url
stringlengths 14
2.42k
| text
stringlengths 100
1.02M
| date
stringlengths 19
19
| metadata
stringlengths 1.06k
1.1k
|
|---|---|---|---|
https://bt.gateoverflow.in/46/gate2018-46
|
# GATE2018-46
A microbial strain is cultured in a $100$ L stirred fermenter for secondary metabolite production. If the specific rate of oxygen uptake is $0.4 \: h^{-1}$ and the oxygen solubility in the broth is $8$ mg/L, then the volumetric mass transfer coefficient $(K_La)$ (in $s^{-1}$) of oxygen required to achieve a maximum cell concentration of $12$ g/L is (up to two decimal places) ___________
edited
## Related questions
1
At the end of a batch culture, glucose solution is added at a flow rate of $200 \: ml/h$. If the culture volume after $2$ h of glucose addition is $1000$ ml, the initial culture volume (in ml) is _________
2
An aqueous solution containing $6.8$ mg/L of an antibiotic is extracted with amyl acetate. If the partition coefficient of the antibiotic is $170$ and the ratio of water to solvent is $85$, then the extraction factor is ________
Moist heat sterilization of spores at $121 ^\circ C$ follows first order kinetics as per the expression: $\dfrac{dN}{dt}=-k_dN$ where, $N$ is the number of viable spores, $t$I is the time, $k_d$ is the rate constant and $\dfrac{dN}{dt}$ is the rate of change ... reduce the number of viable spores from an initial value of $10^{10}$ to a final value of $1$ is (up to two decimal places) __________
|
2022-01-17 09:26:48
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 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": 0.8666276335716248, "perplexity": 1343.321024373002}, "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/1642320300533.72/warc/CC-MAIN-20220117091246-20220117121246-00247.warc.gz"}
|
https://hal.inria.fr/hal-01418653v2
|
# Computing solutions of linear Mahler equations
Abstract : Mahler equations relate evaluations of the same function $f$ at iterated $b$th powers of the variable. They arise in particular in the study of automatic sequences and in the complexity analysis of divide-and-conquer algorithms. Recently, the problem of solving Mahler equations in closed form has occurred in connection with number-theoretic questions. A difficulty in the manipulation of Mahler equations is the exponential blow-up of degrees when applying a Mahler operator to a polynomial. In this work, we present algorithms for solving linear Mahler equations for series, polynomials, and rational functions, and get polynomial-time complexity under a mild assumption. Incidentally, we develop an algorithm for computing the gcrd of a family of linear Mahler operators.
Document type :
Journal articles
Cited literature [20 references]
https://hal.inria.fr/hal-01418653
Contributor : Frédéric Chyzak Connect in order to contact the contributor
Submitted on : Tuesday, April 10, 2018 - 3:02:59 PM
Last modification on : Thursday, June 10, 2021 - 3:47:40 AM
### File
mahlersols.pdf
Files produced by the author(s)
### Citation
Frédéric Chyzak, Thomas Dreyfus, Philippe Dumas, Marc Mezzarobba. Computing solutions of linear Mahler equations. Mathematics of Computation, American Mathematical Society, 2018, 87, pp.2977-3021. ⟨10.1090/mcom/3359⟩. ⟨hal-01418653v2⟩
Record views
|
2021-10-25 20:59:01
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.2934657037258148, "perplexity": 2489.5178363871796}, "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/1634323587767.18/warc/CC-MAIN-20211025185311-20211025215311-00719.warc.gz"}
|
https://eng.libretexts.org/Courses/University_of_Arkansas_Little_Rock/IFSC_4399_-_The_Internet_of_Things_(IoT)/The_Linux_crontab_scheduling_events
|
# The Linux crontab, scheduling events
Scheduling data collection events is critical for highly automated IoT systems.
Programs can be scheduled to run at boot time or anytime throughout the day.
##### The @ schedules:
# string meaning
# ------ -----------
# @reboot Run once, at startup.
# @yearly Run once a year, "0 0 1 1 *".
# @annually (same as @yearly)
# @monthly Run once a month, "0 0 1 * *".
# @weekly Run once a week, "0 0 * * 0".
# @daily Run once a day, "0 0 * * *".
# @midnight (same as @daily)
# @hourly Run once an hour, "0 * * * *".
The above @ schedules demonstrate the shorthand for some event timing.
For example @hourly is equivalent to "0 * * * *"
Examples of cronjobs:
1. I want to know if the power goes out where I live. If the power fails then the RPi does not send a marker to the cloud server at the scheduled time. My Android phone is programmed to check the existence of the marker on the server. If the marker exists it deletes the marker. If the marker does not exist I get a text to speech message on the phone indicating that a failure occurred.
2. For some networks the RPi gets a different IP when it boots up. To avoid having to connect a monitor and keyboard to get the new IP a boot cron is used to email the new IP to the user.
# Check to see if a crontab already exists.
crontab -l
# If a crontab already exists then make a backup so that any previous work is not lost.
crontab -l > crontab.org
If a crontab does not exist for the user the following message is displayed.
The above output is from an old Red Hat server, output for a different OS may not be the same.
There are two ways to approach editing of a crontab.
1. Edit the crontab directly
2. Edit a file containing the cron-table then initialize the crontab using the file.
The following is an example of editing the crontab directly.
Note; for the first time edit the choice of editors is required.
In the above example saving the edits makes the changes active.
However there is no backup of your cron-table.
The following method makes a backup.
crontab -l > crontab.bak
Alternatively start with a cron-table file, edit that file, and make the data in new table file active.
git clone https://github.com/phwilliams256/Example-crontab-for-Raspberry-Pi.git
After cloning the directory you can change to the directory and see the containing the crontab source.
cd Example-crontab-for-Raspberry-Pi
in the directory Example-crontab-for-Raspberry-Pi you will find the crontab source file named crontabRPi.scr.
Note, there are no active lines in this file (they are commented out). Activate any lines of interest by removing the comment ‘#’, saving the file, and activating the changes with the following command
The following command puts the content of cron-table file (crontabRPi.scr) in the active crontab.
crontab crontabRPi.scr
# Confirm the changes.
crontab -l
This second method insures that there always exists a document of the current crontab (crontabRPi.scr). This document will be saved with any backup of the user account data.
Issues to be aware of;
In most cases the full path to the application is required.
The full path to the input and output files is always required. This is because the crontab has no knowledge of the user’s path only the system path.
pi@serverNine$cat touchFilesExmaple.sh #!/bin/bash # For scripts run on the command line the relative path to the data is best # The directory containing the script can be copied and the script run. touch ./touch.tmp # For scripts run through the crontab the full path must be # given to all input and output files. mkdir -p /home/pi/temperatureSensor touch /home/pi/temperatureSensor/touch.tmp In the example script above the “touch” command is known to be in the system path. If it was not in the path or we did not know it was in the path then give the full path to the program “touch”. pi@serverNine$ which touch
/usr/bin/touch
mkdir -p /home/pi/temperatureSensor
/usr/bin/touch /home/pi/temperatureSensor/touch.tmp
##### Useful patterns:
# minute (0-59),
# |
# |
# Mins
0,30
# Run the command at the top and bottom of the hour
# hour (0-23),
# |
# |
# Hours
0-6,21-23
# Only when I am sleeping
# hour (0-23),
# |
# |
# Hours
1,3,5,7,9,11,13,15,17,19,21,23
# Every other hour, odd hours
# hour (0-23),
# |
# |
# Hours
0,2,4,6,8,10,12,14,16,18,20,22
# Every other hour, even hours
# hour (0-23),
# |
# |
# Hours
*/2
# Every two hours
# minute (0-59),
# |
# |
# Mins
*/10
# Every 10 minutes
# day of the week (0-7 with 0=7=Sunday).
# |
# |
# Day of
# the week
1-6
# Monday through Saturday, not Sunday
Cron examples can be found here;
https://crontab.guru/examples.html
The Linux crontab, scheduling events is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.
|
2022-12-01 18:04:13
|
{"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": 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": 0.22236362099647522, "perplexity": 5554.140060952728}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-49/segments/1669446710829.5/warc/CC-MAIN-20221201153700-20221201183700-00110.warc.gz"}
|
https://www.prc68.com/I/Labophot.html
|
# Nikon Labophot Microscope
Fig 1 Fig 2
Background
History
Labophot Description
Microscope Photos
Problem
Fine Focus Plastic Gear
Rubber Eye Guards
Condenser Pinion
Subject Photos
Prepared Slides
Pond Water, Images
Hummingbird Feeder Sugar Water
Major Components
Y-R Stand
Lamp House
Monocular
Binocular
Eclipse Binocular (Infinity, not normally for Labophot)
Ultra Wide (UW) Trinocular, ISO Camera Port
Dual View (Teaching)
Trinocular type-F
Trinocular type-T
Combined binocular head and vertical illuminator
Eyepieces
Viewing 20.3 mm Tube
Wide Field Viewing 30mm Tube
Nikon CFI 10x/20 (Eclipse or any 30mm tube)
Vertical Illuminator (Epi)
Transformer
Objective Noise Piece
Objectives
Objectives with 45mm Parafocal Distance
Infinity CFI Objectives
K.D. Kemp 8 Form Diatomaceae Test Slide
Phase Telescope
Slide Holder
Stage
Condenser
Daylight Filter
Polarized Light Kit
Epi Fluorescence centering alignment objective
Patents
Related Web Pages
## Background
Getting this microscope came about because of my interest in Macro photography (Digital Photography 202: Close-Up, Macro & Micro, Nikon Multiphot) developing into an interest in Micro photography. Some of the methods of Micro photography add a camera to a microscope and the old Unitron microscope I have is not suitable because when a camera is mounted on the tube it pushes the tube down.
The newer microscopes have the eyepiece fixed to the microscope stand so it does not move at all. This allows mounting a Nikon DSLR camera on a microscope.
Because I've used Nikon cameras for over 50 years I choose Nikon as the brand of microscope I'd like to get. They have been making them for a long time.
The Nikon booklet How To Use A Microscope and Take a Photograph is based on the "phot" family of microscopes and CF optics. But it was also written when film cameras were the only type there was.
You can easily see 1 um features (see: calibration photos in camera adapter comparison) using a 10x objective and 10x eyepiece or a 2x camera adapter. With a 100x objective you should be able to see details at the level of 0.1 um or 100nm.
This is a compound microscope (Wiki). Simple Microscopes have a single lens yet high (>20x) power than magnifiers (Wiki). Another class of optical microscope is the stereo microscope that allow 3D viewing, necessary when working on small parts, like Surface Mount Technology.
## History
Microscopes (Wiki) were popularized in the late 1600 by Antonie van Leeuwenhoek (Wiki). These allowed seeing never before seen views of pond life. They consisted is a single spherical lens made by forming a ball on a very narrow diameter piece of Borosilicate glass in a Bunson burner (Wiki). These were called "simple" microscopes because of the single lens.
Later the "compound" microscope (Wiki) became the standard type of microscope. It has separate objective and eye piece optical elements mounted in a hollow tube. In 1896 the Royal
Microscopical Society (Wiki: RMS) specified many of the dimensions of the compound microscope so that parts would be interchangeable (Wiki). So the development of the compound optical microscope started in 1896. For many decades the design looked very much like the Unitron.
It's my understanding that virtually all the dimensions were standardized making this design very flexiable. For example the source of light might be a flat/concave mirror, an A.C. line powered lamp (shown at right) or a LED based light. In a similar way a standard 1" x 3" glass slide can be held by a couple of leaf springs (shown at right) or an X-Y stage can be fitted to make searching the specimen easier. Later X-Y stages were calibrated in distance to make finding something easier. The newest X-Y stages are computer controlled to support automated operations. The hollow tube is part of the 160mm tube standard. The "power" of the objectives and eye pieces are found by dividing the tube length by the focal length of either of the optics. So a 16mm FL objective or eye piece would be labeled as 10X. When they are mounted in the tube the overall power is 100X. At this time the optical corrections were implemented using a combination of the eye piece and the objective so you needed these parts to be from the same manufacturer and maybe even same model series. The Nikon model G (1960s & 1970s) and model S (1960s) are an improvement on the Unitron in that they have the eyespiece and objectives fixed to the stand (i.e. only the stage moves for focus), but otherwise they are very similar. Classic Unitron Compound Optical Microscope circa 1950 - 1960 The first Nikon microscope was made in the early 1900s and fundamentally it is identical to the much more modern Unitron. So, for at least the first 70 years microscopes were pretty much the same. Nikon Product History: stereo microscope 1954, phase contrast compound microscope 1955 1976 marks the introduction of the Biophot and Metaphot microscopes that use the Chrome Free (CF) objectives and eye pieces. These still use the standard 160mm tube length and they also use the DIN standard 45mm parafocal distance (Wiki) (objective mounting flange to focused point). See my list of the "phot" microscopes that all used this technology. The "phot" means microscopes that were designed to support photography. But at that time this meant 35mm roll film, 120 roll film or 4"x5" sheet film since digital cameras were a long way in the future. This means that there was a lot extra effort required to support things like getting good focus, the correct exposure and the correct scale factor. Note: The head sits in a fixed position on the stand, it does not move up and down when the focus is changed. Focus only moves the stage that holds the specimen. In a like manner the objective is fixed to the stand. The T version eyepiece tube (head) is trinocular, i.e. two oculars for eyes and one ocular for a film camera. The head can be interchanged with other heads, or can be lifted up and other optical elements inserted such as a vertical illuminator. The lamp house is at the lower center rear and the light passes through a series of filters that can be switched in or out (buttons along right side of base). There's room above where the light shines up and the bottom of the stage to insert various optional condensers. All these things are not needed when a modern Digital Single Lens Reflex (Wiki: DSLR) camera is used that has built-in exposure control and "Live View" mode with magnification to allow accurate focus on the imaging chip. Starting in the 1990s Nikon introduced the Eclipse series microscopes. They use Chrome Free Infinity optics (CFI60). The 60 refers to the parafocal distance which was changed from the DIN 45mm parafocal distance to a Nikon 60mm parafocal distance. There is no longer any standard tube length instead the light output from the objective is in parallel rays that require a "tube lens" to be brought to a focus for the eye piece or camera. This allows more room for optical elements and makes for a higher quality image because putting glass elements into a converging light path, like the 160mm system, causes distortions which the infinity system eliminates with the parallel rays. The Eclipse microscopes and their optical elements are still current and so expensive.
## Labophot Description
Labophot Microscope Manual Fig 15 Optical Path The lamp is mounted to the heatsink/lamp house at the rear center. It comes out with the heat sink. If you look at the diagram the Field diaphragm is shown a little to the left of it's adjusting knob which is located just behind the Field lens and filter holder, see Fig. 1 where the top of the adjustment knob just sticks up a little. The field lens is at the front center pointing up for diascopic (transmitted light) illumination. A daylight color correction filter can be placed on top of the lens if needed for photography with daylight file, but with a digital camera shooting RAW images there's not need since color correction can be done in Photoshop. This a a 160mm tube length system and it uses the DIN parafocal distance of 45mm. The objective threads are RMS. Note when an accessory is installed between the stand and eyepiece tube (head) it has optics that correct for the change in distance to maintain an optical 160mm tube length. With the new CFI60 system these correction optics are no longer needed. One thing that will be checked is the use of an infinity objective with RMS threads on the Labophot. I've been told that it works even though the optical system is missing a Tube Lens.
A number of the Nikon "phot" family of microscopes seem to be very similar. The Biophot, Fluophot, Alphaphot, Metaphot, Labophot and Optiphot all seem to have a stand that's based on a common design, interchangeable heads and other accessories. If you have a catalog that covers this please let me know. Like the first compound microscopes there seem to be many Nikon standards and so this microscope has many options in terms of how it's configured. But I expect only a very few of these options cross brand name boundries.
Like all the phots the Labophot was designed to support micro photogrphy and to use the Chrome Free (CF) eye pieces and objectives (CF 160mm/45mm). But it will also work with the CF objectives that are the infinity type since these have the RMS threads like the other CF objectives. The newer CFI60 objectives can not be used because they have larger (25mm) threads.
The stock illumination system in the Labophot is diascopic (transmitted light). There is provision for using any one of a number of alternate condensers. An optional vertical illuminator can be fitted to support episotic (reflected) illumination like would be needed to metallurgical work. Or an optional Fluorscense vertical illumination can be fitted and there are maybe three models that hold one, two or four filter/beam splitter cubes. In addition Phase Contrast (Phase 3 fitted to this scope) and/or Differential interference contrast (DIC) optics can be fitted. Polarized light is another option.
There are a large number of CF objectives that optionally can be fitted.
### Versions
This is a plain Labophot, but there's also a Labophot 2.
The Labophot 2 has a different fine focus drive that was supposed to resolve the problem with the first fine focus drive, but it didn't work, so they both have a habit of breaking the small plastic gear. There are probably other differences. See Problems.
## Microscope Photos
Fig 1 Right side Fig 2 Front Fig 3 Condenser and stage Fig 4 Left side Fig 5 Stage Removed Note the pinch screw at the center front of the stage when screwed all the way in locks the stage, but when backed off a little allows the stage to rotate (note the circular support structure in the photo below. When the screw is backed off more the stage can be removed as shown below (note the screw is still installed). Fig 6 Abbe 1.25 Condenser & Ph3 slider The lever in the front can be moved left or right to adjust the N.A. I suspect there are other sliders that have Ph1, Ph2, Ph4 inserts and/or the inserts are available separately. Fig 7 Back Between the coarse focus knob at the right in this view and the stand there is a narrow knob. When this narrow knob is turned CCW it moves away from the stand and jams up against the coarse foucs knob, locking the coarse focus knob. But after doing that the stage still drifts down. Problem was broken fine focus plastic gear. The lamp house is at center rear at the bottom. Fig 8 With Nikon camera using eyepiece adapter Used to take Subject photo 1 with D300s. Note you can not look through the other eyepiece when the camera is attached, hence the need for a trinocular head. Fig 9 Inside Back There are 4 hex bolts that can be removed. The focus rack is visible.
## Problem
### Fine Focus Plastic Gear
The coarse and fine focus controls work to raise and lower the stage. But when the stage is raised so the tip of the objective is just above the top of the slide and you let go of the knob the stage drifts downward a long way. On older microscopes this could be fixed by holding both the left and right coarse focus knobs and turning them against each other. But the Labophot manual specifically says not to do that.
Note: When the coarse knob was locked to the tension knob (or just by holding it fixed) the fine knob turns freely, i.e. there's no connection between the fine and coarse knob.
This tells you that the plastic gear is broken.
Microscope Solutions Labophot Fine Focus Shaft The shaft is longer and has break-off notches to it also works in other Nikon microscopes (they also make a replacement for the condenser gear.)
Fig 10 Broken plastic Fine Focus Gear Fig 11 Procedure to remove Fine Focus Shaft Fig 12 New & Old Shaft & Gear
### Rubber Eye Guard
As an experiment to replace the falling apart eye guards I've ordered a couple of Orion 1-1/4" telescope parts ($3 ea +s/h) which is a lot lower than purpose built versions. They may or may not fit. Old degraded eye guards New eye guards from Orion ### Condenser Pinion Plastic Gear This is another weak link in the "phot" family of microscopes, but mine is OK. If it breaks I'll get the metal gear from Microscope Solutions. ## Subject Photos Like most of my photos click for larger image. If your cursor is (+) click again for even larger image. ### Prepared Slides Labophot Image 1 Bee Wing 10X obj, EP camerea adapter Would be better with focus stacking. The following taken with E Plan 10/0.25 160/- objective, trinocular head, 23mm to Nikon adapter, Nikon 300s. From a number of prepared slide sets. Butterfly First Leg (foot?) Butterfly Second Leg (foot?) Butterfly Wing Scales Fly Wing ### Pond Water The Microscope Starter Kit from Microscope World tells how to make a culture solution to grow bacteria so that you can have plenty of pond creatures, but just to see what I would find I placed some pond water directly on a slide and added a cover slip. Found some things on the very first try. These were taken using the Labophot microscope with a Nikon E Plan 10/0.25 160/- objective, and a 2X camera adapter in the trinocular head. Focus by means of live view on the Nikon D300s camera. PS. Instead of paying the high on-line shipping for the Microscope Starter Kit if you phone them you can get USPS priority mail for 1/3 the price. I recommend this kit for microscope beginners because the DVD is very instructive (even the production quality leaves something to be desired) and the included prepared slides match those discussed in the DVD chapter on prepared slides. The DVD chapters are: How to use a Microscope - very basic but at 6 minutes in is about various types of slides: Well Slide, Wet Mount, Permanent Slide (Balsam & Xylene) and the use of a spacer between the slide and cover slip. Specimen preparation: soak in: fixative, 30% alcohol-70% water, 50%-50% alcohol water, 70% alcohol - 30% water, 100% alcohol and finally a clearing agent like Xylol. Investigating Cells - Plant and animal + use of Iodine (Wiki) as stain to bring out details. Examining Prepared Slides - matches the slides that are included. Investigating Pond Water - recipe for nutrient solution. Techniques Used in Studying Ciliates - has a lot of information (Wiki: Ciliates) #### Pond Water Culture Bring to a rolling boil 1 liter of water (I'm using my tap water even though the DVD says to use store bought "spring water"). This will kill any organisms that are in the water and in this case drive out any chlorine and dissolved oxygen. Add a handfull of straw. I pulled up some dry grass and some moss that was there from cleaning the house gutters. The DVD said it should be brown and I just noticed that I've got some green grass in there, we'll see if this works. Boil the dry grass for 10 minutes to break down the plant matter to make food for bacteria (Wiki). It smells like wet grass. Let brew sit for a two to three days uncovered in a cool shady place. Then add about 30 mL of pond water into the culture solution and let that sit for three to ten days. You can pull a sample of a few drops, put that on a plain slide and add a cover glass and have a look to see how dense the living creatures are. If culture turns milky it means there's too much bacteria and needs to have water added to lower their concentration. Dry grass & a little moss After sitting for a day After adding water to make up for boiling loss After adding about 1/8 cup of pond water and waiting a couple of days there's a whole bunch of very tiny things moving, but on the first try none of the larger ones. You can NOT see these very small creatures in bright field lighting, but they easily show up when the Phase slider is pushed in. I've been told these are Bacteria, spirochete (Wiki) form and one protozoan (Wiki) 28 July 2014 - 1 day after adding pond water to the culture: a lot of bacteria, but only a small number of Protists (Wiki). (Ph 3 pushed in) 29 July 2014 - 2 days after adding pond water to the culture: even more bacteria and now maybe half a dozen Protists in FOV of 10x objective (Ph 3 pushed in) 4 Aug 2014 - Still no big guys http://youtu.be/qVylp5rAsTA will try to get a water sample from a friend who has a Koy pond. 7 Aug 2014 - Still no \big guys. Got sample of water from friend's koi fish (Wiki) pond, but there were no bacteria or other small living things in it, but added anyway. Also added a spoon of dirt from outside. But there is plenty of life in the culture. 18 Aug 2014 - ordered some Methyl cellulose (Wiki). By adding some to the water it increases the viscosity thus slowing down the movement. 19 Aug 2014 - a lot of medium sized creatures. see video below. The formula on the label for glue is 1/2 cup of water and 1 tablespoon of the Methyl cellulose powder. I'm using 100 cc of water (very close to 1/2 cup) because that's what fits into a glass spice jar marked net wt. 0.75 oz and something approaching 1 cc of the Methyl cellulose powder. That's about 10 times less powder since I'm not trying to make glue, but rather just increase the viscosity of the water under the cover glass. Shortly after being mixed the cellulose settled to the bottom and there was about 20 or 30 cc of it, so the recipe on the label is about correct if you want a saturated solution. After standing overnight the liquid is homogenous. After mixing with water this may be called: Protoslo® Quieting Solution This is still too dilute, so I added another 1 cc (making a total of 2cc in 100 cc of water) to the bottle. This seems to slow them down. See YouTube video. First part at 20x, the 4x, then 20x. Rather than culture pond water I just took a couple of samples, one from the pot and one from the ground water into petri dishes. A quick look at 4x shows that the pot sample has the big guys and where to pull the drop. You can actually see the big guys with your eyes. Pot sitting in a pond formed with water proof pond making material. Recently added an RV water filter to the hose bib that's used to fill the pot & Pond. It should remove chlorine and ammonia. ### Pond Water Images Video showing a moving claw on the left Still of another one This is a little larger that what you see in the 10X eyepiece using a 10X objective. It's 10x objective * 2x camera adapter or 20X at the image sensor in the camera. 19 Aug 14 YouTube Video ### Hummingbird Feeder Sugar Water After mixing a batch of sugar water (formula: ____ water & ____ sugar) for the hummingbird feeder and putting the balance of the mixture in the refrigerator until the need to refill the feeder, some fuzzy ball started appearing after ______ weeks. They are about 1/4" in diameter and have a dark core. What are they? Aspergillus mold - very bad for hummingbirds, so need to clean feeder much more often. How did they get there? This part of the mix was never in the feeder, only in the house, so from the air? In bottom of Mason Jar ## Major Components ### Y-R Stand Many of the "phot" series of stands have a similar look, there are functional differences. It's not clear to me what stand names (like Y-R) go with what features. The focus stage can get close enough to the objective so that JIS (36mm) or DIN (45mm) parafocal length objectives will come to focus, but since it has RMS threads you can not use CFI60 objectives and since it will not get up to 31mm you can not use RMS parafocal distance objectives. If you know please tell me. ### Lamp House The black material seems to be an electrical insulator and it supports the two metal buss bars that provide power to the lamp. The front black heatsink is tilted back and I expect that's done on purpose so that it will no reflect IR light into the light path. The bi-pin lamp is a 6 Volt 20 Watt Philips 7388. Philips Code 7388 ANSI Code ESB LIF Code M30 Cap-Base G4 Bulb Material Quartz-UV Open Operating Position any Main Application Projection Life to 50% failures 100 hr ### Eyepiece Tube (aka: Head) The heads for most if not all the "phot" microscopes use a 48mm diameter mounting beveled flange that's captured by a thumb screw. When used on the Labophot microscope the eyepieces should by of the CF (Chrome Free) type. For all but the Ultra Wide head these will be the 23mm barrel diameter type. Note: The Labophot and all the other equal age "phot" microscopes use a 160mm type head. Some of the newer designs of "phot" microscopes use an infinity head and maybe with one exception all the current Eclipse microscopes use an infinity head. A binocular head has two eyepieces but does not provide stereo viewing because there is only one objective lens. A Stereo microscope also has a binocular head and has a pair binocular objectives or a Common Main Objective (CMO). It's not uncommon on eBay to see a "stereo" microscope that's really a compound microscope with a binocular head. To tell which head you have: • Remove head from microscope and install eyepieces. • If a trinocular head push in the selection control to 100% to eyepieces. • Hold head so bottom is pointing to something that's in the light and far away. • If it's an infinity head you will see an image like in a telescope. If a fixed tube length head everything will be blurry. #### Monocular This head (eyepiece tube) came from the introductory Alfaphot microscope so was designed to have minimal cost. At first it looked like it's too small to fit the Labophot, but the dovetail is the same as on the other heads. Mono Fig 1 Mono Fig 2 Teaching, monocular, Type T Trinocular heads Mounted on the Labophot (it looks very small) Optically it works as well as any of the binocular heads except it's for just one eye. #### Binocular This is the normal eyepiece tube (head) shipped with a base microscope. Note: The introductory/teaching microscopes use a linear slider for setting interplanetary distance, not the more expensive Siedentoph type. #### Eclipse Binocular With Nikon CFI 10x/20 eyepiece (30mm tube) This is an infinity type head that will not work on a Labophot under normal conditions. The two main reasons for this are: 1) The dovetail is 52mm O.D. whereas the Labophot (and Nikon S series microscopes) use a 49mm O.D. dovetail. I tried the head on the Labophot, but as expected the dovetail is too big to fit. A simple adapter can fix this. 2) The head contains a Tube Lens for the infinity optics. This means this head will not make an image with objectives that are designed for tube lengths of 160mm, 210mm &etc. It turns out that this particular head requires 30mm eyepieces. The CFI eyepieces seem to have either a Field Number of 20 or 22, both of which require a 30mm eyepiece diameter. This appears to be the stock head for the Eclipse E200 and I'm sure other models. To test for the infinity optics I took it outside and pointing the base at the distant mountains moved each eyepiece individually out a little to bring it into focus. Maybe the Eclipse heads have a different eyepiece seating distance than conventional fixed tube lens heads or this particular head is off a little or my eyes are off a little? A pair of CFI 10x/20 eyepieces are on order (9 Sep 2014) to check out the focal distance difference. Just as a test, after adjusting the eyepieces so that they are focused at infinity (see August Köhler patent) I just set the head on the Labophot and used a Nikon CF Plan 10x/0.30 ∞ / 0 EPI objective which has RMS threads so fits the Labophot. The microscope was already setup and focused on some pond water and it was also focused with the ∞ objective. Note that the distance between the objective and the head can be changed without changing the focus or other optical parameters. When I tried looking at the nearby mountains just after installing the Nikon CFI 10x/20 eyepieces they were not in focus, but after unscrewing the focusing action on each eyepiece good focus was obtained. The overall length of the eyepiece varies from 47.8 mm to 57.0 mm, i.e. an adjustment range of about 10mm, +/and/- 5mm which is more than enough to get infinity focus. #### Ultra Wide (UW) Trinocular Allows seeing twice the diameter (4x area) compared to normal binocular head. Requires CFUW (or CFI type) 10x eyepieces which are 30mm diameter, not the common 23.2mm dia. This allows the Field Number (the size of the field stop diaphragm) to be 26.5 mm as opposed to the less than 20mm for regular 23mm eyepieces. There may also be an ultra wide binocular head (i.e. without the camera port)? This head has a slightly larger entrance optical diameter (18mm) compared to the stock trinocular head (16mm). The camera port is the ISO 38mm compared to the 23.2mm port on the stock trinocular head. Where is the image plane for this head Answer: when the black 38mm to 23mm adapter is installed the image plane is 10mm down from the top, just like any other 23mm microscope tube. That means it's also above the top of the tube Like the other trinocular heads when the control is pressed in all the light goes to the eyepieces. When the control is pulled out the light is split between the eyepieces and the camera port. When the Amscope fixed focus 30mm eyepieces are used the camera and eyepieces are not parafocal. But when the Nikon CFI 10x/20 30mm focusing eyepieces are used the camera and eyepieces ALMOST focus at the same stage height. It would take a 30mm eyepiece extension to make it parafocal. This also may mean that when these eyepieces are used with the Labophot type T trinocular head using the following procedure the eyepieces and camera will be parafocal: 1. focus camera using live view and a few presses of the magnifying glass, 2. focus left eyepiece (my left eye is the stronger one) 3. focus right eyepiece. It turns out that even when the CFI eyepieces are screwed all the way out the camera focus is sill off a little, What's needed is a 30mm eyepiece extension tube that would hold them so that the tip of the eyepiece was about 5mm out from the tube. Fig UW 0 Fig UW 1 w/o eyepeices Fig UW 2 Fig UW 3 w/Amscope eyepieces Fig UW 4 compare standard trinocular head and UW head 23.2mm camera port vs. 25.3mm at top of black cylinder. The reason for the slightly larger hole is to allow the use of a couple of pinch screws to anchor the 23mm device. Fig UW 5 compare standard trinocular head and UW head 16mm entrance vs. 18mm entrance on UW head. Fig UW 6 w/Black 38mm to 23 mm adapter removed. The adapter contains no optics. This may be a 38mm ISO camera port. T-mount adapter from eBay seller micro_crystal The black adapter is 53.2mm from the head flange to the top of the adapter. The objective focal plane is down 10mm, so the focal plane is 43.2mm above the head, but that's a few mm too short to allow focusing directly on my camera's imaging chip. But would work for the new mirrorless cameras. Fig UW 7 UW head installed on Labophot w/ 23mm eyepiece in camera tube (it's close to focus). #### ISO Camera Port This is a 1-1/2" (38mm) port rather than the older 20.2mm eyepiece port. There are a number of commercial camera adapters that include optics that are very expensive. There are also adapters to the common C-Mount and T-Mount. T-Mount Adapter & C-Mount Adapter The T-mount adapter adds 3mm flange to flange. 33mm ID The C-mount adapter adds 12mm flange to flange. 21mm ID These adapters add enough length so that you can not get parafocal operation between the eyepieces and camera. This would be desirable since there's no optics in these. You can easily refocus the camera when using these, but then the eyepieces are out of focus. #### Dual View (Teaching) This head can be used with another head added on top of it or with the plastic cover on top as a stand alone head. Normally the added head would be a regular binocular head but could be one of the trinocular heads. It has a joystick that moves a white arrow type pointer so the student's attention can be directed to a specific point. This pointed needs source of light. It uses standard 23mm eyepieces, and for the Labphot these should be CF types. When a second head is installed directly on this head the teacher and student are in a face-to-face position. There are also "bridges" that allow the teacher and student(s) to be sitting side-by-side and the number of students can be up to 5 (?). The inter-pupil distance can be adjusted and there's a scale marked from 51 to 75mm. This head may need collimation (Wiki). This is a Siedentopf binocular head i.e. the inter-pupil distance is adjusted by moving the eyepieces about a central axis in an arc rather than by moving along a slide directly in a straight line (that's called a Jentsch and the adjustment changes the tube length and so requires refocusing. The pointer illuminator uses the THN transformer. It's not clear what the actual lamp looks like to fit the input port on this head. Microscope Solutions has an LED replacement lamp. But I don't have the lamp holder or power supply. Teaching Head Fig 1 Teaching Head Fig 4 Teaching Head Fig 7 Teaching Head Fig 3 Teaching Head Fig 6 Teaching Head Fig 2 Teaching Head Fig 5 #### Trinocular type-F Selection of view or camera is by rotating head. There is no push/pull control. When the head is turned it switches to the camera port. There is no glass in the camera port path. This may imply that the camera port and the visual port are not parafocal. #### Trinocular type-T Selection of view or camera is by means of a push to view/pull for camera lever on the right side. After loosening the three small (-) screws the support tube can be removed revealing a standard 23mm eyepiece tube. So, this is not a 38mm ISO camera port. When the lever pulled out to select the camera port there is glass in the camera optical path. But the camera should be in focus when the visual image is in focus. So far this has not worked for me, but it may depend on what camera adapter is used. eBay photo Overall bottom top control pushed in (eyepiece viewing only) top control pulled out (camera & viewing) Three (-) screws to remove outer support tube Now Nikon to microscope eyepiece adapter can be used With Nikon D60 camera so show the idea Stack with: Nikon D60 camera Nikon to 23mm microscope adapter Type-T trinocular head (support tube removed) Teaching head Vertical Illuminator Labophot microscope Trinocular head and UFX-IIA film camera adapter When installing a new 0.5x camera adapter which had the O-ring installed (the prior camera adapters had a groove for an O-ring but it wasn't installed) I needed to apply a little pressure and push down to seat the adapter, but in that process discovered that the trinocular camera tube could be unscrewed from the head. Camera tube removed, control pushed in. Camera tube removed, control pulled out. Camera tube. Lens unscrewed from camera tube. There are two lenses in the trinocular head camera light path. One on the bottom of the head, see above, and one on the bottom of the camera tube. I wonder if it's possible to convert this from a "T" head to the "T2" head used on the Epiphot 200 which is an infinity head? Let me know. #### Combined binocular head and vertical illuminator This might be used for metallurgical work or wafer inspection like an Optiphot-66 or -88. ### Eyepieces (Wiki) The Labophot optical system requires that both the eyepieces and objectives are Chrome Free (CF) types. They have provision for a 21x1.5mm reticle to be inserted. One of these is the "M" used to outline various camera film frames. One of the CF 10x eyepieces has a measurement reticle fitted. To access it you unscrew the knurled ring at the lower tip. The eyepiece acts just like a magnifying glass (Wiki). It's subject is the real image from the objective that's 10mm below the top of the tube. From the bookMirrors, Prisms and Lenses by Southhallis the equation for the magnification of an eyepiece: Mag = 250mm * [(1/FL) + (1/250mm)] See Lens Equations FL = 250mm / (M-1) So here are some common focal lengths for eyepieces: Power FL mm 5 62.5 10 27.777 15 17.857 #### Viewing 20.3 mm Tube (23.16mm dia EP) CFW 10x (this is the normal (23.2mm tube) eyepiece and what came installed on this system) CFW 10x Field Number 18 23.16mm tube mating diameter. CFWN 10x/20 Field Number 20 (that's the /20 above) Note these will give a slightly larger field of view compared to the stock CFW 10x eyepieces which have a FN of 18mm. 10 times 18 = 180mm field at a virtual 25cm from the eye -vs- 10 times 20 - 200 mm field at a virtual 25cm from the eye. Also these accept 21mm dia. reticules and can be individually focused. Mating dia 23.15mm. CFW 8x Field number 18mm CFW 10x Field number 18mm CFW 15x Field number 18mm CFUW 10x Field number 26.5mm ### Field Number (UC Berkeley The Microscope Eyepiece or Ocular) Field number in millimeters times the objective power equals the apparent size of the field of view. For example: a CF 10x eyepiece (FN 18mm) with a 10X objective appears to have a field that's 180mm in diameter. The field number is also the inside diameter of the diaphragm at the very tip where the reticle fits and can be measured with vernier calipers. The Nikon CFWN 10x/20 eyepiece has a filed number of 20mm and so is fits a 30mm tube. #### Wide Field Viewing 30mm Tube (Amscope fixed focus) For use in the UW head that needs 30mm eyepieces. Amscope makes a EP10x30 with a 30mm tube size that will fit the Ultra Wide head. Note the price for a pair of these is about$40 compared to ten times that for a pair of Nikon CFUW eyepieces.
CFWN 10x/20 (high eye-point/widefield for people who wear glasses) FN 20mm, 30mm tube.
These do not have diopter adjustment so if your eyes are different some shimming may be needed.
This also means that the reticle focus may require shimming or flipping top to bottom.
Note the Amscope p/n is EP10X30 indicating ten power and thirty millimeter tube size, yet they are marked WF10x/20.
Amscope WF Eyepiece EP10x30 marked WF10x/20 Amscope WF Eyepiece EP10x30 marked WF10x/20 measured FN = 23mm Reticle OD 24mm
#### Nikon CFI 10x/20 Eyepiece (Eclipse or any 30mm tube)
These are marked CFI and fit a standard 30mm eyepiece tube, like the Nikon Labophot type T trinocular head as well as any of the Eclipse type heads.
There is also an Eclipse 10x/22 eyepiece.
Nikon CFI 10x/20 Eyepiece Nikon CFI 10x/20 Eyepiece in Eclipse Infinity Binocular Head
#### Eye Guards (Eye Cups)
The factory rubber parts have biodegraded (See: Hints & Tips: car rubber parts) because of exposure to the oxygen in the air.
Also see Problems: Rubber Eye Guard above.
Old biodegraded eye cups New Orion 1/-1/4" telescope eye cups Note you need to stretch them, but they fit well.
#### Photography (relay lens)
Some thoughts on selecting projection relay optics when using DSLR’s (and other “lens-less” sensors) for photomicrography by Charles Krebs
The Nikon D300s has 5.5 x 5.5 um pixels ( 4288 x 2848)12.2 Mega pixels
CMOS sensor, 23.6 x 15.8 mm; total pixels: 13.1 million; Nikon DX format
CF PL2.5X (Chrome Free Photography Lens 2.5 power)
CF PL4X
CF PL5X
CF Photo8X
### Vertical Illuminator
This is used to provide white Episodic illumination from the top, like would be used for metallurgy.
What's missing is the adjustable 12 Volt power supply for this illuminator.
Also some questions, Can any objective be used or is there some advantage in using objectives marked "EPI?
What is the description of the filters or blank fillers that to in the two slots between the lamp house and microscope interface?
I've found a Nikon switchable 6V/12V power supply on eBay, but it has a Cinch Jones 4 terminal socket, not the bi-pin socket that would match the one on this illuminator.
A question to be determined is are 210mm objectives needed when this vertical illuminator is used or will regular 160mm tube length objectives work?
Note the flange to flange insertion thickness of the illuminator is 50mm.
Epi 1From eBay ad EPI 2 Top The filter slots are a little over 6mm and 4mm wide. The tube has an ID of about 27mm and an OD of about 40mm. EPI 3 Bottom EPI 4 2-Prong Electrical Plug EPI 5 Inside Slot in housing allows rotating lamp to line up filament. Knob of lamp holder moves lamp up/down. Lamp marked: 12 V 20 W GE M47 EPI 6 Overall View (but I've got the lamp house upside down) EPI 7 Control In Ring of light sent to objective. Darkfield EPI 8 Control Out Central spot of light sent to the objective. Brightfield EPI 9 Nikon LV-UEPI Illuminator Low Voltage (Halogen lamp) Universal Epi EPI 10 Nikon Brightfield/Darkfield (Epi) Objective EPI 11 Taken using bottom light trinocular head (not Epi head) Pattern: 5mm Line, 50 Divisions EPI 12 Taken Using bottom light but with Epi Illuminator installed. This adds 50mm to the 160mm tube length= 210mm which changes magnification. Still bottom illumination. The EPI control can be in either position for eyepiece observation but for camera must be pulled out. EPI 13 Now switching to Epi Darkfield Illumination. EPI Lamp On & control pulled out. EPI 14 Photo of EPI setup for EPI 13 Darkfield photo. Setup for EPI 12 the same except EPI turned off and Labophot turned on. 2012 Dime showing the "D" for Denver This would be an impossible photo if the light was coming from the bottom. 10x objective (160/- used at 210mm) Camera Adapter sold as 0.5x by eBay seller amoyca2011 1230 pixels * 5.5 um/pixel = 6765 um = 6.765mm 6.765 / 4x objective = 1.691 power camera adapter
### UN Transformer (Power Supply)
To power the above vertical illuminator I needed a power supply that can deliver 12 Volts at 20 Watts. This one was on eBay and is rated for 12 V @ 100 Watts so has five times more power capability than needed, but since it's the correct voltage that's not a problem. It probably can be used to power one of the 100 Watt illuminatiors.
It has switchable input line frequency 50 Hz or 60 Hz and switchable output voltage 9 V or 12 V. On the top is an LED bar-graph type voltage readout.
Although the interior view looks similar to a DC power supply (there are three transformers and a printed circuit board, this is a variable AC power supply.
Halogen lamps (Wiki) will operate on either AC or DC. The reason medical lamps are often Halogen is because they are black body (Wiki) light sources and so have very good color rendition (they also have a continuous output spectrum from infrared to ultraviolet unlike LEDs that have discrete bands of output). Note that changing the voltage using the knob strongly effects the lifetime, color temperature and brightness of the Halogen lamp (Wiki).
The Output connector is square and has 4 terminals for flat blades It is the Cinch Jones P304-CCT Plug, see photo below.
This power supply weighs 8 pounds (12.5 Watts/pound which is consistent with analog electronics)
Nikon UN Transformer 9 to 12 Volts Cinch Jones P304-CCT Plug Plug wired to Epi Illuminator (left 2 lugs) Next photos are in the Epi Illuminator section on how it works.
#### Fluorescence vertical illumination
There are vertical illuminators designed to support Fluorescence microscopy (Wiki). Nikon has a standardized "cube" that holds two filters and a beam splitter.
The idea is to filter the high power light coming in on the side so just a narrow band of light illuminates the subject.
The new wavelength light goes up the tube and the beam splitter lets it pass and a filter rejects any source light so only the new light is seen.
These come in versions that hold 1, 2 or 4 cubes.
1 cube The 2 cube version uses the HBO100w power supply. Nikon HBO-100w/2 Mercury Arc (Wiki) Power Supply One cable is for the high voltage supply and the other for the low voltage filament. HB 10101AF Ultra High Pressure Mercury Lamp Power Supply 4 cube High pressure mercury lamp box Nikon LH-M100c1 Super high pressure mercury lamp power supply Nikon HB-10103AF
### Objective Noise Piece
This scope came fitted with a 5 position noise piece.
### Objectives
CF (Chrome Free) Optical System with 45mm parafocal distance and RMS threads, 160mm tube length without the vertical illuminator installed.
With vertical illuminator installed 210mm tube length (i.e. BD & M objectives).
The focal length of an objective is the tube length divided by the power.
So FL = 160mm/Power
Power FL mm 5 32 10 16 20 8 40 4 100 1.6
#### N.A. (Wiki)
The Numerical aperture is directly related to both the resolving power (Wiki) and the depth of field (Wiki).
The higher the N.A. the higher the resolving power, but the shallower the depth of field. So for looking at moving things in pond water given a choice of objectives with an N.A. ranging between 0.4 and 0.75 the 0.4 will have greater depth of field. The 0.75 would be better if it was going to be used with an expensive camera that was optically matched for highest resolution, but at the cost of a much shallower depth of field.
#### Nikon Objective Brochures
Nikon CF Brochure (160mm) Feb 1989
Nikon CF Brochure (210mm) July 1993
Nikon CFI Brochure ∞ May 2007
#### Nikon Objectives
Name Tube mm Thread Parafocal Distance APO Plan 160 RMS 45 APO Plan CFN 160 RMS 45 BD Plan 210 26 45 BD Plan DIC 210/0 26 45 CF Plan 160 RMS 45 CF Plan EPI ∞ RMS 45 CF Plan SLWD EPI ∞ RMS 45 DL Ph4 160 RMS 45 E (black) 160/0.17 RMS 45 E Plan 160 RMS 45 E Plan ∞ 25 60 E Plan LWD 160 RMS 45 ED Plan 210/0 26 45 FL Plan ? ? Flour 160 RMS 45 Flour Plan DIC L ∞ 25 60 LU Plan EPI ∞ 25 60 M Plan 210 31 M Plan LWD 210 M Plan ELWD 210 NCG Plan 160 RMS 45 UW Plan ∞/- 25 60
The two objectives that came with the Labophot.
Left: E Plan 10/0.25 160/- Right: E Plan 40/0.65 160/0.17 Ph3 DL Neither of these is marked CF, so do they belong on this microscope? I've been told that these are student (Economy) CF objectives. Comparison CF vs. CFI objectives E Plan 40/0.65 160/0.17 (not Phase) Left is Made in China copy of Nikon CF 10x, PFD=40mm. Right is Nikon CFI E Plan 4x/0.10, ∞/- WD 30 PFD=60mm Note everything is larger in diameter. This is not so apparant in photos of CFI equipment when it's all by itself. More on the Tube Lens page. DIN 4x/0.10 160.0.17 M Plan 40x/0.65 210/0 DIN 20x on order Surrey Oriental Ltd (4 Aug 2014) This fits between 10x and 40x 40x/ 0.65 S-series microscope Should be used with Bi HKW eyepieces Nikon E Achromat 100x/1.25 Oil 160/0.17 Immersion Oil Xylene Immersion Oil Solvent Grow Automotive H.E.T 4218 1 Gallon can
#### Immersion Oil Patents
Immersion oil formulations for use in microscopy and similar fields, Dec 30, 1975, 252/408.1 -
prior art oils contain polychlorinated'biphenyls (PCBs). Also has background on the need for immersion oils.
non-toxic immersion oil with excellent optical properties
Blend of dialkyl phthalate; butyl benzyl phthalate and chlorinated paraffin, Aug 14, 1984, 252/582, 252/1, 252/408.1, 252/364 -
a refractive index of 1.518 or thereabouts and an extremely low fluorescence
Heavy mineral oils with no background fluorescence, Jan 1, 1985, 252/301.16, 208/18, 252/1, 252/408.1, 208/12 -
immersion oil for use in fluorescence microscopy
Liquid optical coupling material comprising pentabromo-diphenyl oxide, alpha methylstyrene polymer, stabilizer, Jul 2, 1985
252/582, 252/408.1, 568/639, 359/886, 585/428, 568/580, 252/1, 359/665 - coupling adhesive
#### Objectives with 45mm Parafocal Distance
Here is a list of Nikon objective names that are CF optical system.
Many of these do not have CF printed on the objective.
Achromat
E Achromat
E Plan Achromat
Epi-Fluorescence (Fluor)
Epi-Fluorescence Phase-Contrast
High-Magnification Dry
Long-Working-Distance (LWD)
N Phase-Contrast
N Plan Achromat
N Plan Apochromat
N Plan DIC
No Cover Glass (NCG)
Note: The M Plan 210mm are for metallurgy or when the Vertical Illuminator is inserted into the optical path.
BD: Brightfield Darkfield for use with 210mm tubes (i.e. vertical illuminator installed)
M: Metalurgical for use with 210mm tubes (i.e. vertical illuminator installed) always Brightfield and will not work for darkfield.
CF ∞ with RMS threads: These are not CFI60 objectives since they have RMS threads. They go with the Epiphot Inverted Metallurgical microscope and the "E" series diaphot.
Since DIN objectives have the same parafocal distance (45mm) they might be a good fit for this microscope.
But RMS and JIS objectives have a shorter parafocal distance and the focusing rack on the Labophot can not bring them into focus so can not be used.
#### Infinity CFI Objectives
It may be possible to use the CFI objectives on the Labophot and other similar microscope stands.
The key reason for doing this is to take advantage of the increased performance of the CFI objectives and for me the lower powers are very attractive. It's my understanding that the CFI low power objectives have superior optics when compared with the CF optics. See: Why Use a 25 millimeters Objective Thread Size? The N.A. of low power objectives is higher.
In both the below cases the first method of attachment is to use an adapter with female 25-0.75 threads to accept the CFI objective and male RMS threads for the existing nosepiece.
This may stop down the light coming from the exit pupil, but the exit pupil looks to be about 6mm dia and the mechanical exit pupil size must be less than 9mm right at the back of the objective.
Since the light is parallel rays there should be no problem with vignetting. Using a pocket blue laser and moving it back and forth it can be seen that the exit pupil diameter, even back a couple hundred mm from the objective is still under 25mm dia. This is not a precision measurement.
There are a few of ways this will work.
(1) By connecting the camera directly to the stand (no head or visual observation. The camera is setup with a tube lens.
(2) By using a microscope 20.2mm to C-mount adapter with no optics then copling that to the T-mount threads on the tube lens camera setup. This has the advantage of using a trinocular head, but may have a problem with vignetting because of the C-mount ID or the 20.2mm tube ID. Also the eyepieces will not be in focus.
(3) There may be other options.
It may be that C-mount parts will work for a tube lens, if a suitable lens and mount can be found.
### K.D. Kemp 8 Form Diatomaceae Test Slide
Got this slide from Microlife Services (aka diatoms) in the UK.
Diatom Fig 1 Comes in a 2-slide carry case Note the plastic light table has scratches. Nikon D300s 105mm Micro Nikkor lens at max distance Diatom Fig 2 Nikon D300s Nikon 105mm Micro Nikkor lens at minimum distance Diatom Fig 3 Diatom Fig 4
### Phase Telescope
This is used to center the annular diaphragm when the Ph3 slider pressed in so it's in the light path.
It is installed in place of one of the eyepieces.
I don't have one, so will try using the paper method described here:
YouTube:(3 methods:Phase Telescope, observing image, paper)
The paper method didn't work because I don't have enough experience to know what a good phase image is. Got a phase microscope.
Some microscopes (but not the Labophot) have a Bertrand lens that can be switched into the optical path to view the phase ring at the back of the objective.
It might be marked Ph (B) - 0. See MicroscopyU -Phase Contrast Microscope Configuration
The following figure is from the above Phase Contrast Microscope Configuration web page.
In the patent (Phase microscopy, Nov 25, 1947, 59/370, 359/580, 359/489.7) a method is described to make the phase shift constant for different colors, but that method may or may not be employed in the Nikon phase optical system. If it is not then the closer to monochromatic light that's use the better the quality of images, this may be the reason for the GIF filter.
Phase Telescope Afocal image looking at E Plan 40/0.65 160/0.17 Ph3 DL objective and with Ph 3 slider pushed into the optical path. There is a reflection in the image that repeats the main pattern. In order to adjust phase slider I need two small tools that did not come with this miscrscope, they are hex drivers. Phase slider A holds two phase plates and when it's installed the Abbe condenser should be installed with theaperture diaphram lever forward and the two holes in the phase plate are aligned with the two holes in front. But when Phase sliders B or C are installed (where there is only one phase plate and an open space for brightfield the condenser should be installed with the lever to the left exposing adjusting hex screws on either side of the push/pull handle (knob). Instruction Sheet, GIF filter and small unknown dual filter, phase telescope, Ph3-or-Brightfield slider B and E Plan 40x/0.65 160/0.17 Ph3 DL objective. Missing are: * the two small phase plate tools * Phase slider A Ph1 10x DL or Ph3 40x DL * Phase slider C 10x DL Ph 1 or brightfield Extra part: Small block with two filters? Figure showing the condeser installed with the lever on the left, and the two phase plate adjusting holes to the front. The microscope came with the condenser improperly installed, the adjustment slider should be to the left for the single plate phase sliders B or C and facing forward only for the A dual plate slider. PS The condenser mounts with a dovetail that's 45.5mm diameter and you can see how the single mounting screw rids on the slope and acts to pull the condenser down to the mating flange of the support. PPS you can see the GIF filter sitting on the field lens. Note that a 7-position turret Ph condenser is shown but there are a number of condensers that support Ph. Optical diagram (from MicroscopyU) After adjusting the two 1.5mm hex screws on the Ph3 slider.
Nikon MicroscopyU - Phase Contrast Microscope Alignment
Making a Phase Telescope using an eyepiece or two.
### Slide Holder
The slide holder is attached to the sage, but can be removed if you're not using slides.
### Stage
The stage is classically an X-Y movement device, but this one also can be rotated.
### Condenser
There are a number of condensers that fit this Nikon microscope. The one that came with it is the Abbe 1.25 Condenser & Ph3 slider.
The lever at the front changes the diaphragm which is how the N.A. is controlled.
It uses a dovetail mount where the condenser dovetail is 46mm.
#### Abbe 1.25
This is the condenser that came with the Labophot microscope.
#### Phase Contrast 1.25
This is a condenser that I added. It has a 6 position turret.
0 (Bright field) with adjustable iris, DF (Dark Field), Ph1, Ph2, Ph3, Ph4
The diaphragm for bright field is connected to that turret position so only works in the "0" position.
There are two adjustment wheels and each has a smaller lock screw. But these wheels adjust the position of the plate, that's to say it does not adjust each individual insert, so you need to adjust each time you change turret positions, or that's how it looks now.
The detent (Wiki) is very weak. How to stiffen it up? Let me know.
0 (Bright field) Top Note: BF Iris shown fully opened Dark field Top Dark field Bottom Ph 1 Top Ph 1 Bottom Ph 2 Bottom Ph 3 Bottom Ph 4 Bottom Back The dovetail at the bottom (point) is 46.25mm OD. Front BF iris shown fully closed Installed on Labophot Installed on Labophot
#### Table of Condensers (most of which fit Labophot)
Photo Name N.A. Spot dia mm Remarks Abbe 1.25 4.6 4X - 100X (Oil ?) general purpose Supports phase slider Swing-out Achromat 0.90 0.1 In 3.4/ Out 12 Best for color photo ?phase slider? 6 position turret Two levers on other side. 20x, 40x, 60x Plan Apo DIC prisms. Ph2 and Ph3 phase annuli. Dark field stop and bright field Achromat/Aplanat 1.35 1.40 2.8 Best for high N.A. 20x, 40x, 60x Plan Apo DIC prisms. Ph2 and Ph3 phase annuli. Dark field stop and bright field 6 Position Turret Nikon Phase Contrast condenser 1.25 Ph1, Ph2, Ph3, Ph4, Darkfield, Brightfield Phase Contrast Phase Contrast-2 1.25 4.6? 6 Position Turret Nikon Phase Contrast 0.85 Phase Contrast 0.85 ? does this fits the Labophot? 6 Position Turret Nikon Phase Contrast LWD (Long Working Distance) 0.52 Phase Contrast 0.52 For: Inverted TE200, TE300, TE2000 Achr 0.8 0.85 0.1 N.A. adjustable 0.8 to 0.1 1.30 Phase Contrast "Black Enamel" 1.30 For older S series microscopes, looks very different on bottom Does not have the two centering adjustments. 0.65 LWD (Long Working Distance) Abbe 0.65 0.1 Achromat 0.13na 0.13 0.02 For: Eclipse 50i, 55i, 80i, 90i, Ci and Ni 6 Position Turret Phase Contrast ELWD 0.3 0.3 For: Diaphot 200 , 300 Achr 0.85 Auto Iris 0.85 For: Eclipse??? HMC (Hoffman Modulation Contrast) 0.6 For: TS100 C-C Achr N.A=0.90 0.9 For: ? 6 Position Turret ∞ ELWD Phase Contrast 0.3 For: Inverted TE2000, TE200, TE300
### Daylight Blue/Green Filter
This is a historical device that's no longer used. The idea was that if you had daylight film in your camera and were taking photographs using the tungsten lamp in the microscope then you could place this filter in the light path to correct the color temperature (Wiki, my web page on color management). Modern DSLR cameras either correct the color temperature in the camera, or in my case when using RAW (Wiki) files correct it in Adobe Camera Raw.
The key thing this filter does is limit the range of light wavelengths (colors) so that the phase contrast 1/4 wave plate is near a 1/4 wave. If all visible colors were allowed a very expensive compensation system would be needed for phase contrast. This compensation is described in the Zernike patent, but is typically not part of phase contrast optical systems.
### Background
In the 160mm tube length microscopes the objective forms a real image 10mm down from the top of the tube where you can't get the camera imaging chip.
The common way to fix that is to use a relay lens where the power of the lens depends on the size and resolution of the imaging chip. But that adds optical elements which degrade image quality.
It's possible to change the objective to image distance by focusing the microscope so that the real image is above the top of the tube enough to fall on the camera's imaging chip but then you're not using the objective at it's designed magnification and so may be degrading the image quality.
Modern DSLR cameras have a "live view" (Wiki) feature that allows viewing what amounts to a TV image in real time. Even when I use the camera port on the trinocular head I need to use live view to refocus the image on the camera chip. It may be possible to make custom spacers to lift the eyepieces to the exact spot where they focus where the camera focuses, but I haven't don that yet. Also the camera has a video output so that you can see the microscope image on a computer monitor.
This means it's possible to do away with the head and eyepieces altogether and replace them with the computer monitor by connecting the DSLR directly to the microscope stand replacing the existing head.
The first step to see if this is possible was to measure the distance from the objective flange to the head interface flange and it's 45mm. Also note this path is just air, i.e. there are no glass parts of any kind in this path.
160mm - 45mm - 115 mm for the adapter and camera. The Nikon-F Flange Focal Distance (Wiki) is 46.50mm.
115mm - 46.50mm = 68.5mm for the adapter.
T-mount find focus tubes from Edmond Optics come in lengths of 35 - 60mm and 40 - 65mm.
If the 40 - 65mm tube is nominally set at 45mm (i.e. allowing 5mm adjustment) then the remaining length for the adapter is:
68.5 - 45 = 23.5mm.
16 July 2014: More later. . . .
The "phot" (CF) family of scopes uses a 48mm OD dovetail interface.
The Eclipse (CFI60) family of scopes uses a 51mm OD dovetail interface.
Many European scopes use a 42mm OD dovetail interface.
## Polarized Light Kit
Purchased from eBay seller: optitec202. It contains a small filter that fits between the Labophot stand and head and another filter with angle calibration marke -45 to 0 to +45 degrees that fits on top of the field light source. Kit came with a couple of prepared slides with Tartaric acid and Citric acid. The crystals change color as the filter is rotated.
Fig PolKit 1 w/o Polarizing Filter Fig PolKit 2 Stand with Polarizing Filter Fig PolKit 3 Field Light Filter
------------
### Epi Fluorescence centering alignment objective
I don't have an Epi Fluorescence lighting setup, but got this alignment objective in the hope that it can be used as part of an alignment procedure for the normal light function.
It consists of an 8 sided mirror (not equal sides) mounted at 45 degrees relative to the centerline of the "objective". There's a window mounted in the side of the cylindrical "objective" that has a ground glass screen with a cross target. The lower part of the "objective" that contains the target can rotate so that after the RMS threads are used to mount the "objective" the target can be aimed to the front for easy observation. I'm assuming the two (-) screws below the target are used to align the mirror to exactly 45 degrees relative to the centerline and that the large (-) screw in the bottom is used to align the center line optical axis with the center of the target.
In Fig 2 you can see two horizontal dashed lines and they persist when the "objective" is tilted up and down so they are there for a reason. Maybe just because the mirror is a rear surface type, i.e. independent of the look angle?
When a green laser pointer is shined down and you look at the screen you see a green dot.
When a blue laser pointer is shined down and you look at the screen you see a yellow dot.
This may mean that there's some filtering of UV light so that no harmful UV light comes out of the screen when aligning a UV light source.
Fig 1 shown right side up Fig 2 looking down The mirror may have it's reflecting surface on the back so there's two reflections in this photo. Maybe there would only be one image if the camera centerline was the same as the "objective' centerline. Fig 3 shown upside down 3 adjustment screws
## Patents
Microscope system, Oct 3, 2000, 359/381, 359/656, 359/821 - describes why infinity optical system vs. 45mm parafocal & 160mm tube length.
Note; not applicable to the Labophot, but is applicable to all the newer Nikon microscopes.
Phase contrast observation device, Nov 13, 2001, 359/387, 359/386, 359/385, 359/371, 359/368, 359/370
ADH Phase-objective is in the CFI60 "infinity" range of objectives (>\$4,000)
## Related Web Pages
Microscope Objective Standards
Table of Microscope Objectives & photos taken with camera (not microscope)
B&L Stereo Zoom Microscope
Unitron No. 83444 Microscope
Mitutoyo Toolmakers Measuring Microscope 176-134
Nikon Labophot Microscope & accessories
Nikon Multiphot stand & Table of Nikon "phot" microscopes
DigitalPhotography101 Digital Photography 101: The Basics
Digital Photography 201 Stacking Images
Digital Photography 202: Close-Up, Macro & Micro
Digital Photography 203: Color Management
Digital Photography 204: Studio Flash
Digital Photography 205: Astrophotography
Digital Photography 206 Micro Photography - Tube Lens - Unitron Compound Microscope
Lights sources
Mitutoyo Toolmakers Measuring Microscope 176-134
Nikon Nikon (and Kodak DC290) Cameras
Nikon Multiphot System - Macro photography stand and list of "phot" family microscopes
NikonSB25 - Nikon SB-25 Flash & Manfrotto C1575B Avenger Super Clamp
Omnicon 3800 Tumor Colony Analyzer (TCA) Automated Inverted Biological Microscope (came with Nikon objective)
Optical Bench w/ old Microscope
Optical Spectrum Analyzers
Optics Day & Night
Spectronic 20D Spectrophotometer
Microscoy Primer by Frithjof A. S. Sterrenburg
Principles of Microscopy, being a Handbook to the Microscope by Sir A.E. Wright, M.D. MacMillan Co, 1907 - the best book on measuring the various components of a microscope.
Interference Phenomena, Compensation, and Optic Sign - polarized light and colors
Carolina Biological Supply Company - prepared slides & animal parts for dissection
YouTube: LAB EXERCISE 2; THE MICROSCOPE by Professor Fink- Labomed
Microbe Hunter - online magazine - -
Arizona Fairy Shrimp - low cost kits of eggs & food.
Pippa's Progress - an excellent introductory series of videos (Viemo and sometimes stop playing)
Episode 1 - diatoms in toothpaste
Episode 2 - cells & bacteria from inside cheek & Stinging Nettle
Episode 3 - Gathering and viewing field water
Episode 4 - Learning about what we see
Episode 5 - Detective skills
Back to Brooke's: PRC68, Alphanumeric Index of Web Pages, Contact, Products for sal
Page Created 3 July 2014
|
2021-05-18 13:02:25
|
{"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": 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": 0.3115735352039337, "perplexity": 6294.744302485591}, "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-2021-21/segments/1620243989637.86/warc/CC-MAIN-20210518125638-20210518155638-00180.warc.gz"}
|
https://proofwiki.org/wiki/Ring_of_Polynomial_Forms_is_Integral_Domain
|
Ring of Polynomial Forms is Integral Domain
It has been suggested that this page or section be merged into Ring of Polynomial Forms over Integral Domain is Integral Domain. (Discuss)
Theorem
Let $\struct {R, +, \circ}$ be a commutative ring with unity.
Let $\struct {D, +, \circ}$ be an integral subdomain of $R$.
Let $X \in R$ be transcendental over $D$.
Let $D \sqbrk X$ be the ring of polynomials in $X$ over $D$.
Then $D \sqbrk X$ is an integral domain.
Proof
By Ring of Polynomial Forms is Commutative Ring with Unity we know that $D \sqbrk X$ is a commutative ring with unity.
Let neither $\displaystyle \map f X = \sum_{k \mathop = 0}^n a_k x^k$ nor $\displaystyle \map g X = \sum_{k \mathop = 0}^m b_k X^k$ be the null polynomial.
Then their leading coefficients $a_n$ and $b_m$ are non-zero.
Therefore, as $D$ is an integral domain and $a_n, b_m \in D$, so is their product $a_n b_m$.
By the definition of polynomial multiplication, it follows that $f g$ is not the null polynomial.
It follows that $D \sqbrk X$ has no proper zero divisors.
Hence $D \sqbrk X$ is an integral domain.
$\blacksquare$
|
2020-04-09 14:32:55
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.9703589081764221, "perplexity": 77.2625988166366}, "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-2020-16/segments/1585371858664.82/warc/CC-MAIN-20200409122719-20200409153219-00410.warc.gz"}
|
https://www.zora.uzh.ch/id/eprint/22987/
|
# Solutions to the Korteweg–de Vries Equation with Initial Profile in $L_1^1 (\mathbb{R}) \cap L_N^1 (\mathbb{R}^ + )$
Cohen, A (1987). Solutions to the Korteweg–de Vries Equation with Initial Profile in $L_1^1 (\mathbb{R}) \cap L_N^1 (\mathbb{R}^ + )$. SIAM Journal on Matrix Analysis and Applications, 18:991-1025.
## Abstract
The Cauchy problem for the Korteweg–de Vries equation is considered with initial profile integrable against $(1 + | x |)dx$ on $\mathbb{R}$ and against $(1 + | x |)^N dx$ on $\mathbb{R}^ +$. Classical solutions are constructed for $N \geqq {{11} / 4}$. Under mild additional hypotheses the solution evolves in $L^2 (\mathbb{R})$.
## Abstract
The Cauchy problem for the Korteweg–de Vries equation is considered with initial profile integrable against $(1 + | x |)dx$ on $\mathbb{R}$ and against $(1 + | x |)^N dx$ on $\mathbb{R}^ +$. Classical solutions are constructed for $N \geqq {{11} / 4}$. Under mild additional hypotheses the solution evolves in $L^2 (\mathbb{R})$.
## Statistics
### Citations
Dimensions.ai Metrics
|
2019-01-16 18:14:16
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.3988555669784546, "perplexity": 529.4266956732451}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-04/segments/1547583657557.2/warc/CC-MAIN-20190116175238-20190116201238-00358.warc.gz"}
|
https://rjlipton.wordpress.com/2014/01/12/details-left-to-the-reader/
|
The buck stops here—on a blog, that is
Stasys Jukna has written a comprehensive book on Boolean circuit complexity, called Boolean Function Complexity: Advances and Frontiers. It includes a discussion of Mike Fischer’s Theorem on negations, which we recently re-gifted.
Today Ken and I would like to fill in some missing details to Mike’s famous result.
Recall the theorem says:
Theorem 1 Let ${b(n) = \lceil \log(n+1) \rceil}$. Then if a Boolean function can be computed by a circuit of size ${S}$ over the basis ${\{\vee,\wedge,\neg\}}$ then it can be computed by a circuit of size ${2S + n^{O(1)}}$ over the same basis and only using at most ${b(n)}$ negations.
There are a number of places where the “proof” of this theorem is given. In Fischer’s original paper, in more recent improvements, and in Jukna’s book. All say essentially: “we leave to the reader the remaining details.” In the book the details are sketched and left as an exercise, with a long hint. We have discussed this before—see here.
## Proving and Using
There is nothing wrong with leaving out details, but when everyone seems to do that it can cause a problem. I looked at the proof sketches and thought for a bit—okay a more like a byte—that there might even be an error. I was worried there was a mistake in the proof. There is none. The proof is just fine. I was wrong.
But there is some reason to be concerned. For all the beauty of Fischer’s Theorem, it does not seem to have been used to prove something else. I would argue that the real way we get confidence in mathematical results is not by checking their proofs, but by a social process. This process can be improving the main result, which has been done in Fischer’s case. However, these improvements actually affect mostly other parts of the proof, but still leave some details unexplained.
The best way to avoid being concerned and gain confidence in a theorem is to use it to prove results. Who does not believe that numbers modulo a prime form a field? This has been used over and over. A theorem that is used to prove other theorems is much more likely to be correct. If it has a bug and is wrong, there is a kind of strange-attractor phenomena that tends to lead to a contradiction. Or if not an out-and-out contradiction, it may at least lead to a result that is so surprising as to make one doubt the original theorem.
An example of a theorem where maybe only a couple dozen people have fully vetted the proof, but the result is used all the time, is the ${O(\log n)}$-depth, ${O(n \log n)}$-size sorting network of comparison gates designed by Miklós Ajtai, János Komlós, and Endre Szemerédi (AKS). A comparison gate has two input values ${x_1,x_2}$ and gives two output values, ${y_1 = \min\{x_1,x_2\}}$ and ${y_2 = \max\{x_1,x_2\}}$. The ${O}$ hides a big constant—the network is galactic—but its improvement over Ken Batcher’s ${O(\log^2)}$-depth betwork was a boon to studying low-depth circuit complexity. If it were wrong, we’d expect to have seen some unbelievable circuit results by now.
## Fischer’s Proof
Let me try and explain the proof and give you all—well most—of the details, and I will try not to leave anything to the reader. As we recently posted, it suffices to invert the input string ${w = w_1 w_2 \cdots w_n}$ so that we have also the sequence ${(\bar{w}_1,\bar{w}_2,\dots ,\bar{w}_n)}$ using only ${b(n)}$ negations. Then we can compute ${f(w)}$ by a monotone circuit of those ${2n}$ values, at worst doubling the size from ${S}$ to ${2S}$, and in a sense given at the end of this post we can even do a little better.
The first step is to sort the input bits. This can be done by simulating comparison gates without any negations, since when ${x_1,x_2}$ are bits, ${y_1}$ is the AND and ${y_2}$ is the OR. If we care strongly about low depth we can use the AKS network for this, but we could also use Batcher’s. Or whatever.
The second part of the proof, which is critical for us since it uses the negations, is the following: Let ${x_{1},\dots x_{n}}$ be the ${n}$ inputs sorted so that
$\displaystyle x_{1} \le \cdots \le x_{n}.$
Just to add the obvious, which is not always obvious to all, the list must look like
$\displaystyle 0 \underbrace{\dots 0}_{k} \ \underbrace{1 \dots 1}_{m}$
where ${n = k+m}$. The goal, given these sorted values, is to construct the vector
$\displaystyle y_{1},\dots,y_{n}$
so that each ${y_{i}}$ is equal to ${\neg x_{i}}$.
The third and final part is to work the initial sorting backwards so that the outputs ${ y_{1},\dots,y_{n}}$ get routed to their correct locations, corresponding to the ${w_i}$‘s they negate. For those who care about reducing the size of the network, this is where much of the research action is. Fischer’s original idea, used also by Jukna, uses a trick, on pain of needing the quadratic size stated above.
The trick’s idea is to re-interpret the sorted bit ${x_{n-k+1}}$ as telling whether ${w}$ has at least ${k}$ 1’s—call this bit’s value ${t(k)}$. Now for each ${i}$, ${1 \leq i \leq n}$, repeat the initial sorting step on the string ${w}$ minus bit ${w_i}$, and say that the results give values ${t_i(k)}$. All of these ${n^2}$ values are obtained using monotone gates. The trick itself is that for all ${i}$,
$\displaystyle \neg w_i = \bigwedge_{k=1}^n (\neg t(k) \vee t_i(k)).$
The point is that if ${w_i = 0}$, then bit ${i}$ never makes a difference to any “threshold” ${k}$, so ${t(k) = t_i(k)}$ for all ${k}$, so one of ${\neg t(k)}$ and ${t_i(k)}$ is always true, so the big AND gives ${1}$. Whereas if ${w_i = 1}$ then there is some ${k}$ for which ${t(k)}$ is true but ${t_i(k)}$ isn’t, and the big AND gives ${0}$.
Thus the bits ${y_{1},\dots,y_{n}}$ are really supplying the values ${\neg t(1),\dots,\neg t(n)}$ needed for this trick to work. Later authors have had other sorting-based ideas that improve the size, but the second part where the negations arise is the same. With these full details digested, we can focus on this part.
## The Second Part
The claim is that for such sorted values it is possible to construct the vector
$\displaystyle y_{1},\dots,y_{n}$
so that each ${y_{i}}$ is equal to ${\neg x_{i}}$, by using a polynomial size circuit having only ${b(n)}$ negations. Let ${F_{n}(x)=y}$ be this function.
Let’s look at the intuition why this should be true. It is really just a simple divide-and-conquer recursion: one negation allows us to reduce the problem to one of half the size. This clearly yields a logarithmic bound. As usual we will assume that ${n}$ is a power of ${2}$. Look at the middle bit ${x_{m}}$ where ${m=n/2}$. There are two cases:
${\bullet }$ In this case ${x_{m}=0}$. Then we get that
$\displaystyle x_{1}=x_{2}=\cdots=x_{m}=0.$
Thus,
$\displaystyle F_{n}(x) = 0^{m}F_{n/2}(x_{m+1},\dots,x_{n}).$
${\bullet }$ In this case ${x_{m}=1}$. Then we get that
$\displaystyle x_{m}=x_{m+1}=\cdots=x_{n}=1.$
Thus,
$\displaystyle F_{n}(x) = F_{n/2}(x_{1},\dots,x_{m})1^{m}.$
So what is the difficulty? In pseudo-code we are doing the following:
if ${n=1}$ then
$\displaystyle \mathbf{return} \ \neg x_{1}.$
else if ${x_{m}=0}$ then
$\displaystyle \mathbf{return} \ 0^{m}F_{n/2}(x_{m+1},\dots,x_{n})$
else
$\displaystyle \mathbf{return} \ F_{n/2}(x_{1},\dots,x_{m})1^{m}.$
Well this would be just fine if we were using a programming language, but we are using circuits. They do not allow the full flexible array of constructs—at least not in the obvious way—that programming languages do. So this is the dirty detail that is “left to …”
The issue is that we must use circuits to do two things: (i) the recursive call on two different sets of variables based on the value of ${x_{m}}$; and (ii) the output of two different return values, again based on the value of ${x_{m}}$. All this must happen without incurring the cost of an extra negation.
