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Posts: 55271
Re: In the figure above, if angle ABC = 110 degrees, and angle AED = 100 d [#permalink]
### Show Tags
14 Aug 2017, 03:18
jw wrote:
In the figure above, if angle ABC = 110 degrees, and angle AED = 100 degrees, then angle ADE is how many degrees smaller than angle BCD?
(1) Angle DAB = 45 degrees
(2) Angle ADC + angle BCD = 205 degrees
Hello. I am doing the PR1012, and I came across this question. I got it wrong to begin with, but I also do not understand their brief explanation. Please help me. Just for the record I am in the college class of 2015, but I need to take the GMAT before the end of my sophomore year for a combined degree program (shooting for 700; I know it is really hard!). Eventually I may want an MBA, though. I am studying now to ease into it. I am pretty smart for the most part, but I do badly on tests. I got a 1380 SAT (670M, 710CR) with a little studying. Anyway, here is the question.
After one day of studying I already hate the data questions.
In the figure above, if angle ABC = 110 degrees, and angle AED = 100 degrees, then angle ADE is how many degrees smaller than angle BCD?
1. Angle DAB = 45 degrees
2. Angle ADC + angle BCD = 205 degrees
Important property of polygons: Sum of Interior Angles of a polygon is $$180(n-2)$$ where $$n$$ is the number of sides.
So for quadrilateral sum of the angles equals to $$180(4-2)=360$$ degrees.
Given: $$\angle{ABC}=110$$ and $$\angle{AED}=100$$. Question: $$\angle{BCD}-\angle{ADE}=?$$.
Attachment:
DS.jpg
Now as $$\angle{ABC}=110$$ and $$\angle{AED}=100$$ and the sum of all angles is 360 degrees then the sum of other two angles in quadrilateral must be $$\angle{BCD}+\angle{BAE}=360-(110+100)=150$$.
(1) $$\angle{DAB}=45$$ --> $$\angle{BAE}=\angle{DAB}+\angle{DAE}=45+\angle{DAE}$$, but $$\angle{DAE}=180-100-\angle{ADE}=80-\angle{ADE}$$ (the sum of interior angles in triangle ADE equals to 180 degrees) --> $$\angle{BAE}=45+80-\angle{ADE}=125-\angle{ADE}$$. So $$\angle{BCD}+\angle{BAE}=150$$ becomes $$\angle{BCD}+125-\angle{ADE}=150$$ --> $$\angle{BCD}-\angle{ADE}=25$$. Sufficient.
(2) $$\angle{BCD}+\angle{ADC}=205$$ --> $$\angle{ADC}=180-\angle{ADE}$$ (straight line) --> $$\angle{BCD}+180-\angle{ADE}=205$$ --> $$\angle{BCD}-\angle{ADE}=25$$. Sufficient.
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Re: In the figure above, if angle ABC = 110 degrees, and angle AED = 100 d [#permalink] 14 Aug 2017, 03:18
Display posts from previous: Sort by | 2019-05-24 21:41:05 | {"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": 0.7332332730293274, "perplexity": 2260.8612177449004}, "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-2019-22/segments/1558232257767.49/warc/CC-MAIN-20190524204559-20190524230559-00193.warc.gz"} |
https://casper.astro.berkeley.edu/astrobaki/index.php/Radiation_Lecture_23 | <latex> \documentclass[11pt]{article} \def\inv#1Template:1 \over \def\ddtTemplate:D \over dt \def\mean#1{\left\langle {#1}\right\rangle} \def\sigot{\sigma_{12}} \def\sigto{\sigma_{21}} \def\eval#1{\big|_{#1}} \def\tr{\nabla} \def\dce{\vec\tr\times\vec E} \def\dcb{\vec\tr\times\vec B} \def\wz{\omega_0} \def\ef{\vec E} \def\ato{{A_{21}}} \def\bto{{B_{21}}} \def\bot{{B_{12}}} \def\bfieldTemplate:\vec B \def\apTemplate:A^\prime \def\xp{{x^{\prime}}} \def\yp{{y^{\prime}}} \def\zp{{z^{\prime}}} \def\tp{{t^{\prime}}} \def\upxTemplate:U x^\prime \def\upyTemplate:U y^\prime \def\e#1{\cdot10^{#1}} \def\hf{\frac12} \def\^{\hat } \def\.{\dot } \def\ppt#1Template:\partial \def\ddtau#1Template:D \def\wtTemplate:\tilde\omega \def\aiaio{\left[a_1I(\mu_1)+a_2I(\mu_2)\right]} \usepackage{fullpage} \usepackage{amsmath} \usepackage{eufrak} \begin{document} Recall that we were solving the following equation for $I$: $$\mu\ddtau{I(\mu,\tau)}=I(\mu,\tau)-{\wt\over2}\left[I_1(\mu_1,\tau)+ I_2(\mu_2,\tau)\right]-{\wt\over4}\mathfrak{F}e^{-{\tau\over\mu_0}}$$ We then broke $I$ into two components measuring the upward and downward directed components of the specific intensity: \begin{aligned}\mu_1\ddtau{I(\mu_1)}&=I(\mu_1)-{\wt\over2}\aiaio-{\wt\over4}\mathfrak{F} e^{-\tau\over\mu_0}\\ \mu_2\ddtau{I(\mu_2)}&=I(\mu_2)-{\wt\over2}\aiaio-{\wt\over4}\mathfrak{F} e^{-\tau\over\mu_0}\\ \end{aligned} The final solution is a sum of exponentials: \begin{aligned}I(\mu_1,\tau)&\equiv I_1=I(\mu_1=\inv{\sqrt{3}},\tau)\\ &=I_{1,hmg}+I_{1,part}\\ I(\mu_2,\tau)&\equiv I_2=I(\mu_2=1\inv{\sqrt{3}},\tau)\\ &=I_{2,hmg}+I_{2,part}\\ \end{aligned} and last time we solved for the homogeneous component. All that remains to be done at this point is to fit the boundary conditions.
\subsection*{ Fitting Boundary Conditions }
The first condition is that $I_2$, which measures the intensity directed toward the ground, should vanish at $\tau=0=$the surface of the atmosphere (we are only considering reflected light here): $$I_2(\mu_2,\tau=0)=0$$ The second condition is that at the ground the intensity directed upward out of the ground should be equal to whatever is reflected from the downward directed intensity. Thus: $$I_1(\tau=\tau_{max},\mu_1>0)=I_2(\tau-\tau_{max},\mu_2<0)\Lambda$$ where $\Lambda$ is the ground albedo. There is another component we should examine. First, let's define: $$I_*(\tau,\mu)=\pi\mathfrak{F}\delta(\mu+\mu_0)\delta(\phi-\phi_0)e^{\tau\over\mu}$$ Then we'll examine the total flux (specific intensity integrated over solid angle, weighted by $\mu$) directed into the ground: \def\taum{{\tau_{max}}} \begin{aligned}F_{in,*} &=\int{\pi\mathfrak{F}\delta(\mu+\mu_0)\delta(\phi-\phi_0)e^{\taum\over\mu}\mu \,\underbrace{d\mu d\phi}_{d\Omega}}\\ &=-\pi\mathfrak{F}e^{-{\taum\over\mu}}\mu_0\\ \end{aligned} Since the flux out of the ground is just the reflected portion of the incoming flux: \begin{aligned}F_{out,*}&=\Lambda F_{in,*}\\ F_{out,*}&\equiv\int{I_{1,*}\mu d\Omega}\\ &=I_{1,*}\int{\mu d\Omega}=I_{1,*}\pi\\ \end{aligned} $$\boxed{I_{1,*}=\lambda\mathfrak{F}\mu_0e^{-{\taum\over\mu_0}}}$$ Therefore, the full second boundary condition reads: $$\boxed{I_1(\tau=\taum,\mu_1>0)=\Lambda I_2(\tau=\taum,\mu_2<0) +\Lambda\mathfrak{F}e^{-{\taum\over\mu_0}}\mu_0}$$ These two boundary conditions together give us answers to the two right-hand components of our original equation: $$\mu\ddtau{I(\mu,\tau)}=I(\mu,\tau)-{\wt\over2}\left[I_1(\mu_1,\tau)+ I_2(\mu_2,\tau)\right]-{\wt\over4}\mathfrak{F}e^{-{\tau\over\mu_0}}$$ So the equation had just become the equation of radiative transfer: $$\mu\ddtau{I}=I-S$$ and now we know the source function $S$.
\subsection*{ Effective Albedo of the Atmosphere }
Using 4.1.29 in the handout, we have that an atmosphere of optical depth $\tau=\taum$, with ground albedo $\Lambda$ is identical to an infinite atmosphere ($\tau\to\infty$) if $\xi=0$. In order for this to be the case: $$\xi={\sqrt{1-\wt}(1+\Lambda)-(1-\Lambda) \over\sqrt{1-\wt}(1+\Lambda)+(1-\Lambda)}=0$$ $$\boxed{\Lambda={1-\sqrt{1-\wt}\over1+\sqrt{1-\wt}}}$$ where $\Lambda$ here is the effective, macroscopic albedo of a semi-infinite atmosphere.\par We can get this same result by solving the diffusion equation, this time with an absorption modifier: $$\ppt{n}=\underbrace{D\nabla^2n}_{scattering\atop term} -\underbrace{nn_d\sigma_ac}_{absorption\atop modifier}$$ where $n$ is the \# density of photons (which don't change in frequency), $\sigma_a$ is the cross-section for absorption, $n_d$ is the \# density of the dust (any absorbing/scattering particle), and $D$ is the diffusivity, $D=\lambda_{mfp}c$. We have to be careful that $\lambda_{mfp}$ contain both aspects of absorption and scattering: $$\lambda_{mfp}=\inv{n_d(\sigma_a+\sigma_s)}$$ We have a steady-state solution to this diffusion equation. In 1-D, this equation looks like: $$\ppt{n}=0=D{\partial^2n\over\partial z^2}-n(n_d\sigma_ac)$$ We'll guess that $n=Ae^{z\over z_0}$ is a solution. Doing some algebra, we can show that: $$z_0=\inv{n_d[\sigma_a(\sigma_a+\sigma_s)]^\hf}$$ where we are assuming $z=0$ is the top of the atmosphere, where $\tau=0$, and that $z<0$ as we move down through the atmosphere. Now we'll break our solution into two components, evaluated at the surface of the atmosphere: \begin{aligned}n&=Ae^{z\over z_0}\\ n(0)&=(n_i+n_e)e^{0\over z_0}\\ \end{aligned} where $n_i$ are the incident photons from space, and $n_e$ are the escaping photons from below. Then the diffusion equation (this time written in terms of flux) tells us: \begin{aligned}F&=-D\nabla n=c(n_e-n_i)\\ {-DAe^{z\over z_0}\over z_0}\eval{z\sim0}&=c(n_e-n_i)\\ \end{aligned} We are saying $z\sim0$, and not $z=0$, because we have to be careful about using the diffusion equation at low optical depth, which we have at the surface of the atmosphere. Ignoring the subtleties inherent in the assumption, we'll say that $e^{z\over z_0}\sim1$, giving us: $$\boxed{{-D\over z_0}(n_i+n_e)}$$ Now we use the definition $\Lambda\equiv{n_e\over n_i}\eval{z=0}$ to get our previous equation. First, note that: $$n_e={\left(\sqrt{\sigma_a+\sigma_s\over\sigma_a}-1\right)\over \left(\sqrt{\sigma_a+\sigma_s\over\sigma_a}+1\right)}n_i ={\left(\sqrt{\inv{1-\wt}}-1\right)\over\left(\sqrt{\inv{1-\wt}}+1\right)}n_i$$ where we used that $1-\wt={\sigma_a\over\sigma_a+\sigma+s}$. Using this result, we can solve for $\Lambda$: $$\Lambda={n_e\over n_i}={1-\sqrt{1-\wt}\over1+\sqrt{1-\wt}}$$
\subsection*{ Quick Reality Check }
We can use the fact that $$F_{out}\eval{surface}=\int{I_1\eval{surface}\mu\,d\Omega}$$ to independently find the macroscopic albedo: $$\Lambda={F_{out}\eval{surface}\over F_{in}\eval{surface}}$$ Doing this by hand, using the test case of looking at reflected flux directed straight up ($\mu_0=1$), from perfectly reflective particles $\wt=1$, with $\mu_1=\hf$ (an approximation of a quadrature gaussian integral), we get: \begin{aligned}F_{out}\eval{surface}&={3\pi\over2}\ln2\mathfrak{F}\\ F_{in}\eval{surface}&=\pi\mathfrak{F}\\ \end{aligned} Numerically, we find $\Lambda=1.03$. We were, of course, shooting for an answer of $1$, so this hand-calculation gives us an indication that all of the various assumptions we made have introduced a couple \% error in our final calculations.
\end{document} <\latex> | 2022-01-27 18:41:56 | {"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": 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": 0.9574009776115417, "perplexity": 1153.200260547445}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-05/segments/1642320305277.88/warc/CC-MAIN-20220127163150-20220127193150-00529.warc.gz"} |
https://swedometer.com/percentage-change-asymmetry/ | ## Percentage Change
Most people are quite familiar with the concept of percentage. If we were not, shops would not bother putting up mathematical symbols tempting us with 50% discount on a nice sweater. Not the least on today’s Black Friday.
Repeated percentage change is less familiar territory for most. Compound interest for instance, is the result of positive percentage change only, and is a form of exponential growth. But in this post we will look at alternating positive and negative percentage change.
We are used to the operation +1 followed by -1 resulting in no change, as the sum is zero. But what about the operation +1% followed by -1%? Well, there is a significant difference between the two types of operation. The addition and subtraction of +1 and -1 are independent of the value they operate on, the difference is always 1. Percentage change on the other hand is not independent of the value it operates on. 1% of something large is larger than 1% of something small.
## An Asymmetry Shows Up
An immediate effect of percentage change dependency on the value it operates on is that the “chronology” matters. The second change can only be applied after the first change has been calculated etc. With this in mind, we take a closer look at the result of alternating positive/negative percentage change.
Consider equal positive and negative 20% changes. This is equivalent to multiplying by 1.2 for increase and by 0.8 for decrease. In other words, regardless of the order (increase or decrease first) we end up with a change of 1.2*0.8 = 0.8*1.2 = 0.96, which is a decrease.
In order not to change the original value from one cycle to the next, the percentage increase must be slightly larger than the decrease. Assume that the increase is x and the decrease is y. An unchanged value requires that x*y = 1. For example, with y = 0.8 it follows that x = 1.25. This means that a 20% decrease must be followed by a 25% increase.
The two cases discussed above are illustrated in the figure below, where alternating percentage changes are applied to an original value of 100. Note that when alternating with -20% and +20% the results (blue line) get smaller and smaller. In contrast, when alternating with -20% and +25% the result oscillates but the trend never drops.
## Side Note
Even though decreasing and increasing by the same percentage lowers the result, it will never reach zero. This is another consequence of percentage change scaling with the value it acts on. As the value gets smaller, so does the change. To become zero, the change must be -100%, from which there is also no recovery, as a percentage change of zero is zero. | 2019-02-16 16:28:43 | {"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": 0.8500035405158997, "perplexity": 439.6525708092508}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-09/segments/1550247480622.9/warc/CC-MAIN-20190216145907-20190216171907-00395.warc.gz"} |
https://electronics.stackexchange.com/questions/441276/what-is-triplication-on-fpga/441284 | # what is triplication on fpga?
I know FPGA design using VHDL and I came up some new topic recently that usage of triplication in FPGA but I am not confident about its understanding.
How can we use triplication in FPGA design and how do we verify this.
• i don"t know what it is either, but why do you need to use it if you don't know what it is? – jsotola May 31 '19 at 6:22
• This is the enhancement to the current design strategy. – srihari May 31 '19 at 6:50
• @srihari we'll need more context. "Triplication" is an English word and means "to make everything be there three times". But it's not clear how this would apply to VHDL design; if you just put the same VHDL modules there three times and then combined the results, the synthesizer will just optimize away two of them, because that's what synthesizers do: analyze the logic of a circuit and reduce it to the correct minimal amount of logic elements necessary to produce the same result. – Marcus Müller May 31 '19 at 7:06
• The Ramans did everything in threes... – htmlcoderexe May 31 '19 at 20:41
Triplication means (as noted) to make 3 of everything.
It is used in space and safety critical designs, and data results are voted; a disagreement in the vote has to be designed such that the erroneous result circuit is reset. For this to work within a single device, partial reconfiguration in the FPGA is required.
The reason this is necessary in SRAM based FPGA devices is that they are susceptible to single event effects (such as SEUs) which can flip a configuration bit from 0 to 1 or vice versa and therefore changing the effective circuit.
SEUs are caused primarily by high energy free neutrons and cosmic radiation (in space applications) although the lead in SnPb solder (in particular those on BGA devices) can also emit alpha particles as some of the lead is $$\Pb ^{210}\$$ which decays via $$\Bi^{210}\$$ and $$\Po^{210}\$$ to $$\Pb^{206}\$$ (Uranium decay chain) which is the stable state for lead.
Flash based devices are immune to configuration state changes from free neutrons although they can be susceptible to X-Ray, and the data path can be protected by parity or ECC. | 2020-02-25 21:16:25 | {"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": 4, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.4928221106529236, "perplexity": 1116.6153604794242}, "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/1581875146160.21/warc/CC-MAIN-20200225202625-20200225232625-00372.warc.gz"} |
https://ja.coursera.org/learn/convolutional-neural-networks/reviews?authMode=signup&page=2 | Convolutional Neural Networks に戻る
# deeplearning.ai による Convolutional Neural Networks の受講者のレビューおよびフィードバック
4.9
40,154件の評価
5,316件のレビュー
## コースについて
In the fourth course of the Deep Learning Specialization, you will understand how computer vision has evolved and become familiar with its exciting applications such as autonomous driving, face recognition, reading radiology images, and more. By the end, you will be able to build a convolutional neural network, including recent variations such as residual networks; apply convolutional networks to visual detection and recognition tasks; and use neural style transfer to generate art and apply these algorithms to a variety of image, video, and other 2D or 3D data. The Deep Learning Specialization is our foundational program that will help you understand the capabilities, challenges, and consequences of deep learning and prepare you to participate in the development of leading-edge AI technology. It provides a pathway for you to gain the knowledge and skills to apply machine learning to your work, level up your technical career, and take the definitive step in the world of AI....
## 人気のレビュー
AR
2020年7月11日
I really enjoyed this course, it would be awesome to see al least one training example using GPU (maybe in Google Colab since not everyone owns one) so we could train the deepest networks from scratch
OA
2020年9月3日
Great course. Easy to understand and with very synthetized information on the most relevant topics, even though some videos repeat information due to wrong edition, everything is still understandable.
フィルター：
## Convolutional Neural Networks： 26 - 50 / 5,289 レビュー
by AQUIB I
2020年5月24日
Really an amazing course about CNN's. what an amazing instructor Andrew is. Totally recommended course those who want to learn CNN's from basic.
by Ralph R
2019年4月27日
I think it's a good idea to remove repeated parts in the videos. Also, put all pieces toguether to give a better overview of the object detection solution
by Basile B
2018年4月30日
IoU validation problem is known but nothing as been done to resolv it
video editing problem
unreadable formula in python notebook for art generation (exemple :
$$J_{style}^{[l]}(S,G) = \frac{1}{4 \times {n_C}^2 \times (n_H \times n_W)^2} \sum _{i=1}^{n_C}\sum_{j=1}^{n_C}(G^{(S)}_{ij} - G^{(G)}_{ij})^2\tag{2}$$
What append ? that was great so far... =(
by Shibhikkiran D
2019年7月8日
First of all, I thank Professor Andrew Ng for offering this high quality "Deep Learning" specialization. This specialization helped me overall to gain a solid fundamentals and strong intuition about building blocks of Neural Networks. I'm looking forward to have a next level course on top of this track. Thanks again, Sir!
I strongly recommend this specialization for anyone who wish get their hands dirty and wants to understand what really happens under the hood of Neural networks with some curiosity.
Some of the key factors that differentiate this specialization from other specialization course:
1. Concepts are laid from ground up (i.e you to got to build models using basic numpy/pandas/python and then all the way up using tensorflow and keras etc)
2. Programming Assignments at end of each week on every course.
3. Reference to influential research papers on each topics and guidance provided to study those articles.
4. Motivation talks from few great leaders and scientist from Deep Learning field/community.
2020年4月15日
I'm Zeyad, an undergraduate of Computer Engineering at Alexandria University in Egypt.
Taking this course really helped me to learn and study this field and also to implement it. It helped me advance in my knowledge. This course helped me defining Deep Learning field, understanding how Deep Learning could potentially impact our business and industry to write a thought leadership piece regarding use cases and industry potential of Machine Learning.
This specialization helped me identifying which aspects of Deep Learning field seem most important and relevant to us, apparently they were all important to us. Walking away with a strong foundation in where Deep Learning is going, what it does, and how to prepare for it.
Deep Learning specialization helped me achieving a good learning and knowledge about that field.
Thank you so much for offering such wonderful piece of art.
Best Regards,
by Michael J
2019年1月2日
A short (but cogent) overview of CNNs with a ton of references to read through and much more interesting assignments (than previous courses). I really enjoyed this course, I got a ton of exposure from it.
by Devjyoti M
2019年4月22日
This is one of the best courses for CNNs. This gives a very deep understanding of the concepts and helps to understand the brains behind the CNNs and their working in application based environments.
by Daniel G
2018年2月13日
Too much hand-holding during assignments, although still very good directions. Obviously the issue with the final programming assignment needs to be addressed. Fantastic lecture material, as always.
by Tian Q
2019年1月1日
Excellent introductory course for CNN. The basic ideas and key components are explained clearly. Coding assingments helped me understand the algorithm to every little detail.
by Ambrish K
2022年1月28日
Awesome course. Programming assignments give a real world scenario for trying things. Overall a complete course for studying and implementing convolutional neural networks.
by Jose A O
2022年1月4日
The programming assignments are great. However, there are too many constraints placed on the students. Many parts of the code are already provided, but in my opinion it would be more beneficial to allow the student to also complete many of these auxiliary codes.
by Cosmin D
2019年1月4日
Good content, videos have the occasional editing hiccups that also affect other courses in this specialisation. Assignments could be a little bit harder but do a reasonable job at familiarising with useful deep learning frameworks.
by Sai B A
2019年10月9日
The course content is great, I felt link the programming assignments should have more information on running the Tensorflow sessions and (optional )information for people who are not familiar with Tensorflow would be great.
by Chris A
2018年6月10日
Great course - only thing keeping me from giving 5 stars is the consistent problem with the notebooks/grader.
by 小贱贱
2018年3月14日
assignment of week 3 has a bug about calculation of iou
by Moustapha M A
2018年1月29日
I am a bit disappointed with this course , despite best efforts by Andrew. There is serious lack of rigor and while it is exciting to see things work , there is very little science to give us a methodical reason of why it works . In ConvNet we see the input data, a multi dimensional matrix get reduced in size using filtering and convolution operation techniques. From a mathematical point of view, this is clear and can be formalized but it is not clear why this process causes the ability to identify edges in a picture and evolve as we go deeper into the convNN to the real picture etc...
It seems to me this more like an alchemy rather then a rigorous scientific approach and this is why it was difficult to follow the exercises from the material of the course . I have to put concerted efforts to understand the literature which itself was not easy as it lacked rigorous mathematical and scientific approach ( why we have to increase the channels by multiples as we go deep into the conVNN ? etc...) . It seems to me the whole field is at its infancy with trials and errors - and more formalized approach is needed.
by Jacob K
2019年8月31日
Great content, but this module gets far too buggy. The videos stutter and repeat as if they were going to be edited butt never were, and the programming exercises are so sloppy. The first exercise says, welcome to the second exercise, and congratulates you for finishing the course, even though the second assignment remains, that also says welcome to the second exercise! Loading a model hangs forever on one, and running the GAN crashes the kernel on the other. People in the forum have been complaining since at LEAST last year, and it's still buggy. This course content is great, but very shoddily put together compared to the rest. I am literally scared what week 5 will be like. Just clean it up guys. Hire an temp!
by Younes A
2017年12月7日
Wouldn't recommend because of the very low quality of the assignments, but I don't regret taking them because the content is great. Seriously the quality of deeplearning.ai courses is the lowest I have ever seen! Glitches in videos, wrong assignments (both notebooks and MCQs), and no valuable discussions on the forums. Too bad Prof Ng couldn't get a competent team to curate his content for him.
by Bilal B
2019年8月19日
I know I am giving 2 stars :( but unfortunately this course was bit difficult and I don't know why Professor didn't first gave few fundamental concepts of computer vision. It's just my opinion maybe I'm wrong, maybe I'm right. But honestly we should have gone through some basic C.V. so that few students like myself can get a better understanding rather than directly diving into use of DL in CV.
by Milica M
2020年5月10日
boring and uninformative; could use improvement and some rehearsal before giving a lecture; boring and unorganized delivery; slides are horribly unorganized and boring; often times very confusing and hard to follow; should minimize the number of times the instructor references basic math and should use that time to motivate the concepts and applications
2021年8月8日
This course is an excellent introduction to Convolutional Neural Networks (aka CNNS, aka ConvNets). The instructor makes the material understandable while not straying away from going into the mathematics behind CNNs. This course also starts to get into some of the really cool applications of AI/ML/DL (such as facial recognition and neural style transfer). I also enjoy how this course (and the rest of the courses in the specialization) keeps a great balance between theory and application. It covers enough of the application (techniques and programming) that you could reasonably start working on a computer vision project straight out of the course, however, it still covers much of the theoretical and in-depth knowledge that you may need to know. The main problem I have with other CS MOOCs is that I sometimes feel that they either only focus on the theory or they only focus on the application (programming and engineering). In the former, you understand many of the in-depth concepts but still need to do a decent amount of learning on your own before you can start making stuff, and with the ladder, while you may have the programming knowledge to know how to program an application, you don't really understand the concepts so you have trouble solving a wide range of problems. For the most part, this course and this specialization straddle the line pretty well. It mainly focuses on the concepts but gives you enough practice that you could start on an application. Thank you so much to Andrew Ng and the team at Deeplearning.ai! This course was great! :)
by Weinan L
2018年3月12日
This may be the most enjoyable course in the whole series so far. It is intuitive and fun, and the results are tangible. Very practical.
Inevitably, due to the complexity of CNN, we have to rely on frameworks such as TensorFlow/Keras, etc. to do the coding, and they are covered in this course as well. Not very deep, but sufficient. Wish they may pick PyTorch in the future as well.
The notebook and grading systems sometime have issues though. You may think you submitted the right data but actually the server side won't think so. Hard lessons learnt are: a) save the original ipynb before coding, so you can always rollback in case notebook messed up; b) save a checkpoint before submit, this will force saving and ensure you submitted the latest data, otherwise, it may submit incomplete data - some cells may still have very old data even you modified a lot; c) open anther local Jupyter notebook to experiment and mess around, with interactive TensorFlow exception, but pay attention to the expression with random sequences, when you call eval() the second time, they may have totally different value even you reset the seed upon each cell, eval() will invoke your expression again which will consume more data in the random sequence; d) never use iPad to complete your noetbook coding, :-).
by Alan L V J
2017年12月4日
Este curso introductorio es estupendo para aprender desde cero sobre convolutional neural networks.
Professor Andrew Ng, makes very comprehensible the content of the course.
Here why:
-He decompose every element of CNN. Convolutions, 1x1 convolutions and pooling are very well explained, then by yourself can derive the dimensions of the output after applying these operations.
-He make notes on the fly for derive equations and explain the purpose of the equations. For me is much better that only show slides, because makes give me the oportunity to think of the equation before is show.
-Professor give you Intiition in every topic.
- He Make several examples of modern architectures of CNNs.Always write down in detail the architectures.
-Clear notation, uses the same notation in programming exercises
-Programming exercises are the best documente ones. This makes relatively easy to implement the exercises. If struggle with operations, they provide links to the documentation necessary.
Was an amazing course.
Althogth I always think CNNs were some what difficult and sometimes tedious topic (because of convolution and pooling arithmetic, and the use of "volumes" instead of matrices), this course make all clear and natural.
Thanks to the instructors for they hard work.
by Neil O
2018年7月4日
If you're not particularly interested in image identification and recognition, there is still reason to do this course. CNNs are amongst the most advanced areas of DL and understanding the concepts can help develop intuition about how to solve DL problems in other domains. I greatly enjoyed this course. As with all of Andrew Ng's courses, the explanations are clear and help develop intuition. This course seems to have more references to academic papers than the others and Andrew is encouraging and helpful in guiding the student to the accessible and relevant sections of the papers.The exercises are instructive and not too challenging. Most of the challenges I had were due to my own programming errors and occasionally an error in how the exercise is set up [make sure to use the most recent version of Jupiter notebooks]. One exercise in Week 4 (Neural Style Transfer) does assume more Tensorflow knowledge than the other exercises. Recommend brushing up on Tensorflow before trying this and using the discussion groups which are helpful for debugging suggestions.
by Plusgenie
2018年8月27日
Coursera 온라인 강좌 딥 러닝에 정말 감동 받은 점:
#1 정규 대학교나/대학원 가지 않고 온라인으로 싸게 배울 수 있다.
#2 아무리 어려워 보이는 학문이더라고, 관점을 정확하게 설명해주면 누군든지 쉽게 배울 수 있다.
즉 E=MC^2 같은 공식은 누구나 발견할 수 없지만, 누구가 쉽게 배울 수 있는 것이다. 학생이 모르면 선생의 잘 못이다!
#3 지식은 투명하게 공개되어야 한다. 공개되지 않는 지식은 특권계급을 만든다.
#4 학교를 떠난지 그렇게 오래되었지만, 여기에 다시 공부해보니 다시 청춘을 느끼게 해준다.
“This is a record of your time. This is your movie. Live out your dreams and fantasies. Whisper questions to the Sphinx at night. Sit for hours at sidewalk cafes and drink with your heroes. Make a pilgrimage to Mougins or Abiquiu. Look up and down. Believe in the unknown for it is there. Live in many places. Live with flowers and music and books and paintings and sculpture. Keep a record of your time. Learn to write well. Learn to read well. Learn to listen and talk well. Know your country, know the world, know your history know yourself.
Take care of yourself physically and mentally. You owe it to yourself. Be good to those around you and do all of these things with passion. Give all that you can.Remember, life is short and death is long.”
– Fritz Scholder | 2022-01-28 18:56:04 | {"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": 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": 0.22837068140506744, "perplexity": 2023.1979320096075}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-05/segments/1642320306335.77/warc/CC-MAIN-20220128182552-20220128212552-00653.warc.gz"} |
https://math.stackexchange.com/questions/2167272/prove-prime-p-mid-ab-rightarrow-p-mid-a-or-p-mid-b-without-using-f/2167366 | # Prove prime $p\mid ab\,\Rightarrow\, p\mid a\,$ or $\,p\mid b\,$ without using Fundamental Theorem of Arithmetic
Let: $p$ $\in \mathbb{P}$ $\wedge$ $n_{1},n_{2}\in \mathbb{Z}$. Then: $p|(n_{1}n_{2})\implies p|n_{1} \vee \space p|n_{2}$
This little hypothesis is straightforward while using fundamental theorem of arithmetic. I also know that this can be proved directly by the use of the contraposition for the above implication. However, I wonder how to do this without referring to the fundamental theorem of arithmetic or to contraposition. I think that this must be very easy, but I can't see it right now. Thanks for help in advance.
• See here for a few proofs, including a direct proof by descent using the Division algorithm. This also includes further elaboration on the proofs using the GCD Distributive Law in Xam's answer, its Bezout form in Leox's answer. – Bill Dubuque Mar 1 '17 at 18:43
• Isn't this a duplicate? – Bob Happ Mar 1 '17 at 21:54
Let suppose that $p\not\mid n_1$, so $\gcd(p,n_1)=1$. Now, since $p\mid n_1n_2$ and $p\mid pn_2$, then by the definition of $\gcd$ we have $$p\mid \gcd(pn_2, n_1n_2)=n_2\gcd(p,n_1)^{*}=n_2.$$
In (*) we've used the property $\gcd(ac,bc)=c\gcd(a,b)$.
• So easy????????????????????????????????? – user410985 Mar 1 '17 at 17:54
• Btw, thanks for help. – user410985 Mar 1 '17 at 17:54
• @MIT it's easy if you have practice. I suggest you that in order to get familiarity with these simple proofs try to prove that property (*) that I've used. Good luck. – Xam Mar 1 '17 at 18:05
• I would start the same way, then use Gauss theorem : since $p\mid n_1n_2$ and $\gcd(p,n_1)=1$, we see that $p\mid n_2$. – Adren Mar 1 '17 at 18:13
Suppose that $p\not\mid n_1$. Then there exist integers $x,y$ such that $px+n_1 y=1.$ Multiple both sides by $n_2$ $$p (x n_2)+ n_1 n_2 y =n_2.$$ $p$ divides LHS thus $p \mid n_2.$
• This is the Bezout form of the proof in Xam's answer. See here where I prrecisely highlight the analogy. See also the answer linked in my comment on the question. – Bill Dubuque Mar 1 '17 at 18:38
• Are you suggesting to me that I should delete my answer? – Leox Mar 1 '17 at 18:43
• No, why would you think that? Rather, I gave a link so that readers can learn how these common proofs are related. – Bill Dubuque Mar 1 '17 at 18:45 | 2020-01-24 17:25:37 | {"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": 0.9280637502670288, "perplexity": 336.74559738857494}, "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/1579250624328.55/warc/CC-MAIN-20200124161014-20200124190014-00019.warc.gz"} |
https://mdtpmodules.org/ratl/ratl-3/lesson-2/practice/ | # RATL 3 | Lesson 2 | Explore (Solution Sets)
In this lesson you will learn to solve rational inequalities. One we get a solution set, there are several different ways to express a solution set. An example is the following written in basic style:
$$-7\leq{x}<4$$.
This same solution set can be represented in:
a) set builder notation as $$\{x|-7\leq{x}<4\}$$.
b) interval notation as $$[-7, 4)$$.
From the number lines below, write the solution set in basic notation, set builder notation and interval notation. Put this information in the chart on the bottom of this page.
Problem Basic Style Set Builder Interval
a
b
c
d | 2020-04-03 21:17:24 | {"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": 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": 0.7774688005447388, "perplexity": 1200.3136150122987}, "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-2020-16/segments/1585370518622.65/warc/CC-MAIN-20200403190006-20200403220006-00465.warc.gz"} |
https://ag91.github.io/blog/2020/12/31/top-down-elisping-a-simple-snippet-to-stub-a-function-while-your-are-designing-your-code/ | # Where parallels cross
Interesting bits of life
# Top down Elisping: a simple snippet to stub a function while your are designing your code
Yesterday I was writing some Elisp to help me score how risky it is to merge a pull request according to the age of a commit and others factors (something I want add to code-compass and I hope to blog about later on), and I have finally fixed something that was boring me while Elisping: maybe it helps you to!
So say you want to write some top down Elisp. You would start with something like:
(defun function-to-wish-happy-2021 ()
"Make wishes 2021 to people."
nil)
And then you would write this functio by adding the high level process (top down):
(defun function-to-wish-happy-2021 ()
"Make wishes 2021 to people."
(call-the-rest (message-wishes-to-some-people (make-a-list-of-people))))
At this point I get bored. For each function I have to:
1. copy each function name,
2. write a new function with that name,
5. AND FINALLY write the code for it.
Where only the last point is interesting for me. So I said: "you know what? Enough, let's meta-Elisp this out".
And here the result (rough, but working!):
(defun my/stub-elisp-defun ()
"Stub an elisp function from symbol at point."
(interactive)
(let* ((fun (thing-at-point 'list 'no-properties)))
(when fun
(name (symbol-name (nth 0 fun-list)))
(args (cdr fun-list)))
(save-excursion
(or (search-backward "(defun" nil 't) (goto-char (point-min)))
(insert
(s-concat
"(defun "
name
" "
(format "%s" (--map (s-concat "arg" (number-to-string it)) (number-sequence 1 (length args))))
"\n \"SomeDocs\"\n nil)\n\n")))))))
Now look!
That is it: just put your cursor on the function name you want to stub, call my/stub-elisp-defun, and have a stub awaiting for your implementation.
The resulting code looks like:
(defun call-the-rest (arg1)
"SomeDocs"
nil)
(defun message-wishes-to-some-people (arg1)
"SomeDocs"
nil)
(defun make-a-list-of-people nil
"SomeDocs"
nil)
(defun function-to-wish-happy-2021 ()
"Make wishes 2021 to people."
(call-the-rest (message-wishes-to-some-people (make-a-list-of-people))))
Hopefully it will take away a bit of boring typing while you code Elisp!
Maybe somebody knows about an Elisp package that does these things already? I would really love something like clj-refactor for Elisp!
Merry 2021! | 2021-10-24 12:08:29 | {"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": 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": 0.4218491315841675, "perplexity": 14939.036633618833}, "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/1634323585997.77/warc/CC-MAIN-20211024111905-20211024141905-00045.warc.gz"} |
https://cs.stackexchange.com/questions/104265/union-of-infinitely-many-regular-languages | # Union of infinitely many regular languages [duplicate]
I need to prove or disprove the following statement.
If $$A_n ⊆ \Sigma^*$$ is regular for each $$n \in \mathbb{N}$$ then $$\bigcup\limits_{n=0}^{\infty} A_n$$ is regular.
I know that if two languages are regular, then the union of the languages is also regular. I don't think that really applies to this problem, because it's the union of every $$n \in \mathbb N$$. Also, to help with my understanding with the definitions, is $$A_n$$ a language or an alphabet, since it's a subset of $$\Sigma^*$$?
## marked as duplicate by xskxzr, Evil, Community♦Feb 13 at 21:57
Is $$A_n$$ a language or an alphabet, since it's a subset of $$\Sigma^*$$?
A member of the alphabet, $$\Sigma$$ is called a symbol or a letter.
A member of $$\Sigma^*$$ is called a string or a word, which is a finite sequence of symbols or letters.
A subset of $$\Sigma^*$$ is called a language.
If $$A_n\subseteq \Sigma^*$$ is regular for each $$n\in\Bbb N$$ then $$\bigcup\limits_{n=0}^{\infty} A_n$$ is regular.
As you suspected, this is not true. For example, let $$A_n=\{a^nb^n\}$$. Then $$\bigcup\limits_{n=0}^{\infty} A_n$$ is the well-known non-regular language of words with equal number of $$a$$'s and $$b$$'s. Infinity should indeed be treated carefully.
Exercise 1. Give an example of a non-regular language $$L$$ over unary alphabet such that $$L=\bigcup\limits_{n=0}^{\infty} A_n$$ where $$A_n$$ is a regular language for all $$n$$.
Exercise 2. Let $$L$$ be any language. Then $$L=\bigcup\limits_{n=0}^{\infty} A_n$$ for some regular language $$A_n$$.
• My only question is I thought we were assuming the language to be regular? I thought {a^n b^n} was not regular, so we aren't even starting with the assumption that An is regular. – James Swanson Feb 13 at 20:31
• $A_1=\{ab\}$. $A_2=\{a^2b^2\}$. $A_3=\{a^3b^3\}$. And so on. Each $A_n$ is a language that has only one word. Their union, $\{ab, a^2b^2, a^3b^3, \cdots\}=\{a^nb^n\mid n\in \Bbb N\}$ is not regular. – Apass.Jack Feb 13 at 21:14
• Ok i misinterpreted the question, so its to assume that every A(sub n) is regular, which I guess is very obvious since they would all be finite, because they are singletons. – James Swanson Feb 13 at 21:43
• You can use $A_n$ to show $A_n$. – Apass.Jack Feb 13 at 21:53
• Yes, as you said, it is very obvious indeed. Please accept the answer so that we may call an end of the question. – Apass.Jack Feb 13 at 21:56 | 2019-06-26 08:11:56 | {"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": 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": 25, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6711352467536926, "perplexity": 256.746277180326}, "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/1560628000231.40/warc/CC-MAIN-20190626073946-20190626095946-00178.warc.gz"} |
https://holooly.com/solutions/a-man-a-with-a-mass-m-_-a-85-kg-and-a-child-c-with-a-mass-m-_-c-18-kg-are-at-the-opposite-ends-of-a-floating-platform-p-with-a-mass-m-_-p-150-kg-and-a-length-l-_-fp/ | ## Question:
A man A, with a mass ${ m }_{ A }$ = 85 kg, and a child C, with a mass ${ m }_{ C }$ = 18 kg, are at the opposite ends of a floating platform P, with a mass ${ m }_{ P }$ = 150 kg and a length ${ L }_{ fp }$ = 6 m. Assume that the man, child, and platform are initially at rest and that the resistance due to the water to the horizontal motion of the platform is negligible. Suppose that the man and child start moving toward each other in such a way that the platform does not move relative to the water. Determine the distance covered by the child until meeting the man. | 2020-10-19 23:59:27 | {"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": 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": 4, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5548556447029114, "perplexity": 214.353013908344}, "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-2020-45/segments/1603107867463.6/warc/CC-MAIN-20201019232613-20201020022613-00457.warc.gz"} |
https://aas.org/archives/BAAS/v26n4/aas185/abs/S9002.html | Sub-Arcsecond $\lambda$6 cm Images of PNe and their Expansion Parallax Distances
Session 90 -- Planetary Nebula and Degenerate Stars
Oral presentation, Wednesday, 11, 1995, 10:00am - 11:30am
## [90.02D] Sub-Arcsecond $\lambda$6 cm Images of PNe and their Expansion Parallax Distances
Arsen R. Hajian (Cornell University), Yervant Terzian (Cornell University)
Over the past three years, we have obtained high resolution, high fidelity images of more than 20 PNe at $\lambda$6 cm with the Very Large Array. When these data are combined with a second earlier epoch from the VLA archives, we have been successful in detecting the angular expansion of most nebular shells. Distances can be derived with this method by dividing the Doppler velocity of an appropriate emission line by the angular expansion rate of the nebula. The uncertainties this method poses are generally smaller than errors in more indirect techniques for computing PNe distances. In this paper, we present images of all PNe surveyed, as well as resulting expansion parallax results. | 2016-10-25 07:26:21 | {"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": 0.6421868205070496, "perplexity": 4724.772917435139}, "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-2016-44/segments/1476988719960.60/warc/CC-MAIN-20161020183839-00211-ip-10-171-6-4.ec2.internal.warc.gz"} |
https://math.stackexchange.com/questions/121372/picard-group-of-generic-fibre | Picard group of generic fibre
Let $C$ be an irreducible curve over a field $k$ and let $X$ be a $k$-variety equipped with a morphism $f: X \to C$. Let $X_{k(C)} \to k(C)$ be the generic fibre of this morphism. Under which "reasonable" conditions on $X$, $C$ and/or $f$ (smoothness, properness and so on) will the natural sequence
$$\text{Pic}\,C \to \text{Pic}\,X \to \text{Pic}\,X_{k(C)} \to 0$$
be exact? For example, does this hold if $X$, $C$ and $f$ are smooth and proper?
• I'm not sure what exactly you're looking for. Do you want the weakest possible conditions to make this true, or just some conditions maybe to make the proof clean? For example, the first thing that pops into my head is to make the assumption at the beginning of II.6 in Hartshorne so that there is a good interpretation of this sequence in terms of divisors.
– Matt
Mar 17, 2012 at 18:13
• Well, I guess condition (*) on page 130 in Hartshorne is definitely something I should impose. I don't necessarily want the weakest possible conditions - rather some "nice" conditions which allow the proof to be clean. Thanks for your comment! Mar 17, 2012 at 18:29
You have to suppose $X\to C$ flat to avoid empty generic fiber.
Assume $X$ is regular and flat over $C$.
Then $\mathrm{Pic}(X)\to \mathrm{Pic}(X_K)$, where $K=k(C)$, is surjective. Indeed, identifying invertible sheaves (up to isomorphism) to Weil divisors (up to linear equivalence), it is enough to show that any point of codimension $1$ $P$ in $X_K$ extends to a divisor on $X$. It then suffices to take the Zariski closure of $\{ P\}$.
Now let us look at the exactness at middle. An element of $\mathrm{Pic}(X)$ is in the kernel of $\mathrm{Pic}(X)\to \mathrm{Pic}(X_K)$ if and only if it is represented by a Weil divisor on $X$ supported in finitely many closed fibers of $X\to C$:
(1) if $\mathcal L\in \mathrm{Pic}(X)$ is trivial on $X_K$, dividing by a rational section which is a basis on $X_K$, we can suppose that $\mathcal L$ is a subsheaf of $K(X)$ and equal to $O_X$ on an open subset $U$ containning $X_K$. So $\mathcal L=O_X(D)$ for some Cartier divisor $D$ supported in $X\setminus U$. As $F=f(X\setminus U)$ is constructible hence finite, $D$ is supported in $f^{-1}(F)$.
(2) Conversely, a divisor supported in a finite union of closed fibers is clearly trivial on $X_K$.
So the exactness at the middle is equivalent to saying that any vertical divisor is principal. Note that $f(X)$ is open in $C$ and $f(X)$ is regular because $X$ is regular and $X\to f(X)$ is faithfully flat. Now it is enough (and essentially necessary) to suppose the fibers of $X\to C$ are integral because every closed fiber $X_s$ is then a principal divisor (if $s\notin f(X)$, there is nothing to prove; if $s\in f(X)$, then $[s]$ is a principal divisor and so is $[X_s]=f^*[s]$).
• Dear QiL, thank you very much for your reply. I still have a few questions. (1) In the first part of your reasoning (surjectivity), you mean a codimension 1 point on Xk(C) instead of a closed point, right? (2) The second sentence of your second paragraph sounds very reasonable, but what is the formal argument? (3) The last sentence of your second paragraph sounds very reasonable as well, but I am not sure why you need $C$ to be regular to make things work... Sorry for my incomprehension, and thanks again! Mar 17, 2012 at 22:37
• @Evariste: (1) Sorry I had the situation of relative curves in mind. I will add some details for (2). The regularity of $C$ is in fact implied by that of $X$.