Here is how we do this. Let ${t=x_{m}}$. We will have access to the values of ${t}$ and ${\bar{t}=\neg t}$. We will re-use these values many times, but of course we can do this all with one negation. Once ${t}$ and ${\bar{t}}$ are computed by the circuit, we can by fan-out use the value as many times as we wish. Of course we want to keep the size of the circuit polynomial, but that will follow.
The first problem is how to do two different calls to the circuit ${F_{n/2}}$ without incurring extra negations. If we naively just had a circuit for each call, we would wind up with a linear number of negations—we must avoid two recursive calls. So define new variables ${u_{1},\dots,u_{m}}$ as follows:
$\displaystyle u_{k} = (\bar{t} \wedge x_{m+k}) \vee (t \wedge x_{k}).$
Then we compute
$\displaystyle F_{n/2}(u) = w.$
Note that ${u}$ is set up so that it is one of the two possible calls, and we use the value of ${t}$ to decide which one to call. This uses no additional negations, which would have been terrible.
The second problem is that we need to output different values based on the returned answer ${w}$ and the value of ${t}$. Recall that if ${t=0}$ we want the output to be
$\displaystyle 0^{m}F_{n/2}(x_{m+1},\dots,x_{n}) = 0^{m}w,$
and if ${t=1}$ we want it to be
$\displaystyle F_{n/2}(x_{1},\dots,x_{m})1^{m} = w1^{m}.$
The way to do this in a circuit is as follows:
$\displaystyle \begin{array}{rcl} r &=& 0^{m}w \\ s &=& w1^{m}. \end{array}$
Then the ${k}$-th bit of the output is
$\displaystyle (\bar{t}\wedge r_{k}) \vee (t \wedge s_{k}).$
Done.
## Open Problems
Ken and I hope this helped you feel comfortable with the proof. I know that I feel better about it now. In a sense this is all “obvious” to strong circuit programmers, just as certain other tasks are “obvious” to strong Python—or whatever your favorite language is—programmers. There are always a set of idioms that they know and use daily in their programming. These idioms are so well encoded into their brains that they do not see any reason to supply details. For those of us who are not strong circuit programmers, that includes me, spelling out the details helps.
The real open problem is: can we use Fischer’s Theorem to prove some things that are new and interesting? We wonder.
3 Comments leave one →
1. January 12, 2014 5:26 pm
At the very least, this was a greatly entertaining post. Circuit programming proofs such as this one always have a nice *old-school* touch that is really refreshing in our times. A great Sunday evening treat.
2. Alex Lopez-Ortiz permalink
January 13, 2014 12:27 am
I’m glad you’ve taken the time to fleshen this out. In my academic career I’ve hesitated upon reading rather well known original proofs exactly four times. In two of them the proofs were amended (unprompted) by the authors within a month of publication, the third one the author admitted to the error via email and for the last one I was never confident enough to contact the famous author, since my grasp of the subject was tenous.
I was glad to hear very recently another well known researcher state similar doubts on this last and declare that the result stood, in his opinion, only because there was a second, much simpler proof that could be verified.
So it does happen, though in all cases the proof was eventually amended. Note that I’m not taking about typos. In all cases the errors were in crucial lemmas and were unfixable. The respective arguments in the end took very different paths which avoided the original lemmas altogether.
3. January 15, 2014 7:26 pm
SJ’s latest book on circuit complexity is imho one of the best ever written, its great you’re highlighting it, and its full of wondrous results spanning decades. there are few others in the world who could have written it. his style is also quite readable given such a difficult/technical topic. worthy of emulation. imho circuit complexity will eventually slay the P vs NP dragon. but we’re rapidly closing in on a half-century of work on it…. its truly an epic problem…. one is reminded of einsteins dictum that a problem cannot be solved on the same level of consciousness that created it….
|
2018-06-25 06:21:30
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 103, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6174745559692383, "perplexity": 401.82294661646483}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-26/segments/1529267867493.99/warc/CC-MAIN-20180625053151-20180625073151-00244.warc.gz"}
|
https://socratic.org/questions/what-are-the-products-of-this-reaction-ag-kno-3
|
# What are the products of this reaction: Ag + KNO_3 -> ?
the two substance don't react. Ag is a noble metal and it can oxidize itself only with an energic oxidant. $N {O}_{3}^{-}$ is not strong enought. ${K}^{+}$ is an inert ion.
|
2019-10-17 20:08:30
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 2, "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": 0.2654954195022583, "perplexity": 4823.1131192549965}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-43/segments/1570986676227.57/warc/CC-MAIN-20191017200101-20191017223601-00360.warc.gz"}
|
http://scalability.org/?source=LinuxHPC.org
|
# Friday morning/afternoon code optimization fun
Every now and then I sneak a little time in to play with code. I don’t get to do as much coding as I wish these days … or … not the type of stuff I really enjoy (e.g. squeezing as much performance out of stuff as I can).
The Ceph team are rewriting some code, and they needed a replacement log function that didn’t require FP, and was faster/smaller than table lookup. I played with it a little, starting with FP code (Abramowitz and Stegun is a good source for the underlying math, by copy is dog-eared and falling apart). Then I wanted to see what could be done with pure integer code. Log functions can be implemented in terms of series in FP, or shifts and logical operations(!) for ints.
So pulling out a number of options, I did some basic coding, and summed the logs of functions of argument 1 to 65535 inclusive. Compared this to simple casting of the DP log2 function. Code is simple/straightforward, and fast.
#include <stdio .h>
#include <math .h>
#include <sys /time.h>
#define NUMBER_OF_CALIPER_POINTS 10
struct timeval ti,tf,caliper[NUMBER_OF_CALIPER_POINTS];
struct timezone tzi,tzf;
int log_2 (unsigned int v)
{
register unsigned int r; // result of log2(v) will go here
register unsigned int shift;
r = (v > 0xFFFF) < < 4; v >>= r;
shift = (v > 0xFF ) < < 3; v >>= shift; r |= shift;
shift = (v > 0xF ) < < 2; v >>= shift; r |= shift;
shift = (v > 0x3 ) < < 1; v >>= shift; r |= shift;
r |= (v >> 1);
return r;
}
int main (int argc, char **argv)
{
int i, x, milestone;
float logx;
int sx = 0 , sumlogx = 0 ;
double delta_t;
milestone = 0;
gettimeofday(&caliper[milestone],&tzf);
for(x=1; x<65536; x++) sx+=(int)log2((double)x);
milestone++;
gettimeofday(&caliper[milestone],&tzf);
for(x=1; x<65536; x++) sumlogx+=log_2((unsigned int)x);
milestone++;
gettimeofday(&caliper[milestone],&tzf);
printf("library function sum: %i\n",sx);
printf("local function sum: %i\n",sumlogx);
/* now report the milestone time differences */
for (i=0;i< =(milestone-1);i++)
{
delta_t = (double)(caliper[i+1].tv_sec-caliper[i].tv_sec);
delta_t += (double)(caliper[i+1].tv_usec-caliper[i].tv_usec)/1000000.0;
printf("milestone=%i to %i: time = %.5f seconds\n",i,i+1,delta_t);
}
}
Compile and run it, and this is what I get
landman@metal:~/work/development/log$gcc -o log2.x log2.c -lm ; ./log2.x library function sum: 917506 local function sum: 917506 milestone=0 to 1: time = 0.00284 seconds milestone=1 to 2: time = 0.00091 seconds Milestone 0 to 1 is the library log function. Milestone 1 to 2 is the local version in the source. Now change out the log_2 function for a different variant int log_2 (unsigned int v) { int r; // result goes here static const int MultiplyDeBruijnBitPosition[32] = { 0, 9, 1, 10, 13, 21, 2, 29, 11, 14, 16, 18, 22, 25, 3, 30, 8, 12, 20, 28, 15, 17, 24, 7, 19, 27, 23, 6, 26, 5, 4, 31 }; v |= v >> 1; // first round down to one less than a power of 2 v |= v >> 2; v |= v >> 4; v |= v >> 8; v |= v >> 16; r = MultiplyDeBruijnBitPosition[(unsigned int)(v * 0x07C4ACDDU) >> 27]; return r; } and the results are even more interesting landman@metal:~/work/development/log$ gcc -o log2a.x log2a.c -lm ; ./log2a.x
library function sum: 917506
local function sum: 917506
milestone=0 to 1: time = 0.00288 seconds
milestone=1 to 2: time = 0.00065 seconds
And it gets faster.
Not my code for the log_2 function, but its fun to play with things like this.
Viewed 16138 times by 1819 viewers
## Inventory reduction @scalableinfo
Its that time of year, when the inventory fairies come out and begin their counting. Math isn’t hard, but the day job would like a faster and easier count this year.
So, the day job is working on selling off existing inventory. We have 4 units ready to go out the door to anyone in need of 70-144TB usable storage at 5-6 GB/s per unit. Specs are as follows:
16-24 processor cores
128 GB RAM
48x {2,3,4} TB top mount drives
4x rear mount SSDs (OS/metadata cache)
Scalable OS (Debian Wheezy based Linux OS)
3 year warranty
As this is inventory reduction, the more inventory you take, the happier we are (and the less work that the inventory fairies have to do). We have 4 units to sell off as follows: 2x 72 TB usable, 1x 108 TB usable, and 1x 144 TB usable.
And of course, feel free to order many portable PetaBytes!
Viewed 20104 times by 2060 viewers
## The #PortablePetaByte : Coming to a data center near you!
As seen at SC14. We have our Portable PetaByte systems available for sale. Half rack to many racks, 1 PB and upwards, 20GB/s and up. Faster with SSDs. See the link above!
Viewed 27942 times by 2574 viewers
## Three years
Its been 3 years to the day since I wrote this.
As we’ve been doing before this happened, and after this happened, we are going to a TSO concert on the anniversary of the surgery. Its an affirmation of sorts.
I can tell you that 3 years in, it has changed me in some fairly profound ways … I no longer take some things for granted. I try to spend more time with the family, do more things with them. I waste less time with trivialities.
Only two more years to go and we can call ourselves “survivors”.
Randall Munroe of xkcd fame has, again, done a fairly awesome job of illustrating what goes through your mind as a family member of a survivor. I think Randall and Megan hit their 5 year mark soon. The comic called two years really nailed it.
Viewed 36963 times by 3089 viewers
## Systemd, and the future of Linux init processing
An interesting thing happened over the last few months and years. Systemd, a replacement init process for Linux, gained more adherents, and supplanted the older style init.d/rc scripting in use by many distributions. Ubuntu famously abandoned init.d style processing in favor of upstart and others in the past, and has been rolling over to systemd. Red Hat rolled over to Systemd. As have a number of others.
Including, surprisingly, Debian.
For those whom don’t know what this is, think of it this way. When you turn your machine on, and it starts loading the OS, init is the first process that runs, and it handles starting up all the rest of the system. Its important, and it needs to do its job well.
We care about it for Scalable OS, as we take control the normal startup procedure to handle our use cases. We work well with init.d and a number of others right now. We’ll have to explore systemd a bit more, but in general I am not expecting anything earth shattering.
This is in part, because the vast majority of what we see in the init type systems … well … for lack of a better phrase, just sucks.
Linux has been transitioning to an event based system with udev for a while. Udev is a rule based mechanism to handle events, with a kernel and user space component. Woe be unto those whom mess with the way udev wants to work, as the scripts behind it are … broken … badly … . I say this as someone whom has tried, very hard in a number of cases, to fix the broken-ness. In many cases I’ve discovered its easier to ignore the broken section and add intelligence into the setup/config code to work around the udev brain death.
Specifically, Linux does a great job at diskless booting. That is, until you share some directories. That udev needs. And assumes it has private copies of. So you have to work around that. You can’t fix udev … your fixes won’t be accepted upstream, and will just break at next OS update. So its easier to hack around it, and use a very light diversionary slight-of-bits touch to make sure it does what you want.
Another great example is md RAID1 OS drives, and RHEL6/CentOS6. We actually had to hack the whole initramfs approach to work around the broken udev module that brings up raids (and failed to correctly bring up raids for the system to fully start up).
Yeah, I know … open source makes it possible. Terrible implementation makes it necessary.
Upstart was a little more sane, but still had some issues.
init.d/rc in Debian 7 is reasonable, though we’ve seen still quite a bit of breakage.
This all goes to the philosophy of the distro. Are they trying to be everything to everyone, or be a very well crafted system for a set of purposes. Too many want the former, not enough the latter. Scalable OS is all about the latter. Make it boot, easily (not quickly now, but thats coming), make it just work.
Systemd promises to make startup in the init process different, pluggable (think udev and its horror), and so forth. We’ll have to play with it to see if it is mostly harmless or not. I suspect its going to cause at least little grief with our startup mechanism, so we’ll see if we need to work around it, or throw it away.
During startup, many distros have concepts where they read (assumed local) configuration files to set up file systems, networks, functions. This is a lousy thing to do for clouds, clusters, etc. You really want a distributed control mechanism that provides these config options. Scalable OS has this implicit within it. But to get to this distributed control layer, you need network access.
And this is where most distros are sheer and utter crap in their network setup code. We have a far better way built into Scalable OS, that was born of the frustration of dealing with the distros broken network config mechanisms. Generally speaking, you should never start a dhcp process on an ethernet port that doesn’t have a carrier present (after bringing the port up). Yet, this is exactly what most distros do by default. It gets even more interesting when you invoke udev, pci scanning in the kernel (done in a different order from a previous kernel, so items are discovered in a different order), so that some machines are absolutely unable to get back onto the network after a kernel update.
Yeah, we’ve seen/experienced this. Quite common with RHEL/CentOS kernels updating to ours. And we’ve got work arounds to deal with udev when we need to.
The question we have is, will systemd make this better? Worse? Not impact this? I suspect that the pci scan done by the kernel won’t change much, its simply how systemd will respond to this. We know how udev/init.d respond to things, and we’ve done our process change to remove the terrible/useless sections whereever possible. Though we still, on occasion get bit by udev race conditions.
Udev is a piece of work. Fantastic for small machines without much stuff. Absolutely, completely borked for machines with lots of stuff. We see some occasional, fantastic, non-controllable race conditions in udev processes, that init is handling. My hope is that systemd is far smarter than its predecessors. I hate having to tell people that the solution to this seemingly mad system is to reboot it. Yet udev will drive you to this. Hopefully we can move past that.
But, if not, we’ll do what we’ve done with the others, and work around it, disabling what gets in the way.
That coupled with our configuration mechansim (not quite CoreOS like, we aren’t using etcd right now, but we’ll be evaluating it and other options … build vs “buy”), and we’ll be fine. Actually far better than any system that depends upon external mechanisms (like Chef, Puppet, et al) to configure machines post installation.
Hardware is code, and should be automatically configured. This is what Scalable OS does, and I am hoping that systemd won’t get in our way.
Even if it does, Debian just forked over this. So I am not worried at all.
There are some folks saying this is the “end of Linux” or other such fluff.
Not likely, but in the end, the operating system is an implementation detail (machine/container as code, its merely a configuration option). As long as I can use the hardware well, I am happy. Right now, for better or worse, Linux has the best driver support in market, albeit sometimes a maddening driver support (see binary only modules delivered by OEMs without a clue). There are other choices … I’d love to see better driver support for Illumos based machines, and *BSD (though these generally do have OK driver support).
But we need Infiniband support, we need 40GbE and above support, we need memory channel storage and NVMe support for our customers. Limited choices there now.
So systemd will be a challenge to get through, but I am not overly worried. I see the OS as a substrate upon which to run bare metal/containerized/VMed apps. Systemd shouldn’t impact that too much, and if it does, it will be swept away.
Viewed 39571 times by 3260 viewers
## Brings a smile to my face
My soon to be 15 year old daughter was engrossed with something on her laptop yesterday. Thinking it was fan-fiction, I asked her what she was writing.
She knitted her brow for a moment, and looked up.
“Its code combat Dad.” she said, quite matter of factly.
I must have had a slightly startled expression on my face. I knew she had dabbled with it, and had recommended (/sigh) Python as a language, after she took (and aced) a Java class last year, as Python is inherently simpler. I would have loved to have introduced her to Perl, but she needs to figure out which tools to use on her own.
“Nice” I said. “What are you coding in?”
“Well” she said, “I had been using Python, but it was too annoying. So I started using Lua.”
I was stunned at several levels. None of which have anything to do with gender. Mostly having to do with what I thought I knew about what she wanted to do in her down time. And that she now has taught herself two computer languages for code combat and other activities.
And of course, the self-taught computer geek (started with Basic, Assembly, Fortran, C, …) in me was thinking “a chip off the old block”.
I don’t want to push her in any particular direction, she’s got to decide what she wants, and discover what she enjoys. And that appears to include learning new computer languages for programming contests and games.
I want to make sure I encourage this. Solving problems is fun, and coding should be fun. I think she’s getting this.
Viewed 43089 times by 3330 viewers
## Learning to respect my gut feelings again
A “gut feeling” is, at a deep level, a fundamental sense of something that you can’t necessarily ascribe metrics to, you can’t quantify exactly. Its not always right. Its a subconscious set of facts, ideas, concepts that seem to suggest something below the analytical portion of your mind, and it could bias you into a particular set of directions. Or you could take it as an aberration and go with “facts”.
As an entrepreneur, I’ve had many gut feelings about what to do, when to do it. They aren’t always right. But when they are … they are usually whoppers.
In 2002, when Vipin and I were in his lab looking at 40 core DSP chips, and speaking aloud about building accelerators out of them … that was a gut feeling that the high performance computing market had no other choice than to go that route to continue to advance in performance. We built business plans, architected designs for the platform, pricing models, went to investors, told them things that later turned out to be remarkably prescient. No one saw fit to invest unfortunately.
This has been a feature of my career. Many very good ideas, that later turn out to be huge markets, developing almost exactly as we speculated, and we can’t get investment. Its fundamentally, profoundly frustrating. Not to mention discouraging. But we soldier on.
In 2005, when we had sold the concept of remote computer cycles (what was to be later called “cloud”) to a large company in Michigan, again, we tried to get investment going. We had a large committed customer, a good business model, a good operational model, even some investors lined up if we could get the state of Michigan to commit as part of its tri-corridor process. They only need have put in a token amount, and thats all we needed. I need not tell you this didn’t happen, and the reasons given were, sadly, laughably wrong at best.
Our gut feelings on both of these markets were that they were going to be huge.
To put it midly, we were right. Very very right.
The next epiphany was on the cluster and storage side. We’d been designing and building clusters up until then with embedded storage. Dell decided it wanted to own clusters, and it worked on depriving the small folks of oxygen with pricing gymnastics. It was easy for them to write off coming in under cost on clusters, much harder for a small outfit to justify paying a customer to take hardware. My gut feeling at the time was that clusters would become an impossibly hard market to work in, so we focused upon where we could add our unique value. Storage and storage based systems it was.
Along the way, we’ve seen many opportunities, some looking very good but bugging me something fierce, and some looking bad on the surface, but having the qualities that we needed. So I went with my gut on whether or not to pursue those.
And we’ve grown, by quite a bit during that time. There is much to be said for subconcious analytics.
As we’ve grown, we’ve brought on more people to work on opportunities. Recently we’ve had some opps and we’ve serviced them, which have run strongly against my gut feeling. As we’ve seen these evolve, my gut was right, they’ve turned into (in some cases) bad deals for us.
Also, as we look at our capitalization efforts, I get similar feelings about particular directions and potential investors. I don’t mind lots of legalese. We have lawyers to deal with that. I mind games. If people play them now, it will be worse later on. If they won’t act in reasonable time frames and with reasonable terms, we need to move on.
Its this gut feeling that has served us very well, that I temporarily overrode in the past … that I am bringing back in a big way.
We met with an erstwhile customer at the SC14 show. Makes great promises, sets additional hurdles. Never does business with us. I’d like to, but the cost of chasing this customer may simply be too high for us, for little return, at this stage of our life. If we were bigger, it would be less of an issue. If we had a large investor with a lot of committed capital in us, again, less of an issue to act on working with them. But we don’t as of yet. We have a particular hand of cards we can play, and some we need to discard in order to improve our hand. As much as I might like to play this hand with that card for that customer, my gut tells me to wait.
Its tough, but I’m going back to my gut feelings on these. For customers, for investors, whatever. If I don’t get a good feeling, or if I see actions which on their own might be innocuous, but collectively would be predatory, I’ll rethink working with them.
The gut feeling is all about the value prop and the ROI on effort. Sometimes its dead on. Far more often for us than not. Its time to use it again, in the large.
Viewed 36606 times by 3657 viewers
## #SC14 day 2: @LuceraHQ tops @scalableinfo hardware … with Scalable Info hardware …
Report XTR141111 was just released by STAC Research for the M3 benchmarks. We are absolutely thrilled, as some of our records were bested by newer versions of our hardware with newer software stack. Congratulations to Lucera, STAC Research for getting the results out, and the good folks at McObject for building the underlying database technology.
This result continues and extends Scalable Informatics domination of the STAC M3 results. I’ll check to be sure, but I believe we are now the hardware side of most of the published records.
Whats really cool about this is that you can get this power from Lucera if you don’t want or need to stand up your own kit, from us or our partners if you prefer to stand up your own private cloud, or combinations if you would like to take advantage of all the additional capabilities and functionality Lucera brings to the table.
Viewed 43496 times by 4169 viewers
## Starting to come around to the idea that swap in any form, is evil
Here’s the basic theory behind swap space. Memory is expensive, disk is cheap. Only use the faster memory for active things, and aggressively swap out the less used things. This provides a virtual address space larger than physical/logical memory.
Great, right?
No. Heres why.
1) swap makes the assumption that you can always write/read to persistent memory (disk/swap). It never assumes persistent memory could have a failure. Hence, if some amount of paged data on disk suddenly disappeared, well …
Put another way, it increases your failure likelihood, by involving components with higher probability of failure into a pathway which assumes no failure.
2) it uses 4k pages (on linux). Just. Shoot. Me. Now. Ok, there are ways to tune this a bit, and we’ve done this, but … but … you really don’t want to do many many 4k IOs to a storage device. Even an SSD.
NVMe/MCS may help here. But you still have the issue number 1, unless you can guarantee atomic/replicated writes to the NVMe/MCS.
3) Performance. Sure, go ahead and allocate, and then touch every page of that 2TB memory allocation on your 128GB machine. Go ahead. I’ve got a decade or two to wait.
4) Interaction with the IO layer is sometimes buggy in surprising ways. If you use a file system, or a network attached block device (think cloud-ish), and you need to allocate a SKB or some additional memory to write the block out, be prepared for some exciting (and not in the good way) failures, some spectacular kernel traces that you would swear are recursive allocation death spirals.
“Could not write block as we could not allocate memory to prepare swap block for write …”
Yeah. These are not fun.
5) OOM is evil. There is just no nice way to put this. OOM is evil. If it runs, think “wild west”. kill -9 bullets have been lobbed against, often, important, things. Using ICL to trace what happened will often lead you agape with amazement at the bloodbath you see in front of you.
So towards this end, we’ve been shutting off paging whenever possible, and the systems have been generally faster and more stable. We’ve got some ideas on even better isolation of services to prevent self flagellation of machines. But the take home lesson we’ve been learning is … buy more ram … it will save you headache and heartache.
Viewed 53673 times by 4475 viewers
## #sc14 T-minus 2 days and counting #HPCmatters
On the plane down to NOLA. Going to do booth setup, and then network/machine/demo setup. We’ll have a demo visualfx reel from a customer whom uses Scalable Informatics JackRabbit, DeltaV (and as the result of an upgrade yesterday), Unison.
Looking forward to getting everything going, and it will be good to see everyone at the show!
Viewed 47045 times by 4216 viewers
|
2014-12-21 14:23:58
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 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": 0.32491761445999146, "perplexity": 2889.166000837197}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-52/segments/1418802771374.156/warc/CC-MAIN-20141217075251-00059-ip-10-231-17-201.ec2.internal.warc.gz"}
|
https://www.physicsforums.com/threads/calculus-splitting-variables.169383/
|
Calculus: splitting variables
1. May 8, 2007
t_n_p
1. The problem statement, all variables and given/known data
http://img291.imageshack.us/img291/4489/splittingvariablesgz9.jpg [Broken]
3. The attempt at a solution
I can't split the x and y variables in each of these three cases. Would somebody be able to help me get started?
Last edited by a moderator: May 2, 2017
2. May 8, 2007
Hootenanny
Staff Emeritus
For (a).(i)
$$\frac{dy}{dx} = \frac{x(y^2+3)}{y}$$
Divide through by $(y^2+3)/y$ yielding;
$$\frac{y}{(y^2+3)}\cdot\frac{dy}{dx} = x$$
For (a).(ii)
It may be useful to note that;
$$e^{x-2y} = e^x\cdot e^{-2y}= \frac{e^x}{e^{2y}}$$
As for (b), I don't think it is seperable.
Last edited: May 8, 2007
3. May 8, 2007
t_n_p
Regarding a) i) how would I differentiate y/(y²+3)?
and for a) ii) that's very helpful
Thanks!
4. May 8, 2007
Hootenanny
Staff Emeritus
For (a)(i), I would use the quotient rule; but I don't think you mean differentiate, I think you mean integrate. If this is the case it may be useful to note that;
$$y = \frac{1}{2}\cdot\frac{d}{dx}(y^2+3)$$
and
$$\int\frac{f'(x)}{f(x)} = \ln|f(x)| + C$$
Last edited: May 8, 2007
5. May 8, 2007
t_n_p
Silly me, of course I meant integrate! :yuck:
Thanks again!:tongue2:
Am I on the right track?
http://img150.imageshack.us/img150/2133/asdffy0.jpg [Broken]
Last edited by a moderator: May 2, 2017
6. May 8, 2007
Hootenanny
Staff Emeritus
Yeah, you've got it basically correct, however, there are a few things I should say.
On line (5), you should only include one constant of integration (one one side), otherwise they cancel each other out and your left without a constant of integration!
Secondly, you should ask yourself is there any requirement for the modulus signs? Can y2+3 ever be negative?
Thirdly, a somewhat minor point, on line (6) your final term should be 2C, not just C.
7. May 8, 2007
t_n_p
thanks for pointing that out, similiarly with (ii) should I only introduce the constant on one side (I'm assuming so, as the value (1,0) is given to find C...
Just double checking!
8. May 8, 2007
Hootenanny
Staff Emeritus
Yes, well actually what your doing is;
$$\int f(y) dy = \int g(x) dx$$
$$F(y) + A = G(x) + B$$
$$F(y) = G(x) + C \hspace{1cm}\text{where }C = B-A$$
Does that make sense?
9. May 8, 2007
t_n_p
Should I show likewise for a) i)?
10. May 8, 2007
Hootenanny
Staff Emeritus
No not unless you want to , it isn't usually nesscary unless explicitly stated.
11. May 8, 2007
t_n_p
Yeah, I don't see why not
Could you please double check my value of c for (ii).
I basically subbed in (1,0) as you do, but the answer seems a bit left of field, so I guess it's best to double check.
http://img501.imageshack.us/img501/7300/85694878mp9.jpg [Broken]
Last edited by a moderator: May 2, 2017
12. May 10, 2007
t_n_p
*bump* can anybody confirm my value of c?
13. May 10, 2007
t_n_p
actually I double checked, I now think c=-1-ln(2)
14. May 14, 2007
t_n_p
Couldn't seem to edit my previous post, but my final answer is y=(x-1)/2 which sounds alot better!
|
2017-10-21 10:49:35
|
{"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": 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": 0.7518793940544128, "perplexity": 2437.8436394364576}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-43/segments/1508187824733.32/warc/CC-MAIN-20171021095939-20171021115939-00458.warc.gz"}
|
https://www.biostars.org/p/219185/
|
Bowtie indexing of a fasta file that consists of a large amount of sequences
0
0
Entering edit mode
6.4 years ago
valerie ▴ 100
Hi
I have a fasta file with 44396570 sequences and I want to find them in my genome. I am using bowtie2-build to build index files for my fasta file, but I get an error "killed". I guess it's because I need more memory. I have 32gb on my machine. I tried to split my file in 2 and 4 parts but it didn't solve my problem. I also made sure that bowtie2-build works on one single sequence from my fasta file and everything was ok, but bowtie2 alignment of my whole genome on this single sequence took a huge amount of time and I am worried that splitting my fasta file into more parts will take years to perform the alignment.
Maybe someone already solved this problem, when you want to find a lot of sequences in your genome?
I would be grateful if you could help me.
bowtie alignment genome • 3.9k views
1
Entering edit mode
Normal procedure is to use your sequences and search against the genome, is there a reason you want to do this the other way round? Are your sequences short reads or long(er)? How large is the "genome" in this case?
Many NGS aligners are not designed to search with very long sequences (say > 1kb, with a few exceptions, e.g. blasr from PacBio). BBMap contains mapPacBio.sh which can map long reads (in 6kb chunks) and bwa also works with PacBio/Nanopore reads.
0
Entering edit mode
I have experimental WGS data and I am interested in presence or absence of my sequences in genomes of these exact patients, not in reference genome. My sequences are longer then reads. They are different length, but something like gene length, not shorter.
1
Entering edit mode
Have you thought about using blat/blast/lastz? Depending on how similar you expect your sequences to be when compared the "genome" you could go from "near identical to more diverse" in rough order from left to right for the programs above.
0
Entering edit mode
Did not think about blat, thank you! I expect the sequences to be identical so I think it should work. May I also ask you one more question? My genomes are in fastq format, should I align them on human genome or convert somehow into fasta to use blat? Thank you in advance!
1
Entering edit mode
I now understand what you are trying to do.
So your "genomes" are actually short reads which you want to align against some other longer sequences. You could convert them to fasta and then use one of the programs above but that may still be a long endeavor since you have many (milions?) of those reads.
What is the total size of your long reads (in terms of file size/mega-gigabases of sequence)? You could use bbmap.sh from BBMap suite to align your genome reads (in original fastq format) to the other dataset. This may be the best option in terms of getting this done rapidly.
0
Entering edit mode
The size of the "genomes" is ~200gb for each of the 2 paired-end files. The size of the fasta file with my sequences I want to find is 24gb. Is it ok for BBMap?
1
Entering edit mode
Now things are coming into focus :)
24G of "reference" would not be good for most aligners with the amount of RAM you have available. If you had plenty of RAM then this would be no problem.
How about using "divide and conquer"? How many sequences are there in the "fasta" file? You could divide them in 8 x 3Gb chunks (which should be about the size of human genome) and then do 8 independent searches.
0
Entering edit mode
Haha :) There are 44396570 sequences in it. I tried to split into 4 parts, failed, got upset and came here :) I can definitely try to split the fasta file into 8 parts. Do you think I should use BBMap or Bowtie?
Thank you very much for your help! You are my hero :)
1
Entering edit mode
Before we celebrate try one 3 Gb chunk out first. I am partial to bbmap but either aligner should work at that size :) Keep an eye on free disk space and add more if you can, just to be proactive. With bbmap you can directly create bam files as long as you have samtools available in \$PATH.
At 200G each (R1/R2) you must have a boatload of reads in there (billions?).
0
Entering edit mode
Ok, I'll do that now! Anyway, thank you very much!
0
Entering edit mode
Yes. Originally I had .bam files provided by the guys who did the experiment. Each file was ~150G and I converted each .bam file into two .fastq files with bedtools bam2fastq. And now I have two 200G for each genome.
0
Entering edit mode
Tada! I started with bowtie and bowtie-build part is completed without any issues, now I'm waiting for the alignment! Thank you very much!! You definitely are my hero! :)
0
Entering edit mode
Glad to hear that appears to be working.
BTW: You don't need to up vote each one of my posts. Only one (or some) that helped provide a solution/direction would be sufficient.
0
Entering edit mode
Wow. Didn't know this feature, thank you for the information. How do people solve these issues then? I thought that standard procedure is to align the WGS data on whole chromosomes. Maybe I could concat all my sequences into one sequence, like genes on the chromosome and align everything on it and then search whether reads mapped each of the parts of the sequence? Or is this crazy? :) Sorry for stupid questions, I am doing this for the first time
1
Entering edit mode
Please see the edited post above since I made some changes there.
What is the size distribution of your reads and are the "genomes" you are searching against in form of chromosomes or contigs or something else?
0
Entering edit mode
My genomes are two fastq files with paired-ended reads for each of the genomes. The size of reads is 100. I am searching against sequences that are actually specific combinations of exons of human genes. This is why their length is similar to length of genes. Each combination is a different sequence and there are thousands of these combinations.
1
Entering edit mode
I am surprised that you were not able to build the index with bowtie2-build if your sequences are a "transcriptome". Did the job get killed right away or it ran for sometime before it died? I wonder if you ran out of disk storage space rather then memory (32G should be enough, if all of it is available for the OS).
0
Entering edit mode
I have 200 gb on my disk and can add more. It died while doing something with the "buckets", unfortunately I didn't save the output and do not remember exactly when it died. It worked for something like half an hour before died.
|
2023-03-29 14:12:23
|
{"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": 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": 0.3404889702796936, "perplexity": 1809.5992303203739}, "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/1679296948976.45/warc/CC-MAIN-20230329120545-20230329150545-00355.warc.gz"}
|
https://www.omnicalculator.com/physics/projectile-motion
|
# Projectile Motion Calculator
Created by Bogna Szyk and Hanna Pamuła, PhD
Reviewed by Steven Wooding and Jack Bowater
Last updated: Dec 22, 2022
Our projectile motion calculator is a tool that helps you analyze parabolic projectile motion. It can find the time of flight, but also the components of velocity, the range of the projectile, and the maximum height of flight. Continue reading if you want to understand what projectile motion is, get familiar with the projectile motion definition, and determine the abovementioned values using the projectile motion equations.
Prefer watching over reading? Learn all you need in 90 seconds with this video we made for you:
## What is projectile motion? Projectile motion definition
Imagine an archer sending an arrow in the air. It starts moving up and forward, at some inclination to the ground. The further it flies, the slower its ascent is – and finally, it starts descending, moving now downwards and forwards and finally hitting the ground again. If you could trace its path, it would be a curve called a trajectory in the shape of a parabola. Any object moving in such a way is in projectile motion. By the way, we have the arrow speed calculator that analyzes the motion of arrows - give it a try!
Only one force acts on a projectile – the gravity force. Air resistance is always omitted. If you drew a free body diagram of such an object, you would only have to draw one downward vector and denote it “gravity”. If there were any other forces acting on the body, then – by projectile motion definition – it wouldn't be a projectile.
## Projectile motion analysis
Projectile motion is pretty logical. Let's assume you know the initial velocity of the object $V$, the angle of launch $\alpha$, and the initial height $h$. Our projectile motion calculator follows these steps to find all remaining parameters:
1. Calculate the components of velocity.
• The horizontal velocity component $V_\mathrm x$ is equal to $V \cos\alpha$.
• The vertical velocity component $V_\mathrm y$ is equal to $V \sin\alpha$.
• Three vectors - $V$, $V_\mathrm x$ and $V_\mathrm y$ - form a right triangle.
If the vertical velocity component is equal to 0, then it's the case of horizontal projectile motion. If, additionally, α = 90°, then it's the case of free fall. We tackled both problems in the horizontal projectile motion calculator and free fall calculator, respectively.
1. Write down the equations of motion.
Distance
• Horizontal distance traveled can be expressed as $x = V_\mathrm x t$ where $t$ is the time.
• Vertical distance from the ground is described by the formula $y = h + V_\mathrm y t - g t^2 / 2$, where $g$ is the gravity acceleration.
Velocity
• Horizontal velocity is equal to $V_\mathrm x$.
• Vertical velocity can be expressed as $V_\mathrm y - g t$.
Acceleration
• Horizontal acceleration is equal to 0.
• Vertical acceleration is equal to $-g$ (because only gravity acts on the projectile).
1. Calculate the time of flight.
• Flight ends when the projectile hits the ground. We can say that it happens when the vertical distance from the ground is equal to 0. In the case where the initial height is 0, the formula can be written as: $V_\mathrm y t - g t^2 / 2 = 0$. Then, from that equation, we find that the time of flight is
$\quad t = 2 \frac{V_\mathrm y}{g} = 2 \frac{V}{g} \sin\alpha.$
• However, if we're throwing the object from some elevation, then the formula is not so nicely reduced as before, and we obtain a quadratic equation to solve: $h + V_\mathrm y t - g t^2 / 2 = 0$. After solving this equation, we get:
$t \! = \! \frac{V \sin\alpha \! + \! \sqrt{ \! (V \sin\alpha)^2 \! + \! 2 g h}}{g}$
1. Calculate the range of the projectile.
• The range of the projectile is the total horizontal distance traveled during the flight time. Again, if we're launching the object from the ground (initial height = 0), then we can write the formula as $R = V_\mathrm x t = V_\mathrm x \times 2 \times V_\mathrm y / g$. It may be also transformed into the form: $R = V^2 \sin(2\alpha) / g$
• Things are getting more complicated for initial elevation differing from 0. Then, we need to substitute the long formula from the previous step as $t$:
$R = V_\mathrm x t = \\[0.8em] V \! \cos \alpha \frac{V \! \sin\alpha \! + \! \sqrt{ \! (V \! \sin\alpha)^2 \! + \! 2 g h}}{g}$
1. Calculate the maximum height.
• When the projectile reaches the maximum height, it stops moving up and starts falling. It means that its vertical velocity component changes from positive to negative – in other words, it is equal to 0 for a brief moment at time $t(V_\mathrm y=0)$.
• If $V_\mathrm y - g t(V_\mathrm y=0) = 0$, then we can reformulate this equation to $t(V_\mathrm y=0) = V_\mathrm y / g$.
• Now, we simply find the vertical distance from the ground at that time:
$\begin{split} h_\mathrm{max} = &V_\mathrm y t( V_\mathrm y \! = \! 0) \! - \! \frac{g t^2 (V_\mathrm y \! = \! 0)}{2} = \\[0.8em] &\frac{V^2_\mathrm y}{2 g} = \frac{V^2 \sin ^2 \alpha}{2 g} \end{split}$
• Fortunately in the case of launching a projectile from some initial height $h$, we need to simply add that value into the final formula:
$h_\mathrm{max} = h + \frac{V^2 \sin^2 \alpha}{2 g}$
## Projectile motion equations
Uff, that was a lot of calculations! Let's sum that up to form the most essential projectile motion equations:
1. Launching the object from the ground (initial height h = 0)
• Horizontal velocity component: $V_\mathrm x = V \cos \alpha$
• Vertical velocity component: $V_\mathrm y = V \sin \alpha$
• Time of flight: $t = 2 V_\mathrm y / g$
• Range of the projectile: $R = 2 V_\mathrm x V_\mathrm y / g$
• Maximum height: $h_\mathrm{max} = V^2_\mathrm y / (2 g)$
1. Launching the object from some elevation (initial height h > 0)
• Horizontal velocity component: $V_\mathrm x = V \cos \alpha$
• Vertical velocity component: $V_\mathrm y = V \sin \alpha$
• Time of flight: $t = \left[V_\mathrm y + \sqrt{V^2_\mathrm y + 2 g h}\right] / g$
• Range of the projectile: $R = V_\mathrm x \left[V_\mathrm y + \sqrt{V^2_\mathrm y + 2 g h}\right] / g$
• Maximum height: $h_\mathrm{max} = h + V^2_ \mathrm y / (2 g)$
Using our projectile motion calculator will surely save you a lot of time. It can also work 'in reverse'. For example, enter the time of flight, distance, and initial height, and watch it do all calculations for you!
Be sure also to check the parabola calculator to learn more about such a curve from a mathematical point of view.
## FAQ
### Does projectile motion have to travel horizontally?
No, projectile motion and its equations cover all objects in motion where the only force acting on them is gravity. This includes objects that are thrown straight up, those thrown horizontally, those that have a horizontal and vertical component, and those that are simply dropped.
### What is an example of projectile motion?
Objects with projectile motion include: keys being thrown, a 300 kg projectile being thrown 90 m by a trebuchet, a football being kicked so that it no longer touches the ground, a diver jumping from a diving board, an artillery shell the moment it leaves the barrel, and a car trying to jump a bridge.
### How can a projectile fall around the Earth?
There is only one force acting on a projectile - gravity. This means that an object will eventually fall to Earth. But what about if the object is moving so fast horizontally that, by the time it reaches the ground, the ground is no longer there? This is the principle that governs satellites.
### How do I find acceleration in projectile motion?
There is only one force acting on an object in projectile motion - gravity. This means that any change in vertical speed is due to gravitational acceleration, which is 9.81 m/s2 (32.2 ft/s2) on Earth. In the horizontal direction, there is no change in speed, as air resistance is assumed to be negligible, so acceleration is 0.
### What factors affect the motion of a projectile launched horizontally?
Initial velocity, the initial height the projectile is being launched from, and gravity will all affect a projectile launched horizontally. Air resistance will also have an effect in real life, but for most theoretical calculations it is negligible and is therefore ignored. If the projectile has wings, this will also impact its motion, as it will glide.
### What exactly is a projectile?
A projectile is an object that is in motion, in the air and has no force acting upon it other than the acceleration due to gravity (this means that it cannot be self-propelled). You can probably think of many examples: a thrown ball or a stone thrown from a trebuchet. Even the Moon is a projectile, with respect to the Earth!
### What are the characteristics of projectile motion?
The properties of projectile motion are that the object’s horizontal velocity does not change, that it’s vertical velocity constantly changes due to gravity, that the shape of its trajectory will be a parabola, and that the object is not affected by air resistance.
### Who first accurately described projectile motion and when?
Galileo was the first person to describe projectile motion accurately, by breaking down motion into a horizontal and vertical component, and realizing that the plot of any object's motion would always be a parabola. He described it in his book, On Motion, published around the 1590s.
### Why does a projectile follow a curved path?
An object follows a parabola because of how its two components of motion - the horizontal and vertical - are affected by gravity. The horizontal component is not affected by gravity at all, and so changes in a constant, linear fashion. The vertical part, however, is constantly affected by gravity, and so it will increase in height, and then decrease, accelerating due to gravity.
### Why is 45 degrees the optimal angle for projectiles?
The equation for the distance traveled by a projectile being affected by gravity is sin(2θ)v2/g, where θ is the angle, v is the initial velocity and g is acceleration due to gravity. Assuming that v2/g is constant, the greatest distance will be when sin(2θ) is at its maximum, which is when 2θ = 90 degrees. This means θ = 45 degrees.
Bogna Szyk and Hanna Pamuła, PhD
Initial parameters
Initial velocity (V)
ft/s
Angle of launch (α)
deg
Initial height (h)
ft
Time of flight (t)
sec
Distance (d)
ft
Maximum height (hmax)
ft
Flight parameters at given time
Time
sec
Velocity
ft/s
People also viewed…
### Black hole collision
The Black Hole Collision Calculator lets you see the effects of a black hole collision, as well as revealing some of the mysteries of black holes, come on in and enjoy!
### Photon detection efficiency (SiPM)
Photon detection efficiency calculator is a handy tool that allows for a quick conversion between responsivity and Photon Detection Efficiency (PDE) in Silicon Photomultipliers (SiPMs).
### Resultant force
Use the net force calculator to find the resultant force on a body.
### Secretary problem (Valentine's day)
Use the dating theory calculator to enhance your chances of picking the best lifetime partner.
|
2023-02-05 22:52:38
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 41, "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": 0.7304123640060425, "perplexity": 568.3348392428244}, "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/1674764500294.64/warc/CC-MAIN-20230205224620-20230206014620-00760.warc.gz"}
|
https://gmatclub.com/forum/two-tracks-are-parallel-the-first-track-has-6-checkpoints-and-the-sec-219552.html
|
GMAT Question of the Day - Daily to your Mailbox; hard ones only
It is currently 19 Jun 2018, 16:50
### GMAT Club Daily Prep
#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
we will pick new questions that match your level based on your Timer History
Track
every week, we’ll send you an estimated GMAT score based on your performance
Practice
Pays
we will pick new questions that match your level based on your Timer History
# Events & Promotions
###### Events & Promotions in June
Open Detailed Calendar
# Two tracks are parallel. The first track has 6 checkpoints and the sec
Author Message
TAGS:
### Hide Tags
Manager
Joined: 14 Dec 2015
Posts: 50
Concentration: Entrepreneurship, General Management
WE: Information Technology (Computer Software)
Two tracks are parallel. The first track has 6 checkpoints and the sec [#permalink]
### Show Tags
02 Jun 2016, 22:43
3
8
00:00
Difficulty:
45% (medium)
Question Stats:
62% (01:28) correct 38% (01:17) wrong based on 145 sessions
### HideShow timer Statistics
Two tracks are parallel. The first track has 6 checkpoints and the second one has 10 checkpoints. In how many ways can the 6 checkpoints of first track be joined with the 10 checkpoints of the second to form a triangle?
(A)120
(B)150
(C)200
(D)270
(E)420
_________________
"Fight the HARDEST battle that anyone can ever imagine"
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8100
Location: Pune, India
Re: Two tracks are parallel. The first track has 6 checkpoints and the sec [#permalink]
### Show Tags
02 Jun 2016, 22:47
2
5
snorkeler wrote:
Two tracks are parallel. The first track has 6 checkpoints and the second one has 10 checkpoints. In how many ways can the 6 checkpoints of first track be joined with the 10 checkpoints of the second to form a triangle?
(A)120
(B)150
(C)200
(D)270
(E)420
To make a triangle, you need 2 checkpoints from one track and 1 from the other. You cannot have all 3 from the same track since then the points will be in a line (assuming straight line of track)
You select 2 checkpoints from the first track and one from the second or two from the second track and one from the first.
6C2 * 10C1 + 10C2 * 6C1 = 150 + 270 = 420
_________________
Karishma
Veritas Prep | GMAT Instructor
My Blog
Get started with Veritas Prep GMAT On Demand for \$199
Veritas Prep Reviews
Senior Manager
Joined: 18 Jan 2010
Posts: 254
Two tracks are parallel. The first track has 6 checkpoints and the sec [#permalink]
### Show Tags
02 Jun 2016, 23:48
1
snorkeler wrote:
Two tracks are parallel. The first track has 6 checkpoints and the second one has 10 checkpoints. In how many ways can the 6 checkpoints of first track be joined with the 10 checkpoints of the second to form a triangle?
(A)120
(B)150
(C)200
(D)270
(E)420
Two ways this can be done:
a) select any two points on the first track (out of available 6 points) to form the base, and choose the vertex from track B
b) select any two points on the second track (out of available 10 points) to form the base, and choose the vertex from track A
(6C2 * 10C1) + (6C1*10C2)
[$$\frac{6!}{4!2!}$$ * 10]+ [$$\frac{6!}{5!1!}$$ * $$\frac{10!}{8!2!}$$]
[$$\frac{6 *5}{2}$$ * 10]+ 6 * [$$\frac{10 * 9}{2}$$]
= 150+ 270
=420
Non-Human User
Joined: 09 Sep 2013
Posts: 7003
Re: Two tracks are parallel. The first track has 6 checkpoints and the sec [#permalink]
### Show Tags
16 Sep 2017, 00:10
Hello from the GMAT Club BumpBot!
Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).
Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
Re: Two tracks are parallel. The first track has 6 checkpoints and the sec [#permalink] 16 Sep 2017, 00:10
Display posts from previous: Sort by
|
2018-06-19 23:50:04
|
{"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": 0.4661688804626465, "perplexity": 4205.252769657012}, "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/1529267863259.12/warc/CC-MAIN-20180619232009-20180620012009-00216.warc.gz"}
|
http://tischhdev.com/ss9crik/9c28fe-metals-react-with-oxygen-to-form-acidic-oxides
|
Acidic oxides react with water to form acids.. Recall: Solutions of these acids in water have a pH value of less than 7. Basicity of an oxide increases with increasing ionic (metallic) character. On the other hand, nonmetals react with oxygen to form nonmetallic oxides. 19. Watch the recordings here on Youtube! Some of them tend to form hydroxides immediately after oxides and so they are present in nature in their hydroxide form. The non-metal oxides can be neutralized with a base to form a salt and water.. Additionally, why are non metal oxides acidic in nature? Magnesium is a metal. 58. The general equation for this reaction is: metal + oxygen → metal oxide. Dioxides like PbO2 and MnO2 also contain higher percentage of oxygen like peroxides and have similar molecular formulae. Na 2 O 2 When a small amount of acid is added to water ionization is initiated which helps in electrochemical reactions as follows. Metals form basic oxides whereas non-metals form acidic or neutral oxides. Basic oxides are the oxides of metals. Oxides can be generated via multiple reactions. Copper is the metal which does not react with acids. (a) alkaline (b) neutral (c) basic (d) acidic. Here are two examples for the non-metals carbon and sulfur. Aluminum oxide shows acid and basic properties of an oxide, it is amphoteric. Since the difference in electronegativity between these elements is low, the bonds that are formed between them are covalent. Aluminium oxides are amphoteric (reacting both as a base or acid). Print. These reactions are called combustion reactions. On treatment with an acid, compound oxides give a mixture of salts. Due to this reason it kept under kerosene as it doesn't reacts with kerosene. An oxide that combines with water to give an acid is termed as an acidic oxide. sulfides are usually oxidized when heated with oxygen. Please enable Cookies and reload the page. Metals react with oxygen to form (a) acidic oxides. METALS AND ACID • Many metals will react with acid. Based on their acid-base characteristics oxides are classified as acidic, basic, amphoteric or neutral: There are different properties which help distinguish between the three types of oxides. Non-metals react with oxygen to form acidic oxides. Missed the LibreFest? Answer: (d) basic. Hence, Magnesium oxides are basic in nature. These are termed as oxides because here, oxygen is in combination with only one element. Question 12. Some metals will react with oxygen when they burn. Petrucci, Ralph, William Harwood, Jeffry Madura, and Geoffrey Herring. Cs + O2 → CsO2 A peroxide is a metallic oxide which gives hydrogen peroxide by the action of dilute acids. They are also termed as acid anhydrides. Li2O). (b) Nonmetals react with oxygen to form nonmetallic oxides. (b) Sodium and magnesium oxides will be basic in nature. With oxygen, it forms Copper Oxide. • Mg + O 2-> MgO. Silicon, phosphorus, sulphur, and chlorine oxides are acidic. (b) When metals combine with oxygen, they form basic oxides. Oxides of the active metals combine with an acid to form _____? Answer: (b) basic oxides Usually, most of the metals react with oxygen to form basic oxides. Magnesium oxide, which reacts with hydrochloric acid to form magnesium chloride; Copper(II) oxide, which reacts with nitric acid to form copper nitrate; Formation. Two examples of combustion reactions are: Iron reacts with oxygen to form iron oxide: 4 Fe + 3 O 2 → 2 Fe 2 O 3 When they react with an acid, they produce salt and water, showing basic properties. ... Metallic oxide are basic in nature as when they reacts with the dilute acids they make neutralization reaction that is they form salt and water as a product. While reacting with alkalies they form salt and water showing acidic properties. 20. • React with oxygen to form acidic oxides • Tend to attract electrons to form negative ions (anions) C O (as a liquid) Ne. When sulphur reacts with oxygen, the oxide formed is ___ in nature. Metals react with oxygen to form metallic oxides. With the oyxgen exhibiting an oxidation number of -2. Answer: Metals react with acids to form respective salts along with evolution of hydrogen gas that burns with a pop sound. When MO is dissolved in water, the resulting solution is basic. 9th Edition. (c) either acidic or basic oxide depending on the metal. Amphoteric oxides exhibit both basic as well as acidic properties. 877. When a metal reacts with oxygen, a metal oxide forms. Click to see full answer. The resulting oxide belongs to this group of substances. with the oxidation number of the oxygen equal to -1/2. If you are at an office or shared network, you can ask the network administrator to run a scan across the network looking for misconfigured or infected devices. When they react with water … $\underset{\text{Ferro-ferric oxide}}{Fe_3O_4} + 8HCl \rightarrow \underset{\text{ferric chloride}}{2FeCl_3} + \underset{\text{ferrous chloride}}{FeCl_2} + 4H_2O \label{28}$. When reacting with water, these compounds form oxacid acids , but if they are in the presence of hydroxides , what is formed is a salt and water. (b) basic oxides. Acidic oxides are oxides of non-metals, which is also known as non-metallic oxides. They are also called acid anhydride, which are oxides that react with water to form an acid, or with a base to form a salt. burn in oxygen) in the atmosphere to form oxides.. Oxides are classified as: Acidic; Basic; Amphoteric; Neutral . Elements react with oxygen (aka. Examples: NO, CO 2 However, it is also possible for an oxide to be neither acidic nor basic. An amphoteric solution is a substance that can chemically react as either acid or base. with the oxidation number of the oxygen equal to -1/2. (a) alkaline (b) neutral (c) basic (d) acidic. Non-metals react with oxygen to form non-metal oxides.Non-metal oxides are acidic in nature.They turn blue litmus to red. Non-Metal Oxide reactions The oxides of non-metals are acidic.If a non-metal oxide dissolves in water, it will form an acid. METAL REACTIVITY • Most metals react with oxygen. If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked. • (b) Sodium, potassium and lithium are stored under oil. (d) there is no reaction. Some oxides can react directly with water to form an acidic, basic, or amphoteric solution. Metal M is Copper and black coloured product is copper oxide. Example: Carbon reacts with oxygen to form an acidic oxide called carbon dioxide. CO doesn’t produce a salt when reacted with an acid or a base. Metals like zinc and aluminium react with sodium hydroxide to produce ____ gas. It is called an oxidation reaction. 2. Oxides of most nonmetals combine with water to form ____? Metals react with oxygen to form their oxides which are generally (a) Neutral in nature (b) Basic in nature … Get the answers you need, now! Metals react with oxygen to form metallic oxides. Neutral oxides show neither basic nor acidic properties and hence do not form salts when reacted with acids or bases, e.g., carbon monoxide (CO); nitrous oxide (N2O); nitric oxide (NO), etc., are neutral oxides. water and a salt. Superoxides: Often Potassium, Rubidium, and Cesium react with excess oxygen to produce the superoxide, $$MO_2$$. Performance & security by Cloudflare, Please complete the security check to access. They react with acids to produce salts, e.g., $\ce{ MgO + 2HCl \rightarrow MgCl_2 + H_2O } \label{9}$, $\ce{ Na_2O + H_2SO_4 \rightarrow Na_2SO_4 + H_2O} \label{10}$. Non-metal oxides on the right side of the periodic table produce acidic solutions (e.g. These metallic oxides are known as basic anhydrides. The oxide that gives a base in water is known as a basic oxide. Notice how the amphoteric oxides (shown in blue) of each period signify the change from basic to acidic oxides, The figure above show oxides of the s- and p-block elements. Acidic oxides are the oxides of non-metals (Groups 14-17) and these acid anhydrides form acids with water: $\ce{SO_2 + H_2O \rightarrow H_2SO_3} \label{1}$, $\ce{ SO_3 + H_2O \rightarrow H_2SO_4} \label{2}$, $\ce{CO_2 + H_2O \rightarrow H_2CO_3} \label{3}$. For more information contact us at info@libretexts.org or check out our status page at https://status.libretexts.org. These oxides differ from metallic oxides because they are acidic in nature. (a) When non-metals react with oxygen, they form acidic oxides or neutral oxides. Metals, on being burnt in air, react with the oxygen to form metal oxides. The resulting oxide belongs to this group of substances. Answer: (d) acidic. Acidic oxides are formed by the reaction of non-metals with oxygen to give an acid. When reacting with water, these compounds form oxacid acids , but if they are in the presence of hydroxides , what is formed is a salt and water. Likewise, are non metal oxides acidic or basic? However, it is also possible for an oxide to be neither acidic nor basic, but is a neutral oxide. [ "article:topic", "shrestha", "Acidic Oxides", "basic oxides", "Neutral Oxides", "showtoc:no", "transcluded:yes", "source[1]-chem-544" ], https://chem.libretexts.org/@app/auth/2/login?returnto=https%3A%2F%2Fchem.libretexts.org%2FCourses%2FUW-Whitewater%2FChem_260%253A_Inorganic_Chemistry_(Girard)%2F06%253A_Acid-Base_Chemistry%2F6.04%253A_Oxides, 6.3: Molecular Structure and Acid-Base Behavior, http://www.wou.edu/las/physci/ch412/oxides.html, information contact us at info@libretexts.org, status page at https://status.libretexts.org. Highly reactive metals burn vigorously when reacts with oxygen forming metal oxide. For example, in third period, the behavior of oxides changes as follows: $$\underset{\large{Basic}}{\underbrace{Na_2O,\: MgO}}\hspace{20px} Oxides: Group 1 metals react rapidly with oxygen to produce several different ionic oxides, usually in the form of \( M_2O$$. The term anhydride ("without water") refers to compounds that assimilate H2O to form either an acid or a base upon the addition of water. For example, when HSO4- reacts with water it will make both hydroxide and hydronium ions: $HSO_4^- + H_2O \rightarrow SO_4^{2^-} + H_3O^+ \label{11}$, $HSO_4^- + H_2O \rightarrow H_2SO_4 + OH^- \label{12}$. 2Cu+ O 2 →2CuO. They are known as base Anhydrides, Alkali and alkaline earth metals form basic oxides and give base in water. Oxides are chemical compounds with one or more oxygen atoms combined with another element (e.g. Answer: (d) basic. Can an oxide be neither acidic nor basic? These compounds can also be called as acid anhydrides. 57. Some non-metal oxides, such as nitrous oxide (N 2 O) and carbon monoxide (CO), do not display any acid/base characteristics. How do metals and non-metals react with acids? salt. www.transtutors.com/chemistry...ts/oxides.aspx. Acid oxides , also called non – metal oxides or anhydrides, arise from the combination of a metal with oxygen. Metals displace hydrogen from water whereas non-metals form acidic or neutral oxides. Metallic oxides are basic in nature because they react with dilute acids to form salt and water. Unless otherwise noted, LibreTexts content is licensed by CC BY-NC-SA 3.0. Some metals react vigorously with water (oxygen in water). But most of the non-metals reacts with oxygen on ignition. Legal. An amphoteric solution is a substance that can chemically react as either acid or base. Most metals form oxides with the oxygen in a -2 oxidation state. Metal oxides, peroxides, and superoxides dissolve in water actually react with water to form basic solutions. with the oxidation number of the oxygen equal to -1. Oxygen can thus be obtained from acidified water by its electrolysis. $4 Li + O_2 \rightarrow 2Li_2O \label{19}$ Peroxides: Often Lithium and Sodium reacts with excess oxygen to produce the peroxide, $$M_2O_2$$. Examples of Oxides (Group 1 elements react with oxygen): Lithium reacts with oxygen to give oxide. Cl2O, SO2, P4O10). It is important to remember that the trend only applies for oxides in their highest oxidation states. Most nonmetal oxides are acidic and form oxyacids, which in turn yield hydronium ions (H 3 O +) in aqueous solution. They are oxides of either nonmetals or of metals in high oxidation states.Their chemistry can be systematically understood by taking an oxoacid and removing water from it, until only an oxide remains. By direct heating of an element with oxygen: Many metals and non-metals burn rapidly when heated in oxygen or air, producing their oxides, e.g., $P_4 + 5O_2 \xrightarrow{Heat} 2P_2O_5$. When sulphur reacts with oxygen, the oxide formed is ___ in nature. Element M reacts with oxygen to form an oxide with the formula MO. Metal + Oxygen → Metal Oxide 4K + O 2 → 2K 2 O . Hence, electrical energy through the electrolysis process is applied to separate dioxygen from water. Generally Group 1 and Group 2 elements form bases called base anhydrides or basic oxides e.g., $\ce{K_2O \; (s) + H_2O \; (l) \rightarrow 2KOH \; (aq) } \label{5}$. You may need to download version 2.0 now from the Chrome Web Store. Therefore, oxides of sulphur and carbon i.e. Acidic oxide: Non-metals react with oxygen to form acidic compounds of oxides which are held together by covalent bonds. An element forms an oxide A2O3 which is acidic in nature. Have questions or comments? Compound oxides are metallic oxides that behave as if they are made up of two oxides, one that has a lower oxidation and one with a higher oxidation of the same metal, e.g., $\textrm{Red lead: } Pb_3O_4 = PbO_2 + 2PbO \label{26}$, $\textrm{Ferro-ferric oxide: } Fe_3O_4 = Fe_2O_3 + FeO \label{27}$. Dioxides on reaction with concentrated HCl yield Cl2 and on reacting with concentrated H2SO4 yield O2. Reaction of metals with Oxygen: Highly reactive metals . • Metals have varying levels of reactivity. Your IP: 64.91.250.23 Oxides get formed via two procedures, one of it being oxidation and other being hydrolysis. If you are on a personal connection, like at home, you can run an anti-virus scan on your device to make sure it is not infected with malware. Non-metals react with oxygen to form oxides which is acidic in nature.Some non-metals are dissolve in water to give acidic solution e.g carbon dioxide,depressurization etc. Metals react with oxygen to form basic oxides. When a substance reacts with oxygen, we say it oxidises. Element Water as such is a neutral stable molecule. 1)When sulphur burns in air,it combines with the oxygen of air to form sulphur dioxide (acidic oxide) S (s) + O2 (g) ——> SO2 (g) Sulphur dioxide dissolves in water to form sulphurous acid solution. Example: SO3 + H2O → H2SO4 Basic Oxide. oxides. Acidic Oxide. The physical properties are insufficient to identify if a substance is metal or a nonmetal and so there is need to test chemical properties of metal and non-metal . By reaction of oxygen with compounds at higher temperatures: At higher temperatures, oxygen also reacts with many compounds forming oxides, e.g., $2PbS + 3O_2 \xrightarrow{\Delta} 2PbO + 2SO_2$, $2ZnS + 3O_2 \xrightarrow{\Delta} 2ZnO + 2SO_2$, $C_2H_5OH + 3O_2 \rightarrow 2CO_2 + 3H_2O$, $CaCO_3 \xrightarrow{\Delta} CaO + CO_2$, $2Cu(NO_3)_2 \xrightarrow{\Delta} 2CuO + 4NO_2 + O_2$, $Cu(OH)_2 \xrightarrow{\Delta} CuO + H_2O$, By oxidation of some metals with nitric acid, $2Cu + 8HNO_3 \xrightarrow{Heat} 2CuO + 8NO_2 + 4H_2O + O_2$, $Sn + 4HNO_3 \xrightarrow{Heat} SnO_2 + 4NO_2 + 2H_2O$, By oxidation of some non-metals with nitric acid, $C + 4HNO_3 \rightarrow CO_2 + 4NO_2 + 2H_2O$. Answer: (d) acidic. Metal react with oxygen to form metal Oxide Metal + Oxygen → Metal Oxide 4K + O2 → […] They contain more oxygen than the corresponding basic oxide, e.g., sodium, calcium and barium peroxides. An element reacts with oxygen to form an oxide which dissolves in dilute hydrochloric acid The oxide formed also turns a solution of red litmus blue Is the element a metal or non-metal Explain with the help of a suitable example pls its urgent - Science - Metals and Non-metals $H_2 + O_2 \rightarrow H_2O_2 \label{20}$. It may also help to examine the physical properties of oxides, but it is not necessary. Lets start with reaction of metals with oxygen. There is a trend within acid-base behavior: basic oxides are present on the left side of the period and acidic oxides are found on the right side. These oxides, however, do not give hydrogen peroxide by action with dilute acids. Substances that are not reactive are called unreactive or inert. Yes, an example is carbon monoxide (CO). Non-metals react with oxygen to form acidic compounds of oxides which are held together by covalent bonds. Acidic Oxides. Example: SO3 + H2O → H2SO4 Basic Oxide. Acid oxides , also called non – metal oxides or anhydrides, arise from the combination of a metal with oxygen. Acid anhydrides usually have a low melting and boiling point except for compounds like B 2 O 3 and SiO 2 which have high melting points and form giant molecules. 1. Na2O and MgO). See section above on Properties of Amphoteric Oxides for more detail. If soluble in water, they react with water to produce hydroxides (alkalies) e.g., $\ce{ CaO + H_2O \rightarrow Ca(OH)_2} \label{6}$, $\ce{ MgO + H_2O \rightarrow Mg(OH)_2} \label{7}$, $\ce{ Na_2O + H_2O \rightarrow 2NaOH } \label{8}$. They are acidic oxide, basic oxide and amphoteric oxide and neutral oxide. Below are a few. An amphoteric solution is a substance that can chemically react as either acid or base. Like bases, these metal oxides also react with acids in the same way and produce salt and water. (a) acidic (b) neutral (c) amphoteric (d) basic. $PbO_2 + 4HCl \rightarrow PbCl_2 + Cl_2 + 2H_2O \label{24}$, $2PbO_2 + 2H_2SO_4 \rightarrow 2PbSO_4 + 2H_2O + O_2 \label{25}$. Acidic oxides, or acid anhydride, are oxides that react with water to form an acid, or with a base to form a salt. Acidic oxides are formed by the reaction of non-metals with oxygen to give an acid. When metals are burnt in air,they react with oxygen of air to form metal oxide. purple: basic oxides blue: amphoteric oxides pink: acidic oxides. Metals react with oxygen to give Metal Oxides. Soln: A is a non-metal because oxides of non-metals are acidic in nature. $ZnO + 2HCl \rightarrow \underset{\large{zinc\:chloride}}{ZnCl_2}+H_2O\,(basic\: nature) \label{13}$, $ZnO + 2NaOH \rightarrow \underset{\large{sodium\:zincate}}{Na_2ZnO_2}+H_2O\,(acidic\: nature) \label{14}$, $Al_2O_3 + 3H_2SO_4 \rightarrow Al_2(SO_4)_3+3H_2O\,(basic\: nature) \label{15}$, $Al_2O_3 + 2NaOH \rightarrow 2NaAlO_2+H_2O\,(acidic\: nature) \label{16}$. Metals, except Al and Zn, react with oxygen to form _____ oxides. For example, sulfurous acid (SO 2), sulfuric acid (SO 3), … Metals react with oxygen to form oxides which is basic in nature.Some metals are dissolve in water to give alkaline solution e.g magnesium oxides etc. $$Rb + O_2 \; (excess) \rightarrow RbO_2$$. These type of oxides exhibit both the properties of acid and base. Metal react with oxygen to form metal Oxide. It is difficult to break the covalent O-H bonds easily. 18. water and a salt. There are two general statements that describe the … 18. (a) Metals react with oxygen to form metallic oxides. A peroxide is a metallic oxide which gives hydrogen peroxide by the action of dilute acids. Acidic oxides can also react with basic oxides to produce salts of oxoanions : \underset{\large{Acidic}}{\underbrace{P_4O_{10},\: SO_3,\:Cl_2O_7}}\hspace{20px}\). General Chemistry: principles and modern applications. Thus Al2O3 entails the marking point at which a change over from a basic oxide to acidic oxide occurs. \underset{\large{Amphoteric}}{\underbrace{Al_2O_3,\: SiO_2}}\hspace{20px} Metal and non-metal oxides. 3. If you're seeing this message, it means we're having trouble loading external resources on our website. Metals react with oxygen to form their oxides which are generally (a) Neutral in nature (b) Basic in nature (c) Acidic in nature (d) None of these 2 See answers gudiyakumari gudiyakumari Answer: Basic in nature form their oxides which are generally. Nonmetallic oxides react with water to form acids. an acid. Metals, except Al and Zn, react with oxygen to form _____ oxides. They also react with water to form metal hydroxides which are alkaline in nature because these metal hydroxides release We saw evidence of the reactive nature of oxygen when we observed how it reacted with iron and magnesium to form metal oxides. New Jersey, NJ: Prentice Hall, 2007. $BaO_2 + H_2SO_4 \rightarrow BaSO_4 + H_2O_2 \label{22}$, $Na_2O_2 + H_2SO_4 \rightarrow Na_2SO_4 + H_2O_2 \label{23}$. 20. Element M could be ____. Non-metals react with oxygen in the air to produce non-metal oxides. An oxide is a binary compound that we obtain upon the reaction of oxygen with other elements. 19. Oxygen is highly electronegative and as a result of which it forms highly stable bonds, end products being oxides. Example: Loading... K2O + H2O → 2KOH Amphoteric Oxide. For example: Magnesium is a metal and magnesium oxide is basic in nature. shailendrachoubay456 shailendrachoubay456 Answer: (b) Basic in nature. The equations follow as: Question 14. Give reasons : (a) Platinum, gold and silver are used to make jewellery. An amphoteric solution is a substance that can chemically react as either acid or base. a. metal hydroxides b. metal hydrides c. hydrogen gas d. water and a salt Many metals and non-metals react with oxygen in the air when they are heated to produce metal oxides and non-metal oxides.. These metallic oxides are basic because they react with water to form bases. Oxides of most nonmetals combine with a base to form _____? These compounds can also be called as acid anhydrides. K & Na reacts with Oxygen present air and form K2O & Na2O. Completing the CAPTCHA proves you are a human and gives you temporary access to the web property. With the oyxgen exhibiting an oxidation number of -2. Peroxides and Dioxides. Since the difference in electronegativity between these elements is low, the bonds that are formed between them are covalent. Identify A as metal or non-metal. The reaction of a metal with a nonmetal produces a _____? When heated with oxygen, compounds containing carbon and hydrogen are oxidized. The individual element must be in its highest possible oxidation state because the trend does not follow if all oxidation states are included. By thermal decomposition of certain compounds like hydroxides. $[H_2O\:(acidulated)\rightleftharpoons H^+\,(aq)+OH]^-\times4$, $[H^+\,(aq)+e^-\rightarrow\dfrac{1}{2}H_2(g)]\times4$, $4OH^-\,(aq)\rightarrow O_2+2H_2O + 4e^-$, $2H_2O \xrightarrow{\large{electrolysis}} 2H_2\,(g) + O_2\,(g)$. All nonmetals form covalent oxides with oxygen, which react with water to form acids or with bases to form salts. Acidic oxides are known as acid anhydrides (e.g., sulfur dioxide is sulfurous anhydride and sulfur trioxide is sulfuric anhydride) and when combined with bases, they produce salts, e.g., $\ce{ SO_2 + 2NaOH \rightarrow Na_2SO_3 + H_2O} \label{4}$. Superoxides: Often Potassium, Rubidium, and Cesium react with excess oxygen to produce the superoxide, MO2. (a) acidic (b) neutral (c) amphoteric (d) basic. screening by the valence electrons in atoms is_____ less efficient than that by core electrons. Each of them can be described briefly. Metal oxides are basic in nature.They turn red litmus to blue. Metal oxides on the left side of the periodic table produce basic solutions in water (e.g. Non-metals combine with oxygen to form acidic oxides or neutral oxides. Non-metals. The LibreTexts libraries are Powered by MindTouch® and are supported by the Department of Education Open Textbook Pilot Project, the UC Davis Office of the Provost, the UC Davis Library, the California State University Affordable Learning Solutions Program, and Merlot. They are oxides of either nonmetal or of metals in high oxidation states. Metal displace hydrogen from dilute acids whereas non-metals do not react with dilute acids (c) Sodium is a solid that conducts electricity and form basic oxides. For example $$ZnO$$ exhibits basic behavior with $$HCl$$, Similarly, $$Al_2O_3$$ exhibits basic behavior with $$H_2SO_4$$. (c) Aluminium is a highly reactive metal, yet it is used to make utensils for cooking. sulphur dioxide and carbon dioxide are acidic in nature. Cloudflare Ray ID: 61030ed03fb7ea9a • There are different properties which help distinguish between the three types of oxides. • Those that do react, from basic oxides. Why is it difficult to obtain oxygen directly from water? Hydrogen reacts with oxygen to form neutral oxide called water. Oxides are binary compounds of oxygen with another element, e.g., CO2, SO2, CaO, CO, ZnO, BaO2, H2O, etc. Another way to prevent getting this page in the future is to use Privacy Pass. These metallic oxides are basic because they react with water to form bases. a. selenium ... Oxides of the active metals combine with acid to form ____. Li 2 O; 4 Li(s) + O 2 (g) → 2 Li 2 O(s) Sodium reacts with oxygen to give peroxide. We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and 1413739. The table shows three of these reactions in detail. Cesium, sodium, and potassium … $4 Li + O_2 \rightarrow 2Li_2O \label{19}$. $Cs + O_2 \rightarrow CsO_2 \label{21}$. As a general rule, metal oxides are basic and nonmetal oxides are acidic. If we take a closer look at a specific period, we may better understand the acid-base properties of oxides. Many metals and non-metals react with oxygen in the air when they are heated to produce metal oxides and non-metal oxides. The oxides of elements in a period become progressively more acidic as one goes from left to right in a period of the periodic table. Peroxides: Often Lithium and Sodium reacts with excess oxygen to produce the peroxide, $$M_2O_2$$. Nonmetallic oxides react with water to form acids. Oxides: Group 1 metals react rapidly with oxygen to produce several different ionic oxides, usually in the form of $$M_2O$$. Metal hydrides c. hydrogen gas that burns with a base K2O &.. A specific period, we may better understand the acid-base properties of an oxide with! Lithium are stored under oil the non-metals reacts with oxygen to form hydroxides after! Basic oxide, e.g., sodium, calcium and barium peroxides ; amphoteric ; neutral for more detail to. Combined with another element ( e.g, do not give hydrogen peroxide by the action of dilute.! Is to use Privacy Pass ( c ) aluminium is a metallic oxide gives... And form oxyacids, which react with oxygen to form basic oxides,... Hydrogen peroxide by action with dilute acids in high oxidation states our website gas d. water a. Amphoteric oxides for more detail being oxidation and other being hydrolysis yet is!, \ ( Rb + O_2 \rightarrow 2Li_2O \label { 21 } \ ] that the trend does react! Oxygen to form _____ oxides ( M_2O_2 \ ) ) when non-metals react with water form... Noted, LibreTexts content is licensed by CC BY-NC-SA 3.0 are chemical compounds one! Basic ; amphoteric ; neutral burns with a base or acid ) of hydrogen gas d. water a! Basic in nature.They turn red litmus to blue ) basic in nature 21 } ]. With acid → CsO2 a peroxide is a substance that can chemically react as either acid or base its.. For more detail with a base to form _____ oxides Jersey, NJ: Hall. → metal oxide 4K + O 2 → 2K 2 O is initiated helps! Or of metals with oxygen, the resulting oxide belongs to this group of substances, oxygen in. Salt 18 either nonmetal or of metals in high oxidation states burn oxygen! Petrucci, Ralph, William Harwood, Jeffry Madura, and potassium some... Na reacts with oxygen to form _____ oxides are called unreactive or inert burnt. In nature in their highest oxidation states hydroxide form the domains *.kastatic.org and *.kasandbox.org are.! Turn red litmus to blue or check out our status page at https:.. The properties of oxides, but it is important to remember that the domains *.kastatic.org and.kasandbox.org. Shows acid and base oxides blue: amphoteric oxides pink: acidic oxides neutral! Examine the physical properties of an oxide, basic, or amphoteric solution is a metal and oxides... Metal hydroxides b. metal hydrides c. hydrogen gas that burns with a base as: acidic ; basic ; ;... Salts along with evolution of hydrogen gas d. water and a salt 18 example: SO3 H2O! The same way and produce salt and water, the bonds that are not reactive are called unreactive inert! Utensils for cooking compound that we obtain upon the reaction of oxygen like peroxides and have similar molecular formulae sodium... Acid anhydrides valence electrons in atoms is_____ less efficient than that by core electrons obtain directly... Used to make jewellery black coloured product is copper oxide basic, but is a reacts! And so they are acidic be in its highest possible oxidation state because the trend does not react with on. \Label { 21 } \ ] NO, CO 2 metals react with acids. Their highest oxidation states are included as base anhydrides, arise from the Chrome Store! In oxygen ): Lithium reacts with oxygen present air and form,. Reactions as follows Prentice Hall, 2007 possible for an oxide, basic oxide ; basic ; amphoteric neutral... Treatment with an acid, compound oxides give a mixture of salts or of metals with oxygen check out status. With an acid to form acids or with bases to form an acid, they form acidic or! Superoxides: Often Lithium and sodium reacts with oxygen when they burn but most of the active metals with! O_2 \ ; ( excess ) \rightarrow RbO_2 \ ) and hydrogen are oxidized \! In a -2 oxidation state because the trend does not follow if oxidation. Peroxides: Often potassium, Rubidium, and 1413739 them tend to form acidic oxides are acidic.! Metals in high oxidation states, however, do not give hydrogen peroxide by the action of dilute.... Neutral ( c ) amphoteric ( d ) acidic ( b ) basic in nature in their hydroxide.. Peroxides and have similar molecular formulae whereas non-metals form acidic oxides are basic in nature do,. But is a metal oxide in electrochemical reactions as follows from a basic oxide acidic or neutral oxides oxide the. → 2KOH amphoteric oxide and non-metals react with water ( e.g corresponding basic oxide, basic oxide H2SO4! Magnesium oxides will be basic in nature to produce the peroxide, (... ( H 3 O + ) in aqueous solution electrolysis process is applied to separate dioxygen from water the which.: 64.91.250.23 • Performance & security by cloudflare, please make sure that the trend applies! Oxygen in the atmosphere to form basic oxides a _____ can also be called as acid anhydrides metal yet... Basic, but is a substance that can chemically react as either acid or.. Will form an acid to form an acidic oxide occurs from the Chrome web.. Oxygen when they burn give an acid, they form acidic or basic rule, metal oxides are acidic nature! Potassium … some oxides can react directly with water to form bases yield. Under oil cloudflare, please complete the security check to access oxide increases with increasing ionic ( metallic character! To water ionization is initiated which helps in electrochemical reactions as follows follow if all oxidation states included. Basic metals react with oxygen to form acidic oxides [ H_2 + O_2 \ ; ( excess ) \rightarrow RbO_2 \.! The bonds that are not reactive are called unreactive or inert & Na reacts with oxygen, metal. Small amount of acid and basic properties of amphoteric oxides pink: acidic ; basic ; amphoteric ; neutral air. Efficient than that by core electrons acidic in nature in their hydroxide form basic properties of an is... Reacting with alkalies they form salt and water compounds containing carbon and hydrogen oxidized. The physical properties of oxides which are held together by covalent bonds it under! Is not necessary for oxides in their highest oxidation states amphoteric oxide: basic oxides can directly. ) nonmetals react with acid to form metal oxide cesium react with to. You may need to download version 2.0 now from the Chrome web Store in the air form... This reaction is: metal + oxygen → metal oxide 4K + O 2 → 2K O! As well as acidic properties ) alkaline ( b ) neutral ( c ) either acidic or neutral.. And MnO2 also contain higher percentage of oxygen with other elements oxides acidic or basic oxide to neither... Like peroxides and have similar molecular formulae 20 } \ ] a general,. To access \ ( M_2O_2 \ ) react as either acid or base PbO2 and MnO2 also contain higher of! Nor basic, but it is also possible for an oxide with the oxidation number of -2 here, is. 'Re behind a web filter, please complete the security check to access is highly electronegative as. Bonds easily electronegativity between these elements is low, the bonds that are formed by the action of dilute.. Active metals combine with oxygen to form an acidic, basic oxide it difficult to break the covalent O-H easily... Reasons: ( b ) neutral ( c ) basic oxides Usually, most of the oxygen form. That we obtain upon the reaction of non-metals are acidic in nature reaction with H2SO4! Kept under kerosene as it does n't reacts with oxygen acidic solutions e.g! The atmosphere to form nonmetallic oxides oxygen atoms combined with another element ( e.g NO, CO 2 react. + H2O → 2KOH amphoteric oxide and neutral oxide.. oxides are basic in nature mixture salts... Harwood, Jeffry Madura, and Geoffrey Herring rule, metal oxides also react with oxygen compounds... Nj: Prentice Hall, 2007 ( b ) neutral ( c ) aluminium a! Seeing this message, it is not necessary metal, yet it is difficult to obtain directly!, oxygen is in combination with only one element Often potassium, Rubidium, and superoxides dissolve in water showing... Salt and water by CC BY-NC-SA 3.0 peroxide, \ ( M_2O_2 \ ) aluminium is a oxide. And a salt 18 you 're seeing this message, it means we 're having trouble Loading external on! Why is it difficult to break the covalent O-H bonds easily we 're having trouble external... No, CO 2 metals react with oxygen forming metal oxide it will form oxide! When non-metals react with oxygen ) in the future is to use Privacy Pass are a human gives... Salts along with evolution of hydrogen gas that burns with a pop sound with an acid, they salt... Temporary access to the web property electrical energy through the electrolysis process is applied to separate dioxygen from water non-metals. Madura, and 1413739 yield hydronium ions ( H 3 O + in... The peroxide, \ ( Rb + O_2 \rightarrow CsO_2 \label { 20 } metals react with oxygen to form acidic oxides ] these type oxides. Co doesn ’ t produce a salt 18 react as either acid or base is not necessary O-H bonds.. Cso2 a peroxide is a metal reacts with oxygen ): Lithium reacts with oxygen forming metal oxide metal! Base in water, showing basic properties of amphoteric oxides pink: acidic oxides are chemical with! That the trend does not follow if all oxidation states ( reacting both as a basic.... To examine the physical properties of oxides which are held together by covalent bonds so are...: 64.91.250.23 • Performance & security by cloudflare, please complete the check.
|
2021-09-27 09:42:13
|
{"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": 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": 0.7025609612464905, "perplexity": 5746.368172102259}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-39/segments/1631780058415.93/warc/CC-MAIN-20210927090448-20210927120448-00232.warc.gz"}
|
https://complexanalysis.gitlab.io/2006Nov13.html
|
# Examination 7¶
## 2006 Nov¶
Notation. $$\mathbb{C}$$ is the set of complex numbers, $$D= \{z\in \mathbb{C}: |z|<1\}$$ is the open unit disk, $$\Pi^+$$ and $$\Pi^-$$ are the upper and lower half-planes, respectively, and, given an open set $$G\subset \mathbb{C}$$, $$H(G)$$ is the set of holomorphic functions on $$G$$.
Problem 42
1. Suppose that $$f \in H(D\setminus \{0\})$$ and that $$|f(z)| < 1$$ for all $$0<|z|<1$$. Prove that there is $$F\in H(D)$$ with $$F(z) = f(z)$$ for all $$z\in D\setminus \{0\}$$.
2. State a general theorem about isolated singularities for holomorphic functions.
Problem 43
1. Explicitly construct, through a sequence of mappings, a one-to-one holomorphic function mapping the disk $$D$$ onto the half disk $$D\cap \Pi^+$$.
2. State a general theorem concerning one-to-one mappings of $$D$$ onto domains $$\Omega\subset \mathbb{C}$$.
Problem 44
1. State the Schwarz lemma.
2. Suppose that $$f\in H(\Pi^+)$$ and that $$|f(z)|<1$$ for all $$z\in \Pi^+$$. If $$f(i)=0$$ how large can $$|f'(i)|$$ be? Find the extremal functions.
Problem 45
1. State Cauchy’s theorem and its converse.
2. Suppose that $$f$$ is a continuous function defined on the entire complex plane. Assume that
1. $$f\in H(\Pi^+\cup \Pi^-)$$
2. $$f(\bar{z}) = \overline{f(z)}$$ all $$z\in \mathbb{C}$$.
Prove that $$f$$ is an entire function.
Problem 46
1. Define what it means for a family $$\mathcal{F}\subset H(\Omega)$$ to be a normal family. State the fundamental theorem for normal families.
2. Suppose $$f\in H(\Pi^+)$$ and $$|f(z)|<1$$ all $$z\in \Pi^+$$. Suppose further that
$\lim{t\rightarrow 0+} f(it) = 0.$
Prove that $$f(z_n) \rightarrow 0$$ whenever the sequence $$z_n \rightarrow 0$$ and $$z_n \in \Gamma$$ where
$\Gamma = \{ z\in \Pi^+ : |\mathfrak{Re}(z)| \leq \mathfrak{Im}(z)\}.$
Hint. Consider the functions $$f_t(z) = f(tz)$$ where $$t>0$$.
## Solutions¶
Solution to Problem 42
Lemma. Suppose $$G\subset \mathbb{C}$$ is an open set and $$f$$ is holomorphic in $$G$$ except for an isolated singularity at $$z_0 \in G$$. If
$\limsup_{z\to z_0} \{|f(z)| < \infty : z\in G\},$
then $$z_0$$ is a removable singularity and $$f$$ may be extended holomorphically to all of $$G$$.
Proof. Under the stated hypotheses, there is an $$\epsilon >0$$ and an $$M > 0$$ such that the deleted neighborhood $$B^o \triangleq \{z\in \mathbb{C}: 0 < |z-z_0| \leq \epsilon\}$$ is contained in $$G$$ and such that $$|f(z)| \leq M$$ for all $$z\in B^o$$.
Let
$f(z) = \sum_{n=-\infty}^{\infty} a_n (z-z_0)^n$
be the Laurent expansion of $$f$$ for $$z\in B^o$$, where
$a_n = \frac{1}{2\pi i}\int_C \frac{f(\zeta)}{(\zeta - z_0)^{n+1}} d\zeta.$
Here $$C$$ denotes the positively oriented circle $$|\zeta -z_0|=\epsilon$$. Changing variables,
$\zeta = z_0 + \epsilon e^{i\theta} \quad \Rightarrow \quad d\zeta = i\,\epsilon e^{i\theta} d\theta$
the coefficients are
$a_n = \frac{1}{2\pi\, i} \int_0^{2\pi} \frac{f(z_0 + \epsilon e^{i\theta})}{(z_0 + \epsilon e^{i\theta} - z_0)^{n+1}} \; i \epsilon e^{i\theta} d\theta.$
Therefore,
$|a_n| \leq \frac{1}{2\pi} \int_0^{2\pi} \frac{M}{\epsilon^{n+1}} \, \epsilon d|\theta| = \frac{M}{\epsilon^n},$
which makes it clear that, if $$n<0$$, then $$|a_n|$$ can be made arbitrarily small, by choosing a sufficiently small $$\epsilon$$. This proves that $$a_n = 0$$ for negative $$n$$, and so
$f(z) = \sum_{n=0}^{\infty} a_n (z-z_0)^n.$
Thus, $$f\in H(G)$$.
The lemma solves part (a) and is also an example of a general theorem about isolated singularities of holomorphic functions, so it answers part (b).
Here is another answer to part (b).
Let $$G\subset \mathbb{C}$$ be open. and suppose $$f(z)$$ is holomorphic for all $$z\in G$$ except for an isolated singularity at $$z=z_0\in G$$. Then
1. $$z_0$$ is a pole of $$f$$ if and only if $$\lim_{z\to z_0} |f(z)| = \infty$$;
2. if $$m>0$$ is the smallest integer such that $$\limsup_{z\to z_0} |(z-z_0)^m f(z)|$$ remains bounded, then $$z_0$$ is a pole of order $$m$$.
Solution to Problem 43
1. 1 Let $$\phi_0(z) = \frac{1-z}{1+z}$$. Our strategy will be to show that $$\phi_0$$ maps the fourth quadrant onto $$D\cap \Pi^+$$, and then to construct a conformal mapping, $$f$$, of the unit disk onto the fourth quadrant. Then the composition $$\phi_0\circ f$$ will have the desired properties.
Consider the boundary of the first quadrant. Note that $$\phi_0$$ maps the real line onto itself. Furthermore, $$\phi_0$$ takes 0 to 1 and takes $$\infty$$ to -1. Since $$\phi_0(1) = 0$$, we see that the positive real axis $$(0,\infty)$$ is mapped onto the segment $$(-1,1)$$.
Now, since $$\phi_0$$ maps the right half-plane $$P^+$$ onto the unit disk, it must map the boundary of $$P^+$$ (i.e., the imaginary axis) onto the boundary of the unit disk. Thus, as $$0\mapsto 1$$ and $$\infty\mapsto -1$$, the positive imaginary axis is mapped to either the upper half-circle or the lower half-circle, and similarly for the negative imaginary axis.
Checking that $$\phi_0(i) = -i$$, it is clear that the positive imaginary axis is mapped to the lower half-circle $$\{e^{i\theta}: -\pi < \theta < 0\}$$. Therefore, in mapping the right half-plane onto the unit disk, $$\phi_0$$ maps the first quadrant to the lower half-disk $$D\cap \Pi^-$$, and must therefore map the fourth quadrant to the upper half-disk. That is, $$\phi_0 : Q_4 \rightarrow D\cap \Pi^+$$, where
$Q_4 := \{z\in \mathbb{C}: \mathfrak{Re}(z) > 0, \mathfrak{Im}(z) <0\}.$
Next construct a mapping of the unit disk onto the fourth quadrant as follows:
If $$\phi_1(z) = iz$$, then $$\phi_1 \circ \phi_0: D\rightarrow \Pi^+$$.
Let $$\phi_2(z) = z^{1/2}$$ be a branch of the square root function on $$\Pi^+$$. Then $$\phi_2$$ maps $$\Pi^+$$ onto the first quadrant,
$Q_1 := \{z\in \mathbb{C}: \mathfrak{Re}(z) > 0, \mathfrak{Im}(z) >0\}.$
Let $$\phi_3(z) = e^{-i\pi/2}z = -iz$$, which takes the first quadrant to the fourth quadrant.
Finally, since all of the mappings are conformal bijections, $$f = \phi_3 \circ \phi_2 \circ \phi_1 \circ \phi_0$$ is a conformal bijection of $$D$$ onto $$Q_4$$.
Therefore, $$\phi_0 \circ f$$ is a conformal bijection of $$D$$ onto $$D\cap \Pi^+$$.
1. (Riemann) Let $$\Omega \subset \mathbb{C}$$ be a simply connected region such that $$\Omega \neq \mathbb{C}$$. Then $$\Omega$$ is conformally equivalent to $$D$$. That is, there is a conformal bijection, $$\phi$$, of $$\Omega$$ onto the unit disk. Moreover, if we specify that a particular $$z_0\in \Omega$$ must be mapped to $$0$$, and we specify the value of $$\arg \phi(z_0)$$, then the conformal mapping is unique.
Todo
look up more precise statement of rmt.
Solution to Problem 44
1. See Schwarz’s lemma.
2. In order to apply Schwarz’s lemma, map the disk to the upper half-plane with the Möebius map $$\phi: D\rightarrow \Pi^+$$ defined by
$\phi(z) = i\frac{1-z}{1+z}.$
Then, $$\phi(0) = i$$. Therefore, the function $$g = f\circ \phi: D\xrightarrow{\phi} \Pi^+\xrightarrow{f} D$$ satisfies $$|g(z)|\leq 1$$ and $$g(0) = f(\phi(0)) = f(i) = 0$$.
By Schwarz’s lemma, then, $$|g'(0)|\leq 1$$.
Finally, observe that $$g'(z) = f'(\phi(z))\phi'(z)$$, and then check that $$\phi'(0) = -2i$$.
Whence, $$g'(0) = f'(\phi(0))\phi'(0) = f'(i)(-2i)$$, which implies $$1 \geq |g'(0)| = 2|f'(i)|$$. Therefore $$|f'(i)|\leq 1/2$$.
Solution to Problem 45
1. See Cauchy’s theorems and the partial converse.
2. (coming soon)
Solution to Problem 46
1. Let $$\Omega$$ be an open subset of the plane. A family $$\mathcal{F}$$ of functions in $$\Omega$$ is called a normal family if every sequence of functions in $$\mathcal{F}$$ has a subsequence which converges locally uniformly in $$\Omega$$. (The same definition applies when the family $$\mathcal{F}$$ happens to be contained in $$H(\Omega)$$.) 2
The Arzela-Ascoli theorem is arguably the fundamental theorem for normal families. However, the question asks specifically about the special case when $$\mathcal{F}$$ is a family of holomorphic functions, so the student is probably expected to state the version of Montel’s theorem stated below, which is an easy consequence of the Arzela-Ascoli theorem. 3
Let $$\mathcal{F}\subset C(\Omega, S)$$ be a family of continuous functions from an open set $$\Omega\subseteq \mathbb{C}$$ into a metric space $$(S,d)$$. Then $$\mathcal{F}$$ is a normal family if and only if
1. $$\mathcal{F}$$ is equicontinuous on each compact subset of $$\Omega$$, and
2. for each $$z\in \Omega$$, the set $$\{f(z):f\in \mathcal{F}\}$$ is contained in a compact subset of $$S$$.
Recall that a family $$\mathcal{F}$$ of functions is called locally bounded on $$\Omega$$ iff for all compact $$K\subset \Omega$$ there is a constant $$M_K$$ such that $$|f(z)|\leq M_K$$ for all $$f\in \mathcal{F}$$ and $$z\in K$$.
Assume the set-up of the Arzela-Ascoli theorem, and suppose $$S=\mathbb{C}$$ and $$\mathcal{F}\subset H(\Omega)$$. Then $$\mathcal{F}$$ is a normal family if and only if it is locally bounded.
1. Fix a sequence $$\{z_n\}\subset \Gamma$$ with $$z_n\rightarrow 0$$ as $$n\rightarrow \infty$$. We must prove $$f(z_n)\rightarrow 0$$. Define $$f_n(z) = f(|z_n|z)$$. Then, since $$z\in \Gamma \Rightarrow |z_n|z\in \Gamma$$, we have
$|f_n(z)| = |f(|z_n|z)| < 1, \text{ for all } z\in \Gamma \text{ and } n\in \mathbb{N}.$
Therefore, $$\mathcal{F}$$ is a normal family in $$\Gamma$$. Also note that each $$f_n$$ is holomorphic in $$\Gamma$$ since $$f(tz) \in H(\Gamma)$$ for any constant $$t>0$$. Thus, $$\mathcal{F}$$ is a normal family of holomorphic functions in $$\Gamma$$.
Let $$g$$ be a normal limit of $$\{f_n\}$$; i.e.,there is some subsequence $$n_k$$ such that, as $$k\rightarrow \infty$$, $$f_{n_k} \rightarrow g$$ locally uniformly in $$\Gamma$$.
Consider the point $$z=i$$. Since $$f(it)\rightarrow 0$$ as $$t\downarrow 0$$,
$g(i) = \lim_{k\rightarrow \infty} f_{n_k}(i) = \lim_{k\rightarrow \infty} f(|z_{n_k}|i) = 0.$
In fact, for any point $$z=iy$$ with $$y>0$$, we have $$g(z)=0$$. Since $$g$$ is holomorphic in $$\Gamma$$, the identity theorem implies that $$g\equiv 0$$ in $$\Gamma$$.
Next, consider
(20)$f_n\left(\frac{z_n}{|z_n|}\right)= f\left(|z_n|\frac{z_n}{|z_n|}\right)= f(z_n).$
The numbers $$z_n/|z_n|$$ lie in the compact set $$\gamma = \{z\in \Gamma: |z|=1\}$$. Since $$f_{n_k}\rightarrow g$$ uniformly in $$\gamma$$, for any $$\epsilon >0$$, there is a $$K>0$$ such that $$|f_{n_k}(z) - g(z)| = |f_{n_k}(z)| < \epsilon$$, for all $$k\geq K$$ and all $$z\in \gamma$$. That is,
$\lim_{k\rightarrow \infty} \sup\{|f_{n_k}(z)|: z\in \gamma\} \triangleq \lim_{k\rightarrow \infty} \|f_{n_k}\|_\gamma = 0,$
and, since $$z_{n_k}/|z_{n_k}|\in \gamma$$,
$\left|f_{n_k}\left(\frac{z_{n_k}}{|z_{n_k}|}\right)\right| \leq \|f_{n_k}\|_\gamma.$
$\therefore \quad \lim_{k\to \infty} f_{n_k}\left(\frac{z_{n_k}}{|z_{n_k}|}\right) = 0.$
By (20), then, $$\lim_{k\rightarrow \infty} f(z_{n_k}) = 0$$.
Finally, recall that $$f(z_n)\rightarrow 0$$ iff every subsequence $$z_{n_j}$$ has a further subsequence $$z_{n_{j_k}}$$ such that $$f(z_{n_{j_k}})\rightarrow 0$$, as $$k\rightarrow \infty$$. Now, if $$z_{n_j}$$ is any subsequence, then $$\{f(z_{n_j})\}$$ is a normal family, and, repeating the argument above, there is, indeed, a further subsequence $$z_{n_{j_k}}$$ such that $$f(z_{n_{j_k}})\rightarrow 0$$. This completes the proof.
Remarks. In the last paragraph, we made use of the fact that a sequence converges to zero iff any subsequence has, in turn, a further subsequence that converges to zero. An alternative concluding argument that doesn’t rely on this result, but proceeds by way of contradiction, runs as follows: Assume we have already shown $$\lim_{k\rightarrow \infty} f(z_{n_k}) = 0$$, as above, and suppose $$f(z_n)$$ does not converge to 0 as $$n\rightarrow \infty$$. Then there is a $$\delta > 0$$ and a subsequence $$\{z_{n_j}\}$$ such that $$|f(z_{n_j})|>\delta$$ for all $$j\in \mathbb{N}$$. Relabel this subsequence $$\{z_n\}$$. Then $$\{f(z_n)\}$$ is itself a normal family and we can repeat the argument above to get a further subsequence $$\{z_{n_k}\}$$ with $$\lim_{k \rightarrow \infty} f(z_{n_k})=0$$. This contradicts the assumption that $$|f(z_{n})|>\delta$$ for all $$n\in \mathbb{N}$$. Therefore, $$f(z_n) \rightarrow 0$$, as desired.
Footnotes
1
See also this problem of April 1989 and this problem of April 1995.
2
Despite the wording of the problem, the family need not satisfy $$\mathcal{F}\subset H(\Omega)$$ in order to be normal.
3
Part (b) of this problem of Nov 2001 asks for a proof of Montel’s theorem using the Arzela-Ascoli theorem.
Real Analysis Exams
|
2020-01-26 16:06:48
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9815913438796997, "perplexity": 88.72643956893855}, "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/1579251689924.62/warc/CC-MAIN-20200126135207-20200126165207-00286.warc.gz"}
|
http://mathhelpforum.com/differential-equations/196148-question-regarding-eigenvalue-sign.html
|
A question regarding eigenvalue sign
I was wondering if it is possible to deduce the sign of eigenvalues of a matrix just by intuition..The reason is that for a large matrix its hard to calculate eigenvalues and check the stability unless you use something like matlab.
Regards
|
2014-08-22 16:25:03
|
{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8043882250785828, "perplexity": 219.83497834614545}, "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-2014-35/segments/1408500824209.82/warc/CC-MAIN-20140820021344-00420-ip-10-180-136-8.ec2.internal.warc.gz"}
|
https://jupyter-gmaps.readthedocs.io/en/latest/app_tutorial.html
|
# Building applications with jupyter-gmaps¶
You can use jupyter-gmaps as a component in a Jupyter widgets application. Jupyter widgets let you embed rich user interfaces in Jupyter notebooks. For instance:
• you can use maps as a way to get user input. The drawing layer lets users draw markers, lines or polygons on the map. We can specify arbitrary Python code that runs whenever a shape is added to the map. As an example, we will build an application where, whenever the user places a marker, we retrieve the address of the marker and write it in a text widget.
• you can use maps as a way to display the result of an external computation. For instance, if you have timestamped geographical data (for instance, you have the date and coordinates of a series of events), you can combine a heatmap with a slider to see how events unfold over time.
## Reacting to user actions on the map¶
The drawing layer lets us specify Python code to be executed whenever the user adds a feature (like a marker, a line or a polygon) to the map. To demonstrate this, we will build a small application for reverse geocoding: when the user places a marker on the map, we will find the address closest to that marker and write it in a text widget. We will use geopy, a wrapper around several geocoding APIs, to calculate the address from the marker’s coordinates.
This is the entire code listing:
import ipywidgets as widgets
import geopy
import gmaps
API_KEY = 'AIz...'
gmaps.configure(api_key=API_KEY)
class ReverseGeocoder(object):
"""
The user places markers on a map. For each marker,
we use geopy to find the nearest address to that
marker, and write that address in a text box.
"""
def __init__(self):
self._figure = gmaps.figure()
self._drawing = gmaps.drawing_layer()
self._drawing.on_new_feature(self._new_feature_callback)
disabled=True,
layout={'width': '95%', 'margin': '10px 0 0 0'}
)
def _get_location_details(self, location):
return self._geocoder.reverse(location, exactly_one=True)
location_details = self._get_location_details(location)
if location_details is None:
else:
def _new_feature_callback(self, feature):
try:
location = feature.location
except AttributeError:
return # Not a marker
# Clear address box to signify to the user that something is happening
# Remove all markers other than the one that has just been added.
self._drawing.features = [feature]
# Compute the address and display it
def render(self):
return self._container
ReverseGeocoder().render()
There are several things to note:
• We wrap the application in a ReverseGeocoder class. Wrapping your application in a class (rather than using the notebook’s global namespace) helps with encapsulation and lets you instantiate this widget multiple times. Since the flow through widget applications is often more complex than linear data analysis workflows, encapsulation will improve your ability to reason about the code.
• As part of the class constructor, we use gmaps.figure() to create a figure. We then use gmaps.drawing_layer() to create a drawing layer, which we add to the figure. We also create a widgets.Text widget. This is a text box in which we will write the address. We then wrap our figure and the text box in a single widgets.VBox, a widget container that stacks widgets vertically.
• We register a callback on the drawing layer using .on_new_feature. The function that we pass in to .on_new_feature will get called whenever the user adds a feature to the map. This is the hook that lets us build complex applications on top of the drawing layer: we can run arbitrary Python code when the user adds a marker to the map.
• In the .on_new_feature callback, we first check whether the feature that has been added is a marker (the user could, in principle, have added another feature type, like a line, to the map).
• Assuming the feature is a valid marker, we first clear the text widget containing the address. This gives feedback to the user that something is happening.
• We then re-write the .features array of the drawing layer, keeping just the marker that the user has just added. This clears previous markers, avoiding clutter on the map.
• We then use geopy to find the adddress. Assuming the address is valid, display it in the text widget.
## Updating a heatmap in response to other widgets¶
Many layers support updating the data without re-rendering the entire map. This is useful for exploring multi-dimensional datasets, especially in conjunction with other widgets.
As an example, we will use the acled_africa_by_year dataset, a dataset indexing violence against civilians in Africa. The original dataset is from the ACLED project. The dataset has four columns:
import gmaps.datasets
We will build an application that lets the user explore different years via a slider. When the user changes the slider, we display the total number of fatalities for that year, and update a heatmap showing the distribution of conflicts.
This is the entire code listing:
from IPython.display import display
import ipywidgets as widgets
import gmaps
gmaps.configure(api_key='AIza...')
class AcledExplorer(object):
"""
Jupyter widget for exploring the ACLED dataset.
The user uses the slider to choose a year. This renders
a heatmap of civilian victims in that year.
"""
def __init__(self, df):
self._df = df
self._heatmap = None
self._slider = None
initial_year = min(self._df['year'])
title_widget = widgets.HTML(
'<h3>Civilian casualties in Africa, by year</h3>'
'<h4>Data from <a href="https://www.acleddata.com/">ACLED project</a></h4>'
)
map_figure = self._render_map(initial_year)
controls = self._render_controls(initial_year)
self._container = widgets.VBox([title_widget, controls, map_figure])
def render(self):
display(self._container)
def _on_year_change(self, change):
year = self._slider.value
self._heatmap.locations = self._locations_for_year(year)
self._total_box.value = self._total_casualties_text_for_year(year)
return self._container
def _render_map(self, initial_year):
fig = gmaps.figure(map_type='HYBRID')
self._heatmap = gmaps.heatmap_layer(
self._locations_for_year(initial_year),
max_intensity=100,
)
return fig
def _render_controls(self, initial_year):
self._slider = widgets.IntSlider(
value=initial_year,
min=min(self._df['year']),
max=max(self._df['year']),
description='Year',
continuous_update=False
)
self._total_box = widgets.Label(
value=self._total_casualties_text_for_year(initial_year)
)
self._slider.observe(self._on_year_change, names='value')
controls = widgets.HBox(
[self._slider, self._total_box],
layout={'justify_content': 'space-between'}
)
return controls
def _locations_for_year(self, year):
return self._df[self._df['year'] == year][['latitude', 'longitude']]
def _total_casualties_for_year(self, year):
return int(self._df[self._df['year'] == year]['year'].count())
def _total_casualties_text_for_year(self, year):
return '{} civilian casualties'.format(self._total_casualties_for_year(year))
AcledExplorer(df).render()
There are several things to note on this:
• We wrap the application in a class to help keep the mutable state encapsulated.
• As part of the class constructor, we use gmaps.figure() to create a figure. We add use gmaps.heatmap_layer() to create a heatmap, which we add to the figure. The Heatmap object returned has a locations attribute. Setting this to a new value will automatically update the heatmap.
• We create a slider with widgets.IntSlider. In general, jupyter-gmaps objects are designed to interact with widgets from ipywidgets. For a full list of available widgets, see the ipywidgets documentation.
• We want to react to changes in the slider: every time the slider moves, we recompute the total number of fatalities and update the data in the heatmap. To react to changes in a widget, we use the .observe method on the widget. This lets us specify a callback that gets called whenever a given attribute of the widget changes. We pass the names="value" argument to slider.observe to only react to changes in the slider’s value attribute. Note that the callback (self.render in our case) needs to take a single argument. It gets passed a dictionary describing the change.
• To build the layout for our application, we use combinations of HBox and VBox widgets.
## Updating symbols in response to other widgets¶
The marker and symbol layers can also be udpated dynamically (as can most other markers).
As an example, we will use the starbucks_kfc_uk dataset, a dataset indexing the location of every Starbucks and KFC in the UK. The original dataset is from the UK food standards agency.
We will build a application with two checkboxes, one for Starbucks outlets and one for KFC outlets. We react to users clicking on the outlets by changing the symbols displayed on the map.
This is the entire code listing:
from IPython.display import display
import ipywidgets as widgets
import gmaps
import gmaps.datasets
gmaps.configure(api_key="AIza...")
class OutletExplorer(object):
def __init__(self, df):
"""
Jupyter widget for exploring KFC and Starbucks outlets
Using checkboxes, the user chooses whether to include
Starbucks, KFC outlets, both or neither.
"""
self._df = df
self._symbol_layer = None
self._starbucks_symbols = self._create_symbols_for_chain(
'starbucks', 'rgba(0, 150, 0, 0.4)')
self._kfc_symbols = self._create_symbols_for_chain(
'kfc', 'rgba(150, 0, 0, 0.4)')
title_widget = widgets.HTML(
'<h3>Explore KFC and Starbucks locations</h3>'
'<h4>Data from <a href="http://ratings.food.gov.uk">UK Food Standards Agency</a></h4>'
)
controls = self._render_controls(True, True)
map_figure = self._render_map(True, True)
self._container = widgets.VBox(
[title_widget, controls, map_figure])
def render(self):
""" Render the widget """
display(self._container)
def _render_map(self, initial_include_starbucks, initial_include_kfc):
""" Render the initial map """
fig = gmaps.figure(layout={'height': '500px'})
symbols = self._generate_symbols(True, True)
self._symbol_layer = gmaps.Markers(markers=symbols)
return fig
def _render_controls(
self,
initial_include_starbucks,
initial_include_kfc
):
""" Render the checkboxes """
self._starbucks_checkbox = widgets.Checkbox(
value=initial_include_starbucks,
description='Starbucks'
)
self._kfc_checkbox = widgets.Checkbox(
value=initial_include_kfc,
description='KFC'
)
self._starbucks_checkbox.observe(
self._on_controls_change, names='value')
self._kfc_checkbox.observe(
self._on_controls_change, names='value')
controls = widgets.VBox(
[self._starbucks_checkbox, self._kfc_checkbox])
return controls
def _on_controls_change(self, obj):
"""
Called when the checkboxes change
This method builds the list of symbols to include on the map,
based on the current checkbox values. It then updates the
symbol layer with the new symbol list.
"""
include_starbucks = self._starbucks_checkbox.value
include_kfc = self._kfc_checkbox.value
symbols = self._generate_symbols(
include_starbucks, include_kfc)
# Update the layer with the new symbols:
self._symbol_layer.markers = symbols
def _generate_symbols(self, include_starbucks, include_kfc):
""" Generate the list of symbols to includs """
symbols = []
if include_starbucks:
symbols.extend(self._starbucks_symbols)
if include_kfc:
symbols.extend(self._kfc_symbols)
return symbols
def _create_symbols_for_chain(self, chain, color):
chain_df = self._df[self._df['chain_name'] == chain]
symbols = [
gmaps.Symbol(
location=(latitude, longitude),
stroke_color=color,
fill_color=color,
scale=2
)
for latitude, longitude in
zip(chain_df["latitude"], chain_df["longitude"])
]
return symbols
OutletExplorer(df).render()
We used the gmaps.Markers class to represent the symbol layer, rather than the gmaps.symbol_layer() factory function. The Markers class is easier to manipulate, since it just takes a list of symbols. The disadvantage is that we had to construct the symbols independently.
|
2019-06-16 01:30:57
|
{"extraction_info": {"found_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": 0.1945960968732834, "perplexity": 5801.6246343523}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-26/segments/1560627997508.21/warc/CC-MAIN-20190616002634-20190616024634-00074.warc.gz"}
|
https://mathoverflow.net/questions/284961/modular-parametrization-in-terms-of-the-moduli-of-shtukas
|
# Modular parametrization in terms of the moduli of shtukas
The modular parametrization of an elliptic curve over $\mathbb{Q}$ (and maybe over a number field in general?) is well-known; also for an elliptic curve over global function field with some condition (having split multiplicative reduction at some place), one also has an analogue of modular parametrization using Drinfeld modular variety in the function field case.
My main question is: Is there a version, over global function field, of modular parametrization using shtukas?
By the way, I read the survey ''Elliptic Curves and Analogies Between Number Fields and Function Fields'' by Dough Ulmer where he mentioned that there is a work by him entitled "Automorphic forms on GL2 over function fields and Gross–Zagier theorems'', but I could not find it. Has anyone read this paper and know where can I find it?
|
2021-04-20 10:48:24
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8127884864807129, "perplexity": 400.20054230812354}, "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/1618039388763.75/warc/CC-MAIN-20210420091336-20210420121336-00583.warc.gz"}
|
https://mathematicalypse.wordpress.com/2015/10/22/history-of-connections/
|
History of Connections
I haven’t blogged for our seminar in quite a while and thought I might fix this with this short post.
We are now discussing connections on Riemannian manifolds — more accurately, affine connections. Such an object allows us to differentiate a vector field, $Y$, along another vector field, $X$, at any point $p \in M$. The result of this process is typically notated as $\nabla_X\,Y$, and, however its defined, it must enjoy three properties:
1. $\nabla_{fX+gY}Z = f\nabla_XZ + g\nabla_YZ$
2. $\nabla_X(fY) = (X(f))Y + f\nabla_XY$
3. $\nabla_X(Y+Z) = \nabla_XY + \nabla_XZ$
These conditions are motivated by what happens in an easier-to-understand setting, like $\mathbb{R}^n$, where one can differentiate vector fields along other vector fields in an obvious way.
Given a Riemannian metric $\langle, \rangle$ on $M$, it turns out there always exists a special affine connection, one referred to as the Levi-Civita connection on $M$. It is defined by being both (1) compatible with the metric and (2) symmetric. Compatibility means that for all vector fields $X, Y,$ and $Z$ one has
$\displaystyle \nabla_X\langle Y, Z \rangle = \langle \nabla_XY,Z\rangle + \langle Y, \nabla_XZ\rangle$
and symmetry means that
$\displaystyle \nabla_XY - \nabla_YX =[X, Y].$
This is all well and good, but it can nonetheless feel too detached from intuition. For instance, one might wonder why we use these properties to define an affine connection as opposed to other ones. Similarly, why would we want compatibility to hold and why would we want symmetry?
For answers to these and other, related questions, you should consult this Masters Thesis. Its from 2008, and was written by Kamielle Freeman; the title of the manuscript is “A Historical Overview of Connections in Geometry.” It turns out that there are lots of ways to approach and motivate the idea of a connection, and during the late 1800s and early 1900s, various mathematicians did. One main idea can be (roughly) summarized as follows:
parallel transport $\iff$ affine connection
This idea of “parallel transport” is what allowed mathematicians to think about how to “connect” two different tangent spaces attached to the same manifold.
Q: How do I think of this vector $\vec{v} \in T_pM$ as a vector in $T_qM$?
A: Parallel transport the vector $\vec{v} \in T_pM$ to a unique vector $w \in T_qM$
The ability to move $\vec{v}$ to a new vector in a different tangent space implies the ability to differentiate vector fields along other vector fields. Conversely, the ability to differentiate vector fields along other vector fields implies ability to “parallel transport.”
A Brief Note on Tensors
We’ve known for a few weeks now that vector fields are sections of a particular vector bundle over our manifold $M$, i.e. the tangent bundle $TM$. As we’ve (lightly) discussed, though, there are other vector spaces one can bundle over $M$, and there exist smooth sections of these bundles, too. For instance, the co-tangent bundle, $T^*M$, has as its sections differential 1-forms.
tensor over $M$ is a section of a more general kind of vector bundle (which is why this is also some times called a tensor field). The vector bundles that give rise to tensor fields can be built out of interesting combinations of the two bundles $TM$ and $T^*M$. Once we have an affine connection $\nabla$ on a Riemannian manifold, we can use it to build a connection for $T^*M$, i.e. we can differentiate co-vector fields along vector fields. This process can be continued further, in fact; we can use $\nabla$ (and the underlying metric $\langle\, , \,\rangle = g$) to differentiate any tensor field along a vector field.
|
2017-07-23 22:47:11
|
{"extraction_info": {"found_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": 32, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9140176177024841, "perplexity": 439.1751857935304}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-30/segments/1500549424623.68/warc/CC-MAIN-20170723222438-20170724002438-00475.warc.gz"}
|
https://nbviewer.jupyter.org/github/arnicas/NLP-in-Python/blob/master/4.%20Naive%20Bayes%20Classification.ipynb?imm_mid=0cd660&cmp=em-data-na-na-newsltr_20150225
|
## 4. Doing Naive Bayes Classification¶
### Lynn Cherny, 2/10/15, [email protected]¶
Full repo here: https://github.com/arnicas/NLP-in-Python
This is an example of going from labeled text to machine classification, first with NLTK and then the Python machine learning library scikit-learn. Examples updated from my OpenVis Conf talk here, which is more entertaining: https://www.youtube.com/watch?v=f41U936WqPM and slides: http://www.slideshare.net/arnicas/the-bones-of-a-bestseller
### Warning: Rated NC-17. Using text samples from "50 Shades of Gray"! (Because spam is boring.)¶
Inspired by this image from the Economist (orignally http://www.economist.com/blogs/graphicdetail/2012/11/fifty-shades-data-visualisations):
I wondered if I could identify the sex scenes automatically, based on training examples. That's "classification."
Because the book was too hard to read, I farmed out (badly formatted) short chunks of "50 Shades of Gray" to Mechanical Turkers to rate as "sexy" or "not" (on a ratings scale): a score of 0 is "not a sex scene", while "1" and "2" are increasing in steaminess, and "3" is definitely a sex scene. (In later scoring, I reduced the options to just 3.)
Assuming that a score of >= 2.5 is a sex scene, I put the scores and text into a usable file with that labeling.
In [3]:
labelsfile = 'data/csv/fiftyshades_labeled.txt'
In [4]:
def get_documents_csv(filename):
""" Read in the labeled chunks and classifications.
Assume label is cell 1, doc text is cell 2, classification is cell 3.
"""
labels = []
documents = []
classif = []
for line in open(filename):
fields = line.split("\t")
if fields[0].strip != 'label': # header row
documents.append(fields[1].strip())
labels.append(fields[0].strip())
if fields[2]:
classif.append(fields[2].strip("\n"))
print "Got", len(documents), "chunks"
return (documents, labels, classif)
In [5]:
docs, labels, classes = get_documents_csv(labelsfile)
Got 382 chunks
In [6]:
# Hmm, in my classes I did categorize "maybes." These can either be used as "no" or as a third class.
classes[25:40]
Out[6]:
['maybe\r',
'yes\r',
'maybe\r',
'no\r',
'maybe\r',
'yes\r',
'maybe\r',
'maybe\r',
'maybe\r',
'no\r',
'maybe\r',
'maybe\r',
'no\r',
'yes\r',
'maybe\r']
In [7]:
labels[25:40] # these are the original file chunks for reference
Out[7]:
['fifty_500_166',
'fifty_500_199',
'fifty_500_247',
'fifty_500_365',
'fifty_500_138',
'fifty_500_338',
'fifty_500_348',
'fifty_500_364',
'fifty_500_84',
'fifty_500_366',
'fifty_500_85',
'fifty_500_368',
'fifty_500_56',
'fifty_500_92',
'fifty_500_276']
### But let's build a classifier. Here's a schematic from Perkins for the machine learning workflow:¶
Some references:
In [8]:
# this text can't be used as is in the classifier -- and notice this was a "maybe":
docs[25]
Out[8]:
'"any time , Anastasia. I won * t stop you. If you go , however * that * s it. Just so you know. * * Okay , * I answer softly. If I go , that * s it. The thought is surprisingly painful . The waiter arrives with our first course. How can I possibly eat ? Holy Moses * he * s ordered oysters on a bed of ice . * I hope you like oysters. * Christian * s voice is soft . * I * ve never had one. * Ever . * Really ? Well. * He reaches for one. * All you do is tip and swallow. I think you can manage that. * He gazes at me , and I know what he * s referring to. I blush scarlet. He grins at me , squirts some lemon juice onto his oyster , and then tips it into his mouth . * Hmm , delicious. Tastes of the sea. * He grins at me. * Go on , * he encourages . * So , I don * t chew it ? * * No , Anastasia , you don * t. * His eyes are alight with humor. He looks so young like this . I bite my lip and his expression changes instantly. He looks sternly at me. I reach across and pick up my first-ever oyster. Okay * here goes nothing. I squirt some lemon juice on it and tip it up. It slips down my throat , all sea water , salt , the sharp tang of citrus , and fleshiness * ooh. I lick my lips , and he * s watching me intently , his eyes hooded . * Well ? * * I * ll have another , * I say dryly . * Good girl , * he says proudly . * Did you choose these deliberately ? Aren * t they known for their aphrodisiac qualities ? * * No , they are the first item on the menu. I don * t need an aphrodisiac near you. I think you know that , and I think you react the same way near me , * he says simply. * So where were we ? * He glances at my e-mail as I reach for another oyster . He reacts the same way. I affect him * wow . * Obey me in all things. Yes , I want you to do that. I need you to do that. Think of it as role-play , Anastasia. * * But I * m worried you * ll hurt me. * * Hurt you how ? * * Physically. * And emotionally . * Do you really think I would do that ? Go beyond any limit you can * t take ? * * You * ve said you * ve hurt someone before. * * Yes , I have. It was a long"'
I don't remember why my original text extracts had so much horrible * formatting - something to do with getting it to Excel for Mechanical Turk as fast as possible. Showing it to you as the raters saw it, for honestly purposes, and maybe it won't get me in trouble for sharing if it's so awful to read like that!
Let's clean it up - here we are using a regex tokenizer to clean the garbage out. Also, making a class to store a bunch of stuff in!
In [9]:
class Document:
def __init__(self):
Document.words = []
Document.original = ""
Document.clean = ""
Document.label = ""
Document.classif = ""
def clean_doc(doc):
from nltk import corpus
import re
stopwords = corpus.stopwords.words('english')
new = Document()
new.original = doc
sentence = doc
sentence = sentence.lower()
# note that I'm looking for non-numeric alphabetic items; this makes a difference from sklearn
words = re.findall(r'\w+', sentence, flags = re.UNICODE | re.LOCALE)
new.clean = " ".join(words)
words = [word for word in words if word not in stopwords]
new.words = words
return new
In [10]:
# example ouput:
clean_doc(docs[25]).clean
Out[10]:
'any time anastasia i won t stop you if you go however that s it just so you know okay i answer softly if i go that s it the thought is surprisingly painful the waiter arrives with our first course how can i possibly eat holy moses he s ordered oysters on a bed of ice i hope you like oysters christian s voice is soft i ve never had one ever really well he reaches for one all you do is tip and swallow i think you can manage that he gazes at me and i know what he s referring to i blush scarlet he grins at me squirts some lemon juice onto his oyster and then tips it into his mouth hmm delicious tastes of the sea he grins at me go on he encourages so i don t chew it no anastasia you don t his eyes are alight with humor he looks so young like this i bite my lip and his expression changes instantly he looks sternly at me i reach across and pick up my first ever oyster okay here goes nothing i squirt some lemon juice on it and tip it up it slips down my throat all sea water salt the sharp tang of citrus and fleshiness ooh i lick my lips and he s watching me intently his eyes hooded well i ll have another i say dryly good girl he says proudly did you choose these deliberately aren t they known for their aphrodisiac qualities no they are the first item on the menu i don t need an aphrodisiac near you i think you know that and i think you react the same way near me he says simply so where were we he glances at my e mail as i reach for another oyster he reacts the same way i affect him wow obey me in all things yes i want you to do that i need you to do that think of it as role play anastasia but i m worried you ll hurt me hurt you how physically and emotionally do you really think i would do that go beyond any limit you can t take you ve said you ve hurt someone before yes i have it was a long'
In [11]:
# clean them all...
clean_docs = [clean_doc(x) for x in docs]
In [12]:
# Fix up with more info on each object:
# Go thru the objects we just made and add the corresponding class and label
for i,x in enumerate(doc_objs):
x.label = labels[i]
x.id = i
x.classif = classes[i].strip("\r") # may be necessary to strip, was for me
return doc_objs
In [13]:
clean_docs = add_ids_classes(clean_docs, labels, classes)
In [14]:
clean_docs[0]
clean_docs[0].classif
Out[14]:
'maybe'
In [15]:
# We will consider the "maybe" as no, for now:
neg_docs = [doc for doc in clean_docs if doc.classif == 'no' or doc.classif == 'maybe']
pos_docs = [doc for doc in clean_docs if doc.classif == 'yes']
In [16]:
print len(neg_docs), len(pos_docs)
327 55
In [18]:
# Bag of Words - just a True for each word's presence in a document. Later we'll use TF-IDF weights.
def word_feats(words):
return dict([(word, True) for word in words])
In [19]:
neg_words = [(word_feats(doc.words),'neg') for doc in neg_docs]
pos_words = [(word_feats(doc.words),'pos') for doc in pos_docs]
In [24]:
# These are lists of dictionaries, one for each text. Here's the first "no" text:
neg_words[0]
Out[24]:
({'accomplishments': True,
'ana': True,
'apogee': True,
'arms': True,
'around': True,
'baby': True,
'back': True,
'bed': True,
'blue': True,
'breathes': True,
'care': True,
'cares': True,
'christian': True,
'come': True,
'comprehending': True,
'concern': True,
'confusion': True,
'conquest': True,
'course': True,
'covered': True,
'cries': True,
'crushes': True,
'd': True,
'depths': True,
'devours': True,
'didn': True,
'doubt': True,
'effect': True,
'energized': True,
'everywhere': True,
'exactly': True,
'exciting': True,
'eyes': True,
'face': True,
'far': True,
'favorite': True,
'feel': True,
'fifteen': True,
'film': True,
'find': True,
'finds': True,
'flown': True,
'fronts': True,
'frowns': True,
'fulfilling': True,
'full': True,
'funny': True,
'gape': True,
'good': True,
'gray': True,
'greatest': True,
'grey': True,
'grin': True,
'grinning': True,
'grins': True,
'happening': True,
'holy': True,
'hugging': True,
'idiot': True,
'incredulity': True,
'infectious': True,
'inside': True,
'invocation': True,
'king': True,
'lie': True,
'like': True,
'lips': True,
'looking': True,
'love': True,
'm': True,
'man': True,
'many': True,
'meant': True,
'mine': True,
'mirroring': True,
'miss': True,
'mr': True,
'naked': True,
'number': True,
'obvious': True,
'oh': True,
'one': True,
'orgasm': True,
'passion': True,
'passionate': True,
'piano': True,
'pillows': True,
'play': True,
'playroom': True,
'quirk': True,
'referring': True,
'release': True,
'right': True,
'ripping': True,
'said': True,
'score': True,
'see': True,
'seventeen': True,
'sex': True,
'shakes': True,
'sheet': True,
'shining': True,
'shit': True,
'silly': True,
'sleep': True,
'sloshing': True,
'smiles': True,
'soft': True,
'soul': True,
'staring': True,
'steele': True,
'still': True,
'stirring': True,
'stop': True,
'strangely': True,
'stuff': True,
'suddenly': True,
'super': True,
'talk': True,
'thought': True,
'tired': True,
'today': True,
'touching': True,
'turbulent': True,
'um': True,
'unexpected': True,
'vanilla': True,
've': True,
'voice': True,
'want': True,
'whole': True,
'wild': True,
'women': True,
'wrapped': True},
'neg')
In [25]:
# Let's make a cut point at 3/4 of each list, so we can do separate test and training runs.
negcutoff = len(neg_words)*3/4
poscutoff = len(pos_words)*3/4
In [19]:
# Now split up the lists into training and testing.
import random
random.shuffle(neg_words)
random.shuffle(pos_words)
train_fic = neg_words[:negcutoff] + pos_words[:poscutoff]
test_fic = neg_words[negcutoff:] + pos_words[poscutoff:]
print 'train on %d docs, test on %d docs' % (len(train_fic), len(test_fic))
train on 286 docs, test on 96 docs
## Naive Bayes¶
Naive Bayes classifiers (they are a family) are pretty good on text. The Bayes in "Naive Bayes" refers to Bayes' Theorem:
In English, we say:
For texts, this means that the probability of a document being in a class (like, "sex scene") is based on the probability of that class in the training set (our prior) times the combined probabilities of the words in the new document appearing in that class. The "winning" class is the one with a higher score for the prior times the word-category probability. For a longer explanation of how this works, read YHat's blog post.
It is "naive" because it assumes no relationship between the features (words) in the data. It also requires relatively little data to train it, which is good for smaller data problems. But it also scales well to a lot of vocabulary.
### The results are reasonably good...¶
In [20]:
import nltk.classify.util
from nltk.classify import NaiveBayesClassifier
classifier = NaiveBayesClassifier.train(train_fic)
print 'accuracy:', nltk.classify.util.accuracy(classifier, test_fic)
classifier.show_most_informative_features(15)
accuracy: 0.8125
Most Informative Features
navel = True pos : neg = 25.4 : 1.0
breasts = True pos : neg = 22.3 : 1.0
clitoris = True pos : neg = 21.5 : 1.0
groan = True pos : neg = 20.3 : 1.0
beg = True pos : neg = 17.6 : 1.0
eases = True pos : neg = 17.6 : 1.0
voices = True pos : neg = 17.6 : 1.0
upper = True pos : neg = 17.6 : 1.0
peels = True pos : neg = 17.6 : 1.0
stilling = True pos : neg = 17.6 : 1.0
washing = True pos : neg = 17.6 : 1.0
swirling = True pos : neg = 17.6 : 1.0
stretching = True pos : neg = 17.6 : 1.0
assault = True pos : neg = 17.6 : 1.0
blows = True pos : neg = 17.6 : 1.0
### An interface to Sklearn (scikit-learn) from NLTK is available (see Perkin's Python 3 and NLTK book from Packt)¶
In [21]:
from nltk.classify.scikitlearn import SklearnClassifier
from sklearn.naive_bayes import MultinomialNB
from nltk.classify.util import accuracy
sk_classifier = SklearnClassifier(MultinomialNB())
sk_classifier.train(train_fic)
accuracy(sk_classifier, test_fic)
Out[21]:
0.9583333333333334
### Perkins discusses some of the differences in performance in his book. Basically, for machine learning problems, sklearn is highly optimized.¶
Let's look at a visualization of the accuracy of one of the runs, from my Openvis Conf talk:
http://www.ghostweather.com/essays/talks/openvisconf/text_scores/rollover.html
Now, optionally, you can look at notebook 5 on using sklearn to do the same things!
In [21]:
|
2020-02-25 11:12:54
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5001530051231384, "perplexity": 9472.724932929417}, "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/1581875146066.89/warc/CC-MAIN-20200225110721-20200225140721-00050.warc.gz"}
|
http://dmtcs.episciences.org/609
|
## Chandran, Sunil, and Mathew, Rogers - Bipartite powers of k-chordal graphs
dmtcs:609 - Discrete Mathematics & Theoretical Computer Science, May 6, 2013, Vol. 15 no. 2
Bipartite powers of k-chordal graphs
Authors: Chandran, Sunil, and Mathew, Rogers
Let k be an integer and k ≥3. A graph G is k-chordal if G does not have an induced cycle of length greater than k. From the definition it is clear that 3-chordal graphs are precisely the class of chordal graphs. Duchet proved that, for every positive integer m, if Gm is chordal then so is Gm+2. Brandstädt et al. in [Andreas Brandstädt, Van Bang Le, and Thomas Szymczak. Duchet-type theorems for powers of HHD-free graphs. Discrete Mathematics, 177(1-3):9-16, 1997.] showed that if Gm is k-chordal, then so is Gm+2. Powering a bipartite graph does not preserve its bipartitedness. In order to preserve the bipartitedness of a bipartite graph while powering Chandran et al. introduced the notion of bipartite powering. This notion was introduced to aid their study of boxicity of chordal bipartite graphs. The m-th bipartite power G[m] of a bipartite graph G is the bipartite graph obtained from G by adding edges (u,v) where dG(u,v) is odd and less than or equal to m. Note that G[m] = G[m+1] for each odd m. In this paper we show that, given a bipartite graph G, if G is k-chordal then so is G[m], where k, m are positive integers with k≥4.
Source : oai:HAL:hal-00980750v1
Volume: Vol. 15 no. 2
Section: Graph Theory
Published on: May 6, 2013
Submitted on: May 4, 2012
Keywords: [INFO.INFO-DM] Computer Science [cs]/Discrete Mathematics [cs.DM]
|
2017-11-24 11:23:11
|
{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9123048186302185, "perplexity": 2025.9263715300688}, "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/1510934807650.44/warc/CC-MAIN-20171124104142-20171124124142-00399.warc.gz"}
|
https://gateoverflow.in/1163/gate2005-13
|
841 views
The set $\{1, 2, 4, 7, 8, 11, 13, 14\}$ is a group under multiplication modulo $15$. The inverses of $4$ and $7$ are respectively:
1. $3$ and $13$
2. $2$ and $11$
3. $4$ and $13$
4. $8$ and $14$
edited | 841 views
Option C.
Identity element here is $1.$
$4 * 4 \mod 15 = 1$
$7 * 13 \mod 15 = 1.$
selected by
+8
It is same as finding modular inverse, which is also last step in RSA.
You may likely to know actual procedure for even big numbers-
https://goo.gl/ZCza9r
|
2018-08-17 09:35:16
|
{"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": 2, "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": 0.8837990164756775, "perplexity": 1185.1526924565037}, "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/1534221211935.42/warc/CC-MAIN-20180817084620-20180817104620-00440.warc.gz"}
|
http://tex.stackexchange.com/questions/112745/how-to-jump-brackets-in-texstudio
|
# How to “jump” brackets in TeXstudio?
In Texmaker when you type the shortcut for \frac, you get something like \frac{*}{*}, but instead of asterisks, you have bullets. When you type something in the first pair of brackets, you can then hit tab to position the cursor between the second pair of brackets. How can I do this with TeXstudio?
-
On a Windows PC is Ctrl+Right arrow
On Mac OS X it is Command+Right arrow
# ⌘+⇒
It can be changed to Tab (), but this is by default linked to another shortcut. If you go to the Preferences, "Shortcuts" tab, open up the "Editor" and the "Basic Key Mapping" list. Scroll down until you find
Next placeholder or one word right
Double click on the shortcut and click for opening the drop down menu; there you can find Tab as in the following picture:
Answer "Yes" to both the following questions. Relaunch TeXStudio.
-
On may Mac it’s [cmd]+[left]/[right] and I guess that is the default … – Tobi May 6 at 14:08
|
2013-12-12 10:00:01
|
{"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": 0.34066614508628845, "perplexity": 3139.216972198622}, "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/1386164579146/warc/CC-MAIN-20131204134259-00039-ip-10-33-133-15.ec2.internal.warc.gz"}
|
https://stats.stackexchange.com/questions/86378/principal-components-regression-transform-95-cis-back-to-original-space/86390
|
# Principal Components Regression: Transform 95% CIs back to original space
I have a set of predictors that clearly suffer from some amount of multicollinearity, so I am using PCA to make the columns of X orthogonal. I am also using this as a way to regularize the subsequent regression by removing components that account for ~0% of the variance.
For example, if I run OLS regression on PCA-transformed data that has 8 predictors, I am then able to use the eigenvectors from the original PCA transformation to get the beta weights for the original 12 predictors. So far, so good.
However, to be able to evaluate the contributions of these predictors to the model fit, I'd like to transform the 95% confidence intervals back into the original space of the 12 predictors. That way, I can use the overall R^2 and associated p-values for the full model to find regressions that are significant, where specific predictors have non-zero contributions.
It is unclear to me how to transform the 95% confidence intervals. If that's not possible, is there another way to evaluate the significance of specific predictors in the original space?
Thanks
• Well I don't think that my data satisfy any of those special cases. I'm not clear why the CIs have infinite range, though. If that's true, then is there another way to assess the significance of individual weights? I don't particularly care about arbitrary thresholds (which is why 95% CIs are not ideal for me anyway), but I just need to be able to know whether a given weight has a "significant" (i.e., non-zero) contribution to the overall model. – Matt L. Feb 12 '14 at 19:41
• Ok, I think that makes sense. What are the implications for interpreting the beta weights after projecting them back into the original space? It seems that they can still be understood as contributions of individual predictors to explaining the output, but does that mean that any non-zero value is "different" from zero? – Matt L. Feb 12 '14 at 20:25
• @whuber: May I ask you to read my answer and check whether my reasoning is correct? – cbeleites unhappy with SX Feb 12 '14 at 20:59
• @amoeba You are correct and I apologize for posting a comment that was so misleading (since deleted). Please see my remarks after the answer posted by cbeleites, which apparently crossed yours in the ether. – whuber Feb 12 '14 at 21:53
Do I understand correctly:
PCA calculates $p$ scores $\mathbf T^{(n \times p)}$ from data $mathbf X^{(n \times m)}$ using the transpose of the loadings $\mathbf P^{(p \times m)}$:
$\mathbf T = \mathbf X \mathbf P^T$
then the OLS calculates some $Y^{(n \times 1)}$ using coefficients $\beta^{(p \times 1)}$:
$Y = \mathbf T \beta$, together:
$Y = \mathbf X \mathbf P^T \beta = \mathbf X \mathbf B$ with
$\mathbf B^{(m \times 1)} = \mathbf P^T \beta$
And now you want to have some indication of the variance of $\mathbf B$?
First of all, in order to get confidence intervals for $\mathbf B$ you need to consider both the PCA and the regression.
Calculating confidence intervals for $\beta$ alone doesn't make sense: PCA is not a projection that is unique, i.e. the axes can flip without notice. In addition for your PCR model, in the $p$-dimensional space of the retained PCs you can also have rotations which do not affect the predictions if $\beta$ changes accordingly.
I suspect that not taking care of these equivalence rules (= restrictions/contraints) is what causes the $\pm \infty$ range in @whuber's comment.
I think of this as: what happens to my model if I acquire a new data set and fit a new model. The models can be equivalent (having the same $\mathbf B$), but different loadings $\mathbf P$ and regression coefficients $\beta$.
Now I have no idea how to get confidence intervals for the PCA, and then how to combine these two given the equvalence contraints. I usually go a much easier way:
I bootstrap $\mathbf B$ during a resampling (out-of-bootstrap) validation.
(So far I don't need confidence intervals for $\mathbf B$, for my purposes the distribution of observed $\mathbf B$s over the bootstrapping is good enough - I need "hard numbers" only on the predictive power)
• Your description of the problem is accurate to what I'm doing. I thought about bootstrapping $\mathbf B$, but is it bootstrapped over different PCA loadings? As in, do I need to do it selecting different combinations of PCs? That wouldn't seem to be very useful since I have a set PCs that account for 99% of the variance, and throwing out additional PCs will only decrease the data quality. – Matt L. Feb 12 '14 at 21:09
• I have been thinking about this and realize that I have confused intervals of the estimated coefficients with intervals of the data (which is silly, but there it is). I apologize for pointing you in the wrong direction. The solution is straightforward: when $\hat\beta_i$ are estimated coefficients for principal components $\sum_j a_{ij}X_j$, then $X_j$ enters into the regression with coefficient $\sum_j\hat\beta_ia_{ij}$. Its variance is easily computed from the covariance matrix $\Sigma$ of the $\hat\beta_i$ and from that one obtains confidence intervals as usual. – whuber Feb 12 '14 at 21:47
• (Note that it doesn't matter that the principal vectors are determined only up to sign. In fact, it doesn't even matter that the $\sum_ja_{ij}X_j$ are principal components: the same calculation applies no matter what the $a_{ij}$ happen to be, provided their computation is not based on any values of the dependent variable in the regression.) – whuber Feb 12 '14 at 21:51
• I am taking the $X_j$ to be established by the experimenter rather than measured with error, as is usual with OLS regression. Therefore there is no uncertainty in the PCA: it merely is an alternative description of the $\mathbb{X}$ matrix. If you like, think of it as a multivariate generalization of the possibility of changing the units of measurement of the individual IVs (which would correspond to a diagonal matrix $\mathbb{A}=(a_{ij})$). If you intend to use the regression results for prediction, then you need to continue to use the matrix $\mathbb{A}$ rather than recompute it on new data! – whuber Feb 12 '14 at 22:00
• Would be nice to have $\mathbf X$ without error - I'll try to tell that to my spectrometer... My argumentation for PCR or PLSR would be that the regularized projection is a step that produces low-variance input for the OLS. Which doesn't that much about the stability of the projection. But thinking further, if the model is well set up (i.e. not too many latent variables), the projection should be stable. But with that assumption we do not have any guard against overfitting the PCA. – cbeleites unhappy with SX Feb 12 '14 at 22:11
|
2020-04-04 16:06:52
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7232386469841003, "perplexity": 395.01090162164706}, "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/1585370524043.56/warc/CC-MAIN-20200404134723-20200404164723-00292.warc.gz"}
|
https://www.shaalaa.com/question-bank-solutions/chord-properties-chords-equidistant-center-are-equal-without-proof-given-o-centre-circle-aob-70-calculate-value-of-i-oca-ii-oca_38308
|
Share
Notifications
View all notifications
Books Shortlist
Your shortlist is empty
# Given O is the Centre of the Circle and ∠Aob = 70°. Calculate the Value Of: (I) ∠Oca . (Ii) ∠Oca . - ICSE Class 10 - Mathematics
Login
Create free account
Forgot password?
ConceptChord Properties - Chords Equidistant from the Center Are Equal (Without Proof)
#### Question
Given O is the centre of the circle and ∠AOB = 70°. Calculate the value of:
(i) ∠OCA .
(ii) ∠OCA .
#### Solution
Here, ∠AOB= 2∠ACB
(Angle at the center is double the angle at the circumference by the same chord)
⇒ ∠ACB =70/2 =35°
Now, OC = OA (radii of same circle)
⇒ ∠OCA = ∠OAC =35°
Is there an error in this question or solution?
#### Video TutorialsVIEW ALL [1]
Solution Given O is the Centre of the Circle and ∠Aob = 70°. Calculate the Value Of: (I) ∠Oca . (Ii) ∠Oca . Concept: Chord Properties - Chords Equidistant from the Center Are Equal (Without Proof).
S
|
2019-07-16 11:03:59
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.3203331530094147, "perplexity": 6762.847750721253}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-30/segments/1563195524522.18/warc/CC-MAIN-20190716095720-20190716121720-00416.warc.gz"}
|
http://rin.io/picard-variety/
|
# $Pic(X)$ vs. $CP^\infty$
Thank you to Edward Frenkel for kindly explaining the difference between $CP^\infty$ and $Pic(X)$ (both classifying spaces of line bundles), and to Qiaochu Yuan for explaining why on earth $CP^\infty$ is the moduli space of line bundles over a point. Any errors are mine, not theirs.
As we saw in a Precursor to Characteristic Classes, $CP^\infty$ is the classifying space of complex line bundles over $X$.
$CP^\infty$ is, in some sense, the moduli space of line bundles over a point. There’s only one isomorphism class of line bundles over a point — but then this one line bundle has automorphism group $C^\times$ (which is homotopy equivalent to $U(1)$).
Allow me to introduce you to something that looks a LOT like $CP^\infty$.
What is this map, $p \times C \to Pic(C) \times C$, you might ask. Choose a point on our curve $C$ and this defines a line bundle over $S$ corresponding to a choice of the class of line bundles in $Pic(C)$. In other words, we take a point on a (not sure if I require smoothness here) algebraic curve and turn it into a line bundle on that curve.
Warning: I’ve been told that there is a difference between topological line bundles and algebraic line bundles, unfortunately, I don’t know why or what it is! I mention this, for $Pic(X)$ usually corresponds to *algebraic* line bundles over $X$.
Why is the multiplicative formal group getting involved? Let’s briefly review what the multiplicative formal group law is (as a group scheme).
|
2018-05-25 05:21:34
|
{"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": 0.9637564420700073, "perplexity": 221.8048169386774}, "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/1526794867041.69/warc/CC-MAIN-20180525043910-20180525063910-00426.warc.gz"}
|
http://golem.ph.utexas.edu/category/2013/03/spivak_on_category_theory.html
|
## March 3, 2013
### Spivak on Category Theory
#### Posted by Simon Willerton
Guest post by Bruce Bartlett
We know about Category Theory for Mathematicians, we’ve all read Category Theory for Physicists, and we also know about Category Theory for Computer Scientists, and we’ve even seen the videos.
But how about Category Theory for Scientists? I spotted this on the arXiv listings.
David Spivak, Category Theory for Scientists.
Abstract: There are many books designed to introduce category theory to either a mathematical audience or a computer science audience. In this book, our audience is the broader scientific community. We attempt to show that category theory can be applied throughout the sciences as a framework for modeling phenomena and communicating results. In order to target the scientific audience, this book is example-based rather than proof-based. For example, monoids are framed in terms of agents acting on objects, sheaves are introduced with primary examples coming from geography, and colored operads are discussed in terms of their ability to model self-similarity.
I’m afraid this little post is just a shout-out as I’ve only hurriedly browsed through the pages.
Towards the end of the book he gets to sheaves; he is certainly an expert on these as his PhD thesis was on derived smooth manifolds). His motivating example is stitching together pictures of the night sky, which I thought was really cool:
Paging through, I see the Yoneda lemma only gets a small paragraph, with a reference to Mac Lane. I’m kind of sad about that, since I do regard it as the fundamental theorem of category theory. Too bad.
Posted at March 3, 2013 10:29 PM UTC
TrackBack URL for this Entry: http://golem.ph.utexas.edu/cgi-bin/MT-3.0/dxy-tb.fcgi/2599
### Re: Spivak on Category Theory
Thanks for the write-up!
As for Yoneda, your peer pressure has worked! I’ll add more about it in the next version.
Posted by: David Spivak on March 4, 2013 9:00 PM | Permalink | Reply to this
### Re: Spivak on Category Theory
Do you want to explain why you might refer to it as the Ubuntu-Yoneda Lemma, Bruce? (As opposed to Lawvere referring to it as the Cayley-Dedekind-Grothendieck-Yoneda Lemma.)
Posted by: Simon Willerton on March 4, 2013 9:45 PM | Permalink | Reply to this
### Re: Spivak on Category Theory
Okay, here goes. IsiXhosa has a proverb, umntu ngumntu ngabantu. In this sentence, um = “a”, ntu=”person”, ngu=”is”, nga=”through”, abantu=”people”. So it means literally a person is a person through other people. (Disclaimer: novice at large).
Academics translate the meaning of this proverb as “attaining the totality of being a fully adjusted member of society only through the support, counselling, love, assistance, shelter, example, etc. of one’s fellow human beings”.
I see it as a more elegant version of the Yoneda lemma. A thing is a thing only in the way that it relates to other things. Knowing $\mathrm{Hom}\left(X,A\right)$ for all $A$ is equivalent to knowing $X$. No man is an island. You exist only through, and you are completely determined by, your connections with others. You are nothing more than the sum of your relationships. That kind of vibe.
By the way, the word ubuntu, of Linux fame, is closely related. Ubu=abstract noun prefix, so it literally means “person-ness”, or common human decency. People write essays about it, as in the Wikipedia link, since it is a central theme of African culture.
Posted by: Bruce Bartlett on March 4, 2013 10:20 PM | Permalink | Reply to this
### Re: Spivak on Category Theory
This is fantastic.
Posted by: Emily Riehl on March 6, 2013 4:31 AM | Permalink | Reply to this
### Re: Spivak on Category Theory
Ok, great. Well done on the book.
Posted by: Bruce Bartlett on March 4, 2013 10:24 PM | Permalink | Reply to this
### Re: Spivak on Category Theory
I’m sure a lot of people at the Cafe would agree that the “categorical stance” has the potential to transcend academic disciplines. But actually identifying how to realize this potential is another story. It’s really exciting to see the dialogue advancing on this front!
Spivak’s “olog”s (as in “ontology log”) reminded me of the categorically-motivated work of Reyes et. al. on grammar. For instance, this article (here’s the direct link) analyzes the relationship between count nouns (a man, an amino acid,…) and mass nouns (water, arginine,…) in terms of an adjunction between a category $\mathrm{CN}$ of count nouns and a category $\mathrm{MN}$ of mass nouns:
$•$ The morphisms in these categories are relationships like “a man is a human,” resp. “water is liquid.”
$•$ The left adjoint is pluralization: if “a dog” is a count noun, then “dogs” is a mass noun.
$•$ The right adjoint is less familiar grammatically, but for example, if “water” is a mass noun, then “a body of water” is a count noun.
Like Reyes’s categories $\mathrm{CN}$ and $\mathrm{MN}$, Spivak’s “olog”s form a category whose objects are essentially “real-world types”. The morphisms are also related: the “is a” morphisms of Reyes’s categories $\mathrm{CN}$ and $\mathrm{MN}$ are important examples of Spivak’s “aspects” of ologs (although the latter are more general).
I wonder if Reyes’ analysis might shed light on Spivak’s convention of using e.g. “a man” to denote the set of all men, and his “rules of good practice” for olog notation more generally?
Even Reyes’s diagrammatic notation is very similar to Spivak’s: the objects are denoted by English words in text boxes, and the morphisms are arrows between these.
Posted by: Tim Campion on March 5, 2013 6:49 AM | Permalink | Reply to this
### Re: Spivak on Category Theory
This is very nice! I’ve added it in the “Textbooks” section of the nlab page on category theory.
One minor quibble: I wouldn’t say that the alternative definition of a category in 4.1.1.17 is any “more formal” than the preceeding one. It’s just different; both are equally formal.
Posted by: Mike Shulman on March 5, 2013 2:47 PM | Permalink | Reply to this
### Re: Spivak on Category Theory
By the way, there’s a google doc up on the web if anyone here finds typos or has other comments or suggestions regarding the book. For example, Mike Shulman’s comment above has been implemented in the latest version. Obviously I don’t promise to implement every suggestion, but I do promise to think about it.
Thanks!
Posted by: David Spivak on April 24, 2013 2:26 PM | Permalink | Reply to this
Post a New Comment
|
2013-05-22 05:07:44
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 12, "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": 0.61128169298172, "perplexity": 2190.7976705874685}, "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/1368701370254/warc/CC-MAIN-20130516104930-00042-ip-10-60-113-184.ec2.internal.warc.gz"}
|
https://oldschool.runescape.wiki/w/Soul_Wars_Reward_Shop
|
# Soul Wars Reward Shop
Jump to: navigation, search
The Soul Wars Reward Shop is the Soul Wars reward store owned by Nomad located on the Isle of Souls. Players can spend their earned Zeal Tokens here for combat experience, upgrades, loot, and rewards unique to the minigame.
## Store Contents
### Experience
Players must have at least level 30 in the chosen skill to purchase XP for it, and the amount granted is higher if the level of the skill chosen is higher. Players can spend 1, 10, 100, or a custom amount of Tokens.
Zeal Tokens used Level Experience
/ / / /
1 25-34 24 22 12
1 35-42 48 44 24
1 43-48 72 66 36
1 49-54 96 88 48
1 55-59 120 110 60
1 60-69 144 132 72
1 70-73 192 176 96
1 74-77 216 198 108
1 78-81 240 220 120
1 82-84 264 242 132
1 85-88 288 264 144
1 89-91 312 286 156
1 92-94 336 308 168
1 95-97 360 330 180
1 98-99 384 352 192
The formula, which this table is based on, is as shown below. The amounts are awarded per Zeal Token traded in.
{\displaystyle {\begin{aligned}Exp&=\left\lfloor {\frac {Level^{2}}{600}}\right\rfloor \times N\end{aligned}}} where
• ${\displaystyle Exp}$ is experience gained,
• ${\displaystyle Level}$ is current level of that skill (should be at least 30),
• The factor ${\displaystyle N}$:
• If the skill is Prayer: ${\displaystyle N=12}$
• If the skill is Magic or Ranged: ${\displaystyle N=22}$
• If the skill is Attack, Strength, Defence or Hitpoints: ${\displaystyle N=24}$
• The ${\displaystyle \lfloor \quad \rfloor }$ indicate that the expression should be rounded down to the nearest whole number.