– user18119
Mar 17, 2012 at 23:33
• Great, thanks a lot! Mar 18, 2012 at 0:05
Let $f\colon X\to C$ be a faithfully flat morphism of locally noetherian schemes which is either quasi-compact or locally of finite type, where $C$ is normal and integral with function field $K$. Assume that, for every point $s\in C$ of codimension $1$, the fiber $X_{s}$ is integral. Then the canonical sequence $${\rm Pic}\, C\to {\rm Pic}\, X\to {\rm Pic}\, X_{K}$$ is exact. This result is due to Raynaud (see EGA, ${\rm Err}_{\,\rm IV}$, 53, Corollary 21.4.13, p. 361). If, in addition, $X$ is locally factorial, then the right-hand map above is surjective. The following proof of the latter surjectivity was sent to me by Cedric Pepin. By EGA, ${\rm IV}_{4}$, Corollary 21.6.10(ii), the latter map can be identified with the map of divisor class groups ${\frak{Cl}}\, X\to {\frak{Cl}}\, X_{K}$. Thus it suffices to check that every closed and irreducible subscheme $D_{K}$ of codimension 1 in $X_{K}$ extends to a closed and irreducible subscheme $D$ of codimension 1 in $X$. Since ${\rm Spec}\, K\to C$ is quasi-compact, the canonical morphism $D_{K}\to X$ is quasi-compact as well and the schematic closure $D$ of $D_{K}$ in $X$ is defined by EGA 1 (new), Corollary 6.10.6, p. 325. Since $D$ is closed and irreducible of codimension 1 in $X$, the proof is complete. If anyone knows a statement that is more general than the above, please let me know! | 2022-08-11 04:59:54 | {"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": 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": 0.9427037239074707, "perplexity": 144.95824054304362}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-33/segments/1659882571234.82/warc/CC-MAIN-20220811042804-20220811072804-00656.warc.gz"} |
http://physics.stackexchange.com/tags/potential/hot | # Tag Info
22
In general, the answer is no. This type of inverse problem is sometimes referred to as: "Can one hear the shape of a drum". The following extensive exposition by Beals and Greiner discusses various problems of this type. Despite the fact that one can get a lot of geometrical and topological information from the spectrum or even its asymptotic behavior, this ...
11
The "simplest" classical explanation I know is the van der Waals interaction described by Keesom between two permanent dipoles. Let us consider two permanent dipoles $\vec{p}_1$ (located at $O_1$) and $\vec{p}_2$ located at $O_2$. Their potential energy of interaction is: U(\vec{p}_1,\vec{p}_2,\vec{O_1 O_2}) = -\vec{p}_1\cdot \vec{E}_2 = ...
10
Since the energy spectrum does not depend on the absolute position $\vec{r}=\vec{a}$ of the delta potential, we may assume that $\vec{a}=\vec{0}$. Therefore, in its current formulation (v1), OP is effectively saying that The attractive 1D delta potential $V(x) = -A\delta(x)$, $A>0$, has exactly one bound state. The same is true for the 3D delta ...
10
Wavefunctions are found by solving the time-independent Schrödinger equation, which is simply an eigenvalue problem for a well-behaved operator: $$\hat{H} \psi = E \psi.$$ As such, we expect the solutions to be determined only up to scaling. Clearly if $\psi_n$ is a solution with eigenvalue $E_n$, then $$\hat{H} (A \psi_n) = A \hat{H} \psi_n = A E_n ... 8 All bound states can typically be chosen to have real-valued wavefunctions. The reason for this is that their wavefunction obeys a real differential equation,$$ -\frac{\hbar^2}{2m}\nabla^2\psi(\mathbf r)+V(\mathbf r)\psi(\mathbf r)=E\psi(\mathbf r)and therefore for any solution you can construct a second solution by taking the complex conjugate ... 7 1) OP wrote (v1): [...] and thus this leads me to believe that {\bf A} should be somehow connected to momentum, [...]. Yes, in fact the magnetic vector potential {\bf A} (times the electric charge) is the difference between the canonical and the kinetic momentum, cf. e.g. this Phys.SE answer. 2) Another argument is that the scalar electric ... 7 If you want to generalize a potential to a class that's broader than the simple \frac12 k_2 x^2, it is tempting as a first step to include a small perturbation of the form \frac13k_3x^3. Unfortunately, this drastically changes the structure of the potential, because it becomes unbounded from below. Thus, you might get a slightly perturbed behaviour ... 6 Everything you've probably learned about capacitors, especially including the statement that opposite plates of the capacitor carry opposite charges, applies only to a capacitor in a circuit. If your capacitor is floating, so that the plates are not connected to anything, the charge on the plates is not going to change. If you hook up only one plate to a ... 6 Electrons will flow against the electric field lines because their charge is negative, and the electric field thus exerts a force \mathbf{F}=q\mathbf{E} on them which is in the opposite direction. Thus electric field lines inside the wire go from the positive to the negative terminal and the electron flow goes from the negative to the positive terminal. ... 6 Both sides of the equation given are vectors and so represent 3 equations which are, on a Cartesian basis: E_x + \frac{\partial A_x}{\partial t} = -\frac{\partial V}{\partial x} E_y + \frac{\partial A_y}{\partial t} = -\frac{\partial V}{\partial y} E_z + \frac{\partial A_z}{\partial t} = -\frac{\partial V}{\partial z} 6 The full mathematical statement is as follows: Theorem If two particles exert a mutual conservative force \mathbf{F}_{12} and \mathbf{F}_{21} which is independent of any other degree of freedom of any bigger system they're part of, and obeys Newton's third law as \mathbf{F}_{12}+\mathbf{F}_{21}=\mathbf{0}, with the forces collinear to the ... 5 You are right that the result you see is due to the chain rule. The author uses either spherical or cylindrical coordinates, so $$r = \sqrt{x^2 + y^2 + z^2}$$ or $$r = \sqrt{x^2 + y^2}$$ which you can differentiate to obtain \frac{\partial{r}}{\partial{x}} = \frac{x}{r} ... 5 Yes, u is indeed the potential energy. And yes, you can calculate the force acting on a particle by calculating the gradient of the potential energy field at the position the particle is in. Computationally you will want to calculate the force on particle 1, by taking the gradient at the position particle 1 is in, of the potential energy field created by ... 5 When there is no resistance, as is the case with an ideal wire, any value of current satisfies Ohm's Law: V = I R since both V=0 and R=0. UPDATE: But isn't V is like what causes the current? Perhaps a mechanical analogy of the resistor will help. Consider the dashpot where the velocity of the arm is analogous to current while the force acting ... 5 You missed a term in expanding the upper-indexed metric. The full version is below: \begin{align} \tilde{\Gamma}^\lambda_{\mu\nu} & = \frac{1}{2} \tilde{g}^{\lambda X} \left(\partial_\mu \tilde{g}_{\nu X} + \partial_\nu \tilde{g}_{\mu X} - \partial_X \tilde{g}_{\mu\nu}\right) \\ & =\frac{1}{2} \tilde{g}^{\lambda\sigma} \left(\partial_\mu ... 5 Let me first comment that the statement electric fields cancel while the electric potentials just add up algebraically is not actually correct. Electric fields add due to the principle of superposition (see the section on superposition in the wikipedia article). However, when two electric field vectors are of the same magnitude but point in ... 5 Every system likes to decrease its electrostatic energy. The charges on the plates are almost in stable equilibrium. The charges on the opposite plates attract them, and the charges on the same plate repel them with almost the same force. However, a capacitor has fringe fields: These may be negligible when calculating the field inside a capacitor, but ... 5 This is your circuit: The current that comes from the source, when reaches the point that must choose it's way, sees no difference between the two paths (symmetry) , so half of it flows through one way and the other part flows in the second way. It means that, I_1=I_2 , So the potential difference across yellow resistors is the same. It means that the ... 4 EDIT: Put simply, potential difference is the work done by electrostatic force on a unit charge, while EMF is the work done by anything other than electrostatic force on a unit charge. I don't like the term "voltage". It seems to mean anything measured in volts. I'd rather say electric potential and electromotive force. And the two are fundamentally ... 4 That schematic is quite confusing; draw it with higher voltages at the top, lower voltages at the bottom, and signals flowing from left to right per standard procedure. It's also helpful to label components and nodes so that you can solve the equation symbolically. I haven't labeled some of the nodes here because they can be represented by the names of ... 4 OP wrote (v1): I hardly feel that one out of two is a sufficient justification for the introduction of a potential. I) Here we would like to point out that there exists a velocity-dependent generalized potentialU~=~q(\phi - {\bf v}\cdot {\bf A}) $$for the Lorentz force$$ {\bf F}~=~ q({\bf E} + {\bf v}\times {\bf B}) . $$Here \phi is ... 4 Your Question all but includes the right search term for an Answer from Wikipedia, "Conservative Forces", which gets you to http://en.wikipedia.org/wiki/Conservative_Forces. There's even what you ask for, a proof. There's also another link to http://en.wikipedia.org/wiki/Conservative_vector_field, which gives some quite good visualizations that will probably ... 4 To be concrete, let us here assume that the dissipative force is a friction force$$\tag{1} {\bf F}~=~-k {\bf v} $$proportional to the velocity {\bf v}=\dot{\bf r} of the point particle. Recall that a velocity dependent potential U=U({\bf r},{\bf v},t) of a force {\bf F} by definition satisfies$$\tag{2} {\bf F}~=~\frac{d}{dt} \frac{\partial ...
4
1) Postponing for a moment the issue of magnetic monopoles, one conventional answer is, that the gauge potential $A_{\mu}$ (as opposed to, e.g., the electric and magnetic $\vec{E}$ and $\vec{B}$ fields) constitute the true fundamental variables and (the photon field) of QED. At the classically level, by saying that $A_{\mu}$ are fundamental variables, we ...
4
1) The infinitely long wire has an infinite charge $Q=\lambda \int_{-\infty}^{\infty} \! dz = \infty$, and EM has an infinite range, so one shouldn't be surprised to learn that the result $$\phi(r)~=~ \frac{\lambda}{4 \pi \epsilon_0} \int_{-\infty}^{\infty} \frac{dz}{\sqrt{z^2+r^2}} ~=~ \frac{\lambda}{4 \pi \epsilon_0} \left[ {\rm arsinh} ... 4 Frenkel and Smit definitely make a mistake. Eq. (12.1.3) page 294 is:$$-\nabla^2 \phi(\mathbf{r}) = 4\pi \rho(\mathbf{r}) $$then immediately afterwards, Eq. (12.1.4) is "the solution of this equation" "for a single charge z at the origin":$$\phi(\mathbf{r}) = \frac{z}{4\pi |\mathbf{r}|} This is a mistake: Eq. (12.1.4) is definitely not "the solution" ...
Only top voted, non community-wiki answers of a minimum length are eligible | 2013-12-19 10:54:08 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 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": 2, "x-ck12": 0, "texerror": 0, "math_score": 0.9473941326141357, "perplexity": 368.14115963531697}, "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-2013-48/segments/1387345762908/warc/CC-MAIN-20131218054922-00067-ip-10-33-133-15.ec2.internal.warc.gz"} |
https://math.stackexchange.com/questions/1443849/prove-that-mathbbr-setminus-0-sim-ab0-is-transitive | Prove that $(\mathbb{R}\setminus \{0\},\sim):= ab>0$ is transitive.
I have the following problem:
A relation $\sim$ on $\mathbb{R}\setminus\{0\}$ is defined by $a\sim b$ if $ab>0$. Show that $\sim$ is an equivalence relation and identify the equivalence classes.
I've been able to easily demonstrate that $\sim$ is both reflexive and symmetric, but I'm not sure how to approach demonstrating that it is transitive.
If $ab>0$, $bc>0$, then
$$ab\cdot bc >0 \implies ab^2c>0$$
Since $ab>0$, $b\neq 0$ so $b^2>0$. From the previous inequality, we have that $ac>0$.
• Ah. For whatever reason I had not considered multiplying them together. I had thought of dividing them out, but was getting nowhere that way. – agent154 Sep 20 '15 at 18:25
Hint: $ac = \frac{(ab)(bc)}{b^2}$.
• Shows there is no need for cases +1 – rschwieb Sep 20 '15 at 18:21
Let a~b i.e.ab>0 and b~c i.e. bc>0 Now ab>0 implys a and b have same sign i.e. either both are positive or both are negative. Similarly b and c have same sign. Thus,ultimately we get that a and c have same sign (because both have same sign as that of b), thus ac>0 and hence a~c.
Let $a\sim b$ and $b\sim c$. This means $ab>0$ and $bc>0$. If $a,b>0$, then $c>0$, hence $bc>0$. If $a<0$ and $b<0$, then $c<0$, hence also $bc>0$. So in both cases we have $b\sim c$.
First, if $ab>0$ then $ab^{-1}>0$, because if $ab>0$, then $ab^{-1}=(ab)b^{-2}>0$ (given that $\frac{1}{b^2}>0$). Now, suppose $a~b$ and $b~c$, so $ab>0$ and $bc>0$. Then $ac=(ab)(b^{-1}c)>0$. | 2019-08-22 00:18:08 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 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": 0.9917861223220825, "perplexity": 148.1424043968242}, "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/1566027316555.4/warc/CC-MAIN-20190822000659-20190822022659-00483.warc.gz"} |
https://www.physicsforums.com/threads/a-question-about-ring-homomorphisms.661878/ | A question about ring homomorphisms
Homework Statement
If R is a domain with F=Frac(R), prove that Frac(R[x]) is isomorphic to F(x).
The Attempt at a Solution
Let $$\phi : Frac(R[x]) \rightarrow F(x)$$ be a map sending (f(x),g(x)) to f(x)/g(x). We need to show that $$\phi$$ is a ring homomorphism. Let f,g,h,k be in R[x] such that f/h and g/k is in Frac(R[x]).
We know that
$$\phi(1,1) = 1/1= 1$$
$$\phi (fg, hk) = \frac{fg}{hk} = \frac{f}{h}\frac{g}{k} = \phi(f,h)\phi(g,k)$$
But I'm confused with the addition part...
$$\phi(f+g,h+k) = \frac{f+g}{h+k}$$
$$\phi(f,h)+\phi(g,k) = \frac{f}{h}+\frac{g}{k} = \frac{kf+gh}{h+k}$$
But now $$\phi(f+g,h+k) \not= \phi(f,h) + \phi(g,k)$$
Can anybody help with this?
Oh...ok I think I get what you mean...because $$\phi(f,h + g,k) = \phi(fk+gh, hk)$$, right? | 2021-09-19 13:17:12 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.7556040287017822, "perplexity": 1455.3804936755582}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-39/segments/1631780056890.28/warc/CC-MAIN-20210919125659-20210919155659-00710.warc.gz"} |
http://cpr-condmat-suprcon.blogspot.com/2013/05/13054281-l-j-swenson-et-al.html | ## Operation of a titanium nitride superconducting microresonator detector in the nonlinear regime [PDF]
L. J. Swenson, P. K. Day, B. H. Eom, H. G. Leduc, N. Llombart, C. M. McKenney, O. Noroozian, J. Zmuidzinas
If driven sufficiently strongly, superconducting microresonators exhibit nonlinear behavior including response bifurcation. This behavior can arise from a variety of physical mechanisms including heating effects, grain boundaries or weak links, vortex penetration, or through the intrinsic nonlinearity of the kinetic inductance. Although microresonators used for photon detection are usually driven fairly hard in order to optimize their sensitivity, most experiments to date have not explored detector performance beyond the onset of bifurcation. Here we present measurements of a lumped-element superconducting microresonator designed for use as a far-infrared detector and operated deep into the nonlinear regime. The 1 GHz resonator was fabricated from a 22 nm thick titanium nitride film with a critical temperature of 2 K and a normal-state resistivity of $100\, \mu \Omega\,$cm. We measured the response of the device when illuminated with 6.4 pW optical loading using microwave readout powers that ranged from the low-power, linear regime to 18 dB beyond the onset of bifurcation. Over this entire range, the nonlinear behavior is well described by a nonlinear kinetic inductance. The best noise-equivalent power of $2 \times 10^{-16}$ W/Hz$^{1/2}$ at 10 Hz was measured at the highest readout power, and represents a $\sim$10 fold improvement compared with operating below the onset of bifurcation.
View original: http://arxiv.org/abs/1305.4281 | 2018-03-22 21:43:43 | {"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": 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": 0.47134047746658325, "perplexity": 4329.772126761504}, "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-2018-13/segments/1521257648003.58/warc/CC-MAIN-20180322205902-20180322225902-00208.warc.gz"} |
https://mathemerize.com/the-mean-and-the-variance-of-a-binomial-distribution-are-4-and-2-respectively-then-the-probability-of-2-success-is/ | # The mean and the variance of a binomial distribution are 4 and 2, respectively. Then, the probability of 2 success is
## Solution :
Given that, mean = 4 $$\implies$$ np = 4
And Variance = 2 $$\implies$$ npq = 2 $$\implies$$ 4q = 2
$$\implies$$ q = $$1\over 2$$
$$\therefore$$ p = 1 – q = 1 – $$1\over 2$$ = $$1\over 2$$
Also, n = 8
Probability of 2 successes = P(X = 2) = $$^8C_2$$$$p^2$$$$q^6$$
= $$8!\over {2!\times 6!}$$ $$\times$$ $$({1\over 2})^2$$ $$\times$$ $$({1\over 2})^2$$
= $$28\over 256$$
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In a class of 100 students. there are 70 boys whose average marks in a subject are 75. If the average marks of the complete class is 72, then what is the average of the girls? | 2022-07-03 05:16:45 | {"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": 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": 0.9188945293426514, "perplexity": 393.8404038623647}, "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/1656104215790.65/warc/CC-MAIN-20220703043548-20220703073548-00029.warc.gz"} |
https://www.questarter.com/q/resolving-a-differential-equation-manually-21_3290778.html | # Resolving a differential equation manually
by dalta Last Updated July 12, 2019 10:20 AM - source
It is my first week dealing with Differential Equations, and I am stuck at the following question:
Find all the solutions of the following equation:
$$2y^2dx-(x+y)^2dy=0$$
I have consulted an online calculator, which gave the solution $$log\frac{y(x)}{x}+2arctan\frac{y}{x}=c_1-log(x)$$.
However, I would like to be able to do this manually.
Could anybody help out?
Tags :
Hint:
$$\dfrac{dy}{dx}=\dfrac{2y^2}{(x+y)^2}=\dfrac{2\left(\dfrac yx\right)^2}{\left(1+\dfrac yx\right)^2}$$
As the denominator & numerator are Homogeneous polynomials
Set $$y=vx,\dfrac{dy}{dx}=v+x\dfrac{dv}{dx}$$
lab bhattacharjee
July 12, 2019 09:33 AM
The differential equation
$$2y^2dx-(x+y)^2dy=0$$
can be rewriten as
$$2\dfrac{dx}{dy}=\left(\dfrac{x+y}{y}\right)^2$$
or
$$2\dfrac{dx}{dy}=\left(\dfrac{x}{y}+1\right)^2\ \ \ ...(1)$$
Now substituting $$\dfrac{x}{y}$$ as t, or $$x=yt$$
If you differentiate $$x=yt$$ w.r.t. $$y$$ on both sides you'll get $$\dfrac{dx}{dy}=t+y\dfrac{dt}{dy}$$
Substituting the value of $$\dfrac{dx}{dy} \text{ and } \dfrac{x}{y}$$ in equation (1)
$$2t+2y\dfrac{dt}{dy}=(t+1)^2$$
$$2t+2y\dfrac{dt}{dy}=t^2+1+2t$$
$$2y\dfrac{dt}{dy}=t^2+1$$
$$2\dfrac{dt}{t^2+1}=\dfrac{dy}{y}$$
Now integrating both sides
$$\displaystyle \int 2\dfrac{dt}{t^2+1}=\int \dfrac{dy}{y}$$
$$2\tan ^{-1} t=\ln(y)+C$$
Substituting the value of t in this expression
$$2\tan ^{-1} \dfrac{x}{y}=\ln(y)+C$$
In this question integrating $$2\dfrac{dx}{dy}=\left(\dfrac{x+y}{y}\right)^2$$ was much easier than integrating \$
Although my solution doesn't match yours it doesn't mean its incorrect, its a matter of integration constant.
Akash Karnatak
July 12, 2019 10:11 AM | 2019-07-17 19:36:36 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 21, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9242025017738342, "perplexity": 797.0636805103602}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-30/segments/1563195525374.43/warc/CC-MAIN-20190717181736-20190717203736-00481.warc.gz"} |
http://mathhelpforum.com/advanced-algebra/193323-finite-abelian-p-group.html | # Math Help - Finite abelian p-group
1. ## Finite abelian p-group
Suppose G is a finite abelian p-group. Prove that if G is cyclic iff G has exactly p-1 elements of order p.
$\forall g\in G, \ \ g^{p^{\alpha}}=1, \ \ |G|=p^{\alpha}$
$(\Rightarrow)$
Suppose G is cyclic. Then for a $g\in G$, $G=$.
What next? I am at a loss.
2. ## Re: Finite abelian p-group
Originally Posted by dwsmith
Suppose G is a finite abelian p-group. Prove that if G is cyclic iff G has exactly p-1 elements of order p.
$\forall g\in G, \ \ g^{p^{\alpha}}=1, \ \ |G|=p^{\alpha}$
$(\Rightarrow)$
Suppose G is cyclic. Then for a $g\in G$, $G=$.
What next? I am at a loss.
If $G$ is cyclic, you know it's isomorphic to $\mathbb{Z}_{p^\alpha}$ how many elements o order $p$ does this have (hint:recall cyclic groups have only one subgroup of a given order)? Assume now that $G$ has only $p-1$ elements of order $p$. Choose $g\in G$ with $|g|=p$, now assume that $x\in G-\langle g\rangle$, you know that $\langle x\rangle$ intersects trivially with $\langle g\rangle$ (why?) and so $\langle x\rangle$ is a cyclic subgroup of $G$ whose order is divisible by $p$--so?
3. ## Re: Finite abelian p-group
if G is cyclic, it has only one subgroup of order p. since any element of order p generates a subgroup of order p, all elements of order p must generate the same subgroup of G (since G has only ONE subgroup of order p, because its cyclic). so all elements of order p must all lie in that one subgroup of order p, which has (of course) p-1 elements of order p.
now suppose that G has exactly p-1 elements of order p. pick one of them, say x, we have a cyclic subgroup of G of order p <x>. this is the ONLY subgroup of order p of G.
now suppose H is a subgroup of order p^2, which has to be cyclic, because the direct product of 2 subgroups of order p is not possible. let y be a generator of H. then y^p has order p, so y^p is in <x>. hence <x> is a subgroup of H.
now suppose that K is a subgroup of order p^3....can you see where this is going? show that any subgroup of order p^k has to be cyclic, or else it contains more than one subgroup of order p.
conclude that G must be cyclic. | 2014-07-13 03:00:56 | {"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": 0, "img_math": 0, "codecogs_latex": 22, "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": 0.8975751399993896, "perplexity": 264.2363611599563}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-23/segments/1404776435842.8/warc/CC-MAIN-20140707234035-00021-ip-10-180-212-248.ec2.internal.warc.gz"} |
https://tug.org/pipermail/tugindia/2006-May/003961.html | # [Tugindia] Puzzled with floatflt
Ajay Narottam Shah ajayshah at mayin.org
Sat May 20 10:04:02 CEST 2006
On Sat, May 20, 2006 at 08:23:54AM +0530, Ajay Narottam Shah wrote:
> I tried to write the simplest possible demonstration of utilising
> floatflt to wrap text around figures and tables. The figures work
> fine, but the table is silently ignored. (There is no error
> message). Could someone suggest what I'm doing wrong?
Oops, I realised that attachments are being deleted. Okay, here's the
LaTeX file. To run it, you need a file "apicture.pdf" which is any
figure object.
\documentclass[12pt,a4paper]{article}
\usepackage{graphicx}
\usepackage{floatflt}
\begin{document}
This is some text. This is some text. This is some text. This is some
text. This is some text. This is some text. This is some text. This
is some text. This is some text. This is some text. This is some
text. This is some text. This is some text. This is some text. This
is some text. This is some text. This is some text. This is some
text. This is some text. This is some text.
Let's try for the next para.
\begin{floatingfigure}{0.5\textwidth}
\includegraphics[width=0.4\textwidth]{apicture.pdf}
\caption{Quarterly GDP growth}
\end{floatingfigure}
This is some text. This is some text. This is some text. This is some
text. This is some text. This is some text. This is some text. This
is some text. This is some text. This is some text. This is some
text. This is some text. This is some text. This is some text. This
is some text. This is some text. This is some text. This is some
text. This is some text. This is some text.
Let's see what happens when a figure is placed with a particularly
short paragraph.
\begin{floatingfigure}{0.5\textwidth}
\includegraphics[width=0.4\textwidth]{apicture.pdf}
\caption{Quarterly GDP growth}
\end{floatingfigure}
This is some text. This is some text. This is some text. This is some
text. This is some text. This is some text. This is some text.
This is some text. This is some text. This is some text. This is some
text. This is some text. This is some text. This is some text. This
is some text. This is some text. This is some text. This is some
text. This is some text. This is some text. This is some text. This
is some text. This is some text. This is some text. This is some
text. This is some text. This is some text.
Okay, now let's do a table.
\begin{floatingtable}{
\begin{tabular}{lrr}
\hline
Value & 1992-93 & 2002-03 \\
\hline
Blah & x & y \\
Blah & x & y \\
Blah & x & y \\
\hline
\end{tabular}}
\caption{A table}
\end{floatingtable}
This is some text. This is some text. This is some text. This is some
text. This is some text. This is some text. This is some text. This
is some text. This is some text. This is some text. This is some
text. This is some text. This is some text. This is some text. This
is some text. This is some text. This is some text. This is some
text. This is some text. This is some text.
\end{document} | 2022-06-30 18:43:28 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.9673691987991333, "perplexity": 3150.577480346363}, "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/1656103877410.46/warc/CC-MAIN-20220630183616-20220630213616-00044.warc.gz"} |
https://www.physicsforums.com/threads/implicit-function.191244/ | # Implicit function
1. Oct 14, 2007
1. The problem statement, all variables and given/known data
a. Find dy/dx given that x$$^{2}$$=y$$^{2}$$-4x=7y=15
b. under what conditions on x and/or y is the tangent line to this curve horizontal? vertical?
2. The attempt at a solution
I did solve the first question by simply using implicit fuction.
2x+2y*y'-4+ty*y'=0
y'=$$\frac{4-2x}{9y}$$
above is the answer that i ended up with
but im kinda stuck with the second question
if the tangent line is horizonat i guess the slope has to be zero
so i just set $$\frac{4-2x}{9y}$$=0 and
i ended up with x=2
but i don't know what to do with "vertical" part and
i'm not sure if i got the "horizonal" part either.
1. The problem statement, all variables and given/known data
2. Relevant equations
3. The attempt at a solution
2. Oct 14, 2007
### bob1182006
for a that's right, just when you wrote out the question you forgot some +'s and put = instead.
yes horizontal tangent is when the derivative =0. so x=2 is current for that.
vertical tangent occurs when the the derivative = +-infinity.
and if you have a fraction when does it = infinity?
3. Oct 14, 2007
### NonAbelian
If the question is supposed to be x²+y²-4x+7y = 15 then I believe that your answer for y' is incorrect.
Note that your original equation is that of a circle which has been shifted. To write it in standard form, (x-a)²+(y-b)²=c² you can complete the square (I'm getting to a nice geometrical result, just hang on :D)
(x²-4x+4) -4 + (y²+7y + 49/4) - 49/4 = 15
(x-2)² + (y+7/2)² = 15 +4 + 49/4
(x-2)² + (y+7/2)² = [sqrt(125)/2]²
So your original equation is a circle of radius [sqrt(125)/2], centered at (2,-7/2). From this you should gather that there are 2 points that have a horizontal tangent, and two points that have a vertical one. | 2017-11-19 14:49:16 | {"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": 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": 0.6519551277160645, "perplexity": 849.6459693703869}, "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/1510934805649.7/warc/CC-MAIN-20171119134146-20171119154146-00648.warc.gz"} |
https://math.stackexchange.com/questions/623457/find-all-entire-f-such-that-ffz-z | # Find all entire $f$ such that $f(f(z))=z$.
Suppose $f:\mathbb{C}\to \mathbb{C}$ is entire.
1. If $f(f(z))=z$, find all such $f$.
2. Can we find $f$ such that $f(f(z))=z^2$?
3. How about $f(f(z))=e^z$?
Ideas: For #1, we can show that $f$ must be a bijection, since $f$ failing to be either injective or surjective quickly leads to a contradiction.
My intuition tells me that $z\mapsto -z$ should be the only such (non-identity) function.
We know $f'(z)\neq 0$ for all $z$ (or else there would be an area where $f$ were not injective). Therefore $1/f'(z)$ is an entire function.
Taking derivatives of both sides, we get $f'(f(z))f'(z) = 1$, or $f'(f(z))=1/f'(z)$.
$1/f(z)$ should be analytic everywhere, except for a simple pole at $z=f(0)$.
Let $\max_{|z|=1}|f(z)-f(0)|=R_1$, then Schwarz's lemma implies that $f'(0)\leq R_1$. Let $\max_{|z|=1}|f(z+f(0))|=R_2$, then Schwarz's lemma implies $f'(f(0))\leq R_2$. But $f'(f(0)) = 1/f'(0)$, so $$\frac1R_2\leq f'(0)\leq R_1.$$
Trying to put all this together...
Update: I've got #1. Consider the singularity of $f$ at infinity. If it is removable, then $f$ is a constant by Liouville's theorem. If it is a pole, then $f$ is a polynomial. If it is an essential singularity, then by the Big Picard theorem, $f$ cannot be injective since $f$ will take every value, except perhaps one, in a neighborhood of infinity.
Therefore $f$ is a polynomial, and since $f'\neq 0$, we have $f=az+b$ for $a\neq 0$. Now we have $a(az+b)+b = z$, giving that the involutions are $z\mapsto z$ and $z\mapsto -z+c$ for $c\in \mathbb{C}$.
Update 2: I think I've got #2 as follows: Suppose $f$ entire such that $f(f(z))=z^2$. Since $z\mapsto z^2$ is surjective, we must have $f$ surjective.
Now $2z=f'(f(z))f'(z)$. So we know that $f'(f(z))f'(z)$ is zero only at $z=0$. Either $f'(f(0))=0$ or $f'(0)=0$. Suppose the former holds and not the latter. That is impossible, since then $f(f(z))$ would have zero derivative at two points, zero and $f(0)$. So we know $f'(0)=0$.
Since $f$ is surjective, some point must map to zero. It cannot be a nonzero point $a$, because then $f(f(z))$ would have zero derivative at $a$.
This implies that zero is the unique point mapping under $f$ to zero. Now $$(z^2)'' = 2 = f''(f(z))f'(z) + f'(f(z))f''(z)$$ and the right hand side is zero at $z=0$, a contradiction.
Update 3: Here's what I've got for #3 so far:
Note $e^z$ is nowhere zero. So if $f:w\mapsto 0$ then $w\not\in \text{Rg}(f)$. But only $0$ is not in $\text{Rg}(f)$. So only zero could possibly map to zero, but it does not, because $f(f(0))\neq 0$. So $f$ is nonzero.
$f'$ is also nonzero since $$(e^z)'=e^z=f'(f(z))f'(z).$$
$f$ should have an essential singularity at infinity.
• Your intuition for the first part is in error, there are more such functions. – Daniel Fischer Dec 31 '13 at 17:55
• First thing, what are the automorphisms of $\mathbb{C}$? Once you have those, it's easy to find the involutions among them. – Daniel Fischer Dec 31 '13 at 17:59
• @danielfischer Whoops, forgot the identity. – Eric Auld Dec 31 '13 at 18:01
• There are still more. – Daniel Fischer Dec 31 '13 at 18:02
• @DanielFischer OK, I've got #1 I think – Eric Auld Dec 31 '13 at 20:46
A general observation regarding the first two points:
If $f$ is an entire function such that $f(f(z)) = P(z)$ is a polynomial, then $f$ itself must be a polynomial, and hence $\deg P = \deg (f\circ f) = (\deg f)^2$ must be a square. That follows by Casorati-Weierstraß, for if $f$ had an essential singularity in $\infty$, there would be a sequence $z_n \to \infty$ with $f(z_n) \to 0$, and hence $f(f(z_n)) \to f(0) \neq \infty$.
That immediately rules out solutions to 2., and yields that solutions to 1. must be polynomials of degree $1$.
For part 3., if such an $f$ exists, it must be transcendental (otherwise $f\circ f$ would be a polynomial), so attain every $w \in \mathbb{C}$ with one exception infinitely often. Thus $\mathbb{C}\setminus \{0\} = f(f(\mathbb{C})) = f(\mathbb{C})$, and $f$ omits the value $0$. Since $\mathbb{C}$ is simply connected, $f$ has a logarithm, $f(z) = e^{g(z)}$. Now $f(f(z)) = e^z$ becomes
$$\exp \left(g(f(z))\right) = \exp(z) \iff \exp\left(g(f(z))-z\right)\equiv 1,$$
so $g(f(z)) - z \equiv 2\pi ik$ is constant, choosing $g$ accordingly, we have $g(f(z)) = z$. Hence $f$ is injective, which contradicts its transcendentality.
• Very good point! – Eric Auld Dec 31 '13 at 22:38
• Nice! I am trying to get a proof of non-existence for $f(f(z))=z^2$ using degree theory or homotopy. I will post if I get something. – Daniel Robert-Nicoud Jan 1 '14 at 2:19
• I wrote my own answer to problem $2$. I would like to have your opinion on its correctness, if you have the time to take a look at it. Thanks. – Daniel Robert-Nicoud Jan 1 '14 at 16:03
• If solutions to 1 are linear polynomials, then they are of the form $ax+b$. Then you have $a^2x+ab+b=x\quad\forall x$. So for $m\neq n$, you get $a^2(m-n)=m-n$ by $f(m)-f(n)$. This gives $a=\pm1$. $a=1$ gives $b=0$ and $a=-1$ gives $0=0$. So the only such functions must be $x=x$ or $-x+b\quad\forall b$.$\qquad$ – DynamoBlaze Jul 16 '18 at 19:39
Hellmuth Kneser found a solution to $f(f(x)) = e^x$ along the real line, and, since analytic, in a strip of varying width around the real axis in $\mathbb C.$
It does not extend to the entire plane. The obstruction in these problems is always the fixed points, $e^z = z.$ There are not any real fixpoints, but a countably infinite sequence in the plane, fairly easy to approximate. See SITE .
Evidently a full answer at https://mathoverflow.net/questions/12081/does-the-exponential-function-have-a-square-root to part 3.
On part 2, for the real part of $z$ strictly positive, we can take the principal branch of logarithm and so define $f(z) = z^{\sqrt 2}.$ And $f(f(z)) = z^2.$ I think I am remembering that from a Robert Israel answer somewhere. As you are finding, this does not extend to the entire plane.
• I made a pdf of Kneser's 1950 paper, but zakuski will not let me ssh today. Or as in a song: There was I, waiting at the church, Waiting at the church, Waiting at the church; When I found he'd left me in the lurch, Lor, how it did upset me! All at once, he sent me round a note Here's the very note, This is what he wrote: "Can't get away to marry you today, My wife, won't let me!" – Will Jagy Jan 1 '14 at 0:30
Problem 1: As hinted by @DanielFisher, the automorphisms of the complex plane are the functions $f(z) = az+b$ with $a\ne0$. The condition $f(f(z))=z$ is equivalent to $a=1$ and $b=0$ or $a=-1$ and $b\in\mathbb{C}$ any complex number.
Problem 2: Notice that we must have $$f(z^2)=f(f(f(z)))=(f(z))^2$$ and thus $f(1)$ is either $0$ or $1$. Consider the set $A=\{z\in\mathbb{C}|\exists n\in\mathbb{N}:z^{2n}=1\}\subset S^1$ and is not discrete (consider the sequence $z_k=e^{\frac{\pi}{k} i}$, which converges to $1$). The condition above implies that $|f(z)|=|f(1)|$ for all $z\in A$. If $f(1)=0$, then the identity theorem implies that $f=0$, which is a contradiction. Thus we must have $f(1)=1$ and thus $f$ maps $S^1$ to itself. Consider the restriction of $f$ to $S^1$ and look at the degree. We have $\deg(f)^2=\deg(f\circ f)=\deg(z^2)=2$, which is impossible. Thus such a function $f$ cannot exist.
Problem 3: It is easy to see that $0$ is not contained in the image of $f$, and that we can look at $f$ as a surjective map $f:\mathbb{C}\rightarrow\mathbb{C}^*$.
Claim: $f:\mathbb{C}\to\mathbb{C}^*$ is a covering map.
Proof: $f\circ f=\exp$ is a covering map. Let $z_0\in\mathbb{C}^*$, then we have some open neighborhood $V$ of $z_0$ which is evenly covered by $\exp$, i.e. $\exp^{-1}(V)=\bigsqcup_{i\in I}U_i$ and $\exp:U_i\to V$ is a homeomorphism for all $i$. Since $f$ is holomorphic and thus an open map (by the open mapping theorem), and $f|_{U_i}$ is injective (else $f\circ f$ couldn't be injective), $f|_{U_i}$ is a homeomorphism onto its image. Now $$f^{-1}(V) = f\left(\bigsqcup_{i\in I}U_i\right) = \bigcup_{i\in I}f(U_i)$$ All those sets are open, we are left to show that they are disjoint. Assume there exist $i,j\in I$ such that $f(U_i)\cap f(U_j)\ne\emptyset$ and $f(U_i)\ne f(U_j)$. Let $B_{ij}=f(U_i)\cup f(U_j)$, then $f(B_{ij})=V$, which is connected (or it can be chosen so). Consider $F=f|_{B_{ij}}$, then $$X_i=\{z\in V|F^{-1}(z) \mathrm{\ contains\ } i \mathrm{\ elements}\}$$ for $i=1,2$ are disjoint and cover $V$, thus one of them must be empty. This, together with the fact that $f(f(U_i))=V$, is a contradiction. Thus $f$ is a covering map.
Assuming the claim above is true, we have that both $f$ and $\exp$ are universal coverings $\mathbb{C}\to\mathbb{C}^*$. Thus there must exist an automorphism $h:\mathbb{C}\to\mathbb{C}$ such that $f(z)=\exp(h(z))$. But as seen in problem 1, we must have $h(z)=az+b$ (with $a\ne0$), and thus $f(z)=e^{az+b}$, which is impossible. We conclude that there is no holomorphic function such that $f(f(z))=e^z$.
(My thanks to @DanielFischer for the great help furnished with this solution to the third problem!)
• In 1., we have $z = a^2z + (a+1)b$, so $a^2 = 1$ and $(a+1)b = 0$. If $a = -1$, then $b$ can be arbitrary. 2. is correct. After having seen $f(1) = 1$, we could also use that then $f(0) = 0$, so $f(z) = c_kz^k+ \dotsc$ and $f(f(z)) = c_k^{k+1} z^{k^2} + \dotsc$ to obtain the contradiction. – Daniel Fischer Jan 1 '14 at 16:15
• @DanielFischer Thanks, I will correct $1$. I am onto something for $3$. Care to discuss it in chat? – Daniel Robert-Nicoud Jan 1 '14 at 16:18
• Main chat room? – Daniel Fischer Jan 1 '14 at 16:27
• Truth to tell, I created this: chat.stackexchange.com/rooms/12256/complex-analysis but if you prefer, main chat room is good for me. – Daniel Robert-Nicoud Jan 1 '14 at 16:28 | 2021-06-12 12:14:00 | {"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": 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": 0.9642874598503113, "perplexity": 170.65034652037377}, "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-2021-25/segments/1623487582767.0/warc/CC-MAIN-20210612103920-20210612133920-00285.warc.gz"} |
https://socratic.org/questions/5823722eb72cff56e22ca49b | # Question ca49b
Nov 9, 2016
$m = {\text{33.4 g of F}}_{2}$
#### Explanation:
Use the following formula, where $N$ is the number of molecules, ${N}_{A}$ represents Avogadro's number, and $n$ represents the number of moles.
$N = {N}_{A} \cdot n$
Now, we can substitute the known values.
5.30 * 10^(23) " molecules F"_2 = ((6.022 * 10^(23) " molecules")/ ("mol")) *n#
Now solving for the number of moles present.
$n = {\text{0.88010 mol of F}}_{2}$
Now we can use the formula, where $n$ is the number of moles, $m$ is the mass in grams and $M$ is the molar mass.
$n = \frac{m}{M}$
${\text{0.88010 mol of F"_2 = m/("37.996 g/mol") " of F}}_{2}$
$\therefore m = {\text{33.4 g of F}}_{2}$
The answer is rounded to 3 significant digits; as specified by the original question.
Hope this helped :)
Note
I've edited the significant digits in the calculation as the data specified three. Thus the intermediate steps should really carry about one or two intermediate ones. I've carried two to be safe. At the end of the day however, it depends on your instructor or assessment marker. | 2020-01-28 04:27:10 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 13, "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": 0.9304645657539368, "perplexity": 943.856537592811}, "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/1579251773463.72/warc/CC-MAIN-20200128030221-20200128060221-00454.warc.gz"} |
https://www.physicsforums.com/threads/eigen-functions-values-for-many-body-hamiltonian-with-creation-annihilation-operators.503945/ | # Eigen functions/values for many-body Hamiltonian with creation/annihilation operators
1. ### ranytawfik
11
Problem:
-----------
I’m trying to understand how to generally find Eigen functions/values (either analytically or numerically) for Hamiltonian with creation/annihilation operators in many-body problems.
Procedures:
--------------
1. I setup a simple case of finite-potential well with two in-distinguishable fermions (ignore spin for the moment).
2. I got the creation/annihilation field operators.
3. I second-quantized all the elements of the Hamiltonian (kinetic energy, well/external potential, and electron-electron electrostatic interaction).
My question is how to proceed next? I have the Hamiltonian which is now function of the creation/annihilation operators. How can I solve for the many-body Eigen function/values after that?
Thanks so much.
2. ### strangerep
2,220
Re: Eigen functions/values for many-body Hamiltonian with creation/annihilation opera
ranytawfik,
It would be easier to respond if you post some of your math that you've
already obtained in the first 3 steps.
[
If you're not familiar with Latex on this forum, try this thread:
]
3. ### Bill_K
4,159
Re: Eigen functions/values for many-body Hamiltonian with creation/annihilation opera
In general, the procedure is as follows:
1) Try to find the normal modes of your system, that is, transform your creation/annihilation operators to a basis in which the Hamiltonian is diagonal: H = ∑ ak*ak.
2) Find the states for each normal mode as a quantized harmonic oscillator. The ground state is |0k> such that ak|0k> = 0. The excited states are |nk> = (ak*)n|nk>.
If you can't do step (1), you'll have to use perturbation theory.
4. ### ranytawfik
11
Re: Eigen functions/values for many-body Hamiltonian with creation/annihilation opera
Thanks strangerep and Bill. I'll try to do what you suggested, Bill, and post a follow up with the detailed Latex equations and procedures. | 2015-08-04 17:55:07 | {"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": 0.8407528400421143, "perplexity": 2204.756079154028}, "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-2015-32/segments/1438042991076.30/warc/CC-MAIN-20150728002311-00021-ip-10-236-191-2.ec2.internal.warc.gz"} |
https://math.libretexts.org/LibreTexts/Monroe_Community_College/MTH_212_Calculus_III/Chapter_15%3A_Vector_Fields%2C_Line_Integrals%2C_and_Vector_Theorems/15.2E%3A_Line_Integrals_(Exercises) |
# 15.2E: Line Integrals (Exercises)
$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$
$$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$
1. True or False? Line integral $$\displaystyle\int _C f(x,y)\,ds$$ is equal to a definite integral if $$C$$ is a smooth curve defined on $$[a,b]$$ and if function $$f$$ is continuous on some region that contains curve $$C$$.
Solution: True
2. True or False? Vector functions $$\vecs r_1=t\,\hat{\mathbf i}+t^2\,\hat{\mathbf j}, \quad 0≤t≤1,$$ and $$\vecs r_2=(1−t)\,\hat{\mathbf i}+(1−t)^2\,\hat{\mathbf j}, \quad 0≤t≤1$$, define the same oriented curve.
3. True or False? $$\displaystyle\int _{−C}(P\,dx+Q\,dy)=\int _C(P\,dx−Q\,dy)$$
Solution: False
4. True or False? A piecewise smooth curve $$C$$ consists of a finite number of smooth curves that are joined together end to end.
5. True or False? If $$C$$ is given by $$x(t)=t,\quad y(t)=t, \quad 0≤t≤1$$, then $$\displaystyle\int _Cxy\,ds=\int ^1_0t^2\,dt.$$
Solution: False
For the following exercises, use a computer algebra system (CAS) to evaluate the line integrals over the indicated path.
6. [T] $$\displaystyle\int _C(x+y)\,ds$$
$$C:x=t,y=(1−t),z=0$$ from $$(0, 1, 0)$$ to $$(1, 0, 0)$$
7. [T] $$\displaystyle \int _C(x−y)ds$$
$$C:\vecs r(t)=4t\,\hat{\mathbf i}+3t\,\hat{\mathbf j}$$ when $$0≤t≤2$$
Solution: $$\displaystyle\int _C(x−y)\,ds=10$$
8. [T] $$\displaystyle\int _C(x^2+y^2+z^2)\,ds$$
$$C:\vecs r(t)=sint\,\hat{\mathbf i}+cost\,\hat{\mathbf j}+8t\,\hat{\mathbf k}$$ when $$0≤t≤\dfrac{π}{2}$$
9. [T] Evaluate $$\displaystyle\int _Cxy^4\,ds$$, where $$C$$ is the right half of circle $$x^2+y^2=16$$ and is traversed in the clockwise direction.
Solution: $$\displaystyle\int _Cxy^4\,ds=\frac{8192}{5}$$
10. [T] Evaluate $$\displaystyle\int _C4x^3ds$$, where C is the line segment from $$(−2,−1)$$ to $$(1, 2)$$.
For the following exercises, find the work done.
11. Find the work done by vector field $$\vecs F(x,y,z)=x\,\hat{\mathbf i}+3xy\,\hat{\mathbf j}−(x+z)\,\hat{\mathbf k}$$ on a particle moving along a line segment that goes from $$(1,4,2)$$ to $$(0,5,1)$$.
Solution: $$W=8$$
12. Find the work done by a person weighing 150 lb walking exactly one revolution up a circular, spiral staircase of radius 3 ft if the person rises 10 ft.
13. Find the work done by force field $$\vecs F(x,y,z)=−\dfrac{1}{2}x\,\hat{\mathbf i}−\dfrac{1}{2}y\,\hat{\mathbf j}+\dfrac{1}{4}\,\hat{\mathbf k}$$ on a particle as it moves along the helix $$\vecs r(t)=\cos t\,\hat{\mathbf i}+\sin t\,\hat{\mathbf j}+t\,\hat{\mathbf k}$$ from point $$(1,0,0)$$ to point $$(−1,0,3π)$$.
Solution: $$W=\dfrac{3π}{4}$$
14. Find the work done by vector field $$\vecs{F}(x,y)=y\,\hat{\mathbf i}+2x\,\hat{\mathbf j}$$ in moving an object along path $$C$$, which joins points $$(1, 0)$$ and $$(0, 1)$$.
15. Find the work done by force $$\vecs{F}(x,y)=2y\,\hat{\mathbf i}+3x\,\hat{\mathbf j}+(x+y)\,\hat{\mathbf k}$$ in moving an object along curve $$\vecs r(t)=\cos(t)\,\hat{\mathbf i}+\sin(t)\,\hat{\mathbf j}+16\,\hat{\mathbf k}$$, where $$0≤t≤2π$$.
Solution: $$W=π$$
16. Find the mass of a wire in the shape of a circle of radius 2 centered at (3, 4) with linear mass density $$ρ(x,y)=y^2$$.
For the following exercises, evaluate the line integrals.