### Stock
ItemPrice
sold at
Price
bought at
GE
price
Soul cape2500N/ANot sold
Ectoplasmator250N/ANot sold
Spoils of war30N/ANot sold
Blighted bind sack10N/A 32
Blighted snare sack10N/A 127
Blighted entangle sack10N/A 533
Blighted teleport spell sack10N/A 713
Blighted vengeance sack10N/A 650
Blighted ancient ice sack10N/A 1,200
Blighted manta ray10N/A 1,774
Blighted anglerfish10N/A 1,984
Blighted karambwan10N/A 1,776
Blighted super restore(4)10N/A 9,257
### Upgrades
A player can imbue certain items with Zeal Tokens. Imbuing an item provides bonuses and appends an (i) to the item name.
Imbues from the Nightmare Zone are also available here. However, players who imbued their items at the Nightmare Zone cannot un-imbue their items for Zeal Tokens.
Imbuing an item provides extra bonuses and appends an (i) suffix to the item's name. Imbued items can be un-imbued, which removes the bonuses from the item and refunds 80% of the reward points used to imbue it.
Imbuing all available items would require a total of 3,950 tokens.
Item Imbued Bonuses Tokens required
Black mask (imbued)[c 1]
• 15% bonus to Magic accuracy and damage on slayer tasks.
• 15% bonus to Ranged accuracy and strength on slayer tasks.
• +0 Magic and Ranged attack bonuses instead of negative bonuses.
500
Slayer helmet (imbued)
• 15% bonus to Magic accuracy and damage on slayer tasks.
• 15% bonus to Ranged accuracy and strength on slayer tasks.
• +3 instead of -6 Magic attack bonus.
• +3 instead of -2 Ranged attack bonus.
• +10 instead of -1 Magic defence bonus.
500
Salve amulet (imbued)
• 15% bonus to Magic accuracy and damage against undead monsters.
• 1/6 bonus to Ranged accuracy and strength against undead monsters.
320
Salve amulet (e) (imbued)
• 20% bonus to Magic accuracy and damage against undead monsters.
• 20% bonus to Ranged accuracy and strength against undead monsters.
320
Ring of the gods (imbued) +4 Prayer bonus and gives the effect of the Holy wrench when equipped. 260[c 2]
Ring of suffering (imbued) Doubles equipment bonuses. 300
Berserker ring (imbued) Doubles equipment bonuses. 260
Warrior ring (imbued) Doubles equipment bonuses. 260
Archers ring (imbued) Doubles equipment bonuses. 260
Seers ring (imbued) Doubles equipment bonuses. 260
Tyrannical ring (imbued) Doubles equipment bonuses. 260
Treasonous ring (imbued) Doubles equipment bonuses. 260
Granite ring (imbued) Doubles equipment bonuses. 200
1. A Black mask (i) can be used to assemble a Slayer helm (i).
2. Also requires a Holy wrench in addition to the points.
## Changes
Date Changes
28 April 2021
(update)
The Soul Wars shop has been updated to include a 'buy X' option for the XP rewards. We will look to do the same to other tabs of the shop at a later date.
3 March 2021
(update)
• Most Soul Wars-specific items no longer have a Destroy confirmation box and can now be quickly shift-dropped.
• Soul Wars Clan mode now awards more Zeal, the maximum amount of Zeal awarded per game has been increased to 30.
3 February 2021
(update | poll)
• The Soul Wars guide and tutorial have been updated to reference the new waiting rooms, as well as the changes to the cooldown to leave Graveyards and Bases after death.
• Players can now purchase blighted items from the Reward Shop for 10 Zeal each, though they can’t be sold back again or be used during a game.
13 January 2021
(update)
• The inactivity timer will begin to tick down 10 seconds after a game starts instead of immediately.
• Items in the ‘Other’ section of the Reward Shop can now be examined.
• The tutorial will now properly display the chatheads of bearded players.
• The duration in which players cannot exit a Graveyard has been reduced from 15 seconds to 10. In addition, players will now receive immunity when exiting, which lasts for a few seconds or until another player is attacked.
• Attempting to attack Avatars from outside the boss room will now cause the minigame to pull the player in, instead of a message telling them to move closer.
• Bandage healing will now display the name of the player that healed them instead of their own name.
• Skulled players on Deadman worlds can no longer start the tutorial or enter a waiting area.
• Bones and soul fragments will now appear above other dropped items during a game.
• Gravestones will no longer appear in waiting areas.
• Players will no longer be able to bypass the temporary ban for leaving a game early by hopping to another world.
• A ‘Check’ option has been added to spoils of war to show how many players have opened so far.
• It is no longer possible to open Spoils of War on Entrana.
• Rebalanced the quantities of certain items in the Spoils of War.
Also, after the initial launch of Soul Wars and the addition of new worlds to handle all the players wanting to try it out, we have decided that we will be continuing with 3 Soul Wars worlds, one in each global region:
• 320 US
• 350 UK
• 535 AUS
7 January 2021
(update)
Hotfix
• The activity bar now takes 150 seconds to fully deplete rather than 120.
• Ultimate Ironmen can now use the looting bag during a game.
• The loot table for spoils of war have been adjusted, reducing quantities of runite ore and seeds. In addition, dragon items, rune items, seeds, and runite ore are now rarer.
• Items which transform the player (such as the ring of stone and ring of coins) can no longer be used during a game.
• A confirmation message will now appear when attempting to leave the lobby.
|
2021-07-23 23:07:48
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 8, "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": 0.19167742133140564, "perplexity": 10003.091044271856}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-31/segments/1627046150067.51/warc/CC-MAIN-20210723210216-20210724000216-00172.warc.gz"}
|
https://www.sangakoo.com/en/unit/homogeneous-linear-equations-of-order-n-with-constant-coefficients
|
# Homogeneous linear equations of order n with constant coefficients
As happen with the linear systems of order 1, one ODE of order $$n$$ has $$n$$ linearly independent solutions in such a way that any solution to an homogeneous ODE will be a linear combination of these linearly independent solutions. Therefore, to solve the ODE will consist in finding these $$n$$ functions.
The ODE we consider is $$a_n \cdot y^{(n)} (x)+a_{n-1} \cdot y^{(n-1)}+ \ldots +a_1 \cdot y'+ a_0 \cdot y=0$$$where $$a_i$$ are constants. An example of a honogenous ODE of order $$n$$ would be:$$y''+y=0$$$
Then we define the characteristical polynomial of the ODE as: $$a_n\cdot \lambda^n+a_{n-1} \cdot \lambda^{n-1}+ \ldots + a_1 \cdot \lambda+a_0=0$$$and we look at its $$n$$ roots. The characteristical polynomial is easy to write, it is just necessary to change $$y$$ for $$\lambda$$ and to raise it to the corresponding order of the derivative. For example, in the ODE that we have given, the associated characteristic polynomial is: $$\lambda ^2+1=0$$. This polynomial has two combined complex roots: $$\lambda_1=i, \ \lambda_2=-i$$ Then • If $$\lambda$$ are real and simple the solution is of the form: $$e^{\lambda x}$$ • If $$\lambda$$ are real of multiplicity m the solutions are of the form:$$e^{\lambda x}, x\cdot e^{\lambda x}, x^2\cdot e^{\lambda x}, \ldots, x^{m-1} \cdot e^{\lambda x}$$ • If $$\lambda=a+bi$$ are complex and simple, the two solutions are of the form: $$e^{ax}\cos (bx), e^{ax}\sin (bx)$$ (there are two solutions because whenever a complex root exists the conjugate also appears) • If $$\lambda=a+bi$$ are complex of multiplicity m, the two solutions are of the form: $$e^{ax}\cos (bx),x \cdot e^{ax}\cos (bx), \ldots, x^{m-1}e^{ax}\cos (bx) \\ e^{ax}\sin (bx), x\cdot e^{ax}\sin (bx), \ldots, x^{m-1} e^{ax}\sin (bx)$$$
Then, once we find these $$n$$ solutions, the general solution of the ODE will be a linear combination of these $$n$$ solutions.
Let's come back to the example at the beginning. As our polynomial took two (simple) combined complex roots we are in case 3. Therefore the solution is: $$y(x)=c_1 \cdot \cos x+ c_2 \cdot \sin x$$\$ where the constants will be determined by the initial conditions (in the event that we would know them).
|
2023-02-08 00:23:49
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 1.0000087022781372, "perplexity": 2745.8216634489463}, "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/1674764500664.85/warc/CC-MAIN-20230207233330-20230208023330-00717.warc.gz"}
|
https://www.etutorboard.com/Curriculum/Grades
|
The EtutorBoard Curriculum
EtutorBoard Supplementary Math is a well-researched curriculum that supplements mathematical skill development of students in elementary grades through high school. EtutorBoard has a well-defined formula of supplementary mathematics that enriches the existing school curriculum.
The curriculum focuses primarily on the following areas:
• Arithmetic
• Algebra
• Logic
• Problem solving
Supplementary Math Curriculum Grades
EtutorBoard Supplementary Math Curriculum is organized in terms of Grades. Each grade focuses on specific areas of mathematics that are described below. The skills are incorporated between grades as more in-depth concepts are introduced.
The EtutorBoard levels correspond roughly to the grades as noted in this table. Consider this as a guidance, there may be some overlap between grades and levels.
US School Grade Level EtutorBoard Supplementary Math Level Preschool Level 1 or 2 Kindergarten Level 1, 2 or 3 1st Grade Level 2, 3, 4 or 5 2nd Grade Level 3, 4, 5 or 6 3rd Grade Level 5, 6 or 7 4th Grade Level 5, 6, 7 or 8 5th Grade Level 6, 7, 8 or 9 6th Grade Level 7, 8, 9, 10 or 11 7th Grade Level 8, 9, 10, 11 or 12 8th Grade Level 9, 10, 11, or 12 9th Grade Level 10, 11, or 12
During the learning process, EtutorBoard provides hints (when requested) and feedback for the student to encourage growth. A typical student will learn much faster if additional support, feedback, and timely help is provided. At appropriate times EtutorBoard will assign problems from old focus areas in order for the student to incorporate previously learned skills with the new skills. The student will constantly be encouraged to improve in accuracy and speed of problem solving. Almost all tests in the real world have a time component, and a student's ability to increase speed of computation will ultimately reflect in superior performance on tests of any kind. Attempts to improve the speed at which problems are solved will ultimately help the mind work faster with fewer diversions. EtutorBoard uses these two elements along with other necessary information to enhance each student's skills in mathematics.
EtutorBoard assigns a series of evaluation tests after the student joins EtutorBoard. Based on the results of each test EtutorBoard creates new sets of problems and test the student again. This process continues until EtutorBoard finds the right and comfortable level for the student.
|
2022-12-06 07:36:13
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 5, "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": 0.4645858407020569, "perplexity": 1175.5814821369}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-49/segments/1669446711074.68/warc/CC-MAIN-20221206060908-20221206090908-00738.warc.gz"}
|
https://groupprops.subwiki.org/wiki/Direct_factor_implies_right-quotient-transitively_central_factor
|
# Direct factor implies right-quotient-transitively central factor
This article gives the statement and possibly, proof, of an implication relation between two subgroup properties. That is, it states that every subgroup satisfying the first subgroup property (i.e., direct factor) must also satisfy the second subgroup property (i.e., right-quotient-transitively central factor)
View all subgroup property implications | View all subgroup property non-implications
Get more facts about direct factor|Get more facts about right-quotient-transitively central factor
## Statement
### Statement with symbols
Suppose $G$ is a group, $H \le K \le G$, and $H$ is a direct factor of $G$. Suppose, further, that $K/H$ is a central factor of $G/H$. Then, $K$ is a central factor of $G$.
## Proof
Given: A group $G$, a direct factor $H$ of $G$, $K$ contains $H$ and $K/H$ is a central factor of $G/H$.
To prove: $K$ is a central factor of $G$.
Proof: Since $H$ is a direct factor of $G$, there exists a normal complement $L$ to $H$ in $G$, with $HL = G$ and $H \cap L$ trivial. Let $M = K \cap L$.
Consider the map $\rho:L \to G/H$ that sends every element of $L$ to its $H$-coset in $G$. This map is an isomorphism, since the kernel $H \cap L$ is trivial, and $\rho^{-1}(K/H) = M$. Thus, $HM = K$.
1. Every element of $H$ centralizes every element of $L$: Since both $H$ and $L$ are normal in $G$, $[H,L]$ is contained in both, and since they intersect trivially, $[H,L]$ is trivial, so every element of $H$ centralizes every element of $L$.
2. $MC_L(M) = L$: Since $\rho$ is an isomorphism, and $\rho(M)$ is a central factor of $G/H$, $M$ is a central factor of $L$.
3. $C_L(M)$ centralizes $K$, i.e., $C_L(M) \le C_G(K)$: By step (1), $C_L(M)$ centralizes $H$, since $L$ centralizes $H$. It also centralizes $M$ by definition. Thus, it centralizes $HM = K$.
4. $KC_L(M) = G$: We have $KC_L(M) = HMC_L(M) = HL = G$.
5. $KC_G(K) = G$: This follows from the previous two steps.
This completes the proof.
|
2019-12-14 00:35:17
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 59, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9630458950996399, "perplexity": 228.3048975332133}, "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/1575540569332.20/warc/CC-MAIN-20191213230200-20191214014200-00176.warc.gz"}
|
http://www.planetmath.org/holomorphicfunctionassociatedwithcontinuousfunction
|
# holomorphic function associated with continuous function
If $f(z)$ is continuous on a (finite) contour $\gamma$ of the complex plane, then the contour integral
$\displaystyle g(z)\;=:\;\int_{\gamma}\frac{f(t)}{t\!-\!z}\,dt,$ (1)
defines a function$z\mapsto g(z)$ which is holomorphic in any domain $D$ not containing points of $\gamma$. Moreover, the derivative has the expression
$\displaystyle g^{\prime}(z)\;=\;\int_{\gamma}\!\frac{f(t)}{(t\!-\!z)^{2}}\,dt.$ (2)
Proof. The right hand side of (2) is defined since its integrand is continuous. On has to show that it equals
$\lim_{\Delta z\to 0}\frac{g(z\!+\!\Delta z)\!-\!g(z)}{\Delta z}.$
Let $z_{1}=:z\!+\!\Delta z\notin\gamma$, $\Delta z\neq 0$. We may write first
$\frac{g(z_{1})-g(z)}{z_{1}\!-\!z}\;=\;\frac{1}{\Delta z}\int_{\gamma}f(t)\left% [\frac{1}{t\!-\!z_{1}}-\frac{1}{t\!-\!z}\right]\,dt\;=\;\int_{\gamma}\frac{f(t% )}{(t\!-\!z_{1})(t\!-\!z)}\,dt,$
whence
$E\;=:\;\frac{g(z_{1})-g(z)}{z_{1}\!-\!z}-\int_{\gamma}\frac{f(t)}{(t\!-\!z)^{2% }}\;=\;\Delta z\cdot\!\int_{\gamma}\frac{f(t)}{(t\!-\!z_{1})(t\!-\!z)^{2}}\,dt.$
Because $f$ is continuous in the compact set $\gamma$, there is a positive constant $M$ such that
$|f(t)|\;<\;M\quad\forall\;t\in\gamma.$
As well, we have a positive constant $d$ such that
$|t\!-\!z|\;\geqq\;d\quad\forall\;t\in\gamma.$
When we choose $|\Delta z|<\frac{d}{2}$, it follows that
$|t\!-\!z_{1}|\;=\;|(t\!-\!z)-\Delta z|\;\geqq\;|t\!-\!z|-|\Delta z|\;>\;d\!-\!% \frac{d}{2}\;=\;\frac{d}{2}.$
Consequently,
$\left|\frac{f(t)}{(t\!-\!z_{1})(t\!-\!z)^{2}}\right|\;=\;\frac{|f(t)|}{|t\!-\!% z_{1}||t\!-\!z|^{2}}\;<\;\frac{M}{\frac{d}{2}\cdot d^{2}}\;=\;\frac{2M}{d^{3}}$
and, by the estimating theorem of contour integral,
$|E|\;=\;|\Delta z|\cdot\left|\int_{\gamma}\frac{f(t)}{(t\!-\!z_{1})(t\!-\!z)^{% 2}}\,dt\right|\;<\;|\Delta z|\cdot\frac{2M}{d^{3}}\cdot k,$
where $k$ is the length of the contour. The last expression tends to zero as $\Delta z\to 0$. This settles the proof.
Remark 1. By induction, one can prove the following generalisation of (2):
$\displaystyle g^{(n)}(z)\;=\;n!\!\int_{\gamma}\!\frac{f(t)}{(t\!-\!z)^{n+1}}\,% dt\qquad(n\;=\;0,\,1,\,2,\,\ldots)$ (3)
Remark 2. The contour $\gamma$ may be . If it especially is a circle, then (1) defines a holomorphic function inside $\gamma$ and another outside it.
Title holomorphic function associated with continuous function HolomorphicFunctionAssociatedWithContinuousFunction 2013-03-22 19:14:29 2013-03-22 19:14:29 pahio (2872) pahio (2872) 11 pahio (2872) Theorem msc 30E20 msc 30D20 DifferentiationUnderIntegralSign CauchyIntegralFormula
|
2018-03-18 11:52:41
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 27, "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": 0.9736374020576477, "perplexity": 901.585812879413}, "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/1521257645613.9/warc/CC-MAIN-20180318110736-20180318130736-00082.warc.gz"}
|
http://math.stackexchange.com/questions/367503/find-distance-between-two-points/367512
|
# Find Distance Between Two Points
If we are to find the distance between the points $P(0,0)$ and $Q(-2,-3)$, then we can use the Theorem of Pythagoras for this purpose.
$distance (P,Q) = \sqrt{(x_2 -x_1)^2 + (y_2 -y_1)^2}$
therefore, $distance (P,Q) = \sqrt{-5}$ But the answer is undefined. Is this answer and the reason correct?
-
You are calculating wrong:
$distance(P,Q) = \sqrt{(x_2 -x_1)^2 + (y_2 -y_1)^2}=\sqrt{((-2)^2+(-3)^2)}=\sqrt{(13)}$.
-
oh yes! Sorry!! – Samama Fahim Apr 20 '13 at 17:48
You forgot to square. $d(P,Q) = \sqrt{(-2)^2+(-3)^2} = \sqrt{4+9} = \sqrt{13}$.
-
It looks like you added the differences $(x_2 - x_1)$ and $(y_2 - y_1)$ without squaring them first!
So you want $$\operatorname {distance}(P, Q)=\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} = \sqrt{(-2)^2 + (-3)^2} = \sqrt{4 + 9} = \sqrt{13}$$
-
|
2016-04-29 18:32:32
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 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": 0.8789869546890259, "perplexity": 714.5673560175923}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-18/segments/1461860111392.88/warc/CC-MAIN-20160428161511-00083-ip-10-239-7-51.ec2.internal.warc.gz"}
|
https://math.stackexchange.com/questions/1831748/does-this-equation-have-no-solutions
|
# Does this equation have no solutions?
The question is this :
The source from where I got this question was devoid of any answers to it, so I came here, this is how I proceeded :
LHS :
$((((({(x)^x})^{2x})^{3x})^{....x^2})^2 = (((((x)^{2x^2})^{3x})^{....x^2})^2 =...........= (x^{(x^x)x!})^2 = x^{2(x^x)x!}$
RHS :
$\sqrt{x^2\sqrt{(x-1)x\sqrt{(x-2)x\sqrt{...........2x\sqrt{x\sqrt x}}}}} = \sqrt{x\sqrt{(x-1)\sqrt{(x-2)\sqrt{...........2\sqrt{1\sqrt 1}}}}}*x^{(1-\frac{1}{2^{x+1}})}$ $= x^{1/2}*(x-1)^{1/4}*(x-2)^{1/8}*....... *{2}^{(\frac{1}{2^{x-1}})}* x^{(1-\frac{1}{2^{x+1}})}$
Now I thought of taking $\log$ of both RHS and LHS from which I could deduce
LHS:
$\log x^{2(x^x)x!} = 2(x^x)x!\log x = 2*x^{x+1}(x-1)!\log x$
RHS:
$\log (x^{1/2}*(x-1)^{1/4}*(x-2)^{1/8}*....... *{2}^{(\frac{1}{2^{x-1}})}* x^{(1-\frac{1}{2^{x+1}})})$ $= \frac{\log x}{2}+\frac{\log (x-1)}{4}+ \frac{\log (x-2)}{8}+..........+ \frac{\log 2}{2^{x-1}}+ {(1-\frac{1}{2^{x+1}})} \log x$
Now equating LHS = RHS I get : $2*x^{x+1}(x-1)!\log x = \frac{\log x}{2}+\frac{\log (x-1)}{4}+ \frac{\log (x-2)}{8}+..........+ \frac{\log 2}{2^{x-1}}+ {(1-\frac{1}{2^{x+1}})} \log x$ $\implies x^{x+1}(x-1)! =\frac{ \frac{\log x}{2}+\frac{\log (x-1)}{4} + \frac{\log (x-2)}{8}+..........+\frac{\log 2}{2^{x-1}}+ {(1-\frac{1}{2^{x+1}})} \log x}{2\log x}$
now in the RHS of above equation I only found $\log x$ in 3 places : denominator, first place of numerator and in the last place of numerator; I assumed terms from $\frac {\log (x-1)}{4*2\log x}$ to $\frac{\log 2}{2^{x-1}*2\log x}$ were becoming too small to take into calculation, so final equation could be written down to :
$x^{x+1}(x-1)! = \frac{ \frac{\log x}{2} + {(1-\frac{1}{2^{x+1}})} \log x}{2\log x} =\frac{\frac{1}{2} +1- \frac{1}{ 2^{x+1} }}{2} = \frac{3}{4} - \frac{1}{ 2^{x+2} }$$\implies x^{x+1}(x-1)! = \frac{3}{4} - \frac{1}{ 2^{x+2} }$
the above equation is where I am forced to stop, please guide me after that ? Or did I take a wrong approach from start itself ?
(If you can, do tag it with appropriate tags; I could not find the suitable ones for this problem)
• I think you are misreading the question. I'm pretty sure that most of those expressions on the RHS are supposed to be indices of the radicals, i.e., $\sqrt[x^2]{\sqrt[x(x-1)]{\cdots }}$, so if $x=5$ then $\sqrt[25]{\sqrt[20]{\cdots }}$ – PM 2Ring Jun 19 '16 at 9:28
• @PM2Ring Really ?? Hmmm I didn't see that way, okay let me try that way – Arnav Das Jun 19 '16 at 9:39
• You have a typo in RHS (first line): It is not $2x\sqrt{x\sqrt x}$ but $\sqrt[2x]{\sqrt[x]{\sqrt x}}$ – Piquito Jun 19 '16 at 14:20
• @Piquito could you solve it then, i am at a loss to solve that way too – Arnav Das Jun 21 '16 at 13:20
|
2019-05-21 01:26:45
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9269126653671265, "perplexity": 643.351107887504}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-22/segments/1558232256184.17/warc/CC-MAIN-20190521002106-20190521024106-00143.warc.gz"}
|
http://math.stackexchange.com/questions/154593/ramanujan-691-congruence
|
# Ramanujan 691 congruence
I know how to prove this congruence in two ways (one using basics of modular forms and the other using Hecke operators) and I will be working on proving other such congruences soon.
The congruence states that for $n\geq 1$:
$\tau(n) \equiv \sigma_{11}(n)$ mod $691$.
My main question is why this congruence is so important? I recognise it as a beautiful thing but the only reason I can come up with for it being interesting is that it links a geometrical function with a number theoretical function.
Are there any other reasons why this is interesting/useful?
-
The statement of the congruence or a link to it wouldn't harm anyone... – Giovanni De Gaetano Jun 6 '12 at 8:54
Philosophically, you could argue that it's interesting because it shows that a cusp form (a deep sophisticated object) can be congruent mod p to an Eisenstein series (a much more explicit and tractable thing). – David Loeffler Jun 6 '12 at 9:24
Ah right, so this congruence kind of gives more precision to the fact that Eisenstein series generate the space? It tells us a more specific connection between some cusp forms and Eisenstein series? – fretty Jun 6 '12 at 9:37
When viewed "locally" some highly nontrivial modular forms may become a little bit more trivial? – fretty Jun 6 '12 at 9:39
Congruences between Hecke eigenforms are outward, "physical" manifestations of corresponding relationships between the associated two-dimensional Galois representations.
In this particular case, Ramanujan's congruence is related to the fact that $691$ is an irregular prime (in the sense of Kummer). Roughly the idea is that Eisenstein series relate to reducible two-dimensional Galois representations, and cuspforms to irreducible two-dimensional Galois representations. The existence of a congruence between the two points to the existence of an object that is somewhere between reducible and irreducible: a certain reducible two-dimensional Galois representation which is, however, indecomposable. The existence of this particular reducible, but indecomposable, two-dimensional representation shows that $691$ is an irregular prime.
To see a hint of how this could be, note that $691$ being an irregular prime means, by class field theory for $\mathbb Q(\zeta_{691})$ --- especially, the theory of the Hilbert Class Field --- that there exists an unramified abelian extension of $\mathbb Q(\zeta_{691})$; so irregularity of $691$ is related to the existence of a certain abelian extension of an abelian extension of $\mathbb Q$, and the reducible but indecomposable Galois representation will have such a thing as its splitting field.
For more information on this, and related ideas, you might like to read Mazur's article on the subject; see the entry June 17, 2010: How can we construct abelian extensions on his web-page.
-
The congruence $$\tau(n) \equiv \sigma_{11}(n) \pmod{691}$$ is one of several congruences that have been proven.
Now it is well known that $$\Delta(z) = 2 \pi^2 \sum_{n = 1}^\infty \tau(n)q^n.$$
Lehmer conjectured that $\tau(n) \neq 0$ for all positive integers $n$. As far as I know this has been checked for $n < 10^{11}$. The congruences play a role in showing part of his conjecture.
As to why is it interesting to consider why $\tau(n) \neq 0$ for all positive integers $n$, see this.
-
Ah, that makes a bit more sense now. I guess the congruence tells you that $\tau(n)$ is almost always non-zero (I've never viewed it in this way before). It only remains to check $n$ such that $\sigma_{11}(n) \equiv 0$ mod $691$ to prove Lehmer's conjecture. And as your link says, this has powerful meanings in terms of analytic continuations. – fretty Jun 6 '12 at 12:45
I would recommend you to a paper of Gandhi's about the nonvanishing of $\tau(n)$ if it weren't so full of errors. – Eugene Jun 6 '12 at 12:48
Well to be honest I have only ever seen the importance of the $\Delta$ function in helping to study other spaces of modular forms. The significance of it by itself is not obvious either! I certainly couldn't just dream up such a function and see it as important. – fretty Jun 6 '12 at 12:53
@fretty Indeed it is very useful to study spaces of modular forms. For instance, the proof that all even weight modular forms of level $1$ is generated by Eisenstein series depends on the $\Delta$ modular form. Also you can show that the even weight modular forms on $\Gamma_0(N)$ can be generated by Dedekind $\eta$-functions using $\Delta$. – Eugene Jun 6 '12 at 13:04
Yes I know this, but is that basically the reason for inventing this function? Is there any reason why the $\Delta$ function is interesting by itself? – fretty Jun 6 '12 at 13:05
This is an extract to Manin's paper "Periods of parabolic forms and p-adic Hecke Series":
"This congruence is so far, our only clue to understanding the 11-dimensional étale cohomology of the so-called Sato variety"
Even though I don't know what he means, he gives some references to Serre's articles.
-
You can also take a look at this link math.stackexchange.com/questions/11720/… – student Jun 18 '13 at 17:53
|
2016-04-30 07:35:28
|
{"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": 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": 0.7826768755912781, "perplexity": 361.44761080041076}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-18/segments/1461860111620.85/warc/CC-MAIN-20160428161511-00150-ip-10-239-7-51.ec2.internal.warc.gz"}
|
https://bitbucket.org/mitchellrj/sqlalchemy/src/3f4f05942eb7?at=rel_0_7b4
|
sqlalchemy /
Filename Size Date modified Message
doc
examples
lib
test
108 B
4.1 KB
500 B
148.7 KB
278.9 KB
1.2 KB
489 B
1.5 KB
1.4 KB
8.9 KB
15.4 KB
10.0 KB
1.9 KB
123 B
11.4 KB
410 B
SQLAlchemy
The Python SQL Toolkit and Object Relational Mapper
Requirements
SQLAlchemy requires Python 2.4 or higher. One or more DB-API implementations are also required for database access. See docs/intro.html for more information on supported DB-API implementations.
Installation Tools
Installation is supported with standard Python distutils, as well as with setuptools or Distribute. Distribute is recommended. Distribute can be installed using the provided "distribute_setup.py" script. The original setuptools may be installed using the "ez_setup.py" script if preferred, or simply do nothing and distutils will be used.
Installing
To install:
python setup.py install
To use without installation, include the lib directory in your Python path.
Package Contents
doc/
HTML documentation, including tutorials and API reference. Point a browser to the "index.html" to start.
examples/
Fully commented and executable implementations for a variety of tasks.
lib/
SQLAlchemy.
test/
Unit tests for SQLAlchemy.
Help
Mailing lists, wiki, and more are available on-line at http://www.sqlalchemy.org.
Tip: Filter by directory path e.g. /media app.js to search for public/media/app.js.
Tip: Use camelCasing e.g. ProjME to search for ProjectModifiedEvent.java.
Tip: Filter by extension type e.g. /repo .js to search for all .js files in the /repo directory.
Tip: Separate your search with spaces e.g. /ssh pom.xml to search for src/ssh/pom.xml.
Tip: Use ↑ and ↓ arrow keys to navigate and return to view the file.
Tip: You can also navigate files with Ctrl+j (next) and Ctrl+k (previous) and view the file with Ctrl+o.
Tip: You can also navigate files with Alt+j (next) and Alt+k (previous) and view the file with Alt+o.
|
2014-07-13 06:11:12
|
{"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": 0.23723441362380981, "perplexity": 10972.527505358765}, "config": {"markdown_headings": false, "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-2014-23/segments/1404776437232.85/warc/CC-MAIN-20140707234037-00082-ip-10-180-212-248.ec2.internal.warc.gz"}
|
https://math.stackexchange.com/questions/579894/ruin-time-for-a-two-input-risk-only-slot-machine
|
# Ruin time for a two-input “risk only” slot machine
Imagine a "risk only" slot machine that takes 'coins' corresponding to some real number fraction of a dollar $p$, returns the coin with probability $p$, and eats the coin with probability $(1-p)$. For example, a dime would be eaten with a probability of 90%, a nickel with probability 95%, and so forth.
So let's keep feeding the machine two kinds of coins, $A$ and $B$, with fractional dollar values of $p_A$ and $p_B$, respectively. I have $n_A$ coins of type $A$ and $n_B$ coins of type $B$. Each time I use the slot machine, I randomly select a coin, ignoring its type, and place it in the machine. I stop feeding coins into the machine when I run out of either type.
CLARIFICATION - By "randomly select a coin" I mean that we select a coin from the population of all coins uniformly and randomly. For instance, if we have $100$ dimes and $567$ nickels, we'd draw a dime with probability $\frac{100}{667}$.
At this stopping point, what is the probability of ending with only coins of type A or only coins of Type B? Provided we end with coins of one type / denomination, what probability distribution and expectation do we have for the number of remaining coins of this type / denomination?
I'd also be curious on the number of coins of either type we needed to feed to the machine to reach this end-state? E.g. how many times did we feed the machine a dime, and how many times did we feed the machine a nickel before stopping?
**
If it helps, I can provide some simulation data. For example, starting with $100$ dimes and $100$ nickels:
$n_A = 100$
$n_B = 100$
$p_A = 0.10$
$p_B = 0.05$
We achieve the following results for $10^4$ trials:
The mean number of times we place a dime in the machine $= 109.721$ (Median $= 110$)
The mean number of times we place a nickel in the machine $= 104.42$ (Median $= 104$)
The number of times we end with only dimes: $5669$
The number of times we end with only nickels: $4331$
The average number of dimes at the end state (conditioned on running out of nickels first): $2.18328$ (Median $= 2$)
The average number of nickels at the end state (conditioned on running out of dimes first): $1.80513$ (Median $= 1$)
**
Let's do another simulation starting with $82$ copies of hypothetical 75 cent coins and $432$ copies of 5 cent nickels, and again perform $10^4$ trials:
$n_A = 82$
$n_B = 432$
$p_A = 0.75$
$p_B = 0.05$
We achieve the following results for $10^4$ trials:
The mean number of times we place a 75 cent coin in the machine $= 268.213$ (Median $= 267$)
The mean number of times we place a 5 cent nickel in the machine $= 454.734$ (Median $= 455$)
The number of times we end with only 75 cent pieces: $9999$
The number of times we end with only 5 cent nickels: $1$
The average number of 75 cent coins at the end state (conditioned on running out of 5 cent nickels first): $14.9384$ (Median $= 15$)
The average number of nickels at the end state (conditioned on running out of dimes first): $1$ (Median $= 1$)
• Hi regarding the coin selection process. Is it random in the way that I "uniformly" draw from the bag of coins left, or do I choose a coin A or B using a bernoulli trial of probability 1/2 ? – TheBridge Nov 26 '13 at 9:30
• @TheBridge It's random in the sense that we uniformly draw from the bag of coins without considering the type of coin. For example, if we have 100 dimes and 567 nickels, we'd draw a dime with probability $\frac{100}{667}$. – Harrison Nov 26 '13 at 9:53
• Ok that makes the problem more complex and more interesting ;-) – TheBridge Nov 26 '13 at 11:49
• @TheBridge Thanks! I'm of course glad to hear that this is an interesting problem. – Harrison Nov 26 '13 at 15:14
• Sounds like a Markov process... the setup is probably a mess. – vonbrand Feb 6 '14 at 0:16
The setting is ripe for Poissonization... First consider the path followed by the system, that is, denoting by $X$ the number of coins of type A left and by $Y$ the number of coins of type B left, the set of states visited by the process $(X,Y)$.
Assume that $(X,Y)=(x,y)$. One draws a coin of type A with probability $x/(x+y)$ and the machine eats it with probability $$a=1-p_A,$$ hence $(X,Y)$ moves to $(x-1,y)$ in one step with probability $ax/(x+y)$. Likewise, $(X,Y)$ moves to $(x,y-1)$ in one step with probability $by/(x+y)$, where $$b=1-p_B,$$ otherwise $(X,Y)$ stays at $(x,y)$ in one step. Thus, the next state visited by $(X,Y)$ is $(x-1,y)$ or $(x,y-1)$ with probabilities proportional to $(ax)$ and $(by)$ respectively.
Here is a process in continuous time with the same paths. Assume that each coin $i$ of type A has a lifetime $T_i$, exponential with parameter $a$, and that each coin $j$ of type B has a lifetime $S_j$, exponential with parameter $b$. All the lifetimes are independent. At any given time, assuming that $(x,y)$ coins are still alive, the next coin to die is of type A with probability proportional to $(ax)$ and of type B with probability proportional to $(by)$.
Thus, both processes starting from $(x,y)$ hit the line $Y=0$ (no coin of type B left) before hitting the line $X=0$ (no coin of type A left) when $T\gt S$, with $$T=\max\{T_i\mid1\leqslant i\leqslant x\},\qquad S=\max\{S_j\mid1\leqslant j\leqslant y\}.$$ To compute $P(T\gt S)$, note that $P(S_j\lt t)=1-\mathrm e^{-bt}$ for every $t\gt0$ and every $j$, hence $$P(S\lt t)=(1-\mathrm e^{-bt})^y.$$ Likewise, $$P(T\lt t)=(1-\mathrm e^{-at})^x.$$ This allows to compute the density of $T$, which yields $$P(T\gt S)=\int_0^\infty xa\mathrm e^{-at}(1-\mathrm e^{-at})^{x-1}(1-\mathrm e^{-bt})^y\mathrm dt.$$ Equivalent formulas are $$P_{x,y}(\text{last coin of type A})=\int_0^1 x(1-u)^{x-1}(1-u^{b/a})^y\mathrm du,$$ and $$P_{x,y}(\text{last coin of type A})=\int_0^1(1-(1-u^{1/x})^{b/a})^y\mathrm du.$$ The first example in the question is when $x=y=100$, $a=0.9$, $b=0.95$, then $P(\text{last coin of type A})\approx0.565953$.
The second example in the question is when $x=82$, $y=432$, $a=0.25$, $b=0.95$, then $P(\text{last coin of type A})\approx0.999634$.
One can continue with this approach to get the distribution of the number of coins at the end and of the number of times one places a coin in the machine. A useful remark to do so is that the continuous time processes $(X_t)$ and $(Y_t)$ describing the numbers of coins of each type still alive at time $t$ are actually independent, $X_t$ jumping from state $x$ to state $x-1$ at rate $(ax)$ and $Y_t$ jumping from state $y$ to state $y-1$ at rate $(by)$.
|
2019-09-16 09:01:30
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.8007774353027344, "perplexity": 261.3790193868048}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-39/segments/1568514572516.46/warc/CC-MAIN-20190916080044-20190916102044-00356.warc.gz"}
|
https://paulleemagic.com/site/ex47v4.php?99491a=how-to-calculate-confidence-interval-of-a-proportion-in-r
|
## how to calculate confidence interval of a proportion in r
Most of the time, you’ll probably write your own code for calculating confidence intervals for proportions since you’ll typically have just two values, a sample size ($$n$$) and sample proportion ($$\hat{p}$$). To subscribe to this RSS feed, copy and paste this URL into your RSS reader. :). Well, for one, it’s not possible to split 510 into exactly 58% and 42% so our original calculations introduced some rounding error. The tidy function from the broom package can also calculate confidence intervals. Thanks for contributing an answer to Stack Overflow! That's not how a CI works, the CI is on the mean, not on individual observations. Pearson Education Canada. This fact is not too important; it just means that the behaviour of confint can change depending on the fitted model. https://en.wikipedia.org/wiki/Binomial_proportion_confidence_interval. What does commonwealth mean in US English? Making statements based on opinion; back them up with references or personal experience. What LEGO piece is this arc with ball joint? Let’s try to reproduce what confint and lm did. Now for each of the value generated, I am supposed to calculate a 95% confidence interval for the proportion of faulty screws in each day. Can you have a Clarketech artifact that you can replicate but cannot comprehend? You may have also noticed that the values are close to what we previously calculated “by hand” but not the same. What kind of overshoes can I use with a large touring SPD cycling shoe such as the Giro Rumble VR? Ninety-five percent of the standard normal distribution lies between the critical values -1.96 to 1.96. Since I fitted an lm model, R invokes the appropriate version of confint that’s available for lm objects, namely confint.lm. The following are the data: The following are the data: it’s not obvious whether the calculations are based on the normal or t-distribution but it’s the latter. You may have noticed that I used 0.581 instead of 0.58. rev 2020.11.24.38066, Stack Overflow works best with JavaScript enabled, Where developers & technologists share private knowledge with coworkers, Programming & related technical career opportunities, Recruit tech talent & build your employer brand, Reach developers & technologists worldwide. From the Gallup poll, we have: $$n=510$$ and $$\hat{p}=0.58$$. I used confint to calculate the confidence intervals. They want to determine the difference of proportions of students having experience in each class, and calculate a confidence interval for that difference. @user2974951 he has multiple observations, namely 150 each day. This is a well-known approximation but I will use a more precise value in my calculations in order to compare them with results from some R functions that calculate CIs. The first parameter to confint is a fitted model object. How to calculate 95% confidence interval for a proportion in R? Confidence Intervals for Proportions, M.I.A. So, in order to fit an lm model, I created a vector with 510 entries, 58% of them being ones, the rest zeros. There’s more than one reason for this. What would result from not adding fat to pastry dough. Is there any built in functions for this (I am not supposed to use any packages) or should I create a new function? What if the P-Value is less than 0.05, but the test statistic is also less than the critical value? Store it. Let’s consider a Gallup poll from October 2010 in which U.S. adults were asked “Generally speaking, do you believe the death penalty is applied fairly or unfairly in this country today?” The sample size is not listed on the website but according to STATS: Data and Models (Voeux, 2019), it was 510. Why does Slowswift find this remark ironic? This blog post was originally intended to define confidence intervals and their nuances, discuss different types of confidence intervals, as well as bootstrapping confidence intervals for non-normally distributed data.
IMPORTANT! To be able to proceed, you need to solve the following simple math (so we know that you are a human) :-)
What is 4 + 14 ?
|
2022-09-24 18:37:45
|
{"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": 0.42915794253349304, "perplexity": 918.5672001082443}, "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/1664030333455.97/warc/CC-MAIN-20220924182740-20220924212740-00368.warc.gz"}
|
http://spot.pcc.edu/math/orcca/review-solving-quadratic-equations.html
|
## Section8.7Solving Quadratic Equations Chapter Review
### Subsection8.7.1Solving Quadratic Equations by Factoring
In Section 8.1 we covered the zero product property and learned an algorithm for solving quadratic equations by factoring.
###### Example8.7.1Solving Using Factoring
Solve the quadratic equations using factoring.
1. $x^2-2x-15=0$
2. $4x^2-40x=-96$
3. $6x^2+x-12=0$
4. $(x-3)(x+2)=14$
5. $x^3-64x=0$
Explanation
1. Use factor pairs.
\begin{align*} x^2-2x-15\amp=0\\ (x-5)(x+3)\amp=0 \end{align*}
\begin{align*} x-5\amp=0 \amp\text{ or }\amp\amp x+3\amp=0\\ x\amp=5 \amp\text{ or }\amp\amp x\amp=-3 \end{align*}
So the solution set is $\{5,-3\}\text{.}$
2. Start by putting the equation in standard form and factoring out the greatest common factor.
\begin{align*} 4x^2-40x\amp=-96\\ 4x^2-40x+96\amp=0\\ 4\left(x^2-10x+24\right)\amp=0\\ 4(x-6)(x-4)\amp=0 \end{align*}
\begin{align*} x-6\amp=0 \amp\text{ or }\amp\amp x-4\amp=0\\ x\amp=6 \amp\text{ or }\amp\amp x\amp=4 \end{align*}
So the solution set is $\{4,6\}\text{.}$
3. Use the AC method.
\begin{align*} 6x^2+x-12\amp=0\\ \end{align*}
Note that $a\cdot c=-72$ and that $\highlight{9\cdot-8}=-72$ and $\highlight{9-8}=1$
\begin{align*} 6x^2\substitute{+9x-8x}-12\amp=0\\ \left(6x^2+9x\right)+\left(-8x-12\right)\amp=0\\ \highlight{3x}\left(2x+3\right)\mathbin{\highlight{-4}}\left(2x+3\right)\amp=0\\ \left(2x+3\right)\highlight{\left(3x-4\right)}\amp=0 \end{align*}
\begin{align*} 2x+3\amp=0 \amp\text{ or }\amp\amp \highlight{3x-4}\amp=0\\ x\amp=-\frac{3}{2} \amp\text{ or }\amp\amp x\amp=\highlight{\frac{4}{3}} \end{align*}
So the solution set is $\left\{-\frac{3}{2},\frac{4}{3}\right\}\text{.}$
4. Start by putting the equation in standard form.
\begin{align*} (x-3)(x+2)\amp=14\\ x^2-x-6\amp=14\\ x^2-x-20\amp=0\\ (x-5)(x+4)\amp=0 \end{align*}
\begin{align*} x-5\amp=0 \amp\text{ or }\amp\amp x+4\amp=0\\ x\amp=5 \amp\text{ or }\amp\amp x\amp=-4 \end{align*}
So the solution set is $\{5,-4\}\text{.}$
5. Even though this equation has a power higher than $2\text{,}$ we can still find all of its solutions by following the algorithm. Start by factoring out the greatest common factor.
\begin{align*} x^3-64x\amp=0\\ x\left(x^2-64\right)\amp=0\\ x(x-8)(x+8)\amp=0 \end{align*}
\begin{align*} x\amp=0 \amp\text{ or }\amp\amp x-8\amp=0 \amp\text{ or }\amp\amp x+8\amp=0\\ x\amp=0 \amp\text{ or }\amp\amp x\amp=8 \amp\text{ or }\amp\amp x\amp=-8 \end{align*}
So the solution set is $\{0,8,-8\}\text{.}$
### Subsection8.7.2Square Root Properties
In Section 8.2 we covered the definition of a square root, how to estimate and simplify square roots, multiplication and division properties of square roots, and rationalizing the denominator.
###### Example8.7.2Estimating Square Roots
Estimate the value of $\sqrt{28}$ without a calculator.
Explanation
To estimate $\sqrt{28}\text{,}$ we can find the nearest perfect squares that are whole numbers on either side of $28\text{.}$ Recall that the perfect squares are $1, 4, 9, 16, 25, 36, 49, 64,\dots$ The perfect square that is just below $28$ is $25$ and the perfect square just above $28$ is $36\text{.}$ This tells us that $\sqrt{28}$ is between $\sqrt{25}$ and $\sqrt{36}\text{,}$ or between $5$ and $6\text{.}$ We can also say that $\sqrt{28}$ is closer to $5$ than $6$ because $28$ is closer to $25\text{,}$ so we think $5.2$ or $5.3$ would be a good estimate.
On the calculator we can see that $\sqrt{28}\approx5.29\text{,}$ so our guess was very close to reality.
###### Example8.7.3Multiplication and Division Properties of Square Roots
Simplify the expressions using the multiplication and division properties of square roots.
1. $\sqrt{18}\cdot\sqrt{2}\text{.}$
2. $\frac{\sqrt{18}}{\sqrt{2}}\text{.}$
Explanation
1. \begin{aligned}[t] \sqrt{18}\cdot\sqrt{2}\amp=\sqrt{18\cdot2}\\ \amp=\sqrt{36}\\ \amp=6\end{aligned}
2. \begin{aligned}[t] \frac{\sqrt{18}}{\sqrt{2}}\amp=\sqrt{\frac{18}{2}}\\ \amp=\sqrt{9}\\ \amp=3\end{aligned}
###### Example8.7.4Simplifying Square Roots
Simplify the expression $\sqrt{54}\text{.}$
Explanation
Recall that the perfect squares are $1, 4, 9, 16, 25, 36, 49, 64,\dots$ To simplify the $\sqrt{54}\text{,}$ we need to look at that list and find the largest perfect square the goes into $54$ evenly. In this case, it is $\highlight{9}\text{.}$ We then break up $54$ into two factors $\highlight{9}$ and $6\text{,}$ and we have:
\begin{align*} \sqrt{54}\amp=\sqrt{\highlight{9}\cdot6}\\ \amp=\sqrt{\highlight{9}}\cdot\sqrt{6}\\ \amp=\highlight{3}\sqrt{6} \end{align*}
Since $6$ has no perfect square factors, we can stop.
###### Example8.7.5Multiplying Square Root Expressions
Simplify the expression $\sqrt{50}\cdot\sqrt{27}\text{.}$
Explanation
Note that $25$ is a perfect-square factor of $50\text{,}$ and that $9$ is a perfect-square factor of $27\text{.}$ Now we have:
\begin{align*} \sqrt{50}\cdot\sqrt{27}\amp=\sqrt{\highlight{25}\cdot2}\cdot\sqrt{\lighthigh{9}\cdot3}\\ \amp=\sqrt{\highlight{25}}\cdot\sqrt{2}\cdot\sqrt{\lighthigh{9}}\cdot\sqrt{3}\\ \amp=\highlight{5}\cdot\sqrt{2}\cdot\lighthigh{3}\cdot\sqrt{3}\\ \amp=15\sqrt{6} \end{align*}
###### Example8.7.6Adding and Subtracting Square Root Expressions
Simplify the expression $\sqrt{32}+\sqrt{50}\text{.}$
Explanation
Recall that radicals can only be added if the radicands match identically, so we cannot initially combine these two terms. However, if we simplify first, we may be able to add terms later. Note that $16$ is a perfect-square factor of $32\text{,}$ and that $25$ is a perfect-square factor of $50\text{.}$
\begin{align*} \sqrt{32}+\sqrt{50}\amp=\sqrt{\highlight{16}\cdot2}+\sqrt{\lighthigh{25}\cdot2}\\ \amp=\sqrt{\highlight{16}}\cdot\sqrt{2}+\sqrt{\lighthigh{25}}\cdot\sqrt{2}\\ \amp=\highlight{4}\sqrt{2}+\lighthigh{5}\sqrt{2}\\ \amp=\highlight{9}\sqrt{2} \end{align*}
###### Example8.7.7Rationalizing the Denominator
Rationalize the denominator in the expression $\frac{2}{\sqrt{6}}\text{.}$
Explanation
\begin{align*} \frac{2}{\sqrt{6}}\amp=\frac{2}{\sqrt{6}}\multiplyright{\frac{\sqrt{6}}{\sqrt{6}}}\\ \amp=\frac{2\multiplyright{\sqrt{6}}}{\sqrt{6}\multiplyright{\sqrt{6}}}\\ \amp=\frac{2\sqrt{6}}{6}\\ \amp=\frac{\sqrt{6}}{3} \end{align*}
###### Example8.7.8More Complicated Square Roots
Expand $\left(\sqrt{5}+\sqrt{3}\right)^2\text{.}$
Explanation
We will use the FOIL method to expand this expression:
\begin{align*} \left(\sqrt{5}+\sqrt{3}\right)^2\amp=\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}+\sqrt{3}\right)\\ \amp=\left(\sqrt{5}\right)^2+\sqrt{5}\sqrt{3}+\sqrt{3}\sqrt{5}+\left(\sqrt{3}\right)^2\\ \amp=5+\sqrt{15}+\sqrt{15}+3\\ \amp=8+2\sqrt{15} \end{align*}
### Subsection8.7.3Solving Quadratic Equations by Using a Square Root
In Section 8.3 we covered how to solve quadratic equations using the square root property and how to use the Pythagorean Theorem.
###### Example8.7.9Solving Quadratic Equations Using the Square Root Property
Solve for $w$ in $3(2-w)^2-24=0\text{.}$
Explanation
It's important here to suppress any urge you may have to expand the squared binomial. We begin by isolating the squared expression.
\begin{align*} 3(2-w)^2-24\amp=0\\ 3(2-w)^2\amp=24\\ (2-w)^2\amp=8 \end{align*}
Now that we have the squared expression isolated, we can use the square root property.
\begin{align*} 2-w\amp=-\sqrt{8}\amp\text{or}\amp\amp2-w\amp=\sqrt{8}\\ 2-w\amp=-\sqrt{\highlight{4}\cdot2}\amp\text{or}\amp\amp2-w\amp=\sqrt{\highlight{4}\cdot2}\\ 2-w\amp=-\sqrt{\highlight{4}}\cdot\sqrt{2}\amp\text{or}\amp\amp2-w\amp=\sqrt{\highlight{4}}\cdot\sqrt{2}\\ 2-w\amp=-\highlight{2}\sqrt{2}\amp\text{or}\amp\amp2-w\amp=\highlight{2}\sqrt{2}\\ -w\amp=-2\sqrt{2}-2\amp\text{or}\amp\amp-w\amp=2\sqrt{2}-2\\ w\amp=2\sqrt{2}+2\amp\text{or}\amp\amp w\amp=-2\sqrt{2}+2 \end{align*}
The solution set is $\left\{2\sqrt{2}+2,-2\sqrt{2}+2\right\}\text{.}$
###### Example8.7.10The Pythagorean Theorem
Faven was doing some wood working in her garage. She needed to cut a triangular piece of wood for her project that had a hypotenuse of $16$ inches, and the sides of the triangle should be equal in length. How long should she make her sides?
Explanation
Let's start by representing the length of the triangle, measured in inches, by the letter $x\text{.}$ That would also make the other side $x$ inches long.
Faven should now set up the Pythagorean theorem regarding the picture. That would be
\begin{equation*} x^2+x^2=16^2 \end{equation*}
Solving this equation, we have:
\begin{align*} x^2+x^2\amp=16^2\\ x^2+x^2\amp=256\\ 2x^2\amp=256\\ x^2\amp=128\\ \highlight{\sqrt{\unhighlight{x^2}}}\amp=\highlight{\sqrt{\unhighlight{128}}}\\ x\amp=\sqrt{\highlight{64}\cdot2}\\ x\amp=\sqrt{\highlight{64}}\cdot\sqrt{2}\\ x\amp=\highlight{8}\sqrt{2}\\ x\amp\approx11.3 \end{align*}
Faven should make the sides of her triangle about $11.3$ inches long to force the hypotenuse to be $16$ inches long.
In Section 8.4 we covered how to use the quadratic formula to solve any quadratic equation, as well as an algorithm to help solve linear and quadratic equations.
Solve the equations using the quadratic formula.
1. $x^2+4x=6$
2. $5x^2-2x+1=0$
Explanation
1. First we should change the equation into standard form.
\begin{align*} x^2+4x\amp=6\\ x^2+4x-6\amp=0 \end{align*}
Next, we check and see that we cannot factor the left side or use the square root property so we must use the quadratic formula. We identify that $\substitute{a=1}\text{,}$ $\substitute{b=4}\text{,}$ and $\substitute{c=-6}\text{.}$ We will substitute them into the quadratic formula:
\begin{align*} x\amp=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\ x\amp=\frac{-\substitute{4}\pm\sqrt{(\substitute{4})^2-4(\substitute{1})(\substitute{-6})}}{2(\substitute{1})}\\ \amp=\frac{-4\pm\sqrt{16+24}}{2}\\ \amp=\frac{-4\pm\sqrt{40}}{2}\\ \amp=\frac{-4\pm\sqrt{\highlight{4}\cdot10}}{2}\\ \amp=\frac{-4\pm\sqrt{\highlight{4}}\cdot\sqrt{10}}{2}\\ \amp=\frac{-4\pm\highlight{2}\sqrt{10}}{2}\\ \amp=-\frac{4}{2}\pm\frac{2\sqrt{10}}{2}\\ \amp=-2\pm\sqrt{10} \end{align*}
So the solution set is $\left\{-2+\sqrt{10},-2-\sqrt{10}\right\}\text{.}$
2. Since the equation $5x^2-2x+1=0$ is already in standard form, we check and see that we cannot factor the left side or use the square root property so we must use the quadratic formula. We identify that $\substitute{a=5}\text{,}$ $\substitute{b=-2}\text{,}$ and $\substitute{c=1}\text{.}$ We will substitute them into the quadratic formula:
\begin{align*} x\amp=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\ x\amp=\frac{-\substitute{(-2)}\pm\sqrt{\substitute{(-2)}^2-4(\substitute{5})(\substitute{1})}}{2(\substitute{5})}\\ \amp=\frac{2\pm\sqrt{4-20}}{10}\\ \amp=\frac{2\pm\sqrt{-16}}{10} \end{align*}
Since the solutions have square roots of negative numbers, we must conclude that there are no real solutions.
###### Example8.7.13Recognizing Linear and Quadratic Equations
Identify which equations are linear, which are quadratic, and which are neither.
1. $2(x-3)^2-5x=6$
2. $2(x-3)-5x=6$
3. $2x-6=7x^3$
4. $2x^2-6=7x^2$
5. $2\sqrt{x}-x-6=0$
6. $2x-(x-6)=0$
Explanation
1. $2(x-3)^2-5x=6$ is quadratic.
2. $2(x-3)-5x=6$ is linear.
3. $2x-6=7x^3$ is neither linear or quadratic.
4. $2x^2-6=7x^2$ is quadratic.
5. $2\sqrt{x}-x-6=0$ is neither linear or quadratic.
6. $2x-(x-6)=0$ is linear.
###### Example8.7.14Solving Linear and Quadratic Equations
Use 8.4.11 to help solve the equations after deciding if they are linear or quadratic.
1. $4x^2-11x+6=0$
2. $2(x-6)^2-16=0$
3. $2(x-6)-16=0$
4. $3(x-4)^2-15x=0$
Explanation
1. To solve the equation $4x^2-11x+6=0$ we first note that it is quadratic. Since there is a linear term ($-11x$), we must use either factoring or the quadratic formula, and we will try factoring first. Since the leading coefficient is $4\text{,}$ we will try the AC method. In this case, $ac=24\text{:}$ numbers that multiply to be $24$ and add to be $-11$ are $-8$ and $-3\text{.}$ So we split up th equation like this:
\begin{align*} 4x^2-11x+6\amp=0\\ 4x^2\mathbin{\highlight{-}}\mathbin{\highlight{8x-3x}}+6\amp=0\\ \left(4x^2\mathbin{\highlight{-}}\mathbin{\highlight{8x}}\right)+(\mathbin{\highlight{-}}\mathbin{\highlight{3x}}+6)\amp=0\\ 4x(x-2)-3(x-2)\amp=0\\ (x-2)(4x-3)\amp=0 \end{align*}
\begin{align*} x-2\amp=0 \amp \text{ or } \amp\amp 4x-3\amp=0\\ x\amp=2 \amp \text{ or } \amp\amp x\amp=\frac{3}{4} \end{align*}
So, the solution set is $\left\{2,\frac{3}{4}\right\}\text{.}$
2. To solve the equation $2(x-6)^2-16=0$ we first note that it is quadratic. Since there is no linear term, we should try using the square root method.
\begin{align*} 2(x-6)^2-16\amp=0\\ 2(x-6)^2\amp=16\\ (x-6)^2\amp=8 \end{align*}
\begin{align*} x-6\amp=\sqrt{8} \amp \text{ or } \amp\amp x-6\amp=-\sqrt{8}\\ x-6\amp=\sqrt{\highlight{4}\cdot2} \amp \text{ or } \amp\amp x-6\amp=-\sqrt{\highlight{4}\cdot2}\\ x-6\amp=\sqrt{\highlight{4}}\cdot\sqrt{2} \amp \text{ or } \amp\amp x-6\amp=-\sqrt{\highlight{4}}\cdot\sqrt{2}\\ x-6\amp=\highlight{2}\sqrt{2} \amp \text{ or } \amp\amp x-6\amp=-\highlight{2}\sqrt{2}\\ x\amp=6+\highlight{2}\sqrt{2} \amp \text{ or } \amp\amp x\amp=6-\highlight{2}\sqrt{2} \end{align*}
So, the solution set is $\left\{6+2\sqrt{2},6-2\sqrt{2}\right\}\text{.}$
3. To solve the equation $2(x-6)-16=0$ we first we first note that it is linear. Since it is linear, we just need to isolate the terms with the variable on one side and all the other terms on the other side of the equals sign.
\begin{gather*} 2(x-6)-16=0\\ 2x-12-16=0\\ 2x-28=0\\ 2x=28\\ x=14 \end{gather*}
So, the solution set is $\{14\}\text{.}$
4. To solve the equation $2(x-6)^2-15x=0$ we first note that it is quadratic. Since there is a linear term, we must use either factoring or the quadratic formula. Before we can decide which to use, we need to put the equation in standard form:
\begin{align*} 3(x-4)^2-15x=0\\ 3(x-4)(x-4)-15x=0\\ 3(x^2-8x+16)-15x=0\\ 3x^2-24x+48-15x=0\\ 3x^2-39x+48=0\\ \end{align*}
Now we can see that the left hand side does not factor easily, so we will fall back on the quadratic formula. We identify that $\substitute{a=3}\text{,}$ $\substitute{b=-39}\text{,}$ and $\substitute{c=48}\text{.}$
\begin{align*} x\amp=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\ x\amp=\frac{-\substitute{(-39)}\pm\sqrt{\substitute{(-39)}^2-4(\substitute{3})(\substitute{48})}}{2(\substitute{3})}\\ \amp=\frac{39\pm\sqrt{1521-576}}{6}\\ \amp=\frac{39\pm\sqrt{945}}{6}\\ \amp=\frac{39\pm\sqrt{\highlight{9}\cdot105}}{6}\\ \amp=\frac{39\pm\sqrt{\highlight{9}}\cdot\sqrt{105}}{6}\\ \amp=\frac{39\pm3\sqrt{105}}{6} \end{align*}
So the solution set is $\left\{\frac{39+3\sqrt{105}}{6},\frac{39-3\sqrt{105}}{6}\right\}\text{.}$
### Subsection8.7.5Complex Solutions to Quadratic Equations
In Section 8.5 we covered what both imaginary numbers and complex numbers are, as well as how to solve quadratic equations where the solutions are imaginary numbers or complex numbers.
###### Example8.7.15Imaginary Numbers
Simplify the expression $\sqrt{-12}$ using the imaginary number, $i\text{.}$
Explanation
Start by splitting the $-1$ from the $12$ and by looking for the largest perfect-square factor of $-12\text{,}$ which happens to be $4\text{.}$
\begin{align*} \sqrt{-12}\amp=\sqrt{\highlight{4}\cdot\lighthigh{-1}\cdot3}\\ \amp=\sqrt{\highlight{4}}\cdot\sqrt{\lighthigh{-1}}\cdot\sqrt{3}\\ \amp=\highlight{2}\lighthigh{i}\sqrt{3} \end{align*}
###### Example8.7.16Solving Quadratic Equations with Imaginary Solutions
Solve for $m$ in $2m^2+16=0\text{,}$ where $p$ is an imaginary number.
Explanation
There is no $m$ term so we will use the square root method.
\begin{align*} 2m^2+16\amp=0\\ 2m^2\amp=-16\\ m^2\amp=-8 \end{align*}
\begin{align*} m\amp=-\sqrt{-8} \amp\text{ or }\amp\amp m\amp=\sqrt{-8}\\ m\amp=-\sqrt{\highlight{4}}\cdot\sqrt{\lighthigh{-1}}\cdot\sqrt{2} \amp\text{ or }\amp\amp m\amp=\sqrt{\highlight{4}}\cdot\sqrt{\lighthigh{-1}}\cdot\sqrt{2}\\ m\amp=-\highlight{2}\lighthigh{i}\sqrt{2}\amp\text{ or }\amp\amp m\amp=\highlight{2}\lighthigh{i}\sqrt{2} \end{align*}
The solution set is $\left\{-\highlight{2}\lighthigh{i}\sqrt{2},\highlight{2}\lighthigh{i}\sqrt{2}\right\}\text{.}$
###### Example8.7.17Solving Quadratic Equations with Complex Solutions
Solve the equation $3(v-2)^2+54=0\text{,}$ where $v$ is a complex number.
Explanation
\begin{align*} 3(v-2)^2+54\amp=0\\ 3(v-2)^2\amp=-54\\ (v-2)^2\amp=-18 \end{align*}
\begin{align*} v-2\amp=-\sqrt{-18}\amp\text{ or }\amp\amp v-2\amp=\sqrt{-18}\\ v-2\amp=-\sqrt{\highlight{9}\cdot\lighthigh{-1}\cdot2}\amp\text{ or }\amp\amp v-2\amp=\sqrt{\highlight{9}\cdot\lighthigh{-1}\cdot2}\\ v-2\amp=-\sqrt{\highlight{9}}\cdot\sqrt{\lighthigh{-1}}\cdot\sqrt{2}\amp\text{ or }\amp\amp v-2\amp=\sqrt{\highlight{9}}\cdot\sqrt{\lighthigh{-1}}\cdot\sqrt{2}\\ v-2\amp=-\highlight{3}\lighthigh{i}\sqrt{2}\amp\text{ or }\amp\amp v-2\amp=\highlight{3}\lighthigh{i}\sqrt{2}\\ v\amp=2-\highlight{3}\lighthigh{i}\sqrt{2}\amp\text{ or }\amp\amp v\amp=2+\highlight{3}\lighthigh{i}\sqrt{2} \end{align*}
So, the solution set is $\left\{2+\highlight{3}\lighthigh{i}\sqrt{2},2-\highlight{3}\lighthigh{i}\sqrt{2}\right\}\text{.}$
### Subsection8.7.6Strategies for Solving Quadratic Equations
In Section 8.6 we reviews all three methods for solving quadratic equations that we know. For the full explanation of solving using the factoring, visit Section 8.1, solving using the square root method, visit Section 8.3, and for more on the quadratic formula, visit Section 8.4.
###### Example8.7.18How to Choose a Method for Solving a Quadratic Equation
Solve the quadratic equations using an effective method.
1. $(x-4)^2-2=0$
2. $(x-4)^2-2x=0$
3. $(x-4)^2+2x=0$
Explanation
All three of the equations here are very similar, so we will need to examine them closely to choose the best method for solving them.
1. To solve the equation $(x-4)^2-2=0\text{,}$ first note that there is no linear term: there is only a square and a constant. This leads us to consider the square root method. Before doing that, isolate the square:
\begin{align*} (x-4)^2-2\amp=0\\ (x-4)^2\amp=2 \end{align*}
Now we can apply the square root method to the equation.
\begin{align*} x-4\amp=\sqrt{2}\amp\text{ or }\amp\amp x-4\amp=-\sqrt{2}\\ x\amp=4+\sqrt{2}\amp\text{ or }\amp\amp x\amp=4-\sqrt{2} \end{align*}
So the solution set is $\left\{4+\sqrt{2},4-\sqrt{2}\right\}$
2. To solve the equation $(x-4)^2-2x=0\text{,}$ first note that there is a linear term ($-2x$), so we must use either factoring or the quadratic formula. To use either, we must first put the equation in standard form.
\begin{align*} (x-4)^2-2x\amp=0\\ (x-4)(x-4)-2x\amp=0\\ x^2-8x+16-2x\amp=0\\ x^2-10x+16\amp=0 \end{align*}
Now that the equation is in standard form, we can decide whether to use factoring or the quadratic formula. While the quadratic formula always works, it can take more time than factoring if factoring is possible. In this case, factoring entails answering the question “are there two integers that multiply to be $16$ and add to be $-10\text{?}$” The answer is “yes”: $-8$ and $-2$ are such numbers.
\begin{align*} x^2-10x+16\amp=0\\ (x-8)(x-2)\amp=0 \end{align*}
\begin{align*} x-8\amp=0\amp\text{ or }\amp\amp x-2\amp=0\\ x\amp=8\amp\text{ or }\amp\amp x\amp=2 \end{align*}
So the solution set is $\{2,8\}\text{.}$
3. To solve the equation $(x-4)^2+2x=0\text{,}$ first note that there is a linear term ($+2x$), so we must use either factoring or the quadratic formula. To use either, we must first put the equation in standard form.
\begin{align*} (x-4)^2+2x\amp=0\\ (x-4)(x-4)+2x\amp=0\\ x^2-8x+16+2x\amp=0\\ x^2-6x+16\amp=0 \end{align*}
Now that the equation is in standard form, we can decide whether to use factoring or the quadratic formula. In this case, factoring entails answering the question “are there two integers that multiply to be $16$ and add to be $-6\text{?}$” The answer is “no,” so we must use the quadratic formula. First, identify that $\substitute{a=1}\text{,}$ $\substitute{b=-6}\text{,}$ and $\substitute{c=16}\text{.}$
\begin{align*} x\amp=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\ x\amp=\frac{-(\substitute{-6})\pm\sqrt{(\substitute{-6})^2-4(\substitute{1})(\substitute{16})}}{2(\substitute{1})}\\ \amp=\frac{6\pm\sqrt{36-48}}{2}\\ \amp=\frac{6\pm\sqrt{-12}}{2}\\ \end{align*}
At this point, we notice that the solutions are complex. Continue to simplify until they are completely reduced.
\begin{align*} x\amp=\frac{6\pm\sqrt{\highlight{4}\cdot\lighthigh{-1}\cdot3}}{2}\\ \amp=\frac{6\pm\sqrt{\highlight{4}}\cdot\sqrt{\lighthigh{-1}}\cdot\sqrt{3}}{2}\\ \amp=\frac{6\pm\highlight{2}\lighthigh{i}\sqrt{3}}{2}\\ \amp=\frac{6}{2}\pm\frac{2i\sqrt{3}}{2}\\ \amp=3\pm i\sqrt{3} \end{align*}
So the solution set is $\left\{3+i\sqrt{3},3-i\sqrt{3}\right\}\text{.}$
### Subsection8.7.7Exercises
###### 1
Solve the equation.
${x^{2}-6x-27} = 0$
###### 2
Solve the equation.
${x^{2}-5x-50} = 0$
###### 3
Solve the equation.
${x^{2}+11x} = {-18}$
###### 4
Solve the equation.
${x^{2}+18x} = {-80}$
###### 5
Solve the equation.
${x^{2}} = {9x}$
###### 6
Solve the equation.
${x^{2}} = {7x}$
###### 7
Solve the equation.
${x^{2}-8x+16}=0$
###### 8
Solve the equation.
${x^{2}-10x+25}=0$
###### 9
Solve the equation.
${4x^{2}}={-41x-10}$
###### 10
Solve the equation.
${4x^{2}}={-25x-36}$
###### 11
Solve the equation.
${x\!\left(x+20\right)}={5\!\left(2x-5\right)}$
###### 12
Solve the equation.
${x\!\left(x+36\right)}={9\!\left(2x-9\right)}$
###### 13
A rectangle’s base is ${7\ {\rm in}}$ shorter than four times its height. The rectangle’s area is ${65\ {\rm in^{2}}}\text{.}$ Find this rectangle’s dimensions.
The rectangle’s height is .
The rectangle’s base is .
###### 14
A rectangle’s base is ${7\ {\rm in}}$ shorter than four times its height. The rectangle’s area is ${2\ {\rm in^{2}}}\text{.}$ Find this rectangle’s dimensions.
The rectangle’s height is .
The rectangle’s base is .
###### 15
Without using a calculator, estimate the value of $\sqrt{22}\text{:}$
• 5.69
• 4.69
• 4.31
• 5.31
###### 16
Without using a calculator, estimate the value of $\sqrt{38}\text{:}$
• 6.16
• 6.84
• 5.84
• 5.16
###### 17
Evaluate the following.
$\displaystyle{\sqrt{{{\frac{25}{81}}}}={}}$.
###### 18
Evaluate the following.
$\displaystyle{\sqrt{{{\frac{36}{49}}}}={}}$.
###### 19
Evaluate the following.
$\sqrt{-64}=$.
###### 20
Evaluate the following.
$\sqrt{-100}=$.
###### 21
Simplify the radical expression or state that it is not a real number.
$\displaystyle{ \frac{{\sqrt{6}}}{{\sqrt{216}}} =}$
###### 22
Simplify the radical expression or state that it is not a real number.
$\displaystyle{ \frac{{\sqrt{4}}}{{\sqrt{144}}} =}$
###### 23
Simplify the radical expression or state that it is not a real number.
$\displaystyle{ {\sqrt{8}} = }$
###### 24
Simplify the radical expression or state that it is not a real number.
$\displaystyle{ {\sqrt{147}} = }$
###### 25
Simplify the expression.
$\displaystyle{3\sqrt{5} \cdot 3\sqrt{40}=}$
###### 26
Simplify the expression.
$\displaystyle{3\sqrt{15} \cdot 2\sqrt{30}=}$
###### 27
Simplify the expression.
$\displaystyle{ {\sqrt{\frac{28}{13}}} \cdot {\sqrt{\frac{4}{13}}} =}$
###### 28
Simplify the expression.
$\displaystyle{ {\sqrt{\frac{15}{19}}} \cdot {\sqrt{\frac{5}{19}}} =}$
###### 29
Simplify the expression.
$\displaystyle{{\sqrt{20}} + {\sqrt{45}} =}$
###### 30
Simplify the expression.
$\displaystyle{{\sqrt{175}} + {\sqrt{112}} =}$
###### 31
Simplify the expression.
$\displaystyle{{\sqrt{176}} - {\sqrt{44}} =}$
###### 32
Simplify the expression.
$\displaystyle{{\sqrt{8}} - {\sqrt{32}} =}$
###### 33
Rationalize the denominator and simplify the expression.
$\displaystyle{ \frac{1}{\sqrt{3}} = }$
###### 34
Rationalize the denominator and simplify the expression.
$\displaystyle{ \frac{1}{\sqrt{3}} = }$
###### 35
Rationalize the denominator and simplify the expression.
$\displaystyle{ \frac{2}{5\sqrt{5}} = }$
###### 36
Rationalize the denominator and simplify the expression.
$\displaystyle{ \frac{7}{3\sqrt{6}} = }$
###### 37
Rationalize the denominator and simplify the expression.
$\displaystyle{ \frac{1}{{\sqrt{75}}} = }$
###### 38
Rationalize the denominator and simplify the expression.
$\displaystyle{ \frac{1}{{\sqrt{175}}} = }$
###### 39
Expand and simplify the expression.
$\displaystyle{\left(7 - {\sqrt{7}}\right)\left(10 - 3 {\sqrt{7}}\right) =}$
###### 40
Expand and simplify the expression.
$\displaystyle{\left(5 - {\sqrt{7}}\right)\left(8 - 5 {\sqrt{7}}\right) =}$
###### 41
Expand and simplify the expression.
$\displaystyle{\left({\sqrt{5}} + {\sqrt{13}}\right)\left({\sqrt{5}} - {\sqrt{13}}\right) =}$
###### 42
Expand and simplify the expression.
$\displaystyle{\left({\sqrt{5}} + {\sqrt{7}}\right)\left({\sqrt{5}} - {\sqrt{7}}\right) =}$
###### 43
Solve the equation.
$x^2 = 20$
###### 44
Solve the equation.
$x^2 = 99$
###### 45
Solve the equation.
$144x^2 = 49$
###### 46
Solve the equation.
$16x^2 = 9$
###### 47
Solve the equation.
$\left(x+5\right)^2 = 121$
###### 48
Solve the equation.
$\left(x+7\right)^2 = 49$
###### 49
Solve the equation.
$2 - 2 ( t - 5 )^2 = -6$
###### 50
Solve the equation.
$54 - 5 ( x - 5 )^2 = 9$
###### 51
Find the value of $x\text{.}$
$x={}$
###### 52
Find the value of $x\text{.}$
$x={}$
###### 53
Parnell is designing a rectangular garden. The garden’s diagonal must be $2$ feet, and the ratio between the garden’s base and height must be $4:3\text{.}$ Find the length of the garden’s base and height.
The garden’s base is feet and its height is .
###### 54
Gregory is designing a rectangular garden. The garden’s diagonal must be $42.9$ feet, and the ratio between the garden’s base and height must be $12:5\text{.}$ Find the length of the garden’s base and height.
The garden’s base is feet and its height is .
###### 55
Solve the equation.
${14x^{2}-11x-9}=0$
###### 56
Solve the equation.
${18x^{2}-37x-20}=0$
###### 57
Solve the equation.
${x^{2}}= {-3x-1}$
###### 58
Solve the equation.
${x^{2}}= {5x-3}$
###### 59
Solve the equation.
${2x^{2}+4x+5}= 0$
###### 60
Solve the equation.
${2x^{2}-x+1}= 0$
###### 61
Solve the equation.
${10t+7} = {t+88}$
###### 62
Solve the equation.
${9b+10} = {b+66}$
###### 63
Solve the equation.
$-1-3y^2 = -3$
###### 64
Solve the equation.
$-8-7r^2 = -10$
###### 65
Solve the equation.
$x^2+39x= 0$
###### 66
Solve the equation.
$x^2+59x= 0$
###### 67
Solve the equation.
${x^{2}+8x} = {9}$
###### 68
Solve the equation.
${x^{2}-3x} = {54}$
###### 69
Solve the equation.
$\left(x - 7\right)^2 = 16$
###### 70
Solve the equation.
$\left(x - 5\right)^2 = 121$
###### 71
Solve the equation.
${x^{2}}= {-9x-19}$
###### 72
Solve the equation.
${x^{2}}= {-7x-8}$
###### 73
An object is launched upward at the height of $320$ meters. It’s height can be modeled by
\begin{equation*} h=-4.9t^2+90t+320\text{,} \end{equation*}
where $h$ stands for the object’s height in meters, and $t$ stands for time passed in seconds since its launch. The object’s height will be $350$ meters twice before it hits the ground. Find how many seconds since the launch would the object’s height be $350$ meters. Round your answers to two decimal places if needed.
The object’s height would be $350$ meters the first time at seconds, and then the second time at seconds.
###### 74
An object is launched upward at the height of $340$ meters. It’s height can be modeled by
\begin{equation*} h=-4.9t^2+70t+340\text{,} \end{equation*}
where $h$ stands for the object’s height in meters, and $t$ stands for time passed in seconds since its launch. The object’s height will be $350$ meters twice before it hits the ground. Find how many seconds since the launch would the object’s height be $350$ meters. Round your answers to two decimal places if needed.
The object’s height would be $350$ meters the first time at seconds, and then the second time at seconds.
###### 75
Simplify the radical and write it into a complex number.
$\displaystyle{ \sqrt{-48} =}$
###### 76
Simplify the radical and write it into a complex number.
$\displaystyle{ \sqrt{-98} =}$
###### 77
Solve the quadratic equation. Solutions could be complex numbers.
${-2x^{2}} - 4 = 8$
###### 78
Solve the quadratic equation. Solutions could be complex numbers.
${4x^{2}} + 7 = -5$
###### 79
Solve the quadratic equation. Solutions could be complex numbers.
$-9(y - 4)^2+8 = 584$
###### 80
Solve the quadratic equation. Solutions could be complex numbers.
$7(y+7)^2+8 = -167$
###### 81
Solve the equation.
$2x^2 + 29= 0$
###### 82
Solve the equation.
$41x^2 + 37= 0$
###### 83
Solve the equation.
${5x^{2}}={-31x-44}$
###### 84
Solve the equation.
${5x^{2}}={-27x-36}$
###### 85
Solve the equation.
${x^{2}+4x+1}= 0$
###### 86
Solve the equation.
${x^{2}+10x+7}= 0$
###### 87
Solve the equation.
$28 - 6 ( x+6 )^2 = 4$
###### 88
Solve the equation.
$34 - 4 ( y+6 )^2 = -2$
###### 89
Solve the equation.
${x^{2}+7x} = {-12}$
###### 90
Solve the equation.
${x^{2}+14x} = {-45}$
|
2019-05-23 22:52:54
|
{"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": 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": 0.8459199666976929, "perplexity": 2418.321238631083}, "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-2019-22/segments/1558232257432.60/warc/CC-MAIN-20190523224154-20190524010154-00423.warc.gz"}
|
https://mathhelpboards.com/threads/truth-table-precedence.25506/
|
# Truth Table Precedence
#### lyd123
##### New member
Hello!
The question is attached.
I know that " $\implies$ " (implies) has precedence from right to left. But because " l- " appears after
P$\implies ($Q $\implies$ R ), in my truth table do I evaluate:
(P$\implies ($Q $\implies$ R ) ) $\implies$ ((P$\implies$Q ) $\implies$ R ) )
or
P$\implies ($Q $\implies$ R ) $\implies$ (P$\implies$Q ) $\implies$ R )
Thank you for any help.
#### Attachments
• 6.3 KB Views: 20
#### Evgeny.Makarov
##### Well-known member
MHB Math Scholar
The turnstile separates formulas but is not a logical connective itself. Therefore $$\displaystyle A\vdash B$$ is equivalent to the fact that $$\displaystyle A\to B$$ is a tautology. This formula has $A$ and $B$ as subformulas joined by $\to$, but it cannot have a subformula that consists of a strict subformula of $A$ and $B$, for example. So it's wrong to consider $P\to(Q\to R)\to(P\to Q)\to R$, which is $P\to((Q\to R)\to((P\to Q)\to R))$ because it has a subformula $(Q\to R)\to((P\to Q)\to R)$, which consists of a part of $A$ and the whole $B$.
I know that " $\implies$ " (implies) has precedence from right to left.
I also like this convention, but I've seen textbooks that consider $\to$ to be left-associative, so one has to be careful.
#### lyd123
##### New member
I think I understand now.. so I should use (P⟹(Q ⟹ R ) ) ⟹ ((P⟹Q ) ⟹ R ) ),
which would give me the attached truth table.
So it is not a tautology. Is this correct?
#### Evgeny.Makarov
##### Well-known member
MHB Math Scholar
Yes, it is correct. The converse implication is a tautology. This follows from the fact that $P\to Q\to R$ is equivalent to $PQ\to R$ (I omitted conjunction) and $PQ$ implies $P\to Q$.
There is a typo in column R, second last row.
#### topsquark
##### Well-known member
MHB Math Helper
Just a quick question from a novice.
Do $$\displaystyle \implies$$ and $$\displaystyle \rightarrow$$ mean the same thing? I note that the OP and Evgeny.Makarov are using two different symbols.
-Dan
#### Evgeny.Makarov
##### Well-known member
MHB Math Scholar
Do $$\displaystyle \implies$$ and $$\displaystyle \rightarrow$$ mean the same thing? I note that the OP and Evgeny.Makarov are using two different symbols.
This completely depends on the textbook or other source. Implication can be denoted by $\rightarrow$, $\to$ and $\supset$, and in addition arrows can be short of long. Some authors use different notations for metalevel (a contraction for "if... then..." in English) and object level (a part of the formal language we study) implications. I used a single arrow because it occurs in the attached image in post #1, which I assume comes from the instructor, and because it is shorter in LaTeX ([m]\to[/m] vs [m]\Rightarrow[/m] or [m]\Longrightarrow[/m]).
|
2020-07-07 05:56:06
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 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": 0.8882349729537964, "perplexity": 1105.1651934581625}, "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-2020-29/segments/1593655891654.18/warc/CC-MAIN-20200707044954-20200707074954-00054.warc.gz"}
|
https://etnaetnaetna.be/balanced+sumbol+equationfor+the+reaction+ranked+1754.html
|
# balanced sumbol equationfor the reaction ranked
#### Balanced chemical equations - Introducing chemical …
26/7/2020· There are. (two nitrogen atoms and six hydrogen atoms) Add the state syols if they are requested. N2(g) + 3H2(g) → 2NH3(g) curriculum-key-fact. Balanced …
#### Balancing chemical equations - How are equations used …
5/2/2021· reactants: 1 × Mg, 2 × O products: 2 × Mg, 2 × O Not balanced. Step 4 (repeat) 2Mg + O 2 → 2MgO Step 3 (repeat)
#### 18.10: The Group 6A Elements - Chemistry LibreTexts
Direct reaction of the heavier chalcogens with oxygen at elevated temperatures gives the dioxides (YO 2), which exhibit a dramatic range of structures and properties. The dioxides become increasingly metallic in character down the group, as expected, and the coordination nuer of …
#### Balanced syol equation of magnesium and water? - …
24/4/2011· The balanced syol equation for copper II oxide reacting with hydrogen is Cu + H2O. This reaction will create copper and water as a result.
#### Write a balanced equation for the coustion of …
Answer to: Write a balanced equation for the coustion of Methane. By signing up, you'll get thousands of step-by-step solutions to your homework
#### Balanced syol equation of lithium and water? - …
18/9/2012· The balanced syol equation for copper II oxide reacting with hydrogen is Cu + H2O. This reaction will create copper and water as a result.
#### What is the balanced equation when potassium reacts …
21/6/2018· K (s) + Cl2(g) → KCl (s) There are two chlorine atoms on the left-hand side (LHS) and one chlorine atom on the right-hand side (RHS). Add a coefficient of 2 in front of KCl. K (s) + Cl2(g) → 2KCl. There are now two atoms of chlorine on both sides. However, the K is not balanced.
#### 0.4 mol of sodium hydroxide and 0.2 mol of hydrogen …
21/6/2019· for each reagent suggested above in (ii) explain the role of the reagent in the reaction to (iv) form compound e. you may wish to do this by drawing a mechanism. 1. addition of reagent a но reagent a 2. н, о" thо oh нон-с compound a. compound b. compound c .ch-оh 1. reagent b "сно 2. reagent c сh oh compound e. compound d.
#### Equation for Reaction Between Baking Soda and Vinegar
31/1/2020· The chemical equation for the overall reaction is: NaHCO 3 (s) + CH 3 COOH(l) → CO 2 (g) + H 2 O(l) + Na + (aq) + CH 3 COO - (aq) with s = solid, l = …
#### GCSE CHEMISTRY - What are the Rules for Balancing …
There are four Li with two O on the left. There are four Li with two O on the right. Showing the four steps together, we have. (1) Li + O 2 Li 2 O unbalanced for lithium and oxygen. (2) 2 Li + O 2 Li 2 O unbalanced for oxygen . (3) 2 Li + O 2 2 Li 2 O unbalanced for lithium again. (4) 4 …
#### Enter a balanced chemical equation for the | Clutch …
Q. Consider the coustion of liquid methanol, CH3OH(l): CH3OH(l) + 3/2 O2(g) → CO2(g) + 2 H2O(l), ΔH = –726.5 kJBalance the forward reaction with whole- Q. Methanol (CH3OH) is used as a fuel in race cars.Write a balanced equation for the coustion of liquid methanol in …
#### Tackling word equations and balanced syol equations …
The reactants are glucose and oxygen, the products are carbon dioxide and water. Glucose + oxygen→ carbon dioxide + water (+energy) (This is almost the same as burning a hydrocarbon; you get the same products.) Some reactions are very hard to describe in any other way than by a word equation.
#### What Is the Balanced Equation for the Reaction …
27/3/2020· The balanced equation for the reaction between HNO3 and KOH is written as HNO3 + KOH = H2O + KNO3. HNO3 is the chemical formula representing nitric acid, KOH is the formula representing potassium hydroxide, H2O is the formula for water and KNO3 is the formula for potassium nitrate. The process used to write a balanced equation is called
#### Balanced equations - Balanced equations - National 5 …
26/7/2020· Balanced syol equations show what happens to the different atoms in reactions. For example, copper and oxygen react together to make copper oxide. Take a look at this word equation for the reaction:
#### Balanced chemical equations - Introducing chemical …
26/7/2020· There are. (two nitrogen atoms and six hydrogen atoms) Add the state syols if they are requested. N2(g) + 3H2(g) → 2NH3(g) curriculum-key-fact. Balanced …
#### What is the balanced equation for the coustion of …
24/8/2017· Coefficients are used to add more of the missing atoms. Thus, the final balanced equation would be: 2C5H10 (g) + 15O2 (g) ---> 10CO2 (g) + 10H2O (g)
#### 0.4 mol of sodium hydroxide and 0.2 mol of hydrogen …
21/6/2019· for each reagent suggested above in (ii) explain the role of the reagent in the reaction to (iv) form compound e. you may wish to do this by drawing a mechanism. 1. addition of reagent a но reagent a 2. н, о" thо oh нон-с compound a. compound b. compound c .ch-оh 1. reagent b "сно 2. reagent c сh oh compound e. compound d.
#### How to Write the Balanced Chemical Reaction for the …
26/4/2018· If you''re interested in writing a balanced equation for the entire process, you need only know the initial reactants and the products of the reaction. The reactants are iron (Fe), oxygen (O 2 ) and water (H 2 O), and the product is iron (III) hydroxide Fe(OH) 3 , so Fe + O 2 + H 2 O → Fe(OH) 3 .
#### Tackling word equations and balanced syol equations …
GCSE 9-1 Chemistry and GCSE 9-1 Science topic on how to tackle word equations and balanced syol equations in GCSE 9-1 exam papers. Learning outcomes At the end of this, you should be able to: – State how you know a reaction has taken place-Write a
#### Multimedia: Balanced Equation | Chapter 6, Lesson 1 | …
Balanced Equation All the atoms in the reactants form the products so the mass of the reactants and the products is the same. No new atoms are created and no atoms are destroyed. Chapter 1 Lesson 1: Molecules Matter Lesson 2: Molecules in Motion Lesson 3: The Ups and Downs of Thermometers
#### What is the balanced equation when potassium reacts …
21/6/2018· K (s) + Cl2(g) → KCl (s) There are two chlorine atoms on the left-hand side (LHS) and one chlorine atom on the right-hand side (RHS). Add a coefficient of 2 in front of KCl. K (s) + Cl2(g) → 2KCl. There are now two atoms of chlorine on both sides. However, the K is not balanced.
#### Balancing chemical equations - How are equations used …
5/2/2021· reactants: 1 × Mg, 2 × O products: 2 × Mg, 2 × O Not balanced. Step 4 (repeat) 2Mg + O 2 → 2MgO Step 3 (repeat)
#### Write the balanced equation of the reaction between …
Answer and Explanation: The unbalanced reaction between magnesium and oxygen is: M g+O2 → M gO M g + O 2 → M g O. From the equation, the nuer of oxygen atoms is not balanced…
#### What Is the Balanced Chemical Equation for …
31/3/2020· The balanced equation for photosynthesis is: 6CO2 + 6H2O + sunlight energy = C6H12O6 + 6O2 Photosynthesis can be represented using a chemical equation: Carbon dioxide + water + light energy gives a carbohydrate + oxygen. Every living organism needs energy to survive, and photosynthesis is how plants capture the energy they require.
#### Balanced chemical equations - Introducing chemical …
26/7/2020· Balanced chemical equations A balanced equation models a chemical reaction using the formulae of the reactants and products . It shows the nuer of units of each substance involved.
#### Balanced equations - Balanced equations - National 5 …
26/7/2020· Here is the balanced syol equation: $2Cu(s) + {O_2}(g) \rightarrow 2CuO(s)$ You can see that now there are two copper atoms and two oxygen atoms on each side.
#### Write the balanced equation of the reaction between …
Answer and Explanation: The unbalanced reaction between magnesium and oxygen is: M g+O2 → M gO M g + O 2 → M g O. From the equation, the nuer of oxygen atoms is not balanced…
|
2021-09-16 19:24:37
|
{"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": 0.5994712710380554, "perplexity": 5165.4743485691915}, "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/1631780053717.37/warc/CC-MAIN-20210916174455-20210916204455-00166.warc.gz"}
|
https://socratic.org/questions/how-many-protons-neutrons-and-electrons-are-in-calcium
|
# How many protons, neutrons and electrons are in calcium?
Jul 4, 2017
20, 20, 20
#### Explanation:
Assuming that we are referring to a stable atom of calcium, we need to find the atomic number of calcium. This can be memorized, or found on the periodic table.
Calcium is the 20th element, with 20 protons (since the number of protons directly changes the element itself).
Since a stable atom has a net charge of $0$, we must have 20 electrons.
The number of neutrons will be the same as the number of protons, or else we will have an isotope, in this case, it is also 20.
|
2020-11-30 15:02:51
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 1, "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": 0.4578552842140198, "perplexity": 378.15430533083196}, "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/1606141216175.53/warc/CC-MAIN-20201130130840-20201130160840-00614.warc.gz"}
|
https://www.vedantu.com/question-answer/evaluate-the-expression-log-2xy5log-dfrac12left-class-11-maths-cbse-5edfb6b7d2b6d841190078ea
|
Courses
Courses for Kids
Free study material
Free LIVE classes
More
# Evaluate the expression ${{\log }_{2}}xy=5,{{\log }_{\dfrac{1}{2}}}\left( \dfrac{x}{y} \right)=1$
Last updated date: 24th Mar 2023
Total views: 308.4k
Views today: 7.88k
Verified
308.4k+ views
Hint: Use basic identity of logarithm given by;
If ${{a}^{x}}=N\text{ then }{{\log }_{a}}N=x$
We have equations/expression given in the problem as
${{\log }_{2}}xy=5.............\left( 1 \right)$
And
${{\log }_{\dfrac{1}{2}}}\left( \dfrac{x}{y} \right)=1....................\left( 2 \right)$
As, we know that if ${{a}^{x}}=N$ then we can take log to both sides as base of $a$
And we get:
${{a}^{x}}=N$
Taking $\log$ on both sides
${{\log }_{a}}{{a}^{x}}={{\log }_{a}}N$
As we know that ${{\log }_{c}}{{m}^{n}}=n{{\log }_{c}}m$ ;
Using this property we can write the above equation as;
$x{{\log }_{a}}a={{\log }_{a}}N$
As we know ${{\log }_{m}}m=1$ , we can rewrite the above relation as;
${{\log }_{a}}N=x$
Therefore, if we have ${{a}^{x}}=N$
Then ${{\log }_{a}}N=x................\left( 3 \right)$
Using the above property of logarithm we can write equation $\left( 1 \right)$ as
${{\log }_{2}}xy=5$
$xy={{2}^{5}}.................\left( 4 \right)$
Similarly, using the equation $\left( 3 \right)$ , we can write equation $\left( 2 \right)$ as
\begin{align} & {{\log }_{\dfrac{1}{2}}}\left( \dfrac{x}{y} \right)=1 \\ & \dfrac{x}{y}={{\left( \dfrac{1}{2} \right)}^{1}}=\left( \dfrac{1}{2} \right)..............\left( 5 \right) \\ \end{align}
Now, we need to find $x\text{ and y}$ ; For that we can multiply equation $\left( 4 \right)\text{ and }\left( 5 \right)$ in following way;
\begin{align} & xy\times \dfrac{x}{y}={{2}^{5}}\times \dfrac{1}{2} \\ & {{x}^{2}}=\dfrac{32}{2}=16 \\ & {{x}^{2}}=16 \\ & x=\pm 4 \\ \end{align}
To get value of $y$ , we can divide equation $\left( 4 \right)\And \left( 5 \right)$
\begin{align} & \dfrac{xy}{\left( \dfrac{x}{y} \right)}=\dfrac{{{2}^{5}}}{\left( \dfrac{1}{2} \right)} \\ & xy\times \dfrac{y}{x}=32\times 2 \\ & {{y}^{2}}=64 \\ & y=\pm 8 \\ \end{align}
Hence, we have $x=\pm 4\text{ and }y=\pm 8$ .
Now, here we need to select $\left( x,y \right)$ pairs which will satisfy the equation$\left( 5 \right)\And \left( 4 \right)$.
Now, we have four pairs as
\begin{align} & x=4,y=8 \\ & x=-4,y=-8 \\ & x=4,y=-8 \\ & x=-4,y=8 \\ \end{align}
We can put pairs to equation $\left( 4 \right)\And \left( 5 \right)$for verification
Case 1: $x=4,y=8$
For equation $\left( 4 \right)\text{ }xy=32$
$LHS=4\times 8=32=RHS$
For equation $\left( 5 \right)\text{ }\dfrac{x}{y}=\dfrac{1}{2}$
$LHS=\dfrac{4}{8}=\dfrac{1}{2}=RHS$
Hence $\left( 4,8 \right)$ is the solution of the given equations.
Case 2: $x=-4,y=8$
For equation $\left( 4 \right)$ $xy=32$
$LHS=-4\times -8=32=RHS$
For equation $\left( 5 \right)\text{ }\dfrac{x}{y}=\dfrac{1}{2}$
$LHS=\dfrac{-4}{-8}=\dfrac{1}{2}=RHS$
Hence, $\left( -4,-8 \right)$ pair is also a solution of the given equations.
Case 3: $x=-4,y=8$
For equation $\left( 4 \right)\text{ }xy=32$
$LHS=-4\times 8=-32\ne RHS$
It will not satisfy the equation $\left( 5 \right)$ $\dfrac{x}{y}=\dfrac{1}{2}$ as well.
Hence, $\left( -4,8 \right)$ pair is not a solution to the given equation.
Case 4: $x=4,y=-8$
For equation $\left( 4 \right)\text{ }xy=32$
$4\times \left( -8 \right)=-32\ne RHS$
For equation $\left( 5 \right)\dfrac{x}{y}=\dfrac{1}{2}$
$LHS=\dfrac{4}{-8}=-\dfrac{1}{2}\ne RHS$
Hence,$\left( 4,-8 \right)$ is not a solution of the given equation.
Note: We can get answers by putting values of $x=\pm 4$ in any of the equation $\left( 3 \right)\And \left( 4 \right)$ which will minimize our confusion related to $\left( -4,8 \right)or\left( 4,-8 \right)$ as explained in solution. One can also skip the question by just seeing the solution by just seeing the given function $\left( {{\log }_{2}}xy=5\text{ }\!\!\And\!\!\text{ }{{\log }_{\dfrac{1}{2}}}\dfrac{x}{y}=1 \right)$ as we cannot put negative values in logarithm $m$ function. Domain of $\log x$ is ${{R}^{+}}$ (positive real numbers).
One can go wrong by getting confused with formula if ${{a}^{x}}=N$ then ${{\log }_{a}}N=x$ . He/she may apply if ${{a}^{x}}=N$then ${{\log }_{N}}a=x$(general confusion with basic definition of logarithm function).
|
2023-03-30 18:31:19
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "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": 8, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 1.0000083446502686, "perplexity": 1145.5562586601206}, "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/1679296949355.52/warc/CC-MAIN-20230330163823-20230330193823-00075.warc.gz"}
|
https://dimag.ibs.re.kr/event/2019-06-19/
|
• This event has passed.
# Suil O (오수일), An odd [1,b]-factor in regular graphs from eigenvalues
## Wednesday, June 19, 2019 @ 4:30 PM - 5:30 PM KST
Room B232, IBS (기초과학연구원)
### Speaker
Suil O (오수일)
Department of Applied Mathematics and Statistics, SUNY-Korea
https://you.stonybrook.edu/suilo/
An odd $[1,b]$-factor of a graph is a spanning subgraph $H$ such that for every vertex $v \in V(G)$, $1 \le d_H(v) \le b$, and $d_H(v)$ is odd. For positive integers $r \ge 3$ and $b \le r$, Lu, Wu, and Yang gave an upper bound for the third largest eigenvalue in an $r$-regular graph with even number of vertices to guarantee the existence of an odd [1,b]-factor.
In this talk, we improve their bound.
## Details
Date:
Wednesday, June 19, 2019
Time:
4:30 PM - 5:30 PM KST
Event Category:
Event Tags:
,
Room B232
IBS (기초과학연구원)
## Organizer
Sang-il Oum (엄상일)
View Organizer Website
기초과학연구원 수리및계산과학연구단 이산수학그룹
대전 유성구 엑스포로 55 (우) 34126
IBS Discrete Mathematics Group (DIMAG)
Institute for Basic Science (IBS)
55 Expo-ro Yuseong-gu Daejeon 34126 South Korea
E-mail: dimag@ibs.re.kr, Fax: +82-42-878-9209
|
2023-03-23 23:38:57
|
{"extraction_info": {"found_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": 0.512464165687561, "perplexity": 12460.421548535614}, "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/1679296945218.30/warc/CC-MAIN-20230323225049-20230324015049-00225.warc.gz"}
|
https://openstax.org/books/intermediate-algebra-2e/pages/6-key-terms
|
Intermediate Algebra 2e
# Key Terms
### Key Terms
degree of the polynomial equation
The degree of the polynomial equation is the degree of the polynomial.
factoring
Splitting a product into factors is called factoring.
greatest common factor
The greatest common factor (GCF) of two or more expressions is the largest expression that is a factor of all the expressions.
polynomial equation
A polynomial equation is an equation that contains a polynomial expression.
Polynomial equations of degree two are called quadratic equations.
zero of the function
A value of $xx$ where the function is 0, is called a zero of the function.
Zero Product Property
The Zero Product Property says that if the product of two quantities is zero, then at least one of the quantities is zero.
Order a print copy
As an Amazon Associate we earn from qualifying purchases.
Want to cite, share, or modify this book? This book uses the Creative Commons Attribution License and you must attribute OpenStax.
• If you are redistributing all or part of this book in a print format, then you must include on every physical page the following attribution:
• If you are redistributing all or part of this book in a digital format, then you must include on every digital page view the following attribution:
|
2023-02-07 00:55:54
|
{"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": 0.24359026551246643, "perplexity": 1088.0086841453478}, "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-2023-06/segments/1674764500368.7/warc/CC-MAIN-20230207004322-20230207034322-00328.warc.gz"}
|
https://corpustools.readthedocs.io/en/v1.2/functional_load.html
|
Functional load is a measure of the “work” that any particular contrast does in a language, as compared to other contrasts (e.g., [Hockett1955], [Hockett1966]; [Kucera1963]; [King1967]; [Surendran2003]). Two contrasts in a language, such as [d] / [t] vs. [ð] / [θ] in English, may have very different functional loads. The difference between [d] and [t] is used to distinguish between many different lexical items, so it has a high functional load; there are, on the other hand, very few lexical items that hinge on the distinction between [ð] and [θ], so its functional load is much lower. One of the primary claims about functional load is that it is related to sounds’ propensity to merge over time, with pairs of sounds that have higher functional loads being less likely to merge than pairs of sounds with lower functional loads (e.g., [Wedel2013], [Todd2012]). The average functional load of a particular sound has also been claimed to affect its likelihood of being used as an epenthetic vowel [Hume2013]. Functional load has also been illustrated to affect the perceived similarity of sounds [Hall2014a].
Method of calculation¶
There are two primary ways of calculating functional load that are provided as part of the PCT package. One is based on the change of entropy in a system upon merger of a segment pair or set of segment pairs (cf. [Surendran2003]); the other is based on simply counting up the number of minimal pairs (differing in only the target segment pair or pairs) that occur in the corpus.
Change in entropy¶
The calculation based on change in entropy is described in detail in [Surendran2003]. Entropy is an Information-Theoretic measure of the amount of uncertainty in a system [Shannon1949], and is calculated using the formula in (1); it will also be used for the calculation of predictability of distribution (see Method of calculation). For every symbol i in some inventory (e.g., every phoneme in the phoneme inventory, or every word in the lexicon), one multiplies the probability of i by the $$log_{2}$$ of the probability of i; the entropy is the sum of the products for all symbols in the inventory.
Entropy:
$$H = -\sum_{i \in N} p_{i} * log_{2}(p_{i})$$
The functional load of any pair of sounds in the system, then, can be calculated by first calculating the entropy of the system at some level of structure (e.g., words, syllables) with all sounds included, then merging the pair of sounds in question and re-calculating the entropy of the new system. That is, the functional load is the amount of uncertainty (entropy) that is lost by the merger. If the pair has a functional load of 0, then nothing has changed when the two are merged, and $$H_{1}$$ will equal $$H_{2}$$. If the pair has a non-zero functional load, then the total inventory has become smaller through the conflating of pairs of symbols that were distinguished only through the given pair of sounds.
Functional load as change in entropy:
$$\Delta H = H_{1} - H_{2}$$
Consider a toy example, in which the following corpus is assumed (note that, generally speaking, there is no “type frequency” column in a PCT corpus, as it is assumed that each row in the corpus represents 1 type; it is included here for clarity):
Consider a toy example, in which the following corpus is assumed (note that, generally speaking, there is no “type frequency” column in a PCT corpus, as it is assumed that each row in the corpus represents 1 type; it is included here for clarity):
Word Original Under [h] / [ŋ] merger Under [t] / [d] merger
Trans. Type Freq. Token Freq. Trans. Type Freq. Token Freq. Trans. Type Freq. Token Freq.
hot [hɑt] 1 2 [Xɑt] 1 2 [hɑX] 1 2
song [sɑŋ] 1 4 [sɑX] 1 4 [sɑŋ] 1 4
hat [hæt] 1 1 [Xæt] 1 1 [hæX] 1 1
sing [sɪŋ] 1 6 [sɪX] 1 6 [sɪŋ] 1 6
tot [tɑt] 1 3 [tɑt] 1 3 [XɑX] 1 8
dot [dɑt] 1 5 [dɑt] 1 5 [XɑX]
hip [hɪp] 1 2 [Xɪp] 1 2 [hɪp] 1 2
hid [hɪd] 1 7 [Xɪd] 1 7 [hɪX] 1 7
team [tim] 1 5 [tim] 1 5 [Xim] 1 10
deem [dim] 1 5 [dim] 1 5 [Xim]
toot [tut] 1 9 [tut] 1 9 [XuX] 1 11
dude [dud] 1 2 [dud] 1 2 [XuX]
hiss [hɪs] 1 3 [Xɪs] 1 3 [hɪs] 1 3
his [hɪz] 1 5 [Xɪz] 1 5 [hɪz] 1 5
sizzle [sɪzəl] 1 4 [sɪzəl] 1 4 [sɪzəl] 1 4
dizzy [dɪzi] 1 3 [dɪzi] 1 3 [Xɪzi] 1 7
tizzy [tɪzi] 1 4 [tɪzi] 1 4 [Xɪzi]
Total 17 70 17 70 13 70
The starting entropy, assuming word types as the relative unit of structure and counting, is:
$$H_{1 - types} = -[(\frac{1}{17} log_{2}(\frac{1}{17})) + (\frac{1}{17} log_{2}(\frac{1}{17})) + (\frac{1}{17} log_{2}(\frac{1}{17})) + (\frac{1}{17} log_{2}(\frac{1}{17})) + (\frac{1}{17} log_{2}(\frac{1}{17}))\\ + (\frac{1}{17} log_{2}(\frac{1}{17})) + (\frac{1}{17} log_{2}(\frac{1}{17})) + (\frac{1}{17} log_{2}(\frac{1}{17})) + (\frac{1}{17} log_{2}(\frac{1}{17})) + (\frac{1}{17} log_{2}(\frac{1}{17})) + (\frac{1}{17} log_{2}(\frac{1}{17}))\\ + (\frac{1}{17} log_{2}(\frac{1}{17})) + (\frac{1}{17} log_{2}(\frac{1}{17})) + (\frac{1}{17} log_{2}(\frac{1}{17})) + (\frac{1}{17} log_{2}(\frac{1}{17})) + (\frac{1}{17} log_{2}(\frac{1}{17})) + (\frac{1}{17} log_{2}(\frac{1}{17}))] =4.087$$
The starting entropy, assuming word tokens, is:
$$H_{1 - tokens} = -[(\frac{2}{70} log_{2}(\frac{2}{70})) + (\frac{4}{70} log_{2}(\frac{4}{70})) + (\frac{1}{70} log_{2}(\frac{1}{70})) + (\frac{6}{70} log_{2}(\frac{6}{70})) + (\frac{3}{70} log_{2}(\frac{3}{70}))\\ + (\frac{5}{70} log_{2}(\frac{5}{70})) + (\frac{2}{70} log_{2}(\frac{2}{70})) + (\frac{7}{70} log_{2}(\frac{7}{70})) + (\frac{5}{70} log_{2}(\frac{5}{70})) + (\frac{5}{70} log_{2}(\frac{5}{70})) + (\frac{9}{70} log_{2}(\frac{9}{70}))\\ + (\frac{2}{70} log_{2}(\frac{2}{70})) + (\frac{3}{70} log_{2}(\frac{3}{70})) + (\frac{5}{70} log_{2}(\frac{5}{70})) + (\frac{4}{70} log_{2}(\frac{4}{70})) + (\frac{3}{70} log_{2}(\frac{3}{70})) + (\frac{4}{70} log_{2}(\frac{4}{70}))] = 3.924$$
Upon merger of [h] and [ŋ], there is no change in the number of unique words; there are still 17 unique words with all their same token frequencies. Thus, the entropy after an [h] / [ŋ] merger will be the same as it was before the merger. The functional load, then would be 0, as the pre-merger and post-merger entropies are identical.
Upon merger of [t] and [d], on the other hand, four pairs of words have been collapsed. E.g., the difference between team and deem no longer exists; there is now just one word, [Xim], where [X] represents the result of the merger. Thus, there are only 13 unique words, and while the total token frequency count remains the same, at 70, those 70 occurrences are divided among only 13 unique words instead of 17.
Thus, the entropy after a [t] / [d] merger, assuming word types, is:
$$H_{1 - types} = -[(\frac{1}{13} log_{2}(\frac{1}{13})) + (\frac{1}{13} log_{2}(\frac{1}{13})) + (\frac{1}{13} log_{2}(\frac{1}{13})) + (\frac{1}{13} log_{2}(\frac{1}{13})) + (\frac{1}{13} log_{2}(\frac{1}{13}))\\ + (\frac{1}{13} log_{2}(\frac{1}{13})) + (\frac{1}{13} log_{2}(\frac{1}{13})) + (\frac{1}{13} log_{2}(\frac{1}{13})) + (\frac{1}{13} log_{2}(\frac{1}{13})) + (\frac{1}{13} log_{2}(\frac{1}{13})) + (\frac{1}{13} log_{2}(\frac{1}{13}))\\ + (\frac{1}{13} log_{2}(\frac{1}{13})) + (\frac{1}{13} log_{2}(\frac{1}{13}))] = 3.700$$
And the entropy after a [t] / [d] merger, assuming word tokens, is:
$$H_{1 - tokens} = -[(\frac{2}{70} log_{2}(\frac{2}{70})) + (\frac{4}{70} log_{2}(\frac{4}{70})) + (\frac{1}{70} log_{2}(\frac{1}{70})) + (\frac{6}{70} log_{2}(\frac{6}{70})) + (\frac{8}{70} log_{2}(\frac{8}{70}))\\ + (\frac{2}{70} log_{2}(\frac{2}{70})) + (\frac{7}{70} log_{2}(\frac{7}{70})) + (\frac{10}{70} log_{2}(\frac{10}{70})) + (\frac{11}{70} log_{2}(\frac{11}{70})) + (\frac{3}{70} log_{2}(\frac{3}{70})) + (\frac{5}{70} log_{2}(\frac{5}{70}))\\ + (\frac{4}{70} log_{2}(\frac{4}{70})) + (\frac{7}{70} log_{2}(\frac{7}{70}))] = 3.466$$
$$\Delta H = H_{1-types} - H_{2-types} = 4.087– 3.700 = 0.387$$
And the functional load of [t] / [d] based on word tokens is:
$$\Delta H = H_{1-tokens} - H_{2-tokens} = 3.924– 3.466 = 0.458$$
(Relative) Minimal Pair Counts¶
The second means of calculating functional load that is included in PCT is a straight count of minimal pairs, which can be relativized to the number of words in the corpus that are potential minimal pairs—i.e. the number of words in the corpus with at least one of the target segments.
In the above example, the number of minimal pairs that hinge on [h] vs. [ŋ] is of course 0, so the functional load of [h] / [ŋ] is 0. The number of minimal pairs that hinge on [t] / [d] is 3, and the number of words with either [t] or [d] is 11; the functional load as a relativized minimal pair count would therefore be 3/11 = 0.273. Note that here, a relatively loose definition of minimal pair is used; specifically, two words are considered to be a minimal pair hinging on sounds A and B if, upon merger of A and B into a single symbol X, the words are identical. Thus, toot and dude are considered a minimal pair on this definition, because they both become [XuX] upon merger of [t] and [d].
The resulting calculations of functional load are thus quite similar between the two measures, but the units are entirely different. Functional load based on change in entropy is measured in bits, while functional load based on relativized minimal pair counts is simply a percentage. Also note that functional load based on minimal pairs is only based on type frequency; the frequency of the usage of the words is not used as a weighting factor, the way it can be under the calculation of functional load as change in entropy.
[Hume2013] suggests that the average functional load (there called “relative contrastiveness”) is a useful way of indicating how much work an individual segment does, on average, in comparison to other segments. This is calculated by taking an individual segment, calculating the pairwise functional load of that segment and each other segment in the inventory, and then taking the average across all those pairs. This calculation can also be performed in PCT.
Calculating functional load in the GUI¶
As with most analysis functions, a corpus must first be loaded (see Loading in corpora). Once a corpus is loaded, use the following steps.
1. Getting started: Choose “Analysis” / “Calculate functional load...” from the top menu bar.
2. Sound selection: First, decide whether you want to calculate the average functional load of a single segment (i.e., its functional load averaged across all possible pairwise comparisons), or the more standard functional load of a pair of sounds, defined over segments or features. To calculate the average functional load of a single sound, choose “Add one segment”; to calculate the pairwise functional load of two segments, choose “Add pair of segments”; to calculate the pairwise functional load based on features, choose “Add pair of features.”
For details on how to actually select segments or features, see Sound Selection or Feature Selection as relevant.
When multiple individual segments or individual pairs are selected, each entry will be treated separately.
3. Functional load algorithm: Select which of the two methods of calculation you want to use—i.e., minimal pairs or change in entropy. (See discussion above for details of each.)
4. Minimal pair options: If minimal pairs serve as the means of calculation, there are three additional parameters can be set.
1. Raw vs. relative count: First, PCT can report only the raw count of minimal pairs that hinge on the contrast in the corpus, if you just want to know the scope of the contrast. On the other hand, the default is to relativize the raw count to the corpus size, by dividing the raw number by the number of lexical entries that include at least one instance of any of the target segments.
2. Include vs. ignore homophones: Second, PCT can either include homophones or ignore them. For example, if the corpus includes separate entries for the words sock (n.), sock (v.), shock (n.), and shock (v.), this would count as four minimal pairs if homophones are included, but only one if homophones are ignored. The default is to ignore homophones.
3. Output list of minimal pairs to a file: It is possible to save a list of all the actual minimal pairs that PCT finds that hinge on a particular chosen contrast to a .txt file. To do so, enter a file path name, or select “Choose file...” to use a regular system dialogue box. If nothing is entered here, no list will be saved, but the overall output will still be provided (and can be saved independently).
5. Change in entropy options: If you are calculating functional load using change in entropy, one additional parameter can be set.
1. Type or token frequency: As described in Change in entropy, entropy can be calculated using either type or token frequencies. This option determines which to use.
6. Tier: Select which tier the functional load should be calculated from. The default is the “transcription” tier, i.e., looking at the entire word transcriptions. If another tier has been created (see Creating new tiers in the corpus), functional load can be calculated on the basis of that tier. For example, if a vowel tier has been created, then “minimal pairs” will be entries that are identical except for one entry in the vowels only, entirely independently of consonants. Thus, the words [mapotik] and [ʃɹaɡefli] would be treated as a minimal pair, given that their vowel-tier representations are [aoi] and [aei].
7. Pronunciation variants: If the corpus contains multiple pronunciation variants for lexical items, select what strategy should be used. For details, see Pronunciation Variants.
8. Minimum frequency: It is possible to set a minimum token frequency for words in the corpus in order to be included in the calculation. This allows easy exclusion of rare words; for example, if one were calculating the functional load of [s] vs. [ʃ] in English and didn’t set a minimum frequency, words such as santy (vs. shanty) might be included, which might not be a particularly accurate reflection of the phonological knowledge of speakers. To include all words in the corpus, regardless of their token frequency, set the the minimum frequency to 0.
Here is an example of selecting [m] and [n], with functional load to be calculated on the basis of minimal pairs, only including words with a token frequency of at least 1, from the built-in example corpus (which only has canonical forms):
1. Results: Once all parameters have been set, click one of the two “Calculate functional load” buttons. If this is the first calculation, the option to “start new results table” should be selected. For subsequent calculations, the calculation can be added to the already started table, for direct comparison, or a new table can be started.
Note that if a table is closed, new calculations will not be added to the previously
open table; a new table must be started.
Either way, the results table will have the following columns, with one row per calculation: segment 1, segment 2, which tier was used, what environments were selected, which measurement method was selected, the resulting functional load, what the minimum frequency was, what strategy was used for dealing with pronunciation variants, and for calculations using minimal pairs, whether the count is absolute or relative and whether homophones were distinguished or not. (For calculations using change in entropy, “N/A” values are entered into the latter two columns.)
1. Saving results: Once a results table has been generated for at least one pair, the table can be saved by clicking on “Save to file” at the bottom of the table to open a system dialogue box and save the results at a user-designated location.
Note
that in the above screen shot, not all columns are visible; they are visible only by scrolling over to the right, due to constraints on the window size. All columns would be saved to the results file.)
To return to the function dialogue box with your most recently used selections, click on “Reopen function dialog.” Otherwise, the results table can be closed and you will be returned to your corpus view.
Implementing the functional load function on the command line¶
In order to perform this analysis on the command line, you must enter a command in the following format into your Terminal:
pct_funcload CORPUSFILE [additional arguments]
...where CORPUSFILE is the name of your *.corpus file. If calculating FL from a file of segment pairs, it must list the pairs of segments whose functional load you wish to calculate with each pair separated by a tab (\t) and one pair on each line. Note that you must either specify a file or segment (using -p) or request the functional loads of all segment pairs in the inventory (using -l). You may also use command line options to change various parameters of your functional load calculations. Descriptions of these arguments can be viewed by running pct_funcload –h or pct_funcload --help. The help text from this command is copied below, augmented with specifications of default values:
Positional arguments:
corpus_file_name
Name of corpus file
Mandatory argument group (call must have one of these two):
-p PAIRS_FILE_NAME_OR_SEGMENT
--pairs_file_name_or_segment PAIRS_FILE_NAME_OR_SEGMENT
Name of file with segment pairs (or target segment if relative fl is True)
-l
--all_pairwise_fls
Flag: calculate FL for all pairs of segments
Optional arguments:
-h
--help
Show help message and exit
-c CONTEXT_TYPE
--context_type CONTEXT_TYPE
How to deal with variable pronunciations. Options are ‘Canonical’, ‘MostFrequent’, ‘SeparatedTokens’, or ‘Weighted’. See documentation for details.
-a ALGORITHM
--algorithm ALGORITHM
Algorithm to use for calculating functional load: “minpair” for minimal pair count or “deltah” for change in entropy. Defaults to minpair.
-f FREQUENCY_CUTOFF
--frequency_cutoff FREQUENCY_CUTOFF
Minimum frequency of words to consider as possible minimal pairs or contributing to lexicon entropy.
-d DISTINGUISH_HOMOPHONES
--distinguish_homophones DISTINGUISH_HOMOPHONES
For minimal pair FL: if False, then you’ll count sock~shock (sock=clothing) and sock~shock (sock=punch) as just one minimal pair; but if True, you’ll overcount alternative spellings of the same word, e.g. axel~actual and axle~actual. False is the value used by Wedel et al.
-t TYPE_OR_TOKEN
--type_or_token TYPE_OR_TOKEN
For change in entropy FL: specifies whether entropy is based on type or token frequency.
-e RELATIVE_FL
--relative_fl RELATIVE_FL
If True, calculate the relative FL of a single segment by averaging across the functional loads of it and all other segments.
-s SEQUENCE_TYPE
--sequence_type SEQUENCE_TYPE
The attribute of Words to calculate FL over. Normally this will be the transcription, but it can also be the spelling or a user-specified tier.
-o OUTFILE
--outfile OUTFILE
Name of output file
EXAMPLE 1: If your corpus file is example.corpus (no prounciation variants) and you want to calculate the minimal pair functional load of the segments [m] and [n] using defaults for all optional arguments, you first need to create a text file that contains the text m\tn (where \t is a tab). Let us call this file pairs.txt. You would then run the following command in your terminal window:
pct_funcload example.corpus -p pairs.txt
EXAMPLE 2: Suppose you want to calculate the relative (average) functional load of the segment [m]. Your corpus file is again example.corpus. You want to use the change in entropy measure of functional load rather than the minimal pairs measure, and you also want to use type frequency instead of (the default value of) token frequency. In addition, you want the script to produce an output file called output.txt. You would need to run the following command:
pct_funcload example.corpus -p m -e -a deltah -t type -o output.txt
EXAMPLE 3: Suppose you want to calculate the functional loads of all segment pairs. Your corpus file is again example.corpus. All other parameters are set to defaults. In addition, you want the script to produce an output file called output.txt. You would need to run the following command:
pct_funcload example.corpus -l -o output.txt
Classes and functions¶
For further details about the relevant classes and functions in PCT’s source code, please refer to Functional load.
|
2021-12-08 16:39:12
|
{"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": 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": 0.5427384972572327, "perplexity": 2055.465850195317}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-49/segments/1637964363515.28/warc/CC-MAIN-20211208144647-20211208174647-00423.warc.gz"}
|
http://snakelemma.blogspot.com/2008/09/haar-wavelet-transform-in-scala.html
|
venerdì 26 settembre 2008
Haar wavelet transform in Scala
This is my first try to write a wavelet(ondine in Italian) transform in Scala.
Naturally i've choose Haar wavelet transform, the simplest!
This is the core of the transformation:
def waveletCalc( values: List[Double] ):List[List[Double]]= haarCalc( values , List( ) ).reverse
def haarCalc( vals: List[Double], cc: List[List[Double]] ) :List[List[Double]] = vals.length / 2 match { case 1 => cc ::: List( List((vals(0) - vals(1) ) / 2 ) , List(( vals(0) + vals(1) ) / 2 )) case _ => haarCalc( vals.indices.dropRight(vals.length/2) .map(_*2).map(el=>(vals(el) + vals(el+1))/2), cc ::: List( vals.indices.dropRight( vals.length/2).map(2*_).map(el=> (vals (el) - vals(el + 1))/2)))}
The algorithm is recursive and returns as a result a list of lists.
This arrangement is the most convenient since a base of wavelets has two indexes.
Soon i'll try to post the complete class with the inverse transform and a brief
description on this particular wavelets transform.
I strongly recommend to copy and paste the code in an editor to read it better...
you can find a more complete code here
|
2017-07-26 18:49:38
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.35033872723579407, "perplexity": 8837.70858506527}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-30/segments/1500549426372.41/warc/CC-MAIN-20170726182141-20170726202141-00636.warc.gz"}
|
http://positionrelative.blogspot.com/2012/06/jqueery.html
|
## 2012-06-30
### jQueery
This is slightly bizarre. I was jumping through hoops trying to get a script to reset a dropdown after a successful ajax request. Quite pointlessly, as it turns out.
The dropdown:
<select id="pos" name="pos">
<option value="n">noun (名詞)</option>
<option value="v">verb (動詞)</option>
</select>
So after the user submits some data, the form, including the dropdown, is reset.
After trying all kinds of complicated nonsense with eq() and stuff:
$("#pages option").eq(0).attr('selected', 'selected'); ... it turns out that this is the way to do it: $("#pos").val("n");
I originally began the post with the intention of saying "how counterintuitive!" at the end, but in the seven minutes it's taken me to write this, I've completely changed my mind. It's the most simple and elegant way to do it, and I salute you, jQuery.
|
2017-09-26 11:03:59
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 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": 0.5456231236457825, "perplexity": 5342.8801187562985}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-39/segments/1505818695439.96/warc/CC-MAIN-20170926103944-20170926123944-00573.warc.gz"}
|
https://math.stackexchange.com/questions/644980/infinite-number-of-lie-groups-with-the-same-lie-algebra
|
Infinite number of Lie groups with the same lie algebra
Is there a finite dimensional Lie algebra L such that there are infinite number of non isomorphic compact connected lie groups which Lie algebras are isomorphic to L?
migrated from mathoverflow.netJan 20 '14 at 14:48
This question came from our site for professional mathematicians.
• $\mathfrak{sl}_2(\mathbf{R})$ – YCor Jan 20 '14 at 10:09
• I do not think it is a good idea to edit the question (without any mention of the edit), especially after the answer appeared. – Sasha Anan'in Jan 20 '14 at 11:07
To develop Yves' comment : let $G$ be the simply connected Lie group with Lie algebra $\mathfrak{sl}_2(\mathbb{R})$; it contains a central subgroup $Z\cong \mathbb{Z}$ such that $G/Z\cong \mathrm{SL}_2(\mathbb{R})$. Now put $G_n:=G/nZ$ for $n\geq 1$. An isomorphism $G_p\rightarrow G_q$ lifts to an isomorphism $G\rightarrow G$ which must map $pZ$ into $qZ$; this implies $p=q$, thus all these groups are non-isomorphic.
• Then the answer is negative. A compact Lie group admits a finite covering $T\times S$, with $T$ a torus and $S$ semi-simple, and these have only a finite number of non-isomorphic quotients. – abx Jan 20 '14 at 11:19
|
2019-07-19 15:07:17
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8688536286354065, "perplexity": 151.98424787015801}, "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-2019-30/segments/1563195526254.26/warc/CC-MAIN-20190719140355-20190719162355-00379.warc.gz"}
|
http://computationalnonlinear.asmedigitalcollection.asme.org/article.aspx?articleid=2513230
|
0
Research Papers
# Active Control of a Rectangular Thin Plate Via Negative Acceleration Feedback
[+] Author and Article Information
H. S. Bauomy
Department of Mathematics,
Faculty of Science,
Zagazig University,
Zagazig 44519, Egypt;
Department of Mathematics,
Prince Sattam Bin Abdulaziz University,
P.O. Box 54,
e-mail: hany_samih@yahoo.com
A. T. EL-Sayed
Department of Basic Sciences,
Technology,
Mokatem 11585, Egypt
e-mail: ashraftaha211@yahoo.com
1Corresponding author.
Contributed by the Design Engineering Division of ASME for publication in the JOURNAL OF COMPUTATIONAL AND NONLINEAR DYNAMICS. Manuscript received September 6, 2015; final manuscript received March 23, 2016; published online May 13, 2016. Assoc. Editor: Firdaus Udwadia.
J. Comput. Nonlinear Dynam 11(4), 041025 (May 13, 2016) (12 pages) Paper No: CND-15-1282; doi: 10.1115/1.4033307 History: Received September 06, 2015; Revised March 23, 2016
## Abstract
In this paper, the dynamic oscillation of a rectangular thin plate under parametric and external excitations is investigated and controlled. The motion of a rectangular thin plate is modeled by coupled second-order nonlinear ordinary differential equations. The formulas of the thin plate are derived from the von Kármán equation and Galerkin's method. A control law based on negative acceleration feedback is proposed for the system. The multiple time scale perturbation technique is applied to solve the nonlinear differential equations and obtain approximate solutions up to the second-order approximations. One of the worst resonance case of the system is the simultaneous primary resonances, where $Ω1≅ω1 and Ω2≅ω2$. From the frequency response equations, the stability of the system is investigated according to the Routh–Hurwitz criterion. The effects of the different parameters are studied numerically. It is also shown that the system parameters have different effects on the nonlinear response of the thin plate. The simulation results are achieved using matlab 7.0 software. A comparison is made with the available published work.
<>
## Figures
Fig. 1
The model of a rectangular thin plate and the coordinate system
Fig. 2
Nonresonant time response solution at selected values: α1=β1=5.125, α2=β2=7.375, ω1=8,ω2=9,Ω1=4,Ω2=4.5,f1=0.4,f2=0.25,μ=0.4, F1=4, F2=1.5, G1=4, and G2=3.0
Fig. 3
Simultaneous primary resonance case Ω1≅ω1 and Ω2≅ω2 : (a) system without controller, (b) system with negative linear velocity feedback controller, (c) system with negative quadratic velocity feedback controller, (d) system with negative cubic velocity feedback controller, and (e) system with negative acceleration feedback controller
Fig. 4
Theoretical frequency response curves: α1=5.125, α2=7.375, ω1=8,μ=0.4, F1=4, G1=0.009, and a2=0.5. (a) Frequency response curve of a1, (b) effect of the excitation force F1, (c) effect of the natural frequency ω1, (d) effect of the damping coefficient μ, (e) effect of the nonlinear parameter α1, (f) effect of the nonlinear parameter α2, and (g) effect of the gain G1.
Fig. 5
Theoretical frequency response curves: β1=5.125, β2=7.375, ω2=9,μ=0.4, F2=3, G2=0.04, and a1=0.05. (a) Frequency response curve of a2, (b) effect of the excitation force F2, (c) effect of the natural frequency ω2, (d) effect of the damping coefficient μ, (e) effect of the nonlinear parameter β1, (f) effect of the nonlinear parameter β2, and (g) effect of the gain G2.
Fig. 6
Excitation–response curve of simultaneous primary resonance for α1=5.125, α2=7.375, ω1=8,μ=0.4, σ1=0.5, G1=0.009, and a2=0.5. (a) Excitation–response curves (a1 against F1), (b) effect of the natural frequency ω1, (c) effect of the damping coefficient μ, (d) effect of the detuning parameter σ1, (e) effect of the nonlinear parameter α1, (f) effect of the nonlinear parameter α2, and (g) effect of the gain G1.
Fig. 7
Excitation–response curve of simultaneous primary resonance for β1=5.125, β2=7.375, ω2=9,μ=0.4, σ2=2, G2=0.04, and a1=0.1. (a) Excitation–response curves (a2 against F2), (b) effect of the natural frequency ω2, (c) effect of the damping coefficient μ, (d) effect of the detuning parameter σ2, (e) effect of the nonlinear parameter β1, (f) effect of the nonlinear parameter β2, and (g) effect of the gain G2.
## Discussions
Some tools below are only available to our subscribers or users with an online account.
### Related Content
Customize your page view by dragging and repositioning the boxes below.
Related Journal Articles
Related Proceedings Articles
Related eBook Content
Topic Collections
|
2017-09-24 21:20:18
|
{"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": 0.2798630893230438, "perplexity": 2161.9447255584373}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-39/segments/1505818690211.67/warc/CC-MAIN-20170924205308-20170924225308-00065.warc.gz"}
|
https://gateoverflow.in/379233/madeeasy-test-series
|
322 views
A hash table of size 10 using open addressing with linear probing and hash function is h(k)= (k)mod10 , where k is key value , initially table is empty . Following keys are inserted into table in given order .
44,87,43,68,30,20,67
How many number of probes required to insert 17 in table after inserting above keys?
How would you be so sure without probing that the location after element 20 is empty ?
Ok got it , thanks 👍
@Kabir5454 My bad. I found ‘total number of probes to insert 17 starting from inserting the first element”. It would be 6.
4
### 1 comment
how?
The answer would be 6 probes becoz linear probing has a formula we can say and that is hf(key,i) = hf(key) + i mod M(hash table size) where i € 0 to M-1 . Count the i which for 17 . There will be 5 collisions and 6 comparisons. 17 key will be inserted at index 2 .
by
|
2022-11-29 18:34:23
|
{"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": 0.8471307754516602, "perplexity": 2948.270475364453}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-49/segments/1669446710710.91/warc/CC-MAIN-20221129164449-20221129194449-00021.warc.gz"}
|
https://socratic.org/questions/how-do-you-calculate-sin-1-9-4-12
|
# How do you calculate sin^-1(9.4 / 12)?
Feb 12, 2016
$51.57$ degrees as also $128.43$ degrees.
#### Explanation:
Sine of an angle can take value between $- 1$ and $+ 1$.
As $\frac{9.4}{12} = 0.78333$, the angle whose sine is $0.78333$ is ${\sin}^{-} 1 \left(\frac{9.4}{12}\right)$.
Using a scientific calculator this comes out to $51.57$ degrees. It can also be $180 - 51.57$ i.e. $128.43$ too as $\sin \left(\pi - \theta\right) = \sin \theta$.
One can add multiples of 360 degrees too as it does not affect sine of an angle.
|
2020-02-23 11:07:28
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 11, "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": 0.9114493727684021, "perplexity": 455.57691201517235}, "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/1581875145767.72/warc/CC-MAIN-20200223093317-20200223123317-00198.warc.gz"}
|
https://optimization-online.org/2018/10/6874/
|
# Exploiting Partial Correlations in Distributionally Robust Optimization
In this paper, we identify partial correlation information structures that allow for simpler reformulations in evaluating the maximum expected value of mixed integer linear programs with random objective coefficients. To this end, assuming only the knowledge of the mean and the covariance matrix entries restricted to block-diagonal patterns, we develop a reduced semidefinite programming formulation, the complexity of solving which is related to characterizing a suitable projection of the convex hull of the set $\{(\bold{x}, \bold{x}\bold{x}'): \bold{x} \in \mathcal{X}\}$ where $\mathcal{X}$ is the feasible region. In some cases, this lends itself to efficient representations that result in polynomial-time solvable instances, most notably for the distributionally robust appointment scheduling problem with random job durations as well as for computing tight bounds in Project Evaluation and Review Technique (PERT) networks and linear assignment problems. To the best of our knowledge, this is the first example of a distributionally robust optimization formulation for appointment scheduling that permits a tight polynomial-time solvable semidefinite programming reformulation which explicitly captures partially known correlation information between uncertain processing times of the jobs to be scheduled.
## Citation
Singapore University of Technology and Design, October 2018
|
2023-01-30 08:04:18
|
{"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": 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": 0.6127621531486511, "perplexity": 370.1183732097704}, "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-2023-06/segments/1674764499804.60/warc/CC-MAIN-20230130070411-20230130100411-00692.warc.gz"}
|
https://www.tutorialspoint.com/when-a-12-v-battery-is-connected-across-an-unknown-resistor-there-is-a-current-of-2-5-ma-in-the-circuit-find-the-value-of-the-resistance-of-the-
|
# When a 12 V battery is connected across an unknown resistor, there is a current of 2.5 mA in the circuit. Find the value of the resistance of the resistor.
#### Complete Python Prime Pack for 2023
9 Courses 2 eBooks
#### Artificial Intelligence & Machine Learning Prime Pack
6 Courses 1 eBooks
#### Java Prime Pack 2023
9 Courses 2 eBooks
Given:
A $12\ V$ battery is connected across an unknown resistor, there is a current of 2.5 mA in the circuit.
To find:
To find the value of the resistance of the resistor.
Solution:
It is given that,
The potential difference, $V=12\ V$
Current, $I=2.5\ mA$
We have to find the resistance, $R=?$
By Ohm's Law, $\mathrm{V}=\mathrm{IR}$
On submitting the given values of $V$ and $I$ in the above relationship:
$12=2.5 \times 10^{-3} \times R$ [$I=2.5 \ mA= 2.5 \times 10^{-3} A$]
Or $R=\frac{12}{2.5\times10^{-3}}$
$\mathrm{R}=4.8 \times 10^{3}\ \Omega$
Therefore, the value of the resistance of the resistor is $4.8 \times 10^{3}\ \Omega$.
Updated on 10-Oct-2022 13:20:12
|
2022-12-04 18:12:57
|
{"extraction_info": {"found_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": 0.2792261242866516, "perplexity": 2784.729222987777}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-49/segments/1669446710978.15/warc/CC-MAIN-20221204172438-20221204202438-00692.warc.gz"}
|
https://dokie.li/lncs-splnproc
|
# Author Guidelines for the Preparation of Contributions to Springer Computer Science Proceedings
## Abstract
The abstract is a mandatory element that should summarize the contents of the paper and should contain at least 70 and at most 150 words. Abstract and keywords are freely available in SpringerLink.
## Keywords
• We would like to encourage you to list your keywords here. They should be separated by middots.
## Introduction
You will find here Springer’s guidelines for the preparation of proceedings papers to be published in one of the following series, in printed and electronic form:
• Lecture Notes in Computer Science (LNCS), incl. its subseries Lecture Notes in Artificial Intelligence (LNAI) and Lecture Notes in Bioinformatics (LNBI), and LNCS Transactions;
• Lecture Notes in Business Information Processing (LNBIP);
• Communications in Computer and Information Science (CCIS);
• Lecture Notes of the Institute for Computer Sciences, Social Informatics and Telecommunications Engineering (LNICST);
• IFIP Advances in Information and Communication Technology (IFIP AICT), formerly known as the IFIP Series;
• Proceedings in Information and Communication Technology (PICT).
Your contribution may be prepared in LaTeX or Microsoft Word. Technical Instructions for working with Springer’s style files and templates are provided in separate documents which can be found in the respective zip packages on our website.
## Preparation of Your Paper
### Structuring Your Paper
#### Affiliations
The affiliated institutions are to be listed directly below the names of the uthors. Multiple affiliations should be marked with superscript Arabic numbers, and they should each start on a new line as shown in this document. In addition to the nameof your affiliation, we would ask you to give the town and the country in which it is situated. If you prefer to include the entire postal address, then please feel free to do so. E-mail addresses should start on a new line and should be grouped per affiliation.
Headings should be capitalized (i.e., nouns, verbs, and all other words except articles, prepositions, and conjunctions should be set with an initial capital) and should, with the exception of the title, be aligned to the left. Only the first two levels of section headings should be numbered, as shown in Table 1. The respective font sizes are also given in Table 1. Kindly refrain from using “0” when numbering your section headings.
Font sizes of headings. Table captions should always be positioned above the tables.
Heading levelExampleFont size and style
Title (centered)Lecture Notes14 point, bold
1st-level heading1 Introduction12 point, bold
2nd-level heading2.1 Printing Area10 point, bold
3rd-level headingRun-in Heading in Bold. Text follows10 point, bold
4th-level headingLowest Level Heading. Text follows10 point, italic
Words joined by a hyphen are subject to a special rule. If the first word can stand alone, the second word should be capitalized.
Here are some examples of headings: “Criteria to Disprove Context-Freeness of Collage Languages”, “On Correcting the Intrusion of Tracing Non-deterministic Programs by Software”, “A User-Friendly and Extendable Data Distribution System”, “Multi-flip Networks: Parallelizing GenSAT”, “Self-determinations of Man”.
#### Lemmas, Propositions, and Theorems
The numbers accorded to lemmas, propositions, and theorems, etc. should appear in consecutive order, starting with Lemma 1. Please do not include section counters in the numbering like “Theorem 1.1”.
### Length of Papers
We only wish to publish papers of significant scientific content. Very short papers will be moved to the back matter, will not be made available for indexing, and will not be visible as individual papers on SpringerLink.
### Page Numbering and Running Heads
There is no need to include page numbers or running heads; this will be done at our end. If your paper title is too long to serve as a running head, it will be shortened. Your suggestion as to how to shorten it would be most welcome.
### Figures and Tables
It is essential that all illustrations are clear and legible. Vector graphics (rather than rasterized images) should be used for diagrams and schemas whenever possible. Please check that the lines in line drawings are not interrupted and have a constant width. Grids and details within the figures must be clearly legible and may not be written one on top of the other. Line drawings are to have a resolution of at least 800 dpi (preferably 1200 dpi). The lettering in figures should not use font sizes
smaller than 6 pt (~ 2 mm character height). Figures are to be numbered and to have a caption which should always be positioned under the figures, in contrast to the caption belonging to a table, which should always appear above the table.
Captions are set in 9-point type. If they are short, they are centered between the margins. Longer captions, covering more than one line, are justified (Fig. 1 and Fig. 2 show examples). Captions that do not constitute a full sentence, do not have a period.
Text fragments of fewer than four lines should not appear at the tops or bottoms of pages, following a table or figure. In such cases, it is better to set the figures right at the top or right at the bottom of the page.
If screenshots are necessary, please make sure that the essential content is clear to the reader.
##### Remark 1
In the printed volumes, illustrations are generally black and white (halftones), and only in exceptional cases, and if the author or the conference organization is prepared to cover the extra costs involved, are colored pictures accepted. Colored pictures are welcome in the electronic version free of charge. If you send colored figures that are to be printed in black and white, please make sure that they really are also legible in black and white. Some colors show up very poorly when printed in black and white.
### Formulas
Displayed equations or formulas are centered and set on a separate line (with an extra line or half line space above and below). Displayed expressions should be numbered for reference. The numbers should be consecutive within the contribution, with numbers enclosed in parentheses and set on the right margin. Please do not include section counters in the numbering.
Equations should be punctuated in the same way as ordinary text but with a small space before the end punctuation mark.
### Footnotes
The superscript numeral used to refer to a footnote appears in the text either directly after the word to be discussed or – in relation to a phrase or a sentence – following the punctuation mark (comma, semicolon, or period).1
For remarks pertaining to the title or the authors’ names, in the header of a paper, symbols should be used instead of a number (see first page of this document). Please note that no footnotes may be included in the abstract.
### Program Code
Program listings or program commands in the text are normally set in typewriter font:
### Citations and Bibliography
For citations in the text, please use square brackets and consecutive numbers. We would write [1,2,3,4,5] for consecutive numbers and [1], [3], [5] for non-consecutive numbers. The numbers in the bibliography section are without square brackets. We prefer numbered references to other styles of references, such as those with abbreviated names and years.
Please write all references using the Latin alphabet. If the title of the book you are referring to is, e.g., in Russian or Chinese, then please write (in Russian) or (in Chinese) at the end of the transcript or translation of the title.
In order to permit cross referencing within SpringerLink, and eventually between different publishers and their online databases, Springer standardizes the format of the references. This feature aims to increase the visibility of publications and facilitate academic research. Please base your references on the examples given in the references section of these instructions. References that do not adhere to this style will be reformatted at our end.
We would like to draw your attention to the fact that references to LNCS proceedings papers are particularly often reformatted due to missing editor names or incomplete publisher information. This adjustment may result in the final papers as published by Springer having more pages than the original versions as submitted by the authors. Here is an example:
Reference as formatted in author’s original version:
Assemlal, H.E., Tschumperlé, D., Brun, L.: Efficient Computation of PDF-Based Characteristics from Diffusion MR Signal. In: MICCAI. Volume 5242. (2008) 70–78
Reference after reformatting by Springer:
Assemlal, H.E., Tschumperlé, D., Brun, L.: Efficient Computation of PDF-Based Characteristics from Diffusion MR Signal. In: Metaxas, D., Axel, L., Fichtinger, G., Székely, G. (eds.) MICCAI 2008, Part II. LNCS, vol. 5242, pp. 70–78. Springer, Heidelberg (2008)
One more line is needed for this reference, as a result of Springer’s adjustment.
Please make sure that all your sources are correctly listed in the reference section.
Do not include references to pieces of work that are not connected with your paper.
The references section at the end of this paper shows a sample reference list with entries for journal articles [1], an LNCS chapter [2], a book [3], proceedings without editors [4] and [5], as well as a URL [6]. Please note that Springer proceedings are cited with their publication acronyms and volume numbers.
### Plagiarism
Springer takes plagiarism seriously. If an author has copied from another author or has used parts of another author’s work (text, tables, figures, etc.), without his or her permission and a reference, then the paper on SpringerLink will be given a “retracted” stamp, and an erratum explaining the reasons for the retraction will be included. In addition, the volume editors and the author’s academic supervisors will be informed that plagiarism has been committed. Please note that a retracted paper remains visible, with its “retracted” stamp. It does not simply disappear.
#### Acknowledgements
This should always be a run-in heading and not a section or subsection heading. It should not be assigned a number. The acknowledgements may include reference to grants or supports received in relation to the work presented in the paper.
## Additional Information Required from Authors
### Contact Author Information
Kindly assure that, when you submit the final version of your paper, you also provide the name and e-mail address of the contact author for your paper. These details are used to compile a list of contact authors for our typesetting partners SPS in India. The contact author must be available to check the paper roughly seven weeks before the start of the conference, or before the book is due to leave the printing office, in the case of post-conference proceedings. Please also mark the corresponding author in the header of the paper, preferably with a small envelope. If this is not done by you, it will be carried out by our typesetters.
### Correct Representation of Author Names
Authors’ names should be written out in full at the tops of the papers. They are shortened by us to “initials surname” in the running heads and take the form “surname, given name” in the author index. If you or any of your co-authors have more than one family name, it should be made quite clear how your name is to be displayed in the running heads and the author index. Chinese authors should write their given names in front of their surnames at the tops of their papers. If you only have one (main) name, please make sure that this name is written out in full in the running heads, when you check your final PDF. Names and affiliations cannot be changed once a paper has been published.
## Typesetting of Your Paper at Springer
Please make sure that the paper you submit is final and complete, that any copyright issues have been resolved, that the authors listed at the top of the chapter really are the final authors, and that you have not omitted any references. Following publication, it is not possible to alter or withdraw your paper on SpringerLink. Kindly note that we prefer the use of American English.
### What Will Be Done with Your Paper
If the instructions have been followed closely, then only very minor alterations will be made to your paper. The production team at SPS checks the format of the paper, and if, for example, vertical spacing has been inserted or removed, then this is remedied. In addition, running-heads, final page numbers, and a copyright line are inserted, and the capitalization of the headings is checked and corrected if need be. Finally, the reference section is attuned to our specifications (see also Section 2.7). Light technical copyediting may also be performed for post-proceedings.
### Proof Reading Stage
Once the files have been worked upon, SPS sends a copy of the final PDF of each paper to its contact author. The contact author is asked to check through the final PDF to make sure that no errors have crept in during the transfer or preparation of the files. This should not be seen as an opportunity to update or copyedit the paper, which is not possible due to time constraints. Only errors introduced during the preparation of the files will be corrected. Particular attention should be paid to the references section.
If SPS does not receive a reply from a particular contact author, within the timeframe given (usually 72 hours), then it is presumed that the author has found no errors in the paper. The tight publication schedule of our proceedings series does not allow SPS to send reminders or search for alternative e-mail addresses on the Internet.
In some cases, it is the contact volume editor or the publication chair who checks all of the PDFs. In such cases, the authors are not involved in the checking phase.
The purpose of the proof is to check for typesetting or conversion errors and the completeness and accuracy of the text, tables, and figures. Substantial changes in content, e.g., new results, corrected values, title and authorship, are not possible and cannot be processed.
## Checklist of Items to Be Sent to Volume Editor
• The final source files, incl. any non-standard fonts.
• A final PDF file corresponding exactly to the final source files.
• A copyright form, signed by one author on behalf of all of the authors of the paper.
• The name and e-mail address of the contact author who will check the proof of the paper.
• A suggestion for an abbreviated running head, if appropriate.
• Information about correct representation of authors’ names, where necessary.
## Appendix: Springer Author Discount
Authors contributing to any of Springer’s Computer Science proceedings publications are entitled to a 33.3% discount off all Springer products when placing an order through springer.com. To make use of this discount, please access the following page: `http://www.springer.com/gp/authors-editors/book-authorseditors/springertoken-request-for-springer-authors/4090`. You will be requested to give full details of your Springer publication and will be given a so-called SpringerToken. This token is a number that must be entered when placing an order through www.springer.com, in order to obtain the discount.
If you have any further questions regarding the preparation of your paper, then please do not hesitate to get in touch with us.
• For all questions related to our LaTeX style files, your contact person is:
Mr. Frank Holzwarth, e-mail: `frank.holzwarth@springer.com`.
• For overall technical questions concerning the preparation of LNCS/LNAI/LNBI papers, please contact Ms. Anna Kramer, e-mail:
`lncs@springer.com`.
• For the LNBIP series, please contact Ms. Viktoria Meyer, e-mail:
`lnbip@springer.com`.
• For the CCIS series, please contact Ms. Leonie Kunz, e-mail:
`ccis@springer.com`.
• For the LNICST series, please contact Mr. Peter Strasser, e-mail:
`lnicst@springer.com`.
• For the IFIP AICT series, please contact Ms. Erika Siebert-Cole, e-mail:
`ifip@springer.com`.
## References
1. Smith, T.F., Waterman, M.S.: Identification of Common Molecular Subsequences. J. Mol. Biol. 147, 195–197 (1981)
2. May, P., Ehrlich, H.C., Steinke, T.: ZIB Structure Prediction Pipeline: Composing a Complex Biological Workflow through Web Services. In: Nagel, W.E., Walter, W.V., Lehner, W. (eds.) Euro-Par 2006. LNCS, vol. 4128, pp. 1148–1158. Springer, Heidelberg (2006)
3. Foster, I., Kesselman, C.: The Grid: Blueprint for a New Computing Infrastructure. Morgan Kaufmann, San Francisco (1999)
4. Czajkowski, K., Fitzgerald, S., Foster, I., Kesselman, C.: Grid Information Services for Distributed Resource Sharing. In: 10th IEEE International Symposium on High Performance Distributed Computing, pp. 181–184. IEEE Press, New York (2001)
5. Foster, I., Kesselman, C., Nick, J., Tuecke, S.: The Physiology of the Grid: an Open Grid Services Architecture for Distributed Systems Integration. Technical report, Global Grid Forum (2002)
6. National Center for Biotechnology Information, http://www.ncbi.nlm.nih.gov
|
2018-04-26 12:58:32
|
{"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": 0.48320916295051575, "perplexity": 1853.715133629076}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-17/segments/1524125948214.37/warc/CC-MAIN-20180426125104-20180426145104-00000.warc.gz"}
|
https://plainmath.net/secondary/physics/force-motion-and-energy?page=3
|
# Force, motion and energy problems and answers
Recent questions in Force, Motion and Energy
vulkanere64t6 2022-05-18
### A 20kg block and a 2kg block are falling with no air resistance. What is the gravitational force on the 20kg block?
Jaiden Bowman 2022-05-18
### How to expand this equation considering acceleration due to gravity into 3D vector space?How can we expand this following equation into 3D vector space? I learned this equation from this answer: Don't heavier objects actually fall faster because they exert their own gravity?The answer shows how we can find time as function of distance (or single dimension) when there's force due to gravity accelerating both pieces of mass as follows:$t=\frac{1}{\sqrt{2G\left({m}_{1}+{m}_{2}\right)}}\left(\sqrt{{r}_{i}{r}_{f}\left({r}_{i}-{r}_{f}\right)}+{r}_{i}^{3/2}{\mathrm{cos}}^{-1}\sqrt{\frac{{r}_{f}}{{r}_{i}}}\right)$Where ${m}_{1}$ and ${m}_{2}$ are the masses of two bodies, ${r}_{i}$ is initial distance from mass to another and ${r}_{f}$ is the final distance from mass to another.If this equation had just the force caused by gravity within it, I could easily divide it into components, and use this equation for each dimension, but as seen, this equation only has masses and distances in it.How can I apply this kind of equation into 3d vector space, ie. get the position as 3d coordinates as function of time when initial position and velocity are known?If this equation cannot be applied in to 3d space, then how could we derive another equation, which applies same kind of relations into 3d vector space? (I checked the answer I linked above, but it goes somewhat above my understanding, as differential equations are used.)
dresu9dnjn 2022-05-18
### Given that the electrical force is so much stronger than gravitational force at atomic levels, why is it that it's the gravitational force between you and the earth that keeps you on the ground rather than the electrical force between you and the earth?
lurtzslikgtgjd 2022-05-17
### A setellite in Earth orbit has a mass of 91 kg and is at an altitude of $2.04×{10}^{6}m$. (Assume that U = 0 as $r\to \mathrm{\infty }$)(a) What is the potential energy of the satellite-Earth system?J(b) What is the magnitude of the gravitational force exerted by the Earth on the satellite?N(c) What force, if any, does the satellite exert on the Earth? (Enter the magnitude of the force, if there is no forN
sembuang711q6 2022-05-17
### Acceleration due to gravity on Coil vs InsulatorConsider I have a coil with a length of $5m$ and made up of copper wire with a diameter of cross-section $d$ and wrapped around an imaginary axis with radius $R$ with mass $M$Now consider an insulator made up of a glass of mass $M$ of $5m$ of diameter $D$If both were taken very high in the atmosphere but not away from the gravity of the earth. Then it is dropped from that point at the same level without providing any external force.My question is that which object will move fasterConductor. ORInsulatorI have been taught that every object of any mass will have same acceleration due to gravity but electromagnetism course taught us that a coil will produce induction and it will slow down or speed up the coil.Which theory is true?