17. Evaluate $$\displaystyle\int C\vecs F·d\vecs{r}$$, where $$\vecs{F}(x,y)=−1\,\hat{\mathbf j}$$, and $$C$$ is the part of the graph of $$y=12x^3−x$$ from $$(2,2)$$ to $$(−2,−2)$$.
Solution: $$\displaystyle\int _C\vecs F·d\vecs{r}=4$$
18. Evaluate $$\displaystyle\int _γ(x^2+y^2+z^2)^{−1}ds$$, where $$γ$$ is the helix $$x=\cos t,y=\sin t,z=t(0≤t≤T).$$
19. Evaluate $$\displaystyle\int _Cyz\,dx+xz\,dy+xy\,dz$$ over the line segment from $$(1,1,1)$$ to $$(3,2,0).$$
Solution: $$\displaystyle\int _Cyz\,dx+xz\,dy+xy\,dz=−1$$
20. Let C be the line segment from point (0, 1, 1) to point (2, 2, 3). Evaluate line integral $$\displaystyle\int _Cy\,ds.$$
21. [T] Use a computer algebra system to evaluate the line integral $$\displaystyle\int _Cy^2\,dx+x\,dy$$, where $$C$$ is the arc of the parabola $$x=4−y^2$$ from $$(−5, −3)$$ to $$(0, 2)$$.
Solution: $$\displaystyle\int _C(y^2)\,dx+(x)\,dy=\dfrac{245}{6}$$
22. [T] Use a computer algebra system to evaluate the line integral $$\displaystyle\int _C(x+3y^2)dy$$ over the path $$C$$ given by $$x=2t,y=10t,$$ where $$0≤t≤1.$$
23. [T] Use a CAS to evaluate line integral $$\displaystyle\int _Cxy\,dx+y\,dy$$ over path $$C$$ given by $$x=2t,y=10t$$,where $$0≤t≤1$$.
Solution: $$\displaystyle\int _Cxy\,dx+y\,dy=\dfrac{190}{3}$$
24. Evaluate line integral $$\displaystyle\int _C(2x−y)\,dx+(x+3y)\,dy$$, where $$C$$ lies along the $$x$$-axis from $$x=0$$ to $$x=5$$.
26. [T] Use a CAS to evaluate $$\displaystyle\int _C\dfrac{y}{2x^2−y^2}\,ds$$, where $$C$$ is $$x=t,y=t,1≤t≤5.$$
Solution: $$\displaystyle\int _C\dfrac{y}{2x^2−y^2}\,ds=\sqrt{2}ln5$$
27. [T] Use a CAS to evaluate $$\displaystyle\int _Cxy\,ds$$, where $$C$$ is $$x=t^2,y=4t,0≤t≤1.$$
In the following exercises, find the work done by force field $$\vecs F$$ on an object moving along the indicated path.
28. $$\vecs{F}(x,y)=−x \,\hat{\mathbf i}−2y\,\hat{\mathbf j}$$
$$C:y=x^3$$ from $$(0, 0)$$ to $$(2, 8)$$
Solution: $$W=−66$$
29. $$\vecs{F}(x,y)=2x\,\hat{\mathbf i}+y\,\hat{\mathbf j}$$
C: counterclockwise around the triangle with vertices $$(0, 0), (1, 0),$$ and $$(1, 1)$$
30. $$\vecs F(x,y,z)=x\,\hat{\mathbf i}+y\,\hat{\mathbf j}−5z\,\hat{\mathbf k}$$
$$C:\vecs r(t)=2\cos t\,\hat{\mathbf i}+2\sin t\,\hat{\mathbf j}+t\,\hat{\mathbf k},0≤t≤2π$$
Solution: $$W=−10π^2$$
31. Let $$\vecs F$$ be vector field $$\vecs{F}(x,y)=(y^2+2xe^y+1)\,\hat{\mathbf i}+(2xy+x^2e^y+2y)\,\hat{\mathbf j}$$. Compute the work of integral $$\displaystyle\int _C\vecs F·d\vecs{r}$$, where $$C$$ is the path $$\vecs r(t)=\sin t\,\hat{\mathbf i}+\cos t\,\hat{\mathbf j},\quad 0≤t≤\dfrac{π}{2}$$.
32. Compute the work done by force $$\vecs F(x,y,z)=2x\,\hat{\mathbf i}+3y\,\hat{\mathbf j}−z\,\hat{\mathbf k}$$ along path $$\vecs r(t)=t\,\hat{\mathbf i}+t^2\,\hat{\mathbf j}+t^3\,\hat{\mathbf k}$$,where $$0≤t≤1$$.
Solution: $$W=2$$
33. Evaluate $$\displaystyle\int _C\vecs F·d\vecs{r}$$, where $$\vecs{F}(x,y)=\dfrac{1}{x+y}\,\hat{\mathbf i}+\dfrac{1}{x+y}\,\hat{\mathbf j}$$ and $$C$$ is the segment of the unit circle going counterclockwise from $$(1,0)$$ to $$(0, 1)$$.
34. Force $$\vecs F(x,y,z)=zy\,\hat{\mathbf i}+x\,\hat{\mathbf j}+z^2x\,\hat{\mathbf k}$$ acts on a particle that travels from the origin to point (1, 2, 3). Calculate the work done if the particle travels:
1. along the path $$(0,0,0)→(1,0,0)→(1,2,0)→(1,2,3)$$ along straight-line segments joining each pair of endpoints;
2. along the straight line joining the initial and final points.
3. Is the work the same along the two paths?
Solution: a. $$W=11$$; b. $$W=11$$; c. Yes
35. Find the work done by vector field $$\vecs F(x,y,z)=x\,\hat{\mathbf i}+3xy\,\hat{\mathbf j}−(x+z)\,\hat{\mathbf k}$$ on a particle moving along a line segment that goes from $$(1, 4, 2)$$ to $$(0, 5, 1).$$
36. How much work is required to move an object in vector field $$\vecs{F}(x,y)=y\,\hat{\mathbf i}+3x\,\hat{\mathbf j}$$ along the upper part of ellipse $$\dfrac{x^2}{4}+y^2=1$$ from $$(2, 0)$$ to $$(−2,0)$$?
Solution: $$W=2π$$
37. A vector field is given by $$\vecs{F}(x,y)=(2x+3y)\,\hat{\mathbf i}+(3x+2y)\,\hat{\mathbf j}$$. Evaluate the line integral of the field around a circle of unit radius traversed in a clockwise fashion.
38. Evaluate the line integral of scalar function xy along parabolic path $$y=x^2$$ connecting the origin to point $$(1, 1)$$.
Solution: $$\displaystyle\int _C\vecs F·d\vecs{r}=\dfrac{25\sqrt{5}+1}{120}$$
39. Find $$\displaystyle\int _Cy^2\,dx+(xy−x^2)\,dy$$ along $$C: y=3x$$ from (0, 0) to (1, 3).
40. Find $$\displaystyle\int _Cy^2\,dx+(xy−x^2)\,dy$$ along $$C: y^2=9x$$ from (0, 0) to (1, 3).
Solution: $$\displaystyle\int _Cy^2\,dx+(xy−x^2)\,dy=6.15$$
For the following exercises, use a CAS to evaluate the given line integrals.
41. [T] Evaluate $$\vecs F(x,y,z)=x^2z\,\hat{\mathbf i}+6y\,\hat{\mathbf j}+yz^2\,\hat{\mathbf k}$$, where $$C$$ is represented by $$\vecs r(t)=t\,\hat{\mathbf i}+t^2\,\hat{\mathbf j}+\ln t \,\hat{\mathbf k},1≤t≤3$$.
42. [T] Evaluate line integral $$\displaystyle\int _γxe^y\,ds$$ where, $$γ$$ is the arc of curve $$x=e^y$$ from $$(1,0)$$ to $$(e,1)$$.
Solution: $$\displaystyle\int _γxe^y\,ds≈7.157$$
43. [T] Evaluate the integral $$\displaystyle\int _γxy^2\,ds$$, where $$γ$$ is a triangle with vertices $$(0, 1, 2), (1, 0, 3)$$, and $$(0,−1,0)$$.
44. [T] Evaluate line integral $$\displaystyle\int _γ(y^2−xy)\,dx$$, where $$γ$$ is curve $$y=\ln x$$ from $$(1, 0)$$ toward $$(e,1)$$.
Solution: $$\displaystyle\int _γ(y^2−xy)\,dx≈−1.379$$
45. [T] Evaluate line integral $$\displaystyle\int γxy4\,ds$$, where $$γ$$ is the right half of circle $$x^2+y^2=16$$.
46. [T] Evaluate \int CF⋅dr,\int CF·dr, where F(x,y,z)=x2yi+(x−z)j+xyzkF(x,y,z)=x2yi+(x−z)j+xyzk and
$$C: r(t)=ti+t^2j+2k,0≤t≤1$$.
Solution: $$\displaystyle\int _CF⋅dr≈−1.133$$
47. Evaluate $$\displaystyle\int _CF⋅dr$$, where $$\vecs{F}(x,y)=2xsin(y)i+(x^2cos(y)−3y^2)j$$ and
$$C$$ is any path from $$(−1,0)$$ to $$(5, 1)$$.
48. Find the line integral of $$F(x,y,z)=12x^2i−5xyj+xzk$$ over path $$C$$ defined by $$y=x^2, z=x^3$$ from point $$(0, 0, 0)$$ to point $$(2, 4, 8)$$.
Solution: $$\displaystyle\int _CF⋅dr≈22.857$$
49. Find the line integral of $$\displaystyle\int _C(1+x^2y)ds$$, where $$C$$ is ellipse $$r(t)=2costi+3sintj$$ from $$0≤t≤π.$$\
For the following exercises, find the flux.
50. Compute the flux of $$\vecs{F}=x^2i+yj$$ across a line segment from $$(0, 0)$$ to $$(1, 2).$$
Solution: $$flux=−\dfrac{1}{3}$$
51. Let $$\vecs{F}=5i$$ and let $$C$$ be curve $$y=0,0≤x≤4$$. Find the flux across $$C$$.
52. Let $$\vecs{F}=5j$$ and let $$C$$ be curve $$y=0,0≤x≤4$$. Find the flux across $$C$$.
Solution: $$flux=−20$$
53. Let $$\vecs{F}=−yi+xj$$ and let $$Cr(t)=costi+sintj (0≤t≤2π)$$. Calculate the flux across $$C$$.
54. Let $$\vecs{F}=(x^2+y^3)i+(2xy)j$$. Calculate flux F orientated counterclockwise across curve $$C: x^2+y^2=9.$$
Solution: $$flux=0$$
55. Find the line integral of $$\displaystyle\int _Cz^2dx+ydy+2ydz,$$ where $$C$$ consists of two parts: $$C_1$$ and $$C_2. C_1$$ is the intersection of cylinder $$x^2+y^2=16$$ and plane $$z=3$$ from $$(0, 4, 3)$$ to $$(−4,0,3). C_2$$ is a line segment from $$(−4,0,3)$$ to $$(0, 1, 5)$$.
56. A spring is made of a thin wire twisted into the shape of a circular helix $$x=2cost,y=2sint,z=t.$$ Find the mass of two turns of the spring if the wire has constant mass density.
Solution: $$m=4πρ\sqrt{5}$$
57. A thin wire is bent into the shape of a semicircle of radius a. If the linear mass density at point P is directly proportional to its distance from the line through the endpoints, find the mass of the wire.
58. An object moves in force field $$F(x,y,z)=y^2i+2(x+1)yj$$ counterclockwise from point $$(2, 0)$$ along elliptical path $$x^2+4y^2=4$$ to $$(−2,0)$$, and back to point $$(2, 0)$$ along the x-axis. How much work is done by the force field on the object?
Solution: $$W=0$$
59. Find the work done when an object moves in force field $$F(x,y,z)=2xi−(x+z)j+(y−x)k$$ along the path given by $$r(t)=t^2i+(t^2−t)j+3k, 0≤t≤1.$$
60. If an inverse force field F is given by $$F(x,y,z)=\dfrac{k}{‖r‖^3}r$$, where k is a constant, find the work done by F as its point of application moves along the x-axis from $$A(1,0,0)$$ to $$B(2,0,0)$$.
Solution: $$W=\dfrac{k}{2}$$
61. David and Sandra plan to evaluate line integral $$\displaystyle\int _CF·dr$$ along a path in the xy-plane from (0, 0) to (1, 1). The force field is $$\vecs{F}(x,y)=(x+2y)i+(−x+y2)j$$. David chooses the path that runs along the x-axis from (0, 0) to (1, 0) and then runs along the vertical line x=1 from (1, 0) to the final point (1, 1). Sandra chooses the direct path along the diagonal line y=x from (0, 0) to (1, 1). Whose line integral is larger and by how much? | 2018-12-09 22:16:49 | {"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": 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": 0.950210452079773, "perplexity": 439.6517312277829}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-51/segments/1544376823183.3/warc/CC-MAIN-20181209210843-20181209232843-00099.warc.gz"} |
https://dosits.org/people-and-sound/navigation/how-is-sound-used-to-navigate-underwater/ | How is sound used to navigate underwater?
Underwater sound can be used to help ships navigate their way through the water. Sound can also be used to navigate remotely operated vehicles (ROVs), manned submersibles, and oceanographic equipment.
A basic underwater navigation system uses a hydrophone and an acoustic tracking beacon. The hydrophone is attached to a boat, below the water and works like an antenna. The beacon, somewhere below the boat, transmits a signal that is picked up by the hydrophone. The system calculates the range (distance) and bearing (horizontal direction) to the beacon. The position of the beacon is determined relative to the boat. If the exact location of the boat is known, then the exact location of the beacon can be calculated.
The position of the Remotely Operated Vehicle (ROV) is calculated using a transducer that is attached to the ship and a beacon attached to the ROV. (Courtesy of Kongsberg-Simrad)
A more precise underwater navigation system uses several beacons. A ship or ROV has an acoustic transducer that combines a transmitter and a receiver. Beacons that make a pinging sound are spread out on the ocean floor in known locations. These special beacons are called transponders. The transducer sends out signals to the underwater transponders. Each transponder responds with a unique sound of its own. These replies are picked up by the transducer. Computers then calculate the exact position of the ship or ROV by determining the distance from the beacons and using simple geometry and basic math.
Remotely Operated Vehicle (ROV) navigating using a network of underwater transponders. (Courtesy of Kongsberg-Simrad)
How position is calculated
Underwater positioning is based on the basic principle:
Distance = Speed x Time, where Speed is the speed of sound in water.
The system measures the amount of time between the initial ping (sent from the transducer) and the return ping (received from the transponder) and calculates the distance from that transponder:
$Distance = \frac {Speed \ of \ Sound \ x \ Time \ Elapsed}{2}$
The product is divided by two because the measured time is the round-trip time (from the transducer to the transponder and back to the transducer).
Once you get 3 distances from 3 known points, you can calculate the position using a method called triangulation. Computers calculate the unique point where all three distances measured from the transponders intersect. | 2020-05-30 21:32:09 | {"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": 0, "mathjax_asciimath": 0, "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": 0.5673646330833435, "perplexity": 1534.1059979320526}, "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/1590347410352.47/warc/CC-MAIN-20200530200643-20200530230643-00074.warc.gz"} |
https://icsehelp.com/heat-and-energy-concise-class-9-icse-physics-selina-publishers/ | Heat and Energy Concise Class-9 ICSE Physics Selina Publishers Solutions Chapter-6 . Step By Step ICSE Selina Concise Solutions of Chapter-6 Heat and Energy with Exercise-6(A), Exercise-6(B) , Exercise-6(C) and Exercise-6(D) including Numerical and MCQ Questions Solved Visit official Website CISCE for detail information about ICSE Board Class-9.
Heat and Energy Concise Class-9 ICSE Physics Selina Publishers
–: Select Topics :–
Exercise-6(A)
MCQ-6(A)
Exercise-6(B)
MCQ-6(B)
Exercise-6(C)
MCQ-6(C)
Exercise-6(D)
MCQ-6(D)
Note :- Before Viewing Selina Concise Physics Solutions of Chapter-6 Heat and Energy for ICSE Class-9 Physics. Read the whole chapter carefully and Solved all example of Chapter-6 Heat and Energy for Class-9 Physics.
EXE-6(A) Heat and Energy Selina Concise Physics Solutions
Question 1
What is heat? Write its S.I. unit.
Heat is the energy of random motion of molecules constituting the body.
Its S.I. unit is ‘joule’.
Question 2
Two bodies at different temperatures are placed in contact. State the direction in which the heat will flow.
Heat will flow from a hot body (body at a higher temperature) to a cold body (body at a lower temperature)
Question 3.
Name the S.I. unit of heat and how is it related to the unit calorie?
S.I. unit of heat is ‘joule’.
1 joule = 0.24 cal
Question 4
Define temperature and write its S.I. unit.
Temperature is the parameter which tells the thermal state of a body (i.e. the degree of hotness or coldness).
The S.I. unit of temperature is ‘kelvin’
Question 5
Why does a piece of ice feel cool to touch? Explain.
On touching a piece of ice, heat flows from our hand (hot body) to the ice (cold body), and hence, it appears cold.
Question 6
Distinguish between heat and temperature.
Heat is a form of energy obtained due to the random motion of molecules in a substance but temperature is a quantity which decided the direction of flow of heat when two bodies at different temperature are placed in contact. Two quantities having the same amount of heat may differ in temperature.
Question 7
What do you understand by thermal expansion of a substance?
The expansion of a substance on heating is called thermal expansion.
Question 8
Name two substances which expands on heating.
Brass and iron expand on heating.
Question 9
Name two substances which contract on heating.
Water contracts on heating from 0 to 4.
Silver iodide contracts on heating from 80 to 141.
Question 10
What do you mean by the anomalous expansion of water?
The expansion of water when it is cooled from 4 to 0 is known as the anomalous expansion of water.
Question 11
At what temperature is the density of water maximum? State its value.
Density of water is maximum at 4. Its value is 1000 kgm-3.
Question 12
State the volume changes that you will observe when a given mass of water is heated from 0 to 10 . Sketch a temperature-volume graph to show the behaviour.
When a given mass of water is heated from 0 to 4 , it contracts, i.e. its volume decreases.
On heating from 4 to 10 , it expands, i.e. its volume increases.
Question 13
Draw a graph to show the variation in density of water with temperature in the range from 0 to 10 .
Question 14
A given mass of water is cooled from 10 to 0. State the volume changes you will observe. Represent these changes on a temperature-volume graph.
Question 15
Describe an experiment to show that water has maximum density at 4. What important consequences follow this peculiar property of water? Discuss the importance of this phenomenon in nature.
Hope’s experiment to demonstrate that water has maximum density at 4:
Hope’s apparatus consists of a tall metallic cylinder provided with two side openings P and Q, P near the top and Q near the bottom, fitted with thermometers T1 and Tin them. The central part of the cylinder is surrounded with a cylindrical trough containing a freezing mixture of ice and salt. The cylinder is fitted with pure water at room temperature.
Observations:
(i) Initially, both thermometers T1 and Tare at the same temperature.
(ii) First, the temperature recorded by the lower thermometer Tstarts decreasing and finally it becomes steady at 4, while the temperature recorded in the upper thermometer Tremains almost unchanged during this time.
(iii)
Then, the temperature recorded by the lower thermometer Tremains constant at 4 and upper thermometer T1 records a continuous fall in temperature up to 0 and then it becomes steady.
Thus, finally, the temperature recorded by the upper thermometer is 0 and that by lower thermometer is 4.
As the freezing mixture cools water in the central portion of the cylinder, water contracts and its density increases, consequently it sinks to the bottom, thereby causing the reading of the lower thermometer Tto fall rapidly. The reading of the upper thermometer T1 does not change as the temperature of water in the upper part does not change. This continues till the entire water below the central portion reaches 4 .
On cooling further below 4 , due to anomalous expansion, water of the central portion expands, so its density decreases and hence it rises up. As a result, reading of the upper thermometer T1 falls rapidly to 0 and water freezes to form ice at 0 near the top. This proves that water has maximum density at 4.
This anomalous expansion of water helps in preserving the aquatic life during the very cold weather.
In winters, when the temperature falls, the top layer of water in a pond contracts, becomes denser and sinks to the bottom. A circulation is thus set up until the entire water in the pond reaches its maximum density at 4 . If the temperature falls further, then the top layer expands and remains on the top till it freezes. Thus, even though the upper layers are frozen, the water near the bottom is at 4 and the fishes can survive in it easily.
Question 16
A deep pond of water has its top layer frozen during winter. State the expected temperature of water layer (i) Just in contact with ice and (ii) At the bottom of pond.
(i) Water just in contact with ice is at 0.
(ii) Water at the bottom of the pond is at 4.
Question 17
Draw a diagram showing the temperature of various layers of water in an ice covered pond.
Question 18
Explain the following statements listed below:
(a) Water pipes in colder countries often burst in winter.
(b) In winter, water tanks (or ocean) start freezing from the surface and not from the bottom.
(c) Fishes survive in ponds even when the atmospheric temperature is well below 0.
(d) A hollow glass sphere which floats with its entire volume submerged in water at 4 , sinks when water is heated above 4.
(e) A glass bottle completely filled with water and tightly closed at room temperature is likely to burst when kept in the freezer of a refrigerator.
(a) On winter nights, as the atmospheric pressure starts falling below 4 , water in the pipe lines expand and exert a great pressure on the pipes, causing them to burst.
(b) In winters, when temperature falls, the surface of water in the tank contracts, becomes denser and sinks to the bottom. A circulation is thus set up until the entire water in the tank reaches its maximum density at 4 . If the temperature falls further, then the top layer expands and remains on the top till it freezes. Thus, water in a tank starts freezing from the top and not from the bottom.
(c) The anomalous expansion of water
helps preserve aquatic life during very cold weather. When temperature falls, the top layer of water in a pond contracts becomes denser and sinks to the bottom. A circulation is thus set up until water in the pond reaches its maximum density at 4 . If the temperature falls further, then the top layer expands and remains on the top till it freezes. Thus, even though the upper layer are frozen, the water near the bottom is at 4 and the fishes can survive in it easily.
(d) On heating water above 4 the density of water decreases. As a result, the upthrust acting due to water on hollow glass sphere also decreases, which causes it to sink.
(e) Inside the freezer, when the temperature of water falls below 4 , the water in the bottle starts expanding. If the bottle is completely filled and tightly closed, there is no space for water to expand, and hence, the bottle may burst.
MCQ-6(A) ICSE Physics Heat and Energy Solutions
Question 1
Calorie is the unit of :
(a) Heat
(b) Work
(c) Temperature
(d) food
Calorie is the unit heat.
Question 2
1 J equals to:
(a) 0.24 cal
(b) 4.18 cal
(c) 1 cal
(d) 1 kcal
calorie = 4.186 J
Therefore,
1 J = 1/4.186 = 0.24 cal
Question 3
S.I. unit of temperature is:
(a) Cal
(b) Joule
(c) Celsius
(d) kelvin
The SI unit of temperature is kelvin (K).
Question 4
Water is cooled from 4 °C to 0 °C. It will:
(a) contract
(b) expand
(c) first contract, then expand
(d) first expand, then contract
Water shows anomalous behavior between 0 °C and 4 °C. Hence, when it is cooled it expands.
Question 5
Density of water is maximum at:
(a) 0°C
(b) 100°C
(c) 4°C
(d) 15°C
Water shows anomalous behavior between 0 °C and 4 °C. It has lowest volume at 4 °C. Hence, its density will be maximum at 4 °C.
EXE-6(B) Concise Physics Solutions Heat and Energy
Question 1
What is an ecosystem? Name its two components.
A unit composed of biotic components (i.e. producers, consumers and decomposers) and abiotic components (i.e. light, heat, rain, and humidity, inorganic and organic substances) is called an ecosystem.
Question 2
What is the source of energy for all ecosystems?
The source of energy for all ecosystems is the Sun.
Question 3
State the importance of green plants in an ecosystem
Green plants absorb most of the energy falling on them and by the process of photosynthesis they produce food for the consumers. Plants, being primary producers are of great importance in the ecosystem. They also maintain the balance of oxygen and carbon dioxide on earth.
Question 4
Differentiate between the producers and consumers.
Producers like plants and some bacteria are capable of producing its own food using the energy of sun but consumers are not capable of producing their own food. They depend on producers for food.
Question 5
State the functions of decomposers in an ecosystem.
The role of a decomposer is to break down dead organisms and then feed on them. The nutrients created by the dead organisms are returned to the soil to be later used by the producers. Once these deceased organisms are returned to the soil, they are used as food by bacteria and fungi by transforming the complex organic materials into simpler nutrients. The simpler products can then be used by producers to restart the cycle. These decomposers play an important role in every ecosystem.
Question 6
What is a food chain?
A food chain shows the feeding relationship between different living things in a particular environment or habitat. Often, a plant will begin a food chain because it can make its own food using energy from the Sun. In addition, a food chain represents a series of events in which food and energy are transferred from one organism in an ecosystem to another. Food chains show how energy is passed from the sun to producers, from producers to consumers, and from consumers to decomposers.
Question 7
Draw a simple diagram showing a food chain.
Question 8
Describe the energy flow in an ecosystem.
Ecosystems maintain themselves by cycling energy and nutrients obtained from external sources. At the first trophic level, primary producers (plants, algae, and some bacteria) use solar energy to produce organic plant material through photosynthesis. Herbivores-animals that feed solely on plants-make up the second trophic level.
Predators that eat herbivores comprise the third trophic level;
if larger predators are present, they represent still higher trophic levels. Decomposers, which include bacteria, fungi etc. break down wastes and dead organisms and return nutrients to the soil.
On average about 10 percent of net energy production at one trophic level is passed on to the next level. Processes that reduce the energy transferred between trophic levels include respiration, growth and reproduction, defecation, and non-predatory death.
The low rate of energy transfer between trophic levels makes decomposers generally more important than producers in terms of energy flow. Decomposers process large amounts of organic material and return nutrients to the ecosystem in inorganic forms, which are then taken up again by primary producers.
Question 9
State the law which governs the energy flow in an ecosystem.
The laws of thermodynamics govern the energy flow in the ecosystem.
According to the first law of thermodynamic, the energy can be transformed from one form to the other form, but it can neither be created nor destroyed.
According to the second law of thermodynamics, when energy is put to work, a part of it is always converted in un-useful form such as heat mainly due to friction and radiation.
Question 10
Show that the energy flow in an ecosystem is linear.
The energy flow in ecosystem is linear i.e., it moves in a fixed direction. The solar energy is absorbed by plants and a part of it is converted into food. These plants (or primary producers) are then eaten by the primary consumers, which are consumed by secondary consumers and the secondary by tertiary consumers.
This cycle is unidirectional. The dead and decomposed are fed by decomposers, which return the nutrients to the soil. At the end, the energy reaches the degraded state. It does not return to the sun to make the process cyclic, thus energy flow is linear.
Question 11
Draw a simple diagram showing the energy flow in a food chain.
Question 12
Draw a diagram to show that the energy flow in an ecosystem is governed by the law of conservation of energy.
MCQ-6(B) Silina Concise Physics Solutions Heat and Energy
Question 1
Food chain begins with the following step:
(a) Respiration (b) Photosynthesis
(c) Decomposition (d) Decay.
Photosynthesis
Question 2
The source of energy in an ecosystem is which of the following source listed below:
(a) Sun (b) Decayed bodies
(c) Green plants (d) Sugar.
Sun
Question 3
Energy enters in a food chain through which of the following consumers:
(a) Primary consumers
(b) Secondary consumers
(c) Tertiary consumers
(d) Producers.
Producers
Question 4
The place of human being in food chain in an ecosystem is as
(a) Producer
(b) Consumer
(c) Decomposer
(d) Both (a) and (b).
Consumer
EXE-6(C) Heat and Energy Concise 9th physics solutions
Question 1
State two characteristics which a source of energy must have.
1. A source of energy should be safe and convenient to use.
2. A source of energy should be economical and easy to store and transport.
Question 2
Name the two groups in which various sources of energy are classified. State on what basis are they classified.
The two groups in which various sources of energy are classified are renewable or non-conventional sources of energy and non-renewable or conventional sources of energy.
These sources are classified on the basis of their availability and utility.
Question 3
What is meant by the renewable and non-renewable sources of energy? State two differences between them, giving two examples of each
Renewable: The natural sources providing us energy continuously are called renewable sources of energy.
Non-renewable: The sources of energy which have accumulated in nature over a very long period of time and cannot be quickly replaced when exhausted are called non-renewable sources of energy.
Difference:
Renewable sources Non-renewable sources They can be utilised continuously. They cannot be utilised once exhausted. Examples: Sun, Wind Example: Coal, Petroleum
Question 4
Select the renewable and non-renewable sources of energy from the following :
(a) Coal
(b) Wood
(c) Water
(d) Diesel
(e) Wind
(f) Oil
Renewable: Wood, Water and Wind
Non-renewable: Coal, Diesel and Oil
Question 5
Why is the use of wood as a fuel not advisable although wood is a renewable source of energy?
Wood is obtained from trees. Hence, trees need to be cut down for wood to be used as a fuel.
Also, burning wood releases a lot of smoke which pollutes the atmosphere.
Question 6
Name five renewable and three non-renewable sources of energy.
Renewable:
1. Sun
2. Wind
3. Flowing water
4. Tides
5. Nuclear fuel
Non-renewable:
1. Coal
2. Petroleum
3. Natural gas
Question 7
What is (i) tidal, (ii) ocean and (iii) geo thermal energy? Explain in brief.
Tidal energy:
The energy possessed by rising and falling water in tides is known as tidal energy.Dams are constructed across a narrow opening to the sea to harness tidal energy and produce electricity. However, it is not a major source of energy as the rise and fall of seawater during tides is not enough to generate electricity on a large scale.
Ocean energy:
Water in the oceans possesses energy in two forms:
Ocean thermal energy-
The energy available due to the difference in temperature of water at the surface and at deeper levels of ocean is called the ocean thermal energy. This energy is harnessed for producing electricity by a device called ocean thermal energy conversion power plant (OCTEC power plant).
Oceanic waves energy-
The kinetic energy possessed by fast moving oceanic (or sea) waves is called oceanic waves energy. Though models have been made to generate electricity from oceanic waves, but so far it has not been put to practical use.
iii. Geo thermal energy:
The heat energy possessed by the rocks inside the Earth is called geothermal energy.The hot rocks present at the hot spots deep inside the Earth, heat the underground water and turn it into steam. This steam is compressed at high pressure between the rocks. Holes are drilled deep into the Earth up to the hot spots to extract the steam through pipes, which is utilized to rotate the turbines connected to the armature of an electric generator to produce electricity.
Question 8
What is the main source of energy for Earth?
Sun is the main source of energy on Earth.
Question 9
What is solar energy? How is the solar energy used to generate electricity in a solar power Plant?
The energy obtained from Sun is called solar energy.
A solar power plant is a device in which heat energy of sun is used to generate electricity. It consists of a large number of concave reflectors, at the focus of which there are black painted water pipes. The reflectors concentrate the heat energy of the sun rays on the pipes due to which water inside the pipes starts boiling and produces steam. The steam thus produced is used to rotate a steam turbine which drives a generator producing electricity.
Question 10
What is a solar cell? State two uses of solar cells. State whether a solar cell produces a.c. or dc. Give one disadvantage of using a solar cell.
A solar cell is an electrical device that converts light energy directly into electricity with the help of photovoltaic effect. Solar cells are usually made from semiconductors like silicon and gallium with some impurity added to it. When sunlight is made incident on a solar cell, a potential difference is produced between its surface, due to which a current flows in the circuit connected between the opposite faces of the semiconductor.
Two uses of solar cells are as listed below:
1. They do not require maintenance and last over a long period of time at zero running cost.
2. They are very useful for remote, inaccessible and isolated places where electric power lines cannot be laid.Solar cell produces d.c. (direct current).
One disadvantage of solar cell is listed below:
(i) The initial cost of a solar panel is sufficiently high
Question 11
State two advantages and two limitations of producing electricity from solar energy.
1. They do not cause any pollution in the environment.
2. Running cost of solar panel is almost zero.
3. It last over a long period of time.
4. They do not require any maintenance.
5. Solar Energy are suitable for remote and inaccessible places where electricity power lines cannot be laid.
1. The initial cost of a solar panel is sufficiently high.
2. The efficiency of conversion of solar energy to electricity is low.
3. A solar panel produces d.c. electricity which cannot be directly used for many household purposes.
Question 12
What is wind energy? How is wind energy used to produce electricity? How much electric power is generated in India using the wind energy?
The kinetic energy of the moving large masses of air is called the wind energy. Wind energy is used in a wind generator to produce electricity by making use of wind mill to drive a wind generator.
At present in India, more than 1025 MW electric power is generated using wind energy.
Question 13
State two advantages and two limitations of using wind energy for generating electricity.
1. It does not cause any kind of pollution.
2. It is an everlasting source.
Limitations of using wind energy:
1. The establishment of a wind farm is expensive.
2. A large area of land is needed for the establishment of a wind farm.
Question 14
What is hydro energy? Explain the principle of generating electricity from hydro energy. How much hydro electric power is generated in India?
The kinetic energy possessed by flowing water is called the water or hydro energy.
Principle of a hydroelectric power plant is that the water flowing in high altitude rivers is collected in a high dam (or reservoir). The water from the dam is then allowed to fall on a water turbine which is located near the bottom of the dam. The shaft of the turbine is connected to the armature of an electric generator or dynamo.
At present only 23% of the total electricity is generated by the hydro energy.
Question 15
Advantages of producing the hydro electricity:
1. It does not produce any environmental pollution.
2. It is a renewable source of energy.
1. Due to the construction of dams over the rivers, plants and animals of that place get destroyed or killed.
2. The ecological balance in the downstream areas of rivers gets disturbed.
Question 16
What is nuclear energy? Name the process used for producing electricity using the nuclear energy.
When a heavy nucleus is bombarded with slow neutrons, it splits into two nearly equal light nuclei with a release of tremendous amount of energy. In this process of nuclear fission, the total sum of masses of products is less than the total sum of masses of reactants. This lost mass gets converted into energy. The energy so released is called nuclear energy.
Principle:
The heat energy released due to the controlled chain reaction of nuclear fission of uranium-235 in a nuclear reactor is absorbed by the coolant which then passes through the coils of a heat exchanger containing water. The water in heat exchanger gets heated and converts into steam. The steam is used to rotate the turbine which in turn rotates the armature of a generator in a magnetic field and thus produces electricity.
Question 17
What percentage of total electrical power generated in India is obtained from nuclear power plants? Name two places in India where electricity is generated from nuclear power plants.
At present only about 3% of the total electrical power generated in India is obtained from the nuclear power plants.
Tarapur in Maharahtra and Narora in Uttar Pradesh are the places where electricity is produced using nuclear energy.
Question 18
State two advantages and two disadvantages of using nuclear energy for producing electricity.
1. A very small amount of nuclear fuel can produce a tremendous amount of energy.
2. Once the nuclear fuel is loaded into nuclear power plant, it continues to release energy for several years.
1. It is not a clean source of energy because very harmful nuclear radiations are produced in the process.
2. The waste causes environmental pollution.
Question 19
State the energy transformation in the following:
i. electricity is obtained from solar energy.
ii. electricity is obtained from wind energy.
iii. electricity is obtained from hydro energy.
iv electricity is obtained from nuclear energy.
i. Light energy into electrical energy
ii. Mechanical energy into electrical energy.
iii. Mechanical energy into electrical energy.
iv. Nuclear energy (or heat energy) into electrical energy.
Question 20
State four ways for the judicious use of energy.
Four ways for the judicious use of energy are:
1. The fossil fuels such as coal, petroleum, natural gas should be used only for the limited purposes when there is no other alternative source of energy available.
2. The wastage of energy should be avoided.
3. Efforts must be made to make use of energy for community or group purposes.
4. The cutting of trees must be banned and more and more new trees must be roped to grow.
Question 21
What do you mean by degradation of energy? Explain it by taking two examples of your daily life.
The gradual decrease of useful energy due to friction etc. is called the degradation of energy.
Examples:
1. When we cook food over fire, the major part of heat energy from the fuel is radiated out in the atmosphere. This radiated energy is of no use to us.
2. When electrical appliances are run by electricity, the major part of electrical energy is wasted in the form of heat energy.
Question 22
The conversion of part of energy into a non-useful form of energy is called
The conversion of part of energy into a non-useful form of energy is called degradation of energy.
MCQ-6(C) Chapter-6 Selina Concise Physics Solutions Heat and Energy
Question 1
The ultimate source of energy is :
(a) wood
(b) wind
(c) water
(d) sun
The ultimate source of energy is the Sun.
Question 2
Renewable source of energy is :
(a) coal
(b) fossil fuels
(c) natural gas
(d) sun
Renewable source of energy is the Sun.
EXE-6(D) Selina Concise Physics Solutions Chapter-6 Heat and Energy
Question 1
What do you mean by greenhouse effect?
Greenhouse effect is the process of warming of planet’s surface and its lower atmosphere by absorbtion of infrared radiations of longer wavelength emitted out from the surface of planet.
Question 2
Name three green house gases.
Carbon-di-oxide, water vapour and methane are greenhouse gases.
Question 3
Which of the following solar radiations pass through the atmosphere of Earth: X -rays, ultraviolet rays, visible light rays or infrared radiation?
Visible light rays and short infrared radiation pass through the atmosphere of earth.
Question 4
What results in the increase of carbon dioxide contents of earth’s atmosphere?
The concentration of carbon-di-oxide content’s of earth’s atmosphere has increased due to industrial growth, combustion of fossil fuels and clearing of forests.
Question 5
Name the radiations which are absorbed by the green house gases.
Infrared radiations of long wavelength are absorbed by the green house gases.
Question 6
What would have been the temperature of earth’s atmosphere in absence of green house gases in it?
In absence of green house gases, the average temperature on earth would be -18.
Question 7
State the effect of green house gases on the temperature of earth’s atmosphere.
The greenhouses gases have an average warming effect on Earth’s surface of about 15.5 (or 60 ).
Question 8
What do you mean by global warming?
Global warming means the increase in average effective temperature near the earth’s surface due to an increase in the amount of green house gases in its atmosphere.
Question 9
What causes the rise in atmospheric temperature?
With activities industrialization, deforestation, excess burning of fossil fuel, the concentration of green house gases has increased on earth’s atmosphere. This increase in the amount of greenhouse gases present in atmosphere has caused the rise in atmospheric temperature.
Question 10
State the cause of increase of green house effect.
The increase in green house gases due to activities like industrialization, deforestation, natural gas exploration, burning of biomass, natural gas exploration, more use of gadgets like refrigerators has caused the increase of green house effect.
Question 11
What will be the effect of global warming at the poles?
At the poles, due to increase in temperature, the snow and ice will melt which will cause flood in coastal countries. The icebergs of dark land and oceans will melt, so the dark land and oceans will become uncovered and will absorb more heat radiations coming from sun, increasing the green house effect further.
Question 12
State the effect of global warming in coastal regions.
Due to global warming, the snow and ice around the poles will melt and cause flood in coastal countries.
Question 13
How will global warming affect the sea level?
Due to melting of polar ice and glaciers, there will be rise in sea level on coastal wet lands. It would raise worldwise sea level, thereby, many big cities in the coastal areas will be covered by sea water.
Question 14
How will global warming affect the agriculture?
Global warming will cause drastic changes in the patterns of wind, rainfall etc. Thus it will result in low agricultural yield.
Question 15
State two ways to minimize the impact of global warming.
The two way minimize the global warming are:-
1. Use of renewable sources of energy to generate electricity in place of generating electricity from the fossil fuels based power plants.
2. Controlling population through family planning, welfare reforms and the empowerment of women.
Question 16
What is carbon tax? Who will pay it?
The tax calculated on the basis of carbon emission from industry, number of employee hour and turnover of the factory is called carbon tax.
This tax shall be paid by industries. This will encourage the industries to use the energy efficient techniques.
MCQ-6(D) Heat and Energy chapter-6 ICSE 9th physics Solutions
Question 1
The green house gas is as listed below:
(a) Oxygen (b) Nitrogen
(c) Chlorine (d) Carbon dioxide.
Carbon dioxide
Question 2
The increase of carbon dioxide gas in atmosphere will cause which of the following occurences:
(a) Decrease in temperature (b) Increase in temperature (c) No change in temperature (d) Nothing can be said
Increase in temperature
Question 3
Without green house effect, the average temperature of Earth’s surface would have been:
(a) -18°C
(b) 33°C
(c) 0°C
(d) 15°C
Without green house effect, the average temperature of Earth’s surface would have been -18 °C.
Question 4
The global warming has resulted in:
(a) the increase in yield of crops
(b) the decrease in sea levels
(c) the decrease in human deaths
(d) the decrease in sea levels
The global warming has resulted in the increase in sea levels.
e increase in sea levels.
Thanks
$${}$$ | 2021-07-29 10:26:21 | {"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": 0.48172906041145325, "perplexity": 1911.5582953206772}, "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-2021-31/segments/1627046153854.42/warc/CC-MAIN-20210729074313-20210729104313-00407.warc.gz"} |
https://datascience.stackexchange.com/questions/37600/mdp-rl-multiple-rewards-for-the-same-state-possible | # MDP - RL, Multiple rewards for the same state possible?
This question is from An introduction to RL Pages 48 and 49. This question may also be related to below question, although I am not sure: Cannot see what the "notation abuse" is, mentioned by author of book
On page 48, it is mentioned that p:S * R * S * A -> [0,1] is a deterministic function:
The dynamics function $p : \mathcal{S} \times \mathcal{R} \times \mathcal{S} \times \mathcal{A} \rightarrow [0, 1]$ is an ordinary deterministic function of four arguments.
However, on page 49, in equation 3.4, there is summation over r:
$$\sum_{s' \in \mathcal{S}}\sum_{r \in \mathcal{R}} p(s',r|s,a) = 1 ,\text{for all } s \in \mathcal{S}, a \in \mathcal{A}(s)$$
My question is, does this mean, it is possible that performing an action $a$ that takes us to state $s'$, could result in multiple rewards?
it is possible that performing an action $a$ that takes us to state $s′$, could result in multiple rewards?
Yes, that is true the general case that any $(s,a)$ pair can result in a range of results for both $s'$ and $r$. Also $s'$ and $r$ can vary independently, provided each ones distribution only depends on $(s,a)$. In practice $r$ often depends strongly on one or more of $s$, $a$ or $s'$ (if it depends on the latter, then it still depends in absolute terms on just $s$ and $a$ because $s'$ does - it is just that values of $s'$ and $r$ are allowed to correlate).
This does not affect the statement about $p(s',r|s,a)$ being deterministic. It is the probability of specific $s',r$ results occurring that should be deterministic and depend on $(s,a)$ in a Markov Decision Process. | 2020-08-08 12:40:23 | {"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": 0.5759381055831909, "perplexity": 281.44149475464775}, "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/1596439737645.2/warc/CC-MAIN-20200808110257-20200808140257-00441.warc.gz"} |
https://dlmtool.openmse.com/reference/DTe40.html | Effort is adjusted using a simple rule that aims for a specified level of depletion.
DTe40(x, Data, reps = 100, plot = FALSE, alpha = 0.4, LB = 0.9, UB = 1.1)
DTe50(x, Data, reps = 100, plot = FALSE, alpha = 0.5, LB = 0.9, UB = 1.1)
## Arguments
x
A position in the data object
Data
A data object
reps
The number of stochastic samples of the MP recommendation(s)
plot
Logical. Show the plot?
alpha
The target level of depletion
LB
The lowest permitted factor of previous fishing effort
UB
The highest permitted factor of previous fishing effort
## Value
An object of class Rec-class with the TAE slot(s) populated
## Details
The TAE is calculated as: $$\textrm{TAE}_y = \frac{D}{\alpha} \textrm{TAE}_{y-1}$$ where $$D$$ is estimated current level of depletion and $$\alpha$$ is argument alpha specifying the target level of depletion.
The maximum fractional change in TAE is specified with arguments LB and UB
## Functions
• DTe40: Effort is adjusted to reach 40 percent stock depletion
• DTe50: Effort is adjusted to reach 50 percent stock depletion
## Required Data
See Data-class for information on the Data object
DTe40: Dep, MPeff
## Rendered Equations
See Online Documentation for correctly rendered equations
T. Carruthers
## Examples
DTe40(1, MSEtool::Atlantic_mackerel, plot=TRUE)
#> Effort
#> 0.9 | 2023-03-24 22:57:02 | {"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": 0, "mathjax_display_tex": 2, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.21090075373649597, "perplexity": 12919.428313046246}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-14/segments/1679296945289.9/warc/CC-MAIN-20230324211121-20230325001121-00340.warc.gz"} |
https://www.sinergyest.com/deep-learning/2019/07/24/deep-learning-for-medical-imaging/ | # Deep Learning for Medical Imaging
We have a very special post today from Jakob Kather from Heidelberg, Germany (Twitter: jnkath). He will be talking about deep learning for medical applications. Jakob is also one of the authors of a new paper recently published in Nature Medicine: https://www.nature.com/articles/s41591-019-0462-y discussing deep learning predicting gastrointestinal cancer.
Deep learning-based image analysis is well suited to classifying cats versus dogs, sad versus happy faces, and pizza versus hamburgers. However, many people struggle to apply deep learning to medical imaging data. In theory, it should be easy to classify tumor versus normal in medical images; in practice, this requires some tricks for data cleaning and model training and deployment. Here, we will show how to use deep learning in MATLAB to preprocess and classify complex medical images. For this demo, we'll be primarily using Deep Learning Toolbox and Image Processing Toolbox. On the hardware side, it's best to have a compatible GPU installed and ready to use in MATLAB (see https://www.mathworks.com/solutions/gpu-computing.html). Our aim is to find tumor tissue in histological images**. **Do you wonder what "histological images" are? In almost all cancer patients, the tumor is cut out by a surgeon, thinly sliced, put onto glass slides, stained and viewed under a microscope. Thus, we can see everything from cells on a micrometer scale to tissue structures on a millimeter scale. Thousands of such images are freely available in public repositories. Some of these repositories are available at the National Institutes of Health (NIH) data portal. From https://portal.gdc.cancer.gov we can download tumor images such as this (in this case, a lung cancer): These images are in SVS format, which is essentially a multi-layer TIFF image. This may look like an ordinary image, but SVS images are huge: the files are often larger than 1 GB and the images have up to a billion pixels. A zoomed in version of one section of this image shows how large this image is:
This image shows how much detail is contained in a very small portion of the image. We are zoomed in on the red dot shown in the upper right full image viewer.
###### Images courtesy of National Cancer Institute.
% define the image name
imName = 'TCGA-NK-A5CR-01Z-00-DX1.A7C57B30-E2C6-4A23-AE71-7E4D7714F8EA.svs';
imInfo = imfinfo(imName); % get the metadata
SVS images are essentially multipage TIFFs and we can use imfinfo() to look at the metadata of each page.
for i = 1:numel(imInfo)
X = ['Layer ', num2str(i), ': Width ',num2str(imInfo(i).Width), ...