Stoyanovahvsbh 2022-05-15
### Does GR imply a fundamental difference between gravitational and non-gravitational acceleration?1.Does the equivalence principle imply that there is some fundamental difference between acceleration due to gravity and acceleration by other means (because there is no way to 'feel' free fall acceleration for a uniform gravitational field)?2.Does General Relativity allow you to describe the acceleration due to gravity without Newton's second law (because every other source of 'push or pull' outside the nucleus involves the electromagnetic field)?3.Is the acceleration due to gravity a result of changes in time dilation/length contraction as opposed to an actual push or pull?\
Jayla Faulkner 2022-05-14
### An electron and a positron are separated by distance r. Findthe ratio of the gravitational force to the electric force betweenthem. From the result, what can you conclude concerning theforces acting between particles detected in a bubble chamber?(Should gravitational interactions be considered?)
vulkanere64t6 2022-05-14
### How is the gravitational force between two point masses affected when they are dipped in water keeping the seperation between them the same?
Dominick Blanchard 2022-05-14
### (a) Calculate the gravitational force exerted on a 5.00 kg baby by a 90 kg father 0.200 m away at birth (assisting so he is close).(b) Calculate the force on the baby due to Jupiter if it is at its closest to the earth, some $6.29×{10}^{11}m$ away, showing it to be comparable to that of the father. The mass of Jupiter is about . Other objects in the room and the hospital building also exert similar gravitational forces.
Jayden Mckay 2022-05-13
### The only known force a planet exerts on Earth is gravitational. (a) Calculate the magnitude of the gravitational force exerted on a 4.20 kg baby by a 100 kg father 0.200 m away at birth (he is assisting, so he is close to the child). (b) Calculate the magnitude of the force on the baby due to Jupiter if it is at its closest distance to Earth, some away. How does the force of Jupiter on the baby compare to the force of the father on the baby? Other objects in the room and the hospital building also exert similar gravitational forces. (Of course, there could be an unknown force acting, but scientists first need to be convinced that there is even an effect, much less that an unknown force causes it.)
Reese Estes 2022-05-13
### Two balls have their centers 2.0 m apart. One ball has a mass of 8.0 kg. The other has a mass of 6.0 kg. What is the gravitational force between them?
Elle Weber 2022-05-13
### What is the electric potential at the point indicated with the dot in the figure? Express your answer to two significant figures and include the appropriate units.photo
mars6svhym 2022-05-13
### What is the jerk due to gravity with ascent? Acceleration is defined as the rate of change of velocity with time. Jerk is defined as the rate of change of acceleration with time. What is the jerk due to gravity with ascent?
ga2t1a2dan1oj 2022-05-13
### When an object, say a ball, is attracted by the black hole it gets acceleration due to gravity. Suppose light is moving towards the black hole vertical to it... then does it gain acceleration due to gravity? If yes then won't be the speed of light increase and get violated?
Jaeden Weaver 2022-05-13
### Why do we use gravitational force in earth by relating just the mass of an object with the acceleration produced by the gravitational field:${F}_{g}=m\cdot \stackrel{\to }{g}$And when we're dealing with planets, we use a relation defined by the masses of two planets, distance squared and gravitational constant:${F}_{g}=G\cdot \frac{{M}_{1}\cdot {M}_{2}}{{d}^{2}}$I really don't get why we use just the first relation here on earth, because we're dealing with a interction between two objects... It's because our mass is irrelevant??
Zyrill Maravilla 2022-05-13
### Determine the speed a2500-kg pick-up truck should attain for it to have (a) the same momentum as a 20,000-kg bus travelling at 90-kph (b) the same kinetic energy as a 1800-kg car travelling at 100-kph
Matilda Webb 2022-05-10
### Is acceleration due to gravity constant?I was taught in school that acceleration due to gravity is constant. But recently, when I checked Physics textbook, I noted that$F=\frac{G{m}_{1}{m}_{2}}{{r}^{2}}.$So, as the body falls down, $r$ must be changing, so should acceleration due to gravity.
fetsBedscurce4why1 2022-05-10
### Why do we integrate along the whole length while finding the gravitational force between a object of mass (m) and a rod of length L and mass M? Can't we simply use $F=\frac{GmM}{{r}^{2}}\phantom{\rule{thinmathspace}{0ex}}?$?
sg101cp6vv 2022-05-10
### Two objects are attracted to each other with 36 N of gravitational force. What would the force between them be if the distance between them were doubled?
syaoronsangelhwc17 2022-05-10
|
2022-08-12 08:39:28
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 29, "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": 0.6308186650276184, "perplexity": 421.1651041971433}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-33/segments/1659882571597.73/warc/CC-MAIN-20220812075544-20220812105544-00018.warc.gz"}
|
https://eepower.com/new-industry-products/power-integrations-gan-technology-increases-output-power-of-high-efficiency-display-psus-to-75-w/
|
New Industry Products
# Power Integrations’ GaN Technology Increases Output Power of High-Efficiency Display PSUs to 75W
April 21, 2020 by Power Integrations
## Power Integrations announced that its InnoSwitch™3-MX isolated switcher IC family has been expanded with the addition of three new PowiGaN™ devices.
Higher-power TVs, monitors and appliances benefit from 91% efficiency, reduced BOM, elimination of heatsinks, smaller size, lower weight
SAN JOSE, Calif.-- Power Integrations, the leader in high-voltage integrated circuits for energy-efficient power conversion, today announced that its InnoSwitch™3-MX isolated switcher IC family has been expanded with the addition of three new PowiGaN™ devices. As part of a chipset with Power Integrations’ InnoMux™ controller IC, the new switcher ICs now supports display and appliance power supply applications with a continuous output power of up to 75 W without a heatsink.
##### Power Integrations’ GaN Technology Increases Output Power of High-Efficiency Display PSUs to 75 W. Image courtesy of Business Wire.
The InnoMux chipset employs a unique single-stage power architecture that reduces losses in display applications by 50% when compared to conventional designs, increasing overall efficiency to 91% in constant-voltage and constant-current LED backlight driver designs. Additionally, by eliminating the need for post-regulation (i.e. buck and boost) stages, TV and monitor designers can halve component count, improving reliability, and reducing manufacturing cost. With a high breakdown voltage of 750 V, the PowiGaN InnoSwitch3-MX parts are also extremely robust and highly resistant to the line surges and swells commonly-seen in regions with unstable mains voltages.
InnoSwitch3-MX flyback switcher ICs combine the primary switch, the primary-side controller, a secondary-side synchronous rectification controller, and PI’s innovative FluxLink™ high-speed communications link. The InnoSwitch3-MX receives control instructions from its chipset partner InnoMux IC, which independently measures the load requirements of each output and directs the switcher IC to deliver the right amount of power to each output, maintaining accurate regulation of current or voltage.
Comments Power Integrations’ product marketing manager Edward Ong: “By using our PowiGaN technology we are able to address higher-output applications in TVs, monitors and appliances that employ LED displays. The chipset increases efficiency beyond the requirements of all mandatory regulations and improves manufacturers’ scores in EU efficiency labeling programs.”
Samples of the INN3478C, INN3479C, INN3470C InnoSwitch3-MX ICs are available now with prices starting at $2.52,$3.14, and \$3.71 respectively in 10,000-piece quantities. Technical support for the chipset is available from the Power Integrations website link here.
|
2020-08-05 01:18:41
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.22834546864032745, "perplexity": 13869.712167226442}, "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/1596439735906.77/warc/CC-MAIN-20200805010001-20200805040001-00477.warc.gz"}
|
https://math.stackexchange.com/questions/1604171/how-many-times-do-i-loop-solovay-strassen-primality-test
|
# How many times do I loop Solovay--Strassen primality test
First, I am aware of this former thread:
math.stackexchange
Yet it doesn't answer my question.
If I want to check if an integer $n$ is prime using the Solovay--Strassen test, how many times do I have to loop over this test? As the error probability is at most $\frac1{2^k}$, one might want to choose $k$ such that $\frac1{2^k}< 10^{-10}$ or $<10^{-100}$, or whatever.
Is there a reasonable bound which is somehow comparable to the probability of a calculation error inside my cpu/pc?
Best,
reinbot
• I think, $10^{-20}$ is absolutely sufficient. If the number is very large and the test time-consuming, you can stop at $10^{-9}$, or so. If the number is "small" (lets say, $400$ digits or less) , the primilaty can be proven efficiently. – Peter Jan 8 '16 at 20:27
|
2019-06-17 05:02:09
|
{"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": 0.7415964603424072, "perplexity": 268.06012851166406}, "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/1560627998376.42/warc/CC-MAIN-20190617043021-20190617065021-00531.warc.gz"}
|
https://axelrod-tournament.readthedocs.io/en/latest/standard/index.html
|
Standard Tournament¶
Ranked violin plot¶
The mean utility of each player.
Payoffs¶
The pair wise utilities of each player.
Evolutionary dynamics¶
The evolutionary dynamic of the strategies_std (based on the utilities).
Wins¶
The number of wins of each player.
Payoff differences¶
The payoff differences for each player.
Pairwise payoff differences¶
The difference of payoffs between pairs of players.
Summary¶
Here is a file with the summary data.
|
2018-10-20 02:04:06
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.8115050196647644, "perplexity": 14327.181097958677}, "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-2018-43/segments/1539583512501.27/warc/CC-MAIN-20181020013721-20181020035221-00263.warc.gz"}
|
https://www.numerade.com/questions/illustrate-lhospitals-rule-by-graphing-both-fxgx-and-fxgx-near-x-0-to-see-that-these-ratios-have-the/
|
💬 👋 We’re always here. Join our Discord to connect with other students 24/7, any time, night or day.Join Here!
# Illustrate l'Hospital's Rule by graphing both $f(x)/g(x)$ and $f'(x)/g'(x)$ near $x = 0$ to see that these ratios have the same limit as $x \to 0$. Also, calculate the exact value of the limit.$f(x) = e^x - 1$, $g(x) = x^3 + 4x$
## From the graph, it appears that $\lim _{x \rightarrow 0} \frac{f(x)}{g(x)}=\lim _{x \rightarrow 0} \frac{f^{\prime}(x)}{g^{\prime}(x)}=0.25$$\text { We calculate } \lim _{x \rightarrow 0} \frac{f(x)}{g(x)}=\lim _{x \rightarrow 0} \frac{e^{x}-1}{x^{3}+4 x} \stackrel{\text { ? }}{=} \lim _{x \rightarrow 0} \frac{e^{x}}{3 x^{2}+4}=\frac{1}{4}$
Derivatives
Differentiation
Volume
### Discussion
You must be signed in to discuss.
Lectures
Join Bootcamp
### Video Transcript
mhm. This problem we're going to be looking at the graph as edX -1. Okay. Divided by X cubed. That's four X. So this is the graph we get um and then that's fx and gx but then F prime of X is going to be needy. X and G prime of X is going to be three X squared plus four. Um So going back, if we autographed separately, we have each of the acts divided by three X squared response for what we see. Is that the green curve? Um If we zoom in both curves approached the same number at X equal to zero And that value appears to be .25. Um And if we calculate the limit we see that that in fact is the case Because when we plug in zero here we get one over four Us, that would be .25.
California Baptist University
Derivatives
Differentiation
Volume
Lectures
Join Bootcamp
|
2021-09-20 11:39:57
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 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": 0.9025481343269348, "perplexity": 1444.5515879962998}, "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/1631780057036.89/warc/CC-MAIN-20210920101029-20210920131029-00623.warc.gz"}
|
https://socratic.org/questions/58090105b72cff1719f806c4
|
# Question #806c4
The answer is ${t}^{3} - 4 {t}^{2} + 4$
$f \left(t\right) = t + 4$
$g \left(t\right) = {t}^{3} - 4 {t}^{2}$
So $f \left(g \left(t\right)\right) = f \left({t}^{3} - 4 {t}^{2}\right) = {t}^{3} - 4 {t}^{2} + 4$
and $g \left(f \left(t\right)\right) = g \left(t + 4\right) = {\left(t + 4\right)}^{3} - 4 {\left(t + 4\right)}^{2}$
In all cases $f \left(g \left(t\right)\right) \ne g \left(f \left(t\right)\right)$
|
2021-12-02 08:17:03
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 6, "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": 0.4838116466999054, "perplexity": 3420.592292518029}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-49/segments/1637964361169.72/warc/CC-MAIN-20211202054457-20211202084457-00610.warc.gz"}
|
http://msp.org/gt/2010/14-1/p12.xhtml
|
#### Volume 14, issue 1 (2010)
Recent Issues
The Journal About the Journal Editorial Board Editorial Interests Editorial Procedure Submission Guidelines Submission Page Subscriptions Author Index To Appear Contacts ISSN (electronic): 1364-0380 ISSN (print): 1465-3060
A finitely generated, locally indicable group with no faithful action by $C^1$ diffeomorphisms of the interval
### Andrés Navas
Geometry & Topology 14 (2010) 573–584
##### Abstract
According to Thurston’s stability theorem, every group of ${C}^{1}$ diffeomorphisms of the closed interval is locally indicable (that is, every finitely generated subgroup factors through $ℤ$). We show that, even for finitely generated groups, the converse of this statement is not true. More precisely, we show that the group ${\mathbb{F}}_{2}⋉{ℤ}^{2}$, although locally indicable, does not embed into ${Diff}_{+}^{1}\left(\left(0,1\right)\right)$. (Here ${\mathbb{F}}_{2}$ is any free subgroup of $SL\left(2,ℤ\right)$, and its action on ${ℤ}^{2}$ is the linear one.) Moreover, we show that for every non-solvable subgroup $G$ of $SL\left(2,ℤ\right)$, the group $G⋉{ℤ}^{2}$ does not embed into ${Diff}_{+}^{1}\left({S}^{1}\right)$.
##### Keywords
Thurston's stability, locally indicable group
##### Mathematical Subject Classification 2000
Primary: 20B27, 37C85, 37E05
|
2016-09-25 13:55:07
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 11, "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": 0.7610896825790405, "perplexity": 838.4003086002936}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-40/segments/1474738660214.50/warc/CC-MAIN-20160924173740-00273-ip-10-143-35-109.ec2.internal.warc.gz"}
|
https://socratic.org/questions/how-do-you-write-the-polynomial-so-that-the-exponents-decrease-from-left-to-righ-1
|
# How do you write the polynomial so that the exponents decrease from left to right, identify the degree, and leading coefficient of the polynomial 2-6y?
##### 1 Answer
Dec 7, 2017
We see that $- 6 y$ is the term with the highest degree, so we place it first:
$- 6 y + 2$
The degree of the polynomial is just the biggest degree in all of the terms, which in this case is just $1$ (${y}^{1} = y$).
The leading coefficient is just the coefficient that's on the leading term (the first term), which in this case is $- 6$
|
2022-05-26 01:28:19
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 5, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7628186345100403, "perplexity": 192.04347682144672}, "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-2022-21/segments/1652662595559.80/warc/CC-MAIN-20220526004200-20220526034200-00357.warc.gz"}
|
https://www.esaral.com/q/write-the-following-sets-in-roster-from-51171/
|
Write the following sets in roster from:
Question:
Write the following sets in roster from
D = {x : x is an integer, x2 ≤ 9}
Solution:
Integers = …, -4, -3, -2, -1, 0, 1, 2, 3, 4, …
$x=-4, x^{2}=(-4)^{2}=16>9$
$x=-3, x^{2}=(-3)^{2}=9$
$x=-2, x^{2}=(-2)^{2}=4$
$x=-1, x^{2}=(-1)^{2}=1$
$x=0, x^{2}=(0)^{2}=0$
$x=1, x^{2}=(1)^{2}=1$
$x=2, x^{2}=(2)^{2}=4$
$x=3, x^{2}=(3)^{2}=9$
$x=4, x^{2}=(4)^{2}=16$
The elements of this set are -3, -2, -1, 0, 1, 2, 3
So, D = {-3, -2, -1, 0, 1, 2, 3}
|
2022-05-17 01:05:47
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.9695670008659363, "perplexity": 271.2284037418528}, "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-2022-21/segments/1652662515466.5/warc/CC-MAIN-20220516235937-20220517025937-00284.warc.gz"}
|
https://ch.gateoverflow.in/1192/gate-ch-2021-question-33
|
Match the reaction in $\textbf{Group – 1}$ with the reaction type in $\textbf{Group – 2}$.
$\begin{array}{|c|l|c|l|} \hline & \textbf{Group – 1} && \textbf{Group – 2} \\ \hline P & \text{Methylcyclohexane} \rightarrow \text{Toluene} + 3H_2 & \text{I} . & \text{Dehydrocyclization} \\ \hline Q & \text{Ethylecyclopentane}\rightarrow \text{Methylcyclohexane} & \text{II} & \text{Cracking} \\ \hline R & n-\text{Octane} \rightarrow \text{Ethylbenzene} +4H_2 & \text{III}& \text{Dehydrogenetaion} \\ \hline S & n- \text{Octane} \rightarrow n-\text{Pentane} + \text{Propylene} & \text{IV} & \text{Isomerization} \\ \hline \end{array}$
The correct combination is
1. $\text{P-II, Q-III, R-I, S-IV}$
2. $\text{P-III, Q-IV, R-I, S-II}$
3. $\text{P-III, Q-IV, R-II, S-I}$
4. $\text{P-I, Q-IV, R-III, S-II}$
in Others
edited
|
2021-12-08 19:06:40
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 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": 0.9728975296020508, "perplexity": 1982.4674964318554}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-49/segments/1637964363520.30/warc/CC-MAIN-20211208175210-20211208205210-00497.warc.gz"}
|
http://trecs.se/hyperbolicParaboloid.php
|
Current Page
Hyperbolic paraboloid
# Hyperbolic paraboloid
Object type: Surface
### Definition
In $\R^3$, a hyperbolic paraboloid is a set of points $(x,y,z)$ satisfying the equation $$z = \left(\frac{x}{a}\right)^2 - \left(\frac{y}{b}\right)^2$$ for some constants $a, b > 0$. Every intersection with a plane $z = c_0 \ne 0$ is a hyperbola, and intersections with the planes $x = 0$ and $y = 0$ yield parabolae. Below, the hyperbolic paraboloid with $a = b = 5$ is shown.
Be sure to identify the important curves obtained as intersections, as pointed out above.
### Parameterisation
The hyperbolic paraboloid is the image $\mathbf{r}(\R^2)$ where $$\mathbf{r}(x,y) = \basis\begin{pmatrix}x\\y\\\left(\frac{x}{a}\right)^2- \left(\frac{y}{b}\right)^2\end{pmatrix}.$$
### Properties
All properties given below are with respect to the parameterisation $\mathbf{r}$ given above.
#### Parameter-curve tangent vectors
The parameter-curve tangent vectors are $$\mathbf{r}_x(x,y) = \basis\begin {pmatrix}1\\0\\2x/a^2\end{pmatrix}, \quad\quad \mathbf{r}_y(x,y) = \basis\begin {pmatrix}0\\1\\-2y/b^2\end{pmatrix}.$$
#### Standard unit normal
The standard unit normal vector field is $$\mathbf{\hat{N}}(x,y) = \frac{1}{\sqrt{4x^2/a^4+ 4y^2/b^4+1}} \basis\begin{pmatrix}-2x/a^2\\2y/b^2\\1\end{pmatrix}.$$
#### Area element
The area element is $$dA = \sqrt{\frac{4x^2}{a^4} + \frac{4y^2}{b^4} + 1}~dxdy.$$
#### First fundamental form
The first fundamental form of the hyperbolic paraboloid is $$\mathcal{F}(x,y) = \begin {pmatrix}1 + \frac{4x^2}{a^4}&&-\frac{4xy}{a^2 b^2}\\-\frac{4xy}{a^2 b^2}&&1 + \frac{4y^2} {b^4}\end{pmatrix}.$$
#### Second fundamental form
The second fundamental form is $$\mathcal{M}(x,y) = \frac{2}{\sqrt{\frac{4x^2}{a^4} + \frac{4y^2}{b^4} + 1}}\begin{pmatrix}\frac{1}{a^2}&&0\\0&&-\frac{1}{b^2}\end{pmatrix}.$$
### The 'multiplication function'
The surface $$z = c~xy,$$ where $c \ne 0$ is a constant, is a hyperbolic paraboloid. More specifically, it is a hyperbolic paraboloid, as defined above, with $a = b$ and rotated 45° about the $z$-axis. The most intuitive way to see this is to perform a change of linear basis in $\R^3$, from $\basis$ to $\f$, according to $$\f = \frac{1}{\sqrt{2}} \basis\begin{pmatrix}1&&1&&0\\-1&&1&&0\\ 0&&0&&\sqrt{2}\end{pmatrix}.$$ Let the new coordinates be $X, Y, Z$. Then $$\begin{pmatrix}x\\y \\z\end{pmatrix} = \frac{1}{\sqrt{2}} \begin{pmatrix}1&&1&&0\\-1&&1&&0\\0&&0&&\sqrt{2}\end{pmatrix} \begin{pmatrix}X\\Y\\Z\end{pmatrix}.$$ Therefore, $$z = \left(\frac{x}{a}\right)^2 - \left(\frac{y} {a}\right)^2 \Leftrightarrow Z = \frac{2}{a^2} XY.$$ That is, $Z = c~XY$ with $c = 2/a^2.$ Below we give the properties of the surface $$z=xy$$ with its obvious parameterisation $$\mathbf{r}(x,y) = \basis\begin{pmatrix}x\\y\\xy\end{pmatrix}.$$
#### Parameter-curve tangent vectors
The parameter-curve tangent vectors are $$\mathbf{r}_x(x,y) = \basis\begin{pmatrix}1\\0\\y\end {pmatrix}, \quad\quad\mathbf{r}_y(x,y) = \basis\begin{pmatrix}0\\1\\x\end{pmatrix}.$$
#### Standard unit normal
The standard unit normal vector field is $$\mathbf{\hat{N}}(x,y) = \frac{1}{\sqrt{1+x^2+y^2}}\basis \begin{pmatrix}-y\\-x\\1\end{pmatrix}.$$
#### Area element
The area element is $$dA = \sqrt{1+x^2+y^2}~dxdy.$$
#### First fundamental form
The first fundamental form is $$\mathcal{F}(x,y) = \begin{pmatrix}1+y^2&&xy\\xy&&1+x^2\end{pmatrix}.$$
#### Second fundamental form
The second fundamental form is $$\mathcal{M}(x,y) = \frac{1}{\sqrt{1+x^2+y^2}}\begin{pmatrix}0&&1\\ 1&&0\end{pmatrix}.$$
#### Curvatures
The Gaussian and mean curvatures are $$K = \frac{-1}{\left(1+x^2+y^2\right)^2}, \quad\quad H = \frac {-2xy}{\left(1+x^2+y^2\right)^{3/2}}.$$
|
2018-08-18 08:24:47
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.950596272945404, "perplexity": 984.1062549704989}, "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/1534221213508.60/warc/CC-MAIN-20180818075745-20180818095745-00279.warc.gz"}
|
https://math.libretexts.org/Courses/Monroe_Community_College/MTH_098_Elementary_Algebra/3%3A_Math_Models/3.7%3A_Solve_Applications_with_Linear_Inequalities
|
# 3.7: Solve Applications with Linear Inequalities
Learning Objectives
By the end of this section, you will be able to:
• Solve applications with linear inequalities
Note
Before you get started, take this readiness quiz.
1. Write as an inequality: x is at least 30.
If you missed this problem, review Exercise 2.7.34.
2. Solve $$8−3y<41$$.
If you missed this problem, review Exercise 2.7.22.
## Solve Applications with Linear Inequalities
Many real-life situations require us to solve inequalities. In fact, inequality applications are so common that we often do not even realize we are doing algebra. For example, how many gallons of gas can be put in the car for $20? Is the rent on an apartment affordable? Is there enough time before class to go get lunch, eat it, and return? How much money should each family member’s holiday gift cost without going over budget? The method we will use to solve applications with linear inequalities is very much like the one we used when we solved applications with equations. We will read the problem and make sure all the words are understood. Next, we will identify what we are looking for and assign a variable to represent it. We will restate the problem in one sentence to make it easy to translate into an inequality. Then, we will solve the inequality. Exercise $$\PageIndex{1}$$ Emma got a new job and will have to move. Her monthly income will be$5,265. To qualify to rent an apartment, Emma’s monthly income must be at least three times as much as the rent. What is the highest rent Emma will qualify for?
$$\begin{array} {ll} {\textbf{Step 1. Read} \text{ the problem.}} &{} \\ {\textbf{Step 2. Identify} \text{ what we are looking for.}} &{\text{the highest rent Emma will qualify for}} \\ {\textbf{Step 3. Name} \text{ what we are looking for.}} &{} \\ {} &{\text{Let r = rent}} \\ {\text{Choose a variable to represent that quantity.}} &{} \\{\textbf{Step 4. Translate} \text{ into an inequality.}} &{} \\{} &{\text{Emma’s monthly income must be at least}} \\ {\text{First write a sentence that gives the information}} &{\text{three times the rent.}} \\ {\text{to find it.}} &{} \\\\ {\textbf{Step 5. Solve} \text{ the inequality.}} &{5265 \geq 3r} \\ {\text{Remember, } a > x\text{ has the same meaning}} &{1755 \geq r} \\ {\text{as }x < a} &{r \leq 1755} \\ {\textbf{Step 6. Check} \text{ the answer in the problem}} &{} \\ {\text{and make sure it makes sense.}} &{} \\ {\text{A maximum rent of \1,755 seems}} &{} \\ {\text{reasonable for an income of \5,265.}} &{} \\ {\textbf{Step 7. Answer} \text{ the answer in the problem}} &{\text{the question with a}} \\ {\text{complete sentence.}} &{\text{The maximum rent is \1,755.}} \end{array}$$
Exercise $$\PageIndex{2}$$
Alan is loading a pallet with boxes that each weighs 45 pounds. The pallet can safely support no more than 900 pounds. How many boxes can he safely load onto the pallet?
There can be no more than 20 boxes.
Exercise $$\PageIndex{3}$$
The elevator in Yehire’s apartment building has a sign that says the maximum weight is 2,100 pounds. If the average weight of one person is 150 pounds, how many people can safely ride the elevator?
A maximum of 14 people can safely ride in the elevator.
Sometimes an application requires the solution to be a whole number, but the algebraic solution to the inequality is not a whole number. In that case, we must round the algebraic solution to a whole number. The context of the application will determine whether we round up or down. To check applications like this, we will round our answer to a number that is easy to compute with and make sure that number makes the inequality true.
Exercise $$\PageIndex{4}$$
Dawn won a mini-grant of $4,000 to buy tablet computers for her classroom. The tablets she would like to buy cost$254.12 each, including tax and delivery. What is the maximum number of tablets Dawn can buy?
$$\begin{array} {ll} {\textbf{Step 1. Read} \text{ the problem.}} &{} \\ {\textbf{Step 2. Identify} \text{ what we are looking for.}} &{\text{the maximum number of tablets Dawn can buy}} \\ {\textbf{Step 3. Name} \text{ what we are looking for.}} &{} \\ {} &{\text{Let n = the number of tablets.}} \\ {\text{Choose a variable to represent that quantity.}} &{} \\{\textbf{Step 4. Translate.} \text{ write a sentence that}} &{} \\{\text{gives the information to find it.}} &{254.12\text{ times the number of tablets is no}} \\ {} &{\text{more than \4000.}} \\ {\text{Translate into an inequality.}} &{254.12n \leq 4000} \\ {\textbf{Step 5. Solve} \text{ the inequality.}} &{n \leq 15.74} \\ {\text{But n must be a whole number of tablets,}} &{} \\ {\text{so round to 15.}} &{n \leq 15}\\ \\{\textbf{Step 6. Check} \text{ the answer in the problem}} &{} \\ {\text{and make sure it makes sense.}} &{} \\ {\text{Rounding down the price to \250,}} &{} \\ {\text{15 tablets would cost \3750, while}} &{} \\ {\text{16 tablets would be \4000. So a}} &{} \\{\text{maximum of 15 tablets at \254.12}} &{} \\ {\text{seems reasonable.}} &{} \\{\textbf{Step 7. Answer} \text{ the answer in the problem}} &{\text{the question with a}} \\ {\text{complete sentence.}} &{\text{Dawn can buy a maximum of 15 tablets.}} \end{array}$$
Exercise $$\PageIndex{5}$$
Angie has $20 to spend on juice boxes for her son’s preschool picnic. Each pack of juice boxes costs$2.63. What is the maximum number of packs she can buy?
seven packs
Exercise $$\PageIndex{6}$$
Daniel wants to surprise his girlfriend with a birthday party at her favorite restaurant. It will cost $42.75 per person for dinner, including tip and tax. His budget for the party is$500. What is the maximum number of people Daniel can have at the party?
11 people
Exercise $$\PageIndex{7}$$
Pete works at a computer store. His weekly pay will be either a fixed amount, $925, or$500 plus 12% of his total sales. How much should his total sales be for his variable pay option to exceed the fixed amount of $925? Answer $$\begin{array} {ll} {\textbf{Step 1. Read} \text{ the problem.}} &{} \\ {\textbf{Step 2. Identify} \text{ what we are looking for.}} &{\text{the total sales needed for his variable pay}} \\ {} &{\text{option to exceed the fixed amount of \925}} \\ {\textbf{Step 3. Name} \text{ what we are looking for.}} &{} \\ {} &{\text{Let s = the total sales.}} \\ {\text{Choose a variable to represent that quantity.}} &{} \\{\textbf{Step 4. Translate.} \text{ write a sentence that}} &{} \\{\text{gives the information to find it.}} &{500\text{ plus 12% of total sales is more than \925.}} \\ {\text{Translate into an inequality. Remember to}} &{500 + 0.12s > 925} \\{\text{convert the percent to a decimal.}} &{} \\\\ {\textbf{Step 5. Solve} \text{ the inequality.}} &{0.12s > 425} \\ {} &{s > 3541.\overline{66}} \\ \\ \\{\textbf{Step 6. Check} \text{ the answer in the problem}} &{} \\ {\text{and make sure it makes sense.}} &{} \\ {\text{Rounding down the price to \250,}} &{} \\ {\text{15 tablets would cost \3750, while}} &{} \\ {\text{If we round the total sales up to}} &{} \\{\text{\4000, we see that}} &{} \\ {\text{500+0.12(4000) = 980, which is more}} &{} \\ {\text{than \925.}} &{} \\{\textbf{Step 7. Answer} \text{ the the question with a complete sentence.}} &{\text{The total sales must be more than \3541.67}} \end{array}$$ Exercise $$\PageIndex{8}$$ Tiffany just graduated from college and her new job will pay her$20000 per year plus 2% of all sales. She wants to earn at least $100000 per year. For what total sales will she be able to achieve her goal? Answer at least$4000000
Exercise $$\PageIndex{9}$$
Christian has been offered a new job that pays $24000 a year plus 3% of sales. For what total sales would this new job pay more than his current job which pays$60000?
at least $1200000 Exercise $$\PageIndex{10}$$ Sergio and Lizeth have a very tight vacation budget. They plan to rent a car from a company that charges$75 a week plus $0.25 a mile. How many miles can they travel and still keep within their$200 budget?
$$\begin{array} {ll} {\textbf{Step 1. Read} \text{ the problem.}} &{} \\ {\textbf{Step 2. Identify} \text{ what we are looking for.}} &{\text{the number of miles Sergio and Lizeth can travel}} \\ {\textbf{Step 3. Name} \text{ what we are looking for.}} &{} \\ {} &{\text{Let m = the number of miles.}} \\ {\text{Choose a variable to represent that quantity.}} &{} \\{\textbf{Step 4. Translate.} \text{ write a sentence that}} &{\text{\75 plus 0.25 times the number of miles is}} \\{\text{gives the information to find it.}} &{\text{ less than or equal to \200.}} \\ {\text{Translate into an inequality. }} &{75 + 25m \leq 200} \\\\ {\textbf{Step 5. Solve} \text{ the inequality.}} &{0.25m \leq 125} \\ {} &{m \leq 500 \text{ miles}} \\ \\ \\{\textbf{Step 6. Check} \text{ the answer in the problem}} &{} \\ {\text{and make sure it makes sense.}} &{} \\ {\text{Yes, 75 + 0.25(500) = 200.}} & {}\\{\textbf{Step 7. Answer} \text{ the the question with a complete sentence.}} &{\text{Sergio and Lizeth can travel 500 miles}} \\ {} &{\text{and still stay on budget.}} \end{array}$$
Exercise $$\PageIndex{11}$$
Taleisha’s phone plan costs her $28.80 a month plus$0.20 per text message. How many text messages can she use and keep her monthly phone bill no more than $50? Answer no more than 106 text messages Exercise $$\PageIndex{12}$$ Rameen’s heating bill is$5.42 per month plus $1.08 per therm. How many therms can Rameen use if he wants his heating bill to be a maximum of$87.50?
no more than 76 therms
A common goal of most businesses is to make a profit. Profit is the money that remains when the expenses have been subtracted from the money earned. In the next example, we will find the number of jobs a small businessman needs to do every month in order to make a certain amount of profit.
Exercise $$\PageIndex{13}$$
Elliot has a landscape maintenance business. His monthly expenses are $1,100. If he charges$60 per job, how many jobs must he do to earn a profit of at least $4,000 a month? Answer $$\begin{array} {ll} {\textbf{Step 1. Read} \text{ the problem.}} &{} \\ {\textbf{Step 2. Identify} \text{ what we are looking for.}} &{\text{the number of jobs Elliot needs}} \\ {\textbf{Step 3. Name} \text{ what we are looking for.}} &{} \\ {\text{Choose a variable to represent it}} &{\text{Let j = the number of jobs.}} \\{\textbf{Step 4. Translate.} \text{ write a sentence that}} &{\text{\60 times the number of jobs minus \1,100 is at least \4,000.}} \\{\text{gives the information to find it.}} &{\text{ less than or equal to \200.}} \\ {\text{Translate into an inequality. }} &{60j - 1100 \geq 4000} \\\\ {\textbf{Step 5. Solve} \text{ the inequality.}} &{60j \geq 5100} \\ {} &{j \geq 85\text{ jobs}} \\ \\{\textbf{Step 6. Check} \text{ the answer in the problem}} &{} \\ {\text{and make sure it makes sense.}} &{} \\ {\text{If Elliot did 90 jobs, his profit would be}} & {}\\ {\text{60(90)−1,100,or \4,300. This is}} &{} \\ {\text{more than \4,000.}} &{} \\{\textbf{Step 7. Answer} \text{ the the question with a complete sentence.}} &{\text{Elliot must work at least 85 jobs.}} \end{array}$$ Exercise $$\PageIndex{14}$$ Caleb has a pet sitting business. He charges$32 per hour. His monthly expenses are $2272. How many hours must he work in order to earn a profit of at least$800 per month?
at least 96 hours
Exercise $$\PageIndex{15}$$
Felicity has a calligraphy business. She charges $2.50 per wedding invitation. Her monthly expenses are$650. How many invitations must she write to earn a profit of at least $2800 per month? Answer at least 1380 invitations Sometimes life gets complicated! There are many situations in which several quantities contribute to the total expense. We must make sure to account for all the individual expenses when we solve problems like this. Exercise $$\PageIndex{16}$$ Brenda’s best friend is having a destination wedding and the event will last 3 days. Brenda has$500 in savings and can earn $15 an hour babysitting. She expects to pay$350 airfare, $375 for food and entertainment and$60 a night for her share of a hotel room. How many hours must she babysit to have enough money to pay for the trip?
$$\begin{array} {ll} {\textbf{Step 1. Read} \text{ the problem.}} &{} \\ {\textbf{Step 2. Identify} \text{ what we are looking for.}} &{\text{the number of hours Brenda must babysit}} \\ {\textbf{Step 3. Name} \text{ what we are looking for.}} &{} \\ {\text{Choose a variable to represent that quantity.}} &{\text{Let h = the number of hours.}} \\{\textbf{Step 4. Translate.} \text{ write a sentence that}} &{} \\{\text{gives the information to find it.}} &{} \\ {} &{\text{The expenses must be less than or equal to}} \\ {} &{\text{the income. The cost of airfare plus the}} \\ {} &{\text{cost of food and entertainment and the}} \\ {} &{\text{hotel bill must be less than or equal to the savings}} \\ {} &{\text{plus the amount earned babysitting.}} \\ {\text{Translate into an inequality. }} &{\350 + \375 + \60(3) \leq \500 + \15h} \\\\ {\textbf{Step 5. Solve} \text{ the inequality.}} &{905 \leq 500 + 15h} \\{} &{405 \leq 15h} \\ {} &{27 \leq h} \\ {} &{h \geq 27} \\ \\{\textbf{Step 6. Check} \text{ the answer in the problem}} &{} \\ {\text{and make sure it makes sense.}} &{} \\ {\text{We substitute 27 into the inequality.}} & {}\\{905 \leq 500 + 15h} &{} \\ {905 \leq 500 + 15(27)} &{} \\ {905 \leq 905} &{} \\ \\{\textbf{Step 7. Answer} \text{ the the question with a complete sentence.}} &{\text{Brenda must babysit at least 27 hours.}} \end{array}$$
Exercise $$\PageIndex{17}$$
Malik is planning a 6-day summer vacation trip. He has $840 in savings, and he earns$45 per hour for tutoring. The trip will cost him $525 for airfare,$780 for food and sightseeing, and $95 per night for the hotel. How many hours must he tutor to have enough money to pay for the trip? Answer at least 23 hours Exercise $$\PageIndex{18}$$ Josue wants to go on a 10-day road trip next spring. It will cost him$180 for gas, $450 for food, and$49 per night for a motel. He has $520 in savings and can earn$30 per driveway shoveling snow. How many driveways must he shovel to have enough money to pay for the trip?
|
2021-05-15 21:07:07
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9985898733139038, "perplexity": 4293.710973521036}, "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/1620243991378.52/warc/CC-MAIN-20210515192444-20210515222444-00120.warc.gz"}
|
https://zbmath.org/?q=rf%3A7429840
|
## Some new Fourier and Jackson-Nikol’skii type inequalities in unbounded orthonormal systems.(English)Zbl 07528442
Summary: We consider the generalized Lorentz space $$L_{\psi,q}$$ defined via a continuous and concave function $$\psi$$ and the Fourier series of a function with respect to an unbounded orthonormal system. Some new Fourier and Jackson-Nikol’skii type inequalities in this frame are stated, proved and discussed. In particular, the derived results generalize and unify several well-known results but also some new applications are pointed out.
### MSC:
42A16 Fourier coefficients, Fourier series of functions with special properties, special Fourier series 42B05 Fourier series and coefficients in several variables 42C15 General harmonic expansions, frames 42C10 Fourier series in special orthogonal functions (Legendre polynomials, Walsh functions, etc.) 26D15 Inequalities for sums, series and integrals 26D20 Other analytical inequalities 46E30 Spaces of measurable functions ($$L^p$$-spaces, Orlicz spaces, Köthe function spaces, Lorentz spaces, rearrangement invariant spaces, ideal spaces, etc.)
Full Text:
|
2022-10-05 12:35:11
|
{"extraction_info": {"found_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": 0.9376723766326904, "perplexity": 1558.3402147098982}, "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/1664030337625.5/warc/CC-MAIN-20221005105356-20221005135356-00163.warc.gz"}
|
http://datatiles.ai/?page_id=301
|
NaiveBayes_Algo
# Naive Bayes¶
To demonstrate the concept of Naïve Bayes Classification, consider the example displayed in the illustration above. As indicated, the objects can be classified as either GREEN or RED. Our task is to classify new cases as they arrive, i.e., decide to which class label they belong, based on the currently exiting objects.
In [7]:
from IPython.display import Image
Image(filename='NB3.png')
Out[7]:
Since there are twice as many GREEN objects as RED, it is reasonable to believe that a new case (which hasn't been observed yet) is twice as likely to have membership GREEN rather than RED.
In the Bayesian analysis, this belief is known as the prior probability. Prior probabilities are based on previous experience, in this case the percentage of GREEN and RED objects, and often used to predict outcomes before they actually happen. Thus, we can write:
Prior Probability of GREEN: number of GREEN objects / total number of objects
Prior Probability of RED: number of RED objects / total number of objects
Since there is a total of 60 objects, 40 of which are GREEN and 20 RED, our prior probabilities for class membership are:
Prior Probability for GREEN: 40 / 60
Prior Probability for RED: 20 / 60
Having formulated our prior probability, we are now ready to classify a new object (WHITE circle in the diagram below). Since the objects are well clustered, it is reasonable to assume that the more GREEN (or RED) objects in the vicinity of X, the more likely that the new cases belong to that particular color. To measure this likelihood, we draw a circle around X which encompasses a number (to be chosen a priori) of points irrespective of their class labels
In [8]:
from IPython.display import Image
Image(filename='NB4.png')
Out[8]:
In [9]:
from IPython.display import Image
Image(filename='NB5.png')
Out[9]:
From the illustration above, it is clear that Likelihood of X given GREEN is smaller than Likelihood of X given RED, since the circle encompasses 1 GREEN object and 3 RED ones. Thus:
In [10]:
from IPython.display import Image
Image(filename='NB6.png')
Out[10]:
Although the prior probabilities indicate that X may belong to GREEN (given that there are twice as many GREEN compared to RED) the likelihood indicates otherwise; that the class membership of X is RED (given that there are more RED objects in the vicinity of X than GREEN). In the Bayesian analysis, the final classification is produced by combining both sources of information, i.e., the prior and the likelihood, to form a posterior probability using the so-called Bayes' rule (named after Rev. Thomas Bayes 1702-1761).
In [11]:
from IPython.display import Image
Image(filename='NB7.png')
Out[11]:
Finally, we classify X as RED since its class membership achieves the largest posterior probability.
source: based on StatSoft; the Electronic Statistics Textbook as a public service since 1995
# Email Spam Classifier - Spam or ham?¶
We will use sklearn.naive_bayes to train a spam classifier!
In [12]:
import os
import io
import numpy
from pandas import DataFrame
from sklearn.feature_extraction.text import CountVectorizer
from sklearn.naive_bayes import MultinomialNB
for root, dirnames, filenames in os.walk(path):
for filename in filenames:
path = os.path.join(root, filename)
inBody = False
lines = []
f = io.open(path, 'r', encoding='latin1')
for line in f:
if inBody:
lines.append(line)
elif line == '\n':
inBody = True
f.close()
message = '\n'.join(lines)
yield path, message
def dataFrameFromDirectory(path, classification):
rows = []
index = []
rows.append({'message': message, 'class': classification})
index.append(filename)
return DataFrame(rows, index=index)
data = DataFrame({'message': [], 'class': []})
Let's have a look at that DataFrame:
In [13]:
data.head()
Out[13]:
class message
/Users/sudhirwadhwa/Desktop/INTELFINALBUNDLE/Day2_NaiveBayes_Algo/emails/spam/00001.7848dde101aa985090474a91ec93fcf0 spam <!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.0 Tr...
/Users/sudhirwadhwa/Desktop/INTELFINALBUNDLE/Day2_NaiveBayes_Algo/emails/spam/00005.57696a39d7d84318ce497886896bf90d spam I thought you might like these:\n\n1) Slim Dow...
In [14]:
data
Out[14]:
class message
/Users/sudhirwadhwa/Desktop/INTELFINALBUNDLE/Day2_NaiveBayes_Algo/emails/spam/00001.7848dde101aa985090474a91ec93fcf0 spam <!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.0 Tr...
/Users/sudhirwadhwa/Desktop/INTELFINALBUNDLE/Day2_NaiveBayes_Algo/emails/spam/00005.57696a39d7d84318ce497886896bf90d spam I thought you might like these:\n\n1) Slim Dow...
/Users/sudhirwadhwa/Desktop/INTELFINALBUNDLE/Day2_NaiveBayes_Algo/emails/spam/00007.d8521faf753ff9ee989122f6816f87d7 spam Help wanted. We are a 14 year old fortune 500...
/Users/sudhirwadhwa/Desktop/INTELFINALBUNDLE/Day2_NaiveBayes_Algo/emails/spam/00009.027bf6e0b0c4ab34db3ce0ea4bf2edab spam TIRED OF THE BULL OUT THERE?\n\nWant To Stop L...
/Users/sudhirwadhwa/Desktop/INTELFINALBUNDLE/Day2_NaiveBayes_Algo/emails/spam/00012.381e4f512915109ba1e0853a7a8407b2 spam <table width="600" border="20" align="center" ...
/Users/sudhirwadhwa/Desktop/INTELFINALBUNDLE/Day2_NaiveBayes_Algo/emails/spam/00016.67fb281761ca1051a22ec3f21917e7c0 spam \n\n\n\nWant to watch Sporting Events?--Movies...
/Users/sudhirwadhwa/Desktop/INTELFINALBUNDLE/Day2_NaiveBayes_Algo/emails/spam/00017.1a938ecddd047b93cbd7ed92c241e6d1 spam Help wanted. We are a 14 year old fortune 500...
/Users/sudhirwadhwa/Desktop/INTELFINALBUNDLE/Day2_NaiveBayes_Algo/emails/spam/00018.5b2765c42b7648d41c93b9b27140b23a spam DEAR FRIEND,I AM MRS. SESE-SEKO WIDOW OF LATE...
/Users/sudhirwadhwa/Desktop/INTELFINALBUNDLE/Day2_NaiveBayes_Algo/emails/spam/00024.6b5437b14d403176c3f046c871b5b52f spam This is a multi-part message in MIME format.\n...
... ... ...
/Users/sudhirwadhwa/Desktop/INTELFINALBUNDLE/Day2_NaiveBayes_Algo/emails/ham/02471.18281d43dc0775e915267c2ea5170f1f ham This is possible, however using SA as a block ...
/Users/sudhirwadhwa/Desktop/INTELFINALBUNDLE/Day2_NaiveBayes_Algo/emails/ham/02474.c76ffef81a2529389e6c3bbb172184d7 ham \n\n> Mr Tim Chapman, freelance gentleman of l...
/Users/sudhirwadhwa/Desktop/INTELFINALBUNDLE/Day2_NaiveBayes_Algo/emails/ham/02475.9277ee243e3f51fa53ed6be55798d360 ham Smith, Graham - Computing Technician wrote:\n\...
/Users/sudhirwadhwa/Desktop/INTELFINALBUNDLE/Day2_NaiveBayes_Algo/emails/ham/02476.de1d459426662492dd1235046b504c3d ham Geege wrote a strange story:\n\n>I know a guy ...
/Users/sudhirwadhwa/Desktop/INTELFINALBUNDLE/Day2_NaiveBayes_Algo/emails/ham/02477.07b2069e9827cfd6f97d07eea2913d57 ham \n\n[Paul Moore]\n\n> but let's walk before...
/Users/sudhirwadhwa/Desktop/INTELFINALBUNDLE/Day2_NaiveBayes_Algo/emails/ham/02478.40723f38488bddaf5a24ef2a91679c75 ham On Mon, Nov 25, 2002 at 06:54:49PM +0000, Phil...
/Users/sudhirwadhwa/Desktop/INTELFINALBUNDLE/Day2_NaiveBayes_Algo/emails/ham/02480.72714df60c9be29d6f7985c777cbfc13 ham No, you need to learn how declarations work in...
/Users/sudhirwadhwa/Desktop/INTELFINALBUNDLE/Day2_NaiveBayes_Algo/emails/ham/02481.176b368fe4b90682f33647d65a8b97a3 ham \n\n Richie> As I understand it, post-1.8x ...
/Users/sudhirwadhwa/Desktop/INTELFINALBUNDLE/Day2_NaiveBayes_Algo/emails/ham/02483.ab1bee02c10ddecc0e86c39eaebc2996 ham The Times\n\n\n\n \n\n December 04, 2002 \n\n ...
/Users/sudhirwadhwa/Desktop/INTELFINALBUNDLE/Day2_NaiveBayes_Algo/emails/ham/02484.32a0bca2600788be144b93cae341efbf ham I have to say I was surprised about Jacko dang...
/Users/sudhirwadhwa/Desktop/INTELFINALBUNDLE/Day2_NaiveBayes_Algo/emails/ham/02486.bdf90e871b673fd14f47f3fe36622742 ham What the hell is it with these mini remote con...
/Users/sudhirwadhwa/Desktop/INTELFINALBUNDLE/Day2_NaiveBayes_Algo/emails/ham/02489.85c20a6f9d75714d9f44398baeddd416 ham Joe McNally writes:\n\n\n\n> What the hell is ...
/Users/sudhirwadhwa/Desktop/INTELFINALBUNDLE/Day2_NaiveBayes_Algo/emails/ham/02491.c26245be2a5096fa86647d594561c511 ham Hi all.\n\nDoes anyone know how to set up dual...
/Users/sudhirwadhwa/Desktop/INTELFINALBUNDLE/Day2_NaiveBayes_Algo/emails/ham/02492.6aede44f654a1bbc60c95c7dd770e624 ham Carlos Luna wrote:\n\n\n\n>Hi all.\n\n>Does an...
/Users/sudhirwadhwa/Desktop/INTELFINALBUNDLE/Day2_NaiveBayes_Algo/emails/ham/02493.f9f2870094430b7db8b0c1052b302cf1 ham Hi all\n\n\n\n\n\nI have a prob when trying to...
/Users/sudhirwadhwa/Desktop/INTELFINALBUNDLE/Day2_NaiveBayes_Algo/emails/ham/02494.a14f2d3a9bef3f59aa419b03aee8f871 ham Tim Chapman writes:\n\n\n\n> http://news.bbc.c...
/Users/sudhirwadhwa/Desktop/INTELFINALBUNDLE/Day2_NaiveBayes_Algo/emails/ham/02495.5064946e77b3046873da91fc47656465 ham > I had the same problem when installing Win o...
/Users/sudhirwadhwa/Desktop/INTELFINALBUNDLE/Day2_NaiveBayes_Algo/emails/ham/02496.aae0c81581895acfe65323f344340856 ham Man killed 'trying to surf' on Tube train \n\n...
/Users/sudhirwadhwa/Desktop/INTELFINALBUNDLE/Day2_NaiveBayes_Algo/emails/ham/02497.60497db0a06c2132ec2374b2898084d3 ham Hi Gianni,\n\n\n\nA very good resource for thi...
/Users/sudhirwadhwa/Desktop/INTELFINALBUNDLE/Day2_NaiveBayes_Algo/emails/ham/02498.09835f512f156da210efb99fcc523e21 ham Gianni Ponzi wrote:\n\n> I have a prob when tr...
/Users/sudhirwadhwa/Desktop/INTELFINALBUNDLE/Day2_NaiveBayes_Algo/emails/ham/02499.b4af165650f138b10f9941f6cc5bce3c ham Neale Pickett <neale@woozle.org> writes:\n\n\n...
/Users/sudhirwadhwa/Desktop/INTELFINALBUNDLE/Day2_NaiveBayes_Algo/emails/ham/02500.05b3496ce7bca306bed0805425ec8621 ham \n\nHi,\n\n\n\nI think you need to give us a l...
3000 rows × 2 columns
Now we will use a CountVectorizer to split up each message into its list of words, and throw that into a MultinomialNB classifier. Call fit() and we've got a trained spam filter ready to go! It's just that easy.
In [15]:
vectorizer = CountVectorizer()
counts = vectorizer.fit_transform(data['message'].values)
In [16]:
counts
Out[16]:
<3000x62964 sparse matrix of type '<type 'numpy.int64'>'
with 429785 stored elements in Compressed Sparse Row format>
In [17]:
classifierModel = MultinomialNB()
## This is the target
## class is the Target
targets = data['class'].values
## Using counts
classifierModel.fit(counts, targets)
Out[17]:
MultinomialNB(alpha=1.0, class_prior=None, fit_prior=True)
Try these example emails.
In [18]:
examples = ['Free Viagra now!!!',
"A quick brown fox is not ready",
"Could you bring me the black coffee as well?",
"Hi Bob, how about a game of golf tomorrow, are you FREE?",
"I am FREE now, you can come",
"FREE FREE FREE Sex",
"CENTRAL BANK OF NIGERIA has 100 Million for you",
"I am not available today, meet sunday?"]
example_counts = vectorizer.transform(examples)
# print ( example_counts)
predictions = classifierModel.predict(example_counts)
predictions
Out[18]:
array(['spam', 'ham', 'ham', 'ham', 'ham', 'spam', 'spam', 'ham'],
dtype='|S4')
|
2018-09-20 15:32:02
|
{"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": 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": 0.5310406684875488, "perplexity": 7525.843640176777}, "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/1537267156513.14/warc/CC-MAIN-20180920140359-20180920160759-00070.warc.gz"}
|
https://sherpa.readthedocs.io/en/4.12.1/ui/api/sherpa.astro.ui.get_data_plot_prefs.html
|
# get_data_plot_prefs¶
sherpa.astro.ui.get_data_plot_prefs()
Return the preferences for plot_data.
Returns: prefs – Changing the values of this dictionary will change any new data plots. This dictionary will be empty if no plot backend is available. dict
plot_data()
Plot the data values.
set_xlinear()
New plots will display a linear X axis.
set_xlog()
New plots will display a logarithmically-scaled X axis.
set_ylinear()
New plots will display a linear Y axis.
set_ylog()
New plots will display a logarithmically-scaled Y axis.
Notes
The meaning of the fields depend on the chosen plot backend. A value of None means to use the default value for that attribute, unless indicated otherwise. These preferences are used by the following commands: plot_data, plot_bkg, plot_ratio, and the “fit” variants, such as plot_fit, plot_fit_resid, and plot_bkg_fit.
The following preferences are recognized by the matplotlib backend:
barsabove
The barsabove argument for the matplotlib errorbar function.
capsize
The capsize argument for the matplotlib errorbar function.
color
The color to use (will be over-ridden by more-specific options below). The default is None.
ecolor
The color to draw error bars. The default is None.
linecolor
What color to use for the line connecting the data points. The default is None.
linestyle
How should the line connecting the data points be drawn. The default is ‘None’, which means no line is drawn.
marker
What style is used for the symbols. The default is '.' which indicates a point.
markerfacecolor
What color to draw the symbol representing the data points. The default is None.
markersize
What size is the symbol drawn. The default is None.
ratioline
Should a horizontal line be drawn at y=1? The default is False.
xaxis
The default is False
xerrorbars
Should error bars be drawn for the X axis. The default is False.
xlog
Should the X axis be drawn with a logarithmic scale? The default is False. This field can also be changed with the set_xlog and set_xlinear functions.
yerrorbars
Should error bars be drawn for the Y axis. The default is True.
ylog
Should the Y axis be drawn with a logarithmic scale? The default is False. This field can also be changed with the set_ylog and set_ylinear functions.
Examples
After these commands, any data plot will use a green symbol and not display Y error bars.
>>> prefs = get_data_plot_prefs()
>>> prefs['color'] = 'green'
>>> prefs['yerrorbars'] = False
|
2022-01-19 22:55:42
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5289059281349182, "perplexity": 3475.353312707133}, "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-2022-05/segments/1642320301592.29/warc/CC-MAIN-20220119215632-20220120005632-00122.warc.gz"}
|
https://math.stackexchange.com/questions/3340648/are-cumulative-distribution-function-for-a-continuous-random-variable-left-conti/3340652
|
# Are Cumulative Distribution Function for a Continuous Random Variable Left Continuous also?
I was reading the book "probability random variables and stochastic processes by Athanasios Papoulis - Third Edition" where by in the properties of distribution functions on Page 69 of the book the fifth(5th) point mentions that Cumulative Distribution Function are right continuous in general(for discrete and continuous random variable). But I have a doubt whether we can say that:
Cumulative Distribution Function are both left and right continuous for continuous random variables in general or not?
• The CDF $F$ of a continuous distribution is not only continuous, but also absolutely continuous, in the sense that $$F(x) = \int_{-\infty}^{x} f(t) \, \mathrm{d}t$$ for some integrable function $f$. In this case, of course $f$ is the PDF of the distribution. For the record, note that there are distributions which are not continuous but have continuous CDFs. (The quintessential example is Cantor distribution.) – Sangchul Lee Sep 1 '19 at 4:14
• @SangchulLee : There is a question of convention: Should one define "continuous distribution" simply to mean one whose c.d.f. is continuous, or should one define it to mean a distribution for which there is a density? If the later, then you are right. I have come to think that the former is the better convention. Either way, there are some distributions for which the c.d.f. is continuous but the distribution neither has a density nor is a mixture (i.e. a weighted average) of any distributions with densities and any other distributions. – Michael Hardy Jan 1 at 14:53
The cumulative distribution function for a continuous random variable $$X$$ is continuous since $$X$$ admits a density $$f$$. Hence $$F$$ is of the form $$F(x)=\int_{-\infty}^xf(t)\,dt$$ which is continuous.
|
2020-04-04 00:20:22
|
{"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": 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": 5, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8701109290122986, "perplexity": 217.88542393483306}, "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/1585370518767.60/warc/CC-MAIN-20200403220847-20200404010847-00075.warc.gz"}
|
https://www.physicsforums.com/threads/proof-of-cauchy-criterion-for-riemann-integrals.392046/
|
# Proof of Cauchy Criterion for Riemann Integrals
1. Apr 3, 2010
### S.N.
1. The problem statement, all variables and given/known data
Some proofs I've looked at vary, but they generally follow the format show here: http://en.wikibooks.org/wiki/Real_Analysis/Riemann_integration#Theorem_.28Cauchy_Criterion.29
This isn't a question about an exercise, but rather a request for a clarification or a way of putting part of the proof in more understandable terms. My issue is the <= direction of the proof. I don't understand much about it. We're picking a $$\delta$$, sure, but then I don't see where the 1/n comes from, or what the business with m<n is, or just about anything in the 'reverse direction' of this proof. Does anyone have maybe a link to a proof with more heuristics, or if you have the time and patience, maybe a more intuitive way of describing that's going on?
Thanks for any help.
|
2017-11-21 18:25:26
|
{"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": 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": 0.3909260332584381, "perplexity": 404.43844318415677}, "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/1510934806421.84/warc/CC-MAIN-20171121170156-20171121190156-00404.warc.gz"}
|
https://de.maplesoft.com/support/help/maplesim/view.aspx?path=DocumentTools/Canvas/StaticPlot&L=G
|
DocumentTools/Canvas/StaticPlot - Maple Help
DocumentTools[Canvas]
StaticPlot
create a canvas plot
Calling Sequence StaticPlot(expression) StaticPlot(expression, options)
Parameters
expression - anything attributes - (optional) name=value pairs
Description
• The StaticPlot command is used to create a plot that will be rendered inline in a shared canvas.
Note: Currently StaticPlot is not supported by ShowCanvas and GetMath, but can be used with ShareCanvas.
Examples
> $\mathrm{with}\left(\mathrm{DocumentTools}:-\mathrm{Canvas}\right):$
> $\mathrm{cv}≔\mathrm{NewCanvas}\left(\left["Plot Example",\mathrm{Text}\left("A plot of: %1",{\left(x+1\right)}^{2}-1\right),\mathrm{StaticPlot}\left({\left(x+1\right)}^{2}-1\right)\right]\right):$
> $\mathrm{ShareCanvas}\left(\mathrm{cv}\right)$
Compatibility
• The DocumentTools[Canvas][StaticPlot] command was introduced in Maple 2021.
|
2022-12-03 11:39:50
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 3, "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": 0.9854300618171692, "perplexity": 7890.938070650943}, "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-2022-49/segments/1669446710931.81/warc/CC-MAIN-20221203111902-20221203141902-00260.warc.gz"}
|
https://aviation.stackexchange.com/questions/33694/do-propeller-coefficients-go-to-zero-at-the-same-value-of-advance-ratio
|
# Do propeller coefficients go to zero at the same value of advance ratio?
Graphs of power coefficient ($C_P$), thrust coefficient ($C_T$), and efficiency ($\eta=JC_T/C_P$) of the 5868-R6 propeller as functions of advance ratio ($J=V/(nD)$) and propeller pitch (blade angle at 0.75R?) are found in many textbooks (such as McCormic):
From these curves it seems that for a given propeller pitch $C_P$, $C_T$, and $\eta$ go to zero at the same value of $J$. At also seems that the slopes of the $C_T$ vs $J$ curves for higher value of $J$ (where $C_T$ approaches zero) are the same.
I have two questions:
1. Do $C_P$, $C_T$, and $\eta$ always go to zero simultaneously at the same value of $J$ for all propellers? Why?
2. Are the slopes of the $C_T$ vs $J$ curves where $C_T\rightarrow 0$ the same for all values of propeller pitch?
• If $C_T$ is 0, then by the definition of $\eta=JC_T/C_P$, $\eta=0$, irrespective of the value of $J$. – ROIMaison Dec 6 '16 at 15:38
• Yes @ROIMaison, that answers one part of the first question. If $C_T=0$ at $J=J_0$, then $\eta=0$ at $0$ and at $J_0$. :-) – Christo Dec 6 '16 at 16:20
Do $C_P$, $C_T$, and $\eta$ always go to zero simultaneously at the same value of $J$ for all propellers? Why?
Yes; Talking in dimensional quantities, This speed, at which all coefficients goes to zero is called pitch speed.
At pitch speed thrust goes to zero due to the fact that the incoming airflow drives the apparent angle of attack seen by the blades to zero.
Power = Thrust * velocity , As thrust goes to zero power goes to zero as well.
Efficiency $\eta = \frac{P_{out}}{P_{in}}$, $P_{out}$ goes to zero as shown above hence $\eta$ goes to zero.
HTH
• Thank you for the answer to the first part of my question, @ABCD. Any thoughts on the second part? – Christo Mar 28 '18 at 12:14
• Hi @Christo, sorry Im not clear what do you mean by the second part. Ct will always be some finite value. it doesnt go to infinity – ABCD Apr 2 '18 at 5:09
• Sorry, @ABCD! It was a typo; I edited the question. It should be the slope of CT vs J as CT decreases to zero. – Christo Apr 2 '18 at 18:17
• HI @Christo, Not really, the slope depends on the pitch/diameter ratio. You can see this trend in your second image attached. The reason is for a given RPM and diameter, higher pitch propeller has higher pitch speed and hence shallower slope. HTH – ABCD Apr 3 '18 at 10:54
|
2019-12-10 11:06:26
|
{"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": 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": 0.9054209589958191, "perplexity": 589.9330058228276}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-51/segments/1575540527205.81/warc/CC-MAIN-20191210095118-20191210123118-00178.warc.gz"}
|
https://pos.sissa.it/332/010/
|
Volume 332 - XIV International Conference on Heavy Quarks and Leptons (HQL2018) - CKM Matrix and Semileptonic Decays
Study of Leptonic and Semi-leptonic B Decays at Belle
K. Varvell* On behalf of the BELLE Collaboration
*corresponding author
Full text: pdf
Pre-published on: 2018 December 05
Published on: 2018 December 11
Abstract
Recent results from the Belle experiment on leptonic and semileptonic decays of $B$ mesons are presented. The measurements are based on an integrated luminosity of 711 fb$^{-1}$ collected at the $\Upsilon(4S)$ resonance at the KEKB $e^{+} e^{-}$ asymmetric collider.
DOI: https://doi.org/10.22323/1.332.0010
Open Access
|
2019-03-26 00:41:54
|
{"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": 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": 0.576026439666748, "perplexity": 10510.268211903405}, "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-2019-13/segments/1552912204736.6/warc/CC-MAIN-20190325234449-20190326020449-00292.warc.gz"}
|
https://tex.stackexchange.com/questions/340895/column-width-specifying
|
# column width specifying
I have the following table. I need to have the same width for all columns except the first one. Could you help me, please?
\begin{minipage}{\linewidth}
\centering
\captionof{table}{label}
\label{table}
\begin{tabular}{|c|c|c|c|c|c|c|}
\hline
Name & 1& 2& 3& 4& 5& 6 \\ \hline
USA & 1 & 1 & 1 & 2 & 22 & 6 \\
Italy & 2 & 6 & 6 & 4 & 11 & 10 \\
UK & 3 & 30 & 3 & 1 & 9 & 9 \\
Canada & 4 & 10 & 7 & 12 & 74 & 37 \\
Spain & 5 & 3 & 2 & 3 & 1 & 1 \\
Switzerland & 6 & 5 & 5 & 6 & 35 & 44 \\
Netherlands & 7 & 4 & 8 & 11 & 17 & 14 \\
Romania & 10 & 14 & 17 & 5 & 2 & 2 \\
Germany & 11 & 37 & 10 & 7 & 12 & 4 \\
New Zealand & 12 & 8 & 4 & 14 & 5 & 23 \\
France & 13 & 36 & 15 & 8 & 7 & 3 \\
Poland & 23 & 44 & 20 & 28 & 4 & 8 \\
India & 32 & 24 & 26 & 32 & 3 & 5 \\
Mexico & 45 & 2 & 56 & 40 & 49 & 40
\\
\hline
\end{tabular}
\end{minipage}
• The data columns currently are all equally wide already. (Each data column has at least one two-digit integer.) How wide do you want the data columns to be? – Mico Nov 25 '16 at 13:10
## 1 Answer
Like this?
\documentclass{article}
\usepackage{siunitx}
\usepackage{caption}
\begin{document}
\begin{minipage}{\linewidth}
\centering
\renewcommand\arraystretch{1.2}
\captionof{table}{label}
\label{table}
\begin{tabular}{|l|*{6}{S[table-format=3.1]|}}
\hline
Name & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline
USA & 1 & 1 & 1 & 2 & 22 & 6 \\
Italy & 2 & 6 & 6 & 4 & 11 & 10 \\
UK & 3 & 30 & 3 & 1 & 9 & 9 \\
Canada & 4 & 10 & 7 & 12 & 74 & 37 \\
Spain & 5 & 3 & 2 & 3 & 1 & 1 \\
Switzerland & 6 & 5 & 5 & 6 & 35 & 44 \\
Netherlands & 7 & 4 & 8 & 11 & 17 & 14 \\
Romania & 10 & 14 & 17 & 5 & 2 & 2 \\
Germany & 11 & 37 & 10 & 7 & 12 & 4 \\
New Zealand & 12 & 8 & 4 & 14 & 5 & 23 \\
France & 13 & 36 & 15 & 8 & 7 & 3 \\
Poland & 23 & 44 & 20 & 28 & 4 & 8 \\
India & 32 & 24 & 26 & 32 & 3 & 5 \\
Mexico & 45 & 2 & 56 & 40 & 49 & 40
\\
\hline
\end{tabular}
\end{minipage}
\end{document}
• Yes, like this! But what if the first row are words and not numbers? Name & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline – Anna Nov 25 '16 at 15:25
• words had to enclosed with curly braces like Name & {word 1} & {word 2} & {word 3} & {word 4} & {word 5} & {word 6} \ – Zarko Nov 25 '16 at 18:30
|
2021-06-14 22:05:06
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.7955493330955505, "perplexity": 57.91413705333174}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-25/segments/1623487613453.9/warc/CC-MAIN-20210614201339-20210614231339-00255.warc.gz"}
|
https://math.stackexchange.com/questions/1432698/when-is-a-vector-field-hamiltonian-with-respect-to-some-symplectic-form
|
When is a vector field hamiltonian with respect to some symplectic form?
Given a vector field $v$ on a $2n$-dimensional manifold, how many symplectic forms are there on $M$ that make $v$ a hamiltonian vector field? Alternatively, take the set of all $(H,\omega)$ pairs, mod adding constants to $H$ and doing $(H,\omega)\to (\alpha H,\omega/\alpha)$. Each element of this set may be associated to a vector field via the map $v^i=\omega^{ij}\partial_j H$ (call the map $F$ for later reference). The question is: given a vector field $v$, what does the preimage of $v$ under this map look like (is it empty/one element/large/etc)?
This question is inspired by this answer, where Qmechanic gives this local existence theorem:
Let there be given a $2n$-dimensional manifold $M$ with a non-vanishing vector field $X\in\Gamma(TM)$. Does there locally exists a symplectic two-form $\omega$, such that $X$ is a Hamiltonian vector field?
My question differs in that I am interested in uniqueness/non-uniqueness in addition to existence, and I am interested in any obstructions to the existence of a global two-form satisfying the quoted result (aside from the manifold itself not supporting a symplectic structure).
For completeness, here are my (haphazard and muddled) attempts at answering this question:
The zeroth thing I did was make the following (probably very flawed) degree of freedom counting argument, forgetting about topological obstructions etc. It takes $2n$ functions on $M$ to specify $v$, $2n-1$ functions on $M$ to specify $\omega$ (since we can add gradients to the tautological one form without changing $\omega$), and $1$ function for the hamiltonian. Thus the very naive reasoning suggests that vectors $v$ should be in correspondence with pairs $(H,\omega)$. As we see below this fails to happen (even for well-behaved examples), so I would be interested in how to make this sort of argument properly.
The next thing I tried was to work out some specific cases. The hamiltonian flow for the harmonic oscillator, which on $\mathbb{R}^2$ with coordinates $(q,p)$ looks something like $(p,-q)$, seems to be hamiltonian with respect to any symplectic form which is a function of $p^2+q^2$ (which seems like a lot of freedom). In particular, let $d\equiv (p^2+q^2)/2$ and consider the hamiltonian $H=f(d)$ and the symplectic form $\omega=f'(d) dq\wedge dp$. Given any function $f'(d)$ which is nonzero everywhere, $\omega$ so defined will be a globally defined two-form, and $H$ will be a globally defined hamiltonian, such that the image of $(H,\omega)$ under $F$ is $v$.
Now let us consider a slightly different situation. Let's work on $\mathbb{R}^2\setminus \{0\}$, and define $v$ to be the flow $(-q,-p)$ (i.e. a radially inward flowing vector field). This is obviously not hamiltonian under the usual form on $\mathbb{R}^2$, but if I did the Mathematica right, any symplectic form of the form
$$\omega = \frac{f'(p/q)}{q^2} dq\wedge dp,$$
with a hamiltonian of the form $H=f(p/q)$, will reproduce this vector field. In this case, unlike before, not any positive $f'$ will give us a globally defined symplectic form, since for instance $f(x)=x$ would produce singularities at $q=0$. Thus, even though locally we have the same freedom to pick $\omega$, globally we do not. However, the choice:
$$f(x)=\frac{x^2}{1+x^2}$$
(which I basically found by guess-and-check) seems to produce a globally defined symplectic form and hamiltonian, so there still exists a global symplectic form on our space with respect to which this vector field is hamiltonian.
Evidently according to this comment, we can also find vector fields on $\mathbb{R}^2\setminus \{0\}$ that don't admit any $(H,\omega)$ pairs globally, although presumably there is still a lot of freedom in picking those pairs locally.
Is there any obvious geometric feature/property that we can use to distinguish these three cases just from looking at $v$?
Lastly, I wanted to briefly note the following. On some manifolds we can choose vector fields that are locally hamiltonian but not hamiltonian (like the last example before; see, for instance, Exercise 7 here). In this situation, we can find a globally defined symplectic form $\omega$, and a collection of locally defined hamiltonians, such that on each patch $v$ arises via hamilton's equations. However, one could imagine a situation where no one global symplectic form will do, no matter how "multivalued" we allow $H$ to be. Is there one? I couldn't think of how to come up with one.
migrated from physics.stackexchange.comSep 12 '15 at 21:01
This question came from our site for active researchers, academics and students of physics.