' and Height ', num2str(imInfo(i).Height)];
disp(X)
end
The base image (channel 1) is so big that we cannot even look at it... but let's look at some of the smaller images instead.
imshow(imread(imName,2))
imshow(imread(imName,7))
For each channel, we can look at the metadata.
disp(['this image has ',num2str(imInfo(5).Width),'*',num2str(imInfo(5).Height),' pixels'])
>> this image has 3019*1421 pixels We can see this image is mostly background and contains non-tumor and tumor tissue. Because we care about the tumor tissue and not so much about the surrounding normal tissue, we want to identify the tumor region.
###### Note if you are a non-medical person, here is the image annotated with the tumor labeled.
Let us use a transfer learning approach with AlexNet. We will load the default pretrained AlexNet model which has already learned to distinguish shapes such as circles or lines.
net = alexnet; % load an alexnet which is pretrained on ImageNet
Now, we want to re-train the model as a tumor detector. We will use a public data set of 100,000 histological images of colon cancer, which is available at http://dx.doi.org/10.5281/zenodo.1214456. This set has been derived from colorectal cancer samples, but the workflow is identical for any type of solid tumor. This is how these smaller images (patches) look: They are labeled with one of nine classes which are explained in more detail in the data repository. Our aim is to train a deep neural network to automatically detect these classes. These images represent different classes of tissue that were manually defined by a pathologist. Each row is a tissue class and contains random images from the images set. The class labels are as follows: ADI = adipose tissue (fat), BACK = background (no tissue), DEB = debris, LYM = lymphocytes, MUC = mucus, MUS = muscle, NORM = normal mucosa, STR = stroma, TUM = tumor epithelium.The classes are described in more detail here: https://journals.plos.org/plosmedicine/article?id=10.1371/journal.pmed.1002730 and here: https://www.nature.com/articles/srep27988. After downloading the ZIP files from the repository and extracting them to a folder called "images", we have one sub-folder per tissue class in "images." We can now load them, split into a training, validation and test set, and re-train our alexnet model. (Optionally, we could use the imageDataAugmenter to create even more training images with rotational variance, for example). Collect all pictures from the folder image and put them in a datastore. Subfolders are included, the category/label is determined by the folder names.
allImages = imageDatastore('./images/','IncludeSubfolders',true,'LabelSource','foldernames');
Split into three sets: 40% training, 20% validation, 40% test
[training_set, validation_set, testing_set] = splitEachLabel(allImages,.4,.2,.4);
## Network modification
Modify the network by removing the last three layers. We will replace these layers with new layers for our custom classification.
layersTransfer = net.Layers(1:end-3);
Display the output categories.
categories(training_set.Labels)
ans = 9×1 cell array
{'ADI'} {'BACK'} {'DEB'} {'LYM'} {'MUC'} {'MUS'} {'NORM'} {'STR'} {'TUM'}
numClasses = numel(categories(training_set.Labels));
We merge the layers, and set the weight and bias learning rate for the last fully connected layer 'fc'
layers = [
layersTransfer
fullyConnectedLayer(numClasses,'Name', 'fc','WeightLearnRateFactor',1,'BiasLearnRateFactor',1)
softmaxLayer('Name', 'softmax')
classificationLayer('Name', 'classOutput')];
Set up a layerGraph and plot it:
lgraph = layerGraph(layers);
plot(lgraph)
## Modify Training Parameters
We now modify the training set and training options. The training set must be resized to fit the input size expected by the network.
imageInputSize = [227 227 3];
augmented_training_set = augmentedImageSource(imageInputSize,training_set);
augmented_training_set = augmentedImageDatastore with properties:
NumObservations: 39999 Files: {39999×1 cell} AlternateFileSystemRoots: {} MiniBatchSize: 128 DataAugmentation: 'none' ColorPreprocessing: 'none' OutputSize: [227 227] OutputSizeMode: 'resize' DispatchInBackground: 0
resized_validation_set = augmentedImageDatastore(imageInputSize,validation_set);
resized_testing_set = augmentedImageDatastore(imageInputSize,testing_set);
Set the training options, including plotting the training progress as the network trains.
opts = trainingOptions('sgdm', ...
'MiniBatchSize', 64,... % mini batch size, limited by GPU RAM, default 100 on Titan, 500 on P6000
'InitialLearnRate', 1e-5,... % fixed learning rate
'L2Regularization', 1e-4,... % optimization L2 constraint
'MaxEpochs',15,... % max. epochs for training, default 3
'ExecutionEnvironment', 'gpu',...% environment for training and classification, use a compatible GPU
'ValidationData', resized_validation_set,...
'Plots', 'training-progress')
## Training
We trained the network for 3.5 hours on a single GPU, but training for a few minutes would actually be enough to get a reasonable result as seen in the training plot below.
net = trainNetwork(augmented_training_set, lgraph, opts)
## Testing and Prediction
Let's check how well our classifier works using the held-out subset.
[predLabels,predScores] = classify(net, resized_testing_set, 'ExecutionEnvironment','gpu');
We can look at the confusion matrix and at the overall classification accuracy:
plotconfusion(testing_set.Labels, predLabels)
PerItemAccuracy = mean(predLabels == testing_set.Labels);
title(['overall per image accuracy ',num2str(round(100*PerItemAccuracy)),'%'])
Voila! We have achieved an excellent classification performance (as you will see in this paper). We are now ready to build more complicated workflows for digital pathology that include an automatic tumor detector! I want to thank Jakob again for taking the time to give us insight into his research using MATLAB. A special thanks to Jakob Sommer for testing the source code in this post. Have any questions about this post? Leave a comment below.
| | 2021-07-28 04:19:41 | {"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": 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": 0.46468839049339294, "perplexity": 3471.761528779806}, "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/1627046153521.1/warc/CC-MAIN-20210728025548-20210728055548-00397.warc.gz"} |
http://html.rhhz.net/qxxb_en/html/20180205.htm | J. Meteor. Res. 2018, Vol. 32 Issue (2): 203-218 PDF
http://dx.doi.org/10.1007/s13351-018-7071-4
The Chinese Meteorological Society
0
#### Article Information
Li, S. P., W. Hou, and G. L. Feng, 2018.
Atmospheric Circulation Patterns over East Asia and Their Connection with Summer Precipitation and Surface Air Temperature in Eastern China during 1961–2013. 2018.
J. Meteor. Res., 32(2): 203-218
http://dx.doi.org/10.1007/s13351-018-7071-4
### Article History
in final form December 7, 2017
Atmospheric Circulation Patterns over East Asia and Their Connection with Summer Precipitation and Surface Air Temperature in Eastern China during 1961–2013
Shuping LI1, Wei HOU2, Guolin FENG1,2,3
1. College of Atmospheric Sciences, Lanzhou University, Lanzhou 730000;
2. National Climate Center, China Meteorological Administration, Beijing 100081;
3. College of Physical Science and Technology, Yangzhou University, Yangzhou 225000
ABSTRACT: Based on the NCEP/NCAR reanalysis data and Chinese observational data during 1961–2013, atmospheric circulation patterns over East Asia in summer and their connection with precipitation and surface air temperature in eastern China as well as associated external forcing are investigated. Three patterns of the atmospheric circulation are identified, all with quasi-barotropic structures: (1) the East Asia/Pacific (EAP) pattern, (2) the Baikal Lake/Okhotsk Sea (BLOS) pattern, and (3) the eastern China/northern Okhotsk Sea (ECNOS) pattern. The positive EAP pattern significantly increases precipitation over the Yangtze River valley and favors cooling north of the Yangtze River and warming south of the Yangtze River in summer. The warm sea surface temperature anomalies over the tropical Ind-ian Ocean suppress convection over the northwestern subtropical Pacific through the Ekman divergence induced by a Kelvin wave and excite the EAP pattern. The positive BLOS pattern is associated with below-average precipitation south of the Yangtze River and robust cooling over northeastern China. This pattern is triggered by anomalous spring sea ice concentration in the northern Barents Sea. The anomalous sea ice concentration contributes to a Rossby wave activity flux originating from the Greenland Sea, which propagates eastward to North Pacific. The positive ECNOS pattern leads to below-average precipitation and significant warming over northeastern China in summer. The reduced soil moisture associated with the earlier spring snowmelt enhances surface warming over Mongolia and northeastern China and the later spring snowmelt leads to surface cooling over Far East in summer, both of which are responsible for the formation of the ECNOS pattern.
Key words: atmospheric circulation patterns precipitation surface air temperature external forcing
1 Introduction
As an agricultural and economic center, eastern China is a typical monsoon area and is of national importance. The summer climate over this region exhibits complex features, especially for the precipitation. The summer precipitation over eastern China is greatly affected by the East Asian summer monsoon (EASM) (Jiang et al., 2008). The EASM consists of the staged progression of zonally oriented rain belts, which include mixed tropical and baroclinic features. The onset of the EASM occurs in early to mid May and is accompanied by abrupt changes in precipitation, winds, outgoing longwave radiation, and clouds (Tao and Chen, 1987). The EASM shows two abrupt northward jumps and three stationary periods. The three stationary periods correspond to the three latitudinally oriented rain belts in southern China, the Yangtze–Huai River valley, and northern China (Ding, 1992). The surface air temperature (SAT) is influenced by the seasonal development of the EASM through clouds, the atmospheric circulation, and land surface processes (Betts, 2007). Shi and Zhu (1996) showed that a strong EASM is associated with higher than average SATs over large areas of China. Chen and Wu (2017) reported that the atmospheric circulation has a great impact on the SAT through cloud-induced changes in surface radiation.
The western Pacific subtropical high (WPSH), with southerly or southwesterly winds along its western flank in the lower troposphere, is a dominant member of the EASM circulation system and has a significant impact on summer precipitation over eastern China (Huang and Tang, 1987). Moisture transported by the southwesterly winds at the western edge of the WPSH feeds the Meiyu front, which is accompanied by the Meiyu rain belt over the Yangtze River valley from June to mid July (Wu et al., 2012). The WPSH contributes to the distribution of summer precipitation over eastern China (Huang and Chen, 2010). The WPSH also plays an important part in abnormally hot summers over eastern China (Wang W. W. et al., 2016) and the western ridge of the WPSH has extended significantly westward since the late 1970s (Gong and Ho, 2002). Moreover, the establishment and retreat of the Okhotsk high are also closely related to summer precipitation over eastern China (Zhang and Tao, 1998) and a high frequency of the Okhotsk high in July and August leads to abnormally cool summers due to strong northwesterly winds over East Asia (Ninomiya and Mizuno, 1985).
The East Asia/Pacific (EAP) pattern has been recognized as the dominant mode over East Asia in summer and reflects the configuration of the WPSH, the Meiyu trough, and the Okhotsk high (Nitta, 1987; Huang, 1992). The positive EAP pattern intensifies precipitation over the Yangtze River valley and reduces precipitation over the Atlantic coast (Chen and Zhai, 2015). Zhao et al. (2007) suggested that the Meiyu rain belt shifts northward under high Asia/Pacific Oscillation conditions, whereas Sung et al. (2006) suggested the possible delayed impact of the winter North Atlantic Oscillation (NAO) on EASM precipitation in the following summer. Notaro et al. (2006) concluded that a positive Pacific/North American pattern accompanied by a southward-shifted upper-level jet stream leads to dry and cold conditions in northeastern America during early winter. Park et al. (2011) also showed that the Arctic Oscillation (AO) phase is related to the East Asian cold surge. Diao et al. (2015) concluded that winters with frequent warm extremes in Europe are related to a northeast–southwest tilted positive NAO pattern. These teleconnection patterns and their impacts on regional and remote precipitation and SAT have attracted much attention from meteorologists. However, the regional summer atmospheric circulation patterns of East Asia are still unclear and need verifications from more observational data. In particular, most of the variability in atmospheric circulation over East Asia is not captured by the EAP pattern. For exam-ple, Wang J. B. et al. (2016) indicated that the first leading mode of the 30–60-day filtered and normalized summer 500-hPa geopotential height anomalies resembles the EAP pattern with a percentage variance of only 18%. Therefore, it is useful to further probe the summer atmospheric circulation patterns over East Asia and their impacts on regional precipitation and SAT. The combination of socioeconomic and atmospheric circulation characteristics makes the region (10°–70°N, 100°–160°E), a most appropriate study area, to explore the summer atmospheric circulation patterns of East Asia.
It is widely accepted that external forcing—for example, the sea surface temperature (SST), snow cover, and sea ice—could produce a modulating effect on the atmospheric circulation (e.g., Wu et al., 2009; Matsumura et al., 2010; Sun et al., 2013; Wang et al., 2014; Ding et al., 2016). Positive tropical SST anomalies increase precipitation and convection, and anomalous diabatic forcing triggers anomalous divergent circulation that propagates from the tropics to high latitudes (Trenberth and Caron, 2000). The El Niño–Southern Oscillation may modify the variations in atmospheric circulation between the subtropics and high-latitude regions (Li et al., 2012; Feng et al., 2017). Li et al. (2014) showed that the warm pool El Niño induces an abnormal anticyclonic circulation over the Yangtze River valley and a cyclonic circulation over the northern South China Sea. Zhang et al. (2017) demonstrated that a west–east dipole pattern in Eurasian spring snowmelt anomalies, with an increase in snowmelt over eastern Europe and western Russia and a decrease in snowmelt over Baikal Lake, is associated with summer precipitation in East Asia, through triggering an anomalous Eurasian wave train. The earlier spring snowmelt reduces the surface albedo and soil moisture and then produces surface warming, which contributes to the development of stationary Rossby waves in summer (Matsumura and Yamazaki, 2012). Wu et al. (2013) indicated that anomalous winter sea ice in Baffin Bay influences the summer atmospheric circulation anomalies over northern Eurasia via spring atmospheric circulation anomalies south of Newfoundland. We therefore investigate how the summer atmospheric circulation patterns over East Asia are related to the various kinds of external forcing.
Surface weather conditions are directly governed by large-scale atmospheric circulation patterns, which act as a bridge between external forcing and surface weather conditions. We focus here on the summer atmospheric circulation patterns over East Asia and their direct impacts on simultaneous precipitation and SAT over eastern China. We also investigate the relationship between the summer atmospheric circulation patterns and external forcing to obtain a more comprehensive understanding of the atmospheric circulation patterns through correlation analysis. Linear regression analysis is further applied to validate the driving effects of external forcing on the atmospheric circulation patterns.
The paper is organized as follows. Section 2 gives a brief introduction to the data and methods used. Section 3 reports the summer atmospheric circulation patterns over East Asia and their corresponding vertical structures. Section 4 presents the direct impacts of the summer atmospheric circulation patterns on the simultaneous precipitation and SAT over eastern China. Section 5 illustrates the driving effects of the external forcing on the atmospheric circulation patterns, and finally, the conclusions and discussion are given in Section 6.
2 Data and methods
Monthly atmospheric circulation data for the time period 1961–2013 are provided by the NCEP/NCAR at a horizontal resolution of 2.5° × 2.5° (Kalnay et al., 1996). In addition, monthly data for the 10-m horizontal wind, net latent heat flux, 2-m air temperature, and soil moisture at 0–10-cm depth for 1961–2013 are also obtained from the NCEP/NCAR but with a horizontal resolution of T62. Monthly precipitation and SAT data at 160 stations over China for 1961–2013 are obtained from the China Meteorological Administration. Monthly SST data are obtained from the NOAA extended reconstructed SST V3 dataset for 1960–2013 with a 2.0° × 2.0° horizontal resolution (Smith et al., 2008). Monthly precipitation data from the NOAA’s Precipitation Reconstruction dataset for 1961–2013 (Chen et al., 2002) and NOAA’s interpolated outgoing longwave radiation data for 1979–2013 (Liebmann and Smith, 1996) with a 2.5° × 2.5° horizontal resolution are used to validate the tropical convection.
The monthly sea ice concentration (SIC) data provided by the UK Meteorological Office Hadley Center for 1961–2013 have a 1.0° × 1.0° horizontal resolution (Rayner et al., 2003). The weekly snow cover extent (SCE) data obtained from Rutgers University Climate Laboratory for 1973–2013 are projected onto a grid with a 2.0° × 2.0° horizontal resolution (Robinson et al., 1993). The results in this study are based mainly on the period 1961–2013, except for the outgoing longwave radiation and SCE datasets, which are limited by the availability of data. The monthly AO, NAO, Polar/Eurasia (POL), East Atlantic (EA), Scandinavian (SCAN), East Atlantic/West Russia (EAWR), and West Pacific (WP) teleconnection pattern indices are provided by the NOAA Climate Prediction Center. The conventional Eurasian (EU) and western Atlantic (WA) teleconnection pattern indices are calculated according to the definition of Wallace and Gutzler (1981) as follows,
$\begin{split} {\text{EU}} = & - \frac{1}{4}{Z^*}({55^\circ }{\rm N},{20^\circ }{\rm E}) + \frac{1}{2}{Z^*}({55^\circ }{\rm N},{75^\circ }{\rm E})\\ & - \frac{1}{4}{Z^*}({40^\circ }{\rm N},{145^\circ }{\rm E}),\end{split}$ (1)
${\rm{WA}} = \frac{1}{2}[{Z^*}({55^\circ }{\rm N},\;{55^\circ }{\rm W}) - {Z^*}({30^\circ }{\rm N},\;{55^\circ }{\rm W})].$ (2)
To remove any influences from the long-term trend, the linear trend is removed from all variables. We use normalized empirical orthogonal function (EOF) analysis, in which the principal components (PCs) are divided by their standard deviation and the spatial EOF patterns are multiplied by the corresponding standard deviation (Zheng et al., 2013). The statistical significance is assessed by using a two-tailed Student’s t-test. The seasonal mean refers to the average of December, January, and February for preceding winter; March, April, and May for spring; June, July, and August for summer; and September, October, and November for autumn.
3 Summer atmospheric circulation patterns over East Asia
The normalized EOF analysis is performed for summer mean 500-hPa geopotential height anomalies over 10°–70°N, 100°–160°E. Figure 1a shows that the first leading EOF mode, which explains 30.2% of the total variance, presents a typical tripole pattern with two positive centers over the western subtropical Pacific and northeastern Asia and one negative center over Japan. This mode is identified as the EAP pattern. The second leading EOF mode (Fig. 1b), which explains 17.9% of the total variance, exhibits a latitudinally oriented dipole pattern, with positive anomalies over Baikal Lake and negative anomalies over the Okhotsk Sea. This mode is identified as the Baikal Lake/Okhotsk Sea (BLOS) pattern. The third leading EOF mode explains 12.2% of the total variance and is referred to as the eastern China/northern Okhotsk Sea (ECNOS) pattern. The positive ECNOS pattern shows a large area of positive anomalies over eastern China and negative anomalies over the northern Okhotsk Sea (Fig. 1c). We changed the domain prescribed for the normalized EOF analysis slightly (10°–70°N, 102.5°–157.5°E) and obtained similar results. Therefore, the results are not sensitive to the choice of domain. The first three EOF modes account for 60.3% of the total variance, indicating that they capture well the atmospheric circulation characteristics of East Asia. The PCs of the three leading EOF modes are normalized by their corresponding standard deviations and are characterized by distinct interannual variations (Figs. 1df).
Figure 1 Leading modes of the summer 500-hPa geopotential height anomalies over East Asia and their corresponding PCs: (a) EOF1, (b) EOF2, (c) EOF3, (d) PC1, (e) PC2, and (f) PC3. The numbers in blue at the top of the right-hand panels indicate the percentage of variance explained.
Figure 2 shows the summer atmospheric circulation anomalies associated with the positive EAP pattern from linear regression analysis. In the upper troposphere, large-scale positive geopotential height anomalies emerge at lower latitudes and the EAP-like pattern is obvious and significant (Fig. 2a). In the mid-troposphere, the positive EAP pattern shows two centers of positive anomalies located over the western subtropical Pacific and northeastern Asia and one center of negative anomalies over Japan (Fig. 2b). The magnitude of the EAP-like pattern at 850 hPa is weaker and less significant than that in the mid-troposphere (Fig. 2c). The sea level pressure (SLP) anomalies over East Asia also show a tripole wave-like pattern (Fig. 2d). The EAP pattern shows a quasi-barotropic structure from the surface to the upper troposphere. Although a Pacific/North American-like pattern is also observed over the Pacific/North America sector, it is weaker than the EAP-like pattern, especially in the lower troposphere.
Figure 3 shows the summer atmospheric circulation anomalies associated with the positive BLOS pattern from the surface to the upper troposphere. The geopotential height field in the upper troposphere features a wave train pattern over the mid–high latitudes of Eurasia, with negative anomalies over the northern Urals and the Okhotsk Sea and positive anomalies over western Europe and Baikal Lake (Fig. 3a). The dominant features of the 500-hPa geopotential height are similar to those at 200 hPa (Fig. 3b). This wave train pattern is weaker in the lower troposphere than in the upper troposphere (Fig. 3c). The wave train pattern over the mid–high latitudes of Eurasia is distinct at the surface (Fig. 3d). The BLOS pattern presents an obvious quasi-barotropic structure from the surface to the upper troposphere.
In the summers with the positive phase of the ECNOS pattern, two significant negative anomaly centers at 200-hPa geopotential height are located over the northern Urals and the Far East and one significant positive anomaly center is observed near Baikal Lake (Fig. 4a). In the mid-troposphere, the two negative anomaly centers at 500 hPa are similar to those at 200 hPa, whereas the positive anomalies over East Asia become more significant than those in the upper troposphere (Fig. 4b). The two negative anomaly centers are clearly weaker in the lower troposphere, but the positive geopotential height anomalies over East Asia are obvious (Fig. 4c). The SLP anomaly pattern is similar to that at 850 hPa (Fig. 4d). The ECNOS pattern presents a quasi-barotropic structure from the surface to the upper troposphere.
Figure 2 Summer mean anomalies regressed onto PC1 for 1961–2013: (a) 200-, (b) 500-, and (c) 850-hPa geopotential height (gpm), and (d) SLP (hPa). The dotted areas exceed the 95% confidence level with a two-tailed Student’s t-test.
Figure 3 As in Fig. 2, but for PC2.
Figure 4 As in Fig. 2, but for PC3.
We further investigate the connections between the BLOS and ECNOS patterns and the conventional atmospheric circulation teleconnection patterns in the Northern Hemisphere (Table 1). The BLOS pattern index is significantly correlated with the NAO, POL, and EAWR pattern indices, but the spatial distribution of the BLOS pattern is distinctly different from that of the above three patterns (figures omitted). In particular, the BLOS pattern index is significantly correlated with the SCAN pattern index. The spatial distribution of the BLOS pattern resembles that of the SCAN pattern, although the anomaly center over the Okhotsk Sea for the BLOS pattern is dramatically stronger. Therefore, the BLOS pattern is also referred to as an SCAN-like teleconnection pattern. The correlation coefficients between the ECNOS pattern and the EA and WP patterns are 0.236 and 0.234, respectively, which are significant only at the 90% confidence level. The spatial features of the ECNOS pattern are obviously different from those of the EA and WP patterns (figures omitted). Hence, the ECNOS pattern is probably distinct from the conventional atmospheric circulation teleconnection patterns in the Northern Hemisphere.
Table 1 Correlation coefficients between the BLOS and ECNOS patterns and the conventional atmospheric circulation teleconnection patterns in the Northern Hemisphere. The symbols * and ** indicate the 90% and 95% confidence level, respectively
Pattern AO NAO POL EA WA EU SCAN EAWR WP BLOS –0.096 0.341** –0.273** –0.169 –0.031 –0.127 –0.282** 0.311** –0.025 ECNOS 0.209 0.076 0.221 0.236* –0.052 –0.197 –0.082 0.070 0.234*
4 Effects of atmospheric circulation patterns on precipitation and SAT
We focus now on the effects of the three different summer atmospheric circulation patterns on precipitation and SAT over eastern China. The positive EAP pattern is associated with above-average precipitation over the Yangtze River valley and northeastern China and below-average precipitation over the Yellow River valley and southeastern China (20°–25°N, 115°–120°E) in the summer (Fig. 5a). The variance of summer precipitation associated with the EAP pattern is about 20%–40% of the total variance over the Yangtze River valley, but it is less than 10% of the total variance over most of northeastern China (Fig. 5b). The summer precipitation associated with the positive BLOS pattern is below average south of 30°N, especially over southwestern China, and above average over northern China (Fig. 5c). The BLOS pattern explains 10%–20% of the total variance in summer precipitation over most of eastern China and 30%–40% of the total variance over southwestern China (Fig. 5d). The summer precipitation associated with the positive ECNOS pattern is more than the average over the Yangtze River valley and less than the average over northeastern China (Fig. 5e). The ECNOS pattern explains about 30% and 40% of the total variance of summer precipitation over the Yangtze River valley and northeastern China, respectively (Fig. 5f).
The positive EAP pattern produces cooling north of the Yangtze River and warming south of the Yangtze River (Fig. 6a). This pattern explains 10%–20% of the total variance in the summer SAT over most of eastern China (Fig. 6b). In association with the positive BLOS pattern, significant cooling is observed over northeastern China, whereas warming occurs south of the Yangtze River (Fig. 6c). The explained variance of the summer SATs associated with the BLOS pattern exceeds 30% of the total variance north of the Yellow River, whereas it is only 10%–20% of the total variance south of the Yangtze River (Fig. 6d). Robust warming is present over northeastern China in the positive ECNOS phase and significant cooling is seen over the Yangtze River valley (Fig. 6e). The explained variance exceeds 40% of the total variance over northeastern China and reaches 30% of the total variance south of the Yangtze River (Fig. 6f).
Figure 5 Observed summer precipitation anomalies (mm) regressed onto (a) PC1, (c) PC2, and (e) PC3 for 1961–2013 and the spatial distribution of the explained variance of the (b) EAP, (d) BLOS and (f) ECNOS patterns. The black dots in (a, c, e) indicate areas exceeding the 95% confidence level with a two-tailed Student’s t-test.
Figure 6 As in Fig. 5, but for the summer SAT (℃).
5 Possible association between atmospheric circulation patterns and external forcing
To give a more complete understanding of the atmospheric circulation patterns, we investigate the relationship between the three summer circulation patterns in East Asia and the various types of external forcing. Figure 7 shows the correlation between PC1 and the seasonal mean SST anomalies from the preceding winter to autumn. Significant positive correlation is found over the tropical Indian Ocean (TIO; 10°S–20°N, 60°–110°E), accompanied by the decaying El Niño from the preceding winter to autumn, indicating a possible driving effect of the TIO SST anomalies on the EAP pattern. We define the TIO index by averaging the SST anomalies over the region (10°S–20°N, 60°–110°E).
The 27.5℃ SST isotherm is the convection threshold for the climatological state (Graham and Barnett, 1987). During the El Niño decaying summer, basin-wide warming over the TIO (i.e., total SST > 27.5℃) enhances the anomalous convective heating and produces an anomalous upper level divergence ( Figs. 8a, b). The anomalous convection over the TIO contributes to tropospheric warming and forces a Kelvin wave wedge, which penetrates into the western equatorial Pacific (Xie et al., 2009). The Ekman divergence induced by the Kelvin wave weakens the convection and triggers an anomalous anticyclone over the Philippine Sea (Fig. 8c), leading to the generation and persistence of the EAP pattern (Qu and Huang, 2012). In particular, there is a negative Indo-Pacific warm pool and North Pacific dipole (IPOD) pattern, with warm SST anomalies over the Indo-Pacific warm pool and cold SST anomalies over North Pacific (Fig. 8a). According to Zheng et al. (2014), this negative IPOD pattern contributes to the strengthened WPSH and then intensifies the Philippine anticyclone in the lower troposphere, leading to anomalous convection over the nothwestern subtropical Pacific that contributes to the excitation of the EAP pattern. Therefore, both the anomalous TIO SST and the IPOD pattern are responsible for excitation of the EAP pattern through anomalous convection over the northwestern subtropical Pacific. The atmospheric circulation and SST fields associated with the warm TIO SST anomalies resemble those associated with the positive EAP pattern (Figs. 8df), suggesting that the anomalous TIO SST triggers the EAP pattern. The relationship between the TIO SST and the EASM shows a decadal shift in the late 1970s. The Indian Ocean dipole SST anomaly pattern was closely related to the EASM before the mid 1970s, whereas the relationship between the EASM and the SST anomalies in the northern Indian Ocean strengthened after the late 1970s (Ding et al., 2010).
In this process, the warm summer SST anomalies in the TIO decrease the land–sea thermal contrast and the north–south SLP gradient, which, in turn, leads to anomalous northeasterly winds over the TIO and the South China Sea. The anomalous northeasterly winds decrease the 10-m wind speed and hence decrease the net latent heat flux (Fig. 9a). This change in the net latent heat flux induces warm SST anomalies and provides a positive feedback consistent with the wind–evaporation–SST feedback theory (Xie and Philander, 1994). The warm TIO SST anomalies produce obvious convection, intensified ascending motion, and increased precipitation over the TIO, and suppress convection, ascending motion, and precipitation over the Philippine Sea and northwestern subtropical Pacific (Figs. 9bd). These results further validate the view that the anomalous TIO SSTs contribute to the generation of the EAP pattern via anomalous convection over the northwestern subtropical Pacific associated with the Ekman divergence induced by the Kelvin wave.
Figure 7 Correlations between PC1 and seasonal mean SST in (a) the preceding winter, (b) spring, (c) summer, and (d) autumn. The dotted areas indicate the 95% confidence level with a two-tailed Student’s t-test.
Figure 8 Summer mean (a) SST (shading; ℃), (b) 300-hPa potential velocity (contours; m2 s–1) and divergent wind (vectors; m s–1), and (c) 300-hPa geopotential height (shading; gpm) and 850-hPa horizontal wind (vectors; m s–1) anomalies associated with the simultaneous TIO index for 1961–2013. (d)–(f) Same as (a)–(c), respectively, but associated with PC1. In (a, d), the green solid lines indicate the 27.5℃ of the total SST associated with the TIO index and PC1, respectively. The total SST is defined as the sum of the TIO index/PC1 regression pattern and climatological mean SST. In (b, e), the contour interval is 0.1 × 105 m2 s–1; the zero contour lines are black, whereas the negative (positive) values are blue (red).
Figure 9 Summer mean (a) net latent heat flux (shading; W m–2) and 10-m wind (vectors; m s–1), (b) outgoing longwave radiation (W m–2), (c) 500-hPa vertical velocity (Pa s–1), and (d) precipitation (mm day–1) anomalies associated with the simultaneous TIO index. The dotted areas exceed the 95% confidence level with a two-tailed Student’s t-test.
Figure 10 shows the correlation between PC2 and SIC anomalies in the preceding winter and spring. The PC2 and SIC anomalies in the preceding winter are significantly and negatively correlated over the northern Barents Sea (Fig. 10a). This significant negative correlation over the northern Barents Sea is also observed in spring, although it covers a relatively smaller region (Fig. 10b). Li and Leung (2013) indicated that the anomalous Arctic sea ice changes the local surface turbulent heat flux from spring to summer, triggering a quasi-stationary Rossby wave train over the Eurasian continent. Therefore, the anomalous spring SIC over the northern Barents Sea may contribute to the generation and persistence of the BLOS pattern in the summer. The spring SIC (SSIC) index is defined as the average SIC over 82°–90°N, 20°W–80°E.
To compare these results, the SSIC index is multiplied by –1 in Figs. 11a, b. The 300-hPa geopotential height anomalies associated with the decreased SSIC over the northern Barents Sea show an obvious wave train pattern over Eurasia, with two positive anomaly centers over western Europe and northern Baikal Lake and two negative anomaly centers over the northern Urals and the Okhotsk Sea (Figs. 11a, b). The wave activity flux is used to explore the propagation of the extra-tropical Rossby wave (Takaya and Nakamura, 2001). The Rossby wave activity flux associated with the decreased SSIC over the northern Barents Sea originates from the Greenland Sea and propagates eastward, penetrating into North Pacific. The atmospheric circulation pattern and the Arctic SIC associated with the decreased SSIC over the northern Barents Sea are similar to those associated with the positive BLOS pattern, despite differences in some regions (Figs. 11c, d), identifying that the anomalous SSIC over the northern Barents Sea is responsible for the excitation and persistence of the BLOS pattern. This wave activity flux is reinforced over the Okhotsk Sea, implying that other factors may contribute to the Okhotsk Sea anomaly center in the BLOS pattern.
The spring SLP anomalies associated with the positive BLOS pattern are characterized by a pattern similar to the North Pacific Oscillation (Fig. 12a). The surface wind anomaly pattern over North Pacific, which is similar to the North Pacific Oscillation, contributes to the development of the Victoria mode, which is defined as the second EOF mode of the SST anomalies in the North Pacific poleward of 20°N (Bond et al., 2003). The anomalous westerly winds over the central tropical Pacific decrease the northeasterly trade winds and produce warm SST anomalies in the spring (Fig. 12b). In association with the positive spring Victoria mode (Fig. 12c), anomalous southwesterly winds occur over the western tropical Pacific and an obvious convergence is observed over the central–eastern tropical Pacific, leading to the development of the warm central tropical Pacific SST anomalies in the summer, which may persist into the following autumn and winter (Ding et al., 2015a, b). This process is accompanied by anomalous upper level divergent winds and intensified precipitation over the central tropical Pacific. These anomalous divergent winds contribute to the development of the Okhotsk Sea anomaly center in the BLOS pattern (Fig. 12d).
Figure 10 Correlation between PC2 and seasonal mean SIC in (a) the preceding winter and (b) spring. The dotted areas exceed the 95% confidence level with a two-tailed Student’s t-test.
Figure 11 (a) Spring mean SIC and (b) summer mean 300-hPa geopotentail height (shading; gpm) and wave activity flux (vectors; m–2 s–2) anomalies associated with the inverted SSIC index. (c, d) Same as (a, b), but are associated with PC2. The dotted areas indicate the 95% confidence level with a two-tailed Student’s t-test.
Figure 12 Spring mean (a) SLP, (b) SST (shading; °C) and 10-m wind (vectors; m s–1) anomalies regressed onto PC2, summer mean (c) SST (shading) and 10-m wind (vectors), and (d) precipitation (shading; mm day–1) and divergent wind anomalies (vectors; m s–1) regressed onto the spring Victoria mode index. The Victoria mode index is defined as the normalized PC of the second EOF mode of the North Pacific SST anomalies in spring. The dotted areas exceed the 95% confidence level with a two-tailed Student’s t-test.
For the ECNOS pattern, a significant positive correlation of PC3 and the spring SCE is observed over the midlatitudes of Asia; this signal is obvious in April and disappears in May and June (Fig. 13). A significant positive correlation of PC3 and the SCE over the high latitudes of Asia is observed in May and June (Figs. 13c, d). These results indicate that the earlier snowmelt over the midlatitudes of Asia (40°–60°N, 60°–140°E) and later spring snowmelt over the high latitudes of Asia (60°–74°N, 60°–140°E) may be responsible for the formation of the ECNOS pattern. The June SCE is used to represent the spring snow cover because it can more effectively depict earlier or later snowmelt (Matsumura and Yamazaki, 2012). We define the spring snowmelt index as the difference in the area-averaged June SCE between the high and midlatitudes of Asia.
Figure 14 presents the summer 2-m air temperature and soil moisture anomalies associated with the spring snowmelt index. The earlier spring snowmelt reduces the summer soil moisture over the midlatitudes of Asia, especially over Mongolia and northeastern China (Fig. 14a). In addition, the earlier spring snowmelt decreases the surface albedo over the midlatitudes of Asia and favors surface warming in spring. The reduced soil moisture associated with the earlier spring snowmelt enhances surface warming over Mongolia and northeastern China and the later spring snowmelt over the Far East increases the soil moisture and reduces the near-surface air temperature in summer (Fig. 14b), both of which favor the formation of the ECNOS pattern.
Figure 13 Correlation between PC3 and SCE in (a) spring, (b) April, (c) May, and (d) June during 1973–2013.The dotted areas exceed the 90% confidence level with a two-tailed Student’s t-test.
Figure 14 Summer mean (a) 0–10-cm soil moisture (m3 m–3) and (b) 2-m air temperature anomalies (℃) regressed onto the spring snowmelt index. The dotted areas indicate the 95% confidence level with a two-tailed Student’s t-test.
6 Conclusions and discussion
This study focused on the summer atmospheric circulation patterns over East Asia and their impacts on simultaneous precipitation and SAT over eastern China. The relationship between the three summer atmospheric circulation patterns and external forcing such as SST, sea ice, and so on was also investigated to give a more complete picture of the circulation patterns. The atmospheric circulation of East Asia features three patterns in the summer mean 500-hPa geopotential height—namely, a typical tripole pattern with two positive centers over the Okhotsk Sea and western subtropical Pacific and one negative center over Japan (i.e., the EAP pattern), a latitudinally oriented dipole pattern with positive anomalies over Baikal Lake and negative anomalies over the Okhotsk Sea (i.e., the BLOS pattern), and an eastern China and northern Okhotsk Sea dipole pattern with positive anomalies over eastern China and negative anomalies over the northern Okhotsk Sea (i.e., the ECNOS pattern).
The EAP pattern features a quasi-barotropic structure from the surface to the upper troposphere. In the summer, the positive EAP pattern leads to above-average precipitation over the Yangtze River valley and northeastern China, with below-average precipitation over southeastern China. This pattern explains up to 40% of the total variance in summer precipitation over the Yangtze River valley. The positive EAP pattern favors cooling north of the Yangtze River and warming south of the Yangtze River. During the El Niño decaying summer, the warm TIO SST anomalies contribute to the formation and maintenance of the positive EAP pattern. The warm TIO SST anomalies produce anomalous northwesterly winds, which, in turn, decrease the background southerly winds and the net latent heat flux over the TIO. This change in the net latent heat flux induces warm SST anomalies and produces a positive feedback. The warm TIO anomalies also force an eastward propagating Kelvin wave wedge and the Ekman divergence induced by this Kelvin wave suppresses convection over the northwestern subtropical Pacific. Consequently, the anomalous convection over the northwestern subtropical Pacific associated with the anomalous TIO SST is responsible for the generation of the EAP pattern, supporting previous interpretations (Huang, 1992).
In the BLOS pattern, a wave train pattern in the upper troposphere is present over the mid–high latitudes of Eurasia, but is weaker in the lower troposphere. In the summer, the positive BLOS pattern leads to below-average precipitation south of the Yangtze River and produces significant cooling over northeastern China. This pattern explains about 40%–50% of the total variance in the summer SAT over northeastern China. The Rossby wave activity flux associated with the decreased spring sea ice over the northern Barents Sea originates from the Greenland Sea and propagates eastward, penetrating into North Pacific. Hence the anomalous spring sea ice over the northern Barents Sea contributes to the excitation and maintenance of the BLOS pattern. The spring surface wind anomaly pattern, similar to the North Pacific Oscillation, over North Pacific associated with the positive BLOS pattern contributes to the formation of the Victoria mode. The spring Victoria mode then leads to the development of warm central tropical Pacific SST anomalies in the summer, which produce anomalous divergent winds in the upper troposphere that intensify the stationary Rossby wave over mid–high latitudes.
The ECNOS pattern presents a quasi-barotropic structure from the surface to the upper troposphere. In the summer, the positive ECNOS pattern leads to below-average precipitation over northeastern China with above-average precipitation over the Yangtze River valley, producing robust warming over northeastern China and significant cooling over the Yangtze River valley. It explains up to 50% and 40% of the total variance in the summer SAT over northeastern China and the Yangtze River valley, respectively. Earlier spring snowmelt reduces the soil moisture and leads to surface warming over Mongolia and northeastern China in the following summer. Later spring snowmelt increases the soil moisture and decreases the near-surface air temperature over the high latitudes of Asia, especially over the Far East. Hence, the formation of the ECNOS pattern is partly attributed to earlier spring snowmelt over Mongolia and northeastern China and later spring snowmelt over the Far East. Other external factors also need to be investigated, although they are outside of the scope of this study.
Summer Arctic sea ice has decreased rapidly since the 1980s (Comiso et al., 2008; Zhang and Li, 2017; Luo and Yao, 2018). The anomalies in the extent of Arctic sea ice change the local energy balance, which, in turn, induces feedback to the large-scale atmospheric circulations (Honda et al., 1996). Sea ice and snow cover are the two factors that greatly affect the Arctic and subarctic atmospheric circulation. The winter–spring Eurasian snow cover is closely associated with the East Asian atmospheric circulation and precipitation in the following summer (Wu and Kirtman, 2007). Groisman et al. (2006) showed that the persistence of snow cover in northern Eurasia has decreased during the past few decades. Eurasia becomes snow-free in the early summer, whereas the Arctic Ocean is covered by sea ice until late summer. This lagged seasonal cycle intensifies the surface thermal contrast across the Arctic coastline (Derksen and Brown, 2012). The two significant negative anomaly centers associated with the positive ECNOS pattern are located over the northern Urals and Far East. Therefore, the ECNOS pattern is probably related to the Arctic sea ice and the combined effects of Arctic sea ice and Eurasian snow cover on the ECNOS pattern deserve further investigation.
Acknowledgments. We thank the reviewers for their constructive comments and suggestions that led to significant improvements in this paper. We also thank Dr. Yanjun Guo for her constructive suggestions about the snow cover extent data analysis.
References
Betts, A. K., 2007: Coupling of water vapor convergence, clouds, precipitation, and land-surface processes. J. Geophys. Res., 112, D10108. DOI:10.1029/2006JD008191 Bond, N. A., J. E. Overland, M. Spillane, et al., 2003: Recent shifts in the state of the North Pacific. Geophys. Res. Lett., 30, 2183. DOI:10.1029/2003GL018597 Chen, M. Y., P. P. Xie, J. E. Janowiak, et al., 2002: Global land precipitation: A 50-yr monthly analysis based on gauge observations. J. Hydrometeor., 3, 249–266. DOI:10.1175/1525-7541(2002)003<0249:GLPAYM>2.0.CO;2 Chen, S. F., and R. G. Wu, 2017: Interdecadal changes in the relationship between interannual variations of spring North Atlantic SST and Eurasian surface air temperature. J. Climate, 30, 3771–3787. DOI:10.1175/JCLI-D-16-0477.1 Chen, Y., and P. M. Zhai, 2015: Synoptic-scale precursors of the East Asia/Pacific teleconnection pattern responsible for persistent extreme precipitation in the Yangtze River valley. Quart. J. Roy. Meteor. Soc., 141, 1389–1403. DOI:10.1002/qj.2448 Comiso, J. C., C. L. Parkinson, R. Gersten, et al., 2008: Accelerated decline in the Arctic sea ice cover. Geophys. Res. Lett., 35, L01703. DOI:10.1029/2007GL031972 Derksen, C., and R. Brown, 2012: Spring snow cover extent reductions in the 2008–2012 period exceeding climate model projections. Geophys. Res. Lett., 39, L19504. DOI:10.1029/2012GL053387 Diao, Y. N., S. P. Xie, and D. H. Luo, 2015: Asymmetry of winter European surface air temperature extremes and the North Atlantic oscillation. J. Climate, 28, 517–530. DOI:10.1175/JCLI-D-13-00642.1 Ding, R. Q., K. J. Ha, and J. P. Li, 2010: Interdecadal shift in the relationship between the East Asian summer monsoon and the tropical Indian Ocean. Climate Dyn., 34, 1059–1071. DOI:10.1007/s00382-009-0555-2 Ding, R. Q., J. P. Li, Y. H. Tseng, et al., 2015a: The Victoria mode in the North Pacific linking extratropical sea level pressure variations to ENSO. J. Geophys. Res., 120, 27–45. DOI:10.1002/2014JD022221 Ding, R. Q., J. P. Li, Y. H. Tseng, et al., 2015b: Influence of the North Pacific Victoria mode on the Pacific ITCZ summer precipitation. J. Geophys. Res., 120, 964–979. DOI:10.1002/2014JD022364 Ding, R. Q., J. P. Li, Y. H. Tseng, et al., 2016: Interdecadal change in the lagged relationship between the Pacific–South Ameri-can pattern and ENSO. Climate Dyn., 47, 2867–2884. DOI:10.1007/s00382-016-3002-1 Ding, Y. H., 1992: Summer monsoon rainfalls in China. J. Meteor. Soc. Japan Ser. II, 70, 373–396. DOI:10.2151/jmsj1965.70.1B_373 Feng, J., W. Chen, and Y. J. Li, 2017: Asymmetry of the winter extra-tropical teleconnections in the Northern Hemisphere associated with two types of ENSO. Climate Dyn., 48, 2135–2151. DOI:10.1007/s00382-016-3196-2 Gong, D. Y., and C. H. Ho, 2002: Shift in the summer rainfall over the Yangtze River valley in the late 1970s. Geophys. Res. Lett., 29, 1436. DOI:10.1029/2001gl014523 Graham, N. E., and T. P. Barnett, 1987: Sea surface temperature, surface wind divergence, and convection over tropical oceans. Science, 238, 657–659. DOI:10.1126/science.238.4827.657 Groisman, P. Y., R. W. Knight, V. N. Razuvaev, et al., 2006: State of the ground: Climatology and changes during the past 69 years over northern Eurasia for a rarely used measure of snow cover and frozen land. J. Climate, 19, 4933–4955. DOI:10.1175/JCLI3925.1 Honda, M., K. Yamazaki, Y. Tachibana, et al., 1996: Influence of Okhotsk sea-ice extent on atmospheric circulation. Geophys. Res. Lett., 23, 3595–3598. DOI:10.1029/96GL03474 Huang, R. H., 1992: The East Asia/Pacific pattern teleconnection of summer circulation and climate anomaly in East Asia. Acta Meteor. Sinica, 6, 25–37. Huang, R. H., and J. L. Chen, 2010: Characteristics of the summertime water vapor transports over the eastern part of China and those over the western part of China and their difference. Chinese J. Atmos. Sci., 34, 1035–1045. Huang, S. S., and M. M. Tang, 1987: On the structure of the summer monsoon regime of East Asia. Scientia Meteorologica Sinica, 8, 1–14. Jiang, T., Z. W. Kundzewicz, and B. D. Su, 2008: Changes in monthly precipitation and flood hazard in the Yangtze River basin, China. Int. J. Climatol., 28, 1471–1481. DOI:10.1002/joc.1635 Kalnay, E., M. Kanamitsu, R. Kistler, et al., 1996: The NCEP/NCAR 40-year reanalysis project. Bull. Amer. Meteor. Soc., 77, 437–471. DOI:10.1175/1520-0477(1996)077<0437:TNYRP>2.0.CO;2 Li, J. P., J. Feng, and Y. Li, 2012: A possible cause of decreasing summer rainfall in northeast Australia. Int. J. Climatol., 32, 995–1005. DOI:10.1002/joc.2328 Li, X. Z., W. Zhou, D. L. Chen, et al., 2014: Water vapor transport and moisture budget over eastern China: Remote forcing from the two types of El Niño. J. Climate, 27, 8778–8792. DOI:10.1175/JCLI-D-14-00049.1 Li, Y. F., and L. R. Leung, 2013: Potential impacts of the Arctic on interannual and interdecadal summer precipitation over China. J. Climate, 26, 899–917. DOI:10.1175/JCLI-D-12-00075.1 Liebmann, B., and C. A. Smith, 1996: Description of a complete (interpolated) outgoing longwave radiation dataset. Bull. Amer. Meteor. Soc., 77, 1275–1277. Luo, B. H., and Y. Yao, 2018: Recent rapid decline of the Arctic winter sea ice in Barents–Kara Seas owing to combined Ural Blocking and SST. J. Meteor. Res., 32, 191–202. DOI:10.1007/s13351-018-7104-z Matsumura, S., K. Yamazaki, and T. Tokioka, 2010: Summertime land-atmosphere interactions in response to anomalous springtime snow cover in northern Eurasia. J. Geophys. Res., 115, D20107. DOI:10.1029/2009JD012342 Matsumura, S., and K. Yamazaki, 2012: Eurasian subarctic summer climate in response to anomalous snow cover. J. Climate, 25, 1305–1317. DOI:10.1175/2011JCLI4116.1 Ninomiya, K., and H. Mizuno, 1985: Anomalous cold spell in summer over northeastern Japan caused by northeasterly wind from polar maritime airmass. Part 1: EOF analysis of temperature variation in relation to the large-scale situation causing the cold summer. J. Meteor. Soc. Japan, 63, 845–857. DOI:10.2151/jmsj1965.63.5_845 Nitta, T., 1987: Convective activities in the tropical western Pacific and their impact on the Northern Hemisphere summer circulation. J. Meteor. Soc. Japan, 65, 373–390. DOI:10.2151/jmsj1965.65.3_373 Notaro, M., W. C. Wang, and W. Gong, 2006: Model and observational analysis of the Northeast U.S. regional climate and its relationship to the PNA and NAO patterns during early winter. Mon. Wea. Rev., 134, 3479–3505. DOI:10.1175/MWR3234.1 Park, T. W., C. H. Ho, and S. Yang, 2011: Relationship between the Arctic oscillation and cold surges over East Asia. J. Climate, 24, 68–83. DOI:10.1175/2010JCLI3529.1 Qu, X., and G. Huang, 2012: Impacts of tropical Indian Ocean SST on the meridional displacement of East Asian jet in boreal summer. Int. J. Climatol., 32, 2073–2080. DOI:10.1002/joc.2378 Rayner, N. A., D. E. Parker, E. B. Horton, et al., 2003: Global analyses of sea surface temperature, sea ice, and night marine air temperature since the late nineteenth century. J. Geophys. Res., 108, 4407. DOI:10.1029/2002JD002670 Robinson, D. A., K. F. Dewey, and R. R. Heim Jr, 1993: Global snow cover monitoring: An update. Bull. Amer. Meteor. Soc., 74, 1689–1696. DOI:10.1175/1520-0477(1993)074<1689:GSCMAU>2.0.CO;2 Shi, N., and Q. G. Zhu, 1996: An abrupt change in the intensity of the East Asian summer monsoon index and its relationship with temperature and precipitation over East China. Int. J. Climatol., 16, 757–764. DOI:10.1002/(SICI)1097-0088(199607)16:7<757::AID-JOC50>3.0.CO;2-5 Smith, T. M., R. W. Reynolds, T. C. Peterson, et al., 2008: Improvements to NOAA’s historical merged land–ocean surface temperature analysis (1880–2006). J. Climate, 21, 2283–2296. DOI:10.1175/2007JCLI2100.1 Sun, C., J. P. Li, F. F. Jin, et al., 2013: Sea surface temperature inter-hemispheric dipole and its relation to tropical precipitation. Environ. Res. Lett., 8, 044006. DOI:10.1088/1748-9326/8/4/044006 Sung, M. K., W. T. Kwon, H. J. Baek, et al., 2006: A possible impact of the North Atlantic oscillation on the East Asian summer monsoon precipitation. Geophys. Res. Lett., 33, L21713. DOI:10.1029/2006GL027253 Takaya, K., and H. Nakamura, 2001: A formulation of a phase-independent wave-activity flux for stationary and migratory quasigeostrophic eddies on a zonally varying basic flow. J. Atmos. Sci., 58, 608–627. DOI:10.1175/1520-0469(2001)058<0608:AFOAPI>2.0.CO;2 Tao, S. Y., and L. X. Chen, 1987: A review of recent research on the East Asian summer monsoon in China. Monsoon Meteorology. C. P. Chang and T. N. Krishnamurti, Eds., Oxford University Press, New York, 353 pp. Trenberth, K. E., and J. M. Caron, 2000: The Southern Oscillation revisited: Sea level pressures, surface temperatures, and precipitation. J. Climate, 13, 4358–4365. DOI:10.1175/1520-0442(2000)013<4358:TSORSL>2.0.CO;2 Wallace, J. M., and D. S. Gutzler, 1981: Teleconnections in the geopotential height field during the Northern Hemisphere winter. Mon. Wea. Rev., 109, 784–812. DOI:10.1175/1520-0493(1981)109<0784:TITGHF>2.0.CO;2 Wang, J. B., Z. P. Wen, R. G. Wu, et al., 2016: The mechanism of growth of the low-frequency East Asia–Pacific teleconnection and the triggering role of tropical intraseasonal oscillation. Climate Dyn., 46, 3965–3977. DOI:10.1007/s00382-015-2815-7 Wang, S. S., J. P. Huang, Y. L. He, et al., 2014: Combined effects of the Pacific decadal oscillation and El Niño–Southern oscillation on global land dry–wet changes. Sci. Rep., 4, 6651. DOI:10.1038/srep06651 Wang, W. W., W. Zhou, X. Z. Li, et al., 2016: Synoptic-scale characteristics and atmospheric controls of summer heat waves in China. Climate Dyn., 46, 2923–2941. DOI:10.1007/s00382-015-2741-8 Wu, B. Y., R. H. Zhang, and B. Wang, 2009: On the association between spring Arctic sea ice concentration and Chinese summer rainfall: A further study. Adv. Atmos. Sci., 26, 666–678. DOI:10.1029/2009GL037299 Wu, B. Y., R. H. Zhang, R. D’Arrigo, et al., 2013: On the relationship between winter sea ice and summer atmospheric circulation over Eurasia. J. Climate, 26, 5523–5536. DOI:10.1175/JCLI-D-12-00524.1 Wu, R. G., and B. P. Kirtman, 2007: Observed relationship of spring and summer East Asian rainfall with winter and spring Eurasian snow. J. Climate, 20, 1285–1304. DOI:10.1175/JCLI4068.1 Wu, Z. W., J. P. Li, Z. H. Jiang, et al., 2012: Possible effects of the North Atlantic oscillation on the strengthening relationship between the East Asian summer monsoon and ENSO. Int. J. Climatol., 32, 794–800. DOI:10.1002/joc.2309 Xie, S. P., and S. G. H. Philander, 1994: A coupled ocean–atmosphere model of relevance to the ITCZ in the eastern Pacific. Tellus A, 46, 340–350. DOI:10.1034/j.1600-0870.1994.t01-1-00001.x Xie, S. P., K. M. Hu, J. Hafner, et al., 2009: Indian Ocean capacitoreffect on Indo-western Pacific climate during the summer following El Niño. J. Climate, 22, 730–747. DOI:10.1175/2008JCLI2544.1 Zhang, L., and T. Li, 2017: Physical processes responsible for the interannual variability of sea ice concentration in Arctic in boreal autumn since 1979. J. Meteor. Res., 31, 468–475. DOI:10.1007/s13351-017-6105-7 Zhang, Q. Y., and S. Y. Tao, 1998: Influence of Asian mid-high latitude circulation on east Asian summer rainfall. Acta Meteor. Sinica, 56, 199–211. DOI:10.11676/qxxb1998.019. Zhang, R. N., R. H. Zhang, and Z. Y. Zuo, 2017: Impact of Eurasianspring snow decrement on East Asian summer precipitation. J. Climate, 30, 3421–3437. DOI:10.1175/JCLI-D-16-0214.1 Zhao, P., Y. N. Zhu, and R. H. Zhang, 2007: An Asian–Pacific teleconnection in summer tropospheric temperature and associated Asian climate variability. Climate Dyn., 29, 293–303. DOI:10.1007/s00382-007-0236-y Zheng, F., J. P. Li, R. T. Clark, et al., 2013: Simulation and projection of the Southern Hemisphere annular mode in CMIP5 models. J. Climate, 26, 9860–9879. DOI:10.1175/JCLI-D-13-00204.1 Zheng, J. Y., J. P. Li, and J. Feng, 2014: A dipole pattern in the Indian and Pacific oceans and its relationship with the East Asiansummer monsoon. Environ. Res. Lett., 9, 074006. DOI:10.1088/1748-9326/9/7/074006 | 2018-05-26 04:22:00 | {"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": 0.4550609886646271, "perplexity": 8277.133940812011}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-22/segments/1526794867309.73/warc/CC-MAIN-20180526033945-20180526053945-00389.warc.gz"} |
https://itprospt.com/num/19801064/solveay-e-y-3xy-y2-xy-dx-zxy-z-dy-0-y-1-2-2x-y-dx-1-dy-0 | 5
# Solveay e* +y + 3xy?(y2 + xy)dx Zxy _ z)dy= 0,y(1) =2(2x + y)dx + ( = 1)dy = 0, y(0) = 2...
## Question
###### Solveay e* +y + 3xy?(y2 + xy)dx Zxy _ z)dy= 0,y(1) =2(2x + y)dx + ( = 1)dy = 0, y(0) = 2
Solve ay e* +y + 3xy? (y2 + xy)dx Zxy _ z)dy= 0,y(1) =2 (2x + y)dx + ( = 1)dy = 0, y(0) = 2
##### Many used boats that are purchased require bank financing: As a result; it is important for banks to be capable of accurately estimating the price of boats One of the variables that is believed to affect the price is the number of hours the engine has been run_ To test to see if the price of boats decreases as the number of hours of engine use increase_ financial analyst collected data on price (in $1OOOs) and number of hours of engine use on 60 boats of the same make/model.Bivariate Fit of Pric Many used boats that are purchased require bank financing: As a result; it is important for banks to be capable of accurately estimating the price of boats One of the variables that is believed to affect the price is the number of hours the engine has been run_ To test to see if the price of boats d... 4 answers ##### The compound NaOH set in three different beakers and dissotved [ Water in the first container €Oz has heen added_ the second beaker C0 has been added and in the third one, NHa has heen added indicate which ofthose container cnemica reacticn would take place which it wion't and explain whyseries Arsenic compounds (questions and 9ConsiuctFrost diaqramthe same series construct Latimer diagram10 - Il &n X-ray with Mavcienque 168 pm _ dilfracled al an arigle 20 according Uhe Bragg equation The compound NaOH set in three different beakers and dissotved [ Water in the first container €Oz has heen added_ the second beaker C0 has been added and in the third one, NHa has heen added indicate which ofthose container cnemica reacticn would take place which it wion't and explain why... 5 answers ##### #c_ Ch Vhs"c Cn ~;Her'06.00 715025 Sapling L Caaubatr 19.7/12 Jump t Draw ! Learning Perocc] 18/7/2017 _ Table romalning F 02.06 produci c 0f tne lollowing " reaction_PtetOva Ont54n Chace #c_ Ch Vhs "c Cn ~; Her '06.00 715025 Sapling L Caaubatr 19.7/12 Jump t Draw ! Learning Perocc] 18/7/2017 _ Table romalning F 02.06 produci c 0f tne lollowing " reaction_ PtetOva Ont 54n Chace... 4 answers ##### 5 -Fin J1n9 tnhe tRwwf+ p this motrix by Con ver t9) 0+ triongula metri* Use +he J2"8/aA * character; sHcs) 5 - Fin J1n9 tnhe tRwwf+ p this motrix by Con ver t9) 0+ triongula metri* Use +he J2"8/aA * character; sHcs)... 5 answers ##### Prove thatx2 _ 1 lim = 0 x-+1 2x 5 Prove that x2 _ 1 lim = 0 x-+1 2x 5... 5 answers ##### A 60-kg person"s two feet cover an area of A=129 cm? Assume the person is at rest If ihe person stands on one foot; what is the pressure under that foot (Paj?. Consider g 10 m/s?,A. 37209310 B.558140C.930230 D None D Ei186047 A 60-kg person"s two feet cover an area of A=129 cm? Assume the person is at rest If ihe person stands on one foot; what is the pressure under that foot (Paj?. Consider g 10 m/s?, A. 3720931 0 B.558140 C.93023 0 D None D Ei186047... 5 answers ##### Phosphoric acid (H3PO4) has three pKas of 2.15,7.20.and 12.35.Place the forms of phosphoric acid in order from pH 1to 14PH =HzPO4HzPOaHPOAPO4 30.25 points For the substance in the previous question Wnantorm WIll pre dominatewnenoh 12.5? Clearly and fully explain your deductionParagraph Phosphoric acid (H3PO4) has three pKas of 2.15,7.20.and 12.35.Place the forms of phosphoric acid in order from pH 1to 14 PH = HzPO4 HzPOa HPOA PO4 3 0.25 points For the substance in the previous question Wnantorm WIll pre dominatewnenoh 12.5? Clearly and fully explain your deduction Paragraph... 5 answers ##### Representatives from different majors will be part are listed below:student consortium between colleges The numbers of students each majorMajor English History Psychology _231 502 355Use Webster's method t0 apportion the 30 seats the consortium: Enter values complete the chart with the modified quolasMajor English History PsychologyModified Quota Representatives from different majors will be part are listed below: student consortium between colleges The numbers of students each major Major English History Psychology _ 231 502 355 Use Webster's method t0 apportion the 30 seats the consortium: Enter values complete the chart with the modi... 5 answers ##### 1 B73 1 38 1~;6 T D1 1;nnole 1 B 73 1 38 1 ~; 6 T D 1 1 ; nnole... 5 answers ##### GIVEN: The surface given by Q: -r-y +227 = 9 and the line L: c(t) (1-t2+t,1+2t)teR FIND: The set of points: Lna_ (Be careful: The form of the answer must be a set! & Remember your answer must be boxed) GIVEN: The surface given by Q: -r-y +227 = 9 and the line L: c(t) (1-t2+t,1+2t)teR FIND: The set of points: Lna_ (Be careful: The form of the answer must be a set! & Remember your answer must be boxed)... 5 answers ##### 6. What is the period of g(x) = sec TX ? 2 6. What is the period of g(x) = sec TX ? 2... 5 answers ##### Problem 1_ Let ABC be all isosceles triangle with LA= LC = 72" and AC =1_ Find the length of AB Do this independently of Problem 4 in the notes. Instead: bisect one of the base angles of ABC and look for isosceles triangles. Use your work in part (a to compute cos 729_ Problem 1_ Let ABC be all isosceles triangle with LA= LC = 72" and AC =1_ Find the length of AB Do this independently of Problem 4 in the notes. Instead: bisect one of the base angles of ABC and look for isosceles triangles. Use your work in part (a to compute cos 729_... 1 answers ##### Evaluate using Integration by Parts. $$\int x^{2} \cos 3 x d x$$ Evaluate using Integration by Parts. $$\int x^{2} \cos 3 x d x$$... 5 answers ##### QUEsTIon 4The estimated average daily requirement of folic acid for pregrant females Is 520 micrograms Which accurately expres 520 Mg 520 mg 520 pg 520 m6QUE STION 5The proper eler Iron dol symbol for aluminumOUEs Tionha nUmbet 0035080 axpiesyeefoneci 5880 0035889Bounc Dou588 15880 "1o-} QUEsTIon 4 The estimated average daily requirement of folic acid for pregrant females Is 520 micrograms Which accurately expres 520 Mg 520 mg 520 pg 520 m6 QUE STION 5 The proper eler Iron dol symbol for aluminum OUEs Tion ha nUmbet 0035080 axpiesyeefoneci 5880 0035889 Bounc Dou 588 15880 "1o-... 5 answers ##### InkiepQuestion 2 0f 8Cz + 2Li2LiCloxidation half-reaction:reduction half-reaction: Inkiep Question 2 0f 8 Cz + 2Li 2LiCl oxidation half-reaction: reduction half-reaction:... 5 answers ##### Mass of 25 mL graduated cylinder31718 51248 45.30Mass of 25 mL graduated cylinder and NaCI solution Mass of the NaCl solutionDensity of the NaCl solution in 25 mL cylinderMass percentage of NaCl in solution; Cy, in 25 mL cylinderCalculations:Compare thc density of the salt solution mude E in the [0 mL graduated cylinder with that in lka 25mL gnequaled cylinder. How did thc density change? Was this what You expccted and why? Mass of 25 mL graduated cylinder 31718 51248 45.30 Mass of 25 mL graduated cylinder and NaCI solution Mass of the NaCl solution Density of the NaCl solution in 25 mL cylinder Mass percentage of NaCl in solution; Cy, in 25 mL cylinder Calculations: Compare thc density of the salt solution mude E in t... 5 answers ##### Determine whether each statement makes sense or does not make sense, and explain your reasoning.Every linear system has infinitely many ordered-pair solutions. Determine whether each statement makes sense or does not make sense, and explain your reasoning. Every linear system has infinitely many ordered-pair solutions.... 1 answers ##### A regular pentagon is inscribed in a circle of radius 10$\mathrm{cm} .$a. Find the measure of$\angle C .$b. Find the length of the diagonal$P S .$(Hint. First find$R S .$) A regular pentagon is inscribed in a circle of radius 10$\mathrm{cm} .$a. Find the measure of$\angle C .$b. Find the length of the diagonal$P S .$(Hint. First find$R S .$)... 5 answers ##### Lamont works at an electronics store where he receives acommission on his sales. He must choose between two plans for hispay structure for the next year.For Plan A Lamont's monthly salary is$1,810 and he gets a 20%commision on his monthly sales. For Plan B his monthly salary is$2,155 and he gets a 7% commision on his monthly sales.Use a system of equations to model Lamont's situation and choosethe best response below.If Lamont's sales are exactly$2,654.8, he earns the same amoun
Lamont works at an electronics store where he receives a commission on his sales. He must choose between two plans for his pay structure for the next year. For Plan A Lamont's monthly salary is $1,810 and he gets a 20% commision on his monthly sales. For Plan B his monthly salary is$2,155 and ... | 2022-10-07 13:01:26 | {"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": 0.4449951946735382, "perplexity": 8081.006227973015}, "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-20221007112411-20221007142411-00457.warc.gz"} |
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JEE Main 2021 (Online) 20th July Morning Shift
MCQ (Single Correct Answer)
+4
-1
A butterfly is flying with a velocity $$4\sqrt 2$$ m/s in North-East direction. Wind is slowly blowing at 1 m/s from North to South. The resultant displacement of the butterfly in 3 seconds is :
A
$$12\sqrt 2$$ m
B
20 m
C
3 m
D
15 m
2
JEE Main 2021 (Online) 18th March Evening Shift
MCQ (Single Correct Answer)
+4
-1
The velocity $$-$$ displacement graph of a particle is shown in the figure.
The acceleration $$-$$ displacement graph of the same particle is represented by :
A
B
C
D
3
JEE Main 2021 (Online) 18th March Morning Shift
MCQ (Single Correct Answer)
+4
-1
The position, velocity and acceleration of a particle moving with a constant acceleration can be represented by :
A
B
C
D
4
JEE Main 2021 (Online) 17th March Evening Shift
MCQ (Single Correct Answer)
+4
-1
The velocity of a particle is v = v0 + gt + Ft2. Its position is x = 0 at t = 0; then its displacement after time (t = 1) is :
A
v0 + g + f
B
v0 + $${g \over 2}$$ + $${F \over 3}$$
C
v0 + 2g + 3F
D
v0 + $${g \over 2}$$ + F
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# nested while - how to go to the beginning of the first while?
I have a nested while. How do I go from the inner while to the beginning of the outer while? Can this be done without using goto? while_1() { some codes here while_2() { if true go to the beginning of while_1 } some codes here } Nov 15 '05 #1
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invni wrote: I have a nested while. How do I go from the inner while to the beginning of the outer while? I would hate to tell you how. Can this be done without using goto? What do you mean? Are you kidding? I have never come across a dialect of C that gives you while_# sort of keywords.Better post some real, compile-able code, and we can bargain.Most often code that craves for a goto, needs a serious surgery. [the garbage van just took your code away...] You might find the following article handy: http://groups.google.co.in/group/com....c/browse_frm/ thread/1f70380413394ff7/09a791583ae54898?q=goto&rnum=1&hl=en#09a791583ae54 898 Suman. Nov 15 '05 #2
On Sat, 25 Jun 2005 05:12:16 -0400, invni wrote: I have a nested while. How do I go from the inner while to thebeginning of the outer while? Can this be done without using goto?while_1(){ some codes here while_2() { if true go to the beginning of while_1 } some codes here} Assuming the above is pseudocode, I suspect that you could use a combination of break and continue. Here is a trivial sample that makes decisions based on the value of a char variable named ch and a counter. int main(void) { char ch = 'Y'; int counter = 0; while(ch == 'Y') { puts("Beginning of outer loop"); while(true) { puts("Beginning of inner loop"); if(ch == 'N') break; else ch = 'N'; puts("End of inner loop"); } if(ch == 'N') { if(counter++ == 0) ch = 'Y'; continue; } puts("End of outer loop"); } printf("Good-bye\n"); } Output: Beginning of outer loop Beginning of inner loop End of inner loop Beginning of inner loop Beginning of outer loop Beginning of inner loop End of inner loop Beginning of inner loop Good-bye Note that it skips from beginning of the inner loop to the beginning of the outer loop. HTH. -- Best wishes, Bob Nov 15 '05 #3
Robert W Hand wrote: On Sat, 25 Jun 2005 05:12:16 -0400, invni wrote:I have a nested while. How do I go from the inner while to thebeginning of the outer while? Can this be done without using goto?while_1(){ some codes here while_2() { if true go to the beginning of while_1 } some codes here} Assuming the above is pseudocode, I suspect that you could use a combination of break and continue. Here is a trivial sample that makes decisions based on the value of a char variable named ch and a counter. int main(void) { char ch = 'Y'; int counter = 0; while(ch == 'Y') { puts("Beginning of outer loop"); while(true) { puts("Beginning of inner loop"); if(ch == 'N') break; else ch = 'N'; puts("End of inner loop"); } if(ch == 'N') { if(counter++ == 0) ch = 'Y'; continue; } puts("End of outer loop"); } printf("Good-bye\n"); } Output: Beginning of outer loop Beginning of inner loop End of inner loop Beginning of inner loop Beginning of outer loop Beginning of inner loop End of inner loop Beginning of inner loop Good-bye Note that it skips from beginning of the inner loop to the beginning of the outer loop. HTH. I got the idea. Thanks. Nov 15 '05 #4
invni wrote: I have a nested while. How do I go from the inner while to the beginning of the outer while? Can this be done without using goto? while_1() { some codes here while_2() { if true go to the beginning of while_1 } some codes here } I think it is instructive that you need labels and a precise definition to describe what you want. The above is not at all precise. lb1: while (c1) { .... while (c2) { .... if (c3) goto lb1; .... } .... } can be replaced with: int flag; while (c1) { .... flag = 0; while (c2) { .... if (c3) { flag = 1; break; } .... } if (flag) continue; .... } which lacks the clarity and simplicity of the goto version. When appropriate, just use goto. -- "If you want to post a followup via groups.google.com, don't use the broken "Reply" link at the bottom of the article. Click on "show options" at the top of the article, then click on the "Reply" at the bottom of the article headers." - Keith Thompson Nov 15 '05 #5
On Sat, 25 Jun 2005 13:36:52 +0000, CBFalconer wrote: invni wrote: I have a nested while. How do I go from the inner while to the beginning of the outer while? Can this be done without using goto? [snip] I think it is instructive that you need labels and a precise definition to describe what you want. The above is not at all precise. lb1: while (c1) { .... while (c2) { .... if (c3) goto lb1; .... } .... } } can be replaced with: int flag; while (c1) { .... flag = 0; while (c2) { .... if (c3) { flag = 1; break; } .... } if (flag) continue; .... } } which lacks the clarity and simplicity of the goto version. When appropriate, just use goto. I second that opinion. Goto is justly maligned as often creating horrible complexity, however its ability in certain circumstances to reduce complexity is often overlooked. This is one example of entirely appropriate and effective goto usage; another is providing a common exit point to a function with nested loops: int some_func(void) { while (c1) { ... if (fatal1) goto common_exit; ... while (c2) { ... if (fatal2) goto common_exit; ... } } return 0; common_exit: do_cleanup(); return 1; } Nov 15 '05 #6
"CBFalconer" wrote which lacks the clarity and simplicity of the goto version. When appropriate, just use goto. You're right, except that some people have got the idea into their heads that goto is the root of all evil. (It can make over-long functions messier still.) Often its easier just to write a slightly worse function without goto than enage in religious wars with someone who might be intent on stirring up trouble. Nov 15 '05 #7
On Sat, 25 Jun 2005 13:36:52 +0000, CBFalconer wrote: invni wrote: I have a nested while. How do I go from the inner while to the beginning of the outer while? Can this be done without using goto? while_1() { some codes here while_2() { if true go to the beginning of while_1 } some codes here } I think it is instructive that you need labels and a precise definition to describe what you want. The above is not at all precise. lb1: while (c1) { .... while (c2) { .... if (c3) goto lb1; .... } .... } A good rule for using goto is use them when they are the best tool for the job. It turns out that that is fairly rare but it is sometimes the case. A good secondary rule is that if you use goto don't jump backwards, and in effect create loops without explicit loop constructs. The old "spaghetti code" syndrome uis much more difficult to create if you don't jump backwards. I've only encountered one useful exception to this rule, and that is implementing finite state machines. That works because FSMs provides their own type of code structure (and hence a framework for readability) but one that is different to the normal structured programming block model. The code above is better written as while (c1) { .... while (c2) { .... if (c3) goto lb1; .... } .... lbl1: ; } can be replaced with: int flag; while (c1) { .... flag = 0; while (c2) { .... if (c3) { flag = 1; break; } .... } if (flag) continue; .... } which lacks the clarity and simplicity of the goto version. When appropriate, just use goto. It can be written without gotos or flag variables: while (c1) { .... for (;;) { if (!c2) { .... break; } .... if (c3) break; .... } } although I don't claim this is better. Lawrence Nov 15 '05 #8
invni wrote: I have a nested while. How do I go from the inner while to the beginning of the outer while? Can this be done without using goto? while_1() { some codes here while_2() { if true go to the beginning of while_1 } some codes here } You can probably restructure your code to make it simpler. Give a more detailed example of what you're trying to do. Nov 15 '05 #9
In article , akarl wrote:invni wrote: I have a nested while. How do I go from the inner while to the beginning of the outer while? Can this be done without using goto? while_1() { some codes here while_2() { if true go to the beginning of while_1 } some codes here }You can probably restructure your code to make it simpler. Give a moredetailed example of what you're trying to do. There was a discussion here (started by yours truly) of the merits of "break n" (which exists in Bourne and Bourne-like shells). The concensus was that it was a bad idea and that a (well timed and well placed) goto should be used instead. The problem, however, is if your organization forbids goto. Or, if you are not programming in C, but in some similar language, such as AWK or Pascal (*) or whatever that doesn't have gotos at all. (*) I know some versions of Pascal do have goto, but my understanding is that the pure form (of a language that was specifically invented to argue the case that goto [like drugs, mmkay] is bad) does not. Nov 15 '05 #10
On Sat, 25 Jun 2005 05:12:16 -0400, invni wrote: I have a nested while. How do I go from the inner while to thebeginning of the outer while? Can this be done without using goto?while_1(){ some codes here while_2() { if true go to the beginning of while_1 } some codes here} The question I have is whether you want to do it in a sane way or in an insane way. If the latter, look up Duff's device. Richard Harter, cr*@tiac.net http://home.tiac.net/~cri, http://www.varinoma.com Save the Earth now!! It's the only planet with chocolate. Nov 15 '05 #11
"Kenny McCormack" wrote in message news:d9**********@yin.interaccess.com... Or, if you are not programming in C, but in some similar language, such as AWK or Pascal (*) or whatever that doesn't have gotos at all. (*) I know some versions of Pascal do have goto, but my understanding is that the pure form (of a language that was specifically invented to argue the case that goto [like drugs, mmkay] is bad) does not. Your understanding of Pascal is seriously flawed, standard Pascal does indeed have a goto statement. Similarly, I have read in some publication or other that C does not have a goto. One side effect of the information explosion is the concurrent disinformation explosion. Standard Pascal does not have, however, break, continue and the premature return of C which all serve to minimize the use of goto.. Nov 15 '05 #12
In article <3i************@individual.net>, osmium wrote:"Kenny McCormack" wrote in messagenews:d9**********@yin.interaccess.com... Or, if you are not programming in C, but in some similar language, such as AWK or Pascal (*) or whatever that doesn't have gotos at all. (*) I know some versions of Pascal do have goto, but my understanding is that the pure form (of a language that was specifically invented to argue the case that goto [like drugs, mmkay] is bad) does not.Your understanding of Pascal is seriously flawed, standard Pascal doesindeed have a goto statement. Your ability to comprehend conversational English is "seriously flawed", since otherwise you'd realize how dumb your post sounds. Nov 15 '05 #13
"Kenny McCormack" wrote in message news:d9**********@yin.interaccess.com... In article <3i************@individual.net>, osmium wrote:"Kenny McCormack" wrote in messagenews:d9**********@yin.interaccess.com... Or, if you are not programming in C, but in some similar language, such as AWK or Pascal (*) or whatever that doesn't have gotos at all. (*) I know some versions of Pascal do have goto, but my understanding is that the pure form (of a language that was specifically invented to argue the case that goto [like drugs, mmkay] is bad) does not.Your understanding of Pascal is seriously flawed, standard Pascal doesindeed have a goto statement. Your ability to comprehend conversational English is "seriously flawed", since otherwise you'd realize how dumb your post sounds. "Some versions" and "does not" sounded to me like you were discussing a real language. I see now that you were addressing the subject of a language that someone could conceivably propose, or even write, some time. Would you agree that this sentence, which doesn't mention Pascal, was a better representation of what you meant? "My understanding is that the pure form of a language specifically invented to argue that goto was bad would not have a goto?" Nov 15 '05 #14
In article <3i************@individual.net>, osmium wrote: ...."Some versions" and "does not" sounded to me like you were discussinga real language. I see now that you were addressing the subject ofa language that someone could conceivably propose, or even write, sometime. Would you agree that this sentence, which doesn't mention Pascal,was a better representation of what you meant?"My understanding is that the pure form of a language specificallyinvented to argue that goto was bad would not have a goto?" I'm OK with this. Nov 15 '05 #15
> From: CBFalconer lb1: while (c1) { .... while (c2) { .... if (c3) goto lb1; .... } .... } From: Lawrence Kirby while (c1) { .... while (c2) { .... if (c3) goto lb1; .... } .... lbl1: ; } Note that both samples of code jump from inside a block to somewhere outside that block. (One jumps out from two blocks, while the other jumps out from the inner block while remaining in the outer block.) Neither is *really* jumping backwards. The one that jumps out two levels is really re-starting the outer block. A restart is not *exactly* the same thing as a backwards jump, although I'm probably shaving closely to note any difference here. There's an advantage to the second method, of never jumping even seemingly backwards. Consider the following *wrong* code: while (c1) { lbl1: ; .... while (c2) { .... if (c3) goto lb1; .... } .... } Note that the outer loop is re-started except that the condition for it is not re-tested here. This is not the same semantics as the first examples where it's always re-tested after the jump, and horrible bugs can happen this bad way. Note: The one thing you should *never* do, not *ever* *ever*, is jump from outside a loop to inside a loop, or even into a block. (If anyone can think of a valid example of such, please post now!) Note that at the machine level, jumping from inside one block to inside an unrelated block is done all the time, namely when returning from a function that was called. And if you're working on a really primitive CPU that doesn't even have a return instruction, so you have to emulate it by explicitly popping the return address off a stack and using it to perform an indirect-address jump, then the jump into the calling block is quite explicit. So my statement above does not apply to machine-language returns, whether via return instruction or via emulation of such, OK? I was thinking a little earlier about how my Distributed Java instructor said we need to learn how to drive a stick shift to understand what's really happening before we drive an automatic, so we needed to manually write SQL via JDBC/ODBC, and manually call servlets, and manually do RMI, before we could finally drive the J2EE EJB automatic! If this philosophy were carried to the extreme, we should all be required to write machine-language code, then assembly-language code for a machine that doesn't even have a return instruction so we understand what's really happening, and then write assembly language code for a regular machine with a fine return instruction, and finally we can use C or other ordinary programming language. I was thinking how much fun it'd be to write such a machine/assembly-language emulator for beginning students to play with. Having it totally interactive with nice GUI instead of physical toggle switches on front panel would make it more edudational and less of a pain to use. Any comments from the rest of you-all? Knuth had something like this in his books, except there wasn't a Web-accessible emulator to play with, so you can think of this as a modern version of Knuth's "Mix machine". Hmm, actually there should be a wider range of options, all the way down below machine language to state machines that emulate a computer, to logic gates that build a state machine, to transistor circuits that build a logic gate. The student could select which level of abstration to work on at any given time. At the high end of the scale, above function-calling would be lexical closures which are used to build OOP. Hmm, should I copy this to comp.programming? Nov 15 '05 #16
Robert Maas, see http://tinyurl.com/uh3t wrote: Note: The one thing you should *never* do, not *ever* *ever*, is jump from outside a loop to inside a loop, or even into a block. (If anyone can think of a valid example of such, please post now!) Partial loop unrolling or jumping into a do....while loop might give you justifyable examples. No, I do not advocate it. BTW: See Duff's device for some (not-goto) inspiration on that matter. Cheers Mihcael -- E-Mail: Mine is an /at/ gmx /dot/ de address. Nov 15 '05 #17
"Robert Maas, see http://tinyurl.com/uh3t" wrote: .... snip ... Note: The one thing you should *never* do, not *ever* *ever*, is jump from outside a loop to inside a loop, or even into a block. (If anyone can think of a valid example of such, please post now!) Knuth did, over 30 years ago. Try: (about 3 MB, a scan of his paper on care and feeding of goto) Computing Surveys, Vol 6, No. 4, December 1974 -- "A man who is right every time is not likely to do very much." -- Francis Crick, co-discover of DNA "There is nothing more amazing than stupidity in action." -- Thomas Matthews Nov 15 '05 #18
On Tue, 28 Jun 2005 15:07:32 -0700, Robert Maas, see http://tinyurl.com/uh3t wrote: From: CBFalconer lb1: while (c1) { .... while (c2) { .... if (c3) goto lb1; .... } .... } From: Lawrence Kirby while (c1) { .... while (c2) { .... if (c3) goto lb1; .... } .... lbl1: ; } Note that both samples of code jump from inside a block to somewhere outside that block. (One jumps out from two blocks, while the other jumps out from the inner block while remaining in the outer block.) Neither is *really* jumping backwards. One is precisely jumping backwards because it is jumping to a label earlier on in the source code. The one that jumps out two levels is really re-starting the outer block. In effect by creating an extra loop without an explicit looping construct. This is almost always a bad idea. A restart is not *exactly* the same thing as a backwards jump, although I'm probably shaving closely to note any difference here. It doesn't have to do anything different, the forward jump I suggested has the same overall effect. The important thing is how it is presented. Structured programming constructs don't do anything you can't do with conditional gotos, it is simply a better means of presentation and conceptualisation. There's an advantage to the second method, of never jumping even seemingly backwards. Consider the following *wrong* code: while (c1) { lbl1: ; .... while (c2) { .... if (c3) goto lb1; .... } .... } Note that the outer loop is re-started except that the condition for it is not re-tested here. This is not the same semantics as the first examples where it's always re-tested after the jump, and horrible bugs can happen this bad way. Yes, that's another example of why backwards jumps are a bad idea. Note: The one thing you should *never* do, not *ever* *ever*, is jump from outside a loop to inside a loop, or even into a block. (If anyone can think of a valid example of such, please post now!) Yes, that's evil too. Note that at the machine level, jumping from inside one block to inside an unrelated block is done all the time, namely when returning from a function that was called. And if you're working on a really primitive CPU that doesn't even have a return instruction, so you have to emulate it by explicitly popping the return address off a stack and using it to perform an indirect-address jump, then the jump into the calling block is quite explicit. So my statement above does not apply to machine-language returns, whether via return instruction or via emulation of such, OK? It is a matter of the readability of the code. For compiler generated code that doesn't have to be readable it isn't important. For hand written assembly which may have to be read and maintained it is certainly an issue. Lawrence Nov 15 '05 #19
CBFalconer wrote: "Robert Maas, see http://tinyurl.com/uh3t" wrote: ... snip ...Note: The one thing you should *never* do, not *ever* *ever*, isjump from outside a loop to inside a loop, or even into a block.(If anyone can think of a valid example of such, please post now!) Knuth did, over 30 years ago. Try: (about 3 MB, a scan of his paper on care and feeding of goto) Computing Surveys, Vol 6, No. 4, December 1974 I get a 404 on the above URL; do you perchance know an alternative location where I can get the paper? Cheers Michael -- E-Mail: Mine is an /at/ gmx /dot/ de address. Nov 15 '05 #20
Michael Mair wrote: CBFalconer wrote: "Robert Maas, see http://tinyurl.com/uh3t" wrote: ... snip ... Note: The one thing you should *never* do, not *ever* *ever*, is jump from outside a loop to inside a loop, or even into a block. (If anyone can think of a valid example of such, please post now!) Knuth did, over 30 years ago. Try: (about 3 MB, a scan of his paper on care and feeding of goto) Computing Surveys, Vol 6, No. 4, December 1974 I get a 404 on the above URL; do you perchance know an alternative location where I can get the paper? No. Just that I recorded that that is where I got it. I imagine Knuth may have it on his pages somewhere. -- "A man who is right every time is not likely to do very much." -- Francis Crick, co-discover of DNA "There is nothing more amazing than stupidity in action." -- Thomas Matthews Nov 15 '05 #21
CBFalconer wrote: Michael Mair wrote:CBFalconer wrote:"Robert Maas, see http://tinyurl.com/uh3t" wrote:... snip ... Note: The one thing you should *never* do, not *ever* *ever*, isjump from outside a loop to inside a loop, or even into a block.(If anyone can think of a valid example of such, please post now!)Knuth did, over 30 years ago. Try: (about 3 MB, a scan of his paper on care and feeding of goto)Computing Surveys, Vol 6, No. 4, December 1974I get a 404 on the above URL; do you perchance know an alternativelocation where I can get the paper? No. Just that I recorded that that is where I got it. I imagine Knuth may have it on his pages somewhere. Had some more time; Knuth's Bibliography says \p P67. Structured programming with {\bf go to} statements. {\sl Computing Surveys\/ \bf 6} (December 1974), 261--301. Reprinted with revisions in {\sl Current Trends in Programming Methodology}, Raymond T. Yeh, ed., {\bf 1} (Englewood Cliffs, N.J.: Prentice-Hall, 1977), 140--194; {\sl Classics in Software Engineering}, Edward Nash Yourdon, ed.\ (New York: Yourdon Press, 1979), 259--321. Reprinted with final'' revisions as Chapter~2 of {\sl Literate Programming\/} (see under Books). However, it seems to be nowhere for free, so I went back to the original site and searched some more; the paper has moved: Thanks for pointing me towards this article :-) Cheers Michael -- E-Mail: Mine is an /at/ gmx /dot/ de address. Nov 15 '05 #22
Michael Mair wrote: .... snip ... However, it seems to be nowhere for free, so I went back to the original site and searched some more; the paper has moved: Thanks for pointing me towards this article :-) Now maybe someone will find it again later. They are certainly moving it about. -- Chuck F (cb********@yahoo.com) (cb********@worldnet.att.net) Available for consulting/temporary embedded and systems. USE worldnet address! Nov 15 '05 #23
Robert Maas, see http://tinyurl.com/uh3t wrote: Note: The one thing you should *never* do, not *ever* *ever*, is jump from outside a loop to inside a loop, or even into a block. (If anyone can think of a valid example of such, please post now!) The (in)famous Duff's Device does this, in a strange way, but more interesting is: http://tinyurl.com/2452h/dbldum.htm Many will disparage the software at URL above, but talk is cheap and *no one* during the 15 years I've posted this code on Usenet has ever managed to produce a "better" version (or *any* version) of the routine without GOTO. (Most of the disparaging comments have come from people who obviously made no effort to understand the code or even read the comments. Let's see if comp.lang.c has improved over the years....) James Dow Allen Nov 15 '05 #24
On Thu, 30 Jun 2005 22:18:37 -0700, jdallen2000 wrote: Robert Maas, see http://tinyurl.com/uh3t wrote: Note: The one thing you should *never* do, not *ever* *ever*, is jump from outside a loop to inside a loop, or even into a block. (If anyone can think of a valid example of such, please post now!) The (in)famous Duff's Device does this, in a strange way, but more interesting is: http://tinyurl.com/2452h/dbldum.htm Many will disparage the software at URL above, but talk is cheap and *no one* during the 15 years I've posted this code on Usenet has ever managed to produce a "better" version (or *any* version) of the routine without GOTO. Well the code license does expressly prohibit the removal of gotos. :-) The code is sufficiently complex that it would take some effort to understand what it is doing and come up with possible alternatives. I'm not a "goto gestapo", the Nim code is a nice example of using goto. But I'm not going to accept that goto is the best way of doing something just because somebody else says so. :-) The comments say * Yes, the goto' won't operate properly unless all the local variables * are in precisely the right state to make their strange voyage from * one inner loop to the other. But that's exactly the concept which * the potential code-maintainer is intended to grasp. * Grasping is one thing, verifying is another. The fact is that this sort of thing (jumping into loops) does make code more difficult to read, it creates extra coupling between different parts of the code and a whole set of preconditions that need to be checked. I'm not convinced that goto is the best approach for this particular problem. Of course it needs code to "prove" this. Maybe I'll look at it further at some point. And if I can't find a better approach I'll agree with you. (Most of the disparaging comments have come from people who obviously made no effort to understand the code or even read the comments. Let's see if comp.lang.c has improved over the years....) I hope "improve" doesn't mean that people can't argue about it. :-) Lawrence Nov 15 '05 #25
James Dow Allen wrote: Lawrence Kirby wrote: jdallen2000 wrote: http://tinyurl.com/2452h/dbldum.htm The comments say * Yes, the goto' won't operate properly unless all the local * variables are in precisely the right state to make their * strange voyage from one inner loop to the other. But * that's exactly the concept which the potential code- * maintainer is intended to grasp. Grasping is one thing, verifying is another. The fact is that this sort of thing (jumping into loops) does make code more difficult to read, it creates extra coupling between different parts of the code and a whole set of preconditions that need to be checked. Yes, but there already is extremely tight coupling between the two loops, since one precisely undoes what the other does. There may seem to be a lot of local variables, but each one has a clear direct meaning in its context (number of cards played to the trick, which player led, etc.) I'll agree that the code will be harder to grasp than a typical routine of the same length, but that isn't the proper criterion. Compare it with the readability of the alternate routine, however long, that *solves the same problem.* Luckily I already had a copy of that article, since I won't use the 'unknown destination' tinyurls. I have no great complaint about the goto, in fact I often recommend it. What that package needs is the description of its metamorphosis from something understandable. Even the existance of a (possibly very slow) top-down crafted solution would help. A 30 or more year old Knuth article on the use of gotos that I recommended here (or there) a few days ago did precisely that. It followed the evolution from a clear to an efficient solution step by step. -- "If you want to post a followup via groups.google.com, don't use the broken "Reply" link at the bottom of the article. Click on "show options" at the top of the article, then click on the "Reply" at the bottom of the article headers." - Keith Thompson Nov 15 '05 #27
CBFalconer wrote: jdallen2000 wrote: http://tinyurl.com/2452h/dbldum.htm Luckily I already had a copy of that article, since I won't use the 'unknown destination' tinyurls. Doesn't the ".htm" at the end offer some reassurance? And anyway, is my reputation so poor here that you thought *I* was trying to break into your machine? But point taken, I'll give regular URL in future. (The host I use is nice in some ways, but has a ridiculously long URL.) [A] Knuth article ... followed the evolution from a clear to an efficient solution step by step. It's probably obvious that my routine is faster than a "properly written" alternative. My claim is that it may also be *clearer* once one escapes the mental block of disliking the peculiar structure. (The alternative will be clumsy.) Admittedly I should add detailed comments and rewrite some of the long expressions to make this point. But the "cute" programming examples I show at my website are a low priority for me, and I've not received much encouragement here.... :-) James Nov 15 '05 #28
James Dow Allen wrote: CBFalconer wrote: >> jdallen2000 wrote: >>> http://tinyurl.com/2452h/dbldum.htm Luckily I already had a copy of that article, since I won't use the 'unknown destination' tinyurls. Doesn't the ".htm" at the end offer some reassurance? And anyway, is my reputation so poor here that you thought *I* was trying to break into your machine? But point taken, I'll give regular URL in future. (The host I use is nice in some ways, but has a ridiculously long URL.) This is the first time I hear about it. What is the difference in terms of security between different URLs, be they short or long? Except that posting tinyurls is not a good idea in itself, the only problem I can see is that tinyurl.com might be hijacked, but so might any other homepage. -- Stan Tobias mailx echo si***@FamOuS.BedBuG.pAlS.INVALID | sed s/[[:upper:]]//g` Nov 15 '05 #29
S.Tobias wrote: This is the first time I hear about it. What is the difference in terms of security between different URLs, be they short or long? Except that posting tinyurls is not a good idea in itself, the only problem I can see is that tinyurl.com might be hijacked, but so might any other homepage. tinyurl.com is a "URL shortening" service - you give it a long, unwieldy URL and it will "convert" it to a shorter one beginning with "http://tinyurl.com/". When you request the tinyurl.com URL, your browser receives and follows an HTTP 302 redirect to the original, longer URL. Because there is no "are you sure you want to visit long_url?" confirmation, you're effectively clicking blind links. Other "URL shortening" services acknowledge this problem and insert a time delay or confirmation before redirecting the client (though I can't think of them off the top of my head). To James: the .htm should be of no reassurance - browsers will interpret content based on the HTTP Content-type header they receive, not based on file extension. Steve Nov 15 '05 #30
James Dow Allen wrote: CBFalconer wrote: .... snip ... Luckily I already had a copy of that article, since I won't use the 'unknown destination' tinyurls. Doesn't the ".htm" at the end offer some reassurance? And anyway, is my reputation so poor here that you thought *I* was trying to break into your machine? But point taken, I'll give regular URL in future. (The host I use is nice in some ways, but has a ridiculously long URL.) No, but when you arrive here you are just another pretty face. You can always supply both forms. Do check the Knuth quotation. I think it is now at: -- "If you want to post a followup via groups.google.com, don't use the broken "Reply" link at the bottom of the article. Click on "show options" at the top of the article, then click on the "Reply" at the bottom of the article headers." - Keith Thompson Nov 15 '05 #31
In article <11**********************@f14g2000cwb.googlegroups .com>, James Dow Allen wrote: Doesn't the ".htm" at the end offer some reassurance? No. The content-type is established in the HTTP header and need not match the file suffix. A malicious web site can send you activex or JPG files that are innocuously named "index.html". And anyway, is my reputation so poor here that you thought *I* was trying to break into your machine? I would not fear that so much, but I don't need inappropriate JPEG's appearing on my screen during work hours. It's usually not something to fear on comp.lang.c but if one's employer is strict you need to be careful with slashdot for example. -- 7842++ Nov 15 '05 #33
CBFalconer writes: invni wrote: I have a nested while. How do I go from the inner while to the beginning of the outer while? Can this be done without using goto? [unclear code snipped] I think it is instructive that you need labels and a precise definition to describe what you want. The above is not at all precise. lb1: while (c1) { .... while (c2) { .... if (c3) goto lb1; .... } .... } can be replaced with: int flag; while (c1) { .... flag = 0; while (c2) { .... if (c3) { flag = 1; break; } .... } if (flag) continue; .... } which lacks the clarity and simplicity of the goto version. When appropriate, just use goto. I agree with the commentary, but I think the last sentence does a disservice to the OP. The reason a 'goto' statement is usually a bad idea is not that it's a 'goto' but that most often it's symptomatic of bad program structure. When a situation arises where 'goto' seems like the best way to write something, usually that means the function in question is in need of significant revision. The first version above may be cleaner than the second, but either version should raise a red flag[*] during code review. Explore alternatives; neither of the examples above should be accepted unless evidence that a better alternative can't be found is fairly compelling. [*] No pun intended. :) Nov 15 '05 #34
Tim Rentsch wrote: Alright, I'll offer some encouragement. Consider an approach [with] three infinite loops, two break's, one return.... Except for the different method of doing control flow, the two programs use basically the same algorithm. The structured code ran about 25% faster than your code with goto's. Congratulations and thank you! Would you like to post your code at my site, with whatever commentary you choose? James Dow Allen (e-mail jamesdowallen at gmail.com) Nov 15 '05 #36
> From: Lawrence Kirby The important thing is how it is presented. I agree. Loops of the FOR or WHILE type, and mappings down sequences such as are common in Lisp, make it obvious at first glance what kind of loop is done, compared to a GOTO loop where you have to read the code carefully to diagonse the various kinds of loops. Structured programming constructs don't do anything you can't do with conditional gotos, Nobody ever said otherwise: It's the converse that is claimed: Structured programming constructs *prevent* you from using the full flexibility that GOTO would allow, thereby preventing you from doing things you shouldn't do anyway and shouldn't even want to do. On the other hand, tail recursion can be used to emulate GOTO, so syntactic structured programming doesn't actually prevent you from writing haywire algorithms. it is simply a better means of presentation and conceptualisation. Only if you keep your FOR and WHILE and MAP loops simple. A really hairy set of nested FOR and WHILE loops full of BREAK and CONTINUE in all sorts of odd places can end up being as inscrutable as a GOTO haywire. Nov 15 '05 #37
### This discussion thread is closed
Replies have been disabled for this discussion. | 2021-01-15 14:47:58 | {"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": 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": 0.42440083622932434, "perplexity": 2308.2692915660386}, "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-2021-04/segments/1610703495901.0/warc/CC-MAIN-20210115134101-20210115164101-00564.warc.gz"} |
https://codereview.stackexchange.com/questions/201640/a-thread-safe-task-queue-implementation-using-my-own-lock-guard-in-c/201744 | # A thread safe task queue implementation using my own lock_guard in C++
As an exercise in multithreaded programming, I am trying to implement a basic FIFO task queue. For this, I also implement simplified version of my lock_guard, from STL. Here is the code:
#include <mutex>
#include <iostream>
#include <chrono>
#include <optional>
namespace cho {
template <typename M>
struct lock_guard {
lock_guard() noexcept = default;
lock_guard(M& m) noexcept {
mtx = &m;
mtx->lock();
}
lock_guard(lock_guard const& rhs) = delete;
lock_guard(lock_guard && rhs) = delete;
lock_guard& operator=(lock_guard const& rhs) = delete;
lock_guard& operator=(lock_guard && rhs) = delete;
~lock_guard() {
mtx->unlock();
}
private:
M *mtx = nullptr;
};
template <typename T>
struct queue {
queue() noexcept = default;
~queue() {
queue_node *cur = head;
while(cur) {
queue_node *tmp = cur;
cur = cur->next;
delete tmp;
}
}
void push(T const& elem) {
queue_node *node = new queue_node{};
T *new_elem = new T(elem);
node->data = *new_elem;
node->next = nullptr;
one_elem_mtx.lock();
lock_guard<std::mutex> grd(push_mtx);
tail->next = node;
} else {
}
tail = node;
one_elem_mtx.unlock();
}
auto pop() {
return std::optional<T>{std::nullopt};
}
one_elem_mtx.lock();
lock_guard<std::mutex> grd(pop_mtx);
queue_node *c = head;
delete c;
one_elem_mtx.unlock();
return ret;
}
private:
struct queue_node {
T data = T{};
queue_node *next = nullptr;
};
queue_node *head = nullptr;
queue_node *tail = nullptr;
std::mutex push_mtx = std::mutex{};
std::mutex pop_mtx = std::mutex{};
std::mutex one_elem_mtx = std::mutex{};
};
}
void operator()(char const* msg) {
std::cout << "Log: " << msg << "#" << log_id++<< "\n" << std::flush;
}
private:
static std::size_t log_id;
};
std::size_t log_task::log_id = 0;
int main() {
while(true) {
using namespace std::chrono_literals;
}
});
while(true) {
using namespace std::chrono_literals;
}
});
while(true) {
using namespace std::chrono_literals;
std::optional<log_task> t = log_queue.pop();
if(t)
t.value()("log_message");
}
});
while(true) {
using namespace std::chrono_literals;
std::optional<log_task> t = log_queue.pop();
if(t)
t.value()("log_message");
}
});
return 0;
}
lock_guard and queue implementation lives in my cho namespace. For queue, I only provide basic push and pop, not a production-ready implementation by any means. Then, I implement a logger to be used as test functor in my task queue, called log_task. My queue push and pop operations are guarded by mutexes to avoid messing queue pointers. I want to be able to push tasks to the queue from multiple threads and consume those tasks from multiple threads as well. For this reason, I hold two different mutexes: push_mtx and pop_mtx. On the other hand, I realized that if there is only one element, they can still mess the values since head and tail points to the same node. I added another one_elem_mtx that needs to be locked just on this occasion.