• Perhaps this question is better suited for Mathematics – Danu Sep 12 '15 at 18:10
• Ah I asked it here because it seemed like a generalization of the linked physics.SE questions. Can I move the question myself? Should I delete it and re-ask it at math.SE? – commutatertot Sep 12 '15 at 20:10
• No, simply flag it and describe your request in a custom flag. – Danu Sep 12 '15 at 20:17
• Any progress on answering the original question or on a given symplectic manifold, when is a vector field Hamiltonian for some function? – Jim Stasheff Jan 27 '17 at 20:43
• @JimStasheff I haven't been working on this lately; any suggestions you might have would be very welcome! – commutatertot Jan 27 '17 at 22:08
|
2019-05-26 11:25:54
|
{"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": 0.8813098073005676, "perplexity": 245.60876581849053}, "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-2019-22/segments/1558232259126.83/warc/CC-MAIN-20190526105248-20190526131248-00014.warc.gz"}
|
https://s138143.gridserver.com/16p3ldp/page.php?page=eta-symbol-physics-244f87
|
(Revised to conform to ANSI Y32.2-1975 and IEEE-Std. A full account of the physics of friction (known as tribology) can be found, for example, in Friction and Wear of Materials, by Ernest Rabinowicz, second Edition, Wiley, 1995. This answer is developed from the principles in Luttwak's Coup de E’tat: A Practical Handbook, which, as J.I. Learn about 24 letters including alpha, beta, gamma and others. Initialism of estimated time approximately. 1 Greek letters; 2 Unary operators; 3 Relation operators; 4 Binary operators; 5 Negated binary relations; 6 Set and/or logic notation; 7 Geometry; 8 Delimiters; 9 Arrows; 10 Other symbols; 11 Trigonometric functions; 12 Notes; 13 External links; Greek letters. The reason is quite simple: there are only so many symbols in the Greek and Latin alphabets, and scientists and mathematicians generally do not use symbols from other languages. Comments: 41 pages; v2: typos corrected, clarifications added; v3: typos corrected, to appear in PTEP; v4: typos corrected in Eq. Contents. First, there’s no such thing as static friction between a solid and a liquid. 153k 25 25 gold badges 384 384 silver badges 433 433 bronze badges. Fraternity, Physics, and Calculus pronunciation. The Greek letter η (Pronounce: “èta”) is the symbol used for Efficiency in physics. Conversions . Curious and quirky videos describing the various symbols used in physics and astronomy Common pronunciations (in British English - Gimson,1981) of mathematical and scientific symbols are given in the list below. Sumber: Karya sorangan: Pangarang: Omegatron: i The source code of this SVG is valid. Title: Exceptional M-brane sigma models and $η$-symbols. Morgan, indicates, is a very informative book on coups. η = F/A ∆v x /∆z: or. Then I explain how to get summation and integration, how to put one thing above another, and, finally, There are a couple of special characters that will combine symbols. Initialism of Electronic Travel Authority. English: The circuit diagram symbol for a Zener diode. La viscosité diminue la liberté d'écoulement du fluide et dissipe son énergie. Depuis 1967, on doit utiliser μm. Chemical Reactions Chemical Properties. Step-by-Step Calculator Solve problems from Pre Algebra to Calculus step-by-step . This is the way your physics teacher spoke Greek, and he learned this pronunciation in his fraternity. More symbols are available from extra packages. Liquid Friction . Mechanics. Modules in Mechanics of Materials List of Symbols A area,freeenergy,Madelungconstant A transformationmatrix A plateextensionalsti ness a length,transformationmatrix,cracklength The significance of Dedekind Eta function in physics that it ensures modular invariance of certain quantities which is an important physical constraint. List of Greek alphabet letters and symbols. What happens if instead of two solid surfaces in contact, we have a solid in contact with a liquid? Stack Exchange Network. Lisénsi. In physics, the nu symbol denotes the frequency of waves, wave-numbers and spatial frequency while in fluid mechanics, it represents kinematic viscosity and in statistics, it represents the degree of freedom. Taip pat „Twitter“, „Instagram“ bei kituose socialiniuose tinkluose ir mobiliose programėlėse. Decimal to Fraction Fraction to Decimal Hexadecimal Scientific Notation Distance Weight Time. Physics. Eta (@) − tipografijos simbolis, naudojamas el. Chez l’homme, les globules rouges sont discoïdes ; leur dimension est de 7 à 8 μ. If a boat is at rest i Eta meson, a physics particle: see list of mesons; Eta (letter in Greek alphabet, lower case only), symbol for efficiency in scientific discourse; Abbreviations: Estimated time of arrival; Estimated Time Amount when using μTorrent, Azureus, and other BitTorrent clients. (all the pages in this section need a unicode font installed - e.g. If you see utf-8, then your system supports unicode characters.To print any character in the Python interpreter, use a \u to denote a unicode character and then follow with the character code. For instance, the code for β is 03B2, so to print β the command is print('\u03B2').. MS-Word File with Mathematical Symbols First I give a list of symbols for both MS-Word and Powerpoint. ETA (plural ETAs) Initialism of estimated time of arrival. Letter & Symbols Includes: Greek alphabet Mathematical symbols Mathematical constants Typographical symbols The Greek alphabet is widely used to demote various constants and values … Contents. In particular, in the SL(5) EFT, we explicitly show the equivalence to the known linear section equation. Symbole [modifier le wikicode] μ (Métrologie) Symbole du préfixe micro-du Système international (×10 −6). answered Mar 14 '11 at 2:44. When you create a presentation for a graduate thesis or dissertation, often, you need to insert Greek letters into the slide. In physics, the square function often rears its head in the context of equations the describe the intensity of some physical quantity as a function of distance. Eta. High Energy Physics - Theory. About the Common uses in Physics While these are indeed common usages, it should be pointed out that there are many other usages and that other letters are used for the same purpose. Importance dans la culture grecque antique. Mathematical and scientific symbols. Due to the 3-D geometry of space, the intensity of any physical quantity that radiates outward in a sphere around the source is inversely proportional to the square of the distance from the source. Xi. Finance. (Australia) Noun . Greek Alphabet Symbols & Characters: mathematical uses The Greek alphabet is widely used in mathematical and scientific equations with letters like psi, rho, eta, mu, omega and many more being widely used. La viscosité (du latin viscum, gui, glu) peut être définie comme l'ensemble des phénomènes de résistance au mouvement d'un fluide pour un écoulement avec ou sans turbulence. Detexify gives me only text mode symbol for LaTeX. 315-1975.) Furthermore, using the $\eta$-symbols, we propose a universal form of the linear section equation. Introducing $\eta$-forms that are defined with the $\eta$-symbols, … Not everyone uses the same symbols. \eta \rho: See Shortcut keys for inserting symbols and templates into the equation to find other frequently used symbols. pašto sistemose. Find out what is the full meaning of ETA on Abbreviations.com! Looking for the definition of ETA? Translations I need to insert an at symbol, @ , which I know how to do in text, but can't figure out how to do in mathmode. Of course, as with all the other Greek symbols, the use of nu symbol in scientific areas goes beyond the ones mentioned here. Next time you hear a physics teacher pronounce Greek, laugh and look superior. Ramnath Ramnath. (Métrologie) Symbole du micron, micromètre. What is the symbol of viscosity? For complete documentation on the various symbols that are available when doing this and how to use them, see ?plotmath. Simple Interest Compound Interest Present Value Future Value. Acronym of Euskadi Ta Askatasuna. Dedekind eta function arises in the partition functions of string theory. share | improve this answer | follow | edited Dec 31 '15 at 18:09. joran. The eta ( η) and eta prime meson ( η′) are isosinglet mesons made of a mixture of up, down and strange quarks and their antiquarks.The charmed eta meson ( η c) and bottom eta meson ( η b) are similar forms of quarkonium; they have the same spin and parity as the (light) η defined, but are made of charm quarks and bottom quarks respectively. How to type Greek letters symbols on PC. ETA. La lettre epsilon tenait une place particulière dans la culture grecque en raison notamment de sa présence dans le sanctuaire de Delphes.Ainsi dans le temple d'Apollon, un epsilon couché était placé au-dessus de la porte à l'entrée du naos.Par ailleurs, un epsilon était gravé sur l'omphalos, symbole du centre du monde [10]. Formally, viscosity (represented by the symbol η "eta") is the ratio of the shearing stress (F/A) to the velocity gradient (∆v x /∆z or dv x /dz) in a fluid. 'Estimated Time of Arrival' is one option -- get in to view more @ The Web's largest and most authoritative acronyms and abbreviations resource. Can anybody please let me know what the ideality factor (greek symbol eta) in a standard current voltage equation of a p-n diode represents? Initialism of Elvis tribute artist. When used in a circuit diagram, the words "Anode" and "Cathode" are not included with the graphic symbol. Chemistry. Mathematical symbols and Greek letters are pervasive today and used everywhere, from physics to social science. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. 49.7k 13 13 gold badges 112 112 silver badges 147 147 bronze badges. Authors: Yuho Sakatani, Shozo Uehara (Submitted on 29 Dec 2017 , last revised 8 Mar 2018 (this version, v4)) Abstract: We develop the M-brane actions proposed in arXiv:1607.04265 by using $\eta$-symbols determined in arXiv:1708.06342. It is a common trap to … A professional impersonator who performs as Elvis Presley. Note that some of the symbols require loading of the amssymb package. Liberté eta symbol physics du fluide et dissipe son énergie pronunciation in his Fraternity videos... In Luttwak 's Coup de E ’ tat: a Practical Handbook, which, J.I... When doing this and how to use them, See? plotmath Cathode '' not. Alpha, beta, gamma and others dimension est de 7 à 8 μ Zener.!: Omegatron: i the source code of this SVG is valid a solid a... Anode '' and Cathode '' are not included with the graphic.. From physics to social science we have a solid and a liquid or dissertation,,. 384 384 silver badges 433 433 bronze badges: a Practical Handbook, which, as J.I |! Omegatron: i the source code of this SVG is valid de 7 à 8.. Tinkluose ir mobiliose programėlėse eta symbol physics linear section equation a couple of special characters that will combine symbols edited Dec '15... Symbols used in a circuit diagram symbol for LaTeX thesis or dissertation, often, you need to Greek! Estimated time of arrival show the equivalence to the known linear section equation Initialism estimated... Command is print ( '\u03B2 ' ) préfixe micro-du Système international ( ×10 −6 ) a... Documentation on the various symbols that are available when doing this and how use..., in the list below to insert Greek letters into the slide time of arrival in Luttwak 's de. Arises in the partition functions of string theory a solid in contact with a liquid explicitly the... Calculator Solve problems from Pre Algebra to Calculus step-by-step for a Zener diode: “ èta ” ) the! Practical Handbook, which, as J.I Dec 31 '15 at 18:09... −6 ) η ( Pronounce: “ èta ” ) is the full meaning of eta on!. Twitter “, „ Instagram “ bei kituose socialiniuose tinkluose ir mobiliose programėlėse in... To Calculus step-by-step there are a couple of special characters that will combine symbols a! “ èta ” ) is the full meaning of eta on Abbreviations.com for β is,!, physics, and he learned this pronunciation in his Fraternity find other frequently used symbols morgan indicates. '' are not included with the graphic symbol Fraternity, physics, he! Coup de E ’ tat: a Practical Handbook, which, as J.I s no thing. ) symbole du préfixe micro-du Système international ( ×10 −6 ) for LaTeX, is a very informative on! Eft, we explicitly show the equivalence to the known linear section equation teacher Greek! This is the full meaning of eta on Abbreviations.com Initialism of estimated of... Practical Handbook, which, as J.I ( Revised to conform to ANSI Y32.2-1975 and IEEE-Std Greek laugh! Pronunciations ( in British English - Gimson,1981 ) of mathematical and scientific symbols are given the! Static friction between a solid in contact, we explicitly show the equivalence to the known linear section.... Of special characters that will combine symbols du fluide et dissipe son énergie are., beta, gamma and others Greek letter η ( Pronounce: “ èta )!, naudojamas el “ èta ” ) is the symbol used for Efficiency in physics and astronomy what the! Furthermore, using the $\eta$ -symbols and how to use them,?... Pre Algebra to Calculus step-by-step way your physics teacher spoke Greek, laugh and look superior physics and what! Teacher spoke Greek, laugh and look superior to the known linear section equation ) of! ) − tipografijos simbolis, naudojamas el are not included with the graphic symbol 24 letters alpha... Models and $η$ -symbols, we propose a universal form of the section. ( Revised to conform to ANSI Y32.2-1975 and IEEE-Std between a solid in contact a... Find other frequently used symbols source code of this SVG is valid the $\eta -symbols... Share | improve this answer | follow | edited Dec 31 '15 at 18:09. joran if a boat is rest. - e.g la liberté d'écoulement du fluide et dissipe son énergie used everywhere, from to! 31 eta symbol physics at 18:09. joran on coups ; leur dimension est de 7 8. On Abbreviations.com is the symbol of viscosity list below du fluide et dissipe son énergie share improve... I the source code of this SVG is valid diagram symbol for.! Letters into the equation to find other frequently used symbols and look superior Instagram. In British English - Gimson,1981 ) of mathematical and scientific symbols are given in the partition functions string! De E ’ tat: a Practical Handbook, which, as J.I we have a in. And used everywhere, from physics to social science a circuit diagram symbol for LaTeX physics teacher Greek. The way your physics teacher eta symbol physics Greek, laugh and look superior code... Particular, in the SL ( 5 ) EFT, we propose a universal form of the linear equation. ; leur dimension est de 7 à 8 μ find out what is the of. Partition functions of string theory: Omegatron: i the source code this! At 18:09. joran “ èta ” ) is the symbol used for Efficiency in physics Calculator Solve from. “ èta ” ) is the way your physics teacher spoke Greek, and... Kituose socialiniuose tinkluose ir mobiliose programėlėse, See? plotmath a Zener diode in British English Gimson,1981. Scientific Notation Distance Weight time instead of two solid surfaces in contact a... Keys for inserting symbols and Greek eta symbol physics into the equation to find other frequently used symbols et son... To use them, See? plotmath SVG is valid equation to find other frequently used symbols a of. 7 à 8 μ this answer is developed from the principles in Luttwak 's Coup de ’... To ANSI Y32.2-1975 and IEEE-Std font installed - e.g ( '\u03B2 ' ) Fraction to! ( Pronounce: “ èta eta symbol physics ) is the way your physics teacher spoke,... Frequently used symbols as J.I only text mode symbol for LaTeX English - )... Graduate thesis or dissertation, often, you need to insert Greek letters are today... 433 433 bronze badges boat is at rest i Fraternity, physics, he... 8 μ Cathode '' are not included with the graphic symbol for complete documentation the! To insert Greek letters into the equation to find other frequently used.... Including alpha, beta, gamma and others 147 147 bronze badges?! Which, as J.I 7 à 8 μ of two solid surfaces in contact, we show. Symbole du préfixe micro-du Système international ( ×10 −6 ) you need to insert Greek letters are today! Conform to ANSI Y32.2-1975 and IEEE-Std time you hear a physics teacher Pronounce Greek laugh... ] μ ( Métrologie ) symbole du préfixe micro-du Système international ( ×10 −6 eta symbol physics source code of SVG... Letter η ( Pronounce: “ èta ” ) is the symbol of?. Section equation between a solid in contact with a liquid: i source... Next time you hear a physics teacher Pronounce Greek, and he learned this pronunciation in his.. L ’ homme, les globules rouges sont discoïdes ; leur dimension est de 7 à 8 μ Omegatron!, which, as J.I and astronomy what is the symbol of viscosity full meaning of eta on!... Instagram “ bei kituose socialiniuose tinkluose ir mobiliose programėlėse 112 112 silver badges 147! In his Fraternity | follow | edited Dec 31 '15 at 18:09. joran solid contact. A couple of special characters that eta symbol physics combine symbols functions of string theory 8 μ badges! '' and Cathode '' are not included with the graphic symbol: sorangan., there ’ s no such thing as static friction between a solid and a liquid sumber: Karya:... And look superior the way your physics teacher spoke Greek, laugh and look.. Liberté d'écoulement du fluide et dissipe son énergie 25 gold badges 384 384 silver badges 433! Gamma and others, naudojamas el your physics teacher Pronounce Greek, and he learned this pronunciation his... Is developed from the principles in Luttwak 's Coup de E ’:. ( Revised to conform to ANSI Y32.2-1975 and IEEE-Std that are available when doing this and how to use,. Linear section equation: Pangarang: Omegatron: i the source code this! Contact, we have a solid in contact, we propose a universal form of the section. -Symbols, we have a solid in contact, we propose a universal form of the linear section equation and. Used for Efficiency in physics Weight time spoke Greek eta symbol physics laugh and superior. Tat: a Practical Handbook, which, as J.I ANSI Y32.2-1975 and IEEE-Std ( −6. \Rho: See Shortcut keys for inserting symbols and Greek letters into equation! Partition functions of string theory Dec 31 '15 at 18:09. joran when you a! Is print ( '\u03B2 ' ) la liberté d'écoulement du fluide et dissipe son énergie how to use them See... Symbols are given in the list below symbol used for Efficiency in physics and astronomy is! Scientific symbols are given in the partition functions of string theory la d'écoulement!$ \eta eta symbol physics -symbols, we explicitly show the equivalence to the known linear equation... Pronounce Greek, and Calculus pronunciation Anode '' and Cathode '' are not included with the graphic....
2020 eta symbol physics
|
2021-06-19 03:53:07
|
{"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": 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": 0.739152729511261, "perplexity": 5585.164204994583}, "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/1623487643380.40/warc/CC-MAIN-20210619020602-20210619050602-00554.warc.gz"}
|
https://punkrockor.com/2013/09/23/how-to-win-at-russian-roulette/
|
# how to win at Russian Roulette
Every time I teach stochastic processes, we discuss whether to play Russian Roulette (don’t!). In the off chance one absolutely has to play, we determine the best time to take a turn and whether it is best to spin the barrel.
I believe in teaching important life lessons in class along with operations research. But in this case, I seriously hope none of my students consider this example on Russian roulette an important life lesson. This example is good for exploring how we can quantify probabilities to confirm our intuition.
Just for fun, here are the relevant probabilities of death based on order and spinning strategy. (I like to mix it up with some dark humor in my classes).
First, consider the odds with not spinning the barrel.
Let Ei = the event that the ith person survives (based on order of play). The first person has a 5/6 chance of survival:
$P(E1) = 5/6$
If we condition on the first outcome, the second person also has a 5/6 chance of survival:
$P(E2) = P(E2|E1)P(E1) + P(E2|E1')P(E1') = (4/5)(5/6) + (1)(1/6) = 5/6$
If we condition on the first two outcomes, the third person has a 5/6 chance of survival:
$P(E3) = P(E3|E1,E2)P(E2|E1)P(E1) + P(E3|E1')P(E1') + P(E3|E1,E2')P(E2'|E1)P(E1) = (3/4)(4/5)(5/6) + (1)(1/6) + (1)(1/5)(5/6) = 5/6$
This makes intuitive sense. The bullet goes into one chamber where it is “preassigned” to one player of the game.
Next, consider the odds with spinning the barrel.
Let Ei = the event that the ith person survives (based on order of play). The first person has a 5/6 chance of survival:
$P(E1) = 5/6$
If we condition on the first outcome, the second person also has improved chance of survival:
$P(E2) = P(E2|E1)P(E1) + P(E2|E1')P(E1') = (5/6)(5/6) + (1)(1/6) = 31/36$
If we condition on the first two outcomes, the third person has an even more improved chance of survival:
$P(E3) = P(E3|E1,E2)P(E2|E1)P(E1) + P(E3|E1')P(E1') + P(E3|E1,E2')P(E2'|E1)P(E1) = (5/6)(5/6)(5/6) + (1)(1/6) + (1)(1/6)(5/6) = 191/216$
Continuing in this way, we can compute the odds of death based on each player’s order.
Without spinning the barrel, someone will lose. If every players spins the barrel prior to his/her turn, there is a 33.5% chance that everyone will walk away from the game. Such a small action greatly affects the outcome of the game, especially for those who are among the last to go.
Do not spin the barrel:
Order P(die) 1 0.1667 2 0.1667 3 0.1667 4 0.1667 5 0.1667 6 0.1667 P(someone dies) 1.0
Spin the barrel:
Order P(die) 1 0.1667 2 0.1389 3 0.1157 4 0.0965 5 0.0904 6 0.0670 P(someone dies) 0.665
We can see here that there is no guaranteed way to win at Russian roulette. However, going last after everyone spins the barrel lowers your probability of losing by 60%.
#### 7 responses to “how to win at Russian Roulette”
• @j_rosenbe
Not entirely staying on topic …
(I like to mix it up with some dark humor in my classes).
I find that including more jokes (even really bad ones) improves my teaching evaluations. But, I have no idea how dark humor would go over with the students.
• David K Smith
In a related way, you could consider the strategy for a three-way duel (truel!) between A, B and C – with A more accurate than B and B more accurate than C. They take it in turns to shoot (i.e not all at once). If C shoots first what should they do? (Thanks to Martin Gardner and others for this problem.)
• Luis Eduardo
Nice problem, altough I always tought spinning the barrel resets conditions so that conditioning does not apply.
• Laura McLay
@Luis, it’s true that we start over fresh with a revolver that has been spun each time. The difference is that the further you are in line, the greater chance that you won’t even have a turn (one of the previous players doesn’t survive). We have to take that into account, which gives us a conditional probability (that is the same for everyone when we reset the barrel conditions by spinning).
• Austin
The barrel on a revolver does not spin. The part that spins is the cylinder.
• johnny
It depends on who spins the cylinder. A bullet will off balance the cylinder, so with a little practice the one who spins it can decide if the gun fires or not fairly easily.
• How To Win At Russian Roulette Against Putin? | Narativ
[…] winning Russian roulette are reduced every time the trigger is pulled. Laura McClay explains why in this post (although it pre-supposes more players). If McClay is correct, then playing last is an […]
|
2019-03-19 19:37:02
|
{"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": 0, "mathjax_asciimath": 0, "img_math": 6, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.4947262704372406, "perplexity": 1006.5563459141225}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-13/segments/1552912202125.41/warc/CC-MAIN-20190319183735-20190319205735-00216.warc.gz"}
|
http://www.ipam.ucla.edu/abstract/?tid=12019&pcode=CCG2014
|
## On a geometric Ramsey number
#### Imre BaranyRenyi Institute of Mathematics
Organizers: Igor Pak and Greta Panova; Location: MS 7608; Abstract: A partial result: if a planar curve intersects every line in at most 3 points, then it can be partitioned into 4 convex curves. This result can be extended to R^d, and the extension implies a good, asymptotically precise, lower bound on a geometric Ramsey number. Joint result with Jiri Matousek and Attila Por.
Back to Algebraic Techniques for Combinatorial and Computational Geometry
|
2017-10-20 21:37:53
|
{"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": 0.8914356231689453, "perplexity": 1575.1833992098266}, "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/1508187824357.3/warc/CC-MAIN-20171020211313-20171020231313-00728.warc.gz"}
|
https://answers.ros.org/question/228750/turtlebot-odometry-calibration-with-and-without-imu-observationsquestions/
|
# Turtlebot Odometry Calibration (with and without IMU) Observations/Questions
After adding the Razor 9DOF IMU from Sparkfun, the resulting Odometry actually seems to have gotten worse as shown in the two videos below:
1. Wheels-only Odometry, IMU disabled, analog Gyro disabled: https://youtu.be/ibY_HCBGm6U
2. Wheels + IMU, analog Gyro disabled: https://youtu.be/mPiKbtP5szM
Both show the robot's position with respect to the /odom frame as calculated by robot_pose_ekf. If the odometry was perfect (i.e. no error) then the position of the walls as seen by the Neato XV-11 LIDAR would remain in the same place as the robot is moved around.
The wheels-only odometry has a long-term drift as the TurtleBot is rotated continuously (via joystick telop). In the short term, though, it is pretty accurate and can probably be improved further by fine tuning the scaling parameters.
The wheel+IMU odometric position gyrates pretty significantly per turn but there is no long term drift.
When localizing with AMCL, adding the IMU will likely produce worse results because Localization can compensate for long term odometric drift but the short term gyrations will throw it off.
I did calibrate the IMU as described in the wiki page, including the Magnetometer (Section 7.1.3). The calibration_magn_use_extended parameter is set to true.
I'm looking for suggestions on how the improve the results of the wheels+IMU odometry. Thanks.
Also, here's the Matlab Analysis of the Magnetometer configuration data. I'm not sure how this is supposed to be interpreted. Please let me know if something's wrong.
edit retag close merge delete
Sort by » oldest newest most voted
So the problem was in the Mag calibration. I redid the calibration procedure while the IMU was mounted on the Turtlebot. I picked up the Turtlebot and moved it around in all directions as in the instructions. The "Ellipsoid Center" values came out significantly different:
Standalone Calibration Results:
const float magn_ellipsoid_center[3] = {138.402, -65.8241, -12.1968};
const float magn_ellipsoid_transform[3][3] = {{0.847013, -0.00538074, 0.00712010}, {-0.00538074, 0.865221, 0.00961725}, {0.00712010, 0.00961725, 0.999017}}
Results when mounted on the Turtlebot:
const float magn_ellipsoid_center[3] = {135.215, 4.24598, 5.24594};
const float magn_ellipsoid_transform[3][3] = {{0.820361, -0.0188837, 0.0166510}, {-0.0188837, 0.831140, 0.0388830}, {0.0166510, 0.0388830, 0.990194}};
Notice the changes in the Y and Z values of the Ellipsoid Center.
After this, the combined results of Wheels + IMU odometry were much better than wheels-only.
Other things to note:
• Experiment with positioning the IMU to minimize interference.
• Verify the ellipsoid center values -- the IMU node prints out the values on startup -- at least twice they were not correct (missing one digit). Not sure why that happened.
• The instructions say to modify the Arduino firmware with these values. I found that it was sufficient to change the parameters in the launch file (my_razor.yaml) -- no need to modify the firmware.
Edited to add: I had noted above that the ellipsoid_center or ellipsoid_transform values were getting corrupted on the readback from the IMU (after calibration configuration was sent to the IMU). Suspecting that it may be a serial line buffer overflow type of problem I added a few rospy.sleep(1) while writing these values from imu_node.py. Happy to report that that seems to have done the trick. The read-back values are no longer corrupted. Most definitely a hack, and perhaps specific to my particular IMU but it works!
more
|
2021-07-30 23:54:40
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.2407081425189972, "perplexity": 4318.137607050529}, "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/1627046154032.75/warc/CC-MAIN-20210730220317-20210731010317-00682.warc.gz"}
|
https://computergraphics.stackexchange.com/questions/7705/variance-of-estimator-monte-carlo-integration
|
# Variance of estimator (Monte Carlo Integration)
So I was reading this paper by Lafortune, "Mathematical Models and Monte Carlo algorithms" and in it he writes.
We have a function or integrand I we want to estimate given as,
$I = \int f(x) dx$
We then have a primary estimator for this as,
$\hat{I_p} = f(\xi)$
where $\xi$ is a random number generated in the interval of the integrand $I$.
The secondary estimator is defined as,
$\hat{I_{sec}} = \frac{1}{N} \sum f(\xi_i)$
Then there is an explanation that goes to show how the expected value of the secondary estimator is equal to the function/integrand I. i.e,
$E(\hat{I_{sec}}) = I$
All fair. The problem comes when he tries to find the variance of this secondary estimator as follows. Im posting a screenshot of the paper here since it's getting a little complex.
I don't understand how did step 2 follow from step 1 of Equation 3.5 in the variance calculation. Note that he has assumed $PDF = 1$ for now
I was trying to calculate the variance using a more standard approach and ignoring the multiple integrals. We know the variance of an estimator $\hat{\theta}$ is given as
$Var(\hat{\theta}) = E[(\hat{\theta} - E(\hat{\theta}))^2]$
If we apply this to above we get
$Var(\hat{I_{sec}}) = E[(\hat{I_{sec}} - E(\hat{I_{sec}}))^2]$
$\hspace{20mm} = E[(\hat{I_{sec}} - I)^2]$
$\hspace{20mm} = E[(\hat{I_{sec}})^2 - 2* I *(\hat{I_{sec}}) + I^2 ]$
$\hspace{20mm} = E[(\hat{I_{sec}})^2] - 2* I *E[\hat{I_{sec}}] + I^2$
$\hspace{20mm} = E[(\hat{I_{sec}})^2] - 2* I^2 + I^2$
$\hspace{20mm} = E[(\hat{I_{sec}})^2] - I^2$
$\hspace{20mm} = E[(\frac{1}{N} \sum\limits_{i=1}^{N} f(\xi_i))^2] - I^2$
Now I am stuck here, I could do this,
$\hspace{20mm} = E[\frac{1}{N^2} \sum\limits_{i=1}^{N} f^2(\xi_i) + \frac{1}{N^2}\sum\limits_{i\neq j}^{N} f(\xi_i) f(\xi_j) ] - I^2 \hspace{10mm} \because [\sum\limits_{i=1}^{N} f(x_i) ]^2 = \sum\limits_{i=1}^{N} f^2(x_i) + \sum\limits_{i\neq j}^{N} f(\xi_i) f(\xi_j)$
$\hspace{20mm} = \frac{1}{N^2} \sum\limits_{i=1}^{N} E[f^2(\xi_i)] + \frac{1}{N^2}\sum\limits_{i\neq j}^{N} E[f(\xi_i) f(\xi_j) ] - I^2$
$\hspace{20mm} = \frac{1}{N^2} \sum\limits_{i=1}^{N} \int f^2(x)dx + \frac{1}{N^2}\sum\limits_{i\neq j}^{N} E[f(\xi_i) f(\xi_j) ] - I^2$
$\hspace{20mm} \because E[f(X)] = \int f(X) p(X) dx$
Note $p(X) = 1$,
$\hspace{20mm} = \frac{1}{N} \int f^2(x)dx + \frac{1}{N^2}\sum\limits_{i\neq j}^{N} E[f(\xi_i) f(\xi_j) ] - I^2$
Don't know what to do next from here, any tips? Or did I do something wrong?
Quoting from "Advanced global illumination" book:
The Monte Carlo estimator is defined as:
And its variance according to the above definition(continuous random variable) would be:
Edit:
An example for how the variance of an estimator G is calculated:
(note how the summation is written as N times the integrand and is cancelled out later.)
• Hmm how did you remove the summation, I would appreciate though if you will tell what's wrong in my calculation rather than proving from a different way. – gallickgunner Jun 13 '18 at 11:22
• the summation is written as N times the f(x)/p(x) which cancels out one of the N factor in variance. I think you missed this on your last proof. Have look at the edit. – ali Jun 13 '18 at 12:20
• Yes I know that, but it can only be done if we are taking the $\sum E[f(x_i)]$. That is when we are taking the expected value inside the summation, that reduces $f(x_i)$ to $f(x)$ and then we can write it as $N*f(x)$. However in my proof I am having problem taking the Expected value of the whole square of the summation. You can't take the E operator inside of the whole square term. example, $E [ ( \sum f(x_i) )^2]$ – gallickgunner Jun 13 '18 at 19:57
So someone at mathstackexchange posted the answer I was satisfied with. Here is the link
And here is the quoted answer.
This computation really has nothing to do with the Monte Carlo integration context. It is a general fact that if $X_i$ are iid random variables with variance $\sigma^2$ then $\frac{1}{n} \sum_{i=1}^n X_i$ has variance $\sigma^2/n$. This is because of two related facts. First, if you sum two independent random variables, you add their variances. This is because:
$$\sigma_{X+Y}^2=E[(X+Y-\mu_X-\mu_Y)^2] \\=E[((X-\mu_X)+(Y-\mu_Y))^2] > \\ > =E[(X-\mu_X)^2]+E[(Y-\mu_Y)^2]+2E[(X-\mu_X)(Y-\mu_Y)] \\=\sigma_X^2+\sigma_Y^2.$$
In the last step you use that independent random variables are uncorrelated. This follows from the fact that $E[XY]=E[X]E[Y]$ if $X$ and $Y$ are independent.
Second, if you multiply a random variable by a constant, you multiply its variance by the square of that constant. This is obvious from the definition and linearity of expectation.
So in the context of $\frac{1}{n} \sum_{i=1}^n X_i$, the summation multiplies the variance by $n$ while the division by $n$ multiplies the variance by $1/n^2$, giving the result.
So if we apply this $E[XY]=E[X]E[Y]$, we get the final result.
|
2019-10-17 15:51:47
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.9436168074607849, "perplexity": 243.5816186960557}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-43/segments/1570986675409.61/warc/CC-MAIN-20191017145741-20191017173241-00472.warc.gz"}
|
https://archive-ouverte.unige.ch/unige:11877
|
Title
Ergodic properties of boundary actions and Nielsen--Schreier theory
Authors
Kaimanovich, Vadim A.
Year 2009
Abstract We study the basic ergodic properties (ergodicity and conservativity) of the action of a subgroup $H$ of a free group $F$ on the boundary $\pt F$ with respect to the uniform measure. Our approach is geometrical and combinatorial, and it is based on choosing a system of Nielsen--Schreier generators in $H$ associated with a geodesic spanning tree in the Schreier graph $X=H\bs F$. We give several (mod 0) equivalent descriptions of the Hopf decomposition of the boundary into the conservative and the dissipative parts. Further we relate conservativity and dissipativity of the action with the growth of the Schreier graph $X$ and of the subgroup $H$ ($\equiv$ cogrowth of $X$), respectively. On the other hand, our approach sheds a new light on entirely algebraic properties of subgroups of a free group. We also construct numerous examples illustrating the connections between various relevant notions.
Identifiers
Note minor editorial changes, added references
Full text
Preprint (484 Kb) - Free access
Structures
Citation
(ISO format)
GRIGORCHUK, Rostislav, KAIMANOVICH, Vadim A., SMIRNOVA-NAGNIBEDA, Tatiana. Ergodic properties of boundary actions and Nielsen--Schreier theory. 2009. https://archive-ouverte.unige.ch/unige:11877
556 hits
266 downloads
Update
Deposited on : 2010-09-23
Export document
Format : HTML citation Plain text citation EndNote format Citation style : APA 6th - American Psychological Association, 6th Edition BibTeX Chicago 15th Edition (Author-Date System) Harvard MLA 6th Edition NLM - National Library of Medicine Turabian (Reference List) 6th Edition Vancouver
|
2023-02-02 08:46:17
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5839626789093018, "perplexity": 2375.090818305617}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-06/segments/1674764499967.46/warc/CC-MAIN-20230202070522-20230202100522-00637.warc.gz"}
|
https://www.vedantu.com/question-answer/two-cars-start-towards-each-other-from-two-class-7-maths-cbse-5ee350f46067f248d16aef45
|
Question
Two cars start towards each other, from two places A and B which are at a distance of 160 km. They start at the same time at 8:10 AM. If the speeds of the cars are 50 km per hour and 30 km per hour respectively, they will meet each other at${\text{A}}{\text{. 10:10 AM}} \\ {\text{B}}{\text{. 10:30 AM}} \\ {\text{C}}{\text{. 11:10 AM}} \\ {\text{D}}{\text{. 11:20 AM}} \\$
Verified
156.6k+ views
Hint: Here, we will proceed by finding out the relative speed between the two cars and then using the formula i.e., Time taken$= \dfrac{{{\text{Distance travelled}}}}{{{\text{Speed}}}}$ to determine after how many hours these cars will meet each other.
As we know that, Time taken$= \dfrac{{{\text{Distance travelled}}}}{{{\text{Speed}}}}$
So, time when both the cars will meet=$\dfrac{{{\text{Initial distance between both the cars}}}}{{{\text{Relative speed between the cars}}}} = \dfrac{{\text{d}}}{{\text{r}}} = \dfrac{{160}}{{80}} = 2$ hours.
|
2021-12-04 14:10:53
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7841399312019348, "perplexity": 314.85736826858096}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-49/segments/1637964362992.98/warc/CC-MAIN-20211204124328-20211204154328-00161.warc.gz"}
|
https://socratic.org/questions/how-do-you-factor-x-3-6x-2-3x-10
|
# How do you factor x^3 - 6x^2 + 3x - 10?
May 9, 2016
See explanation...
#### Explanation:
Use Cardano's method.
First a simple Tschirnhaus transformation, to eliminate the term of degree $2$...
$0 = {x}^{3} - 6 {x}^{2} + 3 x - 10 = {\left(x - 2\right)}^{3} - 9 \left(x - 2\right) - 20$
Let $t = x - 2$ to get the simplified cubic equation:
${t}^{3} - 9 t - 20 = 0$
Next substitute $t = u + v$ to get:
${u}^{3} + {v}^{3} + 3 \left(u v - 3\right) \left(u + v\right) - 20 = 0$
Add the constraint $v = \frac{3}{u}$ to eliminate the term in $\left(u + v\right)$ and get:
${u}^{3} + {\left(\frac{3}{u}\right)}^{3} - 20 = 0$
Multiply through by ${u}^{3}$ to get:
${\left({u}^{3}\right)}^{2} - 20 \left({u}^{3}\right) + 27 = 0$
Use the quadratic formula to find:
${u}^{3} = \frac{20 \pm \sqrt{{20}^{2} - \left(4 \cdot 1 \cdot 27\right)}}{2 \cdot 1}$
$= 10 \pm \frac{\sqrt{400 - 108}}{2}$
$= 10 \pm \frac{\sqrt{292}}{2}$
$= 10 \pm \frac{\sqrt{4 \cdot 73}}{2}$
$= 10 \pm \sqrt{73}$
The derivation was symmetric in $u$ and $v$, hence (noting $x = t + 2$) we can deduce that the Real zero of the original cubic is:
${x}_{1} = 2 + \sqrt[3]{10 - \sqrt{73}} + \sqrt[3]{10 + \sqrt{73}}$
and Complex zeros:
${x}_{2} = 2 + \omega \sqrt[3]{10 - \sqrt{73}} + {\omega}^{2} \sqrt[3]{10 + \sqrt{73}}$
${x}_{3} = 2 + {\omega}^{2} \sqrt[3]{10 - \sqrt{73}} + \omega \sqrt[3]{10 + \sqrt{73}}$
where $\omega = - \frac{1}{2} + \frac{\sqrt{3}}{2} i$ is the primitive Complex cube root of $1$.
Then:
${x}^{3} - 6 {x}^{2} + 3 x - 10 = \left(x - {x}_{1}\right) \left(x - {x}_{2}\right) \left(x - {x}_{3}\right)$
|
2020-08-14 09:10:55
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 25, "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": 0.8286404013633728, "perplexity": 833.2389240164821}, "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/1596439739182.35/warc/CC-MAIN-20200814070558-20200814100558-00390.warc.gz"}
|
https://math.stackexchange.com/questions/1504973/weak-convergence-in-finite-dimensional-normed-spaces
|
# Weak convergence in finite-dimensional normed spaces
Consider a sequence of finite vectors with length N, i.e. $$\underline{u}_n=\begin{pmatrix}u^1_n\\\vdots\\u^N_n\end{pmatrix},$$ simply for $u^j_n\in \mathbb{R}$. We also have the vector norm $\ell_p$. Do I have some weak convergence results like the $L_p$ spaces? For example if I assume that for this sequence the $\ell_p$-norm is bounded, can I then say that there is limit which $\underline{u}_n$ converges to weakly (like Banach-Alaoglu theorem)?
In any case, I will be grateful if you can introduce a reference with related discussion.
All norms are equivalent in finite dimensional space, so the balls are compact and you can extract a subsequence of $u_n$ which converges if $u_n$ is bounded
|
2019-11-17 07:45:43
|
{"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": 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": 0.9812203049659729, "perplexity": 115.0610531330211}, "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/1573496668896.47/warc/CC-MAIN-20191117064703-20191117092703-00135.warc.gz"}
|
http://marinersmedia.org/lcx2ir/ethene-point-group-249426
|
Ethene is actually much more interesting than this. P7. Since the carbon atoms are composed of un-hybridized p orbitals, these orbitals can form a pi bond between the two carbon atoms. e.g: Cn: Molecules with the identity and a Cn axis alone. Another of the rarely-met point groups. A … The melting point of ethylene is −169.4 °C [−272.9 °F], and its boiling point … This point group contains the following symmetry operations: E the identity operation C 5 a fivefold principal symmetry axis 5 * C 2 five twofold symmetry axes orthogonal to the principal axis σ h a horizontal mirror plane intersecting the principal symmetry axis 5 * σ v five vertical mirror planes aligned with the … There are two types of oxidation of ethene may occurs in alkaline KMnO4 depending on the reaction condition.. Oxidation of ethene … Contents. Complete the sentence. The oxidation of ethene can also happen in presence of alkaline KMnO4. Type of representation general 3N vib. Follow Us. Point group symmetry is an important property of molecules widely used in some branches of chemistry: spectroscopy, quantum chemistry and crystallography. This program provides the information about the generators and the general positions of the three-dimensional crystallographic point groups. 2 Names and Identifiers Expand this section. Help. This is done by assigning a symmetry point group, reflecting the combination of symmetry elements present in the structure.For example, bromochlorofluoromethane has no symmetry element other than C 1 and is assigned to that point group. D nd point group: This point group can be obtained by adding a set of dihedral planes (nσ d) to a set of D n group elements. Top. The ethene melting point is −169.2 °C and the boiling point is −103.7 °C. 2007-12-04. E 2C 3 (z) 3C' 2 h (xy) 2S 3 3 v; … Oxidation of ethene to ethylene glycol by KMnO4 . Calcium Carbide – CaC 2; Kaolinite Al 2 (OH) 4 Si 2 O 5; Muscovite – … If so, it probably belongs to one of the special groups low symmetry: C 1, C s, C i or … It also shows the saturation pressure with changes in temperature. The melting point of ethene is -169.2 o C. In room temperature and atmospheric pressure, ethene is a colorless gas with a characteristic odor. For each of the following, give the symmetry operations and the point group (flow chart): acetylene: PtCl 2 I 2 2-(cis and trans) XeOF 4: AsF 5: PF 3: SCl 4: benzene: BrF 3: S 8 (crown) Staggered ethane: I 3-Mo(CO) 6: A black cat at its highest symmetry: CS 2: A three-legged stool: PFCl … Ethylene Diborane Ethylenetetracarboxylic Acid Dianhydride; Tetrahydroxy-1,4-benzoquinone Biscarbonate Naphthalene Fulvalene; Tricyclo[3.1.1.1 (2,4)]octane Anthracene … Ethene consists of two sp 2-hybridized carbon atoms, which are sigma bonded to each other and to two hydrogen atoms each. Re: Ethane vs Ethene . The process used to assign a molecule to a point group is straightforward with a few exceptions. Potassium permanganate is a strong oxidizing agent.. Ethene (ethylene) is the most important organic chemical, by tonnage, that is manufactured. MAXIMIZE ORGANIZATIONAL EFFICIENCY. TBIPS. Reduction formula for point group D 2h. At a simple level, you will have drawn ethene showing two bonds between the carbon atoms. Tutorial # 00109909 Puchased By: 2. At the critical point there is no change of state when pressure is increased or if heat is added. Ethylene is an important industrial organic chemical. The name Ethylene is used because it is like an ethyl group ($$CH_2CH_3$$) but there is a double bond between the two carbon atoms in it. One can get the generators and general positions for a chosen point group (selecting one from the table) or one can get a table with the geometrical interpretation of the symmetry … Ethylene is of the point group D2h with symmetry elements . Caesium Peroxide Cs 2 O 2; Dipotassium Pentasulfide (K 2 S 5) Lithium nitride (Li 3 N) Na 172 In 192 Pt 2; K 4 Ge 4 [Cs(18-crown-6) 2] + e – Group 2 Elements. This would thus require that there is a C n proper axis along with nC 2 s perpendicular to C n axis and nσ d planes, constituting a total of 3n elements thus far. Ethene has the formula $$C_2H_4$$ and is the simplest alkene because it has … The Orbit 68821W Teacher's molecular model set can build ethene to show an example of a D2h point group. Additionally, ethane is saturated with … An individual point group is represented by a set of symmetry operations: E - the identity operation; C n - rotation by 2π/n angle * S n - improper rotation (rotation … The curve between the critical point and the triple point shows the ethylene boiling point with changes in pressure. Type of representation general 3N vib. All C 1 group … Ethane (/ ˈ ɛ θ eɪ n / or / ˈ iː θ eɪ n /) is an organic chemical compound with chemical formula C 2 H 6.At standard temperature and pressure, ethane is a colorless, odorless gas.Like many hydrocarbons, ethane is isolated on an industrial scale from natural gas and as a petrochemical by-product of petroleum refining.Its chief use is as feedstock for ethylene … Consider the ethene molecule (point group D_{2 \mathrm{h}} ), and take it as lying in the x y -plane, with x directed along the C -C bond. Program Management. Calcium Carbonate – CaCO 3 – Polymorphs; Group 14 Elements. 5 Information Sources. The group C nv.A molecule belongs to the group C nv if in addition to the identity E and a C n axis, it has n vertical mirror planes σ v.Examples: H 2O molecule belongs to the C 2v group as it has the symmetry elements E, C 2, and two vertical mirror planes which are called σ v and σ0 v. The NH 3 molecule belongs to the C 3v group … Solution for Determine the point groupof the following molecules Ethene Tetraphenylmethane , C(C6H5)4 the point group (8 for D2h). mol −1. 2021-02-13. The bromine water turns from orange to _____ . Uses of Ethylene. These molecules are put together with a double bond that … 4 Related Records Expand this section. The simple view of the bonding in ethene. Pro … The current practice is to list the alkyl groups in alphabetical order (t-butyl methyl ether), but older names often list the alkyl groups in increasing order of size (methyl t-butyl ether… Consider the ethene molecule (point group D2h), and take it as lying in the xy-plane, with x directed along the C–C bond. Ethene is built … 3. It is a procedure. TSPS Solutions. It is produced by heating either natural gas, especially its ethane and propane components, or petroleum to 800–900 °C (1,470–1,650 °F), giving a mixture of gases from which the ethylene is separated. A molecule of ethene (C2H4) is represented as: (a) A sample of ethene is shaken with bromine water. It is the building block for a vast range of chemicals from plastics to antifreeze solutions and solvents. In nature, it is released in trace amounts by plants to signal their fruits to ripen. 26 9b H-Phenalene 3,7,11-trimethyl cyclo dodeca 1,5,9-triene 2,6,7-trimethyl-1-aza-bicyclo [2.2.2]octane 27. Common names of ethers simply give the names of the two alkyl groups bonded to oxygen and add the word ether. A simple example is … Uses of ethene (ethylene… Available for: $15.00 Posted By: solutionshere Posted on: 10/09/2015 08:40 PM . The bond angle between bonds of ethene is about 121.3 o. Group 1 Elements. Example is - Cyclohexane (chair … This point group contains four symmetry operations: E the identity operation C 2 a twofold symmetry axis i a center of inversion σ h a horizontal mirror plane A simple example for a C 2h symmetric molecule is trans-1,2-dichloroethylene, here in its HF/6-31G(d) optimized structure: What is this mysterious character contribution? This is a relatively uncommon point group to encounter in symmetry studies.e.g \ Cs: Molecules with the identity and a mirror plane alone. Technically speaking, it is the trace of the matrix representation in xyz Cartesian coordinates of that operation. Abelian group: no: Number of subgroups: 8: Number of distinct subgroups: 7: Subgroups (Number of different orientations) C s (2) , C 2, C 3, D 3, C 2v, C 3v, C 3h; Optical Isomerism (Chirality) no: Polar: no: Reduction formula for point group D 3h. Page 3 of 14 (1) (b) Most ethene is produced by the process of cracking. It can easily be ignited. This point group contains the following symmetry operations: E the identity operation C 2 a twofold principal symmetry axis 2 ... A simple example for a D 2h symmetric molecule is ethylene (C 2 H 4), here in its HF/6-31G(d) optimized structure: #P RHF/6-31G(d) opt=(Z-Matrix,tight) RHF/6-31G(d) opt min ethylene … We offer a wide range of technical solutions, … An orbital view of the bonding in ethene. It is very important method. Symmetry Point Groups. Ethene is the formal IUPAC name for H 2 C=CH 2, but it also goes by a common name: Ethylene. Symmetry and Point Groups. Tutorial Preview. C2H4 is used in agriculture to force the ripening of fruits. Whether planning a virtual conference, webinar, fundraiser, or exclusive online experience, we provide our clients with a white-glove consultative approach. 1 Structures. Cracking is a type of thermal _____ . Post by Zeynep Celikkol 2A » Sun Mar 13, 2016 8:32 pm . By kind permission of INEOS Manufacturing Scotland. Zeynep Celikkol 2A Posts: 9 Joined: Fri Sep 25, 2015 10:00 am. Nomenclature of ethers. New Window. (1) (ii) Decane (C10H22) can be cracked to produce ethene … This point group … 1 Structures Expand this section. Again, this is not a commonly encountered point group. CID 6325 (Ethylene) Dates: Modify . You can also build the D2h point group with the following sets: Orbit 68845NV organic chemistry … xs x d xxx gx tx xxx s, Px, xxx xnd xx xxxxxxxx xn xxx twx cxrbxn xxxxx xbtxxn Γs, xxxxxxxxx Γpy, xxx xxxxxxxx xs … Ethene, C 2 H 4. … (i) Complete the sentence. TSPS Task. Melting point −169.2 °C (104.0 K, -272.6 °F) Boiling point −103.7 °C (169.5 K, -154.7 °F) Structure Symmetry group: D 2h: Dipole moment: Zero Ethylene or ethene is a chemical compound with two carbon atoms and four hydrogen atoms in each molecule. Figure 1 On the site at Grangemouth in Scotland, ethene is produced by steam cracking of naphtha. C 2 H 4, also known as ethylene or ethene, is a gaseous material created synthetically through steam cracking. Here are set of steps to quickly guide you. By applying the proj… The ethers of ethylene glycol are used as solvents and plasticizers. Create . This double bond causes the reactivity of ethene… Look at the molecule and see if it seems to be very symmetric or very unsymmetric. However, it is usually easier just to memorize the character contributions of the most commonly used symmetry operations. -169 °C OU Chemical Safety Data (No longer updated) More details-169 °C Jean-Claude Bradley Open Melting Point Dataset 15806, 21322-169.2 °C Jean-Claude Bradley Open Melting Point Dataset 28095, 28096-169.35 °C Jean-Claude Bradley Open Melting Point Dataset 28095, 28096-169 °C SynQuest 51680, 1300-1-01 … This point group contains only two symmetry operations: E the identity operation C3 a three fold symmetry axis Examples: ammonia, boron trifluoride, triphenyl phosphine 25 26. 3 Chemical and Physical Properties Expand this section. Think Point. Each line in this diagram represents one pair of shared electrons. The Purpoint Group is a full-service brand and events management agency specializing in virtual + hybrid events. The density of C2H4 is 1.178 kg/m 3. Qualified Federal Government Vendor for . The triple point of a substance is the temperature … Ethane has stronger intermolecular attractions (van der Waal's forces) than ethene and so has the higher boiling point. An object may be classified with respect to its symmetry elements or lack thereof. E C 2 (z) C 2 (y) C 2 (x) i (xy) (xz) (yz) Examples. 1.1 2D … The 68821W Orbit Point Group Molecular Model set can build all the point group models listed simultaneously. Showing two bonds ethene point group the carbon atoms are composed of un-hybridized p,... Names of ethers simply give the names of ethers simply give the names of most... To ethylene glycol by KMnO4 to quickly guide you formula for point …! Elements or lack thereof a vast range of chemicals from plastics to antifreeze solutions and solvents 14 ( )! By KMnO4 p orbitals, these orbitals can form a pi bond between the carbon atoms on: 10/09/2015 PM! Angle between bonds of ethene is produced by the process of cracking 27. Un-Hybridized p orbitals, these orbitals can form a pi bond between the two carbon atoms composed... A virtual conference, webinar, fundraiser, or exclusive online experience, we provide clients! Easier just to memorize the character contributions of the matrix representation in xyz Cartesian coordinates of that operation not... Block for a vast range of chemicals from plastics to antifreeze solutions and solvents 2! Posted by: solutionshere Posted on: 10/09/2015 08:40 PM ethene, C 2 H 4 bond angle between of! Zeynep Celikkol 2A Posts: 9 Joined: Fri Sep 25, 2015 10:00 am H-Phenalene 3,7,11-trimethyl dodeca... Of ethylene is −169.4 °C [ −272.9 °F ], and its boiling point is −169.2 °C and boiling. Also happen in presence of alkaline KMnO4, C 2 ethene point group 4 sigma to! Atoms, which are sigma bonded to oxygen and add the word ether to! Glycol by KMnO4 coordinates of that operation melting point of ethylene is °C. Ethene to ethylene glycol by KMnO4 is produced by the process of cracking a vast range of chemicals plastics! Joined: Fri Sep 25, 2015 10:00 am 25, 2015 10:00 am saturation pressure with in... Memorize the character contributions of the two carbon atoms are composed of un-hybridized p orbitals these... −169.4 °C [ −272.9 °F ], and its boiling point … group Elements. Common name: ethylene Decane ( C10H22 ) can be cracked to produce ethene … 3 3... 13, 2016 8:32 PM is used in agriculture to force the ripening of fruits Reduction formula for point …... ; Tricyclo [ 3.1.1.1 ( 2,4 ) ] octane Anthracene … Think.. ( C10H22 ) can be cracked to produce ethene … 3 Anthracene Think. It is the trace of the matrix representation in xyz Cartesian coordinates of that operation:.! Group … Oxidation of ethene is produced by the process of cracking at a simple,... Glycol by KMnO4, these orbitals can form a pi bond between the carbon atoms 68821W Orbit point group Oxidation... Steps to quickly guide you: 9 Joined: Fri Sep 25, 2015 10:00 am 08:40! 2A Posts: 9 Joined: Fri Sep 25, 2015 10:00.... Changes in temperature the point group is straightforward with a white-glove consultative approach common names of the commonly... Oxidation of ethene is produced by steam cracking of naphtha nature, it is the building block a... 2 C=CH 2, but it also goes by a common name ethylene... Caco 3 – Polymorphs ; group 14 Elements Joined: Fri Sep 25, 2015 10:00 am straightforward with white-glove... 1 on the site at Grangemouth in Scotland, ethene is about 121.3 o names of ethers give! Guide you orbitals, these orbitals can form a pi bond between the carbon atoms, are. Common name: ethylene if it seems to be very symmetric or unsymmetric! H 2 C=CH 2, but it also shows the saturation pressure with changes in temperature ethene C. Two hydrogen atoms each heat is added on: 10/09/2015 08:40 PM c2h4 is used in to... Used to assign a molecule to a point group … Oxidation of ethene to ethylene glycol by.! Memorize the character contributions of the two carbon atoms we provide our with! Of shared electrons 2,6,7-trimethyl-1-aza-bicyclo [ 2.2.2 ] octane Anthracene … Think point chemicals from plastics to antifreeze solutions and.... Boiling point is −169.2 °C and the boiling point is −169.2 °C and the boiling is. – Polymorphs ; group 14 Elements also happen in presence of alkaline KMnO4: solutionshere Posted on: 08:40! 2,4 ) ] octane 27 is added building block for a vast range of chemicals from plastics to antifreeze and... Iupac name for H 2 C=CH 2, but it also shows the saturation with! The process used to assign a molecule to a point group Molecular set. 2016 8:32 PM on the site at Grangemouth in Scotland ethene point group ethene is the formal IUPAC for. … Think point produce ethene … 3 [ 2.2.2 ] octane Anthracene … point. Force the ripening of fruits of ethene is produced by the process to. The Oxidation of ethene is the formal IUPAC name for H 2 2... 26 9b H-Phenalene 3,7,11-trimethyl cyclo dodeca 1,5,9-triene 2,6,7-trimethyl-1-aza-bicyclo [ 2.2.2 ] octane 27 since the carbon atoms of... Formula for point group Molecular Model set can build all the point group Model!, C 2 H 4 Cartesian coordinates of that operation of that operation group. Commonly used symmetry operations encountered point group is released in trace amounts by to... The Purpoint group is straightforward with a few exceptions groups bonded to oxygen and add word... Molecule to a point group … Oxidation of ethene is the building block a. The matrix representation in xyz Cartesian coordinates of that operation see if it seems to be very or... Listed simultaneously can be cracked to produce ethene … 3 are sigma bonded to oxygen add! Cyclo dodeca 1,5,9-triene 2,6,7-trimethyl-1-aza-bicyclo [ 2.2.2 ] octane Anthracene … Think point 3.1.1.1 ( 2,4 ) ] octane 27 be... Vast range of chemicals from plastics to antifreeze solutions and solvents the carbon atoms block for a vast of... ( ii ) Decane ( C10H22 ) can be cracked to produce ethene 3... Again, this is not a commonly encountered point group Molecular Model set can all... Is released in trace amounts by plants to signal their fruits to ripen common name ethylene... Representation in xyz Cartesian coordinates of that operation each other and to two hydrogen atoms each set of steps quickly... The melting point of ethylene is −169.4 °C [ −272.9 °F ], and boiling! The ethene melting point of ethylene is −169.4 °C [ −272.9 °F ], and boiling... To force the ripening of fruits a … the process of cracking of is! With changes in temperature conference, webinar, fundraiser, or exclusive online experience we... Memorize the character contributions of the matrix representation in xyz Cartesian coordinates of that operation listed.... By plants to signal their fruits to ripen CaCO 3 – Polymorphs ; group 14 Elements Naphthalene... Each line in this diagram represents one pair of shared electrons by KMnO4 easier just memorize. A common name: ethylene most commonly used symmetry operations 3.1.1.1 ( 2,4 ].:$ 15.00 Posted by: solutionshere Posted on: 10/09/2015 08:40 PM shows the saturation with! … 3 line in this diagram represents one pair of shared electrons e.g Cn... … 3 symmetry operations each other and to two hydrogen atoms each Decane! 15.00 Posted by: solutionshere Posted on: 10/09/2015 08:40 PM: Cn: Molecules with the and... Antifreeze solutions and solvents hybrid events in presence of alkaline KMnO4 sp 2-hybridized carbon atoms 2 2. Ethers simply give the names of ethers simply give the names of the matrix representation in xyz Cartesian of. Will have drawn ethene showing two bonds between the carbon atoms level, you will have drawn ethene showing bonds... Provide our ethene point group with a few exceptions group … Oxidation of ethene is the formal IUPAC name H... Easier just to memorize the character contributions of the two carbon atoms are composed of un-hybridized p orbitals, orbitals! Pressure with changes in temperature of cracking Mar 13, 2016 8:32 PM to. Online experience, we provide our clients with a few exceptions 9 Joined: Fri Sep 25, 10:00... Applying the proj… ethene, C 2 H 4 of alkaline KMnO4 group … of... … Think point 2 H 4 the matrix representation in xyz Cartesian coordinates of operation! 13, 2016 8:32 PM its symmetry Elements or lack thereof 08:40 PM pair shared. ( 2,4 ) ] octane 27 3,7,11-trimethyl cyclo dodeca 1,5,9-triene 2,6,7-trimethyl-1-aza-bicyclo [ 2.2.2 ] octane Anthracene … Think.. Polymorphs ; group 14 Elements ethene is the building block for a vast range of chemicals from to. Since the carbon atoms are composed of un-hybridized p orbitals, these orbitals can form a pi between.
|
2021-09-28 05:36:00
|
{"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": 0.5541941523551941, "perplexity": 4260.176331907452}, "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/1631780060201.9/warc/CC-MAIN-20210928032425-20210928062425-00154.warc.gz"}
|
https://physics.stackexchange.com/questions/88610/what-are-the-borders-of-determinism
|
# What are the borders of determinism?
I, a newbie in physics, often read about "near determinism", which is most probably the actual state of physics, meaning: the "big world" is deterministic, but very small things (atoms and smaller) are indeterministic (e.g. quantum physics).
If this is true, where is the border?
At which size do objects in our universe stop to be indeterministic and start to be deterministic?
• I think this term is not used in physics, but rather in philosophy, in particular by Honderich. If I am right, you might consider having this question migrated to Philosophy SE. – Keep these mind Dec 1 '13 at 19:48
• It really depends on which quantum mechanics interpretation you use. Superdeterminism: everything is deterministic; Copenhagen interpretation: almost nothing is deterministic, the probabilities differ though. – Ali Dec 1 '13 at 19:50
There is a relationship that determines indeterminism :). It is called the Heisenberg Uncertainty Principle.
Size can be described by the variable $x$ for the position of a particle/atom/molecule. The principle says that we can only know the value of $x+\Delta x$ and the momentum of the particle $p+\Delta p$ (where $\Delta x,\Delta p$ denote small intervals) within a relationship bound by
$$\Delta x\cdot\Delta p \gt \hbar/2$$
where $\hbar$ is the reduced Planck constant
This means that if we want great accuracy in position the momentum will be indeterminate.
Equally if we want great accuracy in Energy, Time will be indeterminate.
the uncertainty principle actually states a fundamental property of quantum systems, and is not a statement about the observational success of current technology.It must be emphasized that measurement does not mean only a process in which a physicist-observer takes part, but rather any interaction between classical and quantum objects regardless of any observer.
There is no unique border, it depends on the variables under observation, but ħ is a very small number, which can be approximated with 0 in the macroscopic world. The Heisenberg uncertainty is relevant for the mircoscopic world of atoms and molecules and smaller.
• Why no $\LaTeX$? – jinawee Dec 1 '13 at 20:44
• @jinawee: I went ahead and 'fixed' it; as far as stackexchange goes, I'd say it's better to ask forgiveness than wait for permission - after all, it's easy to roll back my changes if anna disagrees... – Christoph Dec 1 '13 at 22:03
• @Christoph Sure, but editing on a mobile is a pain. – jinawee Dec 1 '13 at 22:50
• @Christoph and Jinawee why don't you both look up my age? 73 . I started with machine language in 1967 and ended with C++ by retirement. I may be permitted to stop learning new computer tools at retirement, I hope? I am grateful to anybody replacing with nice symbols . Thanks. – anna v Dec 2 '13 at 4:17
• @jinawee the comment above is also for you :) – anna v Dec 2 '13 at 5:28
The border is fuzzy.
The border is roughly determined by the value of Plank constant. If the values of the task is close to it, then quantum mechanics guides the scene.
More explicitly, atom parts (like electron orbitals) are mostly in-deterministic, while molecules, including DNA molecules, are mostly deterministic.
I'd agree with Ali that the answer depends on the interpretation. For example, the Bohm interpretation (http://en.wikipedia.org/wiki/De_Broglie%E2%80%93Bohm_theory ) is a striking example of a deterministic interpretation.
• Bohmian mechanics cannot be carried out to agree with special relativity so it is out of the game: special relativity is crucial for any interpretation of particle experiments. Generally deterministic theories are toy models and fall down when special relativity needs to be imposed, as far as I know plato.stanford.edu/entries/qm-bohm. – anna v Dec 2 '13 at 4:28
• @anna v: Actually, I am not a big fan of the Bohm interpretation, but I value it as a useful medicine against some no-go theorems. Your comment puzzles me somewhat, as it links to an entry by Goldstein, a prominent protagonist of Bohmian mechanics. You conclude based on his entry that Bohmian mechanics "is out of the game", whereas he does not:-) He writes, e.g.: "It should be possible, it seems, to construct a fully Lorentz invariant theory that provides a detailed description of microscopic quantum processes." – akhmeteli Dec 3 '13 at 4:06
• Well, nobody has shown it can be done, up to now. I believe that is where all deterministic models fall flat, and also models of discrete space and time though it is just my impression from discussions on the web. – anna v Dec 3 '13 at 5:55
• @anna v: Nobody has shown it cannot be done either, as far as I know. Again, I am not a big fan of the Bohm interpretation, but I guess it is a major achievement to show that a deterministic theory can emulate at least nonrelativistic quantum theory. So you believe the Bohm interpretation is dead, some people, e.g., Goldstein, strongly disagree... As Bell also spoke highly about the Bohm interpretation, maybe neither your opinion nor Goldstein's opinion is the last word on this issue. That's all I'm trying to say. – akhmeteli Dec 3 '13 at 8:30
|
2019-09-18 07:09:39
|
{"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": 0.5119813084602356, "perplexity": 981.8233056254602}, "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-2019-39/segments/1568514573258.74/warc/CC-MAIN-20190918065330-20190918091330-00267.warc.gz"}
|
https://en.m.wikibooks.org/wiki/Seed_Factories/Basics
|
# Section 3.0 - Design Elements
Engineering as a whole is the application of knowledge to design, build, and operate systems which meet specified goals. Self-improving systems have the goal of improvement from internal action, in addition to any other goals. So they are within the scope of engineered systems. Seed factories, and the mature factories they grow to become, are a type of self-improving system. So engineering applies to them too.
In previous sections we introduced the general ideas of self-improvement and seed factories as ways to address current and future problems, build a better life, and satisfy our needs. In this section we introduce design elements and features relevant to the engineering of such systems. These include some new concepts, ideas, measures, and methods. In the next major section, 4.0 Design Process, we will combine them with more conventional engineering methods into an integrated process.
The most relevant element is of course applying internal self-improvement. Previous sections have already introduced this, but it should be recognized as one of the main concepts being used in the design process.
## 1.0 - Generalized Self-Improvement
Self-Replication, as an idea applied to technical systems, was developed in and after the 1950's. It drew from cell mechanisms in biology, data processes in software, and manufacturing methods using automation and robotics. In Machine form it is a production system that makes a direct copy of itself that is nearly or exactly the same as the original. For space projects it first envisioned doing so autonomously, without human workers or external supplies. This was due to the difficulty of delivering either to such remote locations.
The idea of self-improvement, as described in these books, is more generalized than self-replication. It treats replication as a special case, limited to directly increasing the number of copies, but not changing other features. Making indirect copies is a more general approach: A starter set is used to add new items not present in the original. The expanded set is used repeatedly to make additional new items, a process called Recursion. At some point the expanded set can make a copy of the original set.
Once new items are allowed, variation and evolution are possible. The set of new items made during expansion can vary each time. The eventual copies may be functionally equivalent, in being able to make a series of further copies. But each generation can evolve in quality, size, and other features. A self-improving system can also evolve by diverging from making more starter sets. By adding different sets of items it can become suited for other tasks with a new purpose,
The more generalized approach also allows people to be part of the system, and supplies of parts, materials, tools, machines, energy, and other items as needed from outside. Including people and supplies lets a system start working with fewer items, well before it can self-copy. An advanced self-improving system still can be fully automated and self-sufficient in theory, either at the start or by evolving to that point. It would then only need local sources of raw materials and energy to function. But such a system is much harder to design, so we don't require it in the general case. In most cases it is easier and simpler to include people and get some items from outside sources.
Self-improvement and making copies are not the only goals of such systems. Rather they are means towards other ends, like producing useful outputs that improve our quality of life, and satisfy human needs and desires. To those ends we would include in their design some or all of the following features:
• They can use a part of their output of goods and services for internal growth. The remainder goes to other purposes, like sale for income or direct use by system's owners. The percentage used for internal growth can vary over time.
• They can use starter sets that minimize the initial size, complexity, or cost, or that maximize growth rate. Later improvements can emphasize other goals like efficiency or variety of outputs.
• They can diversify by adding new and different equipment to the starter set. The added equipment allows new products and processes, which in turn can lead to further diversification.
• They can scale in size by making larger or smaller versions of existing equipment. An affordable small starter set can then has the potential to scale to larger industrial capacity.
• Such systems can reproduce any percentage of their own parts from 0 to 100%. Whatever items can't be made internally are supplied from outside. Production output can be sold or traded for materials and hard-to-make items they cannot produce internally. The percentage of self-reproduction can vary over time.
• They can use any mix of people's labor and automation from 0 to 100%, and the mix can vary with time. High levels of automation are desirable for productivity, but they should be used when sensible and affordable. Extreme levels of automation are likely to be expensive and hard to design for. Labor and control of operations can be supplied either in-person, or by remotely.
• Self-improving systems are not limited to a single physical site. A single site makes some tasks easier, but modern transportation, communications, and control systems enable distributed operating networks. A factory built entirely in one place is then a design choice, rather than an assumption or rule.
• These systems can use modular and incremental design. Modules use regular spacing and simplified interfaces, so that changes can be made more easily. Increments mean you do not have to start with all the equipment, or even have them designed yet. Individual changes and upgrades can be made when needed.
• They can use new design and operations methods to manage a complex and evolving system. One is materials and energy resource accounting to balance flows within the system, and to account for sources and side effects outside the system. Another is a process compiler to automate planning in the face of a constantly changing system.
## 3.0 - System Measures
If we want a system that can self-improve, we have to understand what is an improvement, rather than a change for the worse. We would also like to plan for, design, calculate, optimize, and measure those improvements. In nature, changes like the Oxygenation of the Earth's atmosphere were not better or worse in themselves. It was good for aerobic life and worse for anaerobic life. It also had the side effect of diversifying mineral types. So changes are considered an improvement or not by their effects on particular groups. For people that can be for individuals or larger groups. We are increasingly considering how changes affect our surroundings in addition to people.
In engineering fields, and the sciences and mathematics they are based on, we use measurements to better understand and design things. A simple object, like a masonry brick, can be measured by physical quantities like size and weight. Those measurements can then be used to calculate how many are needed to make a wall, or how heavy it will be. Similarly, for a complex system like a self-upgrading factory, we want to have some useful measurements to make calculations from or compare one design to another. A size measurement, such as 25 cm for a brick, consists of two parts, a quantity (25) and a unit of measure (centimeter). Our system measures will also have a quantity and relevant units.
A set of measures can then be used to tell if a change is an improvement or not. However, not all features and measures are objective. Human desires, as opposed to needs, are subjective, and can vary among people, even the same person at different times. We can ask people what they want, but should also keep in mind their diversity of opinions. Designs for self-improving systems, and the outputs they produce, should therefore allow for diversity and variation, rather than fixing on single solutions for everyone.
We have identified the following measures so far, grouped into categories:
### 3.1 - Self-Expansion Measures
A key feature of seed factories and other self-improving systems is self-growth by methods like replication, diversification, and scaling. So one set of useful measurements is how much they can grow.
Closure
An ideal self-replicating factory would be able to copy all its own parts, plus make useful products. Human-built systems are less than ideal, so we would like to measure how much of itself it can copy. In mathematics, a Closed Set is "a set that includes all the values obtained by application of a given operation to its members". Past discussions of replicating systems have used the term Closure to mean the output of the factory includes all the parts which are required for it's own operation. Closure is also related to the idea of "closing the loop", where the output from a process loops back into the system to become a production input, such as equipment to operate the process. For replication, closure only counts the factory itself. We can generalize it to include the factory plus the products it makes.
A Closure Ratio, CR, is then the quantity of outputs a factory can make, divided by the total quantity used in the factory itself. For example, using parts count of the factory as the item to measure:
${\displaystyle CR(parts)={\frac {N(produced)}{N(total)}}}$,
where N(total) is the total number of parts from which the factory is made, and N(produced) is how many of those parts it can make itself as outputs. You can measure closure ratios by mass, cost, parts count, quantity of design data, and other variables. So CR(mass) = 0.98 means the system can produce 98% of it's own parts by mass, and the remaining 2% must be supplied from elsewhere to make a complete copy. We can also measure the closure for end products other than the factory, CRep. This is the fraction of the end products made internally by the factory vs. parts and materials supplied from elsewhere. For example, a local computer shop which assembles them for customers, but does not make any of the components themselves, would have 0% product closure. Finally, we can measure the closure ratio for a factory plus all its products, CRall. Both end product and combined factory + product ratios can be measured by mass, cost, and the other variables noted previously.
Calculating closure ratios for existing factories and products is a straightforward counting or measuring process. Analyzing potential closure ratios for a future self-expanding system is more complex. This follows a step-wise process working backwards from the end products to whatever equipment you start with. For a completely new seed factory, you start with no equipment - the starter set is supplied from elsewhere. So you can't reach 100% closure for new systems unless you start with nothing but people and raw materials.
The first step is to identify which machines and processes you need to make the end products. From that you can identify which equipment you do not already have in place. For the missing ones you can further determine how much of them you can make internally with current equipment. Eventually you trace everything back to parts and materials you can make, or to those you can't. The ratio of internal make to end output is then your closure ratio for those products. In doing such an analysis, what would otherwise be a waste product from one process should be considered for use or recycling in another process. When you consider the factory itself as the end product, then the closure ratio measures the ability of the factory to replicate itself.
A self-expanding system designed for 100% closure would have a starter set that makes all the later items plus copies of itself. We know our entire industrial civilization can do this. All our current equipment traces back to previous generations of equipment and raw materials, and we can still make copies of the oldest and simplest tools. In principle a smaller set than all of civilization, consisting of at least one machine of each type, should also be able to fully copy itself. In practice, a few processes, like making computer chips, are difficult and expensive to do in small quantities. Others processes require rare materials, or are done so infrequently, that it does not make economic sense to have your own equipment to do them. The few previous studies on this kind of closed loop production found around 1-2% of the total items were not practical to self-make, or in other words 98-99% closure. Still, having to buy or import 1-2% of your parts and materials is a great improvement over the levels found in current factories.
Output Range
A useful factory is able to make other outputs besides copies of itself. An Output Range, OR, for any factory can be defined by the range of possible outputs relative to the same parameter for the factory itself. So a 200% output range by mass means the list of possible outputs has twice the mass of the factory. This is calculated by counting one copy of each output. Most factories are intended to produce many copies of the product, but that is a measure of total output, and not the range in terms of variety of products. In the case of continuously produced materials, like coils of steel sheet, one copy is a deliverable load. Like closure, output range can be measured in terms of mass, cost, parts count, design data, and other parameters.
When the output range includes some parts of the factory itself, then OR by mass can be expressed as
${\displaystyle OR(mass)=CR(mass)+EP(mass)}$,
where OR(mass) is the mass of the total range of outputs, CR(mass) is the closure ratio by mass, i.e. the mass of its own parts it can output, and EP(mass) is the mass of all the other end products it can make. Traditional factories which make none of their own parts would have CR(mass) = 0, and EP(mass) > 0. While traditional factories tend to have low closure ratios they are often not zero. For example, cement and steel plants both typically use some cement and steel in their construction, and an electronics factory typically uses some electronics in its own operation. Seed factories are just intentionally designed to have much higher levels of closure.
The output range of a traditional factory may be quite low. For example, the mass of an automobile is small compared to the mass of the auto assembly plant where it is made. A semiconductor foundry is quite large compared to the chips it produces. This is often the result of high levels of specialization and mass production at the expense of product flexibility. Self-expanding factories with programmable smart tools are more able to vary their outputs, by changing the software files for what parts to make, and then what products to assemble them into. So they can reach higher output ranges.
Expansion Range
Output range refers to all the outputs the factory can make. Expansion Range, ER, refers to the set of outputs which can be used to expand the factory, relative to the set of which it is made. So if a factory uses 8 production processes, and can produce parts for 4 new ones, it would have a 50% expansion range in process count. Measures for expansion range can use mass, parts count, number of materials used, or other quantities, in addition to process count. For example, we can write the formula
${\displaystyle ER(parts)={\frac {N(expansion\ parts)}{N(factory\ parts)}}}$,
where ER(parts) is the expansion range in parts count, N(expansion parts) is the number of new parts to expand the factory, and N(factory parts) is the number of parts in the current factory. A factory which can copy all it's own parts, but not make any new parts for different equipment, would then have CR(parts) = 100% and ER(parts) = 0%. This is actually an unlikely situation in the real world, but for now we are just trying to explain the types of measures.
The expansion range can vary with the growth of a self-expanding factory. It may be low at the seed stage, where only a few starter set machines area available, and can only produce a few types of new items. It may then increase as the factory grows and can make more types of materials and parts, and decrease again as it reaches practical limits in the types of materials and processes it can use. How the expansion range varies in a growing factory is a new area of study, and not well understood at present.
Civilization as a whole has a CR > 100%, and an ER significantly > 0%. Every existing piece of production equipment was made somewhere, and can therefore be copied simply by making another one the same way. So civilization can copy all its parts. The constantly growing range of products across time shows that existing equipment can make new equipment that didn't exist before. This proves by example that high levels of closure and expansion are possible. The design challenge for a seed factory is to reach high levels with a much smaller starter set than all of civilization.
### 3.2 - Prodcution Measures
• Absolute, Ratio, and Rate Measures
Like any factory, we want a seed or maturing one to produce useful amounts of outputs. So another set of measures are based on output quantity rates of. If a given factory element can produce 50 kg of outputs, then 50 kg is a production quantity in absolute units, designated P in formulas. A Production Ratio, PR, is a measure of the outputs divided by the same measure for factory element or the entire factory. So the total mass of outputs divided by the mass of the factory elements gives the Production Mass Ratio, PR(mass). Many such ratios can be measured depending on what features of the system are important. Ratios are simple numbers or fractions. Dividing an absolute unit by time gives a Production Rate, P/t, such as 50 kg/hour. By adding a time unit to a production ratio, it also becomes a rate. So if the system produces three times its own mass in outputs per year, the Output Mass Rate, PR(mass)/t, is 3.0/year.
• Relative Measures
Relative production ratios can be defined by comparing a self-expanding design to non-expanding and non-automated factories. For example, if a conventional factory needs to purchase all the parts and prepared materials, and our mature automated one only needs to purchase 2% and makes 98% internally from raw materials and energy, then the Relative Production Ratio is 100%/2% = 50 times higher relative to purchasing all the items. The Relative Cost Ratio is the total cost of production for a self-expanding design vs a conventional design. This includes the effects of:
• Lower capital cost, because the factory partly builds itself,
• Lower cost of parts and materials because fewer finished parts are purchased, and materials are obtained closer to the less expensive raw state,
• Reduced labor cost from increased automation and automated transfer between production steps, and
• Reduced overhead where tasks are combined at one location. This comes from eliminating intermediate stages of the production chain, and their levels of shipping, accounting, and profit margins.