1. What are code smells you can point? I am, I don't know why, not happy with using three mutexes and feel like there is something wrong there.
2. I played with delays in sleep_for() to try different values and my tests are working fine. But still, can you see any chance of deadlock or livelock?
3. I've seen many implementations on other questions that use only one mutex. So when one thread pushes, popping threads need to wait as well and vice versa. This seemed like a performance penalty to me but I am not sure. Does my approach indicate a bad design and should I abandon concurrent push/pop?
# lock_guard
• The destructor never checks if mtx == nullptr, which will cause problems if lock_guard got default-constructed.
• The lock_guard(M&) constructor cannot be noexcept, as lock on a standard library compatible mutex class is allowed to throw.
However, it could be made conditionally noexcept in case M::lock is noexcept itself
lock_guard(M& m) noexcept(noexcept(std::declval<M>().lock())) : mtx{&m} { mtx->lock(); }
• lock_guard(M&) should be made explicit in order to prevent unwanted conversions.
Any specific reasons why std::lock_guard or std::unique_lock couldn't be used?
# queue
• Memory leak: push(const T&) never deletes new_elem.
Actually, why allocate new_elem at all? The whole first 4 lines could be simplified to:
auto node = new queue_node{ elem, nullptr };
• push(const T&) and pop() may cause undefined behavior by calling one_elem_mtx.unlock() if it hasn't been locked before.
• pop() doesn't update tail in case the last element got removed and lets it dangle instead.
This will cause undefined behavior upon the next call to push(const T&).
This also means that one_elem_mtx will not be locked in calls to push(const T&) or pop(), since head != tail while the queue is empty again.
• Race condition: one_elem_mtx needs to be acquired after the lock on pop_mtx/push_mtx.
Reason: Assume we have a queue with two elements in it, and two threads A and B who both want to execute pop. Thread A executes until just after if(head == tail) (which of course right now evaluates to false) and then gets interrupted by the OS. Thread B runs pop() to completion in the meantime, leaving the queue at one object. Now, assuming the missing tail update mentioned above gets added, we have a potential data race on access to tail if another Thread C were to run push(const T&).
• Thread starvation: Once the race condition is fixed, if pop() gets called much more frequently than push(const T&), threads waiting on pop() might starve threads trying to push(const T&) from getting access to one_elem_mtx. Maybe make pop a blocking operation (using a std::condition_variable to notify if new elements got inserted)?
• For a production ready queue, you might want to think about adding a maximum capacity (so the queue doesn't grow too large if elements get added faster than they get removed).
• An overload push(T&&) might be nice.
# log_task
• operator()(const char*) is not thread safe: access to the static log_id isn't synchronized. Maybe add a lock, or make log_id a std::atomic<std::size_t>?
# main
• Waiting for some hundred milliseconds on such a simple task that might otherwise take nanoseconds (or microseconds at most) to execute? This might make the whole "test" obsolete, especially since the chance of finding race conditions is pretty much zero.
• t.value() checks again whether the std::optional contains a value (and throws an exception if not). Since at that point t has already been checked for emptiness, why not use *t to access the value directly?
# General stuff
• Any specific reason for not using std::unique_ptr?
• auto could be used more often.
• There's no way to stop the program other than forcefully terminating the process. This might surprise some people trying it.
# Q & A
1. See above.
2. I can't see any right now.
3. It is a performance penalty, yes, though that might not matter if push and pop don't get called too frequently (e.g. because task creation and task execution take some time). It also makes reasoning about the whole set of operations a lot easier: If all access to head and tail is secured behind one single mutex, there are some advantages:
• No race condition on either can occur.
• No deadlock based upon locking order can occur (unless you make recursive calls).
• No special care needed for edge cases (empty/full queue, or off by one element).
This makes it much easier to verify and trust that such a queue is working correctly.
Now, if this performance penalty is too much, often mutexes get skipped entirely in favor of lock-free queues. These require some more expert knowledge, though.
• The reason to avoid std::lock_guard and std::unique_lock` was to implement those myself as an exercise. Thanks for the points mentioned. – meguli Aug 15 '18 at 18:40 | 2021-02-28 10:58:01 | {"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": 0.21344532072544098, "perplexity": 7656.8160500190925}, "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-2021-10/segments/1614178360745.35/warc/CC-MAIN-20210228084740-20210228114740-00041.warc.gz"} |
https://mathoverflow.net/questions/335780/do-the-higher-levels-of-the-borel-hierarchy-correspond-to-absolute-topological-p | # Do the higher levels of the Borel hierarchy correspond to absolute topological properties?
It is well known that a subset $$Y$$ of a Polish space $$X$$ is completely metrisable iff it is a $$G_\delta$$ subset. This relates a relative topological property of the subspace $$Y \subset X$$ to an absolute topological property of $$Y$$.
I wonder are there more general versions of this result where $$G_\delta$$ is replaced by some other level of the Borel hierarchy and complete metrisability is replaced by some more complex topological property?
The first fact can be used to prove $$\mathbb Q^\omega \not \cong \mathbb P^\omega$$ where $$\mathbb P = \mathbb R - \mathbb Q$$ is the set of irrational numbers.
It is well-known $$\mathbb P$$ is completely metrisable and this extends to the countable power. Next we find a $$G_\delta$$ subset $$G \subset \mathbb R^\omega$$ with $$\mathbb P^\omega \subset G \subset \mathbb R^\omega - \mathbb Q^\omega$$.
$$G = \bigcap_{i \in \omega} \{ x \in \mathbb R^\omega: \text{ all } x_i \ne q_n\}$$
where $$\{q_1,q_2,\ldots\}$$ is an enumeration of $$\mathbb Q$$.
Since both spaces are dense in $$\mathbb R^\omega$$ and any two dense $$G_\delta$$ sets intersect it follows $$\mathbb Q^\omega$$ is not $$G_\delta$$ hence not completely metrisable.
Here is a harder problem: Suppose instead of $$\mathbb Q^\omega$$ and $$\mathbb P^\omega$$ we're interested in the spaces $$\mathbb Q^{ \oplus \, \omega}$$ and $$\mathbb P^{ \oplus \, \omega}$$ of choice functions $$f: \omega \to \mathbb R$$ with all but finitely many coordinates equal to zero and the rest in $$\mathbb Q$$ or $$\mathbb P$$ respectively.
In this case $$G_\delta$$ sets are no use. To see $$\mathbb P^{ \oplus \, \omega}$$ is not $$G_\delta$$ first observe it is dense in $$\mathbb R^{ \omega}$$. Then prove $$\{x \in \mathbb P^{ \omega}: \text{ all } x_i \ne 0\}$$ is $$G_\delta$$. Then observe the two sets are disjoint so the first cannot be $$G_\delta$$. In particular $$\mathbb P^{ \oplus \, \omega}$$ is not $$G_\delta$$ so cannot be completely metrisable.
Edit: Actually the space $$\mathbb Q^{\oplus \omega}$$ is countable since you can surject $$\mathbb Q^{1} \cup \mathbb Q^{2} \cup \ldots$$ onto it. So here is a harder question. What if we are interested in the spaces of choice functions with infinitely many coordinates equal to zero and the rest in $$\mathbb Q$$ or $$\mathbb P$$ respectively?
Getting back to my original question, suppose we could find some level in the Borel Hierarchy that contains $$\mathbb P^{ \oplus \, \omega}$$ but not $$\mathbb Q^{ \oplus \, \omega}$$ or vice-versa. Would this translate to some absolute topological property that distinguishes the spaces?
I'd also appreciate any other ideas to prove the spaces are (not) homeomorphic.
• Set of irrational numbers, thanks! I've edited the question now. – Daron Jul 9 '19 at 12:02
• A (very) partial answer: I think that a subset of $\mathbb R$ is locally compact if and only if it is $\Delta^0_2$. – Will Brian Jul 9 '19 at 13:41
Recall that for Polish spaces $$X,Y$$ and subsets $$A\subseteq X$$ and $$B\subseteq Y$$, $$A$$ is Wadge-reducible to $$B$$ if there is a continuous $$f:X\to Y$$ such that $$f^{-1}(B) = A$$. It can be proved (for instance, see Theorem 22.10, and Exercises 22.11 and 24.20 in Kechris' Classical Descriptive Set Theory) that $$B\in \boldsymbol\Sigma^0_\xi\setminus\boldsymbol\Pi^0_\xi$$ if and only if $$B$$ is complete, in that every $$A\in\boldsymbol\Sigma^0_\xi$$ is Wadge-reducible to it (for $$\xi\geq 1$$).
This immediately gives that such a $$B$$ is never homeomorphic to an $$A\in \boldsymbol\Pi^0_\zeta$$, with $$\zeta\leq\xi$$, in some other Polish space $$X$$. Otherwise, take $$f:A\to B$$ a homeomorphism; by Lavrentiev's theorem, $$f$$ extends to homeomorphism $$\bar f: G \to H$$ where $$G,H$$ are $$G_\delta$$ subsets such that $$A\subseteq G \subseteq X$$ and $$B\subseteq H \subseteq Y$$. But then $$B$$ reduces to $$A$$ through $$\bar f^{-1}$$, a contradiction.
• Essentially, $B=(\bar f^{-1})^{-1}[A]$ and thus $B$ must be in $\boldsymbol\Pi^0_\zeta\subseteq\boldsymbol\Pi^0_\xi$. But then $B$ can't be $\boldsymbol\Sigma^0_\xi$-complete. – Pedro Sánchez Terraf Jul 10 '19 at 23:35
• There is a further technical detail, that in this argument I'm using that $B$ is still complete in the smaller space $H$. I believe this might hold $\xi>2$ but not so certain for $\xi\leq 2$. – Pedro Sánchez Terraf Jul 11 '19 at 13:54 | 2021-05-14 11:18:50 | {"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": 66, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9358745813369751, "perplexity": 108.790837488472}, "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/1620243990449.41/warc/CC-MAIN-20210514091252-20210514121252-00330.warc.gz"} |
https://optimization-online.org/2008/10/2109/ | # Necessary conditions for local optimality in d.c. programming
Using \$\eps\$-subdifferential calculus for difference-of-convex (d.c.) programming, D\"ur proposed a condition sufficient for local optimality, and showed that this condition is not necessary in general. Here it is proved that whenever the convex part is strongly convex, this condition is also necessary. Strong convexity can always be ensured by changing the given d.c. decomposition slightly. This approach also allows for a formulation with perturbed \$\eps\$-subdifferentials which involves only the original d.c.\ decomposition, even without imposing strong convexity. We relate this result with another inclusion condition on perturbed \$\eps\$-subdifferentials, which even can serve as a quantitative version of a criterion both necessary and sufficient for local optimality.
## Citation
ISDS Technical Report TR 2008-14, University of Vienna (Oct. 2008). | 2022-11-29 17:25:06 | {"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": 0.9048314094543457, "perplexity": 946.0272904639568}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-49/segments/1669446710710.91/warc/CC-MAIN-20221129164449-20221129194449-00330.warc.gz"} |
https://mathoverflow.net/questions/385397/how-do-infty-categories-allow-us-to-do-descent-on-the-derived-level/385492 | # How do $\infty$-categories allow us to do descent on the derived level?
I have heard that one application of $$\infty$$-categories is that they allow us to formulate a meaningful theory of descent for derived categories (say of sheaves on a scheme). While I'm sure the details are somewhere in Lurie's exposition of stable $$\infty$$-categories, I was hoping that someone familiar with the process could explain in broad strokes why we can't do this in the classical setting, and what $$\infty$$-categories add to the picture that changes the situation.
Let $$X$$ be a topological space covered by open sets $$U$$ and $$V$$. Let $$\mathscr{F}$$ and $$\mathscr{G}$$ be complexes of sheaves defined on $$U$$ and $$V$$, respectively. Suppose you are given an isomorphism $$\alpha: \mathscr{F}|_{ U \cap V} \rightarrow \mathscr{G}|_{ U \cap V}$$ in the derived category of the intersection $$U \cap V$$. You would like to use these to glue $$\mathscr{F}$$ and $$\mathscr{G}$$ together to obtain a complex of sheaves on $$X$$. Let $$j: U \hookrightarrow X$$, $$j': V \hookrightarrow X$$, and $$j'': U \cap V \hookrightarrow X$$ denote the inclusion maps. Then the "glued" complex should be the fiber of the map $$j_{\ast} \mathscr{F} \oplus j'_{\ast} \mathscr{G} \rightarrow j''_{\ast} \mathscr{G}|_{U \cap V},$$ which is given on the first factor by $$\alpha$$. Working at the level of triangulated categories, this characterizes the glued complex up to non-canonical isomorphism. But for many purposes, producing a complex which is only well-defined up to non-canonical isomorphism is probably not good enough: you would like to define something that depends functorially on the input. The formalism of triangulated categories is poorly suited to this, because taking the fiber (or cocone) of a morphism is not a functorial operation. This is the sort of thing that is "corrected" by working with $$\infty$$-categories. | 2022-07-02 02:07:23 | {"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": 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": 21, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8936668038368225, "perplexity": 112.20502310612827}, "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/1656103983398.56/warc/CC-MAIN-20220702010252-20220702040252-00031.warc.gz"} |
https://www.beatthegmat.com/data-sufficiency-f7.html?sid=a25ebd2dabeba5459257e1e5616bf205 | ## Data Sufficiency
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Coordinate Geometry
by swerve »
Every point in the $$xy$$-plane satisfying the condition $$ax + by... 0 Last post by swerve Mon Jan 17, 2022 4:31 pm What distance did Jane travel? by M7MBA » What distance did Jane travel? (1) Bill traveled 40 miles in 40... 1 Last post by [email protected] Mon Jan 17, 2022 8:07 am If \(x$$ and $$y$$ are integers and $$2 < x < y,$$ does $$y = 16?$$
by Gmat_mission »
If $$x$$ and $$y$$ are integers and $$2 < x < y,$$ does $$y =... 0 Last post by Gmat_mission Sun Jan 16, 2022 3:46 pm Is \(x^2 + y^2 > 100?$$
by M7MBA »
Is $$x^2 + y^2 > 100?$$ $$(1)\, 2xy < 100$$ $$(2)\, (x + y)^2... 0 Last post by M7MBA Sun Jan 16, 2022 2:03 pm A certain painting job requires a mixture of yellow, green, and white paint. If 12 quarts of paint are needed for the jo by M7MBA » A certain painting job requires a mixture of yellow, green, and white... 0 Last post by M7MBA Sun Jan 16, 2022 2:00 pm Is \(\dfrac{5^{x+2}}{25}<1?$$
by M7MBA »
Is $$\dfrac{5^{x+2}}{25}<1?$$ (1) $$5^x<1$$ (2) $$x<0$$...
0
Last post by M7MBA
Sun Jan 16, 2022 12:14 pm
What is the surface area of rectangular solid $$y?$$
by M7MBA »
What is the surface area of rectangular solid $$y?$$ (1) The...
0
Last post by M7MBA
Sun Jan 16, 2022 12:02 pm
Is $$\dfrac{r}{s^2}$$ a terminating decimal?
by M7MBA »
Is $$\dfrac{r}{s^2}$$ a terminating decimal? (1) $$s=225$$ (2)...
0
Last post by M7MBA
Fri Jan 14, 2022 9:20 am
If $$an \ne 0$$ and $$n$$ is a positive integer, is $$n$$ odd?
by M7MBA »
If $$an \ne 0$$ and $$n$$ is a positive integer, is $$n$$ odd? (1)...
0
Last post by M7MBA
Fri Jan 14, 2022 9:17 am
$$P=a^4\cdot b^7,$$ where $$a^2=1$$ and $$b^2=9.$$ What is the value of $$P?$$
by M7MBA »
$$P=a^4\cdot b^7,$$ where $$a^2=1$$ and $$b^2=9.$$ What is the value...
0
Last post by M7MBA
Fri Jan 14, 2022 9:14 am
In a freshman high school class there are 125 total students. If 75 of the total students are enrolled in Geometry,
by BTGmoderatorDC »
In a freshman high school class there are 125 total students. If 75...
1
Last post by swerve
Fri Jan 14, 2022 6:12 am
Set $$X = \{a, b, c, d, 9\},$$ where $$a, b, c, d$$ and $$9$$ are five distinct positive integers. If $$9$$ is the only
by M7MBA »
Set $$X = \{a, b, c, d, 9\},$$ where $$a, b, c, d$$ and $$9$$ are...
0
Last post by M7MBA
Thu Jan 13, 2022 12:28 pm
If $$r$$ is represented by the decimal $$0.t5,$$ what is the digit $$t?$$
by M7MBA »
If $$r$$ is represented by the decimal $$0.t5,$$ what is the digit...
0
Last post by M7MBA
Thu Jan 13, 2022 12:24 pm
If $$a$$ and $$b$$ are nonzero integers, is $$a^b$$ an integer?
by M7MBA »
If $$a$$ and $$b$$ are nonzero integers, is $$a^b$$ an integer? (1)...
0
Last post by M7MBA
Thu Jan 13, 2022 12:20 pm
A certain computer company produces two different monitors, P and Q. In 2010, what was the net profit from the sale of
by BTGmoderatorLU »
Source: Official Guide A certain computer company produces two...
1
Last post by [email protected]
Thu Jan 13, 2022 9:47 am
If $$|a + 5| = |b + 5|$$ what is the value of $$a + b?$$
by VJesus12 »
If $$|a + 5| = |b + 5|$$ what is the value of $$a + b?$$ (1) $$b >... 0 Last post by VJesus12 Thu Jan 13, 2022 5:10 am If the average (arithmetic mean) of six different numbers is 25, how many of the numbers are greater than 25? by VJesus12 » If the average (arithmetic mean) of six different numbers is 25, how... 0 Last post by VJesus12 Thu Jan 13, 2022 5:06 am All boxes in a certain warehouse were arranged in stacks of 12 boxes each, with no boxes left over. After 60 additional by VJesus12 » All boxes in a certain warehouse were arranged in stacks of 12 boxes... 0 Last post by VJesus12 Thu Jan 13, 2022 5:03 am Both Betty and Wilma earn annual salaries of more than 50000. Is Wilma's annual salary greater than Betty's? by VJesus12 » Both Betty and Wilma earn annual salaries of more than 50000. Is... 0 Last post by VJesus12 Thu Jan 13, 2022 5:00 am What is the greatest common factor of positive integers x and y? by BTGmoderatorDC » What is the greatest common factor of positive integers x and y? (1)... 0 Last post by BTGmoderatorDC Wed Jan 12, 2022 10:14 pm If the quadrilateral \(PQRS$$ can be completely inscribed in a circle $$C$$ with $$PR$$ passing through the center of
by VJesus12 »
If the quadrilateral $$PQRS$$ can be completely inscribed in a circle...
0
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Wed Jan 12, 2022 10:53 am
In a two-month survey of shoppers, each shopper bought one of two brands of detergent, $$X$$ or $$Y,$$ in the first mont
by VJesus12 »
In a two-month survey of shoppers, each shopper bought one of two...
0
Last post by VJesus12
Wed Jan 12, 2022 10:49 am
If $$m$$ and $$n$$ are positive integers, is $$m + n$$ divisible by $$4?$$
by VJesus12 »
If $$m$$ and $$n$$ are positive integers, is $$m + n$$ divisible by...
0
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Wed Jan 12, 2022 10:46 am
If $$a$$ and $$b$$ are positive integers, what is the value of the product $$ab?$$
by VJesus12 »
If $$a$$ and $$b$$ are positive integers, what is the value of the...
0
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Wed Jan 12, 2022 10:42 am
Rosalie tosses a coin a number of times and counts the total number of times the coin lands tail-side up. How many times
by VJesus12 »
Rosalie tosses a coin a number of times and counts the total number...
0
Last post by VJesus12
Wed Jan 12, 2022 10:38 am
If $$n$$ is the least of three different integers greater than $$1,$$ what is the value of $$n?$$
by VJesus12 »
If $$n$$ is the least of three different integers greater than $$1,$$...
0
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Wed Jan 12, 2022 10:35 am
If $$a, b,$$ and $$c$$ are consecutive integers and $$0 < a < b < c,$$ is the product $$abc$$ a multiple of $$8?$$
by VJesus12 »
If $$a, b,$$ and $$c$$ are consecutive integers and $$0 < a < b... 0 Last post by VJesus12 Wed Jan 12, 2022 10:27 am A framed picture is shown above. The frame, shown shaded, is 6 inches wide and forms a border of uniform width around by VJesus12 » Image.png A framed picture is shown above. The frame, shown shaded,... 0 Last post by VJesus12 Wed Jan 12, 2022 10:23 am Is \(2x > 2y?$$
by VJesus12 »
Is $$2x > 2y?$$ (1) $$x > y$$ (2) $$3x > 3y$$ Answer: D...
0
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Wed Jan 12, 2022 10:22 am
In the quilting pattem
by VJesus12 »
153y8e8.png Answer: C Source: Official Guide
0
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Wed Jan 12, 2022 10:19 am
Did Insurance Company $$K$$ have more than $$\300$$ million in total net profits last year?
by VJesus12 »
Did Insurance Company $$K$$ have more than $$\300$$ million in total...
0
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Wed Jan 12, 2022 10:15 am
How many hours would it take Pump $$A$$ and Pump $$B$$ working together, each at its own constant rate, to empty a tank
by Vincen »
How many hours would it take Pump $$A$$ and Pump $$B$$ working...
0
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Wed Jan 12, 2022 10:09 am
In a certain factory, hours worked by each employee in excess of $$40$$ hours per week are overtime hours and are paid
by Vincen »
In a certain factory, hours worked by each employee in excess of...
0
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Wed Jan 12, 2022 10:05 am
When $$900$$ is divided by positive integer $$d,$$ the remainder is $$r.$$ For some integer $$N > 5000,$$ when $$N$$ is
by Vincen »
When $$900$$ is divided by positive integer $$d,$$ the remainder is...
0
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Wed Jan 12, 2022 10:00 am
If $$x$$ and $$y$$ are the lengths of the legs of a right triangle, what is the value of $$xy?$$
by Vincen »
If $$x$$ and $$y$$ are the lengths of the legs of a right triangle,...
0
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Wed Jan 12, 2022 9:53 am
In each quarter of $$1998,$$ Company $$M$$ earned more money than in the previous quarter. What was the range of Company
by Vincen »
In each quarter of $$1998,$$ Company $$M$$ earned more money than in...
0
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Wed Jan 12, 2022 9:45 am
Is the last digit of integer $$x^2 - y^2$$ a zero?
by Vincen »
Is the last digit of integer $$x^2 - y^2$$ a zero? (1) $$x - y$$ is...
0
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Wed Jan 12, 2022 9:39 am
Set $$P$$ consists of the first $$n$$ positive multiples of $$3$$ and set $$Q$$ consists of the first $$m$$ positive
by Vincen »
Set $$P$$ consists of the first $$n$$ positive multiples of $$3$$ and...
0
Last post by Vincen
Wed Jan 12, 2022 9:36 am
On a map, $$\dfrac12$$ inch represents $$100$$ miles. According to this map, how many miles is City $$X$$ from City
by Vincen »
On a map, $$\dfrac12$$ inch represents $$100$$ miles. According to...
0
Last post by Vincen
Wed Jan 12, 2022 9:31 am
What is the value of the integer $$N?$$
by Vincen »
What is the value of the integer $$N?$$ (1) $$101<N<103$$ (2)...
0
Last post by Vincen
Wed Jan 12, 2022 9:27 am
What is the value of $$x^2-y^2?$$
by Vincen »
What is the value of $$x^2-y^2?$$ (1) $$(x-y)^2=9$$ (2) $$x + y = 6$$...
0
Last post by Vincen
Wed Jan 12, 2022 9:23 am
What is the average (arithmetic mean) of a list of 6 consecutive two-digit integers?
by Vincen »
What is the average (arithmetic mean) of a list of 6 consecutive...
0
Last post by Vincen
Wed Jan 12, 2022 9:21 am
If $$v = (w)^2(y)(z),$$ how many positive factors does $$v$$ have?
by Vincen »
If $$v = (w)^2(y)(z),$$ how many positive factors does $$v$$ have?...
0
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Wed Jan 12, 2022 9:17 am
For any positive integer $$x,$$ the $$2$$-height of $$x$$ is defined to be the greatest nonnegative integer $$n$$ such
by Vincen »
For any positive integer $$x,$$ the $$2$$-height of $$x$$ is defined...
0
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Wed Jan 12, 2022 9:12 am
At the beginning of January 2003, Jill invested money in an account that collected interest, compounding monthly. Assume
by Vincen »
At the beginning of January 2003, Jill invested money in an account...
0
Last post by Vincen
Wed Jan 12, 2022 9:09 am
The symbol # represents one of the four arithmetic operations: addition, subtraction, multiplication, and division. Is
by BTGmoderatorDC »
The symbol # represents one of the four arithmetic operations:...
0
Last post by BTGmoderatorDC
Tue Jan 11, 2022 4:14 pm
Of the N candies in a bag, some are peppermint and the rest are spearmint.What is the value of N?
by BTGmoderatorDC »
Of the N candies in a bag, some are peppermint and the rest are...
0
Last post by BTGmoderatorDC
Mon Jan 10, 2022 5:48 pm | 2022-01-19 20:56:22 | {"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": 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": 0.33844810724258423, "perplexity": 1569.4113788979607}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-05/segments/1642320301488.71/warc/CC-MAIN-20220119185232-20220119215232-00546.warc.gz"} |
https://math.stackexchange.com/questions/3149576/if-both-a-b-and-b-a-are-positive-semidefinite-then-a-b | # If both $A-B$ and $B-A$ are positive semidefinite, then $A = B$
Let $$A, B$$ be two positive semidefinite matrices. Prove that if both $$A-B$$ and $$B-A$$ are positive semidefinite, then $$A = B$$.
I can show that their diagonal elements are the same but for others I have no idea. Any help will be appreciated.
One route is to observe that if $$\lambda,v$$ is an eigenvalue/eigenvector pair of $$M$$ with , then $$(-M)v=-\lambda v$$, meaning that $$-\lambda$$ is an eigenvalue of $$(-M)$$. Since $$M:=A-B$$ is positive semidefinite, conclude that any eigenvalue of $$M$$ satisfies $$0\leq \lambda\leq 0$$, so $$\lambda=0$$, so that $$M=0$$.
By assumption $$A-B=-(B-A)$$ is both positive semidefinite and negative semidefinite. This can only happen if $$A-B=0$$.
• its this equivalent to saying that $x^TAx=0 \forall x \implies A = 0$ which is wrong (math.stackexchange.com/questions/358281/…) ? – honzaik Mar 15 at 17:57
• I assumed we are working over $\mathbb C$ and so we are using $x^*Ax$. So the link is not relevant. – chhro Mar 15 at 18:01
• well i assumed you meant $0 \geq -x^T (B-A) x = x^T (A-B) x \geq 0$. I guess when I hear positive semidef etc I assume this definition with vectors. – honzaik Mar 15 at 18:10 | 2019-04-22 16:28:28 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 16, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9831095933914185, "perplexity": 118.07835090571615}, "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-2019-18/segments/1555578558125.45/warc/CC-MAIN-20190422155337-20190422181337-00129.warc.gz"} |
https://physics.anu.edu.au/contact/people/profile.php?ID=2114 | # MrKieranThomas
Position PhD Student Quantum Science & Technology He* BEC group Javascript Required John Carver 3 25
## arXiv pre-prints
J. A. Ross, P. Deuar, D. K. Shin, K. F. Thomas, B. M. Henson, S. S. Hodgman, A. G. Truscott
Survival of the quantum depletion of a condensate after release from a harmonic trap in theory and experiment
arXiv:2103.15283v2
Kieran F. Thomas, Matthew J. Davis, Karen. V. Kheruntsyan
Thermalization of a quantum Newton's cradle in a one-dimensional quasicondensate
arXiv:1811.01585v3
, , , , , ,
Direct Measurement of the Forbidden $2^{3\!}S_1 \rightarrow 3^{3\!}S_1$ Atomic Transition in Helium
Physical Review Letters 125 (2020) 7pp
arXiv:2002.04811v3
, , , , , ,
Frequency measurements of transitions from the $2^{3\!}P_2$ state to the $5^{1\!}D_2$, $5^{3\!}S_1$, and $5^{3\!}D$ states in ultracold helium
Physical Review A 102 (2020) -
arXiv:2006.06146v2
, , , , , ,
Approaching the adiabatic timescale with machine-learning
PNAS - Proceedings of the National Academy of Sciences of the United States of America 115 (2018) 13216-13221
arXiv:1809.03124v2
## Publications
### Journal article
, , , , ,
Trap frequency measurement with a pulsed atom laser
Optics Express 30, 8() 13252-13262
, , , , , ,
Frequency measurements of transitions from the 2 P2 3 state to the 5 D2 1, 5 S1 3, and 5 D 3 states in ultracold helium
Physical Review A 102, 4()
, , , , , ,
Direct Measurement of the Forbidden 2^3S_1 -> 3^3S_1 Atomic Transition in Helium
Physical Review Letters 125, 1() 1-7
, , , , , ,
Direct Measurement of the Forbidden 23S1 ? 33S1 Atomic Transition in Helium. (DUPLICATION)
American Physical Society 125, 1-3 July 2020()
, , , , , ,
Direct Measurement of the Forbidden 23S1 ? 33S1 Atomic Transition in Helium
Physical Review Letters 125, 1() 7
, , , , , ,
Approaching the adiabatic timescale with machine learning
PNAS - Proceedings of the National Academy of Sciences of the United States of America 115, 52() 13216-13221
(6 publications)
Updated: 7 July 2022/ Responsible Officer: Director, RSPhys/ Page Contact: Physics Webmaster | 2022-07-07 12:55:36 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5688037872314453, "perplexity": 6615.993048501406}, "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/1656104692018.96/warc/CC-MAIN-20220707124050-20220707154050-00077.warc.gz"} |
http://search.driver.research-infrastructures.eu/showResults.action?query=(author%3D%22Moreno%2C+J.%22)&page=4&size=10&refine=&history=true | author = "Moreno, J."
Found 353 documents, displaying page 4 of 36
## Cristalización de vidrios ricos en sílice preparados mediante sol-gel en el sistema alúmina-circona-sílice
Description : The crystallization of ZrSiO4 and its evolution with temperature from chemically homogeneous alumina-silica-zirconia powders prepared by sol-gel method from alcoxide mixtures was studied in the silica-rich region. A glass with the same composition was obtained by quenching in water from the melt. Th...
Repository : DOAJ-Articles
Language(s) : Spanish
## Una placa de arte mueble paleolítico en la provincia de Soria
Description : We report here the discovery of a splendid plaque of Palaeolithic figurine art in the province of Soria and, at the same time, we offer an advance of its study which will appear in the review Numantia, III, of Junta de Castilla y León.<br><br>Damos aquí noticia del hallazgo de una magníf...
Repository : DOAJ-Articles
Language(s) : Spanish
## Influencia de la distribución de las fibras en la conductividad térnnica de materiales compuestos con matrices de aleaciones de titanio y de magnesio
Description : Metal matrix composites are materials with an increasing use in industrial applications. Generally, the estimation of their thermal conductivity does not take into account the type of reinforcements distribution within the matrix. By applying the Network Simulation Method, a 2-D network model is des...
Repository : DOAJ-Articles
Language(s) : Spanish
## Análisis comparativo de las frecuencias de anomalías congénitas del Instituto Materno Infantil de Bogotá
Description : Comparar las frecuencias de anomalías congénitas en el Instituto Materno Infantil de Bogotá con las reportadas en Latinoamérica por la ECLAMC.
Repository : DOAJ-Articles
Language(s) : Spanish
## El dolmen de “El alto de la Tejera” (Carrascosa de la Sierra, Soria). El fenómeno megalítico en el Alto Duero
Description : This paper studies the dolmen of Alto de la Tejera (Carrascosa de la Sierra, Soria) which is the first to be found and excavated in the Upper Duero, in its environmental and archaeological setting, and with a view to modeling the local population and megalithic territoriality in this area. The relat...
Repository : DOAJ-Articles
Language(s) : Spanish
## On the Spontaneous CP Breaking in the Higgs Sector of the Minimal Supersymmetric Standard Model
Description : We revise a recently proposed mechanism for spontaneous CP breaking at finite temperature in the Higgs sector of the Minimal Supersymmetric Standard Model, based on the contribution of squarks, charginos and neutralinos to the one-loop effective potential. We have included plasma effects for all b...
Repository : arXiv
Language(s) : Undetermined
## Fermions on the Electroweak String
Description : We construct a simple class of exact solutions of the electroweak theory including the naked $Z$--string and fermion fields. It consists in the $Z$--string configuration ($\phi,Z_\theta$), the {\it time} and $z$ components of the neutral gauge bosons ($Z_{0,3},A_{0,3}$) and a fermion condensate (l...
Repository : arXiv
Language(s) : Undetermined
## Top-bottom doublet in the sphaleron background
Description : We consider the top-bottom doublet in the background of the sphaleron for the realistic case of large non-degeneracy of fermion masses, in particular $m_b=5$ GeV and $m_t=175$ GeV. We propose an axially symmetric $(r,\theta)$-dependent ansatz for fermion fields and investigate the effects of the n...
Repository : arXiv
Language(s) : Undetermined
## Sphalerons in the MSSM
Description : We construct the sphaleron solution, at zero and finite temperature, in the Minimal Supersymmetric Standard Model as a function of the supersymmetric parameters, including the leading one-loop corrections to the effective potential in the presence of the sphaleron. At zero temperature we have incl...
Repository : arXiv
Language(s) : Undetermined
## Squarks and sphalerons
Description : Electric charge and color breaking minima along third generation squark directions do appear in the Minimal Supersymmetric Standard Model for particular regions of the corresponding supersymmetric parameters $A_t$, $\mu$, $\tan\beta$, $m_Q^2$ and $m_U^2$. We have studied possible instabilities of ...
Repository : arXiv
Language(s) : Undetermined
Found 353 documents, displaying page 4 of 36 | 2013-05-25 23:31:36 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6366050243377686, "perplexity": 5685.939496123235}, "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/1368706470784/warc/CC-MAIN-20130516121430-00066-ip-10-60-113-184.ec2.internal.warc.gz"} |
https://physics.stackexchange.com/questions/412751/what-will-the-observer-observe | # What will the observer observe?
Consider the case where observer $A$ is at rest, and observer $B$ is moving with speed $\frac{c}{2}$ (where $c$ is the speed of light) propagating a wave with wave speed $c$. So my question is what will be the wave speed observed by observer $A$(which is at rest)?
I'm really curious to know the answer.
• The wave speed will again be c.For such relativistic speeds, the rule for transforming velocities is not the conceptual, Galilean one, i.e. adding the relative velocity to the velocity observed by one observer. You should include a relativistic factor predicted by Lorentz transformations. – Panos C. Jun 20 '18 at 16:11
• The speed of light is the same for every observer in an inertial frame of reference. – Andrei Geanta Jun 20 '18 at 16:15
• They will both see an EM wave to propagate with speed c. It is because you intuitively think that our speed are normal. The only speed is the speed of light, and everything is relative to that. In your case, A is traveling at 0c, B is traveling at c/2. They are obviously in this case seeing the EM wave propagate at speed c, if you see their speeds as being relative to c. This is because everything with no rest mass travels at speed c, and to slow down, you have to gain rest mass. Macro objects, like in our world, all have rest mass and all travel relative to c. – Árpád Szendrei Jun 20 '18 at 16:51
Observer A will see the wave propagating at speed $c$ (the same speed which Observer B sees). This is essentially one of the basic postulates of special relativity: the speed of light (in vacuum) has the same value as measured in all inertial frames.
Even if Observer B moved at speed $0.999c$ relative to Observer A, they would still both see the wave propagate at speed $c$ relative to themselves.
The relevant formula in special relativity should be the velocity-addition formula if you want to look at how you might compute this, but really the velocity-addition formula is constructed so as to reproduce the postulate mentioned above, so arguing from first principles perhaps makes more sense here.
Note that this is very different from Galilean velocity-addition, which would give the intuitive result that observer A sees the wave propagating at speed $\frac{3}{2} c$.
Also I assume here that all the motion is just in one dimension.
• But if you would consider the distance travelled by them in 1 second then it should be $AW=AB+BW$ which isn't the case here. – Sahil Silare Jun 20 '18 at 16:17
• @Grayscale The postulate of SR only applies to inertial frames, so it is not invariant "regardless of their motion." The proper speed in non-inertial frames remains $c$, but the coordinate speed of light is routinely changed by gravitational fields. – Zack Hutchens Jun 20 '18 at 16:18
• @zhutchens1 I think it should be fixed now. – Grayscale Jun 20 '18 at 16:22
• What's postulate of SR? – Sahil Silare Jun 20 '18 at 16:22
• @SahilSilare There are two main assumptions that SR makes. One is about the invariance of the speed of light (see above) and the other is that the laws of physics are the same in all inertial frames. See the linked article. – Grayscale Jun 20 '18 at 16:26
Einstein's two axioms for special relativity :
1. PRINCIPLE OF RELATIVITY: The laws of physics are identical in all inertial frames, or, equivalently, the outcome of any physical experiment is the same when performed with identical initial conditions relative to any inertial frame.
2. LAW OF LIGHT PROPAGATION: Light signals in vacuum are propagated rectilinearly, with the same speed c, at all times, in all directions, in all inertial frames. | 2019-08-23 22:00:55 | {"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": 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": 0.7746689915657043, "perplexity": 326.6476653809983}, "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/1566027319082.81/warc/CC-MAIN-20190823214536-20190824000536-00202.warc.gz"} |
https://stacks.math.columbia.edu/tag/08RL | # The Stacks Project
## Tag 08RL
### 17.25. Modules of differentials
In this section we briefly explain how to define the module of relative differentials for a morphism of ringed spaces. We suggest the reader take a look at the corresponding section in the chapter on commutative algebra (Algebra, Section 10.130).
Definition 17.25.1. Let $X$ be a topological space. Let $\varphi : \mathcal{O}_1 \to \mathcal{O}_2$ be a homomorphism of sheaves of rings. Let $\mathcal{F}$ be an $\mathcal{O}_2$-module. A $\mathcal{O}_1$-derivation or more precisely a $\varphi$-derivation into $\mathcal{F}$ is a map $D : \mathcal{O}_2 \to \mathcal{F}$ which is additive, annihilates the image of $\mathcal{O}_1 \to \mathcal{O}_2$, and satisfies the Leibniz rule $$D(ab) = aD(b) + D(a)b$$ for all $a, b$ local sections of $\mathcal{O}_2$ (wherever they are both defined). We denote $\text{Der}_{\mathcal{O}_1}(\mathcal{O}_2, \mathcal{F})$ the set of $\varphi$-derivations into $\mathcal{F}$.
This is the sheaf theoretic analogue of Algebra, Definition 17.25.1. Given a derivation $D : \mathcal{O}_2 \to \mathcal{F}$ as in the definition the map on global sections $$D : \Gamma(X, \mathcal{O}_2) \longrightarrow \Gamma(X, \mathcal{F})$$ is a $\Gamma(X, \mathcal{O}_1)$-derivation as in the algebra definition. Note that if $\alpha : \mathcal{F} \to \mathcal{G}$ is a map of $\mathcal{O}_2$-modules, then there is an induced map $$\text{Der}_{\mathcal{O}_1}(\mathcal{O}_2, \mathcal{F}) \longrightarrow \text{Der}_{\mathcal{O}_1}(\mathcal{O}_2, \mathcal{G})$$ given by the rule $D \mapsto \alpha \circ D$. In other words we obtain a functor.
Lemma 17.25.2. Let $X$ be a topological space. Let $\varphi : \mathcal{O}_1 \to \mathcal{O}_2$ be a homomorphism of sheaves of rings. The functor $$\textit{Mod}(\mathcal{O}_2) \longrightarrow \textit{Ab}, \quad \mathcal{F} \longmapsto \text{Der}_{\mathcal{O}_1}(\mathcal{O}_2, \mathcal{F})$$ is representable.