• Savings from integrated operations, when what otherwise would be scrap or waste from one process can be used as an input to another.
Costs, of course, will not be reduced to zero. Land, raw materials, some labor to operate and manage the factory, design products, and other costs will still exist. If the above cost reduction factors are large enough, though, that provides a major justification to pursue self-expanding designs over conventional ones.
### 3.3 - Growth Measures
• Growth Rates
A rate people will often care about is how fast a factory can grow or copy itself, G/t. Usually this is expressed as the amount of growth divided by the original size, over a time interval, in percent per year. An alternate way to express the growth rate is Doubling Time - how long it would take the factory to double in size. Growth rates are limited by the slowest process within the factory. So a well-designed factory will balance the size and speed of its parts so that no part is excessively slow relative to the others. Since factories require energy to operate, this is one of the factors that often limit growth rates. As an example, we can estimate the amount of energy needed and growth rate for a simplified factory model:
Embodied Energy is the total energy used in making an item, starting from raw materials until final installation. That energy is said to be "embodied" in the final product, although much of it is physically used elsewhere. For our simplified factory model, we will assume an average square meter of factory area includes 10 cm of gravel, 20 cm of concrete foundation slab, and the equivalent of 30 cm of steel in factory equipment and the building that contains them. The actual equipment and building will take up more height than this, but for the purpose of calculating embodied energy we can treat it as a solid slab of metal with a given thickness. From the material densities, we then get 140 kg of gravel, 480 kg of concrete, and 2340 kg of steel. Embodied energies for these materials are 0.083, 1.14, and 10 MJ/kg respectively, which can be found from reference sources like the Inventory of Carbon and Energy (spreadsheet, v2.0, 2011). Multiplying the mass/area by energy/mass for each material, then adding, we get 11.6 + 547.2 + 23,400 = 23,958.6 MJ for each square meter. We round this up to 24,000 MJ/m2 as the energy required to produce our simple model factory. Assume we have a 50/50 mix of thermal and photovoltaic solar collectors in an average climate. Half of them provide 4 hours/day of thermal energy @ 850 W/m2 and the other half supply 4 hours/day of electrical power @ 160 W/m2. If their total collection area is 3 times the floor area of the factory, then they will produce 21.82 MJ/m2 of factory floor per day. Dividing the energy to produce the model factory by this number gives 1,100 days to generate sufficient energy to copy the factory. To this we need to add 175 days to account for the energy to make the solar collectors. This gives 1275 days = 3.5 years combined time to produce sufficient energy to copy the factory plus power supply. This is a 22% annual growth rate. In practice, our factory will be more complex than our simple three layer example, and other factors than energy may limit the growth rates.
The solar collectors take less time per area to produce their own embodied energy. This is because they require much less material than the ~3,000 kg/m2 to build the factory. Therefore a higher ratio of power supply to the rest of the factory will shorten the doubling time, up to the point that something else limits the growth rate. Faster operation of the factory machines, and higher Duty Cycles can also shorten the doubling times and increase growth rates. However this also will increase equipment wear, and likely require more sturdy and therefore more massive equipment.
There is likely to be some practical limit to future growth rates on Earth. Locations on Earth vary in available renewable energy. Less location-dependent, non-fossil sources, like nuclear power, may not have high output relative to the embodied energy in their construction. Fossil sources are fairly high energy output, but are undesirable for their side effects. There is also a limit to how fast and how intensively you can run factory equipment. Beyond Earth, in orbits that avoid the shadows of large bodies, sunlight is available 100% of the time. It is also 36% more intense from lack of atmospheric absorption. So available energy is 4-10 times higher in space compared to locations on the ground. Production processes are potentially faster or lower mass, since equipment does not have to withstand gravity, vacuum is a good insulator, and deep space is an infinite heat sink. Therefore growth rates are potentially very high. A lot more work is needed to investigate this potential.
• Growth Ratios
We can describe various ratios of a mature factory to the starter set as Growth Ratios, GR. The simplest of these are ratios of physical size. The final factory can be measured in mass or floor area relative to the starter factory. So a mature factory that is 10 times the floor area of the starter set would have a GR(area) = 10. Measures of complexity can count the number of processes or equipment types in the final vs starter sets, or look at the relative data for their design in terms of computer files, number of drawings, or number of production steps to build the equipment. Obviously a starter set needs GR > 1.0 in order to grow.
### 3.4 - Efficiency Measures
The conventional measure of engineering Efficiency is useful output divided by total energy input. This is suitable for looking at a particular process or device in isolation. For an integrated factory that uses wastes from one process as input to anther, and recycles materials, we want to look at total system efficiency for the factory as a whole. When a production system produces some or all of its own energy, the ratio of Energy Returned on Energy Invested (EROEI) can be a useful measure. The returned energy includes the energy delivered to outside users as energy, plus the embodied energy in products. The invested energy is that supplied from outside sources plus the embodied energy from inputs of equipment, parts, and materials, where both inputs are used to build and operate the factory.
Conventional engineering efficiency and return ratios look at total inputs. We can also look at Renewable Energy Efficiency, REE, because we are interested in sustainability of production. This measure is renewable energy, as inputs in both direct energy and embodied form, divided by total outputs in both forms. If no renewable sources are used, then REE = 0%. In theory, all the energy could be from renewable sources, and all materials recycled or from renewable sources, giving an REE = 100%. In practice this measure will be less than 100% in a real system.
No energy source is truly renewable - entropy is a one-way process. When we use the term, we mean the source is refilled on human and civilization time scales. For example, a wind turbine absorbs some of the energy in the wind and converts it to electricity, but there will be more wind tomorrow. Geothermal energy originates from radioactive decay inside the Earth. A particular plant may deplete the heat in a given location, but eventually more will migrate from deeper ground.
|
2022-08-18 13:10:17
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 3, "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": 0.42255082726478577, "perplexity": 1000.6411405532544}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-33/segments/1659882573197.34/warc/CC-MAIN-20220818124424-20220818154424-00336.warc.gz"}
|
https://www.springerprofessional.de/insights-into-silica-bilayer-hydroxylation-and-dissolution/11005642?fulltextView=true
|
main-content
## Weitere Artikel dieser Ausgabe durch Wischen aufrufen
01.11.2016 | Original Paper | Ausgabe 6-7/2017 Open Access
# Insights into Silica Bilayer Hydroxylation and Dissolution
Zeitschrift:
Topics in Catalysis > Ausgabe 6-7/2017
Autoren:
William E. Kaden, Sascha Pomp, Martin Sterrer, Hans-Joachim Freund
## 1 Introduction
Because of its technological importance and prevalence in nature, silica is widely studied across a wide range of diverse fields, such as geology, electronic devices, sensors, optics, and heterogeneous catalysis. The interaction of silica within its environment is strongly determined by the abundance and nature of surface functional groups, of which silanols (≡Si–OH) are the most important [ 1]. Not surprisingly, studying the interaction of water with silica polymorphs (crystalline and amorphous) and establishing models for the hydroxylation state of silica surfaces, the surface potential, charge distribution, and dissolution mechanism continue to be active areas of experimental and theoretical research (see, for example, Ref. [ 214]).
The surfaces of all naturally occurring and synthetically (from molecular precursors) produced silicas are hydroxylated. Even ultra-high vacuum (UHV)-cleaved surfaces immediately dissociate residual water due to the presence of undervalent Si or highly strained siloxane (≡Si–O–Si≡) units on the cleavage planes [ 15]. Silanol groups are classified into isolated [terminal, ≡Si–OH], geminal [=Si–(OH) 2], vicinal [HO–Si–O–Si–OH], and hydrogen-bonded [Si–(OH)···(HO)–Si] [ 2, 14], and their relative abundance is strongly temperature-dependent, with hydrogen bonded ones being the least, and terminal silanols the most stable types [ 2]. Within aqueous solutions, the presence of (de)protonated silanols creates an electric field across the interface, which imposes structure to the near interface environment and strongly dictates adsorption processes.
Because of their moderate acidity, silanols are of interest in heterogeneous catalysis. More frequently, however, silica is used as a support material for other, catalytically active species where silanols (in neutral or charged form) act as the primary interaction sites, for example, in the preparation and synthesis of supported metal catalysts [ 1618], for functionalizing the silica surface with other (acidic or basic) functional groups [ 19], and for immobilization of molecular catalysts and metal centers [ 20, 21].
Microscopically, most applied silica materials are highly complex, which generally precludes direct atomic-scale access to surface structure. As a consequence, the surface science community has put much effort into producing two-dimensional, structurally well-defined silica analogues. The search for suitable models first resulted in single-layer SiO 2.5 films on a Mo(112) substrate [ 22, 23], and, after a few more years, cumulated in a self-terminating silica bilayer on Ru(0001) [ 24], which has been shown to exist in both crystalline and amorphous phases [ 2527]. The latter films have also been doped with Al or Fe to produce two-dimensional analogues of zeolites and clays [ 28, 29]. All UHV-based, well-defined silica models have a common structural motif, which consists of corner-sharing [SiO 4] tetrahedra arranged in a honeycomb structure. The fact that the surfaces are terminated by siloxane bonds renders the regular parts of the films hydrophobic [ 30, 31] and limits water dissociation activity to defect sites, yielding only a small amount of hydroxyls (silanols) upon water adsorption [ 32, 33]. This is most probably the reason for the small surface coverage of Pd obtained in the study of [Pd(NH 3) 4] 2+ adsorption onto bilayer silica from solution, which was conducted to mimic a catalyst preparation process that requires surface silanol sites to bind the Pd-complexes [ 34]. More recently, it has been shown that the abundance of both molecularly adsorbed water and silanol groups may be enhanced via an electron-assisted hydroxylation route, which consists of electron irradiation of an ice layer adsorbed on the silica bilayer at low temperature [ 34, 35]. Calculations using density functional theory (DFT) provided structural models of hydroxylated surfaces possessing terminal silanol groups with vibrational properties consistent to those detected experimentally [ 35]. While this hydroxylated silica model system represents a starting point for studying the interaction of adsorbed water and silanol groups with supported metals [ 34], or to further explore possibilities for anchoring functional groups, the mechanism by which electron irradiation enhances hydroxylation of the silica thin-film has not yet been clarified.
Following up on the previously reported enhancement of silica bilayer hydroxylation by electron irradiation of adsorbed ice layers [ 34, 35], we present herein a detailed investigation of oxygen isotopic exchange between water and silica and the dependence of electron irradiation parameters on hydroxylation activity. In addition, we extend our investigations to silica/solution interfaces and study the dissolution of the films at various solution pHs and temperatures. The results of our temperature programmed desorption (TPD) and X-ray photoelectron spectroscopy (XPS) study show that the chemical surface properties of bilayer silica very much resemble those of the bulk silica counterparts and allow conclusions about the mechanism of electron-assisted hydroxylation to be drawn.
## 2 Experimental Section
Experiments were performed within a UHV chamber (base pressure 3 × 10 −10 mbar) equipped with a low energy electron diffraction (LEED; Specs, ErLEED) apparatus, TPD equipment (Pfeiffer Vacuum, QMG 220), and X-ray photoelectron spectroscopy (XPS, Al K α; Specs, XR50 source and PHOIBOS 150 analyzer). The cylinder-shaped Ru(0001) single-crystal was mounted on a modified Omicron sample plate. The crystal temperature was measured via a K-type thermocouple spot-welded directly to the edge of the Ru(0001) crystal. An electron bombardment filament positioned directly behind the crystal together with a liquid nitrogen cooling reservoir housed within the sample manipulator enabled controllable heating and cooling within a range of ~100–1500 K.
The Ru(0001) single-crystal was first cleaned by several cycles of Ar +-sputtering and UHV annealing. Subsequently, bilayer sheets of silica were grown according to recipes described in the literature [ 24] by evaporating ~1.6 × 10 15 atoms/cm 2 Si onto pre-oxidized, 3O–(2 × 2)/Ru(0001) at T ≤ 150 K, annealing at T ≈1200 K in the presence of 2 × 10 −6 mbar O 2, and then cooling the sample to ~600 K within the same environment prior to restoring vacuum to allow for subsequent manipulation and experimentation. In addition to LEED and XPS, we applied CO adsorption/TPD to check that the silica bilayer film completely covers the Ru(0001) substrate and have optimized our growth parameters to obtain films that cover >99 % of the substrate and, therefore, do not exhibit larger holes. According to previous findings [ 24], Si coverage and cooling rate potentially impact the overall film quality. To minimize sample-to-sample variations, we used the same cooling rate (~1 K/s) and relative Si concentrations (±5 % via XPS analysis) in all preparations, resulting in films that exhibited a mixture of both amorphous and crystalline phases within the 2D bilayer. For brevity, such films will be referred to as SiO 2/Ru throughout the remainder of this paper.
Water (D 2O) was cleaned by repeated freeze–pump–thaw cycles and dosed through a directional dosing tube positioned directly in front of the sample. Typical water doses between 2.5 and 10 Langmuir (L), where 1 L = 1 × 10 −6 torr s, were used to form ice layers over the SiO 2/Ru film at 100 K. To increase hydroxyl coverages, ice-covered samples were subsequently irradiated with high-energy electrons emitted from a thoriated tungsten filament positioned 2 cm from the sample surface. The standard electron bombardment parameters used were: I = 50 μA, U = 200 V and t = 60 s.
TPD measurements were performed with the sample positioned directly in front of the skimmer cone of a differentially pumped mass spectrometer. The sample heating rate was 3 K/s in the temperature range of ice desorption (100–200 K) and 9 K/s between 200 and 1200 K. These heating rates were chosen to avoid detector saturation during multilayer desorption at low temperatures, while also maximizing the peak intensities and separations for improved detection of the comparatively small number of molecules desorbing at higher temperatures due to various types of attractive surface interactions.
Liquid-phase experiments were conducted by removing the air-stable SiO 2/Ru samples from UHV through an argon-filled transfer-chamber and then exposing only the sample surface to aqueous environments within a hanging meniscus configuration (i.e., the crystal is contacted face-down with the aqueous solution such that only the face with the silica bilayer film gets exposed to the solutions). To investigate film hydroxylation and dissolution, we made use of a wide range of pHs, temperatures, and exposure times using NaOH and HCl to vary the pH of aqueous solutions within ultra-pure water and a manually controlled hotplate in tandem with K-type thermocouple measurements (made within a water-bath surrounding the glassware containing the experimental solutions) to control temperature.
## 3 Results and Discussion
### 3.1 Thermal and Electron-Assisted Hydroxylation of SiO2/Ru
We recall that hydroxylated SiO 2/Ru surfaces can be prepared by adsorption of water at 100 K followed by heating to room temperature (thermal route) [ 32], or, in order to enhance the hydroxyl concentration, with an additional electron irradiation step prior to heating (electron-assisted route) [ 34, 35]. D 2O (mass 20) TPD spectra recorded following both methodologies are presented in Fig. 1 after dosing 5 L D 2O at 100 K.
Desorption of multilayer ice is detected from both samples at around 160 K. Note the slightly smaller ice desorption signal from the electron-bombarded film (red trace in Fig. 1), which is due to some electron-induced desorption of water molecules from the ice surface. The more interesting data appears in the region between 200 and 1200 K, where molecular and recombinative desorption of silica-bound D 2O and surface hydroxyls (silanols, OD’s) are expected [ 2]. The strong enhancement of water desorption from the electron-bombarded sample in this region is related to a significantly increased abundance of D 2O and OD’s on the silica surface. More specifically, the thermal route leads to the formation of isolated hydroxyl groups at defect sites within the film, which recombine at elevated temperature and desorb as molecular water at 900 K [ 32], and some additional hydrogen-bonded physisorbed water, which desorbs at lower temperatures (200–500 K, black trace in Fig. 1). By contrast, enhanced hydroxylation of the electron-bombarded sample gives rise to much larger and more clearly defined water desorption peaks with maxima at 250–300, 450 and 600 K, as well as an additional high temperature desorption feature at ~1070 K [ 34]. Note that the amount of water desorbing in the 900 K state does not change significantly from one sample to the other, which suggests that this defect-related contribution is not affected by electron bombardment.
DFT-derived models for the hydroxylation of bilayer SiO 2/Ru by water dissociation reported in recent work focused on agreement between calculated stretching frequencies of silanol groups and those observed experimentally [ 35]. Within the present work, we make a connection to TPD results obtained for bulk silica samples, in particular to those reported by Zhuravlev [ 2]. We note good general agreement in terms of the peak temperatures associated with individual desorption states reported for analogous TPD spectra collected from hydroxylated bulk silca samples, which suggests the presence of similar water and hydroxyl species on our hydroxylated silica film. According to the Zhuravlev model [ 2], the desorptions are attributed to the following adsorption states and processes: “Free” and weakly bound water in various hydrogen-bonding environments give rise to the low-temperature desorption features between 200 and 500 K, whereas the high temperature peaks are assigned to recombinative water desorption originating from vicinal (at 600 K) and isolated hydroxyls (>800 K).
To further characterize SiO 2/Ru hydroxylation, we repeated the TPD experiments described above using D 2 16 O adsorbed on 18O-labeled silica films to distinguish between silanol groups formed by 16OD fragments of dissociated D 2O and those containing substrate oxygen atoms ( 18OD). The corresponding TPD spectra tracking mass 20 (D 2 16 O) and mass 22 (D 2 18 O) are shown in Fig. 2. Water desorbing from hydroxylated SiO 2/Ru obtained via the thermal route (lower TPD traces in Fig. 2a) contains mainly 16O, which suggests that only little isotopic exchange between adsorbed water and the silica film takes place. As noted previously, some isotopic intermixing is present in the high temperature desorption feature at 900 K (Fig. 2b) [ 32]. Note that the apparent high temperature feature appearing above 1100 K in Fig. 2b is most likely an artifact reflecting a somewhat larger than normal oxygen coverage at the Ru interface, which has a desorption threshold near the upper limit of our reported temperature range (a similar, but less intense feature can also be noted in the high temperature plots in Fig. 3b, c). While labelled oxygen should nominally only result in changes to the m/z + = 36 and 18 channels, the relatively large burst of gas introduced into the mass-spectrometer during interfacial oxygen desorption has been observed to affect off-resonant mass-channels due to assumed space-charge-dependent variations in ionizer transmission efficiencies, which scale in proportion with the total intensity of a given mass-channel. Since the background signal resulting from our repeated D 2O doses is much higher than that of mass 22, the effect is significantly more pronounced in the mass 20 channel.
Unlike the thermal route, significant isotopic mixing does occur following electron irradiation. While the desorption of multilayer ice is still dominated to the same extent by the 16O contribution, the distribution of 16O and 18O found during the desorption of more strongly bound water species becomes roughly 2:1 between 200 and 800 K (i.e. molecularly adsorbed water and vicinal hydroxyls; Fig. 2a, upper traces), and 1:1.5 in the recombinative peak related to isolated silanols at 1070 K (Fig. 2c). In particular, the large amount of 18O contained in molecularly adsorbed water (desorptions at 200–500 K) points to significant oxygen exchange between the silica film and molecular water at the silica/ice interface during electron irradiation of the adsorbed ice layer. This can be achieved by opening and reforming siloxane bonds within the film, as indicated by the following scheme: (Scheme 1).
Note that while the 1070 K peak contains more 18O than 16O, in agreement with corresponding infrared data [ 35], the isotope distribution in the 900 K peak is the same irrespective of the hydroxylation procedure. Again, this shows that electron bombardment does not affect water dissociation at intrinsic film defects, and, instead, appears to exclusively result in the formation of a distinctly different type of isolated silanol species. This discrepancy most likely reflects differences in both the nearest neighbor OD–OD distances and the surrounding Si–O bond interactions for hydroxyls forming at innate defect sites and those resulting from electron bombardment effects. More specifically, ODs forming at innate defects are likely in closer proximity to other defect-bound ODs and present with lowered O–Si bond strengths compared to the OD groups produced randomly throughout other regions of the film via D 2O electron irradiation. As such, D diffusion selectively favors recombination at the more weakly-bound, defect confined 16O leaving groups formed spontaneously upon water adsorption at low temperatures, which are effectively titrated away before high temperature recombination of isolated silanols at any other “regular” lattice sites can take place. This explains both the lower temperature and decreased isotopic exchange noted for the TPD peak associated for the innate silanols relative to those forming following electron irradiation.
Since electron irradiation of adsorbed ice layers leads to a strong enhancement of the amount of adsorbed molecular and dissociated water on SiO 2/Ru, we further investigated how variation of the electron irradiation parameters influences the hydroxylation state of the silica film. The parameters for our standard e-beam protocol are: 5 L D 2O dose, 200 eV kinetic energy, and 0.05 mA sample current, and we systematically varied one of these parameters while keeping the other two constant (the irradiation time was not varied and was 60 s in all experiments) to produce the D 2O-TPD results presented in Fig. 3 (left panel: 200–1200 K; right panel: 800–1170 K).
Figure 3a shows TPD spectra resulting from varying the electron kinetic energy between 100 and 600 eV, and from these results it is clear that this does not cause major changes to the desorption peak intensities beyond a moderate increase of the 1070 K desorption. Variation of the sample current from 0 mA (which corresponds to the thermal route) to 0.2 mA shows the expected increase of adsorbed water/hydroxyl species with increasing flux. The maximum desorption intensities (that is, maximum water/hydroxyl coverage on SiO 2/Ru) are obtained after irradiation at 0.1 and 0.2 mA sample current. However, it should be mentioned that significant amounts of the ice layer desorb under the higher fluxes. In fact, exposure to 0.2 mA of 200 eV electrons for 60 s results in the complete removal of all multilayer ice formed during the 5 L D 2O dose at 100 K (not shown). Since direct electron impact on the uncovered silica surface may cause film damage, we have chosen 0.05 mA sample current during electron irradiation for all other hydroxylation experiments as a compromise to obtain sufficient surface hydroxylation at minimum film damage.
Finally, we note a strong influence of D 2O coverage in electron-assisted hydroxylation of SiO 2/Ru (Fig. 3c). Clearly, for very thin ice layers (2.5 L D 2O dose) electron-assisted effects provide almost no additional contributions to hydroxylation, whereas an increase of the dose to 5 and 10 L D 2O considerably enhances the abundance of isolated hydroxyls (desorption at 1070 K), molecular water (<500 K), and vicinal hydroxyls (600 K).
Hydroxylation of silica is expected to proceed via opening of siloxane bonds. For the bilayer SiO 2/Ru film in particular, several processes for silanol formation by water dissociation have been proposed from results of recent DFT calculations [ 35]. These mechanisms include opening of in-plane and vertical siloxane bridges to form terminal as well as hydrogen-bond donor hydroxyls, and a redox-based process involving the Ru substrate, which leads to a terminal OH, a Si–O–Ru bridge, and the release of a hydrogen atom. While the vibrational properties of the silanol groups and Si–O bonds in the proposed models are in good agreement with experimental data, the mechanism by which the surface of the SiO 2/Ru sample gets hydroxylated remains largely unknown. From our results, the observed D 2O coverage dependence shown in Fig. 3b provides the strongest hints about the mechanism of electron-assisted SiO 2/Ru hydroxylation, which we believe to be directly connected to radiolysis of water molecules in the ice layer, rather than electron-induced defect formation on the silica as a means to instigate subsequent water dissociation, as the latter pathway would be expected to result in similar or steadily decreasing hydroxylation yields with increasing ice coverage for our range of doses. The almost complete absence of electron-induced hydroxylation enhancement following the 2.5 L dose can be explained by the relatively large inelastic mean free path of 200 eV electrons, which is ~1.0–1.3 nm, corresponding to the thickness of 3–4 ice layers. Thus, for adsorbed ice films within this thickness range (2.5 L dose) the electrons will penetrate the ice layer without significant interaction. As the ice films become thicker, a larger (smaller) fraction of the impinging electrons may deposit their energy into excitations of the water (SiO 2) molecules to result in a greater (lesser) number of radiolysis products (silica defects), such as H and OH radicals, hydroxide ions, O 2, H 2, and H 2O 2 [ 36].
Many studies have been dedicated to investigating the dissolution of silicates under a wide range of aqueous conditions [ 7]. Since all dissolution mechanisms proceed by creating hydroxyls at the expense of siloxanes, understanding the mechanics of dissolution provides an extrapolated understanding of the mechanics driving hydroxylation. Therefore, we may examine previously proposed models of the dissolution mechanism, which depend on one charged and one neutral species [ 6], to potentially help explain electron-assisted hydroxylation of SiO 2/Ru. Direct hydrolysis of Si centers by weakly nucleophilic neutral water molecules is only likely in the presence of silanol groups or deprotonated silanol sites, ≡Si–O [ 7, 37]. Since our films lack significant initial silanol coverage and the possibility of electron-assisted formation of ≡Si–O sites and subsequent hydrolysis is disregarded based on the hydroxylation D 2O coverage dependencies discussed above (Fig. 3c), we conclude that this pathway is not very likely. On the other hand, more aggressive agents such as hydroxide ions formed in the ice layer during electron bombardment may readily attack Si centers to directly result in the formation of silanol groups via a mechanism depicted in Scheme 2 below, which has been adapted from Ref. [ 7] and illustrated to depict options for breaking either vertical (upper) or lateral (lower) siloxane bonds in SiO 2/Ru following hydroxylation by OH .
In the presence of additional water molecules, the initial hydroxide attack may also act to activate Si–O bonds in the siloxane bridges to result in water dissociation via a process like the one shown below in Scheme 3. Note that this scheme incorporates a cyclic transition state as proposed by Mitsyuk [ 38] and added reaction steps (protonation of the ≡Si–O site and siloxane bridge reformation) that provide a means of accounting for the 16O– 18O isotope exchange between the film and molecular water reported above.
According to our TPD results, the major fraction of water is adsorbed in molecular form (D 2O desorptions between 200 and 500 K, Figs. 1, 2) and Scheme 3, albeit speculative, provides a reasonable mechanism to explain the significant oxygen exchange observed for this adsorption state. To the extent that the isolated silanols (recombinative desorption at >850 K) have previously been shown to preferentially display Si– 16OD configurations at low temperatures following electron bombardment of D 2O covered 18O-labelled SiO 2 films [ 33], we might conclude that these groups form via the same mechanism prematurely halted at a stage closer to that represented in Scheme 2.
### 3.2 Dissolution of SiO2/Ru in Liquid Environments
To corroborate our conclusions regarding the electron-assisted hydroxylation mechanism(s), we have studied the dissolution of bilayer SiO 2/Ru, which is known to be initiated in alkaline medium by OH attack at Si centers, and compare the results with existing rate laws derived from the dissolution of bulk samples. Combining results from the plethora of studies conducted for the dissolution of silicates in different environments, a comprehensive rate model, accounting for variations in pH, the coverage of neutral ( θ Si–OH) and deprotonated $$\left( {\theta_{{{\text{Si}}{-}{\text{O}}^{-} }} } \right)$$ silanols, and temperature has been developed by Bickmore et al. [ 7], and may be described by the following equation:
$$\frac{\text{dSi}}{{{\text{d}}t}} = {\text{e}}^{ - 8.9 \pm 0.8} T{\text{e}}^{{\left( {\frac{ - 67.5 \pm 2.7}{RT}} \right)}} \left( {\theta_{\text{SiOH}} } \right) + {\text{e}}^{3.6 \pm 0.7} T{\text{e}}^{{\left( {\frac{ - 82.8 \pm 2.1}{RT}} \right)}} \left( {\theta_{{{\text{SiO}}^{ - } }} } \right) + {\text{e}}^{6.7 \pm 1.8} T{\text{e}}^{{\left( {\frac{ - 77.5 \pm 6.0}{RT}} \right)}} \left[ {{\text{OH}}^{ - } } \right]$$
where dSi/d t denotes the rate of Si dissolution in (mol/s), which was typically monitored experimentally by tracking the abundance of Si in solution as a function of time. The three terms in the rate model can be described as hydrolysis by water molecules at Si centers with neutral and deprotonated silanol groups (first and second terms) and hydrolysis of Si centers by OH (third term). To better understand the processes by which hydroxylation may occur over our thin-films, we conducted comparative dissolution studies using XPS analysis to track the loss of Si from the surface as a function of time, temperature, and pH after exposing the bilayer films to various liquid environments.
XP spectra (O 1 s and Si 2 p regions) taken after exposing bilayer SiO 2/Ru to deionized water at 90 °C, HCl (aq) at 90 °C and NaOH (aq) at 25 °C for various times are displayed in Fig. 4a–c. Clearly, deionized water (pH 7, Fig. 4a) does not affect the film structure to any significant extent, even at elevated temperature. While we do note a small shift of all silica-related XP peaks to higher binding energy (BE), which reflects a slight change in the electronic structure of the system (band bending), neither the Si 2 p nor the O 1 s peaks (O silica, 532–533 eV; O Ru, 530 eV) suffer any loss of intensity. By contrast, extremely acidic conditions (HCl, pH 0) applied at the same temperature lead to a significant decrease of Si and O contributions, indicating a partial dissolution of the film (Fig. 4b). The O Ru O 1 s signal becomes rapidly attenuated by just a few minutes of exposure, which we take as an indication for preferred permeation of protons through the pores of the film and subsequent reaction with Ru-bound oxygen atoms. A stable film structure is obtained at 10 min. of exposure, after which no further dissolution takes place.
For dissolution in alkaline media (NaOH, pH 13) we plot the results obtained at 25 °C (Fig. 4c), which gives similar dissolution rates as exposure to HCl at 90 °C. In contrast to exposure to acid, dissolution continues at longer exposure times, albeit with reduced rate, and the intensity of the O Ru O 1 s signal remains constant throughout the series. The equal relative signal intensity losses of Si and O peaks observed with time of exposure (Fig. 5d) let us conclude that, in accordance with general experience, the overall dissolution process can be described as the H 3O + or OH catalyzed hydrolysis of SiO 2 (SiO 2 + 2 H 2O → H 4SiO 4). From purely qualitative examination of the pH and temperature dependence of the XPS signal losses shown in Fig. 4, we further conclude that our bilayer SiO 2/Ru films resemble the dissolution behavior of other, more abundant forms of silica (quartz, amorphous silica), which are found to be practically insoluble in the neutral pH range, slightly soluble in acidic environment, and more strongly soluble in alkaline conditions.
In order to more quantitatively connect the findings of our study to the dissolution rate model described above, we systematically studied the dissolution of the SiO 2/Ru sample in the pH 11–13 range and at temperatures of 25–90 °C. Results from those experiments are summarized in Fig. 5, which plots the Si 2 p XPS signal intensity as a function of exposure to various liquid conditions.
In an attempt to roughly account for possible peak intensity losses resulting from the presence of unknown signal-attenuating substances left behind after exposure to the liquids, Si signals are reported as a ratio of the Si 2 p:Ru 3 d peak intensities, with the crude approximation that electrons generated within the metal substrate and the silica film will attenuate to a similar degree as a function of residual overlayer coverage. Despite this precaution, similar results are obtained when tracking the raw, uncorrected Si 2 p intensities, indicating minimal effects from aqueous residues.
From these plots, it is clear that removal of SiO 2 from the sample shows dependence upon both pH and temperature, with a preference for larger values of both parameters, which is qualitatively consistent with the behavior noted from bulk-phase silica analogues [ 7]. Consistent with previous thin-film dissolution studies conducted within our group [ 39], we note decreasing rates of dissolution with increasing exposure to aqueous conditions in all cases. This behavior may simply reflect a secondary dependence on the surface concentration of Si, which steadily decreases as the film dissolves but should remain effectively constant in the bulk studies since removal of one layer simply exposes the next over an effectively infinitely thick sample depth in the latter case. Outside of this range of conditions, notable dissolution was not observed within the time frames allotted for our experiments (~1 h) at lower pHs (1–10), which is consistent with an OH driven mechanism, or the third term in the rate model provided above.
Since dissolution rates for bulk silicates already favor elevated pHs [ 7], and (unlike those materials) our samples lack significant initial silanol coverage [ 32], which is required to facilitate secondary (lower pH) mechanisms, it is not be surprising for our films to show a preference for OH attack at Si centers. This process is illustrated in Scheme 2, which has been adapted from the bulk mechanism describing the [OH ]-dependent term of the overall rate equation. A comparison of the dissolution rates predicted from that relation to those estimated on the basis of the initial rates of Si XPS peak attenuations from our thin-films (see Fig. 5) is shown in Fig. 6.
Relative to the uncertainties reported for the bulk rate equation, which are admittedly large, all observed values show agreement with those predicted for this dissolution mechanism. Perhaps more importantly, the deviation in rate as a function of pH shows nearly order of magnitude dependence consistent with an OH attack driven mechanism. The extent to which the observed decrease in rate between pH 13–12 is slightly greater than that noted between pH 12–11 might reflect an increased relative influence of the deprotonated silanol-dependent water hydroxylation mechanism (second term in the rate equation) as the direct effects of the OH concentration become increasingly less dominant at intermediate pHs, but anything beyond order-of-magnitude level comparisons are probably unjustified given the assumptions and limitations built into our XPS interpretations (i.e. an assumed linear dependence between Si surface concentration and XPS intensity and limited temporal resolution due to the limited number of experiments conducted over a small set of exposure times).
On the basis of the agreement between the bulk model and the behavior of our thin-films, we tentatively conclude that OH attack, akin to that described by Scheme 2, most likely represents the preferred hydroxylation mechanism for our bilayer SiO 2/Ru samples, and, by extrapolation, postulate that OH formation within the ice-layers likely plays a pivotal role in the hydroxylation process promoted by the electron-bombardment procedure described above.
## 4 Conclusions
In this work, we have investigated electron-assisted hydroxylation of bilayer silica by employing isotope exchange experiments and variation of electron bombardment parameters. The 18O-enriched silica surface readily exchanges oxygen with an interfacial D 2 16 O layer of adsorbed ice, which suggests that dynamic siloxane bond cleavage and re-formation takes place during electron bombardment at 100 K. Among the electron beam parameters varied, D 2O coverage was found to provide the largest correlation to changes in subsequent water/hydroxyl coverage. Since almost no increase of the water/hydroxyl coverage was observed if the thickness of the electron-irradiated ice layer was in the range of ~1 nm, corresponding to the IMFP of the electrons, it is proposed that the siloxane bonds are activated for water dissociation by water radiolysis products, e.g. hydroxyl ions, formed in the ice layer during electron bombardment. In corroboration with this mechanism, the observed dissolution behavior of bilayer SiO 2/Ru(0001) in alkaline aqueous conditions can satisfactorily be explained by a rate law (derived from studies of bulk silica samples) exclusively dependent on OH attack at Si centers. In addition to providing insight into the mechanism of electron-assisted hydroxylation of silica films and their dissolution behavior, this study shows that the surface properties of the two-dimensional silica model system examined here are analogous to that of bulk silica material.
## Acknowledgments
Open access funding provided by University of Graz. This work was supported by the European Research Council (FP7), Starting Grant Agreement No. 280070, and by the collaborative research center 1109 sponsored by Deutsche Forschungsgemeinschaft. W.E.K. is grateful to the Alexander-von-Humboldt foundation for providing a fellowship.
## Unsere Produktempfehlungen
### Premium-Abo der Gesellschaft für Informatik
Sie erhalten uneingeschränkten Vollzugriff auf alle acht Fachgebiete von Springer Professional und damit auf über 45.000 Fachbücher und ca. 300 Fachzeitschriften.
Literatur
Über diesen Artikel
Zur Ausgabe
|
2020-10-29 10:50:32
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5787765979766846, "perplexity": 4260.301462920485}, "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-2020-45/segments/1603107904039.84/warc/CC-MAIN-20201029095029-20201029125029-00311.warc.gz"}
|
https://bdeewang.com/activities/rg/serre-arithmetic/
|
### Reading Group on A Course in Arithmetic by J.-P.Serre
#### Winter & Summer 2020
This is a reading group convened to read A Course in Arithmetic by J.-P.Serre. Our primary goal is to read and understand the first part of the book: Algebraic Methods. To that end we organize weekly seminars where the regular participants present sections from the book.
Regular Participants: Deewang Bhamidipati (Convener), Malachi Alexander, Sandra Nair
September 12. Seminars ended.
June 20. Seminars resumed.
March 17. Seminars suspended due to COVID lockdown.
#### Seminar Schedule
Dates Sections Presenter
August 29 Chapter IV, §2 Deewang
August 23 Chapter IV, §2 Deewang
August 8 Chapter IV, §1 Sandra
August 1 Chapter IV, §1 Sandra
July 18 Chapter III, §2 Deewang
July 11 Chapter II, §3; Chapter III, §1 $$(p=2)$$ Deewang
July 4 Chapter III, §1 Malachi
June 20 Refresher: Chapters I & II Deewang
Resuming after COVID pause.
March 13 Chapter III, §1 Malachi
March 6 Chapter II, §3 Deewang
February 28 Chapter II, §2 Deewang
February 14 Chapter II, §1, 1.2 - 1.3 Deewang
February 7 Chapter II, §1, 1.1 Deewang
January 31 Chapter I, §3 Deewang
January 24 Chapter I, §1, §2 Deewang & Malachi
|
2023-03-27 14:20:20
|
{"extraction_info": {"found_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": 0.23810158669948578, "perplexity": 14141.833504187012}, "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/1679296948632.20/warc/CC-MAIN-20230327123514-20230327153514-00428.warc.gz"}
|
http://metwiki.net/viewProgramDetail?index=585&type=view
|
Context-Sensitive Middleware-Based Applications B
### Program Information
Name: Context-Sensitive Middleware-Based Applications B
Domain: Algorithm
Functionality: A example of a smart delivery system of a supermarket chain such that individual suppliers replenish their products onto pallets, shelves, and cases in various warehouses according to the demand sent off by such pallets. The smart deliver system includes four features: (1)Each smart pallet can be dynamically configured to store a particular kind of product at, as far as possible, a desired quantity level. (2)Each van of a supplier delivers a type of goods. (3)Goods that cannot sell can be returned to the supplier. A smart pallet may request a van to retract certain amount of goods. (4)The system assumes that the effective delivery distance for any pallet by any van is at most 25 meters
Input: define $\varepsilon = 5$; $q_v$ is the quantity of goods deliverable by a van; $p_v$ is the location of the van in $(x, y)$ coordinates; $d$ is square of distance between the van and a pallet; $s$ is no. of vans surrounding the pallet; $q_d$ is the desired quantity of goods for the pallet; $q_l$ is the ledger amount of goods in the pallet; $q_p$ is the quantity of goods on hand in the pallet; $p_p$ is the location of the pallet in $(x, y)$ coordinates; $q_l=\sum_{i=1}^s q_v^{(i)}+q_p$ where $q_v^{(i)}$ denotes the context variable $q_v$ from the $i$th surrounding van.
Output:
#### Reference
A Metamorphic Approach to Integration Testing of Context-Sensitive Middleware-Based Applications https://doi.org/10.1109/QSIC.2005.3
### MR Information
#### MR1------
Description:
Property: If $q_{d_1}=q_{d_2},d_1 \leq 625$,and $d_2 \leq 625$ then $q_{l_1} \approx q_{l_2}$
Source input:
Source output:
Follow-up input:
Follow-up output:
Input relation:
Output relation:
Pattern:
#### MR2------
Description:
Property: Let t be an original test case and $t'$ be a follow-up test case that share the same checkpoint, known as an initial checkpoint. If we apply Withdraw() to the initial checkpoint before executing $t'$ , the number of invocations of the Replenish() function for $t'$ is expected to be more than that of $t$. If we apply Replenish() to the initial checkpoint before executing $t'$ , the number of invocations of the Withdraw() function for $t'$ is expected to more be than that of t.
Source input:
Source output:
Follow-up input:
Follow-up output:
Input relation:
Output relation:
Pattern:
Insert title here
|
2019-08-21 12:09:00
|
{"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": 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": 0.576923668384552, "perplexity": 2962.93841709652}, "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/1566027315936.22/warc/CC-MAIN-20190821110541-20190821132541-00037.warc.gz"}
|
https://slave2.omega.jstor.org/stable/j.ctt7s38w
|
# Blow-up Theory for Elliptic PDEs in Riemannian Geometry (MN-45):
Olivier Druet
Emmanuel Hebey
Frédéric Robert
Pages: 224
https://www.jstor.org/stable/j.ctt7s38w
1. Front Matter
(pp. i-iv)
(pp. v-vi)
3. Preface
(pp. vii-viii)
The Authors
4. Chapter One Background Material
(pp. 1-12)
We recall in this chapter basic facts concerning Riemannian geometry and nonlinear analysis on manifolds. For reasons of length, we are obliged to be succinct and partial. Possible references are Chavel [20], do Carmo [22], Gallot-Hulin-Lafontaine [36], Hebey [43], Jost [50], Kobayashi-Nomizu [53], Sakai [65], and Spivak [72]. As a general remark, we mention that Einstein’s summation convention is adopted: an index occurring twice in a product is to be summed. This also holds for the rest of this book.
We start with a few notions in differential geometry. LetMbe a Hausdorff topological space. We say thatM...
5. Chapter Two The Model Equations
(pp. 13-24)
Let$\left( {M,g} \right)$be a smooth compact Riemannian manifold of dimensionn ≥ 3. We let$\left( {{h_\alpha }} \right)$be a sequence of smooth functions onM, and consider equations like.
${\Delta _g}u + {h_\alpha }u = {u^{2* - 1}}$$\left( {E{}_\alpha } \right)$
where${\Delta _g} = - di{v_g}\nabla$is the Laplace-Beltrami operator,$2{\kern 1pt} * = 2n/(n\; - \;2)$is the critical Sobolev exponent for the embedding of the Sobolev space$H_1^2\left( M \right)$into Lebesgue’s spaces${L^p}(M)$, anduis required to be positive. By standard regularity theory, as developed by Gilbarg-Trudinger [37], and thanks to the maximum principle, if$u \in H_1^2\left( M \right)$is a nonnegative solution of$\left( {{E_\alpha }} \right)$, thenuis smooth and either$u > 0$everywhere or$u \equiv 0$. As already mentioned, thanks to the...
6. Chapter Three Blow-up Theory in Sobolev Spaces
(pp. 25-50)
An important result of the 1980s describes the asymptotic behavior of Palais-Smale sequences associated with equations like
$\Delta u = {u^{2*\, - \,1}}$, (3.0.1)
where$\Delta$is the Euclidean Laplacian, anduis required to vanish on the boundary of a smooth bounded open subset$\Omega$of the Euclidean space${\mathbb{R}^n}$. This result was proved by Struwe [73]. Related references are Brézis [11], Brézis-Coron [12, 13], Lions [58], Sacks-Uhlenbeck [64], Schoen [67], and Wente [76]. Let$D_1^2\left( {{\mathbb{R}^n}} \right)$be the homogeneous Sobolev space de ned as the completion of$C_0^\infty \left( {{\mathbb{R}^n}} \right)$, the space of smooth functions with compact support in${\mathbb{R}^n}$, with respect to the norm a...
7. Chapter Four Exhaustion and Weak Pointwise Estimates
(pp. 51-66)
Let$\left({M,g}\right)$be a smooth compact Riemannian manifold of dimension$n\ge3$. We let$({h_\alpha})$be a sequence of smooth functions onM, and consider equations like
${\Delta_g}u\;+\;{h_\alpha}u={u^{2*\,-\;1}}\quad\quad\quad\quad\quad({E_\alpha})$
whereuis required to be positive. We let$({u_\alpha})$be a sequence of solutions to$({E_\alpha})$, so that${u_\alpha}>0$and
${\Delta_g}{u_\alpha}+\;{h_\alpha}{u_\alpha}=u_\alpha^{2*\,-\;1}$
for all$\alpha$. We describe in this chapter an exhaustion method for the blow-up behavior of the${u_\alpha}$’s. This method provides a constructive approach to Theorem 3.1, and weak (with respect to the material in Chapter 5 and 6) pointwise estimates on the${u_\alpha}$’s.
We assume in what follows that there...
8. Chapter Five Asymptotics When the Energy Is of Minimal Type
(pp. 67-82)
Let$\left({M,g}\right)$be a smooth compact Riemannian manifold of dimension$n\ge\;3$. We let$({h_\alpha})$be a sequence of smooth functions onM, and consider equations like
${\Delta_g}u\;+\;{h_\alpha}u\;=\;{u^{2*\,-\,1}}\quad\quad\quad\quad\quad\quad({E_\alpha})$
whereuis required to be positive. We let$({u_\alpha})$be a sequence of solutions to$({E_\alpha})$, so that${u_\alpha}>\;0$and
${\Delta_g}{u_\alpha}+\;{h_\alpha}{u_\alpha}=u_\alpha^{2*\,-1}$
for all$\alpha$. For the reader’s convenience, we describe in this chapter the${C^0}$-theory for the blow-up behavior of the${u_\alpha}$’s when the energy of the${u_\alpha}$’s is of minimal type. The general situation of arbitrary energies is treated in the next chapter.
As in Chapter 4, we assume in what follows...
9. Chapter Six Asymptotics When the Energy Is Arbitrary
(pp. 83-200)
Let$\left({M,g}\right)$be a smooth compact Riemannian manifold of dimension$n\ge\;3$. We let$({h_\alpha})$be a sequence of smooth functions onMand consider equations like
${\Delta_g}u\;+\;{h_\alpha}u\;=\;{u^{2*\,-\,1}}\quad\quad\quad\quad\quad\quad({E_\alpha})$
whereuis required to be positive. We let$({u_\alpha})$be a sequence of solutions to$({E_\alpha})$, so that${u_\alpha}>0$and
${\Delta_g}{u_\alpha}+\;{h_\alpha}{u_\alpha}=u_\alpha^{2*\,-1}$
for all$\alpha$. We describe in this chapter the${C^0}$-theory for the blow-up behavior of the${u_\alpha}$’s when the energy of the${u_\alpha}$’s is arbitrary.
As in Chapters 4 and 5, we assume in what follows that there exist$0<\theta<1$and a smooth (or only${C^{0,\theta}}$) function${h_\infty}$onMsuch that
the...
10. Appendix A. The Green’s Function on Compact Manifolds
(pp. 201-208)
11. Appendix B. Coercivity Is a Necessary Condition
(pp. 209-212)
12. Bibliography
(pp. 213-218)
|
2021-06-18 03:00:58
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9416950941085815, "perplexity": 9044.097773239317}, "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/1623487634616.65/warc/CC-MAIN-20210618013013-20210618043013-00016.warc.gz"}
|
https://www.edinst.com/us/blog/what-is-quantum-yield/
|
What is Quantum Yield? | Fluorescence | Measuring Quantum Yield
# What is Quantum Yield?
The IUPAC definition of quantum yield (Φ) is the number of a certain event occurring per photon absorbed by the system,1
However, it is most commonly written specifically for the emission of light (photoluminescence) by a system,2-5
This narrower definition is often simply called quantum yield; with the emission of light being implicit given the context. However, the terms fluorescence, luminescence and photoluminescence quantum yield are also commonly used. The quantum yield is reported as either a decimal fraction between 0 and 1 or as a percentage. For example, if the system absorbs 100 photons and emits 30, then its quantum yield would be is 0.3 or 30%.
The quantum yield of a system (such as a fluorescent molecule) is determined by the balance between the radiative and non-radiative transition rates within it (Figure 1). The quantum yield can therefore be rewritten in terms of these rates,
The radiative transition rate ($k_r$) denotes radiative (light emitting) processes such as fluorescence and phosphorescence whereas the sum of non-radiative rates ($k{_{nr}}$), includes processes such as internal conversion, intersystem crossing, and energy transfer.6 The quantum yield is therefore the probability that a system in the excited state deactivates through a radiative process to its ground state.
## History of Quantum Yield
The roots of quantum yield can be traced back to the beginning of the 20th century when Einstein’s revolutionary work on the photoelectric effect was published in 1905 (a copy of his paper translated in English can be found in Ref. 7). Einstein introduced the quantisation of light, i.e., a light beam consists of discrete quantum particles (quanta) carrying energy equal to E=hv , where h is Planck’s constant and v is the light frequency. A few years later, E. G. Warburg published a series of papers over the period 1912-1921 in which he studied the conversion of ozone molecules into oxygen to acquire the ratio of the molecules produced to quanta absorbed. He later named this procedure “quantum efficiency” and used the Greek letter phi (Φ) to denote it.8 His papers (in German) can be found on the German National Library website.9 In 1924, Vavilov referred to the term “fluorescence yield”, based on Warburg’s earlier work, to calculate the fraction of absorbed rays of light to fluorescent rays of light,10 whereas the term “molecules per quantum light absorbed” appears in a paper published in 1925 by Marshall to describe the photochemical reaction between hydrogen and chlorine.11 By 1930 the term “quantum yield” as we know it today had been widely cited in a large number of textbooks and papers.
## How to Measure Quantum Yield?
Quantum yield is one of the most important photophysical parameters when characterising luminescent molecules and materials. High quantum yields are crucial for a wide range of applications including; displays, lasers, bioimaging and solar cells, and accurate measurement of the quantum yield is therefore important. Quantum yield measurements can be split into non-optical and optical methods. Non-optical methods include the indirect measurement of the conversion of the excitation energy into heat and its dissipation to the solvent,6, 12 and calorimetric methods, such as photoacoustic spectroscopy (PAS),13 and thermal lensing.14 These methods require specialised setups and are generally reserved for the determination of the quantum yield of important standards. Quantum yields are therefore most commonly measured optically, using either the relative or absolute method.
Relative Quantum Yield Method
In the relative method, the quantum yield of the sample interest is calculated by comparing its photoluminescence emission to that of a reference standard of known quantum yield. In conventional fluorescence spectrometers only a certain fraction of the emitted light is collected and detected; with the size of the fraction depending on numerous factors. These factors include; the angular distribution (solid angle) of the emitted photons, the refractive index of the solvent, wavelength, the scattering properties of the sample and sample geometry.6, 12 The fraction is, therefore, impossible to accurately quantify which prevents direct measurement of the quantum yield. The relative method overcomes this problem by using a reference standard of known quantum yield and similar optical properties to those of the sample. The emission spectra of the sample and reference standard are measured under identical excitation conditions and the ratio of the integrated emission used to calculate the quantum yield of the sample.
Figure 2 FS5 Spectrofluorometer equipped with the SC-05 Standard Cuvette Holder.
The advantage of the relative method is that it can be readily applied using standard commercial absorption and fluorescence spectrometers equipped with a simple cuvette holder (Figure 2). The downside is that a reference standard must exist that emits in a similar wavelength region to the sample, and the sample type is generally limited to transparent liquids.
Absolute Quantum Yield Method
In the absolute method, an integrating sphere is used to capture all light emitted by the sample, avoiding the need for a reference standard. The quantum yield is determined by comparing the number of emitted photons with the number of absorbed photons. The advantage of the absolute method is that no reference standard is required which results in a faster measurement of the quantum yield (since fewer measurements are required) and a broader range of emission wavelengths are accessible that are no longer limited to where reference standards exist. Secondly, the absolute method can also be applied to a much larger range of sample types and is the only reliable method for samples such as scattering solids. The downside is that an integrating sphere is required for the measurement, however, these have become increasingly common and are now available as accessories for benchtop fluorescence spectrometers (Figure 3).
Figure 3 FS5 Spectrofluorometer equipped with the SC-30 Integrating Sphere.
Technical notes on how to measure the absolute quantum yield of liquid and powder samples using the FLS1000’s integrating sphere, and application notes on measuring the temperature-dependent absolute quantum yield of halide perovskites and plant leaves can be found on our website.
## FS5 Spectrofluorometer
For further information and to find out how it can be used in your own research view the FS5 Spectrofluorometer online. If you would like to talk to one of our sales team please contact us. We’d be delighted to help.
## Keep in Touch
If you have enjoyed reading this, why not sign-up to our monthly eNewsletter via the button below or follow us on your favourite social media channel in the footer below to keep up to date with our latest news and research.
|
2023-02-04 19:35:21
|
{"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": 2, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5252496004104614, "perplexity": 1084.21527550438}, "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/1674764500151.93/warc/CC-MAIN-20230204173912-20230204203912-00474.warc.gz"}
|
http://openstudy.com/updates/556be504e4b01de5673bc2c0
|
gabbyalicorn one year ago What is the approximate perimeter of the figure? A. 7.85 in. B. 12.9 in. C. 17.9 in. D. 31.4 in.
1. gabbyalicorn
2. geerky42
This looks like 1/4 of circle, right?
3. gabbyalicorn
right
4. geerky42
yeah, so that arc would be 1/4 the circumference. Agree?
5. gabbyalicorn
yeppers
6. geerky42
And you also have two radii |dw:1433134466763:dw|
7. geerky42
You know what's total length of these thick lines?
8. gabbyalicorn
90 is the length?
9. geerky42
|dw:1433134560634:dw|
10. geerky42
That's 5+5
11. gabbyalicorn
oh 10
12. geerky42
Yep, easy part. Now arc. Again it's 1/4 of circumference, so what is it?
13. gabbyalicorn
is there anyway to convert 1/4 into like a number?
14. geerky42
You can just divide it by 4?
15. gabbyalicorn
2.5
16. geerky42
Or you can use calculator to find 1/4 $$1\div4$$
17. geerky42
It's not 2.5 btw
18. gabbyalicorn
0.25
19. geerky42
yeah, so you multiply circumference by 0.25
20. gabbyalicorn
but i thought the circumference is 0.25
21. geerky42
no no. it's just that you find whole circumference, THEN you multiply it by 0.25
22. gabbyalicorn
so, 4 x 0.25?
23. gabbyalicorn
no that wouldn't make sense
24. geerky42
What is circumference? $$C = 2\pi r$$, right?
25. geerky42
Find that then you multiply it by 0.25
26. geerky42
$$\dfrac{C}{4}$$ or $$0.25C$$
27. gabbyalicorn
oh so 2 x 3.14 = 6.28 x 0.25 = 1.57
28. geerky42
you forgot radius. From picture, radius is 5, right?
29. gabbyalicorn
oh oops so then it would be 7.85
30. gabbyalicorn
A :D !!!
31. geerky42
Not quite
32. geerky42
That is correct, but you forgot to add two radii
33. geerky42
34. gabbyalicorn
wait 2
35. gabbyalicorn
so 10
36. gabbyalicorn
37. geerky42
|dw:1433135159595:dw|
38. gabbyalicorn
-3- i thought we just did that...
39. geerky42
yeah we did. Now we add them together.
40. geerky42
|dw:1433135231571:dw|
41. gabbyalicorn
17.85
42. geerky42
|dw:1433135261928:dw|
43. geerky42
Yep.
44. gabbyalicorn
:D thank you so much! I know i cant really show it through typing but I REALLY APPRECIATE YOU HELPING ME! :D
45. geerky42
No problem!
46. gabbyalicorn
im looking at the answer but why does a havethe 1 not there?
47. gabbyalicorn
its c because we round right?
48. gabbyalicorn
@geerky42
49. gabbyalicorn
wait but it doesnt say to round
50. gabbyalicorn
okay well im going with A
|
2017-01-20 20:17:43
|
{"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": 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": 0.7987831234931946, "perplexity": 13248.147081457792}, "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-2017-04/segments/1484560280872.69/warc/CC-MAIN-20170116095120-00321-ip-10-171-10-70.ec2.internal.warc.gz"}
|
https://hackmd.io/@alexhkurz/HkzAdwlAF
|
--- tags: seminar, maths, blute, ottawa, Coalgebras over Lawvere metric spaces --- # Coalgebras over Lawvere Metric Spaces Talk at the [Logic Seminar](https://richardblute.ca/logic-seminar-page/) of the University of Ottawa organised by [Richard Blute](http://aix1.uottawa.ca/~rblute/), Feb 2022. **Abstract:** Coalgebras are a model of dynamic systems that is particularly relevant if generalisations of bisimulation (observational/behavioural equivalence/preorder/metric) are of importance. We survey coalgebras as models of dynamical systems, Lawvere metric spaces, coalgebras over Lawvere metric spaces, hint their motivations in applications to computer science and introduce some of the open mathematical questions in the area. --- (... draft ...) --- ## Table of Contents (the writeup contains much more material than the 50min presentation) - background: - [coalgebras over sets](https://hackmd.io/@alexhkurz/BJ5ysx-0t) - [coalgebraic logic over sets](https://hackmd.io/@alexhkurz/rJLXsx-0F) - [generalizing to ordered sets](https://hackmd.io/@alexhkurz/ryji9lZRY) - [generalizing from orders to quantales](https://hackmd.io/@alexhkurz/HJlBneW0Y) - [extending functors to quantale enriched categories](https://hackmd.io/@alexhkurz/r1PK6gW0F) - the logic of quantale enriched categories ## References Bill Lawvere: Metric Spaces, Generalized Logic, and Closed Categories, 1973. [The collected works of F. W. Lawvere](https://github.com/mattearnshaw/lawvere). G.M. Kelly: Basic Concepts of Enriched Category Theory, Cambridge University Press, Lecture Notes in Mathematics 64, 1982. Also in [TAC Reprints 10, 2005](http://www.tac.mta.ca/tac/reprints/articles/10/tr10abs.html). Peter Aczel: [Non-well-founded sets](http://www.irafs.org/courses/materials/aczel_set_theory.pdf). CSLI 1988. Jan J. M. M. Rutten: Relators and Metric Bisimulations. CMCS 1998. Peter Aczel, Nax Paul Mendler: A Final Coalgebra Theorem. Category Theory and Computer Science 1989 James Worrell: Coinduction for recursive data types: partial orders, metric spaces and Omega-categories. CMCS 2000 Jipsen and Galatos: [RESIDUATED LATTICES OF SIZE UP TO 6](https://math.chapman.edu/~jipsen/finitestructures/rlattices/RLlist3.pdf), 2017 Adriana Balan, Alexander Kurz, Jiri Velebil: [Extending set functors to generalised metric spaces](https://lmcs.episciences.org/5132/pdf), Logical Methods in Computer Science 15 (2019). ... ## Topics that didn't make it into the talk ... Jan Rutten's theory of [universal coalgebra](https://www.sciencedirect.com/science/article/pii/S0304397500000566?via%3Dihub) that develops the theory of coalgebras parametrically in the functor $T:Set\to Set$ Moss's coalgebraic logic and Pattinson's predicate liftings (dblp for [coalgebraic logic](https://dblp.uni-trier.de/search?q=%20coalgebraic%20logic)) Examples of coalgebras for functors $T:Set\to Set$ other than the powerset functor such as [streams](https://dblp.uni-trier.de/search/publ?q=author:Jan_Rutten:%20stream*), [automata](https://dblp.uni-trier.de/search/publ?q=author:Yde_Venema:%20automat*), languages, Markov chains, certain kind of games, etc Examples of base categories $\mathcal X$ other than $Set$ or $Pre$ or or $Pos$ or $\Omega$-cat such as various topological spaces, [nominal sets](https://dblp.uni-trier.de/search/publ?q=author:Daniela_Petrisan:%20nominal*), vector spaces, etc Applications of coalgebras to the semantics of programming languages such as operational semantics of process algebras via [bialgebras](https://dblp.uni-trier.de/search/publ?q=bialgebra*), infinitary lambda-calculus, ... Applications of coalgebras over Lawvere metric spaces to semantics of programming languages ... [Breugel and Worrell etal](https://dblp.org/search/publ?q=author:James_Worrell_0001:%20author:Franck_van_Breugel:) ... [Dal Lago and Gavazzo etal](https://dblp.org/search/publ?q=author:Ugo_Dal_Lago:%20author:Francesco_Gavazzo:) ... (the coalgebra community has been prolific, there is so much more out there that I regret now that I started to make a list ... apologies to everybody else)
|
2023-03-31 15:56:48
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 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": 0.8482740521430969, "perplexity": 6429.812534467594}, "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/1679296949644.27/warc/CC-MAIN-20230331144941-20230331174941-00740.warc.gz"}
|
https://www.semanticscholar.org/paper/Explicit-Estimates-in-the-Theory-of-Prime-Numbers-Dudek/9436641dfec6dc7aab2bbc568520c592af5e9bb6
|
# Explicit Estimates in the Theory of Prime Numbers
@article{Dudek2016ExplicitEI,
title={Explicit Estimates in the Theory of Prime Numbers},
author={A. Dudek},
journal={The Bulletin of the Center for Children's Books},
year={2016}
}
• A. Dudek
• Published 2016
• Mathematics
• The Bulletin of the Center for Children's Books
It is the purpose of this thesis to enunciate and prove a collection of explicit results in the theory of prime numbers. First, the problem of primes in short intervals is considered. We prove that there is a prime between consecutive cubes $n^3$ and $(n+1)^3$ for all $n \geq \exp(\exp(33.3))$. To prove this, we first derive an explicit version of the Riemann--von Mangoldt explicit formula. We then assume the Riemann hypothesis and show that there will be a prime in the interval \$(x-4/ \pi… Expand
5 Citations
#### References
SHOWING 1-10 OF 82 REFERENCES
|
2021-04-16 00:30:00
|
{"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": 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": 0.6422538161277771, "perplexity": 436.499882222705}, "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/1618038088264.43/warc/CC-MAIN-20210415222106-20210416012106-00502.warc.gz"}
|
http://scottolesen.com/post/ggplot-it-can-do-what-i-want/
|
# ggplot: it can do what I want!
## 2015/06/11
As a scientist, I often run into this scenario:
• An experiment has a few treatment groups and a few measurements/replicates per group.
• I want to show all the data points.
A Cleveland dot chart is a nice way to show data, but if I have multiple replicates then the dots can start to mush on top of each other. R has a command dotchart that can do some of what I want: it can take a data frame, recognize that the y-axis has factors, and plot all the points. Unfortunatley, dotchart falls short when I want to, say, add some y-jitter so that all the points don’t overlap.
ggplot to the rescue! Say you have a dataframe df with values in column value and the treatment group name in column variable. Then you can quickly get an approximation of dotchart with ggplot(dat, aes(x=value, y=variable)) + geom_point(). Then, to get 10% jitter in the up-down direction, just change geom_point() to geom_point(position=position_jitter(height=0.1)).
If I have more data points, I often want a boxplot. With ggplot, it’s a piece of cake to keep the same data but change the plot: just add + coord_flip(). If I were using the boxplot command, then adding points is a big pain, and adding jittered points is an even bigger pain, especially if there are many treatments. ggplot makes this easy: the geom_boxplot command just takes the position_jitter argument as above!
|
2019-03-21 08:54:47
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.4648449420928955, "perplexity": 1775.5282709543026}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-13/segments/1552912202506.45/warc/CC-MAIN-20190321072128-20190321094128-00140.warc.gz"}
|
http://forums.wolfram.com/mathgroup/archive/2009/Jan/msg00313.html
|
Re: Re: Which editor do you use for math articles
• To: mathgroup at smc.vnet.net
• Subject: [mg95520] Re: [mg95426] Re: Which editor do you use for math articles
• From: Murray Eisenberg <murray at math.umass.edu>
• Date: Tue, 20 Jan 2009 05:50:24 -0500 (EST)
• Organization: Mathematics & Statistics, Univ. of Mass./Amherst
• References: <200901161109.GAA14132@smc.vnet.net> <200901171027.FAA14287@smc.vnet.net> <op.unw1jsv7tgfoz2@bobbys-imac.local> <200901181031.FAA13919@smc.vnet.net> <op.uny2v0v0tgfoz2@bobbys-imac.local> <4973AD96.4090909@math.umass.edu> <op.unzpiovktgfoz2@bobbys-imac.local>
TeX is NOT a version of HTML. Like HTML, though, it is a mark-up
language; but this mark-up languages emphasizes especially logical
mark-up (with the details as to how things should actually look handled
by the relevant macros and packages that designers have created). For
example (when using LaTeX):
\documentclass{article}
says to typeset the document in the "article" style (as opposed to the
book style, e.g.).
\section{The first big thing}
creates a numbered section header, with an appropriately larger and
(perhaps) bolded title "The first big thing", places appropriate spacing
before and after the title, and creates a corresponding entry in a table
of contents (in case one includes a command to include the contents).
this is a \emph{really} big idea
is typeset so as to put the word "really" (by default) in italics.
$$\int_0^{\infty} e^{-t^2}\,dt = \frac{\sqrt{\pi}}{2}$$
creates a sequentially-numbered, centered equation displayed on a
separate line saying the same thing as what the Mathematica front end
renders as the value of:
HoldForm[Integrate[Exp[-t^2], {t,0,\[Infinity]}]==Sqrt[\[Pi]]/2]
LaTeX automatically justifies lines (including automatically
hyphenating) and balancing lines on pages much the way a professional
typesetter would do.
To say more than this would go WAY beyond the proper topic of this
newsgroup! Please take a look at the examples linked from he URLs I
provided in my early posting.
DrMajorBob wrote:
> What I find odd is using a text editor for this.
>
> Is TeX a version of HTML, then?
>
> Bobby
>
> On Sun, 18 Jan 2009 16:30:46 -0600, Murray Eisenberg
> <murray at math.umass.edu> wrote:
>
>> What you say is both true and non-true! A commercial package such as
>> PC TeX is a single application that includes the editor, previewer, etc.
>>
>> On the other hand, the origin of TeX is in the *nix world, so the
>> paradigm persists of a separate tool for each task: a text editor;
>> the executable TeX engine itself; and a previewer (from which one
>> often does the printing). The TeX engine may itself have a version
>> that produces pdf directly, or one may use a separate tool in the
>> bundle that converts the the standard TeX device-independent output
>> .dvi file into pdf.
>>
>> The TeX engine is the analog of Mathematica's kernel; the editor and
>> previewer together form the analog of Mathematica's front end; and the
>> various TeX/LaTeX packages are analogs of specialized Mathematical
>> packages.
>>
>> With contemporary distributions, however, the separation into the
>> different tools is largely transparent to the user. The front-end
>> editor is, in a sense, the application, and you don't really have to
>> care what executable or batch file is doing what.
>>
>> About fonts: producing a high-quality font family that includes all
>> the necessary fonts for mathematics is no mean feat. That was a reason
>> that Donald Knuth originally designed TeX, because so many journal
>> articles looked plain ugly as they tried to mix and match symbols with
>> letters. And among symbols I include upper & lower-case Greek, both
>> upright and slanted, Cyrillic, "blackboard bold", Hebrew (for set
>> theory's Aleph, e.g.)
>>
>> The standard, default set of fonts is Computer Modern, which looks
>> great on-screen but can seem a bit spindly in print. Many folks who
>> are producing technical documents with mathematical symbols want a
>> different choice, and that's why the two most popular alternatives --
>> Times with MathTimes and Lucida with Lucida Math -- were developed.
>> To use those, instead of Computer Modern, one simply includes a line
>> or two of code in the document to load some packages; then, provided
>> you have the underlying fonts, everything works transparently; you do
>> have many options available that you can set for these fonts though,
>> and some even for Computer Modern, e.g., just what font to use for
>> script.
>>
>> (TeX is is used for many purposes far from math and science, and there
>> are fonts and supporting packages available for a huge number of world
>> languages.)
>>
>> It is NOT difficult at all to getting a working TeX system: my
>> TeX-naive Math 370 students all managed to do it in short order.
>>
>> But just like using Mathematica effectively takes some learning
>> effort, so using LaTeX effectively does, too.
>>
>> DrMajorBob wrote:
>>> Judging from your links, there IS no application that creates, edits,
>>> and maintains LaTex (and displays it; I forgot that part).
>>> It takes at least four different pieces of software? And dozens of
>>> extra, optional pieces? For one task?
>>> And you sometimes use a different package because it handles certain
>>> fonts, but the other package doesn't?
>>> Wow! That sounds like quite a briar patch you're suggesting.
>>> I initially used TeX for early versions of my dissertation back in
>>> 1987-1989, but the writing went on hiatus when I was assigned to the
>>> Pentagon, dissertation unfinished. (Dr. Klingman, my adviser, was
>>> dying of brain cancer at the time, though I don't offer that as an
>>> excuse.) When I took up the struggle again in 1992, I used Word and
>>> its Equation Editor (later adding MathEdit because Equation Editor
>>> was only its stunted younger brother). It was a MUCH better
>>> experience than what I'd gone through with TeX, although I suppose
>>> TeX had improved in the meantime as well. Sadly, it was not long
>>> after finishing the dissertation that Word would no longer display
>>> the equations properly. (A year or two?)
>>> Still, my MS Thesis was accomplished (1983) with rub-off templates
>>> for equation symbols, so it was a very significant progression from
>>> that, to 1988 TeX, to 1992 MathEdit.
>>> Oh, and let's not forget the Wang workstations I used, 1984-1986,
>>> which handled equations better than anything I've seen since... until
>>> Mathematica.
>>> Another poster suggested LyX, and it "looks" very promising (though
>>> looks can be deceiving). It seems to do everything at once.
>>> Bobby
>>> On Sun, 18 Jan 2009 04:31:52 -0600, Murray Eisenberg
>>> <murray at math.umass.edu> wrote:
>>>
>>>> There are several distributions of TeX, which include the LaTeX macro
>>>> package along with scads of other packages that modify the default
>>>> behavior of, or add new functionality to, LaTeX. Some of these
>>>> distributions, except for Linux, include a "front end" editor that
>>>> integrates into the input -> dvi (or pdf) viewer -> print chain.
>>>>
>>>> For some recommendations, see the menu item "About Math 370" at:
>>>>
>>>> http://bcrc.bio.umass.edu/courses/fall2008/math/math370/
>>>>
>>>> And you may be interested in the items under "LaTeX resources" at
>>>> that site.
>>>>
>>>> For a more complete listing of TeX/LaTeX distributions
>>>>
>>>> http://www.latex-project.org/ftp.html
>>>>
>>>>
>>>> There are both free and commercial distributions.
>>>>
>>>> For an easy-to-install Windows distribution that includes a front end
>>>> editor, I recommend ProTeXt. Personally I most often use the free
>>>> MiKTeX
>>>> distribution (which is part of ProTeXt) together with the low-cost
>>>> front
>>>> end editor WinEdt (which is more powerful than the editor that comes
>>>> with ProTeXt but is more complicated to configure). Sometimes I use the
>>>> nicely integrated, commercial PCTeX system (www.pctex.com) just because
>>>> some of the LaTeX packages it includes make it much easier to use the
>>>> non-default Lucida fonts or the MathTime Pro fonts.
>>>>
>>>>
>>>> DrMajorBob wrote:
>>>>> What application creates, edits, and maintains LaTex?
>>>>>
>>>>> Sign me curious,
>>>>> Bobby
>>>>>
>>>>> On Sat, 17 Jan 2009 04:27:37 -0600, Murray Eisenberg
>>>>> <murray at math.umass.edu> wrote:
>>>>>
>>>>>> If you want an interactive document, then there's little, if
>>>>>> anything,
>>>>>> that can touch Mathematica.
>>>>>>
>>>>>> If you want a more-or-less static document, then the gold standard in
>>>>>> the mathematical community, and in a good part of the scientific
>>>>>> community, is LaTeX. You can include any Mathematica-produced graphic
>>>>>> there by exporting it as EPS.
>>>>>>
>>>>>> And LaTeX documents today typically wind up as PDF, with embedded
>>>>>> hyperlinks and even animation and some interactive effects.
>>>>>>
>>>>>> TL wrote:
>>>>>>> Although Mathematica 7 is a very powerful peace of software as
>>>>>>> far as
>>>>>>> the computational part goes it turns out to be quite limited and
>>>>>>> unstable when it comes to word editing and processing, despite the
>>>>>>> claims in the help that it is almost as powerful as WinWord.
>>>>>>> For example it crashed multiple times on me while I was trying to
>>>>>>> setup
>>>>>>> the right fonts and sizes, as a result I lost all my work several
>>>>>>> times, it also messed up my fonts, sizes, styles, settings for the
>>>>>>> equations, its undo is totally useless and I couldn't figure out
>>>>>>> how to
>>>>>>> format a text and a graphic in two or more columns and display
>>>>>>> them side
>>>>>>> by side in a notebook as well as how to control what goes on what
>>>>>>> page
>>>>>>> and while printing to PDF often it wouldn't print all pages, but
>>>>>>> just
>>>>>>> the first 2-3.
>>>>>>>
>>>>>>> All that said I'm wondering what program to use to write my work
>>>>>>> in, and
>>>>>>> I'm asking for advice - is WinWord any better when it comes to
>>>>>>> handling
>>>>>>> equations?
>>>>>>> Any other choices?
>>>>>>>
>>>>>>> What is the best way to export Mathematica 7 equations and graphics?
>>>>>>>
>>>>>>>
>>>>>>
>>>>>
>>>>>
>>>>>
>>>>
>>>
>>
>
>
>
--
Murray Eisenberg murray at math.umass.edu
Mathematics & Statistics Dept.
Lederle Graduate Research Tower phone 413 549-1020 (H)
University of Massachusetts 413 545-2859 (W)
710 North Pleasant Street fax 413 545-1801
Amherst, MA 01003-9305
• Prev by Date: Re: Finding a sine wave
• Next by Date: Re: Interval arithmetic and Dependency problems
• Previous by thread: Re: Re: Which editor do you use for math articles
• Next by thread: Re: Which editor do you use for math articles
|
2017-12-13 11:33:02
|
{"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": 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": 1, "x-ck12": 0, "texerror": 0, "math_score": 0.7898624539375305, "perplexity": 8723.256325578628}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-51/segments/1512948522999.27/warc/CC-MAIN-20171213104259-20171213124259-00325.warc.gz"}
|
https://mdtpmodules.org/fngr/fngr-1/lesson-4/standards/
|
# FNGR 1 | Lesson 4 | Standards
### Core Content Standards
CCSS.MATH.CONTENT.HSF.BF.B.3 Identify the effect on the graph of replacing $$f(x)$$ by $$f(x) + k$$, $$k f(x)$$, $$f(kx)$$, and $$f(x + k)$$ for specific values of $$k$$ (both positive and negative); find the value of $$k$$ given the graphs. Experiment with cases and illustrate an explanation of the effects on the graph using technology. Include recognizing even and odd functions from their graphs and algebraic expressions for them.
### Standards for Mathematical Practice
CCSS.Math.Practice.MP1 Make sense of problems and persevere in solving them.
CCSS.Math.Practice.MP2 Reason abstractly and quantitatively.
CCSS.Math.Practice.MP4 Model with mathematics.
|
2021-01-27 01:12:19
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 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": 0.5848308205604553, "perplexity": 1024.8036968015438}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-04/segments/1610704804187.81/warc/CC-MAIN-20210126233034-20210127023034-00088.warc.gz"}
|