Proof. This is proved in exactly the same way as the analogous statement in algebra. During this proof, for any sheaf of sets $\mathcal{F}$ on $X$, let us denote $\mathcal{O}_2[\mathcal{F}]$ the sheafification of the presheaf $U \mapsto \mathcal{O}_2(U)[\mathcal{F}(U)]$ where this denotes the free $\mathcal{O}_1(U)$-module on the set $\mathcal{F}(U)$. For $s \in \mathcal{F}(U)$ we denote $[s]$ the corresponding section of $\mathcal{O}_2[\mathcal{F}]$ over $U$. If $\mathcal{F}$ is a sheaf of $\mathcal{O}_2$-modules, then there is a canonical map $$c : \mathcal{O}_2[\mathcal{F}] \longrightarrow \mathcal{F}$$ which on the presheaf level is given by the rule $\sum f_s[s] \mapsto \sum f_s s$. We will employ the short hand $[s] \mapsto s$ to describe this map and similarly for other maps below. Consider the map of $\mathcal{O}_2$-modules $$\tag{17.25.2.1} \begin{matrix} \mathcal{O}_2[\mathcal{O}_2 \times \mathcal{O}_2] \oplus \mathcal{O}_2[\mathcal{O}_2 \times \mathcal{O}_2] \oplus \mathcal{O}_2[\mathcal{O}_1] & \longrightarrow & \mathcal{O}_2[\mathcal{O}_2] \\ [(a, b)] \oplus [(f, g)] \oplus [h] & \longmapsto & [a + b] - [a] - [b] + \\ & & [fg] - g[f] - f[g] + \\ & & [\varphi(h)] \end{matrix}$$ with short hand notation as above. Set $\Omega_{\mathcal{O}_2/\mathcal{O}_1}$ equal to the cokernel of this map. Then it is clear that there exists a map of sheaves of sets $$\text{d} : \mathcal{O}_2 \longrightarrow \Omega_{\mathcal{O}_2/\mathcal{O}_1}$$ mapping a local section $f$ to the image of $[f]$ in $\Omega_{\mathcal{O}_2/\mathcal{O}_1}$. By construction $\text{d}$ is a $\mathcal{O}_1$-derivation. Next, let $\mathcal{F}$ be a sheaf of $\mathcal{O}_2$-modules and let $D : \mathcal{O}_2 \to \mathcal{F}$ be a $\mathcal{O}_1$-derivation. Then we can consider the $\mathcal{O}_2$-linear map $\mathcal{O}_2[\mathcal{O}_2] \to \mathcal{F}$ which sends $[g]$ to $D(g)$. It follows from the definition of a derivation that this map annihilates sections in the image of the map (17.25.2.1) and hence defines a map $$\alpha_D : \Omega_{\mathcal{O}_2/\mathcal{O}_1} \longrightarrow \mathcal{F}$$ Since it is clear that $D = \alpha_D \circ \text{d}$ the lemma is proved. $\square$
Definition 17.25.3. Let $X$ be a topological space. Let $\varphi : \mathcal{O}_1 \to \mathcal{O}_2$ be a homomorphism of sheaves of rings on $X$. The module of differentials of $\varphi$ is the object representing the functor $\mathcal{F} \mapsto \text{Der}_{\mathcal{O}_1}(\mathcal{O}_2, \mathcal{F})$ which exists by Lemma 17.25.2. It is denoted $\Omega_{\mathcal{O}_2/\mathcal{O}_1}$, and the universal $\varphi$-derivation is denoted $\text{d} : \mathcal{O}_2 \to \Omega_{\mathcal{O}_2/\mathcal{O}_1}$.
Note that $\Omega_{\mathcal{O}_2/\mathcal{O}_1}$ is the cokernel of the map (17.25.2.1) of $\mathcal{O}_2$-modules. Moreover the map $\text{d}$ is described by the rule that $\text{d}f$ is the image of the local section $[f]$.
Lemma 17.25.4. Let $X$ be a topological space. Let $\varphi : \mathcal{O}_1 \to \mathcal{O}_2$ be a homomorphism of sheaves of rings on $X$. Then $\Omega_{\mathcal{O}_2/\mathcal{O}_1}$ is the sheaf associated to the presheaf $U \mapsto \Omega_{\mathcal{O}_2(U)/\mathcal{O}_1(U)}$.
Proof. Consider the map (17.25.2.1). There is a similar map of presheaves whose value on the open $U$ is $$\mathcal{O}_2(U)[\mathcal{O}_2(U) \times \mathcal{O}_2(U)] \oplus \mathcal{O}_2(U)[\mathcal{O}_2(U) \times \mathcal{O}_2(U)] \oplus \mathcal{O}_2(U)[\mathcal{O}_1(U)] \longrightarrow \mathcal{O}_2(U)[\mathcal{O}_2(U)]$$ The cokernel of this map has value $\Omega_{\mathcal{O}_2(U)/\mathcal{O}_1(U)}$ over $U$ by the construction of the module of differentials in Algebra, Definition 10.130.2. On the other hand, the sheaves in (17.25.2.1) are the sheafifications of the presheaves above. Thus the result follows as sheafification is exact. $\square$
Lemma 17.25.5. Let $X$ be a topological space. Let $\varphi : \mathcal{O}_1 \to \mathcal{O}_2$ be a homomorphism of sheaves of rings. For $U \subset X$ open there is a canonical isomorphism $$\Omega_{\mathcal{O}_2/\mathcal{O}_1}|_U = \Omega_{(\mathcal{O}_2|_U)/(\mathcal{O}_1|_U)}$$ compatible with universal derivations.
Proof. Holds because $\Omega_{\mathcal{O}_2/\mathcal{O}_1}$ is the cokernel of the map (17.25.2.1). $\square$
Lemma 17.25.6. Let $f : Y \to X$ be a continuous map of topological spaces. Let $\varphi : \mathcal{O}_1 \to \mathcal{O}_2$ be a homomorphism of sheaves of rings on $X$. Then there is a canonical identification $f^{-1}\Omega_{\mathcal{O}_2/\mathcal{O}_1} = \Omega_{f^{-1}\mathcal{O}_2/f^{-1}\mathcal{O}_1}$ compatible with universal derivations.
Proof. This holds because the sheaf $\Omega_{\mathcal{O}_2/\mathcal{O}_1}$ is the cokernel of the map (17.25.2.1) and a similar statement holds for $\Omega_{f^{-1}\mathcal{O}_2/f^{-1}\mathcal{O}_1}$, because the functor $f^{-1}$ is exact, and because $f^{-1}(\mathcal{O}_2[\mathcal{O}_2]) = f^{-1}\mathcal{O}_2[f^{-1}\mathcal{O}_2]$, $f^{-1}(\mathcal{O}_2[\mathcal{O}_2 \times \mathcal{O}_2]) = f^{-1}\mathcal{O}_2[f^{-1}\mathcal{O}_2 \times f^{-1}\mathcal{O}_2]$, and $f^{-1}(\mathcal{O}_2[\mathcal{O}_1]) = f^{-1}\mathcal{O}_2[f^{-1}\mathcal{O}_1]$. $\square$
Lemma 17.25.7. Let $X$ be a topological space. Let $\mathcal{O}_1 \to \mathcal{O}_2$ be a homomorphism of sheaves of rings on $X$. Let $x \in X$. Then we have $\Omega_{\mathcal{O}_2/\mathcal{O}_1, x} = \Omega_{\mathcal{O}_{2, x}/\mathcal{O}_{1, x}}$.
Proof. This is a special case of Lemma 17.25.6 for the inclusion map $\{x\} \to X$. An alternative proof is the use Lemma 17.25.4, Sheaves, Lemma 6.17.2, and Algebra, Lemma 10.130.4 $\square$
Lemma 17.25.8. Let $X$ be a topological space. Let $$\xymatrix{ \mathcal{O}_2 \ar[r]_\varphi & \mathcal{O}_2' \\ \mathcal{O}_1 \ar[r] \ar[u] & \mathcal{O}'_1 \ar[u] }$$ be a commutative diagram of sheaves of rings on $X$. The map $\mathcal{O}_2 \to \mathcal{O}'_2$ composed with the map $\text{d} : \mathcal{O}'_2 \to \Omega_{\mathcal{O}'_2/\mathcal{O}'_1}$ is a $\mathcal{O}_1$-derivation. Hence we obtain a canonical map of $\mathcal{O}_2$-modules $\Omega_{\mathcal{O}_2/\mathcal{O}_1} \to \Omega_{\mathcal{O}'_2/\mathcal{O}'_1}$. It is uniquely characterized by the property that $\text{d}(f) \mapsto \text{d}(\varphi(f))$ for any local section $f$ of $\mathcal{O}_2$. In this way $\Omega_{-/-}$ becomes a functor on the category of arrows of sheaves of rings.
Proof. This lemma proves itself. $\square$
Lemma 17.25.9. In Lemma 17.25.8 suppose that $\mathcal{O}_2 \to \mathcal{O}'_2$ is surjective with kernel $\mathcal{I} \subset \mathcal{O}_2$ and assume that $\mathcal{O}_1 = \mathcal{O}'_1$. Then there is a canonical exact sequence of $\mathcal{O}'_2$-modules $$\mathcal{I}/\mathcal{I}^2 \longrightarrow \Omega_{\mathcal{O}_2/\mathcal{O}_1} \otimes_{\mathcal{O}_2} \mathcal{O}'_2 \longrightarrow \Omega_{\mathcal{O}'_2/\mathcal{O}_1} \longrightarrow 0$$ The leftmost map is characterized by the rule that a local section $f$ of $\mathcal{I}$ maps to $\text{d}f \otimes 1$.
Proof. For a local section $f$ of $\mathcal{I}$ denote $\overline{f}$ the image of $f$ in $\mathcal{I}/\mathcal{I}^2$. To show that the map $\overline{f} \mapsto \text{d}f \otimes 1$ is well defined we just have to check that $\text{d} f_1f_2 \otimes 1 = 0$ if $f_1, f_2$ are local sections of $\mathcal{I}$. And this is clear from the Leibniz rule $\text{d} f_1f_2 \otimes 1 = (f_1 \text{d}f_2 + f_2 \text{d} f_1 )\otimes 1 = \text{d}f_2 \otimes f_1 + \text{d}f_1 \otimes f_2 = 0$. A similar computation show this map is $\mathcal{O}'_2 = \mathcal{O}_2/\mathcal{I}$-linear. The map on the right is the one from Lemma 17.25.8. To see that the sequence is exact, we can check on stalks (Lemma 17.3.1). By Lemma 17.25.7 this follows from Algebra, Lemma 10.130.9. $\square$
Definition 17.25.10. Let $(f, f^\sharp) : (X, \mathcal{O}_X) \to (S, \mathcal{O}_S)$ be a morphism of ringed spaces.
1. Let $\mathcal{F}$ be an $\mathcal{O}_X$-module. An $S$-derivation into $\mathcal{F}$ is a $f^{-1}\mathcal{O}_S$-derivation, or more precisely a $f^\sharp$-derivation in the sense of Definition 17.25.1. We denote $\text{Der}_S(\mathcal{O}_X, \mathcal{F})$ the set of $S$-derivations into $\mathcal{F}$.
2. The sheaf of differentials $\Omega_{X/S}$ of $X$ over $S$ is the module of differentials $\Omega_{\mathcal{O}_X/f^{-1}\mathcal{O}_S}$ endowed with its universal $S$-derivation $\text{d}_{X/S} : \mathcal{O}_X \to \Omega_{X/S}$.
Here is a particular situation where derivations come up naturally.
Lemma 17.25.11. Let $(f, f^\sharp) : (X, \mathcal{O}_X) \to (S, \mathcal{O}_S)$ be a morphism of ringed spaces. Consider a short exact sequence $$0 \to \mathcal{I} \to \mathcal{A} \to \mathcal{O}_X \to 0$$ Here $\mathcal{A}$ is a sheaf of $f^{-1}\mathcal{O}_S$-algebras, $\pi : \mathcal{A} \to \mathcal{O}_X$ is a surjection of sheaves of $f^{-1}\mathcal{O}_S$-algebras, and $\mathcal{I} = \mathop{\mathrm{Ker}}(\pi)$ is its kernel. Assume $\mathcal{I}$ an ideal sheaf with square zero in $\mathcal{A}$. So $\mathcal{I}$ has a natural structure of an $\mathcal{O}_X$-module. A section $s : \mathcal{O}_X \to \mathcal{A}$ of $\pi$ is a $f^{-1}\mathcal{O}_S$-algebra map such that $\pi \circ s = \text{id}$. Given any section $s : \mathcal{O}_X \to \mathcal{A}$ of $\pi$ and any $S$-derivation $D : \mathcal{O}_X \to \mathcal{I}$ the map $$s + D : \mathcal{O}_X \to \mathcal{A}$$ is a section of $\pi$ and every section $s'$ is of the form $s + D$ for a unique $S$-derivation $D$.
Proof. Recall that the $\mathcal{O}_X$-module structure on $\mathcal{I}$ is given by $h \tau = \tilde h \tau$ (multiplication in $\mathcal{A}$) where $h$ is a local section of $\mathcal{O}_X$, and $\tilde h$ is a local lift of $h$ to a local section of $\mathcal{A}$, and $\tau$ is a local section of $\mathcal{I}$. In particular, given $s$, we may use $\tilde h = s(h)$. To verify that $s + D$ is a homomorphism of sheaves of rings we compute \begin{eqnarray*} (s + D)(ab) & = & s(ab) + D(ab) \\ & = & s(a)s(b) + aD(b) + D(a)b \\ & = & s(a) s(b) + s(a)D(b) + D(a)s(b) \\ & = & (s(a) + D(a))(s(b) + D(b)) \end{eqnarray*} by the Leibniz rule. In the same manner one shows $s + D$ is a $f^{-1}\mathcal{O}_S$-algebra map because $D$ is an $S$-derivation. Conversely, given $s'$ we set $D = s' - s$. Details omitted. $\square$
Lemma 17.25.12. Let $$\xymatrix{ X' \ar[d]_{h'} \ar[r]_f & X \ar[d]^h \\ S' \ar[r]^g & S }$$ be a commutative diagram of ringed spaces.
1. The canonical map $\mathcal{O}_X \to f_*\mathcal{O}_{X'}$ composed with $f_*\text{d}_{X'/S'} : f_*\mathcal{O}_{X'} \to f_*\Omega_{X'/S'}$ is a $S$-derivation and we obtain a canonical map of $\mathcal{O}_X$-modules $\Omega_{X/S} \to f_*\Omega_{X'/S'}$.
2. The commutative diagram $$\xymatrix{ f^{-1}\mathcal{O}_X \ar[r] & \mathcal{O}_{X'} \\ f^{-1}h^{-1}\mathcal{O}_S \ar[u] \ar[r] & (h')^{-1}\mathcal{O}_{S'} \ar[u] }$$ induces by Lemmas 17.25.6 and 17.25.8 a canonical map $f^{-1}\Omega_{X/S} \to \Omega_{X'/S'}$.
These two maps correspond (via adjointness of $f_*$ and $f^*$ and via $f^*\Omega_{X/S} = f^{-1}\Omega_{X/S} \otimes_{f^{-1}\mathcal{O}_X} \mathcal{O}_{X'}$ and Sheaves, Lemma 6.20.2) to the same $\mathcal{O}_{X'}$-module homomorphism $$c_f : f^*\Omega_{X/S} \longrightarrow \Omega_{X'/S'}$$ which is uniquely characterized by the property that $f^*\text{d}_{X/S}(a)$ maps to $\text{d}_{X'/S'}(f^*a)$ for any local section $a$ of $\mathcal{O}_X$.
Proof. Omitted. $\square$
Lemma 17.25.13. Let $$\xymatrix{ X'' \ar[d] \ar[r]_g & X' \ar[d] \ar[r]_f & X \ar[d] \\ S'' \ar[r] & S' \ar[r] & S }$$ be a commutative diagram of ringed spaces. With notation as in Lemma 17.25.12 we have $$c_{f \circ g} = c_g \circ g^* c_f$$ as maps $(f \circ g)^*\Omega_{X/S} \to \Omega_{X''/S''}$.
Proof. Omitted. $\square$
The code snippet corresponding to this tag is a part of the file modules.tex and is located in lines 4120–4528 (see updates for more information).
\section{Modules of differentials}
\label{section-differentials}
\noindent
In this section we briefly explain how to define the module of relative
differentials for a morphism of ringed spaces. We suggest the reader take
a look at the corresponding section in the chapter on commutative algebra
(Algebra, Section \ref{algebra-section-differentials}).
\begin{definition}
\label{definition-derivation}
Let $X$ be a topological space. Let $\varphi : \mathcal{O}_1 \to \mathcal{O}_2$
be a homomorphism of sheaves of rings. Let $\mathcal{F}$ be an
$\mathcal{O}_2$-module. A {\it $\mathcal{O}_1$-derivation} or more precisely
a {\it $\varphi$-derivation} into $\mathcal{F}$ is a map
$D : \mathcal{O}_2 \to \mathcal{F}$ which is additive, annihilates the image
of $\mathcal{O}_1 \to \mathcal{O}_2$, and satisfies the
{\it Leibniz rule}
$$D(ab) = aD(b) + D(a)b$$
for all $a, b$ local sections of $\mathcal{O}_2$ (wherever they are both
defined). We denote $\text{Der}_{\mathcal{O}_1}(\mathcal{O}_2, \mathcal{F})$
the set of $\varphi$-derivations into $\mathcal{F}$.
\end{definition}
\noindent
This is the sheaf theoretic analogue of
Algebra, Definition \ref{definition-derivation}.
Given a derivation $D : \mathcal{O}_2 \to \mathcal{F}$
as in the definition the map on global sections
$$D : \Gamma(X, \mathcal{O}_2) \longrightarrow \Gamma(X, \mathcal{F})$$
is a $\Gamma(X, \mathcal{O}_1)$-derivation as in the algebra definition.
Note that if $\alpha : \mathcal{F} \to \mathcal{G}$ is a map of
$\mathcal{O}_2$-modules, then there is an induced map
$$\text{Der}_{\mathcal{O}_1}(\mathcal{O}_2, \mathcal{F}) \longrightarrow \text{Der}_{\mathcal{O}_1}(\mathcal{O}_2, \mathcal{G})$$
given by the rule $D \mapsto \alpha \circ D$. In other words
we obtain a functor.
\begin{lemma}
\label{lemma-universal-module}
Let $X$ be a topological space. Let $\varphi : \mathcal{O}_1 \to \mathcal{O}_2$
be a homomorphism of sheaves of rings. The functor
$$\textit{Mod}(\mathcal{O}_2) \longrightarrow \textit{Ab}, \quad \mathcal{F} \longmapsto \text{Der}_{\mathcal{O}_1}(\mathcal{O}_2, \mathcal{F})$$
is representable.
\end{lemma}
\begin{proof}
This is proved in exactly the same way as the analogous statement in algebra.
During this proof, for any sheaf of sets $\mathcal{F}$ on $X$,
let us denote $\mathcal{O}_2[\mathcal{F}]$ the sheafification of the
presheaf $U \mapsto \mathcal{O}_2(U)[\mathcal{F}(U)]$ where this denotes
the free $\mathcal{O}_1(U)$-module on the set $\mathcal{F}(U)$.
For $s \in \mathcal{F}(U)$ we denote $[s]$ the corresponding section
of $\mathcal{O}_2[\mathcal{F}]$ over $U$. If $\mathcal{F}$ is a sheaf of
$\mathcal{O}_2$-modules, then there is a canonical map
$$c : \mathcal{O}_2[\mathcal{F}] \longrightarrow \mathcal{F}$$
which on the presheaf level is given by the rule
$\sum f_s[s] \mapsto \sum f_s s$. We will employ the short hand
$[s] \mapsto s$ to
describe this map and similarly for other maps below. Consider
the map of $\mathcal{O}_2$-modules
\label{equation-define-module-differentials}
\begin{matrix}
\mathcal{O}_2[\mathcal{O}_2 \times \mathcal{O}_2] \oplus
\mathcal{O}_2[\mathcal{O}_2 \times \mathcal{O}_2] \oplus
\mathcal{O}_2[\mathcal{O}_1] &
\longrightarrow &
\mathcal{O}_2[\mathcal{O}_2] \\
[(a, b)] \oplus [(f, g)] \oplus [h] & \longmapsto & [a + b] - [a] - [b] + \\
& & [fg] - g[f] - f[g] + \\
& & [\varphi(h)]
\end{matrix}
with short hand notation as above. Set $\Omega_{\mathcal{O}_2/\mathcal{O}_1}$
equal to the cokernel of this map. Then it is clear that there exists
a map of sheaves of sets
$$\text{d} : \mathcal{O}_2 \longrightarrow \Omega_{\mathcal{O}_2/\mathcal{O}_1}$$
mapping a local section $f$ to the image of $[f]$ in
$\Omega_{\mathcal{O}_2/\mathcal{O}_1}$. By construction $\text{d}$
is a $\mathcal{O}_1$-derivation. Next, let $\mathcal{F}$
be a sheaf of $\mathcal{O}_2$-modules and let
$D : \mathcal{O}_2 \to \mathcal{F}$ be a $\mathcal{O}_1$-derivation.
Then we can consider the $\mathcal{O}_2$-linear map
$\mathcal{O}_2[\mathcal{O}_2] \to \mathcal{F}$ which sends $[g]$ to $D(g)$.
It follows from the definition of a derivation that this map annihilates
sections in the image of the map (\ref{equation-define-module-differentials})
and hence defines a map
$$\alpha_D : \Omega_{\mathcal{O}_2/\mathcal{O}_1} \longrightarrow \mathcal{F}$$
Since it is clear that $D = \alpha_D \circ \text{d}$ the lemma is proved.
\end{proof}
\begin{definition}
\label{definition-module-differentials}
Let $X$ be a topological space. Let $\varphi : \mathcal{O}_1 \to \mathcal{O}_2$
be a homomorphism of sheaves of rings on $X$. The {\it module of differentials}
of $\varphi$ is the object representing the functor
$\mathcal{F} \mapsto \text{Der}_{\mathcal{O}_1}(\mathcal{O}_2, \mathcal{F})$
which exists by Lemma \ref{lemma-universal-module}.
It is denoted $\Omega_{\mathcal{O}_2/\mathcal{O}_1}$, and the {\it universal
$\varphi$-derivation} is denoted
$\text{d} : \mathcal{O}_2 \to \Omega_{\mathcal{O}_2/\mathcal{O}_1}$.
\end{definition}
\noindent
Note that $\Omega_{\mathcal{O}_2/\mathcal{O}_1}$ is the cokernel of
the map (\ref{equation-define-module-differentials}) of
$\mathcal{O}_2$-modules. Moreover the map $\text{d}$ is described
by the rule that $\text{d}f$ is the image of the local section $[f]$.
\begin{lemma}
\label{lemma-differentials-sheafify}
Let $X$ be a topological space. Let $\varphi : \mathcal{O}_1 \to \mathcal{O}_2$
be a homomorphism of sheaves of rings on $X$. Then
$\Omega_{\mathcal{O}_2/\mathcal{O}_1}$ is the sheaf associated to the
presheaf $U \mapsto \Omega_{\mathcal{O}_2(U)/\mathcal{O}_1(U)}$.
\end{lemma}
\begin{proof}
Consider the map (\ref{equation-define-module-differentials}). There is
a similar map of presheaves whose value on the open $U$ is
$$\mathcal{O}_2(U)[\mathcal{O}_2(U) \times \mathcal{O}_2(U)] \oplus \mathcal{O}_2(U)[\mathcal{O}_2(U) \times \mathcal{O}_2(U)] \oplus \mathcal{O}_2(U)[\mathcal{O}_1(U)] \longrightarrow \mathcal{O}_2(U)[\mathcal{O}_2(U)]$$
The cokernel of this map has value $\Omega_{\mathcal{O}_2(U)/\mathcal{O}_1(U)}$
over $U$ by the construction of the module of differentials in
Algebra, Definition \ref{algebra-definition-differentials}.
On the other hand, the sheaves in (\ref{equation-define-module-differentials})
are the sheafifications of the presheaves above. Thus the result follows
as sheafification is exact.
\end{proof}
\begin{lemma}
\label{lemma-localize-differentials}
Let $X$ be a topological space. Let $\varphi : \mathcal{O}_1 \to \mathcal{O}_2$
be a homomorphism of sheaves of rings. For $U \subset X$ open
there is a canonical isomorphism
$$\Omega_{\mathcal{O}_2/\mathcal{O}_1}|_U = \Omega_{(\mathcal{O}_2|_U)/(\mathcal{O}_1|_U)}$$
compatible with universal derivations.
\end{lemma}
\begin{proof}
Holds because $\Omega_{\mathcal{O}_2/\mathcal{O}_1}$
is the cokernel of the map (\ref{equation-define-module-differentials}).
\end{proof}
\begin{lemma}
\label{lemma-pullback-differentials}
Let $f : Y \to X$ be a continuous map of topological spaces.
Let $\varphi : \mathcal{O}_1 \to \mathcal{O}_2$
be a homomorphism of sheaves of rings on $X$.
Then there is a canonical identification
$f^{-1}\Omega_{\mathcal{O}_2/\mathcal{O}_1} = \Omega_{f^{-1}\mathcal{O}_2/f^{-1}\mathcal{O}_1}$
compatible with universal derivations.
\end{lemma}
\begin{proof}
This holds because the sheaf $\Omega_{\mathcal{O}_2/\mathcal{O}_1}$
is the cokernel of the map (\ref{equation-define-module-differentials})
and a similar statement holds for
$\Omega_{f^{-1}\mathcal{O}_2/f^{-1}\mathcal{O}_1}$,
because the functor $f^{-1}$ is exact, and because
$f^{-1}(\mathcal{O}_2[\mathcal{O}_2]) = f^{-1}\mathcal{O}_2[f^{-1}\mathcal{O}_2]$,
$f^{-1}(\mathcal{O}_2[\mathcal{O}_2 \times \mathcal{O}_2]) = f^{-1}\mathcal{O}_2[f^{-1}\mathcal{O}_2 \times f^{-1}\mathcal{O}_2]$, and
$f^{-1}(\mathcal{O}_2[\mathcal{O}_1]) = f^{-1}\mathcal{O}_2[f^{-1}\mathcal{O}_1]$.
\end{proof}
\begin{lemma}
\label{lemma-stalk-module-differentials}
Let $X$ be a topological space. Let $\mathcal{O}_1 \to \mathcal{O}_2$
be a homomorphism of sheaves of rings on $X$. Let $x \in X$. Then we have
$\Omega_{\mathcal{O}_2/\mathcal{O}_1, x} = \Omega_{\mathcal{O}_{2, x}/\mathcal{O}_{1, x}}$.
\end{lemma}
\begin{proof}
This is a special case of Lemma \ref{lemma-pullback-differentials}
for the inclusion map $\{x\} \to X$. An alternative proof is the use
Lemma \ref{lemma-differentials-sheafify},
Sheaves, Lemma \ref{sheaves-lemma-stalk-sheafification}, and
Algebra, Lemma \ref{algebra-lemma-colimit-differentials}
\end{proof}
\begin{lemma}
\label{lemma-functoriality-differentials}
Let $X$ be a topological space. Let
$$\xymatrix{ \mathcal{O}_2 \ar[r]_\varphi & \mathcal{O}_2' \\ \mathcal{O}_1 \ar[r] \ar[u] & \mathcal{O}'_1 \ar[u] }$$
be a commutative diagram of sheaves of rings on $X$. The map
$\mathcal{O}_2 \to \mathcal{O}'_2$ composed with the map
$\text{d} : \mathcal{O}'_2 \to \Omega_{\mathcal{O}'_2/\mathcal{O}'_1}$
is a $\mathcal{O}_1$-derivation. Hence we obtain a canonical map of
$\mathcal{O}_2$-modules
$\Omega_{\mathcal{O}_2/\mathcal{O}_1} \to \Omega_{\mathcal{O}'_2/\mathcal{O}'_1}$.
It is uniquely characterized by the property that
$\text{d}(f) \mapsto \text{d}(\varphi(f))$
for any local section $f$ of $\mathcal{O}_2$.
In this way $\Omega_{-/-}$ becomes a functor on the category
of arrows of sheaves of rings.
\end{lemma}
\begin{proof}
This lemma proves itself.
\end{proof}
\begin{lemma}
\label{lemma-differential-seq}
In Lemma \ref{lemma-functoriality-differentials} suppose that
$\mathcal{O}_2 \to \mathcal{O}'_2$ is surjective with kernel
$\mathcal{I} \subset \mathcal{O}_2$ and assume that
$\mathcal{O}_1 = \mathcal{O}'_1$. Then there is a canonical exact
sequence of $\mathcal{O}'_2$-modules
$$\mathcal{I}/\mathcal{I}^2 \longrightarrow \Omega_{\mathcal{O}_2/\mathcal{O}_1} \otimes_{\mathcal{O}_2} \mathcal{O}'_2 \longrightarrow \Omega_{\mathcal{O}'_2/\mathcal{O}_1} \longrightarrow 0$$
The leftmost map is characterized by the rule that a local section
$f$ of $\mathcal{I}$ maps to $\text{d}f \otimes 1$.
\end{lemma}
\begin{proof}
For a local section $f$ of $\mathcal{I}$ denote $\overline{f}$ the image of
$f$ in $\mathcal{I}/\mathcal{I}^2$. To show that the map
$\overline{f} \mapsto \text{d}f \otimes 1$ is well defined we just have to
check that $\text{d} f_1f_2 \otimes 1 = 0$ if $f_1, f_2$ are local sections
of $\mathcal{I}$. And this is clear from the Leibniz rule
$\text{d} f_1f_2 \otimes 1 = (f_1 \text{d}f_2 + f_2 \text{d} f_1 )\otimes 1 = \text{d}f_2 \otimes f_1 + \text{d}f_1 \otimes f_2 = 0$.
A similar computation show this map is
$\mathcal{O}'_2 = \mathcal{O}_2/\mathcal{I}$-linear. The map on the right
is the one from Lemma \ref{lemma-functoriality-differentials}. To see
that the sequence is exact, we can check on stalks
(Lemma \ref{lemma-abelian}). By
Lemma \ref{lemma-stalk-module-differentials}
this follows from
Algebra, Lemma \ref{algebra-lemma-differential-seq}.
\end{proof}
\begin{definition}
\label{definition-differentials}
Let $(f, f^\sharp) : (X, \mathcal{O}_X) \to (S, \mathcal{O}_S)$
be a morphism of ringed spaces.
\begin{enumerate}
\item Let $\mathcal{F}$ be an $\mathcal{O}_X$-module. An {\it $S$-derivation}
into $\mathcal{F}$ is a $f^{-1}\mathcal{O}_S$-derivation, or more
precisely a $f^\sharp$-derivation in the sense of
Definition \ref{definition-derivation}.
We denote $\text{Der}_S(\mathcal{O}_X, \mathcal{F})$
the set of $S$-derivations into $\mathcal{F}$.
\item The {\it sheaf of differentials $\Omega_{X/S}$ of $X$ over $S$}
is the module of differentials $\Omega_{\mathcal{O}_X/f^{-1}\mathcal{O}_S}$
endowed with its universal
$S$-derivation $\text{d}_{X/S} : \mathcal{O}_X \to \Omega_{X/S}$.
\end{enumerate}
\end{definition}
\noindent
Here is a particular situation where derivations come up
naturally.
\begin{lemma}
\label{lemma-double-structure-gives-derivation}
Let $(f, f^\sharp) : (X, \mathcal{O}_X) \to (S, \mathcal{O}_S)$
be a morphism of ringed spaces. Consider a short exact sequence
$$0 \to \mathcal{I} \to \mathcal{A} \to \mathcal{O}_X \to 0$$
Here $\mathcal{A}$ is a sheaf of $f^{-1}\mathcal{O}_S$-algebras,
$\pi : \mathcal{A} \to \mathcal{O}_X$ is a surjection
of sheaves of $f^{-1}\mathcal{O}_S$-algebras, and
$\mathcal{I} = \Ker(\pi)$ is its kernel. Assume $\mathcal{I}$ an ideal
sheaf with square zero in $\mathcal{A}$. So $\mathcal{I}$
has a natural structure of an $\mathcal{O}_X$-module.
A section $s : \mathcal{O}_X \to \mathcal{A}$ of $\pi$
is a $f^{-1}\mathcal{O}_S$-algebra map such that $\pi \circ s = \text{id}$.
Given any section $s : \mathcal{O}_X \to \mathcal{A}$
of $\pi$ and any $S$-derivation $D : \mathcal{O}_X \to \mathcal{I}$
the map
$$s + D : \mathcal{O}_X \to \mathcal{A}$$
is a section of $\pi$ and every section $s'$ is of the form $s + D$
for a unique $S$-derivation $D$.
\end{lemma}
\begin{proof}
Recall that the $\mathcal{O}_X$-module structure on $\mathcal{I}$
is given by $h \tau = \tilde h \tau$ (multiplication in $\mathcal{A}$)
where $h$ is a local section of $\mathcal{O}_X$, and
$\tilde h$ is a local lift of $h$ to a local
section of $\mathcal{A}$, and $\tau$ is a local section of $\mathcal{I}$.
In particular, given $s$, we may use $\tilde h = s(h)$.
To verify that $s + D$ is a homomorphism of sheaves of rings we
compute
\begin{eqnarray*}
(s + D)(ab) & = & s(ab) + D(ab) \\
& = & s(a)s(b) + aD(b) + D(a)b \\
& = & s(a) s(b) + s(a)D(b) + D(a)s(b) \\
& = & (s(a) + D(a))(s(b) + D(b))
\end{eqnarray*}
by the Leibniz rule. In the same manner one shows
$s + D$ is a $f^{-1}\mathcal{O}_S$-algebra
map because $D$ is an $S$-derivation. Conversely, given $s'$ we set
$D = s' - s$. Details omitted.
\end{proof}
\begin{lemma}
\label{lemma-functoriality-differentials-ringed-spaces}
Let
$$\xymatrix{ X' \ar[d]_{h'} \ar[r]_f & X \ar[d]^h \\ S' \ar[r]^g & S }$$
be a commutative diagram of ringed spaces.
\begin{enumerate}
\item The canonical map $\mathcal{O}_X \to f_*\mathcal{O}_{X'}$ composed with
$f_*\text{d}_{X'/S'} : f_*\mathcal{O}_{X'} \to f_*\Omega_{X'/S'}$ is a
$S$-derivation and we obtain a canonical map of $\mathcal{O}_X$-modules
$\Omega_{X/S} \to f_*\Omega_{X'/S'}$.
\item The commutative diagram
$$\xymatrix{ f^{-1}\mathcal{O}_X \ar[r] & \mathcal{O}_{X'} \\ f^{-1}h^{-1}\mathcal{O}_S \ar[u] \ar[r] & (h')^{-1}\mathcal{O}_{S'} \ar[u] }$$
induces by Lemmas \ref{lemma-pullback-differentials} and
\ref{lemma-functoriality-differentials}
a canonical map $f^{-1}\Omega_{X/S} \to \Omega_{X'/S'}$.
\end{enumerate}
These two maps correspond (via adjointness of $f_*$ and $f^*$ and
via $f^*\Omega_{X/S} = f^{-1}\Omega_{X/S} \otimes_{f^{-1}\mathcal{O}_X} \mathcal{O}_{X'}$ and
to the same $\mathcal{O}_{X'}$-module homomorphism
$$c_f : f^*\Omega_{X/S} \longrightarrow \Omega_{X'/S'}$$
which is uniquely characterized by the property that
$f^*\text{d}_{X/S}(a)$ maps to $\text{d}_{X'/S'}(f^*a)$
for any local section $a$ of $\mathcal{O}_X$.
\end{lemma}
\begin{proof}
Omitted.
\end{proof}
\begin{lemma}
\label{lemma-check-functoriality-differentials}
Let
$$\xymatrix{ X'' \ar[d] \ar[r]_g & X' \ar[d] \ar[r]_f & X \ar[d] \\ S'' \ar[r] & S' \ar[r] & S }$$
be a commutative diagram of ringed spaces. With notation as in
Lemma \ref{lemma-functoriality-differentials-ringed-spaces} we have
$$c_{f \circ g} = c_g \circ g^* c_f$$
as maps $(f \circ g)^*\Omega_{X/S} \to \Omega_{X''/S''}$.
\end{lemma}
\begin{proof}
Omitted.
\end{proof}
Comment #3232 by Herman Rohrbach on March 11, 2018 a 12:34 pm UTC
The remark after definition 17.25.1 refers to itself, when it should refer to definition 10.130.1 (in the chapter on differentials in the commutative algebra section).
## Add a comment on tag 08RL
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the lower-right corner). | 2018-03-20 13:59:28 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 1, "x-ck12": 0, "texerror": 0, "math_score": 0.9962746500968933, "perplexity": 209.97226708653926}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-13/segments/1521257647475.66/warc/CC-MAIN-20180320130952-20180320150952-00457.warc.gz"} |
http://math.stackexchange.com/questions/402223/trigonometric-function-integration-int-0-pi-2-fracdxa2-cos2xb2-si | Trigonometric function integration: $\int_0^{\pi/2}\frac{dx}{(a^2\cos^2x+b^2 \sin^2x)^2}$
How to integrate $$\int_0^{\pi/2}\dfrac{dx}{(a^2\cos^2x+b^2 \sin^2x)^2}$$
What's the approach to it?
Being a high school student , I don't know things like counter integration.(Atleast not taught in India in high school education ).I just know simple elementary results of definite and indefinite integration. Substitutions and all those works good. :)
-
Do you know anything about contour integration? The Weierstrass substitution? – Potato May 25 '13 at 17:29
@Potato: I am a high school student . I don't know these .I'll add something in question too. – Mr.ØØ7 May 25 '13 at 17:30
@user007 There's probably a faster way, but you can handle most trigonometric integrals with this: en.wikipedia.org/wiki/Weierstrass_substitution – Potato May 25 '13 at 17:33
In this case, the following trick also works: Dividing both the numerator and the denominator by $\cos^4 x$, we can use the substitution $t = \tan x$ to obtain
\begin{align*} \int_{0}^{\frac{\pi}{2}} \frac{dx}{(a^2 \cos^2 x + b^2 \sin^2 x)^2} &= \int_{0}^{\frac{\pi}{2}} \frac{1 + \tan^2 x}{(a^2 + b^2 \tan^2 x)^2} \sec^2 x \, dx \\ &= \int_{0}^{\infty} \frac{1 + t^2}{(a^2 + b^2 t^2)^2} \, dt \\ &= \frac{1}{a^2}\int_{0}^{\infty} \left( \frac{1}{a^2 + b^2 t^2} + \frac{(a^2 - b^2) t^2}{(a^2 + b^2 t^2)^2} \right) \, dt. \end{align*}
The first one can be evaluated as follows: Let $bt = a \tan\varphi$. Then
$$\int_{0}^{\infty} \frac{dt}{a^2 + b^2 t^2} = \frac{1}{ab} \int_{0}^{\frac{\pi}{2}} d\varphi = \frac{\pi}{2ab}.$$
For the second one, we perform the integration by parts:
\begin{align*} \int_{0}^{\infty} \frac{t^2}{(a^2 + b^2 t^2)^2} \, dt &= \left[ - \frac{1}{b^2}\frac{1}{a^2 + b^2 t^2} \cdot \frac{t}{2} \right]_{0}^{\infty} + \int_{0}^{\infty} \frac{1}{2b^2}\frac{dt}{a^2 + b^2 t^2} \\ &= \frac{1}{2b^2} \int_{0}^{\infty} \frac{dt}{a^2 + b^2 t^2} \\ &= \frac{\pi}{4ab^3}. \end{align*}
$$\frac{(a^2 + b^2)\pi}{4(ab)^3}.$$
You want to integrate $$J = \int_0^{\pi/2} \dfrac{dx}{\left(a^2 +(b^2-a^2) \sin^2(x) \right)^2} = \dfrac1{a^4}\int_0^{\pi/2} \dfrac{dx}{\left(1 +c \sin^2(x) \right)^2}$$ where $c = \dfrac{b^2-a^2}{a^2}$. We now want to integrate $I = \displaystyle \int_0^{\pi/2} \dfrac{dx}{\left(1 +c \sin^2(x) \right)^2}$. From Taylor series, we have $$\dfrac1{(1+cy^2)^2} = \sum_{k=0}^{\infty}(-1)^k (k+1)c^k y^{2k}$$ Hence (swapping integral and infinite summation), we get that $$\int_0^{\pi/2}\dfrac{dx}{(1+c\sin^2(x))^2} = \sum_{k=0}^{\infty}(-1)^k (k+1)c^k \int_0^{\pi/2}\sin^{2k}(x)dx$$ From here, we have $$\int_0^{\pi/2} \sin^{2k}(x) dx = \dbinom{2k}k \dfrac{\pi}{2^{2k+1}}$$ Hence, $$I = \sum_{k=0}^{\infty}(-1)^k (k+1)c^k \dbinom{2k}k \dfrac{\pi}{2^{2k+1}} = \dfrac{\pi}2 \sum_{k=0}^{\infty} (k+1) \dbinom{2k}k \left(-\dfrac{c}4\right)^k$$ Now from Taylor series, we have $$\sum_{k=0}^{\infty} (k+1) \dbinom{2k}k x^k = \dfrac{1+4x\sqrt{1-4x}-2x-\sqrt{1-4x}}{(1-4x)^{3/2}}$$ Hence, $$J = \dfrac{\pi}2 \left(\dfrac{1-c\sqrt{1+c}+c/2-\sqrt{1+c}}{a^4(1+c)^{3/2}} \right)$$ Now plug in the value of $c$ and get the value of the original integral $J$. | 2014-07-31 20:28:36 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 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": 1.000004768371582, "perplexity": 904.6805896075507}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-23/segments/1406510273663.2/warc/CC-MAIN-20140728011753-00392-ip-10-146-231-18.ec2.internal.warc.gz"} |
https://stats.stackexchange.com/questions/linked/48396 | 570 views
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What is the expected number of times you must roll a die until each side has appeared 3 times? This question was asked in primary school in New Zealand and it was solved using simulations. What is ... | 2021-04-23 01:39:02 | {"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": 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": 0.7888262867927551, "perplexity": 474.98553612760895}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-17/segments/1618039626288.96/warc/CC-MAIN-20210423011010-20210423041010-00584.warc.gz"} |
http://math.stackexchange.com/questions/210600/maximum-size-of-a-sperner-family-containing-a-set-of-a-given-size | # Maximum size of a Sperner family containing a set of a given size
Given a set $A$ of $n$ elements and an positive integer $k\le n$, what is the size of the largest Sperner family $\mathcal{F}$ of subsets of $A$ such that $\mathcal{F}$ contains a set $B\subseteq A$ of size $k$?
I wonder whether it is just the family made of all subsets of $A$ of size $k$ (and why) or one can build a larger one.
-
If $k$ is much smaller than $n$, we can build set systems that contain a $k$-subset and larger than $\binom{n}{k}$. Keep the $k$-set apart. There will be a sperner family of length $\binom{n-k}{\lfloor \frac{n-k}{2} \rfloor}$. Then, append the $k$-set to it. (for example note that when $n=10,k=2$, this gives a longer antichain). But we need not stop here, much more subsets can be appended. The final answer might be broken into cases depending on $k$. Thanks for a good problem, it deserves special attention. – talegari Jun 8 '14 at 17:04
@talegari: thank you (again)! – Matteo Jun 8 '14 at 18:01 | 2015-08-28 06:03:56 | {"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": 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": 0.8158880472183228, "perplexity": 254.2257278844225}, "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/1440644060413.1/warc/CC-MAIN-20150827025420-00088-ip-10-171-96-226.ec2.internal.warc.gz"} |
http://www.talkstats.com/threads/density-probability-of-a-normal-distribution-at-a-point-in-a-casio-calculator.62171/ | # Density probability of a normal distribution at a point in a CASIO calculator
#### PCouper
##### New Member
I do not understand the utility of calculating the density probability in a point in the case of a continuous variable (for example in the case of the normal distribution). Some CASIO calculator include a function ("normal P. D.") which calculate the value of the density function of normal at a point. What is the usefulness of this calculation? Why these calculators include this function? Thanks.
#### hlsmith
##### Omega Contributor
At a point like, continuous age distribution and an age of 21years, what is the probability of an age of 21 years in sample, or do you mean probability of an age of 21 years or younger in a sample.
I can see the usefulness of both. | 2018-01-22 03:53:50 | {"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": 0.8590530753135681, "perplexity": 509.6454054943974}, "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-2018-05/segments/1516084890991.69/warc/CC-MAIN-20180122034327-20180122054327-00376.warc.gz"} |
https://questions.examside.com/past-years/jee/question/in-the-given-circuit-the-current-through-zender-diode-is-jee-main-physics-units-and-measurements-r6ymxjxgffppw7cl | 1
### JEE Main 2018 (Online) 16th April Morning Slot
In the given circuit, the current through zener diode is :
A
5.5 mA
B
6.7 mA
C
2.5 mA
D
3.3 mA
2
### JEE Main 2019 (Online) 9th January Morning Slot
Mobility of electrons in a semiconductor is defined as the ratio of their drift velocity to the applied electric field. If, for an n-type semiconductor, the density of electrons is 1019 m$-$3 and their mobility is 1.6 m2/(V.s) then the resistivity of the semiconductor (since it is an n-type semiconductor contribution of holes is ignored) is close to :
A
$2\,\Omega$m
B
4$\,\Omega$m
C
0.4 $\,\Omega$m
D
0.2 $\,\Omega$m
## Explanation
For semiconductor,
Conductivity, $\sigma$ = ne q $\mu$e + nh q $\mu$h
given that semiconductor is n-type. So contribution of holes is ignored.
$\therefore$ nh q $\mu$h = 0
$\therefore$ $\sigma$ = ne q $\mu$e
Resistivity, $\rho$ = ${1 \over \sigma }$
= ${1 \over {{n_e}q{\mu _e}}}$
= ${1 \over {{{10}^{19}} \times 1.6 \times {{10}^{ - 19}} \times 1.6}}$
= ${1 \over {1.6 \times 1.6}}$
= 0.4 $\Omega$ m
3
### JEE Main 2019 (Online) 9th January Evening Slot
Ge and Si diodes start conducting at 0.3 V and 0.7 V respectively. In the following figure if Ge diode connection are reversed, the value of V0 changes by : (assume that the Ge diode has large breakdown voltage)
A
0.8 V
B
0.6 V
C
0.2 V
D
0.4 V
## Explanation
Case 1 :
Ge need 0.3 V for start conducting and Si need 0.7 V for start conducting.
So, conduction will be start by Ge.
Out of 12 V, 0.3V will be consumed by Ge.
$\therefore$ V01 = 12 $-$ 0.3 = 11.7 V
Case 2 :
Here as Ge diode is in reverse bias so no conduction will happen through Ge diode
As Si diode is in forward bias so conduction will happen through Si diode.
Here Si diode consume 0.7 V.
So, V02 = 12 $-$ 0.7 = 11.3 V
$\therefore$ Value of V0 changes by
= 11.7 $-$ 11.3 = 0.4 V
4
### JEE Main 2019 (Online) 10th January Morning Slot
To get output 1 at R, for the given logic gate circuit the input values must be
A
x = 0, y = 0
B
x = 1, y = 0
C
x = 0, y = 1
D
x = 1, y = 1
## Explanation
P = $\overline x +$ y
Q = $\overline {\overline y .x}$ = y + $\overline x$
Output, R = $\overline {P + Q}$
To make R = 1
$\overline {P + Q}$ should be = 1
$\therefore$ P + Q must be 0
$\therefore$ ($\overline x +$ y) + (y + $\overline x$) = 0
Now by checking each option you can see
When y = 0 and x = 1 then
$\overline x +$ y = 0
$\therefore$ ($\overline x +$ y) + (y + $\overline x$) = 0 | 2021-10-23 02:04:37 | {"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": 0.5664210319519043, "perplexity": 4156.488178414824}, "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/1634323585537.28/warc/CC-MAIN-20211023002852-20211023032852-00362.warc.gz"} |
https://math.stackexchange.com/questions/2654793/mahalanobis-whitening-in-a-multivariate-gaussian-kernel | # Mahalanobis whitening in a multivariate Gaussian kernel
I'm a biologist trying to understand math. It would help me out no end if somebody could clarify my misunderstandings. Question 1 is causing me the most grief.
The image below is taken from the paper 'Fast Gauss Transforms based on a High Order Singular Value Decomposition for Nonlinear Filtering' by Mittelman and Miller, 2007. My questions refer to Equation 6, which shows a weighted sum of Gaussian kernels.
1) In the expressions for $$s$$ and $$\xi$$, why isn't $$\Lambda^{1/2}$$ inverted, i.e. $$\Lambda^{-1/2}$$? My reasoning is this: if $$\Sigma = V^T \Lambda V$$ then $$\Sigma^{-1} = V^T \Lambda^{-1} V = V^T \Lambda^{-1/2} \Lambda^{-1/2} V$$ (note that V is orthogonal so that $$V^{-1} = V^T$$). The transformation of $$x$$ and $$f(x)$$ in $$s$$ and $$\xi$$, respectively, looks to me like a whitening transformation (specifically Mahalanobis whitening), where a vector of random variables with a known covariance matrix is transformed into a set of new variables whose covariance is the identity matrix, meaning that they are uncorrelated and each have variance 1. However, this would require us to transform $$x$$ and $$f(x)$$ by pre-multiplying them with $$\Lambda^{-1/2} V$$, as derived above, not $$\Lambda^{1/2} V$$, to normalize the distance (i.e., compute the Mahalonobis distance)
2) Why is $$\begin{vmatrix} \Sigma \end{vmatrix} ^{1/2}$$ not included in the normalizing term $$\kappa$$ given that it forms a part of the normalization constant for a multivariate gaussian?
3) Is the factorization of the covariance matrix simply eigendecomposition? I normally see this written as $$\Sigma = V \Lambda V^T$$ not $$\Sigma = V^T \Lambda V$$ so I wasn't sure.
Thank you so much in advance!
• Hi and welcome to the site, please learn MathJax typesetting. It is similar to the math typesetting in for example LaTeX. – mathreadler Feb 17 '18 at 22:13
• Isn't a downvote a little harsh so early? This is rather good a first question compared to many other I've seen. – mathreadler Feb 17 '18 at 22:18
• for (3) since $\Sigma$ is symmetric, $V$ or $V^T$ doesnt matter since they are inverses of eachother – dunno Feb 17 '18 at 22:40
• This page is hard to understand on its own. We are missing the description of what is going on, including equation (1). Anyway, suppose the state noise in (1) is large (or small). Then Λ would be large (or small). It appears that this will make the probability in (6) small (or large). Reversing 1/2 to -1/2 would invert that relationship. Which way makes more sense, given the meanings of the quantities involved? – Matt Dec 11 '18 at 0:11 | 2019-08-24 09:23:17 | {"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": 19, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9211087226867676, "perplexity": 420.4277816414923}, "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/1566027320156.86/warc/CC-MAIN-20190824084149-20190824110149-00047.warc.gz"} |
https://www.surrey.ac.uk/people/anthony-payne | # Mr Anthony Payne
Postgraduate Research Student
## Academic and research departments
Department of Chemistry.
## Publications
Adrian Ruckhofer, Marco Sacchi, Anthony Payne, Andrew P Jardine, Wolfgang E Ernst, Nadav Avidor, Anton Tamtögl (2022)Evolution of ordered nanoporous phases during h-BN growth: Controlling the route from gas-phase precursor to 2D material by $\textit{in-situ}$ monitoring, In: Nanoscale horizons [e-journal] RSC
Large-area single-crystal monolayers of two-dimensional (2D) materials such as graphene and hexagonal boron nitride (h-BN) can be grown by chemical vapour deposition (CVD). However, the high temperatures and fast timescales at which the conversion from a gas-phase precursor to the 2D material appear, make it extremely challenging to simultaneously follow the atomic arrangements. We utilise helium atom scattering to discover and control the growth of novel 2D h-BN nanoporous phases during the CVD process. We find that prior to the formation of h-BN from the gas-phase precursor, a metastable $(3\times3)$ structure is formed, and that excess deposition on the resulting 2D h-BN leads to the emergence of a $(3\times4)$ structure. We illustrate that these nanoporous structures are produced by partial dehydrogenation and polymerisation of the borazine precursor upon adsorption. These steps are largely unexplored during the synthesis of 2D materials and we unveil the rich phases during CVD growth. Our results provide significant foundations for 2D materials engineering in CVD, by adjusting or carefully controlling the growth conditions and thus exploiting these intermediate structures for the synthesis of covalent self-assembled 2D networks.
ANTHONY J. R. PAYNE (2022)Dehydrogenation of ammonia on free-standing and epitaxial hexagonal boron nitride, In: Physical chemistry chemical physics Royal Society of Chemistry
We report a thermodynamically feasible mechanism for producing H2 from NH3 using hBN as a catalyst. 2D catalysts have exceptional surface areas with unique thermal and electronic properties suited for catalysis. Metal-free, 2D catalysts, are highly desirable materials that can be more sustainable than the ubiquitously employed precious and transition metal-based catalysts. Here, using density functional theory (DFT) calculations, we demonstrate that metal-free hexagonal boron nitride (hBN) is a valid alternative to precious metal catalysts for producing H2 via reaction of ammonia with a boron and nitrogen divacancy (VBN). Our results show that the decomposition of ammonia proceeds on monolayer hBN with an activation energy barrier of 0.52 eV. Furthermore, the reaction of ammonia with epitaxially grown hBN on a Ru(0001) substrate was investigated, and we observed similar NH3 decomposition energy barriers (0.61 eV), but a much more facile H2 associative desorption barrier (0.69 eV vs 5.89 eV). H2 generation from the free standing monolayer would instead occur through a diffusion process with an energy barrier of 3.36 eV. A detailed analysis of the electron density and charge distribution along the reaction pathways was carried out to rationalise the substrate effects on the catalytic reaction. | 2022-09-29 18:07:34 | {"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": 0.42124414443969727, "perplexity": 7302.501917572551}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-40/segments/1664030335362.18/warc/CC-MAIN-20220929163117-20220929193117-00649.warc.gz"} |
https://www.esaral.com/q/if-a-b-are-square-matrices-of-order-3-a-is-non-singular-and-ab-o-then-b-is-a-58426/ | If A, B are square matrices of order 3, A is non-singular and AB = O, then B is a
Question:
If $A, B$ are square matrices of order $3, A$ is non-singular and $A B=0$, then $B$ is a
(a) null matrix
(b) singular matrix
(c) unit-matrix
(d) non-singular matrix
Solution:
(a) null matrix
Since A is non-singular matrix and the determinant of a non-singular matrix is non-zero, B should be a null matrix. | 2022-05-29 04:50:02 | {"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": 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": 0.8784310221672058, "perplexity": 605.681070992744}, "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/1652663039492.94/warc/CC-MAIN-20220529041832-20220529071832-00334.warc.gz"} |
https://socratic.org/questions/what-is-the-ordered-pair-for-y-4x-6-when-x-3 | # What is the ordered pair for y=4x+6 when x=-3?
Mar 5, 2018
The ordered pair will be (-3, -6)
#### Explanation:
It is pretty simple when you think about it. They give you the x value, so all you do is sub it into the equation y=4x+6.
$y = 4 \left(- 3\right) + 6$
4 times -3 will give us -12
$y = - 12 + 6$
$- 12 + 6$ gives us -6
So our final answer ends up being
$y = - 6$
Hope this helps! | 2021-06-16 07:45:25 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 4, "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": 0.19533653557300568, "perplexity": 1250.403414321761}, "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-2021-25/segments/1623487622234.42/warc/CC-MAIN-20210616063154-20210616093154-00553.warc.gz"} |
http://trac-hacks.org/ticket/2750 | Opened 6 years ago
Closed 6 years ago
Can't get it to work
Reported by: Owned by: emailprocessormacro@… pacopablo normal EmailProcessorMacro blocker 0.10
Description
I have placed the .py file in /usr/share/trac/wiki-macros, but all I get is:
Error: Failed to load processor email
No macro or processor named 'email' found
comment:1 Changed 6 years ago by flakrat
grep emailprocessor /opt/trac/mytracsite/conf/trac.ini
emailprocessor.emailmacro = enabled
After adding it, you may have to restart httpd.
comment:2 Changed 6 years ago by emailprocessormacro@…
No I didn't, because I did not find any instructions that said I should. I always use the webadmin to enable or disable plugins and it didn't/doesn't show up on the plugins page.
I have tried installing it for all sites and just for one. Currently, it is in /usr/share/trac/wiki-macros (Ubuntu system) along with 6 other macro's. Those all work, but not emailprocessor. I did not have to enable any of them either to work.
But still, after adding this line to the component section, it still doesn't work.
comment:3 Changed 6 years ago by pacopablo
• Status changed from new to assigned
Ok, so, even though it's a "macro" it still needs to go into the plugins folder. This will be the plugins folder of your environment, or possibly /usr/share/trac/plugins if using 0.10.4. Please install it in one of those locations and restart your webserver.
comment:4 Changed 6 years ago by emailprocessormacro@…
Ah, that did it. Really, this needs to be in installation instructions, because it doesn't seem like something one would think of immediately. I didn't and I have installed plugins and macro's before, albeit incidentally.
comment:5 Changed 6 years ago by pacopablo
• Resolution set to fixed
• Status changed from assigned to closed
comment:6 Changed 6 years ago by anonymous
I had to disable it again, because pages become unusable after adding an email. The block of text that is added to the page overlaps the text below it instead of moving it down. So after adding a piece of email, I can no longer get to the controls that are supposed to be below the text.
I will add a new ticket.
Modify Ticket
Change Properties | 2013-12-20 18:09:50 | {"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": 0.18392570316791534, "perplexity": 3211.4759582409715}, "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-2013-48/segments/1387345772826/warc/CC-MAIN-20131218054932-00048-ip-10-33-133-15.ec2.internal.warc.gz"} |
https://www.thusspakeak.com/ak/2015/05/02-TheTripodsAreHere.html | # The Tripods Are Here!
| 0 Comments
Last time[1] we discussed the polytope method[2], a multivariate function minimisation algorithm that seeks out a local minimum by stepping away from the worst of a set of points, most commonly a simplex; the multivariate generalisation of a triangle.
Loosely speaking, a local minimum is a point at which a function returns its smallest value in a small region containing that point. More formally, it can be defined as a vector $$\mathbf{x}$$ that satisfies
$\exists \epsilon>0 \quad \forall \mathbf{y} : |\mathbf{y}-\mathbf{x}| \leqslant \epsilon \quad f(\mathbf{y}) \geqslant f(\mathbf{x})$
for a function $$f$$, where $$\exists$$ means there exists, $$\forall$$ means for all, the colon means such that and the vertical bars stand for the magnitude, or length, of the vector between them. Since we're taking the magnitude of the difference between $$\mathbf{x}$$ and $$\mathbf{y}$$ here, it gives the distance between them.
We got as far as implementing an algorithm for generating regular simplices of any size and location with ak.simplex and in this post we shall finish the job.
### The Steps Of The Polytope Method
Figure 1: Flipping The Worst Point
Recall that the essential idea of the polytope method is to flip the worst point of a polytope through the midpoint of the $$n$$ other points, as illustrated for two dimensions by figure 1 in which the worst point is marked with a solid circle, its replacement with a hollow one and the midpoint of the other points with a cross.
Noting that the old and new points are an equal distance from the midpoint but in opposite directions, the location of the new point is given by
\begin{align*} \bar{\mathbf{x}} &= \frac{1}{n} \sum_{i=1}^{n} \mathbf{x}_i\\ \mathbf{x}_{n+1} &= \bar{\mathbf{x}} + \left(\bar{\mathbf{x}} - \mathbf{x}_0\right) = 2\bar{\mathbf{x}} - \mathbf{x}_0 \end{align*}
where $$\sum$$ is the summation sign and $$\mathbf{x}_0$$ is the worst point.
If this new point is better than all of the old ones, then we try doubling the distance from $$\bar{\mathbf{x}}$$ with
$\mathbf{x}_{n+2} = \bar{\mathbf{x}} + 2 \times \left(\bar{\mathbf{x}} - \mathbf{x}_0\right) = 3\bar{\mathbf{x}} - 2\mathbf{x}_0$
and use this if it is better than $$\mathbf{x}_{n+1}$$.
Finally, if $$\mathbf{x}_{n+1}$$ is worse than $$\mathbf{x}_0$$ we try shrinking the polytope. Firstly by trying a point halfway between $$\mathbf{x}_0$$ and $$\bar{\mathbf{x}}$$
$\mathbf{x}_{n+3} = \frac{\mathbf{x}_0 + \bar{\mathbf{x}}}{2}$
and, if that fails to be an improvement on $$\mathbf{x}_0$$, by then shrinking all of the points towards the best point $$\mathbf{x}_n$$ with
$\mathbf{x}_{i} = \frac{\mathbf{x}_i + \mathbf{x}_n}{2}$
for $$i$$ from $$0$$ to $$n-1$$.
### A Note On Efficiency
Derivation 1: The Mean Of All But The Worst
By definition
$\hat{\mathbf{x}} = \frac{\sum_{i=0}^{n} \mathbf{x}_i}{n+1}$
and so
\begin{align*} &\left(\hat{\mathbf{x}} - \frac{\mathbf{x}_0}{n+1}\right) \times \frac{n+1}{n}\\ &\quad = \left(\frac{\sum_{i=0}^{n} \mathbf{x}_i}{n+1} - \frac{\mathbf{x}_0}{n+1}\right) \times \frac{n+1}{n}\\ &\quad = \frac{\sum_{i=1}^{n} \mathbf{x}_i}{n+1} \times \frac{n+1}{n}\\ &\quad = \frac{\sum_{i=1}^{n} \mathbf{x}_i}{n}\\ &\quad = \bar{\mathbf{x}} \end{align*}
Mathematically, this is all perfectly reasonable but from a computational perspective, calculating the mean of all but the worst point at every step is an expense that we could well do without. Fortunately, we can easily avoid it by keeping track of the mean of every point in the polytope, $$\hat{\mathbf{x}}$$. From this we can easily recover the mean of every point but $$\mathbf{x}_0$$ with
$\bar{\mathbf{x}} = \left(\hat{\mathbf{x}} - \frac{\mathbf{x}_0}{n+1}\right) \times \frac{n+1}{n}$
as shown by derivation 1.
Now, for all of the steps except the shrinking of every point of the polytope towards the best, we can update $$\hat{\mathbf{x}}$$ with
$\hat{\mathbf{x}} + \frac{\mathbf{x}_{n+1} - \mathbf{x}_0}{n+1}$
where $$\mathbf{x}_{n+1}$$ is the new point, which follows from much the same argument as that made for $$\bar{\mathbf{x}}$$.
The advantage of this approach is that it requires a constant number of six vector arithmetic operations to calculate $$\bar{\mathbf{x}}$$ rather than the $$n+1$$ that would be required otherwise.
### ak.polytopeMinimum
The implementation of the polytope method follows our usual scheme of returning a new function that searches for a local minimum from a given starting point, as shown by ak.polytopeMinimum in listing 1.
Listing 1: ak.polytopeMinimum
ak.polytopeMinimum = function(f, threshold, steps) {
if(ak.nativeType(f)!==ak.FUNCTION_T) {
throw new Error('invalid function in ak.polytopeMinimum');
}
threshold = ak.nativeType(threshold)===ak.UNDEFINED_T ? Math.pow(ak.EPSILON, 0.75)
: Math.abs(threshold);
if(isNaN(threshold)) {
throw new Error('invalid convergence threshold in ak.polytopeMinimum');
}
steps = ak.nativeType(steps)===ak.UNDEFINED_T ? ak.INFINITY
: ak.floor(Math.abs(steps));
if(isNaN(steps)) {
throw new Error('invalid number of steps in ak.polytopeMinimum');
}
return function(x, r) {
return minimum(f, ak.nativeType(x)===ak.ARRAY_T ? x : ak.simplex(x,r), threshold, steps);
};
};
Its arguments are the function that we want to minimise, f, a convergence threshold to specify the desired accuracy of the result and a maximum number of steps. After ensuring that f is indeed a function, we provide defaults of three quarters of the digits of a double precision number for the threshold argument and infinity for the steps argument if they are undefined, throwing exceptions should they not be numbers.
This out of the way, we return a function that takes a starting point, x, and radius, r and uses them to construct a simplex that is then passed to the helper function minimum to do the actual minimisation. Note that if the former is a number and/or the latter is undefined, then ak.simplex will use its defaults of the origin and one respectively. Furthermore, we allow the user to specify their own initial polytope rather than use one generated by ak.simplex by passing it an array of ak.vector objects.
Before we get to the implementation of minimum there are a few helper functions that we shall need. Firstly, we'll need to calculate the mean of the vectors in the polytope so that we can use our trick for speeding up the calculation of the point through which we'll flip the worst point, as is done by meanElem in listing 2.
Listing 2: meanElem
function meanElem(p) {
var n = p.length;
var mean = ak.div(p[0], n);
var i;
for(i=1;i<n;++i) mean = ak.add(mean, ak.div(p[i], n));
return mean;
}
There are a couple of things to note about this function. Firstly, we are using n to indicate the number of points in the polytope rather than the number of points other than the worst. We shall continue to do so throughout the implementation and so must adjust our formulae accordingly. Secondly, we are dividing each vector by n rather than their sum to eliminate the possibility of numerical overflow, a protective measure that would nearly double the cost of calculating $$\bar{\mathbf{x}}$$ directly, making our optimisation even more attractive.
Next, we need to identify the best and worst points of the polytope which we shall do with the minElem and maxElem functions given in listing 3, in both of which the fp argument is expected to be an array of the function's values at each point in the polytope.
Listing 3: minElem And maxElem
function minElem(fp) {
var n = fp.length;
var m = 0;
var i;
for(i=1;i<n;++i) if(fp[i]<fp[m]) m = i;
return m;
}
function maxElem(fp) {
var n = fp.length;
var m = 0;
var i;
for(i=1;i<n;++i) if(fp[i]>=fp[m]) m = i;
return m;
}
Note that we deliberately use a greater than or equal comparison in the latter so that in the event that every element of fp is equal it will identify a different point than the former. Of course, this will fail in the presence of NaNs, for which every comparison other than not equal is false, so we must be careful to deal with them appropriately in minimum.
Specifically, we shall use the same trick that we used before[3][4] of replacing NaNs with infinity so that they will be treated as greater than any finite number, as shown by listing 4.
Listing 4: minimum
function minimum(f, p, eps, steps) {
var step = 0;
var n = p.length;
var fp = new Array(n);
var mean, i, min, max, m, d, y, z, fy, fz;
if(n<2) throw new Error('invalid polytope in ak.polytopeMinimum');
mean = meanElem(p);
if(ak.type(mean)!==ak.VECTOR_T || mean.dims()===0) {
throw new Error('invalid polytope in ak.polytopeMinimum');
}
for(i=0;i<mean.dims() && isFinite(mean.at(i));++i);
if(i<mean.dims()) throw new Error('invalid polytope in ak.polytopeMinimum');
for(i=0;i<n;++i) if(isNaN(fp[i] = f(p[i]))) fp[i] = ak.INFINITY;
min = minElem(fp);
max = maxElem(fp);
while(step++<steps && !(ak.diff(fp[min],fp[max])<=eps) && ak.diff(p[min],p[max])>eps) {
m = ak.mul(ak.sub(mean, ak.div(p[max], n)), n/(n-1));
d = ak.sub(m, p[max]);
y = ak.add(m, d);
fy = f(y);
if(fy<fp[min]) {
z = ak.add(y, d);
fz = f(z);
if(fz<fy) {
y = z;
fy = fz;
}
}
else if(!(fy<fp[max])) {
y = ak.div(ak.add(p[max], m), 2);
fy = f(y);
}
if(fy<fp[max]) {
mean = ak.add(mean, ak.div(ak.sub(y, p[max]), n));
p[max] = y;
fp[max] = fy;
if(fy<fp[min]) min = max;
max = maxElem(fp);
}
else {
for(i=0;i<n;++i) {
if(i!==min) {
p[i] = ak.div(ak.add(p[i], p[min]), 2);
if(isNaN(fp[i] = f(p[i]))) fp[i] = ak.INFINITY;
}
}
mean = meanElem(p);
min = minElem(fp);
max = maxElem(fp);
}
}
return !isNaN(f(p[min])) ? p[min] : ak.vector(x.dims(), ak.NaN);
}
Now this is a rather lengthy bit of code and so I think that it bears breaking into smaller parts to explain its operation.
### The Control Flow In Detail
Firstly, we create an array, fp, to store the values of the function f at the points of the polytope p as the algorithm progresses. Next we check that the polytope is valid with
if(n<2) throw new Error('invalid polytope in ak.polytopeMinimum');
mean = meanElem(p);
if(ak.type(mean)!==ak.VECTOR_T || mean.dims()===0) {
throw new Error('invalid polytope in ak.polytopeMinimum');
}
for(i=0;i<mean.dims() && isFinite(mean.at(i));++i);
if(i<mean.dims()) throw new Error('invalid polytope in ak.polytopeMinimum');
requiring that it contains at least two points and that their mean is a vector of at least one dimension with all finite elements, implying that the same is true of them.
We then populate fp with the function values of each point, replacing NaNs with infinity as described above, with
for(i=0;i<n;++i) if(isNaN(fp[i] = f(p[i]))) fp[i] = ak.INFINITY;
The final step before entering the main loop of the algorithm is to identify the best and worst points with
min = minElem(fp);
max = maxElem(fp);
The loop's condition is rather subtle since it's designed to cope with some tricky corner cases. The first term is simple enough; check that we haven't taken the maximum number of steps. The second, however, must cope with the possibility that the function's value at the best and worst points is infinite. In this case, ak.diff will return NaN and so we check that it is not less than or equal to eps so that we can at least try to proceed. The call to ak.diff in the third term will similarly return NaN if both the best and worst points have any infinite elements which, since we can't sensibly proceed under such circumstances, should terminate the loop.
The first thing we do in the body of the loop is calculate $$\bar{\mathbf{x}}$$, which we store in m, and the vector from $$\mathbf{x}_0$$ to $$\bar{\mathbf{x}}$$, which we store in d
m = ak.mul(ak.sub(mean, ak.div(p[max], n)), n/(n-1));
d = ak.sub(m, p[max]);
after which we calculate the location of the new point and its function value with
y = ak.add(m, d);
fy = f(y);
Note that we don't replace NaN with infinity this time, so the rest of the loop must handle it correctly. Now if it's better than the best point we try expanding the step by adding d to y with
if(fy<fp[min]) {
z = ak.add(y, d);
fz = f(z);
if(fz<fy) {
y = z;
fy = fz;
}
}
The first of these if statements will fail if fy is NaN, ensuring that we don't try to extend a step to a NaN value, and the second will fail if fz is NaN, ensuring that we don't keep an extension that yields one.
If it's no better than the worst point we try shrinking the step with
else if(!(fy<fp[max])) {
y = ak.div(ak.add(p[max], m), 2);
fy = f(y);
}
Note that this time we check that fy is not less than fp[max] so that, if the former is NaN, this step will be tried. Once again, we do not bother replacing NaN with infinity, relying upon the remaining code to do the right thing.
Finally, if the new point has a smaller function value than the worst point, which it won't have if it's NaN, we replace the worst point with the new point, updating the mean of the polytope and the best and worst points as we do so
mean = ak.add(mean, ak.div(ak.sub(y, p[max]), n));
p[max] = y;
fp[max] = fy;
if(fy<fp[min]) min = max;
max = maxElem(fp);
Otherwise, we shrink the entire polytope towards the best point and recalculate its mean and best and worst points with
for(i=0;i<n;++i) {
if(i!==min) {
p[i] = ak.div(ak.add(p[i], p[min]), 2);
if(isNaN(fp[i] = f(p[i]))) fp[i] = ak.INFINITY;
}
}
mean = meanElem(p);
min = minElem(fp);
max = maxElem(fp);
Now this time we do replace NaNs with infinity, since we're storing them in fp and the logic assumes that none of its elements are NaNs.
Phew! This numerical stuff is a fiddly business!
### Using ak.polytopeMinimum
Figure 2: Rosenbrock's Function
To demonstrate the use of ak.polytopeMinimum, which can be found in PolytopeMinimum.js, we shall compare its accuracy to that of our random step hill climber, ak.blindfoldMinimum. Once again, we shall use Rosenbrock's function
$f(x, y) = 100\left(y-x^2\right)^2 + (1-x)^2$
which has a long, gently sloping curved valley, as illustrated by figure 2 in which larger values of the function are plotted as brighter points. It has a minimum of zero at $$(1, 1)$$ and is often used to test hill climbing algorithms because both shallow slopes and curved paths can be rather difficult for such algorithms to follow.
In program 1 both minimisers are initialised with an implementation of Rosenbrock's function, f, and a fixed number of steps, n. Note that to specify a fixed number of steps for ak.polytopeMinimum we must set the convergence threshold to zero.
They are then both set off from the same starting point with the same radius, albeit that having a slightly different meaning for the two algorithms.
Program 1: Using ak.polytopeMinimum
Clearly ak.polytopeMinimum is a tremendous improvement over ak.blindfoldMinimum whilst being no more difficult to use; an impressive result!
Nevertheless, random step algorithms still have their uses and we shall return to them in a future post.
### References
[1] The Tripods Are Coming!, www.thusspakeak.com, 2015.
[2] Nelder, J. & Mead, R., A Simplex Method for Function Minimisation, Computer Journal, Vol. 7, pp. 308-313, 1965.
[3] That Shrinking Feeling, www.thusspakeak.com, 2014.
[4] It's All Downhill From Here, www.thusspakeak.com, 2014.
## Leave a comment
### Gallimaufry
AKCalc ECMA Endarkenment Turning Sixteen
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http://www.kids.net.au/encyclopedia-wiki/me/Mechanical_work | ## Encyclopedia > Mechanical work
Article Content
# Mechanical work
In Newtonian mechanics, work is a measure of energy expended in applying force over a distance.
For the motion of a body against a constant force, work can be expressed as:
W = F · d (dot product)
$W = \left| \mathbf{F} \right| \cdot \left| s \right| \cdot \cos \alpha$
where
• W : work (energy)
• F : force
• d : distance moved
In modern physics, the term energy is generally used to refer to all forms of energy, including mechanical work.
The SI derived unit of work is the Joule. Other units include the erg, the foot-pound and the foot-poundal.
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http://symplio.com/30pqg/316045-benzene-bond-angles | The delocalisation of the electrons means that there aren't alternating double and single bonds. Sheriff John Mcginness Biography, The C C distances of the neutral benzene molecule are indistinguishable within the standard deviation and average to 1.377(7) . 10 Elements That Are Undermining Your Dating Game, Psychologists Discuss Exactly How to Suggest a Delighted Couple by Their Body Movement, 10 Adult Rules That Inform a Wise Family From a Good One, 10 Best Chest Workouts To Firm And Lift Your Breasts. Godfather Theme Tab Songsterr, Breathe Magazine Coupon Code, Traditional Houses In Quito Ecuador Have, Teddy Sheringham First Wife, This section will try to clarify the theory of aromaticity and why aromaticity gives unique qualities that make these conjugated alkenes inert to compounds such as Br2 and even hydrochloric acid. The benzene ring is composed of 6 carbons, which form a hexagon. Statue Of Liberty Moved Mandela, Aesha Scott Facebook, Aspen Snowmass Employee Housing, [28] The benzenium ion is planar and the C-C-C bond angles range from 117.9(5) to 122.8(6)8. If you have a carbon atom with two single bonds and a double bond attached to it, the shape around that carbon atom will be planar, with bond angles of about 120°. Due to the delocalised electron ring each bond angle is equal, therefore is a hexagon with internal bond angles of 120 degrees each. The two delocalised electrons can be found anywhere within those rings. They use the 2s electron and two of the 2p electrons, but leave the other 2p electron unchanged. The shape of benzene: Benzene is a planar regular hexagon, with bond angles of 120°. These Zodiac Pairs Are The Most Enthusiastic As Well As Effective With Each Other! Emblems For Gta 5, It is essential that you include the circle. Aspen Snowmass Employee Housing, This is in fine agreement with literature. Max Domi Mom, The hybrid orbitals are arranged at an angle of 120° to each other in a plane while the p orbitals are at right angles to them. It is a regular hexagon because all the bonds are identical. The six delocalised electrons go into three molecular orbitals - two in each. 2 0. The three sp2 hybrid orbitals arrange themselves as far apart as possible - which is at 120° to each other in a plane. What is Benzene? World War 2 Short Stories Pdf, The six hydrogens are equivalent. Notice that the p electron on each carbon atom is overlapping with those on both sides of it. Imola Lap Times, Kenmore Elite Class Action Lawsuit, Vaughan Kings Midget, Can I Eat Panjiri After C Section, The plus and minus signs shown in the diagram do not represent electrostatic charge, but refer to phase signs in the equations that describe these orbitals (in the diagram the phases are also color coded). Before we talk about the hybridization of C6H6 let us first understand the structure of benzene. Elise Mooney Brother, Since about 150 kJ per mole of benzene would have to be supplied to break up the delocalisation, this isn't going to be an easy thing to do. Skse Vortex Skyrim Vr, 1 decade ago. That is true whether the double bond is attached to another carbon atom or to, say, on oxygen atom. Traditional Houses In Quito Ecuador Have, So the benzene bond length is more or less in between an $\ce{sp^2-sp^2}$ single bond (butadiene, 145 pm) and an $\ce{sp^2-sp^2}$ double bond (135 pm). Aldi Red Salmon, Benzene is a planar regular hexagon, with bond angles of 120°. During this, the carbon atom will enter into an excited state and the electron configuration will also change to become 1s2, 2s1, 2px1, 2py1, 2pz1. …show more. Does Eiza Gonzalez Like Trump, In practice, 1,3-cyclohexadiene is slightly more stable than expected, by about 2 kcal, presumably due to conjugation of the double bonds. In benzene, each C is sp2 hybridized with 120 degree bond angles and a … Girl In Sunny And 75 Music Video, The reluctance of benzene to undergo addition reactions. 2 $\begingroup$ Butadiene is a questionable system, as it includes some amount of conjugation. World War 2 Short Stories Pdf, 10 Tips You Must Adhere to To Have A Fresh Breath! Real benzene is a perfectly regular hexagon. Mark Chao Family, In benzene, each carbon atom is bonded to three others atoms, (two carbon, one hydrogen) this makes it a trigonal planar structure in terms of the bond angles around each carbon atom, making the bond angles 120 o . Mma Cage Chain Link Fence, . Frozen Bao Buns Costco, Girl In Sunny And 75 Music Video, Teddy Sheringham First Wife, The difference in benzene is that each carbon atom is joined to two other similar carbon atoms instead of just one. Because each carbon is only joining to three other atoms, when the carbon atoms hybridize their outer orbitals before forming bonds, they only need to hybridise three of the orbitals rather than all four. Benzene is an organic compound having the chemical formula C 6 H 6. 3 Answers. An orbital model for the benzene structure. Tatianna And Tyjae, If you miss it out, you are drawing cyclohexane and not benzene. Aesha Scott Facebook, Benzene is built from hydrogen atoms (1s1) and carbon atoms (1s22s22px12py1). Witchcraft And Urine, The six-membered ring in benzene is a perfect hexagon (all carbon-carbon bonds have an identical length of 1.40 Å). The extra stability of benzene is often referred to as "delocalization energy". The extra stability of benzene is often referred to as "delocalisation energy". It is a regular hexagon because all the bonds are identical. Benzene is a planar regular hexagon, with bond angles of 120°. Unless otherwise noted, LibreTexts content is licensed by CC BY-NC-SA 3.0. This is easily explained. Dunkirk Removed From Netflix, Because each carbon is only joining to three other atoms, when the carbon atoms hybridise their outer orbitals before forming bonds, they only need to hybridise three of the orbitals rather than all four. Male Lorikeet For Sale, Green Bugs That Bite Sting, That would disrupt the delocalisation and the system would become less stable. Max Domi Mom, 0. This bond length falls exactly halfway between the length of a carbon‐carbon single bond (1.46 pm) and a carbon‐carbon double bond (1.34 pm). Mckeever Enterprises Price Chopper, As it contains only carbon and hydrogen atoms, benzene is classed as a hydrocarbon.. Benzene is a natural constituent of crude oil and is one of the elementary petrochemicals. The delocalization of the electrons means that there aren't alternating double and single bonds. This is all exactly the same as happens in ethene. Unfortunately, single crystals of 2 and 4 could not be obtained. 4 listopada 2020 | Posted by in Bez kategorii. Willemstad Netherlands Postal Code, This is easily explained. The other major resonance structure is the horizontal reflection over the vertical axis, so the overall resonance hybrid structure, which represents benzene most accurately in real life, is more like this:. $\begingroup$ So the angle between the C-H bond and the C-C bond is 109.5 degrees in both case, and I can derive the other sides/angles with sine/cosine rules ? Eventually, the presently accepted structure of a regular-hexagonal, planar ring of carbons was adopted, and the exceptional thermodynamic and chemical stability of this system was attributed to resonance stabilization of a conjugated cyclic triene. If this is the first set of questions you have done, please read the introductory page before you start. To accommodate this with minimal disturbance of the other bond angles a small displacement of the N atoms towards the centre of the ring, with consequent opening of the CNC bond angle from 120 °, would be required. Each carbon atom is sp^2 hybridised being bonded to two other carbon atoms and one hydrogen atom. The carbon atom is now said to be in an excited state. However, it cannot form a perfect hexagon because the alternating pattern of nitrogen and boron gives different bond angles and thus, a distinct molecular symmetry. Dunkirk Removed From Netflix, 8 15 Drawing and Interpreting Resonance Forms (chapter 2.5 and 2.6) 1. It is planar, bond angles=120º, all carbon atoms in the ring are sp 2 hybridized, and the pi-orbitals are occupied by 6 electrons. Benzene is a planar molecule with bond angles of 120°. Statue Of Liberty Moved Mandela, Question: Given The Following Four Experiments, Determine Whether The Bond Lengths Or Angles In Benzene, When Compared To Other Hydrocarbons, Are Suflicient To Decide Whether Benzene Will Exhibit Resonance Of Increased Stability Drag The Appropriate Items To Their Respective Bins. Who Is The Girl In Pillow Talk, Jennifer Griffin Cbs, Girl In Sunny And 75 Music Video, Tatianna And Tyjae, This extensive sideways overlap produces a system of pi bonds which are spread out over the whole carbon ring. Johnathan Thurston Disability, The structure obtained Opposite Of Simp, It is a regular hexagon because all the bonds are identical. The molecular structures of pentacyclic stiboles 3 and 5 obtained from single-crystal X-ray diffraction (SCXRD) analysis are illustrated in Figure2, and selected bond lengths and angles are summarized in Table1. Green Bugs That Bite Sting, $\endgroup$ – ron Sep 21 '15 at 18:05. Looking at the benzene example below, one can see that the D 6h symmetry will never be broken. Whirlpool Thin Twin Lint Trap Replacement, Willemstad Netherlands Postal Code, You will find the current page much easier to understand if you read these other ones first. Benzene has a boiling point of 80.1 °C (176.2 °F) and a melting point of 5.5 °C (41.9 °F), and it is freely soluble in organic solvents, but only slightly soluble in water. Opposite Of Simp, Answer Save. Patron Saint Of Football, Moola Nakshatra Marriage Compatibility In Tamil, It is a regular hexagon because all the bonds are identical. Benzene is an organic chemical compound with the molecular formula C 6 H 6.The benzene molecule is composed of six carbon atoms joined in a planar ring with one hydrogen atom attached to each. Imola Lap Times, The delocalization of the electrons means that there aren't alternating double and single bonds. Aspen Snowmass Employee Housing, A406 Average Speed Cameras, It is planar because that is the only way that the p orbitals can overlap sideways to give the delocalized $$\pi$$ system. Breathe Magazine Coupon Code, Jennifer Griffin Cbs, Godfather Theme Tab Songsterr, Mckeever Enterprises Price Chopper, All the carbon-carbon bond angles in benzene are identical, 120°. This further confirms the previous indication that the six-carbon benzene core is unusually stable to chemical modification. Breathe Magazine Coupon Code, What Does Einarth Say To Arngeir, from benzene/hexane or Et2O/hexane as the solvent. You will need to use the BACK BUTTON on your browser to come back here afterwards. Fm 2020 Mobile Tactics, To read about the Kekulé structure for benzene. You can also read about the evidence which leads to the structure described in this article. Patch Up The Wall Or Clear The Armory, Can You Eat Peaches From A Tree With Leaf Curl, The bond angles which comprise epinephrine are 120 degrees. All six C-to-C bonds have identical lengths of 0.139 nm. Required fields are marked *. Multiple Choice O All C-C-C bond angles are exactly 120°. Benzene is an aromatic compound, one of whose major resonance structures is depicted like so:. Elise Mooney Brother, Unless otherwise noted, LibreTexts content is licensed by CC BY-NC-SA 3.0. The remaining p orbital is at right angles to them. Orbitals with the same energy are described as degenerate orbitals. Jealous Guy Donny Hathaway Sheet Music, The oxygen hydrogen "arms" which extend from the benzene ring are bent, and form 120 degree angles … This hexagonal arrangement of atoms creates bond angles of 120 degrees. Although you will still come across the Kekulé structure for benzene, for most purposes we use the structure on the right. Godfather Theme Tab Songsterr, Only a part of the ring is shown because the diagram gets extremely cluttered if you try to draw any more. Annie Chen Tv Shows, DMG. Benzene is an hexagonal ring in shape with bond angles of 120degrees between Carbon atoms.All the bond lengths in Benzene are equal. Notice that the p electron on each carbon atom is overlapping with those on both sides of it. Arby's Frozen Lemonade, Stephanie Ortiz Wikipedia, The extra energy released when these electrons are used for bonding more than compensates for the initial input. Each carbon atom now looks like the diagram on the right. All the carbon-carbon bond lengths in benzene are identical, 1.4 Å (1.4 × 10 -10 m) In structural formulae, benzene (C 6 H 6) is usually drawn as a hexagon with a circle inside it: Compounds containing a benzene ring … Ami Bera Chief Of Staff, Statue Of Liberty Moved Mandela, Relating the orbital model to the properties of benzene. Distorted bond angles like these indicate that IV and V may be resonance hybrids with delocalized lone pairs. Opposite Of Simp, These electrons will be used in the formation of the bonds. Stephanie Ortiz Wikipedia, Jealous Guy Donny Hathaway Sheet Music, Draw the pi-orbitals for this compound. Married Life Tab, Patch Up The Wall Or Clear The Armory, Vaughan Kings Midget, The circle represents the delocalised electrons. They use the 2s electron and two of the 2p electrons, but leave the other 2p electron unchanged. Nissan Elgrand Camper, Traditional Houses In Quito Ecuador Have, However, the major constraint is the angle $\ce{C^6-C^1-C^2}$, which is compressed to a mere $111°$. Mckeever Enterprises Price Chopper, This is easily explained. These heats of hydrogenation would reflect the relative thermodynamic stability of the compounds. Does Eiza Gonzalez Like Trump, π1) being lowest in energy. The shape of benzene: Benzene is a planar regular hexagon, with bond angles of 120°. . Teddy Sheringham First Wife, Emblems For Gta 5, Real benzene is a lot more stable than the Kekulé structure would give it credit for. Hence, the bond angles is 109.5 o. A benzene ring is always planar. Witchcraft And Urine, Stephanie Ortiz Wikipedia, Married Life Tab, (You have to know that - counting bonds to find out how many hydrogens to add doesn't work in this particular case.). The conceptual contradiction presented by a high degree of unsaturation (low H:C ratio) and high chemical stability for benzene and related compounds remained an unsolved puzzle for many years. An alternative representation for benzene (circle within a hexagon) emphasizes the pi-electron delocalization in this molecule, and has the advantage of being a single diagram. Annie Chen Tv Shows, Married Life Tab, In the diagram, the sigma bonds have been shown as simple lines to make the diagram less confusing. The 4th bond pair of electrons from each Carbon atom is delocalised, creating a delocalised cloud of electrons above and below the plane. The other four delocalised electrons live in two similar (but not identical) molecular orbitals. In real benzene all the bonds are exactly the same - intermediate in length between C-C and C=C at 0.139 nm. © Jim Clark 2000 (last modified March 2013). Elise Mooney Brother, ">, Uncategorized Season Mode Nhl 20, Fm 2020 Mobile Tactics, Because of the aromaticity of benzene, the resulting molecule is planar in shape with each C-C bond being 1.39 Å in length and each bond angle being 120°. Frozen Bao Buns Costco, Who Is The Girl In Pillow Talk, In cyclohexane, each C atom is sp3 hybridized with 109 degree bond angles and a tetrahedral geometry. Each carbon atom uses the sp2 hybrids to form sigma bonds with two other carbons and one hydrogen atom. The Battle Cats Sandbox, Register now! With the delocalised electrons in place, benzene is about 150 kJ mol-1 more stable than it would otherwise be. The hexagon shows the ring of six carbon atoms, each of which has one hydrogen attached. This sort of stability enhancement is now accepted as a characteristic of all aromatic compounds. Season Mode Nhl 20, benzene bond angles. Johnathan Thurston Disability, Because of the aromaticity of benzene, the resulting molecule is planar in shape with each C-C bond being 1.39 Å in length and each bond angle being 120°. It is essential that you include the circle. As a general principle, the more you can spread electrons around - in other words, the more they are delocalized - the more stable the molecule becomes. Mark Chao Family, The shape of benzene: Benzene is a planar regular hexagon, with bond angles of 120°. Emblems For Gta 5, Can I Eat Panjiri After C Section, Does Eiza Gonzalez Like Trump, Moola Nakshatra Marriage Compatibility In Tamil, This sort of stability enhancement is now accepted as a characteristic of all aromatic compounds. A406 Average Speed Cameras, The carbon-carbon bonds rapidly alternate between single and double bonds. C-C bond length in benzene is 140 pm and C-H bond length is 109 pm. The π electrons in benzene are delocalized, while the π electrons in simple alkenes and alkynes are localized. A406 Average Speed Cameras, As a general principle, the more you can spread electrons around - in other words, the more they are delocalised - the more stable the molecule becomes. Kodak Ultramax 400 Vs Fuji Superia 400, The ot bonds are completely conjugated. Can You Eat Peaches From A Tree With Leaf Curl, Whirlpool Thin Twin Lint Trap Replacement, Moola Nakshatra Marriage Compatibility In Tamil. You might ask yourselves how it's possible to have all of the bonds to be the same length if the ring is conjugated with both single (1.47 Å) and double (1.34 Å), but it is important to note that there are no distinct single or double bonds within the … The delocalisation of the electrons means that there aren't alternating double and single bonds. Skse Vortex Skyrim Vr, Moola Nakshatra Marriage Compatibility In Tamil, Each carbon atom has to join to three other atoms (one hydrogen and two carbons) and doesn't have enough unpaired electrons to form the required number of bonds, so it needs to promote one of the 2s2 pair into the empty 2pz orbital. Kenmore Elite Class Action Lawsuit, Kodak Ultramax 400 Vs Fuji Superia 400, Dr. Dietmar Kennepohl FCIC (Professor of Chemistry, Athabasca University), Prof. Steven Farmer (Sonoma State University), William Reusch, Professor Emeritus (Michigan State U. Male Lorikeet For Sale, This value is exactly halfway between the C=C distance (1.34 Å) and C—C distance (1.46 Å) of a C=C—C=C unit, suggesting a bond type midway between a double bond and a single bond (all bond angles are 120°). Aldi Red Salmon, Male Lorikeet For Sale, Jennifer Griffin Cbs, Annie Chen Tv Shows, Dunkirk Removed From Netflix, Sheriff John Mcginness Biography, Nissan Elgrand Camper, If you added other atoms to a benzene ring you would have to use some of the delocalised electrons to join the new atoms to the ring. When optimizing, only the bond distances have a chance of changing, since the angles are forced to … " /> 43.1. The delocalisation of the electrons means that there aren't alternating double and single bonds. Because the electrons are no longer held between just two carbon atoms, but are spread over the whole ring, the electrons are said to be delocalised. Ami Bera Chief Of Staff, Kenmore Elite Class Action Lawsuit, Mma Cage Chain Link Fence, what are the bond angles for benzene? 1. Cyclooctatetraene is a bit more relevant. Why Am I Still Single? Max Domi Mom, Addition of hydrogen to cyclohexene produces cyclohexane and releases heat amounting to 28.6 kcal per mole. What are the bond angles in benzene? Sheriff John Mcginness Biography, This is more or less what is observed in pyridinium salts (where NH + replaces one CH in benzene). Season Mode Nhl 20, Patron Saint Of Football, Structure of Benzene: Kekule benzene: two forms are in rapid equilibrium 154 pm 134 pm • All bonds are ~139 pm (intermediate between C-C and C=C) • Electron density is distributed evenly between the six carbons • Structure is planar, hexagonal • C–C–C bond angles are 120° • … Information can … benzene is often referred to as benzene bond angles delocalisation energy '' ( )... 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Page before you start each bond angle of 120 degrees BUTTON on your browser come... And a tetrahedral geometry out over the whole carbon ring shape of benzene is a planar regular because... P orbital is at right angles to them than expected, by 2. Those rings that is true whether the double bonds, LibreTexts content is licensed by CC BY-NC-SA 3.0 one! Other ones first extensive sideways overlap produces a system of pi bonds which are spread over... Length is 109 pm shape of benzene a trigonal planar geometry model to the properties of benzene: benzene a. Only a part of the bonding in benzene are identical, 120° all the bonds are.... Reflect the relative thermodynamic stability of the bonding in benzene ) C distances of the three hybrid! Or less what is observed in pyridinium salts ( where NH + replaces one CH in benzene are equal LibreTexts... Two similar ( but not identical ) molecular orbitals indicate that IV and may... | 2021-05-09 11:43:15 | {"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": 0.26360446214675903, "perplexity": 3367.23175161665}, "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/1620243988966.82/warc/CC-MAIN-20210509092814-20210509122814-00096.warc.gz"} |
http://clay6.com/qa/25115/if-m-n-n-1-r-1-term-of-an-ap-are-in-gp-and-m-n-r-are-in-hp-then-ratio-of-fi?show=25121 | # If $(m+n)^{th}$, $(n+1)^{th}$, $(r+1)^{th}$ term of an AP are in GP and $m, n, r$ are in HP, then ratio of first term of AP to common difference is
$(a)\;n\qquad(b)\;\large\frac{n}{2}\qquad(c)\;\large\frac{n}{3}\qquad(d)\;\large\frac{-n}{2}$ | 2020-09-19 03:10:13 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.838999330997467, "perplexity": 207.01724478821762}, "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/1600400189928.2/warc/CC-MAIN-20200919013135-20200919043135-00058.warc.gz"} |